CHARACTER THEORY OF
FINITE ROUPS
f. MARTIN ISAACS

Character Theory
of Finite Groups

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CHARACTER THEORY
OF FINITE GROUPS
I. MARTIN ISAACS
Department of Mathematics
University of Wisconsin
Madison, Wisconsin
1976
ACADEMIC PRESS New York San Francisco London
A Subsidiary of Harcourt Brace Jovanovich, Publishers

Copyright © 1976, by Academic Press, Inc.
all rights reserved.
no part of this publication may be reproduced or
transmitted in any form or by any means, electronic
or mechanical, including photocopy, recording, or any
information storage and retrieval system, without
permission in writing from the publisher.
ACADEMIC PRESS, INC.
Ill Fifth Avenue, New York, New York 10003
United Kingdom Edition published by
ACADEMIC PRESS, INC. (LONDON) LTD.
24/28 Oval Road, London NW1
Library of Congress Cataloging in Publication Data
Isaacs, Irving Martin, (date)
Character theory of finite groups.
(Pure and applied mathematics, a series of monographs
and textbooks; 69)
Bibliography: p.
1. Finite groups. 2. Characters of groups. I.
Title. II. Series. QA3.P8 [QA171] 510'.8s [512.24]
ISBN 0-12-374550-0 75-19652
AMS (MOS) 1970 Subject Classifications: 20C15, 20C05,
20C25, and 20C20
PRINTED IN THE UNITED STATES OF AMERICA

Contents
PREFACE vii
ACKNOWLEDGMENTS ix
NOTATION xi
1. Algebras, modules, and representations l
Problems • •
2. Group representations and characters 13
Problems 29
3. Characters and integrality 33
Problems 44
4. Products of characters 47
Problems 59
5. Induced characters 62
Problems 73
6. Normal subgroups 78
Problems 95
7. T.I. sets and exceptional characters
Problems
99
121

vi Contents
8. Brauer's theorem 126
Problems 141
9. Changing the field 144
Problems 156
10. The Schur index 160
Problems 171
11. Projective representations 174
Problems 194
12. Character degrees 198
Problems 216
13. Character correspondence 219
Problems 237
14. Linear groups 240
Problems 260
15. Changing the characteristic 262
Problems 285
Appendix Some character tables 287
Bibliographic notes 292
References 295
INDEX
297

Preface
Character theory provides a powerful tool for proving theorems about
finite groups. In fact, there are some important results, such as Frobenius'
theorem, for which no proof without characters is known. (Until fairly
recently, Burnside's p"<f theorem was another outstanding example of this.)
Although a significant part of this book deals with techniques for applying
characters to "pure" group theory, an even larger part is devoted to the
properties of characters themselves and how these properties reflect and are
reflected in the structure of the group.
The reader will need to know some basic finite group theory: the Sylow
theorems and how to use them and some elementary properties of
permutation groups and solvable and nilpotent groups. A knowledge of additional
topics such as transfer and the Schur-Zassenhaus theorem would be helpful
at a few points but is not essential. The other prerequisites are Galois theory
and some familiarity with rings. In summary, the content of a first-year
graduate algebra course should provide sufficient preparation.
Chapter 1 consists of ring theoretic preliminaries, and Chapters 2-6 and
8 contain the basic material of character theory. Chapter 7 is concerned with
one of the more important techniques for the application of characters to
group theory.
The emphasis in all chapters except 1,9, 10, and 15 is on characters over
the complex numbers rather than on modules and representations over
other fields. In Chapter 9, irreducible representations over arbitrary fields
are considered; and in Chapter 10, this is specialized to subfields of the
complex numbers. Chapter 15 is an introduction (and only that) to Brauer's
Vll

vm
Preface
theory of blocks and " modular characters " and so is concerned with
connections between complex characters and characteristic ^-representations.
The remaining chapters concern more specialized topics. Chapter 12
deals with the connections between the set of degrees of the irreducible
characters and the structure of a group. Chapter 13 is essentially an
exposition of a paper of Glauberman, and Chapter 14 contains some "classical"
results on complex linear groups and a small sampling of more recent
developments.
Following each chapter there is a selection of problems. In addition to
some routine exercises, these include examples, further results, and
extensions and variations of theorems in the text. It is hoped that the reader will
find that these problems enhance his understanding (and enjoyment) of
character theory.

Acknowledgments
At various times during the preparation of this book I was supported by
research grants from the National Science Foundation and a research
fellowship from the Sloan Foundation. These organizations deserve (and
hereby receive) my sincere thanks. Finally, I thank my teacher, Richard
Brauer, both for introducing me to character theory and for his large part
in developing the subject which I find so fascinating.
IX

This Page Intentionally Left Blank

Notation
Mn(F)
Xy
Ay
A°
+.!■
M(V)
ndV)
J(A)
C\(g)
£„
XT
o(g)
Til)
Xh
det/
cox
[x,y]
v„(z)
cp x 9
9°
9°
Ga
/g(9)
algebra of n x n matrices over F
linear transformation of the A-module V, induced by x e A
{xv\x eA}
regular A-module
internal direct sum
see Definition 1.12
see Lemma 1.13
the Jacobson radical, Problem 1.4
conjugacy class of g
symmetric group of degree n
transpose of the matrix X
the order of g
see Definition 2.26
restriction of / to H
see Problem 2.3
the algebra homomorphism Z(C[G]) -» C induced by /
x~ 1y~ 1xy, the commutator
see Lemma 4.4 and the discussion preceding it
see Definition 4.20
see Definition 5.1
induced character
stabilizer of a in permutation representation
inertia group, Definition 6.10
xi

Xll
Notation
o(x) determinantal order of /, equals the order of det / in group
of linear characters
Qk the field Q(e27"lk)
Op(G), Op (G) minimal normal subgroups of p-power index and index
prime to p
R[£f] ring of /^-linear combinations of y £ Irr(G)
R[_£T\° {9e/?[^]|9(l) = 0}
P(G,je) Z[{lHG|He^}]
9*> 9*>gP> 9P' see the discussion following Lemma 8.18
3E£ see the discussion at beginning of Chapter 9
F(x) the field generated over F by the values of /
rrifix) the Schur index, Definition 10.1
Z(G, A), B(G, A), H(G, A) see the discussion preceding Theorem 11.7
M(G) the Schur multiplier, Definition 11.12
A the group of linear characters of A
Ch(G|9), Irr(G|9) see the discussion preceding Definition 11.23
coreG(H) f] HB for g e G
cd.(G) {z(l)|zeIrr(G)}
\{x) vanishing-off subgroup, <x e G | x(x) ^ 0>
d.l.(G) derived length of G
F(G) Fitting subgroup of G
b(G) max(c.d.(G))
Op(G), Op(G) maximal normal subgroup of order a power of p, prime to p
Irrs(G) the set of S-invariant / e Irr(G)
<J>(G) Frattini subgroup of G
IBr(G) the set of irreducible Brauer characters of G
dXv decomposition numbers, Definition 15.9
<£,, the projective character associated with q> e IBr(G)
B1(G) the set of p-blocks of G
eB, XB the idempotent and algebra homomorphism of Z(F[G])
associated with B e B1(G)
(5( Jf) the set of defect groups for the class Jf
Jf the sum of the elements of Jf in F[G]
a^Jf) coefficient of Jf in eB
8(B) the set of defect groups for B e B1(G)
d(B) defect of BeBl(G)

1 Algebras, modules, and representations
Character theory provides a means of applying ring theoretic techniques
to the study of finite groups. Although much of the theory can be developed
in other ways, it seems more natural to approach characters via rings (or
more accurately, algebras). The purpose of this chapter is to provide the
reader with the ring theoretic prerequisites needed in the rest of the book.
Many of the results in this chapter are true in more general contexts
than those considered here. Nevertheless, an effort has been made to avoid
excess generality and to prove only that which will be needed later.
(1.1) definition Let F be a field and let A be an F-vector space which is
also a ring with 1. Suppose for all c e F and x,ye A, that
(cx)y = c(xy) = x(cy).
Then A is an F-algebra.
We mention some examples of algebras over a field F:
(a) M„(F) is the algebra of n x n matrices over the field F.
(b) Let V be an F-vector space. Then End(l/), the set of F-linear
transformations of V, is an F-algebra under the following conventions. If x,
ye End(l/), then xy is defined by (v)xy = ((v)x)y and if c e F, then ex is
defined by (v)(cx) = (cv)x. Of course, (v)(x + y) = (v)x + (v)y.
This is a good place to digress briefly to discuss some notational
conventions which will be used throughout this book. In writing scalar
multiplication in a vector space, the scalar may be written on either side of the vector
1

2
Chapter 1
to which it is applied. Similarly, functions are written on whichever side is
convenient, but the rule for function composition is always " fg means do /
first, then g." Because of this rule, in those situations where function
composition is important [as in example (b)], it will usually be convenient to write
functions on the right.
We now return to another example of an algebra, the one of primary
importance for us: the group algebra.
(c) Let G be a finite group. Then F[G] is the set of "formal" sums
{Z»eG«99 \ag e F}- The structure of an F-vector space is given to F[G] in the
obvious way and the element of F[G] for which ag = 1 and ah = 0 if h j= g is
identified with g. This identification embeds G into F[G] and in fact G is a
basis for F[G]. One result of this identification is to give a new meaning for
£ agg. We may now view this expression not only as a formal sum, but also as
an actual sum, a linear combination of the basis vectors. Finally, to define
multiplication on F[G], we multiply the basis vectors according to their
group multiplication and extend linearly to all of F[G]. It is routine to check
that this defines the structure of an F-algebra on F\_G]-
The construction of F\_G] suggests a general method of constructing
algebras which should be mentioned. Let A be a finite dimensional F-algebra
with F-basis vu ..., v„. We have then v^j = £ cijkvk, where cijk e F are the
multiplication constants of A with respect to the basis {vt}. It is clear that these
constants determine the algebra, so that any n-dimensional algebra may be
specified by prescribing n3 constants cljk e F. Of course, only a small subset
of all possible sets of constants define an algebra since most sets of constants
define multiplications that turn out to be nonassociative.
From now on, the word "algebra" in this book will mean a finite
dimensional algebra. We make a few observations and definitions before going
on to prove anything.
Let A be an F-algebra. Then F • 1 = {c\ \c e F} is a subalgebra of A
contained in the center Z(y4) since (c\)x = c(\x) = c(x\) = x(c\) for x e A.
It is sometimes convenient to identify F with F • 1 and thus to view F as a
subalgebra of A. If / is a left or right ideal of A as a ring and xe I and c e F,
we have ex = (cl)x = x(cl)e I and / is a subspace. (If we had not required
algebras to contain 1, then ideals would not automatically be subspaces.)
If / is an ideal (this means two-sided), then A/I has the structure of an F-
algebra in a natural manner.
If A and B are F-algebras and q> is a ring homomorphism from A to B
with <p(l) = 1, it is not necessarily true that cp is an F-linear transformation.
(1.2) definition Let A and B be F-algebras. Suppose that <p: A -» B
satisfies

Algebras, modules, and representations
3
(a) <p(xy) = <p(x)<p(y) for all x, y e A;
(b) <p(l)=l;
(c) cp is an F-linear transformation.
Then <p is an algebra homomorphism or an F-homomorphism.
(1.3) definition Let A be an F-algebra and let V be a finite dimensional
F-vector space. Suppose for every v e V and x e A that a unique vx e V is
defined. Assume for all x, y e A, v, w e V, and ceF that
(a) (d + w)x = vx + wx,
(b) u(x + y) = vx + vy,
(c) (wc)>> = v(xy),
(d) (cu)x = c(tuc) = v(cx),
(e) t>l = v.
Then F is an A-module.
Let F be an A-module. Each x e A defines a map xv: V -» K by t> i—> dx.
By (a) and part of (d) of the definition of a module, xve End(F). By (b), (c),
(e), and part of (d), the map x \-* xv is an algebra homomorphism A -» End( V).
Its image is denoted by Av.
Some important examples of modules are the following. If A £ End(K),
then 7 is an A-module in a natural way. If A = M„(F\ then the row space of
dimension n over F is an A-module under matrix multiplication. If A is any
algebra, then A itself is an A-module under right multiplication. This module
is called the regular A-module and is denoted by A°.
If V is an A-module and W £ V is an A-invariant subspace, then W is a
submodule of V. Thus the right ideals of A are exactly the submodules of A°.
If W is a submodule of V, then the space J^W becomes an A-module in the
usual manner. Note that if / is a proper ideal of A, then the objects A/I,
A°/I, and (A/If are all defined and all different, being respectively an algebra,
an A-module, and the regular (A//)-module. However, A°/l is an A-module
which is annihilated by / (i.e., if v e A°/I and xel, then vx = 0), and so it
may be viewed as an (A//)-module. As such it becomes (A/If.
If V and W are A-modules, a linear transformation cp: V -» W is an
A-homomorphism if (p(vx) = (p(v)x for all v e V and x e A. An A-isomorphism
is an A-homomorphism which is one-to-one and onto and, as is the usual
situation, if V and W are A-isomorphic, they are exactly the same "as seen
by A." For instance, in this situation they are annihilated by exactly the same
set of elements of A.
The set Hom^K W) of A-homomorphisms from V to W has the structure
of an F-space by (ccp)(v) = c((pv) for ce F and (<p + 9)(t>) = (p(v) + 9(d).
In addition Hom/1(K V) is a ring [remember, <p9 is defined by (v)(p& =
(u<p)9], and in fact Hom^K V) is an F-algebra. It is exactly the centralizer

4
Chapter 1
of Av in End(F) and is denoted EA(V). If E = EA(V), then V is an E-module
and Ef/iV) 3 Av. Later in this chapter we shall find a sufficient condition
for equality here.
(1.4) definition Let V be a nonzero ,4-module. Then V is irreducible
if its only submodules are 0 and V.
It is obvious that if cp e Hom^CV, W) then ker cp and im (p are sub-
modules of V and W, respectively. The following important lemma is now
immediate.
(1.5) lemma (Schur) If V and W are irreducible A -modules, then every
nonzero element of Hom/1(K W) has an inverse in Hom/1(VK V).
An immediate consequence of Schur's lemma is that if V is an irreducible
A-module, then EA(V) is a division algebra, i.e., every nonzero element is
invertible.
(1.6) corollary Let F be algebraically closed, A an F-algebra, and V an
irreducible A-module. Then EA(V) = F-1, the set of scalar multiplications
on V.
Proof Clearly, F ■ 1 £ EA(V). Let 9 e EA(V). Then 9 is a linear
transformation of a finite dimensional vector space V over F and so has an
eigenvalue L Then 9 — 11 e EA(V) and is not invertible. Thus 9 — 11 = 0 and
9 = 11 e F • 1 as claimed. |
To justify the study of irreducible modules we remark that for certain
algebras, every module is a direct sum of irreducible ones, and thus to know
all irreducibles is to know all modules for these algebras. It happens that
group algebras over fields of characteristic zero are among these fortunate
algebras. This will follow from Maschke's theorem which will be proved
shortly.
(1.7) definition Let V be an A-module. Suppose for every submodule
W £ V, there exists another submodule U £ V such that V = W + U
(the dot indicating that the sum is direct). Then V is a completely reducible
module.
Observe that irreducible modules are completely reducible as are all
modules over fields (i.e., vector spaces).
(1.8) definition An algebra A is semisimple if its regular module, A0 is
completely reducible.
(1.9) theorem (Maschke) Let G be a finite group and F a field whose
characteristic does not divide \G\. Then every F[G]-module is completely
reducible.

Algebras, modules, and representations
S
Proof Let V be an F[G]-module with submodule W. Let U0 be an
F-subspace of V which is complementary to W, i.e., V = W + U0. Let cp be
the projection map of V onto W with respect to U0. Now cp is a linear
transformation which is not necessarily an F[G]-homomorphism. The object
of the remainder of the proof is to modify (p in order to create, an F[G]-homo-
morphism.
Define 9: V -+ W by
Clearly, 9 is F-linear. To show that 9 is an F[G]-homomorphism, we compute
|G| 9EG I O | g
since hg runs over G as g does for fixed /j.
If w e H< we have wg eW for all g e G and thus (p(wg) = wg. It follows
that 9(w) = w. Now let U = ker 9, an F[G]-submodule of V. We have
9(d) e VK so that 9(9)d)) = 9(d) and 9(d - 9(d)) = 9(d) - 9(d) = 0. Thus
v = 9(d) + (d - 9(d)) eW+U and V = W + U. Finally, if w e W n U,
we have w = S(w) = 0, so that V = W + V and the proof is complete. |
A consequence of Maschke's theorem is that F[G~] is semisimple if
char(F)^|G|. The converse of this statement is also true and the reader is
referred to the problems for a proof.
(1.10) theorem Let V be an ^module. Then V is completely reducible
iff it is a sum of irreducible submodules.
Proof Suppose V = Y,K where the Va are irreducible. Let W £ V.
By finite dimensionality, choose U £ V maximal such that W n U = 0.
We claim that W + U = V. Otherwise, we may choose Va <£. W + U and
thus (W+U)nVa = 0 by the irreducibility of Fa. It follows that
W r\ (U + Va) = 0 and this violates the maximality of U. Thus V = W + U
and V is completely reducible.
Conversely, suppose V is completely reducible and let S be the sum of all
of the irreducible submodules of V. If S < V, we may write V = S + T with
T t6 0. By finite dimensionality, T contains an irreducible submodule which
is a contradiction since T n, S = 0. |
(1.11) lemma Let V be an ,4-module and suppose V = Y,K where the Va
are irreducible submodules. Then V is the direct sum of some of the V^s.
Proof Choose W £ V maximal with the property that W is the direct
sum of some Vx's. If W < V, then Vx £ W for some a. Since Va is irreducible,

6
Chapter 1
we have W r\ Va = 0 and Wi = W + Vx > W. This violates the maximality
of W and we conclude that W = V as desired. |
We see from the previous two results that the completely reducible
modules are exactly the direct sums of irreducible modules. It follows that
in order to know all modules for a group algebra over a field of
characteristic 0, it suffices to know all irreducible modules.
(1.12) definition Let V be a completely reducible ,4-module and let M
be an irreducible A-module. The M-homogeneous part of V, denoted M(V),
is the sum of all those submodules of V which are isomorphic to M.
Observe that if M s N, then M(V) = N(V). As will be shown shortly, if
M and N are nonisomorphic irreducible /4-modules, then M(V) n, N(V) = 0.
If V has no submodules isomorphic to M, then M(V) = 0.
(1.13) lemma Let V = £• Wi be a direct sum of ,4-modules with W(
irreducible for all i. Let M be any irreducible ,4-module. Then
(a) M(HisanE/1(F)-submoduleof V;
(b) M(V) = li{Wt\Wl*Mh
(c) The number nM(V) of Wt which are isomorphic to M is an invariant
of K independent of the given direct sum decomposition.
Proof (a) Let 9 eEA(V). We need to show M(V)$ £ M(V). It suffices
to show that if W £ V and W s M, then W& £ M(V). This is sufficient
because M(V) is the sum of such W. If W9 = 0, there is nothing to prove; if
W9 t6 0, then since W9 is a homomorphic image of the irreducible module W,
we have W& S W S M. Thus W9 £ M(7).
(b) Clearly, £ {V^ | Wt = M} £ M(V). Let ^ be the projection map of F
onto Wt. Now let W£ K W^ M. If nj(W) ± 0, then nfW) = Wj and
W £ W). Thus TtjiW) £ X {Wil Wt s M} for all j. However, W £ £, n/W)
and hence W £ ^{H^| Wt S M}. It follows that M(V) £ £ Ml Wt S M}.
(c) By (b), we have dim M(F) = nM(V) dim M and it is immediate that
nM(V) is an invariant. |
By part (b) of Lemma 1.13 it follows that a completely reducible module is
the direct sum of its M-homogeneous components for distinct M. In
particular, M(V) n N(V) = 0 if M ik N.
We wish to discuss "all irreducible /4-modules." This poses some
difficulties. First, we really mean "all isomorphism classes of irreducible
A -modules." It would be convenient to have a representative set of /4-modules.
By this we mean a set Jl{A) of irreducible /4-modules with the property that
every irreducible y4-module is isomorphic to exactly one element of Jt{A\
Since modules are external to the algebra, it is not clear how one can find a

Algebras, modules, and representations
7
representative set for a given algebra A. To do this we need to produce a
set of A -modules large enough to contain copies of all irreducibles. The next
lemma shows that the set of homomorphic images of A° suffices.
(1.14) lemma Let A be an F-algebra. Then every irreducible ,4-module is
isomorphic to a factor module of A°. \{A is semisimple, then every irreducible
A -module is isomorphic to a submodule oi A0.
Proof Let V be an irreducible A-module, choose 0 ^ v e V, and define
ft A -» V by 9(x) = vx. Then 9 is clearly F-linear and 9(xy) = vxy = 9(x)y
so that 9 e HomA(A°, V). Now, v e im 9 £ V and hence im 9 = V since V
is irreducible. Let W = ker 9. Then V = A°/W. If A is semisimple, then
A0 = W + U and /T/W s I/. The proof is complete. |
Now fix a representative set M{A) of irreducible A-modules. By part (b)
of Lemma 1.13, we have V = £• MejnA)M(V) for every completely
reducible A-module V. Suppose A is a semisimple algebra. We can then
apply the above to A0 and write A° = £ • M(A°). It turns out that M(A°) is
actually a two-sided ideal of A. For notational convenience, we write M(A)
for M(A°) in what follows.
(1.15) theorem (Wedderburn) Let A be a semisimple algebra and let
M be an irreducible A-module. Then
(a) M(A) is a minimal ideal of A;
(b) if W is irreducible, then it is annihilated by M(A) unless W s M,
(c) the map x i-> xM is one-to-one from M(A) onto y4M £ End(M);
(d) Jt{A) is a finite set.
Proof If x e A, the map 9*: y i—> x_y satisfies 9* e E^A0). Therefore, by
Lemma 1.13(a), xM(A) = M(A)$X £ M(A) and M(A) is a left ideal. Since
M(A) is a submodule of A0, it follows that it is an ideal of A. Minimality will
follow after (c) is proved.
If W is an irreducible A-module with W yk M, then W(A) r\ M(A) = 0 by
Lemma 1.13(b). Since W(A) and M(A) are ideals, we have W(A)M(A) = 0.
By Lemma 1.14, A0 has a submodule W0 s W and W0 £ W(A), so that
M(A) annihilates WQ. Since W = W0, they have the same annihilator in A
and (b) follows.
By (b), it follows that xw = 0 if x e M(A) and M ^ W.lt now follows from
the direct sum decomposition A = £ • M€^u)W(^)tnat f°r y e ^> we have
^ = xM, where x is the component of y in M(A). We conclude that the map
xbxm maps M(A) onto AM. If x e M(A) and xM = 0, then it follows from (b)
that x annihilates every irreducible, and hence every completely reducible
A -module. Thus x= lx£y4°x = 0 and (c) is proved.

8
Chapter 1
To show the minimality of M(A), let I < M(A) be an ideal of A. Now
M(A) is a sum of submodules isomorphic to M and thus there exists M0 £
M(A), M0 = M,M0^/.AsM0n/< M0 and M0 = M is irreducible, we
have M0 n 7 = 0. Thus M07 £ M0 n J = 0 and 7 annihilates M0 and hence
also M. Therefore, if x e 7, we have xM = 0. However, x h-» xm is one-to-one
for x e M(y4) and thus x = 0. We conclude that 7 = 0 and the minimality of
M(A) is proved.
Finally, M(A) =£ 0 for every M by Lemma 1.14, and yet
A = £• M€.*u)W(^4) is finite dimensional. It follows that |^(X)| is finite
and the proof is complete. |
Observe that each M(A) is actually an algebra, its unit element being the
component of 1 in M(A) under the decomposition A = £ • M(A). Since the
map x i—> xM is an algebra homomorphism from A to y4M, it follows from (c)
of the theorem that the restriction of this map to M(A) is an algebra
isomorphism from M(A) onto Au.
Since M1(A)M2(A) = 0 for Mj % M2, it follows that every ideal of the
algebra M(A) is in fact an ideal of A. We conclude from the minimality of
M(A) as an ideal of A, that M(A) is a simple algebra, i.e., it has no nontrivial
proper ideals. Thus the preceding theorem asserts (among other things)
that a semisimple algebra is a direct sum of simple algebras. This is a
somewhat more usual statement of Wedderburn's theorem. (It is also true that
every simple algebra is semisimple. This follows from Problem 1.5.)
To review the situation now: group algebras over fields of characteristic
zero are semisimple; semisimple algebras are direct sums of ideals M(A)
and M(A) is naturally isomorphic to Au. What remains is to study the simple
algebras Au. This is the purpose of the "double centralizer" theorem which
follows. The proof we give here is based on an idea of M. Rieffel.
It should be remarked that the hypothesis that A is semisimple is actually
superfluous since the theorem is really about AM which is automatically
semisimple when M is completely reducible. See Problem 1.6.
(1.16) theorem (Double Centralizer) Let A be a semisimple algebra and
let M be an irreducible A-module. Let D = EA(M). Then ED(M) = AM.
Proof It is no loss to replace M by an isomorphic module, and so by
Lemma 1.14, we may assume that M £ A°. Let I = M{A) so that M £ 7.
It is clear that AM £ ED(M) so we prove the reverse inclusion. Let
9 e ED(M) so that (ma)9 = (m9)a for a e D. If m e M, define am: M -» A by
(x)am = mx. Since m e M is a right ideal of A, we have mx e M and am: M -> M.
If a e A and x e M, we have (xa)oim = m(xa) = (mx)a = (xa.m)a. It follows
that am e EA(M) = D. Thus for m,ne M, we have
(*) (mn)S = (nam)9 = (n9)am = m(n9>).

Algebras, modules, and representations
9
Now fix n e M with n#0 and let e be the unit element of /. We have
AnA £ / and by the minimality of the ideal /, we have e e I = AnA and
e = £ ainbi for suitable ah bt e A. Urns M, we have
m = me = m £ «,-«£>,• = X(ma;)(nfc,).
Since mat e M and nbt e M, Equation (*) yields for all m e M that
m9 = X {{mad(nbd)9 = I NIW = m £ a,.((nfc;)9).
Thus 9 = mm e y4M where u = £ aj((nfcj)9). The proof is complete. |
(1.17) corollary Let A be a semisimple algebra over an algebraically
closed field F and let M be an irreducible A-module. Then
(a) AM = End(M);
(b) dim(AM) = dim(M(A)) = dim(M)2;
(c) nM(A°) = dim(M).
Furthermore, if M{A) is a representative set of irreducible ,4-modules, then
(d) dim(A) = XM€^M>dim(M)2,
(e) dim(Z(A)) = |uT(X)|.
Proof By Corollary 1.6, EA(M) = F ■ 1 for irreducible M and thus y4M =
EF.i(M) = End(M). By elementary linear algebra, End(M) s M^F), the
algebra of d x d matrices, where d = dim M. Thus dim AM = dim(End(M)) = d1.
Since M(A) s Au, we have (b). Now M(A) is the direct sum of nM(A°)
isomorphic copies of M, thus d2 = dim M(A) = nu(A°) dim M = dnM(y4°),
and (c) follows. Since A = £ • Me.*M)Af(/4), (d) is immediate from (b).
Finally, let ZM = HM(A)\ Now Z{AM) = AM n E^M) = Au n F • 1
= F • 1. Thus dim ZM = dim(ZUM)) = 1. Clearly, £ • M,M(A)ZM <= Z(A)
and dim(£ ZM) = l-^(4)|. However, if z e Z{A), write z = £ aM, aM e M(A).
If w e M(y4), then waM = wz = zu = aMu since the distinct M(A)'s annihilate
each other. Thus aM e ZM and £ ZM = Z{A). The proof is complete. |
Although the word "representations" constitutes one third of the title
of this chapter, so far nothing has been said about them. In fact, that isn't
really true, because as we shall see, representations are just a different way
of looking at modules.
(1.18) definition Let A be an F-algebra. A representation of A is an
algebra homomorphism 3E: A -» M„(F). The integer n is the degree of 3E.
Two representations 3E, 9) of degree n are similar if there exists a nonsingular
n x n matrix P, such that 3E(a) = P~l<!0(a)P for all a e A.

10
Chapter 1
Clearly, similarity is an equivalence relation among representations. Also,
if 9) is a representation of degree n and P is any nonsingular n x n matrix,
then the formula 3E(a) = P~ ll$)(a)P defines a new representation 3E.
It is easy to build modules from representations and representations from
modules. If 3E is a representation of degree n of the F-algebra A, let V be the
n-dimensional row vector space over F. If v e Fand X is any n x n matrix
over F, then vX e V. Define va = vH{a) for aeA.lt is routine to check that
this gives the structure of an ,4-module to V.
Conversely, if M is an A-module, choose an F-basis for M and let 3E(a)
be the matrix of au with respect to this basis. It is now easy to check that 3E
is a representation. Note that a different choice of basis might give a different
representation (and usually does).
Starting with a representation 3E, constructing the module V as above,
and then choosing the appropriate basis for V and constructing the
corresponding representation will result in the original representation 3E.
Suppose V and W are A-modules and 9e Hom^K W). How does this
situation look from the representation point of view? Choose bases 6Y and 6W
for V and W. Since 9 is a linear transformation from V to W, it has a matrix
P with respect to the given bases. Note that P is m x n where m = dim V
and n = dim W. Now, the fact that (va)S = (v$)a for all veV and a e A
yields 3E(a)P = Ptyla), where 3E and 9) are the representations given by V
and W with respect to the bases 6V and 6W. If 9 is a module isomorphism, then
P is nonsingular and the above matrix equation yields that 3E and 9) are
similar representations. In particular, the different representations arising
from a given module with different choices of basis are all similar.
The above reasoning may be reversed to show that if the representations
arising from V and W with respect to bases 6V and Sw are similar, then V
and W are in fact isomorphic modules. It follows that there is a natural
one-to-one correspondence between isomorphism classes of A-modules and
similarity classes of representations of A.
If V is an /4-module and W < V is a proper nonzero submodule, choose a
basis 6W for W and extend this to 6V, a basis for V. Number 6V so that the last
m vectors are £w, where m = dim W. Let 3E be the representation of A
corresponding to V with respect to the basis 6V and let 9) be the
representation corresponding to W with respect to the basis 6W. It is then easy to see for
a e A that 3E(a) has the form
(3(a) U(aj
Furthermore, 3 is a representation corresponding to V/W. Note that U is a
function from A into (n — m) x m matrices (where n = dim V), but U is not
a representation.

Problems
11
The representation 3E is said to be in reduced form and one similar to 3E
isreducible. Thus the irreducible representations correspond to the irreducible
modules.
If there exists a submodule U £ V in the above situation, with V =
W + U, then the basis for W may be extended to V by adjoining to it a
basis for U. When this is done, the result is that U(a) = 0 for all aeA.lt
follows from this discussion that if 3E is any representation corresponding to
a completely reducible module, then 3E is similar to a representation in block
diagonal form, where each of the blocks is an irreducible representation.
Problems
(1.1) Let V be an A-module. Show that V is completely reducible iff the
intersection of all of the maximal submodules of V is trivial.
Hint To prove "if," embed V into a sum of irreducible modules. Recall
that our definition of "module" requires finite dimensionality.
Note This problem is "dual" to Theorem 1.10 which implies that V
is completely reducible iff it is the sum of its minimal submodules. Theorem
1.10 is true much more generally than we have proved it. It holds for arbitrary
modules over rings. Problem 1.1, however, is not valid in this greater
generality. A counterexample is the regular module of the ring of integers Z.
(1.2) Let 3E and 9) be representations of an F-algebra, A. A nonzero matrix
P is said to intertwine 3E and 9) if PH(a) = ^(afP for all a e A. Assume 3E and
9) are irreducible.
(a) If P intertwines 3E and 9), show that P is square and nonsingular.
(b) Assume that F is algebraically closed and that P and Q both
intertwine 3E and 9). Show that Q = XP for some X e F.
(1.3) Show that an algebra A is semisimple iff every A-module is completely
reducible.
(1.4) Let A be an algebra. For A-module V, let d{V) = {a e A\ Va = 0}.
Let J(A) = C\mzjha)<s4(M), where Jt(A) is a representative set of irreducible
A-modules. Show
(a) s#(V) is an ideal of A for all V.
(b) VJ(A) < V for every nonzero A-module V.
(c) J(A)n = 0 for some integer n.
(d) If / is a right ideal of A and Im = 0 for some m, then / £ J(A).
Note The ideal J(A) is called the Jacobson radical of A; jtf(V) is the
annihilator of V and a right ideal / with I" = 0 is said to be nilpotent.

12
Chapter 1
(1.5) Prove that the following are equivalent for the algebra A.
(a) J(A) = 0.
(b) A has no nonzero nilpotent right ideals.
(c) A has no nonzero nilpotent ideals.
(d) A is semisimple.
Hint If V is irreducible, then sf(V) is an intersection of maximal right
ideals of A.
(1.6) Let A be an algebra and V a completely reducible A-module. Show that
the algebra Av is semisimple.
Note A consequence of Problem 1.6 is that the hypothesis that A is
semisimple in the Double Centralizer Theorem 1.16 may be dropped since this
theorem is really about AM for an irreducible module M.
(1.7) Let A be an algebra and V an irreducible A-module. Show that
\Jf(Ay)\ = 1.
(1.8) Let G be a group, HsGa subgroup and F a field with characteristic
prime to | G: H |. Let V be an F[G]-module with submodule W. Suppose that
there exists U0 £ V, an F[H]-submodule such that V = W + U0. Show
that there exists an F[G]-submodule U £ V with V = W + U.
Note This generalization of Maschke's Theorem 1.9 is due to D. G.
Higman.
(1.9) Let G be a group and F a field of characteristic p. Suppose p\\G\ and
show that F[G] is not semisimple.
Hint (I9eG0)2=O.
(1.10) Let M be an A-module. Show that M is completely reducible iff
MJ(A) = 0.
Hint A/J(A) is semisimple.

Group representations and characters
Let G be a finite group and let F be a field. Suppose 3E is a representation
of F[G] with degree n. Since 3E is an algebra homomorphism, X(l) = /, the
identity matrix. It follows for g e G that 3E(g) is nonsingular and 3E(g)_1 =
X(g~'). If we restrict the function 3E to G £ F[G], we obtain a group
homomorphism from G into the general linear group GL(n, F), that is, the
multiplicative group of nonsingular n x n matrices over F.
(2.1) definition Let F be a field and G a group. Then an F-representation
of G is a homomorphism 3E: G -» GL(n, F) for some integer n.
We have seen that a representation of F[_G] determines an
F-representation of G by restriction. Conversely, an F-representation 3E0 of G determines a
representation 3E of F[G] by linear extension. That is,
We shall usually use the same symbol to denote both an F-representation
of G and the corresponding representation of F[G~\. Also, the adjectives
"similar" and "irreducible" will be applied to F-representations of G as if
they were the corresponding representations of F[G~\. Some caution is
necessary here since if F £ E, a larger field, and 3E is an F-representation of G,
then 3E is automatically an E-representation. It is entirely possible, however,
that 3E is irreducible as an F-representation, and yet is reducible as an
E-representation. We will explore this situation in some depth in Chapter 9.
One further triviality which should be mentioned now is the following.
If JV o G and 3E is an F-representation of G with JV £ ker 3E, then there is a
13

14
Chapter 2
unique F-representation X of G/N defined by X(Ng) = X(g). This formula can
also be used to define the representation 3E if J is given. Note that 3E is
irreducible iff J is. We shall often fail to distinguish between 3E and X.
The trouble with representations is that they contain too much
information. If 3E is an F-representation of G of degree n, then for each element of G
we have n2 entries in X(g). Some of this data is clearly redundant because it
distinguishes between similar representations. The idea behind character
theory is to throw away most of the information and to save just enough to be
useful. This is done by calculating the traces (that is, the sums of the diagonal
entries) of the matrices in question. Recall that if A and B are any two n x n
matrices over a field, then tr(y4B) = tr(BA).
(2.2) definition Let 3E be an F-representation of G. Then the F-character
X of G afforded by 3E is the function given by 1(g) = tr X(g).
As is the case with F-representations of G, we may view F-characters as
functions on all of F[G]. Note that if the characteristic char(F) ± 0, then the
constant function 0 is an F-character. On the other hand, if char(F) = 0,
then 0 is definitely not an F-character because x(l) = deg 3E, where 3E is an
F-representation of G which affords /. In this case, we say that x(l) is the
degree of /.
Most F-characters of a group G are not homomorphisms of any kind.
However, if 1 is a homomorphism from G into the multiplicative group of F,
then 3E(g) = (1(g)) is an F-representation of G of degree 1 which affords X
as its character. Characters of degree 1 are called linear characters. In
particular, the function lc with constant value 1 on G is a linear F-character.
It is called the principal F-character.
(2.3) lemma (a) Similar F-representations of G afford equal characters,
(b) Characters are constant on the conjugacy classes of a group.
Proof UP is nonsingular then tr(P~ 1AP) = tr(P ■P~1A) = tr(A). Both
(a) and (b) follow from this observation. To see (b), observe that X(h~ lgh) =
X(h)~ lX(g)X(h) if 3E is a representation of G, and hence tr(X(h~1gh)) =
tr(X(g)). |
We make one further general observation. If 3E and 9) are
F-representations of G, then
is also an F-representation. Since tr 3(#) = tr X(g) + tr ?)(#), it follows that
the set of F-characters of G is closed under addition.

Group representations and characters
IS
We now restrict our attention to the special case that the field F = C,
the complex numbers. We emphasize, however, that the subfield consisting
of the algebraic elements in C would work exactly as well, and in fact with
only minor modifications, most of what follows works for any algebraically
closed field of characteristic not dividing \G\.
Let us establish some notation. Fix a finite group G and choose a
representative set of irreducible C[G]-modules, Jt(C[G]) = {Mu ..., MJ.
Choose a basis in each M, and let 3E; be the resulting representation of
C[G]. Let Xt be the character afforded by 3Ej. It follows that the set Irr(G) =
{*i> • • • > Xk) is the set of all irreducible C-characters of G (that is, characters
afforded by irreducible representations). Henceforth, the word "character"
will mean C-character unless otherwise stated.
Since sums of characters are characters, it follows that / = £?= j n,x, is a
character whenever the n, are nonnegative integers which are not all zero.
Conversely, if / is any character of G afforded by a representation 3E
corresponding to a module K we can decompose V into a direct sum of irreducible
modules. It follows that / is the sum of the corresponding irreducible
characters. We have, in fact, / = £ nMi(V)x,.
Corollary 1.17(d) asserts that dim(C[G]) = X?=1(dim M,-)2. Since
dim C[G] = \G\ and dim M, = deg £, = x,(l), we obtain the fundamental
formula
|G|= £>,-(l)2.
i= 1
It seems natural at this point to ask how we can determine the integer k
purely group theoretically, without looking at representations. By Corollary
1.17(e), we have k = |^(C[G])| = dim Z(C[G]).
(2.4) theorem Let Jf i, jf2»• • •■» ^r be the conjugacy classes of a group
G. Let X, = £«*-, * e C[G]. Then the X, form a basis for Z(C[G]) and if
KtKj = Y,fl,jv^v> then the multiplication constants a,JV are nonnegative
integers.
Proof It is clear that the K; lie in Z(C[G]). Moreover, they are linearly
independent because they are sums of disjoint sets of elements. If z =
£ agg e Z(C[G]) and h e G, we have z = h~ 1zh = £ aggh. Comparing the
coefficients of gh on both sides, we obtain agh = ag. In other words, the
coefficients ag have the constant value a, for all g e if;. It follows that z =
X a(Ki and thus the X, span Z(C[G]).
To find aiJV, pick g e Jfv. Then aijv is the coefficient of g in KtKj. From
the definition of multiplication in a group algebra, this is | {(x, y)\x e Jf h
y e Jfy, xy = g}\. Since ai]v is the cardinality of a set, it is a nonnegative
integer. |

16
Chapter 2
(2.5) corollary The number k of similarity classes of irreducible
representations of G is equal to the number of conjugacy classes of G.
(2.6) corollary The group G is abelian iff every irreducible character
is linear.
Proof Let k be the number of classes of G. Then k = \ G | iff G is abelian.
Now \G\ = Yj=i Z;(l)2 and x,-(l) > 1 for all i. It follows that k = \G\ iff
Xi(l) = 1 for all i. The proof is complete. |
We have not yet proved that the & are distinct. To see this, we introduce
a little more notation. From the results of Chapter 1 we have the direct sum
C[G]= i-M;(C[G]).
i=i
Let 1 = £ et with et e M;(C[G]). Since M/C[G]) annihilates the module
Mi if i t6 j, we have 3E;(e;) = 0 in this case. It follows that 3£,<e,) = 3E;(1) = /.
Therefore, ^(e,) = 0 if i ^ j and x,(e,) = /;(1) =£ 0 and we conclude that the /,-
are distinct as functions on C[G]. Thus the /,- are also distinct as functions
onG.
We may now restate two of the earlier results in a slightly more convenient
form.
(2.7) corollary Let G be a group. Then | Irr(G) | equals the number of
conjugacy classes of G and
I Z(D2 = |G|.
Xelrr(G)
For certain very small groups, the information contained in Corollary 2.7
is sufficient to determine the irreducible character degrees. For instance, if
G = Z3, the symmetric group on three symbols, then G has exactly three
conjugacy classes and \G\ = 6. It follows that the /;(1) are 1,1, and 2.
Actually, the preceding argument with the e,'s yields more than the fact
that the Xis are distinct. A class function on a group G is a function, cp: G -» C
which is constant on conjugacy classes. All characters are class functions.
(2.8) theorem Every class function <p of G can be uniquely expressed in
the form
9= Z axK
Xelrr(G)
where ax e C. Furthermore, q> is a character iff all of the ax are nonnegative
integers and (^#0.

Group representations and characters
17
Proof The set of class functions of G forms a vector space over C whose
dimension is the number of classes of G. We claim that Irr(G) is a basis for this
space. Since |Irr(G)| = k = number of classes, it suffices to show that if
£ atXi = 0, then each a; = 0. This is immediate by evaluation at e;.
The second statement has already been proved. |
If i = Yj= i «iZ; is a character, then those & with nt > 0 are called the
irreducible constituents of /. In general, if ip is a character such that / — \p
is also a character or is zero, then \p is called a constituent of /. An important
consequence of Theorem 2.8 is the following.
(2.9) corollary Let 3E and 9) be C-representations of a group G. Then
3E and 9) are similar iff they afford equal characters.
Proof We already know that similar representations afford equal
characters.
Let V and W be C[G]-modules corresponding to 3E and 9) respectively.
Then 3E affords the character £ nMi(V)x,- and 9) affords £ nMi(W)Xi- If these
characters are equal, it follows that nMi{V) = nMi(W) for all i. Therefore,
V and W are isomorphic to identical direct sums of irreducible C[G]-
modules and thus are isomorphic and 3E and 9) are similar as required. |
For a particular group G, the irreducible characters are usually presented
in a character table: a (square) array of complex numbers whose rows
correspond to the /, and whose columns correspond to the classes Jft.
An example of a character table is the accompanying one for the symmetric
group Z4 on four symbols.
g: 1 (12) (1 2)(3 4) (12 3 4) (12 3)
icto)i
\Cl(g)\
Xi
ll
X.3
n
Xs
24
1
1
1
2
3
3
In this table, the classes are denoted by writing a representative element
g in its cyclic notation. (The reader is reminded that in a symmetric group,
two elements are conjugate iff they have the same cycle structure.) The sizes
of the centralizer C(g) and of the corresponding conjugacy class Cl(g) are

18
Chapter 2
given for convenience, although they are not, properly speaking, a part of
the table. Further character tables are given in the Appendix (page 287).
It is rather difficult to describe the process by which these tables are
constructed. Usually, various combinations of ad hoc arguments and general
theorems are necessary. As the student learns some of these theorems, he is
urged to try to construct some character tables on his own. The important
point is that it is very much easier to construct a character table than it is to
construct representations.
Theorem 2.8 tells us that every class function is a linear combination of
irreducible characters. For example, if G = E4 and (p(g) is defined to be the
number of points moved by g e G, then cp is a class function and is a linear
combination of %lt /2, • • •, Xs • Thus the row (0,2,4,4, 3) is a linear
combination of the five rows of the given table. It turns out that there is an easy method
for computing the coefficients, practically by inspection from the table.
This is done by using the so called "orthogonality relations," which we are
about to derive. These relations are also extremely useful in the construction
of character tables.
Before we leave Z4 we should comment on the fact that all of the character
values, which a priori are only known to be complex numbers, in fact turn out
to be integers. (Of course the xf(l), being the degrees of representations, are
always positive integers.) It is not always true that all character values are
ordinary integers, although frequently a large fraction of them are. It is true,
however, for all symmetric groups. All of this should become clear later.
The key to the orthogonality relations is to compute explicitly the
coefficients of the group elements in the e,'s in terms of the characters. To
do this we use the character p of G afforded by a representation corresponding
to the regular module C[G]°. This regular character p will be computed
in two ways.
(2.10) lemma IfgeGandg ^ 1, then p(g) = 0. Also p(\)= \G\.
Proof We must choose a basis for C[G]° to obtain a corresponding
representation. We simply take G, in some ordering, as the basis and let 5R be
the corresponding representation. If 5R(g) = (ay), then ay = 0 unlessgtg = gr
in which case ay = 1. Since p(g) is the number of gt satisfying g,g = gh the
lemma follows. |
Since p is a character of G, it may be expressed as an integer linear
combination of the Xi- We do this explicitly.
(2.11) LEMMA P = JJ-iZl(l)Zl-

Group representations and characters
19
Proof If V is any C[G]-module, it may be decomposed as a direct sum of
irreducibles. The character afforded by a corresponding representation
is£nM,(P0Xi- Now by Corollary 1.17(c) we know that nMi(C[G]°) =
dim M; = x,(l). The result follows. |
It is suggested that the reader use Lemma 2.11 and the character table of
Z4 to compute p explicitly for this group, and check the result against
Lemma 2.10.
(2.12) theorem e, = (1/IGI) L,eG U\)l,\g-l)g.
Proof Write et = Y,ag9- By Lemma 2.10, we have p(etg~l) = ag\G\.
Lemma 2.11 thus yields
ag\G\ = Y.xmj{eig-1).
j
Since
(0 if i*j,
wehavex/e.g-1) = xt(g~ '^y, where the Kronecker (5,7is0or 1 depending on
whether i and j are unequal or equal. We now have
aB\G\ = xi(l)Xi(g-1)
and the result follows. |
(2.13) theorem (Generalized Orthogonality Relation) The following
holds for every he G.
^Ixi(gh)xj(9~1) = ziM.
Proof The e{ lie in trivially intersecting ideals of C[G] and thus etej = 0
if i t6 j. Since 1 = £ e-p multiplication by et yields et2 = et. We now
substitute the formula of Theorem 2.12 into the equation e,ej = (5l7e; and
compare the coefficients of the group elements on both sides.
The coefficient of a fixed h e G on the right hand side is (d 0/| G |)/j( 1 )x;(h ~1)
and that on the left-hand side is
|G| gsG
The result now follows by equating these expressions and substituting h
for/T1. |

20
Chapter 2
By taking h = 1 in Theorem 2.13 we obtain
(2.14) corollary (First Orthogonality Relation)
rprr Exited-1) = <V
|G| 9€G
Since the expression %(g~') has come up several times, we digress for a
while to discuss its connection with %(g) and some related questions.
(2.15) lemma Let 3E be a representation of G affording the character / and
let g e G. Let n = o(g\ the order of g. Then
(a) 3E(g) is similar to a diagonal matrix diagta,..., ef);
(b) e," = 1;
(c) X(0) = le,and|x(g)| <;*(l);
(d) X(g~1) = M-
Proof The restriction of 3E to the cyclic group <g> is a representation of
<g> and hence it is no loss to assume G = <g>. By Maschke's theorem and
other results of Chapter 1, it follows that 3E is similar to a representation in
block diagonal form, with irreducible representations of G appearing as the
diagonal blocks. Since G = <g> is abelian, Corollary 2.6 asserts that its
irreducible representations have degree 1, and thus 3E is similar to a diagonal
representation. Now (a) follows, and we may assume that 3E is diagonal. We
have / = 3E(g") = X(gf = diag(e1",..., e/). Therefore (b) is proved. It follows
that |e,| = 1 and |£e,| ^ Yje,-| = / = *0). It is clear that x(g) = £ e, so
that (c) follows. Now 3E(gf_1) = 3E(g)_1 = diag(e1_1,..., Ej-'1) _so that
Xig"1) = Ze._1- Since |e,| = 1, we have e,"1 = e^ and i(g~l) = i(g). The
proof is complete. |
Combining Corollary 2.14 and Lemma 2.15(d), we obtain
I"! 06G
This suggests the following definition.
(2.16) definition Let (p and 9 be class functions on a group G. Then
0,9] = -!- 2>(3)%)
|G| gsG
is the inner product of q> and 9.
Some of the obvious properties of this "inner product" are
(a) [>,8] = E5T£I;
(b) [<p, <p] > 0 unless <p = 0;

Group representations and characters
21
(c) [£-!<?! + c2(p2, 9] = CjE^j, 9] + c2[<p2, 9];
(d) O, CjSj + c292] = c7[<p, 8,] + c2[_q>, S2].
Therefore, [ , ] has all of the properties usually used to define an inner product
in linear algebra and analysis. (In fact, this makes the space of class functions
into a finite dimensional Hilbert space.)
We know that Irr(G) is a basis for the space of class functions and it is
the content of the orthogonality relation that it is, in fact, an orthonormal
basis, that is,
[Zi.Zy] = <V
This yields the promised method for expressing an arbitrary class function in
terms of the irreducible characters; for if [<p, /,] = q, then (p = £ c,Xj.
Another application of the inner product is to determine instantaneously
whether or not a given character is irreducible.
(2.17) corollary Let x and \ji be (not necessarily irreducible) characters
of G. Then [/, i/f] = [i/f, /] is a nonnegative integer. Also / is irreducible iff
[*. X] = 1-
Proof We have / = £ n,-/,- and \p = £ »>,•/,-, with all n{ and m; non-
negative integers. Then [/, t/f] = £ n,mj = |>, /] and [*, *] = £ n;2. The
result is now immediate, since / is irreducible exactly when one nt = 1 and
all other nt = 0. |
The following "second orthogonality relation" is derived from the first
and so imposes no new necessary condition for an array of complex numbers
to be a character table. Nevertheless, it is often extremely useful in the
construction of character tables and in the extraction of information from them.
(2.18) theorem (Second Orthogonality Relation) Let g,heG. Then
I x(g)W) = o
XeIrr<G1
if g is not conjugate to h in G. Otherwise, the sum is equal to |C(g)|.
Proof Let gu g2,..., gk be representatives of the conjugacy classes of
G. Let X be the k x k matrix whose (i,j) entry is Xi(9j)- (In other words, X is
the character table, viewed as a matrix.) Let D be the diagonal matrix .with
entries (5iy | Jf; | where Jf,- is the conjugacy class Cl(g,). The first orthogonality
relation asserts that
\g\s,]= Y.xte)xjg)= El^vlzitev)JW
gsG v=l

22
Chapter 2
This system of k2 equations may be replaced by the single matrix equation
|G|/ = XDXT,
where / is the identity matrix and the superscript denotes transpose.
Since a right inverse for a square matrix is necessarily also a left inverse,
this yields
|G|/ = DXTX.
We now write this as a system of equations and obtain
\G\8„ = l.\jrl\xJgdx*(gj)-
V
Since | G | /1 Jf; | = | Qsi) I, this yields
I xtiX)=\Qg,)\&i},
*eIrr(G)
which is the desired result. |
In the character table for Z4 that was given earlier, the size of each con-
jugacy class was given. We see now that this information is derivable from the
body of the table. It was given only for ease of computation of inner products.
As a check on our results so far, observe what happens if we take h = 1 in
the second orthogonality relation. The reader should recognize what results
as a combination of Lemmas 2.10 and 2.11.
Let £ £ C be the field of algebraic numbers. By Lemma 2.15, all of the
character values %(g) e E for / e Irr(G) and g e G. How is Irr(G), which is a set
of functions G -» E related to the set of irreducible £-characters of G, which
we denote by Irr£(G)? In fact these sets are equal.
Suppose 3E is an irreducible £-representation of G which affords / e Irr£(G).
Now X may be viewed as a C-representation and thus / is a character (that is,
a C-character) of G. Since the entire development of character theory up to
this point would have been the same over £ as over C and since / e Irr£(G), it
follows that [/, /] = 1. However / is a C-character of G and it follows from
Corollary 2.17 that / elrr(G). Thus Irr£(G) £ Irr(G).
We also have that both Irr^G) and Irr(G) have the same cardinality,
namely that of the set of conjugacy classes of G. Therefore Irr^G) = Irr(G).
The point of this digression is to suggest that there is something
"absolute" about a character table. It is not entirely an artifact of our choice of
the particular field C.
Another consequence of this argument which is sometimes useful is that
if i e Irr(G), then / is afforded by an £-representation of G. This type of
consideration will be discussed much more fully in Chapter 9. Until then, we
resume our convention that "character" means "C-character."

Group representations and characters
23
A great deal of information about a group can be recovered from its
character table. In particular, all of the normal subgroups of G can be found.
A normal subgroup is a union of conjugacy classes and the word "found"
in the preceding sentence means that those sets of conjugacy classes whose
unions form subgroups can be listed. In particular, the orders of all normal
subgroups and the inclusion relations among them can be determined.
(2.19) lemma Let 3E be a C-representation of G which affords the
character /. Then g e ker 3E iff %{g) = x(l).
Proof If g e ker 3E, then X(g) = I = 3E(1) and %(g) = /(l). Conversely, by
Lemma 2.15, xdd) = gj + • • • + ef, where e; is a root of unity and/ = /(l).
Since \e-t\ = 1, the equation %(g) =f forces e; = 1 for all i. Now 3E(g) is similar
to diagta,... ,ef) = I and therefore X(g) = I and the proof is complete. |
(2.20) definition Let x be a character of G. Then ker / = {g e G \ %(g) =
(2.21) lemma Let x be a character of G with / = £ n,Z; for /, e Irr(G).
Then ker / = f]{ker Xi\nt > 0}. Also f){ker Xi\ I < i < k} = I.
Proof Since \xiig)\ ^ Z,(l) by Lemma 2.15, xig) = x(l) forces Xi(g) =
X,(l) whenever n-t ^ 0. The reverse inclusion is trivial and the first assertion
follows.
To prove the second statement, consider the regular character p. By
Lemma 2.10, ker p = 1 and the result follows. |
The normal subgroups N{ = ker /; can be found by inspection from the
character table of a group G. We claim that every normal subgroup is the
intersection of some of the JV,'s and thus can be found from the character
table. To see this, let N o G and let 5R be the regular representation of the
group G/N so that ker 5R = N/N. Now view 5R as a representation of G with
kernel N and let / be the corresponding character of G. Then N = ker 5R =
kwZ=n{tfil[X.Z.]*0}.
Given a normal subgroup NofG (where "given" means listing the classes
Jf; which it contains), we may calculate | N \ from the character table using
|AM = £{|Jf,||jr,£N}. (Recall that |Cl(g)| = |G: Qg)\ and |C(g)| =
Xx€irr<G) \x(g)\2 so tnat \$r-,\ is determined from the character table.)
It follows from this discussion that G is simple iff ker / = 1 for all non-
principal x e Irr(G) and therefore simplicity of a group can be easily
determined from its character table.

24
Chapter 2
The group G is solvable iff it has a chain of normal subgroups, 1 = M0 £
M( £ • • • £ M„ = G such that | M,: Mf_ , | is a prime power for all i, 1 <
i < n. Since the M, can be located and their orders determined from the
character table of G, it follows that the table determines solvability or non-
solvability of G.
This may be a good place to remark that the character table of G does not
determine G up to isomorphism. In fact if p is any prime, there are two non-
isomorphic nonabelian groups of order p3. These two groups have identical
character tables.
Let N o G. It seems natural to ask whether the character tables of N and
of G/N can be calculated from that of G. The answer is "no" for N but "yes"
for G/N.
We have already observed that there is a one-to-one correspondence
between representations of G/N and representations of G with kernel
containing N. Furthermore, under this correspondence, irreducible
representations correspond to irreducible representations. This situation may be
interpreted in terms of characters as follows.
(2.22) lemma Let N <i G.
(a) If x is a character of G and N c ker /, then / is constant on cosets of
N in G and the function x on G/N defined by x(Ng) = x(0) is a character of
GIN.
(b) If x is a character of G/N, then the function / defined by x(g) = t{Ng)
is a character of G.
(c) In both (a) and (b), / e Irr(G) iff x e Irr(G/N).
Usually, we shall identify / and %■ Under this identification, we have
\xx(G/N) = {x e Irr(G)| N £ ker /}. To demonstrate what is happening here,
let us consider the example G = Z4 and N the normal subgroup of order 4.
The classes of G which are contained in N are the identity and Cl(( 1 2)(3 4)).
Referring to the character table on p. 17, we see that the irreducible characters
/,• of G with N £ ker /,- are / (, Xi ■, and X3 and so under above identification, we
haveIrr(G/JV)={x„x2,x3}.
In order to write the character table for G/N we need to know its con-
jugacy classes. If Jf is a class of G, then Jf, its image in G/N, is a class of G/N.
However, distinct classes of G may have equal images in G/N. This poses no
problem since if g, hsG, then g and h are conjugate in G = G/N iff x(g) = x(h)
for all / e Irr(G/./V). (The second orthogonality relation, applied to G/N
proves this.)
The part of the character table of G = E4 corresponding to the characters
of G/N is

Group representations and characters
25
xr-
Xi-
Xi-
■*\
l
l
2
.w\
l
-l
0
■*\
l
l
2
■W\
1
-1
0
■^'s
1
1
-1
Observe that columns 1 and 3 are identical, as are columns 2 and 4. Deleting
repeats, we obtain the character table
1 1 1
1-1 1
2 0-1
for G/N. (In this case G/N sl3.|
The preceding discussion provides a second method for computing | N \
from the character table of G. Namely, by using the fact that
|G:A/| = X{*(l)2|zelrr(G/JV)}
= I{/(D2|/elrr(G) and JVsker*}.
By Corollary 2.7, a group is abelian iff all of its irreducible characters are
linear. It follows that given N o G, the character table of G determines
whether or not G/N is abelian. There is no known way to determine from the
table whether or not N is abelian.
(2.23) corollary Let G be a group with commutator subgroup G'. Then
(a) G'= fliker/.|/eIrr(G),/(l) = 1};
(b) | G: G' | = the number of linear characters of G.
Proof If / is a linear character of G, then /. is a homomorphism
into the abelian multiplicative group of C. It follows that G' £ ker X.
Since G/G' is abelian, all / e Irr(G/G') are linear and thus Irr(G/G') =
{/eIrr(G)|/(l) = 1}. (This equality, of course, depends on the
identification of characters of G/G' with characters of G.) Finally, for any JV <i G,
we have N = f){ker %\% e Irr(G) and N £ ker /} and hence (a) follows.
The number of linear characters of G is equal to the total number of
irreducible characters of the abelian group G/G' and hence equals |G/G'|.
The proof is now complete. |
The information in Corollary 2.23(b) is useful for finding the set of
character degrees of a group G. For instance, if G is a nonabelian group of
order 27, then \G: G'\ = 9 and G has exactly 11 conjugacy classes. By the

26
Chapter 2
preceding, G has exactly nine linear characters and two nonlinear irreducible
characters / and \p. We have
27 = |G| = 9 + z(l)2 + ^(l)2.
Since the only way that 27 — 9 = 18 can be written as a sum of two squares is
32 + 32, it follows that jrfl) = 3 = i/<l).
Before leaving the discussion of character tables and factor groups, we
mention an amusing result. The following could be proved without
characters but it is somewhat tricky to do so.
(2.24) corollary Let g e G and N <i G. Then | CG/N(Ng) \ < \ CG(g) \.
Proof From the second orthogonality relation, we have
\CGIN(Ng)\= £ |z(Ng)|2 = £{lx(0)l2|xeIrr(G),JVckerz}
XeIrr(G/N)
< E \X(9)\2 = \CG(g)\. |
Xelrr(G)
We now discuss the connections between characters and the center of a
group.
(2.25) lemma Let 3E be an irreducible C-representation of G of degree n.
Suppose A is an n x n matrix over C which commutes with 3E(g) for all g e G.
Then A = al for some a e C.
Proof Let M be the n-dimensional row space over C so that M is an
irreducible C[G]-module via m-a = m3E(a) for meM and aeC[G]. Let
9: M -» M be defined by m9 = mA. Then
9eEC[G](M)=Cl
and 9 = a • 1 for some a e C. The result follows. |
(2.26) definition Let x be a character of G. Then Z(/) = {g e G11 x(g) | =
Z(l)}-
If H £ G and 3E is a representation of G, then its restriction to H, denoted
3EH, is a representation of H. Similarly, the restriction Xh °f a character / of G
to H is a character of H and we can write
Xh = Z n^
l/(€lrr(H)
for suitable integers n^. Note that if Xh e Irr(H), then / e Irr(G). Of course, the
converse of this statement is false.

Group representations and characters
27
(2.27) lemma Let x be a character of G and let Z = Z(x) and/ = x(l). Let
3E be a representation of G which affords /. Then
(a) Z = {geG\X(g) = el for some eeC};
(b) Z is a subgroup of G;
(c) Xz = ft- f°r some linear character k of Z;
(d) Z/ker / is cyclic;
(e) Z/ker / c Z(GAer z).
Furthermore, if x £ Irr(G), then
(f) Z/ker z = Z(GAer z).
Proof By Lemma 2.15, H(g) is similar to diag^,..., ef), with |e;| = 1,
1 < i < f. Since x(g) = X fy, it follows that |/(g)| =/ iff all £j are equal. Since
the only matrix similar to el is el itself, conclusion (a) follows.
Define the function k: Z -> C by H{g) = 1(g)/ for geZ. It follows for
g,heZ that 3E(g/i) = k(g)k(h)I and hence Z is a subgroup and k is a homomor-
phism (linear character) of Z. We have that xio) = P\o) for g e Z and (b) and
(c) have been proved.
Clearly, ker x = ker k and thus Z/ker x is isomorphic to the image of k,
a finite multiplicative subgroup of the field C. This subgroup is necessarily
cyclic and (d) follows. Also, ker x = ker 3E and 3E(Z) £ Z(3E(G)) and (e) is an
immediate consequence.
Finally, if g(ker x) e Z(G/ker xl then X(g) e Z(3E(G)). If x e Irr(G), then by
Lemma 2.25, we conclude that X(g) = el for some e e C. Now (f) follows from
(a) and the proof is complete. |
(2.28) corollary Let G be a group. Then
Z(G)=f){Z(z)lzeIrr(G)}.
Proof Since (TAG) ker z)/ker x £ Z(G/ker /), it follows from Lemma
2.27(f) that Z(G) £ Z(/). Conversely, suppose g e Z(/) for every / e Irr(G).
It follows that g(ker(/)) e Z(G/ker /) and thus for any xeG, the commutator
[0,*] = g'^-tgxekerx-
Thus [#, x] e f){ker x\xe !"•((?)} = i and g commutes with x. Since xeG
was arbitrary, we have geZ^G). |
It is apparent from Corollary 2.28 that Z(G) can be located from the
character table of G. It follows that it can be determined from the table
whether or not G is nilpotent. This is done by finding Z(G), then finding the
character table of G/Z(G), and iterating this process. The sequence of
subgroups of G which results is the upper central series and G is nilpotent iff this
sequence reaches G.

28
Chapter 2
Some information about character degrees can be obtained using
Lemma 2.27(c). We need a lemma first.
(2.29) lemma Let H c G and let / be a character of G. Then
[Zh,Zh]£|G:H|[x,Z]
with equality iff %(g) = 0 for all g e G — H.
Proof We have
\H\[Xb,Xh]= I\x(h)\2< l\x(9)\2 = \G\h,Xl
hell (EC
since \x(g)\2 Sr 0 for geG — H. Equality thus holds iff x(g) = 0 for all
geG — H. The result follows. |
(2.30) corollary Let x e Irr(G). Then /(l)2 < | G: Z(/)|. Equality occurs
iff x vanishes on G — Z(/).
Proof By Lemma 2.27(c), we have XzM = X(W and thus [_XzM, Xuxd =
X(lKl A] = z(l)2. Therefore
X(l)2<|G:Zto|[z,Z] = |G:Z(Z)|
with equality iff / vanishes on G — Z(/). |
We already knew, of course, that /(I)2 < | G | for / e Irr(G). We now have
the slight improvement that /(l)2 < \G: Z(G)|. Equality can occur here, and
when it does, TXy) = Z(G) and / vanishes on G — Z(G). It has been
conjectured that only in a solvable group is it possible to have x(l)2 = \G: Z(G)|
with x £ Irr(G). As of this writing, the question is still open. Observe that the
nonabelian groups of order 27 which were discussed earlier give examples
where equality occurs.
(2.31) theorem Suppose that / e Irr(G) and that G/Z(/) is abelian. Then
|G:Z(z)| = z(l)2.
Proof It suffices to prove that / vanishes on G — Z(/). Let geG — Z(/).
Then by Lemma 2.27(f), we have that there exists h e G with g~1h~1gh$ ker /.
However, since GjTXx)is abelian, we haveg~ lh~lgh = ze 2Ij). Now, if 3E is
a representation of G which affords /, then 3E(z) = el and e j= 1 since z £ ker 3E.
We have X(gz) = X(g)X(z) = eX(g) and thus /(gz) = e/(g). However, gz =
h~ lgh and so x(gz) = x(g\ Since ex(g) = x(d) and e ^ 1, we have x(g) = 0 as
desired. |
A character / of G is said to be faithful if ker / = 1. Every group has a
faithful character, namely its regular character p; but not every group has a
faithful irreducible character.

Problems
29
(2.32) theorem (a) If G has a faithful irreducible character, then Z(G) is
cyclic.
(b) If G is a p-group and Z(G) is cyclic, then G has a faithful irreducible
character.
Proof (a) Let / e Irr(G) be faithful. By Lemma 2.27(f), Z(G) = Zfa)
and by part (d) of that lemma, Z(/) is cyclic.
(b) Since G is a p-group, it follows that if 1 ^ N <i G, then N n Z(G)
t^ 1. Now let Z be the unique subgroup of order p in the cyclic group Z(G),
so that Z £ N for every nontrivial normal subgroup iV of G. Since
P){ker xl/eIrr(G)} = 1, it follows that Z £ ker / for some /elrr(G). We
conclude that ker / = 1 and the proof is complete. |
Problem 2.19 provides an example to show that the full converse of
Theorem 2.32(a) is not true.
Problems
(2.1) (a) Let 3E be an irreducible F-representation of G over an arbitrary
field. Show that Y^gea %(g) = 0 unless 3E is the principal representation.
(b) Let H £ G and g e G be such that all elements of the coset Hg are
conjugate in G. Let / be a C-character of G such that [/H, 1H] = 0. Show
that X(g) = 0.
Hint (b) Compute the trace of £,,€ H H(hg), where 3E affords /.
In the following, all characters are over C.
(2.2) (a) Let % be a character of G. Show that / is afforded by a
representation 3E such that all entries of 3E(g) for all g e G lie in some field F £ C with
|F: 0| < oo.
(b) Let e = e2nil", where n= \G\ and let / be a character of G. (Note that
X(g)eQ[e] for all ge G by Lemma 2.15.) Let a be an automorphism of the
field Q[e] and define /": G -» C by /"(g) = /(g)". Show that /" is a character
and that f e Irr(G) iff / e Irr(G).
(2.3) Let i be a character of G. Define det /: G -» C as follows. Choose 3E
affording / and set
(det X)(g) = det X(g).
Show that det / is a uniquely defined linear character of G.
(2.4) (a) Let G be a nonabelian group of order 8. Show that G has a unique
nonlinear irreducible character /. Show that x(l) = 2, x(z) = —2, and
X(x) = 0, where z e G - {1} and x e G - G'.

30
Chapter 2
(b) If G S £>8, show that det / ¥= 1G-
(c) If G S £>8, show that det z = 1G.
Hint Show that ker(det /) contains all elements of order 4. Use Lemma
2.15.
Note Although D8 and Q8 have identical character tables, the map
det: Irr(G) -» Irr(G) is not the same for both groups.
(2.5) (a) Find a real representation of Ds which affords the charaoter / of
Problem 2.4(a).
(b) Show that this cannot be done for the group Q8.
(2.6) Let x, i> be characters of G. Define/xtft/G--♦ 'C'"byi(^)(gr) = .jigMg).
(a) If 1^(1) = 1, show that xi> is a character.
(b) If tA(l) = 1, show that rfi e Irr(G) iff % e'Irr(G).
(c) If ip = x (that is, ip(g) = /[^.and^l) > U,^how'that # £ irr(G).
Note In Chapter 4 we will showtthat. x^ is always a character.
(2.7) Let G be abelian and write G -= Hrr(G).
(a) Show that G is an abelian ..group under the multiplication of
Problem 2.6.
(b) If H s G, let H1 = {1 e G|H^iker 1}. Show<that 1 is a bijection
from the set of subgroups of G onto the-setof subgroupstof((j.
(c) Show that G ^G.
Hints There is a natural isomorphism of;G onto G. lUsetthis for(b).
For (c), use the fundamental theorem of abelian1 groups.
(2.8) Let x be a faithful character of G. Show that H £ G is abelian: iff every
irreducible constituent of Xh is linear.
(2.9) (a) Let / be a character of an abelian group A. Show
E IzWi2 > \a\x{\\
xeA
(b) Let A £ G with A abelian and \G:A\ = n. Show that *(1) <, n for
allzelrr(G).
(2.10) Suppose G = (J"= j Ah where the y4; are abelian subgroups of G and
A{ n Aj = 1 if i t6 y.
(a) Let/eIrr(G). Show that if/(l) > l,then/(l) > \G\/(n - 1).
(b) IfGisnonabelian^henl/ljl <, n — 1 for each i and n - 1 ^ (|G|)1/2.
Hints For (a), bound £9€G |/(g)|2 using Problem 2.9(a). For (b), use
Problem 2.9(b).

Problems
31
(2.11) Let g e G. Show that g is conjugate to g l in G iff %(g) is real for all
characters / of G.
Note An element of a group which is conjugate to its inverse is called a
real element. If G has any real elements other than 1, then G must necessarily
have even order.
(2.12) Let | G | = n and let g e G. Show that i{g) is rational for every
character i of G iff g is conjugate to gm for every integer m with (m, n) = 1.
Hints Let e be a primitive nth root of 1 in C and let E = Q[e]. Let <§ be
the Galois group of E over Q. Given (m, n) = 1, show that there exists a e 'S
with %(gm) = /(g)" for all g e G and all characters /. Conversely, for every
er e 'S, there is an m such that this formula holds.
(2.13) Let | G' | = p, a prime. Assume that G' £ Z(G). Show that
X(1)2 = |G:Z(G)|
for every nonlinear / e Irr(G).
(2.14) Let H c G' n Z(G) be cyclic of order n and let m be the maximum of
the orders of the elements of G/H. Assume that n is a prime power and show
that | G | > n2m.
Hints Choose /elrr(G) with Hnker / = 1. Let k = det / (as in
Problem 2.3). We have /H = #(l)/z with fi e Irr(H). Using k and fi, show that
Z(l) ^ n. Finish the proof using Problem 2.9(b).
Note The assumption on n can be removed using Corollary 5.4.
(2.15) Let x £ Irr(G) be faithful and suppose H £ G and /H e Irr(H). Show
that
CG(H) = Z(G).
(2.16) Let H £ G and let / be a (possibly reducible) character of G which
vanishes on G — H. Assume either that H = 1 or that G is abelian. Show
that |G:H| divides/(l).
Hint Let 1 be an irreducible constituent of %H. Under either hypothesis,
find fi e Irr(G) with fiH = k. Compute \j, /*] and conclude | G: H11 {jH, 1].
./Vote A natural common generalization of the situations H = 1 and G
is abelian is H c Z(G). Is the result true under the hypothesis H £ Z(G)?
(2.17) Let A < G be abelian and assume there exists / e Irr(G) with x(l) =
| G: A |. Show that G has a nontrivial normal abelian subgroup.
Hint Show that / vanishes on G — A.

32
Chapter 2
(2.18) Let A o G and suppose A = CG(a) for every a j= I, aeA. Assume
further that G/A is abelian. Show that G has exactly (\A\ - 1)/|G: X|
nonlinear irreducible characters and that these all have degree equal to
\G:A\ and vanish on G - A.
Hints Let k = | Irr(G)|. By counting classes, show that
k£l+(\A\- l)/\G:A\ + (\G\-\A\)/\A\.
Using characters, show that
k>\G:A\ + (\G\-\G:A\)/\G:A\2.
Use Problem 2.9(b).
Note The group G of Problem 2.18 is a special case of a Frobenius
group. The character theory of such groups will be discussed more fully later.
Observe that although the hypotheses of 2.18 are very special, this situation
does arise frequently. Some examples are AA and nonabelian groups of order
pq where p and q are primes with p\(q — 1).
(2.19) Let£= <x1,x2,x3,x4> be an elementary abelian group of order 16.
Let P = <y> be cyclic of order 3. Let P act on E by
Xi = *2i x2 = *1*2> -*3 = x4i x4 = *3*4-
Let G be the semidirect product E x P. Show that Z(G) = 1 but that G does
not have a faithful irreducible character.
Hint The smallest possible degree for a faithful character of E is 4.
(2.20) Let 3E and 9) be irreducible C-representations of G and define the
functions a,J(g) and b^g) by H{g) = (a,jg)) and 9)(gf) = {bi}{g)). Write
SPqrs= ZflP«(^«(9-1)-
Show that Sp,„ = 0 if 3E and 9) are not similar. If 3E = ?), show that Spqrs = 0
unless p = s and q = r in which case Sp„p = | G |/deg 3E.
Hint Let P„ = £,.<; X(g)£,r?)(g~') where Ev is the (deg X) x (deg 9))
matrix with all entries zero except the (q, r) entry which equals one. Note that
XPqr = P,r9). Use Schur's lemma. For the last statement use Lemma 2.25
and compute tr(P,r).
Note The results of this problem are called the Schur relations. They
can be used to give another proof of the orthogonality relations.

Characters and integrality
One of the most celebrated applications of character theory to pure group
theory is Burnside's theorem which asserts that a group with order divisible
by at most two primes is solvable. The proof of this theorem (and much of the
rest of character theory) depends on properties of algebraic integers. We begin
by establishing some of the most basic of these properties.
(3.1) definition An algebraic integer is a complex number which is a root
of a polynomial of the form
x" + an-.!*"-1 + ■■■ + a0,
where af e Z for 0 < i < n — 1.
(3.2) lemma The rational algebraic integers are precisely the elements
ofZ.
Proof If a e Z, then a is a root of the polynomial x — a and thus is an
algebraic integer. Conversely, let r/s be an algebraic integer with r,seZ. We
may assume that (r, s) = 1. We have
(r/sf + a„^(r/sf-1 +-.. + ao = 0.
Now multiply by s" and rearrange terms to obtain
r" = -s(a„_1r""1 + a„_2sr""2 + ••• + aos""1).
We conclude that s\r". However, since (r, s) = 1, this yields s = +1 and
r/s e Z as desired. |
33

34
Chapter 3
Frequently, the word "integer" is used to mean an algebraic integer, and
the elements of Z are referred to as " rational integers." One of the most
important properties of the set of algebraic integers is that it is a ring. In other
words, sums and products of integers are integers. This fact seems surprising
from the definition, but it is not hard to prove indirectly. This we proceed
to do.
(3.3) lemma Let X = {<*!,..., at} be a finite set of algebraic integers.
Then there exists a ring S satisfying
(a) ZzSz C:
(b) XsS;
(c) there exists a finite subset, Y of S such that every element of S is a
Z-linear combination of elements of Y.
Proof The integer a,- satisfies an equation of the form
«"' = ».
where f is a polynomial of degree n{ — 1 with coefficients in Z. Let Y =
{ai'otj2 • • • afcfc|0 <, rt < nt - 1} and let S be the set of all Z-linear
combinations of elements of Y.
Using the equation a."' = /((a,), any power of a.{ may be written as a Z-
linear combination of 1, a,-, a,2,..., a"'~'. It follows from this that the product
of any two elements of Y lies in S and hence S is a ring. All of the properties
claimed for S are now clear. |
Condition (c) of the above lemma may be paraphrased by saying that S is
finitely generated as a Z-module. We now prove a strong converse to Lemma
3.3.
(3.4) theorem Let S be a ring with Z £ S £ C. Suppose that S is finitely
generated as a Z-module. Then every element of S is an algebraic integer.
Proof Let se S and let Y = {yu ...,y„}zS have the property that
every element of S is a Z-linear combination of elements of Y. We then have
Wi = Z auyj
j
for all i, with ayeZ. Let A be the matrix (ay) and let v be the column,
colfj!,..., y„). Then
Av = sv
and thus s is a root of the polynomial
f(x) = det(x/ - A).
It follows that s is an algebraic integer and the proof is complete. |

Characters and integrality
3S
(3.5) corollary Sums and products of algebraic integers are algebraic
integers.
Proof Let a and /? be algebraic integers. By Lemma 3.3, there exists a
ring S with Z z S £ C such that a, /? e S and S is finitely generated as a
Z-module. Since a + /? and a/? e S, it follows from Theorem 3.4 that they are
algebraic integers. |
(3.6) corollary Let / be a character of a group G. Then %(g) is an
algebraic integer for all g e G.
Proof By Lemma 2.15, we know that %(g) = gj + ■■■ + ef, where the
gj are roots of a polynomial of the form x" — 1, and therefore are algebraic
integers. The result now follows. |
We can now see the reason for the assertion made in Chapter 2 that all of
the entries in the character table of the symmetric group £„ lie in Z.lfge £„
and m is relatively prime to o(g), then gm and g have identical cycle structures
and therefore these elements are conjugate in £„. It follows from Problem 2.12
that %(g) is rational for all / e Irr(£„). Since %(g) is an algebraic integer,
Lemma 3.2 yields that %(g) e Z as claimed.
Let G be a group and / e Irr(G). We wish to define a function co depending
on /, from the center of the group algebra C[G] into C. Let 3E be any
representation which affords /. If z e Z(C[G]), then we may conclude from
Lemma 2.25 that X(z) = el for some g e C. Observe that since the only
matrix similar to el is el itself, the complex number g does not depend on the
choice of the particular representation affording /. We now define co by
setting co(z) = g. In other words
3E(z) = co(z)I
for all z e Z(C[G]). We shall often write co = cox in order to emphasize the
dependence of co on /.
Since 3E is an algebra homomorphism, it is easy to see that co is also a
homomorphism. In particular, co is C-linear and hence to determine co on
Z(C[G]), it suffices to calculate its values on a basis. Such a basis is given by
the class sums for»the conjugacy classes of G. Let Jf be a class with sum
K e C[G] and let g e Jf. Calculation of traces in the equation 3E(X) = co(K)I
yields
xHMK) = X(K) = X x(x) = \^\x(g)

36
Chapter 3
and thus
Zte)|JT|
coJK) =
Z(D
Note that it follows from this formula that the functions a>x are determined by
the character table of G.
(3.7) theorem Let /elrr(G) and let K be a class sum in C[G]. Then
cox(K) is an algebraic integer.
Proof Let Jf u ..., Jf k be the classes of G, with corresponding class
sums Ku...,Kk. By Theorem 2.4, we have KtKj = £v aiJvKv where
a0-v e Z. Since co = cox is an algebra homomorphism from Z(C[G]) to C, we
have
V
Let S be the set of all Z-linear combinations of the co(X(). It follows that S is
closed under multiplication. Since eo(l) = 1, it follows that Z z S z C and
Theorem 3.4 applies. All of the elements of S are therefore algebraic integers
and the proof is complete. |
It should be emphasized that the fact that x(g)|Cl(g)|/x(l) is an algebraic
integer does not follow from the fact that %(g) is integral since division of
an integer by an integer does not usually result in an integer.
We proceed now toward Burnside's solvability theorem. The essence of
the argument is contained in the next result.
(3.8) theorem (Burnside) Let / e Irr(G) and let Jf be a conjugacy class
of G with g e Jf. Suppose that (/(1), | Jf |) = 1. Then either g e Z(/) or else
1(9) = 0.
Proof We know that x(gf)|Jf |/x(l) is an algebraic integer. Since (/(l),
|jf |) = 1, we may choose rational integersu and v so that m/(1) + v\Jf\= 1.
Thus
x(9)(i - «x(i)) = VM\*\
z(D zd)
is an algebraic integer. Since u%(g) is also integral, it follows that a = x(g)/x(l)
is an algebraic integer. Suppose that g£Z(x), so that \iig)\ < z(l) and
|a|<l.
Now let n = o[g) and let E be the splitting field for the polynomial x" — 1
over Q in C so that a e E. Let ^ be the Galois group of £ over Q. Since x(g) is a
sum of x( 1) roots of unity, so is x(g)a for each a e <§. It follows that | /(gf)" | ^ /(1)

Characters and integrality
37
and I a" I <. 1 for a e 0. We have then
n<
< i.
For each a e CS, a." satisfies the same rational polynomials that a satisfies
and hence is integral. Therefore /? = ]~[ a" is an algebraic integer. However, /?
is clearly fixed by all a e <& and therefore /? e Q by elementary Galois theory.
It follows from Lemma 3.2 that /? e Z. Since | /? | < 1, we have /? = 0 and hence
a." = 0 for some er. Therefore 0 = a = i(g)li(V) and /(g) = 0. The proof is
complete. |
(3.9) theorem Let G be a nonabelian simple group. Then {1} is the only
conjugacy class of G which has prime power size.
Proof Suppose geG, \Cl(g)\ = p", and g * 1. Let /elrr(G), x * lc-
Then ker / = 1 since G is simple and Z{/) = Z(G) = 1 since G is nonabelian.
Thus if pJ(x(l), then x(g) = 0 by Theorem 3.8. Now
0 = P(g) = I x(Mg) = i + Z x(Dx(0)-
X€lrr(G) X€lrr(G);pU(l)
We have — 1 = pa, where
the sum being taken over / e Irr(G) where p | /(1). It follows that a = — 1/p is
an algebraic integer and this violates Lemma 3.2. |
(3.10) theorem Let \G\ = p"qb, where p and q are primes. Then G is
solvable.
Proof Use induction on \G\. We may assume |G| > 1 and choose a
maximal proper normal subgroup N. If N > 1, then by the inductive
hypothesis, N and G/N are solvable and thus G is solvable and the result follows.
Suppose then N = 1, so that G is simple. Let P?^ 1 be a Sylow subgroup
of G. We may choose geZ(P), g # 1. Then |C%)| = |G:C(g)| divides
|G: P\, which is a prime power. It now follows from Theorem 3.9 that the
simple group G is abelian and the proof is complete. |
We now obtain some strong results about the degrees of the irreducible
characters of a group G. The fact is that /(1) 11G: Z(/) | for / e Irr(G). We shall
first prove the weaker statement that the irreducible character degrees divide
the group order. This proof is much less complicated and serves to motivate
the stronger proof.

38
Chapter 3
(3.11) theorem Let x e Irr(G). Then /(1) 11G |.
Proof From the first orthogonality relation we have
\G\= ZxigMg'1).
geG
We wish to rewrite this equation in terms of cox. Let Jf 1; Jf 2, • • •» ^k be the
classes of G, with class sums Kt and representative elements gr We have then
k k
\g\= Y\^i\x(g-M9i'l)= IxVMKMgr1),
■=i i=i
where co = cox. This yields
|G|/z(l) = 2>(K,M?f1),
which is an algebraic integer. Since |G|//(1) is rational, it lies in Z and the
result follows. |
(3.12) theorem Let x e Irr(G). Then /(1) 11G: Z(/) |.
Proof Since / may be viewed as a character of G/ker /, it is no loss to
assume that ker / = 1. Under this assumption, Z(G) = Z{y).
For x, yeG, define x = y if there exists zeZ = Z(G) such that x is
conjugate to yz. It is easy to check that = is an equivalence relation and thus
partitions G into equivalence classes. We claim that | x(x) | is constant as x runs
over one of these classes. To see this, observe that Xz = xWh where X is a
faithful linear character of Z, and that x(yz) = Kz)x(y) for z e Z and yeG. If
x = y, then xg = yz for some z and x(x) = X(z)x{y). Since | l(z) | = 1, the claim
follows.
Let #!, #2* • • •, ^r be those ( = )-classes on which / does not vanish. We
have then
|G|= Ilx(9)l2= i^iWxig-M2,
geG i=l
where the gt are representatives for the #,. We claim 1^1 = |Cl(gf,-)||Z|.
Clearly, every x e #,- is of the form yz where y e Cl(gt) and z e Z. It suffices to
show that all of these elements yz are distinct. Suppose that _y1z1 = y2z2,
yu y2 e Cl(g{), and zu z2 e Z. Then
X(.yiM(zi) = Zt^MO^)
and /(yj = x(y2) = x(gf,-) j= 0. Thus l(zt) = l(z2) and hence zt = z2 since 1
is faithful on Z. Thus _yt = y2 and the claim is established.
We have now
\G\ = Zl^llz^)!2 = Y\Cl(gi)\x(9iMgi-l)\Z\
= YJx(iMKi)x(gr1)\z\,

Characters and integrality
39
where Kt = ^eci^)* an(* a = cox. It follows that
|G:Z|/jrfl)= t^K-Mdr1),
■ =i
an algebraic integer which is a rational number. The result now follows. |
As a combined application of Theorems 3.8 and 3.12, we prove the
following.
(3.13) theorem Let G have a faithful irreducible character of degree p",
where p is a prime and suppose that a Sylow p-subgroup of G is abelian. Then
p" is the exact power of p dividing | G: Z(G) |.
Proof Let / be the given faithful character of G. By Theorem 3.12,
pa = Z(l) divides |G:Z(G)|.
Let P e Sylp(G) and let xeP. Thus P £ C(x) and hence (/(l), |Cl(x)|) = 1.
By Theorem 3.8, /(*) = 0 if x i Z(/) = Z(G). Let Z = P n Z(G) so that /
vanishes on P — Z. Now by Problem 2.16, we conclude that |P:Z||p".
Since P/Z s PZ(G)/Z(G), which is a Sylow subgroup of G/Z(G), the result
follows. I
The above result is typical of a number of theorems about "complex
linear groups," that is, groups of nonsingular matrices over C. In these
theorems, one is given the degree n of a finite linear group G and the object is
to control the structure of G in terms of n, often under certain additional
assumptions such as the irreducibility of G. In Theorem 3.13 we are given a
group having a faithful representation of known degree, and this is obviously
equivalent to being given a linear group of that degree.
If G is a complex linear group, let S be the subgroup consisting of the
elements of G which are scalar matrices (that is, of the form el). Observe that if
G is irreducible, then S = Z(G). The group GlS is called the collineation group
associated with G. Frequently, the object of a theorem about linear groups of
given degree is to obtain information about the associated collineation group.
Theorem 3.13 is of this nature.
The reason for this situation may be seen from the following. Let G be a
linear group of degree n and let C be a group of n x n scalar matrices. Let
G* = GC and let S* be the scalar subgroup of G*. The linear group G* may be
much larger than G but G/S s G*/S*.
We have already seen several situations in which a character value is
forced to be zero. We shall now prove that every nonlinear irreducible
character vanishes somewhere. We begin with a preliminary result.

40
Chapter 3
(3.14) lemma Let G be a cyclic group and let / be a (possibly reducible)
character of G. Let S = {g e G | G = <#>} and assume that /(s) ^ 0 for all
seS. Then
Ilx(s)|2>|S|.
seS
Proof Let n = |G| and let £ be the splitting field for the polynomial
x" — 1 over Q in C. Let ^ be the Galois group of £ over Q. If a e <§ and e is an
nth root of 1, then e" = em for some m e Z, (m, n) = 1. Now /(s) = et + • • •
+ ef where e," = 1 and hence /(s)" = e^ + • • • + efm = /(s™).
The group ^ is abelian and therestriction of complex conjugation to £ is
an element of <&. It follows that a" = (a)" for all a e £ and a e 0, and thus
la"!2 = aV = a'a" = (| a I2)". Therefore (|x(s)|2r = |x(sm)|2, where (m, n) = 1
and m depends only on a.
Observe that if s e S and (m, n) = 1, then sm e S. Also, the map x h-» xm is
one-to-one on G and therefore effects a permutation of S. It follows that
Flses I X(s) I2 's invariant under ^ and hence is rational. Since it is an algebraic
integer, it must lie in Z, and since / does not vanish on S, we have
ni*(s)i2*i.
seS
for any positive real numbers ru r2,..., rfc, we
lln>(Un)llk
^Ilx(s)|2>l
I J I seS
and the proof is complete. |
(3.15) theorem (Burnside) Let /elrr(G) with *(1) > 1. Then i(g) = 0
for some g e G.
Proof Partition G into equivalence classes by calling two elements of G
equivalent if they generate the same cyclic subgroup of G. Assume %(g) ^ 0
for all g e G. Then by Lemma 3.14, we have
1\X(S)\2^\S\
seS
for every equivalence class S. Sum this inequality over all equivalence classes
of nonidentity elements to obtain
IIX(0)|2>|G|-1
9*1
Now we use the fact that
have
and we conclude that

Characters and integrality
41
and thus
|G|= Ilx(0)l2>|G|-l+x(l)2.
geG
This forces /(l) <, 1 which contradicts the hypothesis. |
The next topic we shall discuss does not, strictly speaking, depend on
algebraic integers. Nevertheless it seems appropriate to include it here.
Let G and H be finite groups and suppose C[G] = C[H], where this is a
C-algebra isomorphism. What can we infer about the relationship between G
and HI Clearly, \G\ = |H| and there exists a degree-preserving one-to-one
correspondence between Irr(G) and Irr(H). We cannot conclude, however,
that G and H are isomorphic or even that they have identical character tables.
Indeed, if G is abelian, it follows from the results of Chapter 1 that C[G] is the
direct sum of | G | copies of C. Thus if G and H are abelian and \G\ = \H\, then
C[G] s C[H].
The situation becomes more interesting if we make the weaker
assumption that Z[G] = Z[H], where Z[G] represents the group ring of G over Z
and may be identified with the ring of Z-linear combinations of elements
of G in C[G]. It is conjectured that if Z[G] £ Z[H] for finite G and H,
then G = H. The best result in this direction that has been proved as of
this writing is due to A. Whitcomb. It asserts that if G is metabelian and
Z[G] sZ[H],thenG sH.
If Z[G] = Z[H], then it is clear that we may view H as a multiplicative
subgroup of Z[G] £ C[G] and that H spans C[G] over C. In particular,
IH| > |G| and hence by symmetry \H\ = \G\ and H is a basis for C[G].
(3.16) definition Let H £ C[G] be such that H is a multiplicative group
which is a basis for C[G]. Suppose that every element of H is a Z-linear
combination of elements of G. Then H is an integral group basis in C[G].
If H is an integral group basis in C[G], we identify C[G] with C[H] in the
natural manner. An important result used in studying the isomorphism
problem is due to G. Glauberman. Most of the rest of this chapter is devoted
to its proof and some consequences. We use the notation Z Jf to denote the
sum of the elements of the conjugacy class Jf, computed in the group
algebra.
(3.17) theorem (Glauberman) Let H be an integral group basis in
C[G]. Then there exists a one-to-one correspondence between the sets of
conjugacy classes of H and G such that if if corresponds to Jf, then | if | =
|Jf| and Ii?= ±ZjT.

42
Chapter 3
We need a lemma.
(3.18) lemma Let a, /? e Z(C[G]) and view the characters of G as being
defined on all of C[G]. Define
/elrr(G)
Then
(a) <ctat + c2a2, /?> = c^a,, 0> + c2<a2, 0>;
(b) </?,a> = <^>;
(c) <K„ Ky> = 0 if i*j;
(d) <K„K,> = |G||jr,|;
where the jf, are the classes of G and Xf = EjT,.
Proof Statements (a) and (b) are immediate since characters are linear
functions. To prove (c) and (d), let x e jf t and y e Jf j. Then x(Kj) = | Jf f | x(x)
and x(Xj-) = |Jf/|x(}0- The result is now immediate from the second
orthogonality relation. |
Proof of Theorem 3.17 Identify C[H] with C[G] so that k =
dim(Z(C[G])) is the common number of classes of H and G. Let ift, Jz?2,
..., ifk be the classes of H and jf t, Jf2,..., Jf t the classes of G. Write X;
= ZjT, and L,- = Ei?,. Since the X,- are a basis for Z(C[G]), we may write
U = Z c0K;.
j
Since L( is a Z-linear combination of elements of G, we conclude that all
Ctj-eZ.
The form < , > denned in Lemma 3.18 depends only on the algebra
C[G] = C[H] and not on the particular groups G and H. This is because the
irreducible characters, defined on C[G], are simply the traces of the
irreducible representations of the algebra. We therefore have
(1) \H\\X,\ = <LltLi> = S(c|.J.)2|G||JfJ.|.
j
Sum (1) over i to obtain
(2) |H|2 = |G|Xfe(cy)2W.|.
Since the Lt span Z(C[G]) but no proper subset of {Kj} spans this space,
it follows that for each j, there exists i with cis =£ 0. Thus £,- (cy)2 > 1. Since
| H | = | G |, Equation (2) yields
|G| = Z (% (ctA\JTj\ > X |JT;| = \G\

Characters and integrality
43
and thus we have equality and £, (cu)2 = 1 for all j. Therefore, for each j
there is a unique i with ctJ ^ 0. This defines a map{jfj} -» {if,}. Since for
each i there exists 7 with cy ,£ 0, this map is onto and hence is one-to-one.
Because all nonzero c,7 = ± 1, we have Lt = ±Kj when ££, and jf,
correspond. Finally, Equation (1) yields |H|| if f| = I G||jf j I and thus I iff I = \Jfi\
and the proof is complete. |
(3.19) corollary Let H be an integral group basis in C[G]. Then there
exists an integral group basis H* in C[G] such that H = H* and the class
sums of H* equal the class sums of G.
Proof Let 3 be the linear extension of the principal character of G to
C[G]. (This is usually called the augmentation map.) Note that 8(£ agg) =
£ ag and 5 is an algebra homomorphism C[G] -» C.
If/i e H, then c5(/i)e Zand since 5(h)5(h~l) = (5(1) = l,wehavec5(/i) = ±1.
Let /1* = 5(h)h e C[G] for /ieH and note that c5(/i*) = c5(/i)2 = 1. Put
H* = {h*\heH} and note that H* is a group and the map hi->/i* is an
isomorphism. Clearly, H* is an integral group basis in C[G].
Let if be a class of H* so that Y,S£ = ±ZJf for some class Jf of G by
Theorem 3.17. Since 8(h*) = 1 for h* e H*, we have (5(Zif) = |if |. Since
(5(ZJf) = I Jf I, the ambiguous sign above must be positive and the proof is
complete. |
(3.20) corollary Let H be an integral group basis in C[G]. Then G and
H have identical character tables.
Proof By Corollary 3.19, we may assume that there exists a one-to-one
correspondence between the classes of H and the classes of G such that if ^£
and Jf correspond, then Y,<£ = £Jf and | if | = | Jf |. Let / be the trace of
an irreducible representation of C[H] = C[G]. Let if and Jf correspond
and let /1 e if and geJf. It suffices to show that /(/i) = /(g). However
tih)\2\ = x(Zif) = z(ZJf) = *(0)|jr|
and since | if | = | Jf |, the result follows. |
We have seen that if Z[G] s Z[H], then C[G] has an integral group
basis isomorphic to H. The converse is true but is less obvious.
(3.21) theorem Let H be an integral group basis in C[G]. Then Z[G] =
Z[H] in C[G].
Proof Here, Z [H] denotes the ring of Z-linear combinations of elements
of H. (It is isomorphic to the abstract integral group ring.) Since H £ Z[G],
we have Z[H] £ Z[G]. We show that G £ Z[if].
WriteG = {g,|l < i <. n}andH = {h(\l < i < n},wheren = \G\ = \H\.

44
Chapter 3
We have h; = £, a^gj for ai} e Z. We shall show that the matrix A = (au)
has an inverse with entries in Z and this will complete the proof. Write hf1 =
Z.' ^ij9i~' and B = (^"'A an integer matrix.
Let p be the character of the regular representation of C[G] = C[H] so
that p(hihf~l) = ndij. Now express /i,/ij~ ' as a linear combination of elements
of G and observe that the coefficient of 1 is £v ahbvj. It follows that
ndfj = P(hihf') = n £ a,vbv>
v
and thus AB = /, the identity matrix. The result now follows. |
Problems
(3.1) Let a be an algebraic integer and suppose that /(a) = 0, where
/(x)e 0[x] is irreducible and monic. Show that/(x) e Z[x].
(3.2) Let G be a group, g e G, and let / be a character of G. Suppose \%(g) | = 1.
Show that %(g) is a root of unity.
Hint Let £ £ C be the splitting field for x" - 1 over Q. For integral
a e E, with | a | = 1, let/a e Z[x] be the polynomial of Problem 3.1. Show that
only finitely many polynomials can arise this way. Do this by bounding the
degree and the coefficients offa.
(3.3) Show that no simple group can have an irreducible character of
degree 2.
Hint Problem 2.3 is relevant.
(3.4) Let G be a simple group and suppose / e Irr(G) with /(l) = p, a prime.
Show that a Sylow p-subgroup of G has order p.
Hint If the Sylow p-subgroup P is nonabelian, then Z(P) s Z(/).
(3.5) Suppose A £ G is abelian and \G:A\ is a prime power. Show that
G' < G.
(3.6) Let G be a p-group and suppose / e Irr(G). Show that x(l)2||G: Z(/)|.
(3.7) Let /elrr(G) be faithful and suppose /(l) = p" for some prime p.
Let P e Sylp(G) and suppose that CC(P) £ P. Show that G' < G.
Hint Let Q £ CC(P) be a p'-subgroup, Q ± 1. Show that Q n Z(x) # 1
and consider det /.
(3.8) Let x be a (possibly reducible) character of G which is constant on
G — {1}. Show that / = a\G + bpG, where a,beZ and p is the regular
character of G. Show that if G ¥= ker /, then x(l) > | G | - 1.

Problems
45
Hint First show that / = a\a + bpG for some a, b e C.
(3.9) Let Jfu ..., Jfk be the conjugacy classes of a group G and let
K,,..., Kk be the corresponding class sums. Choose representatives gt e Jf,
and let a,jv be the integers defined by
V
Show that
ip \G\ X£{G) jrfl) •
Hint Use the eo^ and the second orthogonality relation.
Notes This formula shows that the a1JV can be calculated from the
character table. Therefore, the character table of G can be used to answer questions
such as: Is an element g e jf „ the product of an element of Jf, with one of Jf fl
The fact that the a1JV are nonnegative rational integers imposes another
necessary condition on an array of complex numbers that the array be a
character table for some group.
Since the K{ are a basis for Z(C[G]), it follows that the character table of
G determines Z(C[G]) up to algebra isomorphism.
(3.10) We write [x, y] for the commutator x~ ly~lxy of x and y in a group G.
(a) Let g e G and fix x e G. Show that g is conjugate to [x, y] for some
^ e G iff
y lx(*)|2x(g)/Q
X€lrr(G) Z(l)
(b) Show that g = [x, _y] for some x, y e G iff
X€lrr(G) Z(l)
Note We already knew that the character table of a group determines the
commutator subgroup. Problem 3.10 says more than this, since the
commutator subgroup usually does not consist entirely of commutators.
(3.11) Let g e G be a commutator. Suppose m e Z, (m, o{g)) = 1. Show that
gm is a commutator.
Hint See the hint to Problem 2.12.
(3.12) Let x e Irr(G) and g,heG. Show
Zfo)Z(*) = 7^7 IzW-
|G| 2€C

46
Chapter 3
Hint Let 3E afford / and let X, and K, be the class sums in C[G] which
contain g and h. Use the fact that 3E(X,)3E(XJ) = 3E(X1XJ).
(3.13) {Brauer) Let Ku ..., Kk be the class sums in C[G]. Suppose there
exists c e C such that £ X, = cflK,. Show G = G'.
Hint Let lc#xeIrr(G). Show that oix(Kj) = 0 for some i.
(3.14) (Brauer) In the notation of the previous problem, show that if
G = G', then there exists c e Q such that £ Kt = cf]K.-
Hint To show that two elements a, be Z(C[G]) are equal, it suffices to
prove that a>x(a) = ax(b) for all / e Irr(G).
(3.15) (Thompson) Let £ be a Galois extension of Q with Galois group ^.
Let a e £ be an algebraic integer with the property that a" is real and positive
forallere^. AtheoremofSiegel(^nn.o/Mat/i.46 (1945) p. 303, Theorem III)
asserts that if a =£ 1, then
Use this to show that if / e Irr(G) then %(x) is either zero or a root of unity for
more than a third of the elements xeG.
Hints Mimic the proof of Theorem 3.15. Use Problem 3.2.
(3.16) (Burnside) Let |G| be odd and suppose/elrr(G) is not principal.
Show that x ¥= X-
Hint Using orthogonality, show that if/ =£ lcand^ = x,then/(l) = 2a
for some algebraic integer a.
(3.17) (Burnside) Let |G| be odd and suppose that G has exactly k con-
jugacy classes. Show that
\G\ = kmodl6.
Hint If n is an odd integer, then n2 = 1 mod 8.

4
Products of characters
Let x and \\i be characters of G. The fact that / + \p is a character is a
triviality. We may define a new class function xi> °n G by setting (x^){g) =
x(g#(g). It is true but somewhat less trivial that /i/f is a character. [If either /
or \ji is linear, this is Problem 2.6(a).]
Let V and W be C[G]-modules. We shall construct a new C[G]-module
V ® W called the tensor product of V and W. Choose bases {vu ...,vn} for
7 and {wu ..., wm} for W. Let V ® W be the C-space spanned by the mn
symbols vt (g) vv,-. [More precisely, V ® W is the set of formal sums of the
form £ a^0>j ® wj), with ay e C] U veV and w e W, suppose t> = £ a-^
v ® w = £ a.-b/iv ® Wj) eV ®W.
and w = >, ftj-Wj-. We define
Note that not every element of V ® W has the form v ® w for v e V and
weW (except in the special case that n or m = 1).
We define an action of G on V ® W by setting
(v, ® Wj)g = vtg ® Wjg
and extending this by linearity to all of V ® W. The reader should check that
live V, w e W, and g e G, then (v ® w)g = vg ® wg. It follows that (xg^ =
^192) for x e V ® W and gt e G.
Next we give V ® W the structure of a C[G] -module by extending the
action of G by linearity in C[G]. In other words, for x e V ® W we define
It is routine to check that this really makes V ® W into a C[G]-module.
47

48
Chapter 4
A few words of caution are in order here. If a e C[G], it is not necessarily
true that (v-t ® wjp. = t^a ® Wjtx. It is for this reason that we first defined the
action of G on V ® W and then extended it to C[G]. If A. is an arbitrary
algebra with modules V and W, it is not generally possible to define the
structure of an /1-module on V ® W.
We have not yet shown that the C[G]-module V ® W is determined (up
to isomorphism) by V and W, independently of the choice of bases. This is,
in fact, not hard to prove for group algebras over any field. We shall leave the
general situation to the problems.
(4.1) theorem Let C[G]-modules V and W afford characters / and \ji,
respectively. Choose bases in V and W and construct V ® W. Then V ® W
affords the character x*P and is independent of the choice of bases.
Proof Let {i>,| 1 < i < n} and {wr\ 1 < r <, m} be bases for V and W,
respectively, and let g e G. Write
n m
vi9 = Hauvj and wrg = Y,brsWs,
j=l -S=l
with Ay, brse C.
Then %(g) = YH= i «i,and \ji(g) = JJ*=, b„. Let 9 be the character afforded
by V ® W. Since
(vt ® wr)g = X flyMfj ® vvs),
we have
%) = Z flii ftrr = Z aii Z ftrr = /(g)^).
i, r i r
We now know that the character afforded by V ® W is independent of
the choice of bases and the result follows by Corollary 2.9. |
(4.2) corollary Products of characters are characters.
Corollary 4.2 provides another necessary condition for an array of
complex numbers to be a character table. It asserts that the inner product of any
row with the product of any two rows is a nonnegative integer. In this
connection, the following formula is relevant:
ExiiA, /2] = j-gj Z Zi(s#(0)/2(i) = Lzi, fe]-
In general, products of irreducible characters are not irreducible. For
instance, if / e Irr(G), then te Irr(G); nevertheless, lc is a constituent of XX
since
[XL lc] = [LK\g\ = !•

Products of characters
49
(4.3) theorem (Burnside-Brauer) Let / be a faithful character of G and
suppose %(g) takes on exactly m different values for g e G. Then every \p e Irr(G)
is a constituent of one of the characters (xf for 0 < j < m.
Note In the special case x = p, the regular character of G, we have
m = 2. The theorem is already known to be true in this case.
Proof Let at, a2,..., am be the distinct values taken on by /. Let
Gi = {geG\y>(g) = *i}.
Assume a.^ = /(l) so that Gt = ker / = {1}. Fix \\i elrr(G) and let ft =
ZseGi #?)• Now for./ ^ 0, we have
|G| 1 = 1
If t/^ is not a constituent of xJ for any y", 0 < j < m, we have
E(«^« = o, o<y<m.
i
The determinant of this system of m equations in the m "unknowns" /Sj is the
so-called "Vandermonde determinant" and is equal to ± J^f^</a,- — a,-) ^ 0.
It follows that all ft = 0.
On the other hand, jSt = i^(l) ^ 0 and this contradiction proves the
theorem. |
Let g e G and let n > 0 be an integer. We ask how many nth roots g has
in G. Let
$n(g) = \{heG\h'' = g}\.
Observe that if (| G |, n) = 1, we may choose an integer m such that nm = 1
mod | G |. Thus if h" = k", then h = h"m = A:nm = fc and we have 9„(g) < 1 for
all g eG. Since the map hh-»/i" is one-to-one on G, it must be onto and it
follows that 8n(g) = 1 for all g e G.
The situation becomes considerably more interesting if we drop the
assumption that n is prime to | G |. Since 9„ is clearly a class function on G, we
may write
#n = Z VnMX<
X€lrr(G)
where v„(/) is a uniquely determined complex number.
(4.4) LEMMA Vn(x) = ( 1/| G\)Y,geGX(g").

50
Chapter 4
Proof By the orthogonality relations we have
\V\gzG
Since 9nfef)Hg) = L,tCr)l»=9 H^j, we have
l«l (l€C
Finally, replace hby h'1 to obtain the desired result. |
Let cp be any class function of G and let n be a positive integer. We define
a new function <p<n) by <p(n)(gf) = <p{g"). Note that <p<n) is a class function.
Lemma 4.4 asserts that v„(/) = [x(n), lc] for / e Irr(G). It is not, in general,
true that x(n) is a character for % e Irr(G), however, it is always a difference of
two characters (see problems) and thus v„(/) e Z.
(4.5) theorem (Frobenius-Schur) Let / e Irr(G). Then
(a) x<2) is a difference of characters;
(b) v2(X)= 1,-l,or0;
(c) v2(x) t6 0 iff / is real valued.
Proof Let V be a C[G]-module which affords / and let vu v2,---,v„
be a basis for V. Let W = V (8> V and define the linear map *: W -> W by
(vt ® Vj)* = Vj® v{. Let
Ws= {weW\w* = w} and WA = {weW\w* = -w}.
These subspaces are called the symmetric and antisymmetric parts of W,
respectively.
If w e W, we have w + w*eH's and:w — w* eWA. Since
w + w* w — w*
it follows that W = Ws + WA.
. We claim that if w e W and g e G, then (wg)* = w*g. It suffices to check
this as w runs over the basis v( ® v}, and thus we need to show that
(v,g ® Vjg)* = Vjg ® vtg.
In fact, it is true that
(x ® y)* = y ® x
for all x, yeV. This is seen by expanding x and y in terms of the vt.

Products of characters
51
It now follows that Ws and WA are C[G]-submodules of W and we have
X = Xs + Xa>
where xs and Xa are tne characters afforded by Ws and H^. We compute Xa
using the basis
Wjj = t)f ® Vj - Vj <g> l>„ i < j,
for WA.
Suppose v-,g = Y. Qirvr- We have
">ij9 = Z (airaJs ~ ajrais)Vr ®VS= Y,(airajs - fljrfl«SKs.
r,s r<s
Therefore
n(g)= Hauajj- aaaiy
This yields
ZXaIq) = Ha>>ajj ~ \Zauan = ( Z «»)( Z ajj) ~ Z aiJa)i-
i*j i*j \i J\ j J i,J
Let 3E be the representation corresponding to the basis {vt} of V, so that
X(g) = (Ay) = M. We have
2Zi4(0) = (tr M)2 - tr(M2) = X(g)2 - X(g2)
since M2 = 3E(gr2).
This yields z(2)(gf) = x(g)2 - 2z/1(gr) and thus z<2) = z2 - 2^, a difference
of characters, as claimed.
Now v2(z) = [z<2), lc] = [z2, lc] - 21x4, lc]. If z is not real valued,
then 0 = [z, z] = Lz2, lc] and thus [z^, lc] = 0 since Xa is a constituent of
Z2. It follows that v2(z) = 0 in this case. If z is real valued, then 1 = [z2, lc]
tnus Ix*, lc] = 0 or 1 and v2(z) = 1 or -1. The proof is now complete. |
(4.6) corollary Let G have exactly t involutions. Then
i + f= Z v2(z)z(i),
X€lrr(C)
where v2(z) = 0 if z 7^ Z and v2(z) = ± 1 if z = Z-
Proof This is immediate since 92(1) = 1 + t. |
If N <i G, we have identified Irr(G/iV) with a subset of Irr(G). The
following lemma shows that v„(z) is well defined under this identification.
(4.7) lemma Let N <i G and let z e Irr(G) with N £ ker z- Let v„(x) be as
above and let v„(z) be the corresponding number computed in G/N. Then
v„(z) = i>„(z)-

52
Chapter 4
Proof We have
= i7q Z xte") = vn(xl I
(4.8) lemma Let x £ Irr(G) and let X be a linear character of G with X" = lc.
Then vn(x) = v„(l/).
Proof Recall that Xx e Irr(G) by Problem 2.6(b). Now
v„(ix) = r^ Z to)(0") = r^ Z %"k(0n) = t4 Z xig") = v„(z)
|G|9€C |G| g \G\
since 1(9") = 1(g)" = X"{g) = \. |
As an application of the Frobenius-Schur theorem, we prove the
following.
(4.9) theorem (Alperin-Feit-Thompson) Let G be a 2-group containing
exactly t involutions. If t = 1 mod 4, then either G is cyclic or | G: G' | = 4.
Proof If G is abelian, then it clearly must be cyclic. We suppose G is not
abelian and show that | G: G' | = 4.
If Z(G) is not cyclic, choose K s Z(G) elementary abelian of order 4. The
set {x e G\x2 = 1} is a union of cosets of X and hence 4|(r + 1), a
contradiction. Therefore, Z(G) is cyclic and G contains the unique minimal subgroup
Z of order 2.
Since G' > 1, we have Z s G'. Also G/Z is not cyclic since G is not
abelian. If G/Z satisfies the hypotheses of the theorem, then |G:G'| =
|(G/Z):(G/Z)' | = 4 by induction and we are done. We may therefore assume
that the number of involutions in G/Z is not = 1 mod 4.
We have
Z v2(z)x(l) = t + 1 =2 mod 4
xelrr(G)
and
Z v 2(x)xW # 2 mod 4,
Z€lrr(C);ZSkerz
where the second statement follows via Lemma 4.7. We conclude that
(*) Z v2(Z)Z(l)#0mod4.
Z€lrr(C);Z«kerz
Now let C be the group of linear characters X of G which satisfy X2 = lc.
For x e Irr(G), we have Xx e Irr(G) and since Z £ ker X, we conclude that

Products of characters
S3
Z £ ker x iff Z £ ker(l/). Therefore C permutes {/ e Irr(G)|Z £ ker /} and
partitions this set into orbits 6t. By Lemma 4.8, v2(x) is constant on each orbit
asisx(l).
Since 16X | is a power of 2, we conclude from Equation (*) that there exists
X e Irr(G) such that
(i) Z £ ker /,
(ii) v2(Z) * 0,
(iii) *(1)|0|£2,
where 0 is the orbit containing /.
Since Z £ ker /, we have ker z = 1. Since G is not abelian, z(l) > 1 and
hence z(l) = 2 and 101 = 1. Therefore, Ax = X f°r a^ ^ e C.
Now let F = <I>(G), the Frattini subgroup. If g e G - F, then there exists
k e C for which 1(g) =£ 1 and it follows that xid) = 0. By Lemma 2.29, we now
have
4 = Z(D2>[ZF,ZF] = |G:F|.
Since G is not cyclic, \G:F\ > 4 and we have equality above. This forces
XF = 2//, where fi is a faithful linear character off.
Since v2(z) ¥= 0, we know that / is real valued. Therefore, /z is also, and so
\F\ < 2. Thus \G\ <, 8 and hence G s D8 or Q8. In either case, \G: G'\ = 4
as desired. |
A theorem of O. Taussky [Theorem III 11.9(a) in Huppert] asserts that
the only nonabelian 2-groups G for which | G: G' | = 4 are the dihedral, semi-
dihedral, and generalized quaternion groups. In each of these groups, the
number of involutions is in fact = 1 mod 4.
As another application of Corollary 4.6, we prove some results of Brauer
and Fowler that were originally obtained in a different way.
(4.10) lemma Let a,, a2,..., a„ be arbitrary real numbers. Then
Proof The well-known Schwarz inequality asserts that
(2>&i)2 s: (2><2)(2>2)
for real a, and bt. The lemma follows by setting all bt = 1. |

S4
Chapter 4
(4.11) theorem Assume \G\ = g is even and that G contains exactly t
involutions. Let a = (g — l)/r. Then
(a) There exists xeG,xJ= 1, with | G: C(x) \ < a2.
(b) There exists real / e Irr(G), ](# le, with x(l) < a.
Proof By Corollary 4.6,
0 < (g - l)/a = t < Z Z(l),
where
^ = {zeIrr(G)|z = z and X * U)-
In particular, s = \Sf\ =£ 0. Now
t2<(lx(l)Y<sI>(l)2<S(0-l)
by Lemma 4.10. It follows that
g — 1 < sol2
and thus
sa2>Xx(l)2-
Therefore, /(l) < a for some / e y, proving (b).
Since s < k — 1, where k = |Irr(G)| is the total number of conjugacy
classes of G, we have
(k - l)a2 > g - 1
and hence some nonidentity class of G has size <a2. This proves (a). |
Recall that an element x e G is said to be real if x is conjugate in G to x~'.
It is a fact (which will be proved later) that the number of classes of real
elements of G is equal to the number of real irreducible characters.
Assuming this, it is immediate from the inequality sa2 > g — 1 in the
above proof that statement (a) can be strengthened to guarantee that a real
xeG exists with x ± 1 and |G: C(x)\ <, a2.
(4.12) corollary Let n be a positive integer. There exist at most finitely
many simple groups containing an involution with centralizer of order n.
Proof Let G be such a group with \G\ = g. Then G contains at least g/n
involutions and hence a < n in the notation of Theorem 4.11.
By (a) of that theorem, there exists xeG with 1 < | G: C(x) \ < n2.
Therefore, G is isomorphic to a subgroup of the alternating group A„2_ 1. The result
follows. |

Products of characters
SS
(4.13) corollary Let G have even order, g > 2. Then G contains a
proper subgroup of order > (g)113.
Proof We use induction on \G\. First assume Z(G) = Z > 1. If |G: Z\
is even, then there exists H/Z < G/Z with
\G:H\ = \(G/Z):(H/Z)\<\G/Z\2'3<g2'3
and the result follows. If | G: Z | is odd, then G has a central Sylow 2-subgroup
and thus has a normal 2-complement by a transfer theorem of Burnside.
(Or see Theorem 5.6.) It follows that G has a subgroup of index 2 and the
result follows.
Now assume Z(G) = 1 and let a be as in Theorem 4.11. Then
1 < |G:C(x)| < a2
for some xeG and we are done if a2 < g213. Assume, then, that a > g113 and
let t be an involution in G. Then
l<|G:C(T)|<(0-l)/a<g/a<g2'3
and the result follows. |
We now wish to discuss the question of which real / e Irr(G) satisfy
v2(z) = + 1- The answer is interesting but does not seem to be very important
in the applications of the Frobenius-Schur theorem. The solution to this
problem involves some matrix theory.
Let U and V be C[G]-modules with bases {uu ...,«„} and {vu ..., vm},
respectively. An element w e U ® V is uniquely of the form
W = £ fli/Mi ® Vj)
'J
and this defines the n x m matrix (ay). We write M(w) = (ay). For g e G, we
compute M(wg).
Let 3E and 9) be representations of G corresponding to U and V
respectively, with respect to the given bases. Write
*(0) = (Ks), <D(g) = (cpq).
Then
(«,• ® t>j)0 = u,g ® Vjg = £ feisCj,(Ms ® i>,).
Thus
W9 = Z bisaijCJq(us ® vq)
•', j, s,«

56
Chapter 4
and, therefore,
M(wg) = mT M(wW(g\
(4.14) theorem Let/elrr(G) be real valued and let 3E be a representation
which affords /. Then there exists a nonzero matrix M such that
X(g)TMX(g) = M
for all geG. Furthermore, for any such matrix, MT = v2(/)M.
Proof Let V be a C[G]-module corresponding to 3E and write W =
V ® V. Let M: W -> M„(C) be as previously stated so that
M(wg) = X(0)TM(w)X(0)
for w e W and geG.
Now W affords the character x2 and \j2, lc] = [/, /] = 1 and hence
there is a one-dimensional space of G-fixed points, W0 £ W. The matrices
M satisfying X(g)JMX(g) = M are precisely the M(x) for xeW0.
In the proof of Theorem 4.5, we had a map *: W -» W and we
decomposed W = Ws + W^, where Ws = {we W|w* = w} and H^ =
{weW|w* = — w}. Also, the proof of Theorem 4.5 showed that v2(x)
= 1 iff [Xa< lc] = 0, where /^ is the character afforded by W^.
We must have one of the following two situations:
(i) W0^Wa,[Xa,Ig] = Iv2(x)=-U
(ii) W0 £ Ws, Ixa, 1C] = 0, v2(X) = 1.
Thus if x e W0, we have x* = v2(/)x.
Now, for we W we clearly have M(w*) = M(w)T. Thus for xe W0 we
have
M(x)T = v2(x)M(x)
and the result follows. |
(4.15) corollary Suppose / e Irr(G) is afforded by a real representation.
Then v2(x) = + 1.
Proof Let 3E be an R-representation which affords / and let
M= XX(/i)TX(/i).
heG
It is clear that MJ = M and 3E(gr)T M3E(gr) = M for all geG. All that remains is
to show that M^O. For any n x n real matrix X, is it clear that the diagonal
entries of XJX are ^ 0, and at least one of them is >0 unless X = 0. The
result now follows. |

Products of characters
S7
By Problem 2.5(b), we know that it is possible that a real-valued
irreducible character / is not afforded by any real representation. Such
characters are exactly those for which v2(z) = — 1. The proof of this seems to
require some nontrivial matrix theory.
If M is a square, complex matrix, we write M* = MT. and say M is
unitary if MM* = /and Mis normal if MM* = M*M. A standard theorem of
linear algebra asserts that if M is normal, then there exists a unitary matrix,
U, such that U~ lMU is diagonal.
(4.16) lemma Let D be a diagonal matrix. Then D = E2 for some
diagonal matrix E such that every matrix which commutes with D also
commutes with E.
Proof Suppose a,, ...,a„ are distinct complex numbers and /?,,...,
/?„ e C are arbitrary. Then there exists a polynomial / such that/(a;) = /?,-.
We may therefore choose / such that /(a)2 = a for every diagonal entry
a of D. Then E = f(D) has the desired properties. |
(4.17) theorem Let 3E be a C-representation of a group, G. Then 3E is
similar to a representation 9) such that 9)(g) is unitary for all g e G.
Proof Let M = £«,6C X(g)*X(gl Then M* = M and M is normal. We
may choose a unitary matrix U such that U~ lMU is diagonal. Since U~i =
I/*, we have
(U-iX(g)U)*=U-1X(g)*U and tr'Ml/ = £ *i(ff)*3Eite).
where 3E, = l/~'3El/. It is therefore no loss to assume that M is diagonal.
The diagonal entries of 3E(g)*3E(g) are all real and nonnegative and since
£( 1) = /, it follows that the diagonal entries of M are positive. We may
therefore write M = P2 where P is a nonsingular real diagonal matrix.
Since y.{g)*MH{g) = M for all g e G, we have
(P-1X(&)*P)(PX(g)P-') = /.
Since (PX(g)P-1)* = P_13E(gf)*P, it follows that 9) = P3EP"1 is the desired
representation. |
(4.18) lemma Let P*P = I and PT = P for a square matrix P. Then
P = QQ~' for some matrix Q.
Proof Since P is unitary, it is normal and we may write U~ iPU =
D = E2, where U is unitary, D and E are diagonal, and every matrix which
commutes with D also commutes with E. Now DJ = D, PJ = P, and
UJ = [7 "'.This yields
Z) = DT = (l/-'Pl/)T = Cr'PlJ = TJ-^UDV-^U.

58
Chapter 4
Therefore, U ' U commutes with D and hence also with E. It follows that
E = Cr'l/ElT'l/and
UEU'1 = UEU'1.
From P*P = /, we obtain / = D*D = DD and it follows that the diagonal
entries of D and hence of E have absolute value 1 and E = £_1. Now let
Q = UEU-1. Then
QQ-1 = (0£tr')(t/£t/-1) = (I/EIT1)2 = UDU'1 = P
and the proof is complete. |
(4.19) theorem Let / e Irr(G) and suppose v2(/) = 1. Then / is afforded by
some real representation of G.
Proof By Theorem 4.17, / is afforded by a representation X, such that
X(g) is unitary for all g e G. Since / = /, X and X are similar and we may
choose P such that P~XJP = 3E. Now
Since v2(/) = 1, it follows that PT = P by Theorem 4.14.
From the equation X(g)P = PX(g), we obtain P*X(g)* = 3E(gf)*P* and
since X(g)* = 3E(gr', this yields P*X(g)'1 = 3E(g)-'P*. Thus 3EP* = P*J
and
3EP*P = P*X~P = P*P3E.
By Schur's lemma, it follows that P*P = a.1 for some a. Clearly, a is real and
positive and we may therefore replace P by P/oc1/2 and assume P*P = /.
Now Lemma 4.18 applies to yield P = QQ~' for some Q and we have
QQ1X = IQQ1
or Q- '3EQ = Q~ XXQ. In other words, 9) = Q_ '3EQ is a real representation
which affords / and the proof is complete. |
We may summarize the situation as follows for / e Irr(G):
(a) v2(x) = 0 iff/is not real;
(b) v2(x) = 1 iff x is afforded by a real representation;
(c) v2(z) = — 1 iff/ is real but is not afforded by any real representation.
One further remark should be made. If z e Irr(G) and v2(/) = — 1, then
X(l) must be even. This may be seen as follows. The nonzero matrix M in
Theorem 4.14 is nonsingular by Schur's lemma. If v2(/) = — 1 then MT =
-Mand
det(M) = det(MT) = det(-M) = (-1)*(1) det M.
It follows that (-1)*(1) = 1 and /(l) is even.

Problems
59
We close this chapter with a discussion of the characters of direct products.
(4.20) definition Let G = H x K and let <p and 9 be class functions on
H and K, respectively. Define / = cp x y by %(hk) = (p(h)S(k) for h e H and
keK.
Observe that cp x y is a class function of G. If cp is a character of H, then
under the isomorphism H s G/X, there is a corresponding character q> of G
with X ^ ker 0 and <p(/i/c) = (p(h). Similarly, if y is a character of K, there is a
corresponding character y of G, with §(hk) = y(/c). It follows that q> x y
= $y is a character of G.
(4.21) theorem LetG = H x K. Then the characters <p x y for <p e Irr(H)
and y £ Irr(K) are exactly the irreducible characters of G.
Proof Let <p, <p, e Irr(H) and y, y, e Irr(K). Let / = cp x y and /, =
<p, x y,. Then
[/, Zi] = iTT, E X(0)Z^)" = 777^7 £ (pQWkfaWAk)
\(j\gsG \H\\K\ hBB:kEK
= (j^ £/{h)V^)(\k\ X ^W)) = l>, <P.][S, Si]-
It follows that the <p x y are all distinct and irreducible.
Now
£ (? x y)(i)2 = X<p(i)29(D2 = (X<p(i)2YxW
^elrr(H).,Selrr(K) ^.S \^ /\s
= |H||K| = |G|,
and thus the cp x y are all of Irr(G). |
Problems
(4.1) Let / and \ji be characters of G. Express det(/i/0 in terms of det / and
deti/f.
Hint It suffices to assume that G is cyclic.
(4.2) (Garrison) Let Jf be a class of G which is not contained in any
proper normal subgroup. Let K be the corresponding class sum in C[G] and
let m be the number of distinct values of cox(K) for / e Irr(G). Show that every
element of G can be written as a product of fewer than m elements of Jf.

60
Chapter 4
Hint Mimic the proof of Theorem 4.3 using the second orthogonality
relation.
(4.3) Let G = H x K. Let <p e Irr(H) and 9 e Irr(X) be faithful. Show that
cp x 9 is faithful iff (| Z(H) |, | Z(K) |) = 1.
(4.4) Suppose G = HK with H <= QK).
(a) Let /elrr(G). Show that Xh = S(l)<p and Xk = <p(l)9 for some
<p e Irr(H) and 9 e Irr(X).
(b) Let <p e Irr(H) and 9 e Irr(X) and suppose that q>HnK and 9HnK have
a common constituent. Show that there exists a unique / e Irr(G) with
Xn= 9U)<pandxK = <p(l)9.
Hint G is a homomorphic image of H x K.
(4.5) Let G = H x A where A is abelian and let n > 0 with (|H|, n) = 1.
Show that x(n) e Irr(G) for every / e Irr(G).
(4.6) Let n > 0 and assume that x(n) e Irr(G) for every / e Irr(G). Show that
G = H x A, where A is abelian and (|H|, n) = 1.
Hints (a) Let rf = (| G |, n). Show that it is no loss to assume that
(|G|/d,n)=l.
(b) Let A = f)^irr(G) ker x(n). Show that A = {geG\gn = 1} and
\A\ = d.
(c) Let H = f){ker zlze Irr(G), z(n) = 1G}. Show \G : H\ = d.
(4.7) Let V be a C[G]-module with basis {vu ...,v„} and let p be a prime.
Let W = F ® • • • ® FQ? times) and define a on W so that (xt ® • • • ® xpf =
x2 ® x3 ® • • • ® xp ® Xi for all x, e 7. Let e be a primitive pth root of 1 in C.
Let Wv = {weWlw" = ew}. Show that Wt is a C[G]-submodule of W
which affords the character Xi = (Xp — X<p))/P> where / is the character
afforded by V. Conclude that x(m) is a difference of characters for all m > 1.
Hints Let "W = {t>fl ® • • • ® vip \ not all i3 are equal}. For we if, let
w = w + e~lw* + e~2w"2 + ••• + e~<p-1V"'.
Now <a> permutes niT into orbits of size p. Let 1^0 be a set of representatives
of these orbits. Then {vv| w e "W^\ is a basis for Wt.
(4.8) Let G be a p-group with Frattini factor group of order > p2a~l. Show
that the number of elements of order p in G is = — 1 mod p".
Hint Mimic the proof of Theorem 4.9. Use the fact that vp(x) is an
integer (Problem 4.7).

Problems
61
(4.9) If/e Irr(G), then |v2(/)| <, 1. Show that no absolute bound on v„(x)
exists for any n > 2.
Hint There exist groups of prime exponent p#2 and of exponent 4
with irreducible characters of arbitrarily large degree.
(4.10) Let V, W, Vu and Wt be F[G]-modules for an arbitrary field F. Fix
a particular F-basis in each of these modules and construct V ®W and
Vt ® Wt as was done for C[G]. Now suppose a: V -» Vt and /?: W -* Wx are
module homomorphisms. Show that there exists a module homomorphism
y.V ®W -^ViQWi such that y(v (g) w) = <x(v) (g) 0(w) for t> e 7 and weW.
Conclude that V ® W is determined up to module isomorphism
independently of the choice of bases.
(4.11) Let G be simple and let S e Syl2(G) be elementary abelian of order q.
Suppose S = CG(x) for all x, 1 # x e S. Show that v2(x) = 1 for all / e Irr(G)
and that x(x) = /(l) - q for / ^ 1G and xeS.
Hint First show that G contains exactly | G | /q involutions and that the
remaining elements of G have odd order. This is done without characters
using the fact that if x, y are nonconjugate involutions, then |Z«x, y»| is
even.
Note This problem is continued as Problem 5.20 where it is proved that
\G\=q(q- l)(q+ 1).
(4.12) Let x, «A, 9 e Irr(G). Show that [#, 9] < 9( 1).

Induced characters
Let H £ G be a subgroup. Given a character / of G, we obtained the
character Xh °f # by restriction. In this chapter we study an approximately
dual process, where characters of G are induced from characters of H.
(5.1) definition Let H £ G and let <p be a class function of H. Then <pG,
the induced class function on G, is given by
I" I xeG
where <p° is defined by <p°(h) = q>(h) ifheH and <p°(y) = 0 if y £ H.
Observe that <pG is really a class function on G and that <pG(\) = \G: H\q>(\).
Another useful formula for (pG(g) is obtained by choosing a transversal T
for the right cosets of H in G (that is, a set of representatives for these cosets).
It is then easy to see that
9G(g)= ?,<p0(tgt-1).
(5.2) lemma {Frobenius Reciprocity) Let H £ G and suppose that cp is a
class function on H and that 9 is a class function on G. Then
[>, Sfl] = l>G> 8].
Proof We have
[>G, 9] = t^ E <PG(0)%) = r^rj ^ E E 9°(xgx- l)W\
|G| 9eG |w| \"\ gsG xeG
62

Induced characters
63
Setting y = xgx l and observing that 9(g) = 9(y), we obtain
I O I \H\ yBGxBG \tI\yeB
(5.3) corollary Let H £ G and suppose <p is a character of H. Then
<pG is a character of G.
Proof Let / e Irr(G). Then Xh is a character of H and thus [<pG, /] =
[<P> Z«] is a nonnegative integer. This proves the result since cpG j= 0. |
(5.4) corollary Let H £ G and suppose <p e Irr(H). Then there exists
X e Irr(G) such that <p is a constituent of Xh ■
Proof Take / to be an irreducible constituent of (pG. Then
0 * [>G, z] = L>, Zfl]
and the result follows. |
(5.5) corollary Let G be abelian and suppose H £ G. Then every
1 e Irr(H) is the restriction of some fi e Irr(G).
Induction of characters is closely related to the transfer map from a
group into an abelian factor group of a subgroup. Many of the results
provable by transfer can also be proved using induced characters. We give one
example of this now.
(5.6) theorem Let G have an abelian Sylow p-subgroup. Then
|G'nZ(G)|
is not divisible by p.
Proof Assume the contrary and choose U £ G' n Z(G) of order p,
with U £ P e Sylp(G). Let X e Irr(I/), X * lv, and choose fi e Irr(P) with
/% = 1. Now write
V°= X «**
xelrr(G)
and observe that fiG( 1) = /z( 1) | G: P | = | G: P | is prime to p. Therefore, there
exists x £ Irr(G) such that ax =£ 0 and p^ 1). We have
0 * [>G z] = D*, ZJ
and thus X is a constituent of Xv Since 1/ £ Z(G), we have Xv = z(*M and
(det z)t; = ^*(1)- (See Problem 2.3.) Since U e G', we have (det x)v = lu and
thus XxW = lv. Since pJfxi^X ^ ^ li/> and 1^1 = P> this is a contradiction
and proves the theorem. |

64
Chapter 5
Explicit computation of induced characters is extremely useful for the
construction of character tables, despite the fact that cpG is usually reducible
even if cp is irreducible. [Note that if <p is reducible, then <pG is necessarily
reducible since (cpi + <p2)G = <PiG + (P2°-l
Given H £ G, cp a character of H and g e G, an efficient way to compute
cpa(g) explicitly is to choose representatives, xu ..., xm for the classes of H
contained in Cl(g) in G and to use the formula
cpG(g) = \CG(g)\l j^-
where it is understood that (pG(g) = 0 if H n Cl(g) = 0. This formula is
immediate from the definition of <pG since as x runs over G, xgx~i = Xj for
exactly | CG(g) | values of x.
We shall now construct the character table of G = A5. There are five
conjugacy classes; one each of elements of order 1, 2, and 3 and two of
elements of order 5. It is convenient to label these classes 1, 2, 3, 5l5 and 52 in
the table. Let Xi = 1g- We have:
Class Jf; 1 2 3 5t 52
\C(x)\iorxeJf:
\X\
60 4 3 5 5
1 15 20 12 12
11111
Now let H = y44 £ G and compute (1H)G. We have
(1H)G: 5 12 0 0
using the above formula. [For instance, if o(g) = 3, there are two classes in H
of elements conjugate in G to g. If xt and x2 are representatives of these
classes, we have lH(Xi) = 1 and |CG(g)| = 3 = |CH(xj)| and thus (lH)G(gf) = 2
as listed.]
Now [(1H)G, 1G] = [1h, 1h] = 1 by Frobenius reciprocity and thus
(1H)G — 1G is a character. We call this character /4. Thus
n = iUf- iG: 4 0 i -l -l.
Direct computation now yields [/4, /4] = 1 and we have found an
irreducible character.
Next, let 1 be a nonprincipal linear character of H = A4 so that l(x) = 1
if o(x) = 1 or 2 and l(x) = co = e2nil3 for one class of elements of order 3 in H
and X(x) = a2 for the other class. We get
X5 = 1G: 5 1 -1 0 0
since co + co2 = -1. We compute [/5, x5] = 1 and so Xs e Irr(G).

Induced characters
65
Now let K be a cyclic subgroup of G of order 5 and let /z e Irr(K), fi =£ lK.
The class 5t of G contains a pair of inverse elements of K, say x and x~i.
Choose fi so that fi{x) = e = e2nil5. Since x2 and x~2 lie in the class 52 we have
HG: 12 0 0 e + e4 e2 + e3.
Computation yields [/iG, /5] = 1 and thus
HG - Xs- 7 -1 1 e + e4 e2 + e3
is a character.
Further computation yields that [/iG — Xs, /4] = 1 and hence
Z3 = V° - Xs ~ U- 3 - 1 0 e + e4 + 1 e2 + e3 + 1
is a character and [/3, x3] = 1 so that /3 is irreducible.
Finally, replacing fi above, by v e Irr(X) with v(x) = e2 yields
ip3: 3 - 1 0 e2 + e3 + 1 e + e4 + 1
is irreducible and we have found all five irreducible characters of G.
We now consider the module theoretic interpretation of induced
characters.
(5.7) definition Let V be a F[G]-module. Suppose V = Wf + • • • + Wk
where the Wt are subspaces of V which are transitively permuted by the
action of G. Then V = Wv + ■ ■ ■ + Wk is an imprimitivity decomposition of V.
If V is irreducible and has no proper imprimitivity decomposition (that is,
with k > 1), then V is a primitive F[G]-module.
We emphasize that if V = Wt -j- • • • 4- W* is an imprimitivity
decomposition of V with /c > 1, then the Wt are definitely not F[G]-submodules of
V. In this situation, let H = {g eG\Wth = WJ, then Wt is a F[H]-module.
(5.8) theorem Let V = Wt + ■■ ■ + Wkbe an imprimitivity
decomposition of the F[G]-module V. Assume FgC, Let H be the stabilizer of Wv
Suppose V affords the character x of G and Wv affords the character 9 of H.
Then x = 9G-
Proof Let T be a right transversal for H in G and write W = Wv Then
Wt runs over the Wt as t runs over T and
V = £ • Wt.
Choose a basis, wt,..., wm for W so that {wj 111 < i <, m, t e T} is a basis for
V. Let g e G. We compute xig) using this basis.

66
Chapter 5
For te T, the contribution of the basis vectors wtt to %(g) will be zero
unless Wtg = Wt, that is, tgt~l e H. Assuming tgt~1 = heH, write w-th =
5] a/;w; so that £|"=1 ait = 9(rgr~')- Now, (Wjt)9 = Wj/if = 5] fluw;f an(*
hence the contribution of w{ t to /(g) is a^. Therefore, the total contribution of
the w,t, 1 < i < m, is 9(tgt~'). It follows that x(g) = ^(eT 9°(fgr') = $G(g)
as desired. |
(5.9) theorem Let H c g and let W be an F[H]-module. Then there
exists an F[G]-module Shaving an imprimitivity decomposition V = £• Wt,
where H is the stabilizer of Wt and Wt s W as F[H]-modules.
Proof Let V be the external direct sum of | G: H | copies of W as an
F-vectorspace. Let T be a right transversal for H in G and assume 1 e T. Let
W ® t, for t e T, denote subspaces of V, each isomorphic to W, such that
^ = E"<<et(W® t)- Fix isomorphisms a,: W-> W ® t and let w (g) t
denote at(w) so that W ® t = {w ® t \ w e W}.
Now for weW and g e G, define w ® ge V as follows. Write g = ht,
with fi£fl and te T, and set w ® g = vWi ® t. Note that for we W, xeH,
and g e G, we have wx ® g = w ® xg. (The usual notation for what we have
constructed is V = W ® f[H]F[G].)
We now let G act on V by defining (w ® t)g = w ® tg for g e G. Observe
that (w ® gj^j = w ® g^ for all gu g2e G. Extending this definition by
linearity gives V the structure of an F[G]-module and
V = £ -W®t
is an imprimitivity decomposition.
Clearly the stabilizer of W ® 1 is H and the map at: W -^ W ® 1 is an
isomorphism of F[H]-modules. The proof is complete. |
If s e Ht, then W ® s = W ® t in the above construction and it is
reasonably clear that the module V is independent of the choice of coset
representatives. We write V = WG. Note, however, that we have not yet proved that
WG is the unique (up to F[G]-isomorphism) F[G]-module which satisfies the
conclusion of Theorem 5.9.
In the special case that F = C, let W afford the character 9. Then by
Theorem 5.8, it follows that any C[G]-module which satisfies the conclusion
of 5.9 affords / = 9G. This proves uniqueness in this case. The general case is
left to the problems.
Note that Theorems 5.8 and 5.9 provide an alternate proof of the fact that
induced characters are characters. Another consequence of these results is
that if i e Irr(G), then / is afforded by a primitive module iff / =£ 9G for any
character 9 of a proper subgroup of G. Such / are called primitive characters.

Induced characters
67
(5.10) definition Let x be a character of G. Then / is monomial if / = AG,
where X is a linear character of some (not necessarily proper) subgroup of G.
The group G is an M-group if every / e Irr(G) is monomial.
Note that if / is monomial, then by Theorems 5.8 and 5.9, / is afforded by a
module which has an imprimitivity decomposition into subspaces of
dimension 1. It follows that i is afforded by a representation X with the
property that each row and column of X(g) has exactly one nonzero entry for
each g e G.
(5.11) lemma Let 9 be a character ofHsG. Then
ker(9G) = f] (ker 3)x.
xeG
Proof Let / = 9G. Then g e ker / iff
Yj$°(xgx-1)= £3(1).
xbG xbG
Since |9°(xgx-1)| < 9(1), we conclude that geker / iff ^(xgx"1) = 9(1)
for all x e G. This happens iff g e (ker 9-)* for all x and the proof is complete. |
(5.12) theorem Let G be an M-group and let 1 = ft < f2 < ■ ■ ■ < fk
be the distinct degrees of the irreducible characters of G. Let / e Irr(G) with
Z(l) = fi- Then G(i) £ ker /, where G(i) denotes the ith term of the derived
series of G.
Proof If i = 1, then / is linear and G' £ ker /. Assume i > 1 and work by
induction on i. If ifre Irr(G) and t/f(l) < /(l), then lA(l) = f} for j < i and
G(.--DcGU)£ker^.
Since G is an M-group, we may choose H < G and linear 1 e Irr(H) with
X = 1G. Now [(1H)G 1G] = [1«> 1«] = 1 and thus (1H)G is not irreducible.
If \p is any irreducible constituent of (1H)G, then \p( 1) < (1H)G(1) =\G:H\ =
XG(l) = /(l) and thus G(i_ u £ ker i>. It follows that G(i_ u £ ker((lH)G) £ H
by Lemma 5.11.
Now G<0 £ H' £ ker 1 and since G<0 -a G, we have
G<0 £ f) (ker Xf = ker /
and the proof is complete. |
(5.13) corollary (Taketa) Let G be an M-group. Then G is solvable.
It is not the case that all solvable groups are M-groups. [The smallest
counterexample is SL(2, 3) of order 24.] In Chapter 6 we shall discuss some
sufficient conditions. There is no known characterization of M-groups other
than in terms of characters.

68
Chapter 5
We shall now discuss some of the connections between character theory
and the theory of permutation groups. Suppose the group G acts on the
finite set Q. In other words, for each a e Q and g e G, an element a9 e Q is
defined such that a1 = a and (a9)*1 = a9*1 for a e Q and g,heG. For a e Q, we
shall use the notation Ga = {g e G|a9 = a} so that \G:Ga\ = \6\, where & is
the orbit containing a.
Let G act on Q. For g e G, let %(g) = | {a e Q | a9 = a} |. The nonnegative
integer valued function / is called the permutation character associated with
the action. To see that / really is a character, let V be a C-space with a basis
identified with fi. Let G act on V by permuting the basis. This makes V into
a C[G]-module called a permutation module. It is clear that V affords /.
(5.14) lemma Let G act transitively on Q, let a e Q and let H = Ga. Then
(1H)G is the permutation character of the action.
Proof Let V be the permutation module with basis fi. Then V =
E • 0efiC/? is an imprimitivity decomposition for V. The result is now
immediate from Theorem 5.8. |
(5.15) corollary Let G act on Q with permutation character /. Suppose
Q decomposes into exactly k orbits under the action of G. Then [/, 1G] = k.
Proof Write Q = (J*= t (9{ where the 0j are orbits. Let Xt be the
permutation character of G on &x so that / = £ xf. For aefi);, we have %t =
(lGa)Gby Lemma 5.14. Thus [*„ 1G] = [(1G/, 1G] = [1G„, 1GJ = 1.
Therefore, [/, 1G] = £ [/i, 1G] = /c and the proof is complete. |
Corollary 5.15 can also be proved as follows. Let &x be as above with
ae0».Then|0,| = |G|/|GJ. Now let <f = {(a,g)|aefi,geG, a9 = a} and
compute | y | two ways. We have
Ez(0) = m= Eigj = £ £igj = E i M = £|Gi = *iGi
and the result follows.
(5.16) corollary Let G act transitively on Q with permutation character
/. Suppose that aefl and that Ga has exactly r orbits on Q. (One of these is
{a}.) Then [*, *] = r.
Proof We have
r = LXg., IgJ = [X, (1g„)g] = ih Z\
by 5.15, 5.14 and Frobenius reciprocity. |

Induced characters
69
The integer r in Corollary 5.16 is called the rank of the transitive action.
Recall that G is doubly transitive (or 2-transitve) on Q if |Q| > 2 and Ga is
transitive on Q — {a}, that is, if r = 2.
(5.17) corollary Let G act on Q with permutation character /. Then the
action is 2-transitive iff / = 1G + \ji where \ji e Irr(G) and \\i j= 1G.
Corollary 5.17 is often useful for finding irreducible characters. For
instance, it shows that the symmetric group E„ has an irreducible character of
degree n — 1 for n > 2, and that this character restricts irreducibly to A„
for n > 4.
We now consider the question of finding (not necessarily normal)
subgroups of a group G from information about its characters.
Suppose we wish to prove that G has a subgroup H of fairly small index.
This could be done if the character (1H)G could be recognized as such. Because
(1H)G is a transitive permutation character (of the action on the set of right
cosets of H), there are a number of necessary conditions it must satisfy.
(5.18) theorem Let H £ G and let / = (1H)G. Then
(a) Z(D||G|;
(b) [z,£|£W)for,MIrr(G);
(c) [x, lc] = 1;
(d) x(g) is a nonnegative rational integer for all g e G;
(e) xig) < X(9m) for g e G and integers m;
(f) X(g) = Oifo(g)Jf\G\/x(l);
(g) x(g)\Cl(g)\/xW is integral for all g e G.
Proof Most of these assertions are immediate from the fact that z is a
transitive permutation character. Statement (b) follows since
[X, -A] = E(1h)G, •/'] = [!«, -Ah] ^(l)-
Statement (f) follows since |H| = |G|/z(l) and if o{g)Jf\H\, then no
conjugate of g lies in H and so (lH)G(g) = 0.
To prove (g), let Q be the set of right cosets of H so that x is the character
of the action of G on Q. Let X = Cl(g) and let y = {(a, x)| a e Q, x e Jf,
and ax = a}. Since / is constant on Jf, we have
\Jf\x(9) = \y\= ElXnGJ.
Since all Ga are conjugate in G, | Jf n GJ = m is independent of a and we
have | Jf |xig) = m\0.\ = m/( 1), proving the result. |
The necessary conditions given in Theorem 5.18 are definitely not
sufficient to guarantee that z is a permutation character. For example, the

70
Chapter 5
simple group M22, of order 443,520, has an irreducible character of degree 55.
The character obtained by adding the principal character to it satisfies the
conditions (a)-(g) and yet M22 has no subgroup of index 56.
The following result can often be used to find proper subgroups of a
group.
(5.19) theorem (Brauer) Let / be a character of G with [/, 1G] = 0. Let
A, B £ G and suppose
Then A and B generate a proper subgroup of G.
Proof Let U be a C[G]-module which affords / and let V, W and Y be
the subspaces of fixed points of U under A, B and A r\ B respectively. Then
V ^ Y,W£ 7,and
dim V + dim W = [jA, 1J + [/B, 1B] > [xAnB, lAnB] = dim Y.
It follows that V r\ W ^ 0 and thus {A, B} has nontrivial fixed points on U.
Since [/, 1G] = 0, we conclude that (A, B} < G. |
An example showing how Theorem 5.19 can be used is the following
proof that A6 is the unique simple group ofits order. [Since | A61 = |PSL(2, 9)|,
it follows that A6 s PSL(2,9).] As is often the case when trying to identify a
particular group up to isomorphism, there is a great deal of tedious and
detailed work involved. Since the result hardly seems to be worth that much
effort, the proof which follows is somewhat more sketchy than most in this
book.
(5.20) theorem Let G be simple and suppose \G\ = 360 = 23 • 32 • 5.
ThenG s A6.
Proof Suppose G0 Z G with | G: G01 = k > 1. Then G is isomorphic to a
subgroup of the alternating group Ak. Therefore k > 6 and the result follows
if we can find G0 with k = 6.
The only divisors of 360 which are = 1 mod 3 are 1, 4, 10, and 40. Let
PeSyl3(G) and N = NG(P). We conclude that \G:N\ = 10 or 40. By
Burnside's transfer theorem, N > P and hence | G: N \ = 10.
Now suppose P <=: M <G. We claim M £ N. If not, then 1 <
| M: M n N \ <, 10 and hence | M: M n N \ = 4 or 10 by Sylow's theorem.
Since the Sylow 3-subgroups of G generate all of G, we have | Syl3(M)| < 10
and thus \M: M n N\ = 4 and \G:M\ divides 10. Thus \G:M\ = 10 and
M r\ N = P. Now application of Burnside's theorem to M yields K<M
with | X | = 4. Thus 811 NG(X) | and P £ NG(X) so that | G: NG(X) | < 5. This
contradiction shows M £ N as claimed.

Induced characters
71
Now let P ± Pt e Syl3(G) and set D = P r\ Pv Then Pt £ N(£>) so
that N(£>) £ N. However, P s N(£>) and we conclude that N(£>) = G so
thatD = 1.
Next we claim that G contains no element of order 6. If x is an even
permutation on 10 points and o(x) = 6, then x must contain a 2-cycle and x2
fixes at least two points. Since no element of G of order 3 lies in two distinct
Sylow 3-subgroups, the claim follows. In particular, N/P acts faithfully on P.
Since the automorphism group of a cyclic group of order 9 has order 6, it
follows that P is elementary abelian. Furthermore, each involution in N
inverts all elements of P.
Now P = C(y) for each l#yeP and it follows that N has exactly six
conjugacy classes and thus exactly two nonlinear irreducible characters each
of degree 4. If 1G j= % e Irr(G), then %N must have one of these as a constituent
since N' £ ker /. Therefore x(l) > 4.
Let %e Irr(G) with /(l) odd. Since 4^/(1), it follows that Xn has a linear
constituent X and thus x is a constituent of XG. Therefore, /(l) < 1G(1) = 10
and if / t6 1g we have z(l) = 5 or 9.
We will show that G has an irreducible character of degree 5. From
Zxeirr(G) X(l)2 — 360 = 0 mod 4, it follows that there exist at least three
nonprincipal irreducible characters of odd degree. It suffices to show that
there is at most one of degree 9. In fact we show that if / e Irr(G) and /(l) = 9,
then / is a constituent of (lN)G. Indeed, / is a constituent of Xa for some linear
character X of N. Thus XG = / + y., where /z(l) = 1 and thus fi = 1G. Since
[fiN, 1] t6 0, it follows that X = 1^ as claimed.
Now fix i e Irr(G) with /(l) = 5. Then %N has a linear constituent and so
[Xp> lp] t6 0. Let v be a nonprincipal linear constituent of %P. Since every
element of P is real, %P is real valued and {jP, v] = [/P, v]. Let 1/ = ker v, so
that | U\ = 3. It follows that \jv, 1^] > 3 and hence by Theorem 5.19, any
two conjugates of U generate a proper subgroup of G. Note that we can find
ten conjugates of U such that no two of them centralize each other.
Let V be a conjugate of U which does not centralize U, and let H =
(U, V} < G. If \G: H\ = 6, we are done, so assume that |H| < 60.
Now U e Syl3(H) or else P S H and thus H £ N. This would imply
V £ P which is not the case. Since I/, V e Syl3(H), it follows that 3 x 4 or
3 x 10 divides |H| and thus | H\ = 12, 24, or 30.
If | H | = 30, then there exists S<fl with | S \ = 5. Therefore, | G: N(S) | =
1 mod 5 and divides \G:H\= 12. Thus |G: N(S)| = 6. We may assume that
this does not occur and hence |H| = 12 or 24. The only group of order 24
which is generated by elements of order 3 has a normal subgroup of order 8
and thus contains elements of order 6. We conclude |H| = 12 and there
exists K<H, elementary of order 4.
Next, we claim \jH, 1H] = 2. To see this, we use the fact that the unique

72
Chapter 5
nonlinear irreducible character 9 of H has degree 3 and is real. It follows that
[#i/> It/] = 1 and thus Xh = 9 + lt + 12 where (Xt)D = 1^. It follows that
1; = 1H and the claim is established. Also [xv, lu] = 3.
Now choose W conjugate to U with W £ H and W not commuting with
U. Let L= (U, W). We may assume |L| < 60 and thus \L\ = 12 and
H r\L = U. Since [/H, 1H] = 2 = [/L, 1 J, Theorem 5.19 applies again to
show R = <H, L> < G. Since R £ N, it follows that 9^1/? |. Reasoning as
before, \R\ ^ 24. Since 12||/?|, this forces \R\ = 60 and the proof is
complete. |
It is often convenient to be able to write a character of a group in terms of
characters induced from linear characters of subgroups of a specified type.
For example, an important theorem of Brauer (which we will prove in
Chapter 8) asserts that every character is a Z-linear combination of characters
induced from linear characters of nilpotent subgroups. For now we consider
only rational valued characters and cyclic subgroups. We prove the following.
(5.21) theorem (Artin) Let/be a rational valued character of G. Then
y an , g
*■ ^|N(H):H| " '
where H runs over the cyclic subgroups of G and aH e Z.
The proof of Artin's theorem depends on a well-known result of
algebraic number theory, namely that the cyclotomic polynomials are irreducible
over Q. In other words, for each positive integer n, all of the primitive nth
roots of unity are conjugate over Q. (This fact is also needed for Problem 3.11.)
(5.22) lemma Let i be a rational valued character of G and let x, yeG
with <x> = <y>. Then /(*) = x(y).
Proof Let n = o(x) = o(y) and let e be a primitive nth root of unity. We
have y = xm, with (m, n) = 1 and thus em is a primitive nth root of unity and
em = e" for some automorphism a of Q[e].
Now %(x) = Y.f=l ei' where each e; is an nth root of unity and hence is a
power of e. We have
x(y) = x(xl = E *r = £ e," = x(x)a.
Since x(x) is rational, we have x{x)a = %(x) and the proof is complete. |
Proof of 5.21 Define an equivalence relation = on G by x = y if <x>
and <>>> are conjugate in G. It follows from Lemma 5.22 that / is constant on
the equivalence classes under = and thus x is a Z-linear combination of the
characteristic functions of these classes.

Problems
73
Let #!, ^2,..., (€m be the distinct (= )-classes of G and let 4>, be the
characteristic function of <€t so that <&;(*) = 1 if x e #, and <J)((x) = 0, otherwise.
Choose representatives xte <€t and let Hf = <x(> and nt = \Ht\. We have
(1) |«f,| = |G:N(Hl)|<»(fil).
We prove by induction on nt that
(2) \nnd\^>i = Y.aJ^H1)G
i
for suitable as e Z where a,- = 0 unless H} is conjugate to a subgroup of Hf.
The result will then follow since \N(Ht)\ divides |N(H,)| when a} ¥= 0.
If n, = 1, then ^ = {1} and |G|<D( = (1H()G and (2) holds in this case.
Suppose Hi > 1. Write
(1h,)g = 5>,4>;
and compute the coefficients b3 as follows using Frobenius reciprocity and the
fact that [(D;, <Dk] = <5;k|<^.|/|G|:
b;|^.|/|G| = [(1H()G, <&j] = [lfl„ (*/)*,].
Now (<D;)H. = 0 unless H; is conjugate to a subgroup of Hj. In that case,
(<D;)Hi takes on the value 1 on <p{n}) elements of Ht and vanishes elsewhere. Thus
[1H(, (<bj)nj = <P(nj)/ni and we conclude that
bJ=\G\<p(nj)/nt\Vj\ = \N(HJ)\/nl
using (1).
Therefore,
n,(WG = ElN(H;)|<Di,
where j runs over the set of subscripts for which H3 is conjugate to a
subgroup of Hj. Solving for |N(H()|<I>; and applying the inductive hypothesis
yields (2) and proves the theorem. |
(5.23) corollary Let x be any rational valued character of G. Then
IG | X = Eh an l»°> wnere H runs over tne cyclic subgroups of G and aH e Z.
(5.1) Let H £ X £ G and suppose that <p is a class function of H. Show that
(9Kf = <PG-
(5.2) Let H, K £ G with H/C = G and suppose <p is a class function of H.
Show that (9G)K = (9HnK)K.

74
Chapter 5
(5.3) Let H £ G and suppose <p is a class function of H and i/f is a class
function of G. Show that {(p\liH)G = ip6!/'.
Note Problems (5.1)-(5.3) will be used frequently.
(5.4) Let b(G) = max{x(l)|/e Irr(G)}. If H £ G, show that b(H) < b(G) <
\G:H\b(H).
Note Since H is abelian iff b(H) = 1, the inequality b(G) <\G:H\b(H)
generalizes Problem 2.9(b).
(5.5) Let F be an arbitrary field and let H £ G. Let W be an F[H]-module.
Show that there is a unique (up to F[G]-isomorphism) F[G]-module V
such that V has an imprimitivity decomposition V = £ • Wf, where H is the
stabilizer of Wl and Ws W^ as F[H]-modules.
Notes By Theorem 5.9, V s WG. In the following problem, if V is an
F[G]-module and H s G, we use the notation FH to denote the module V
viewed as an F[H]-module.
(5.6) (Mackey) Let H, K £ G and let T be a set of double coset
representatives so that
G = (J HtK
is a disjoint union. Let W be an F[H]-module for an arbitrary field F. Show
that
(WG)K= l-(WH.nKf.
Note Mackey's theorem generalizes Problem 5.2.
(5.7) Let H s G and suppose \ji is a character of H. Let X £ G and assume
(tAG)K e Irr(X). Show that HK = G.
Note Problem 5.7 is an immediate consequence of Problem 5.6. It can,
however, be done without using modules. The key step is to show that
K^g)k, (ipHnicf] * 0 and then conclude that \G: H\ <\K:K r\ H\.
(5.8) Let / be a monomial character of G and suppose K c g with
Xk £ Irr(X). Show that %K is a monomial character of K.
(5.9) Suppose that F is a subfield of C and let H £ G. Suppose that 9 is a
character of H which is afforded by an F-representation. Show that 9G is
afforded by an F-representation.
Note It follows that every monomial character of G is afforded by a
O [^-representation, where e is a primitive nth root of unity, n being the

Problems
7S
exponent of G (that is, the least common multiple of the orders of the
elements of G). Clearly, an arbitrary character of G has values in Q[e]. In
Chapter 10 we shall prove the theorem of Brauer which asserts that every
character of G is afforded by a Q[e]-representation.
(5.10) Let H s G and view C[H] s C[G]. Let / be a right ideal of C[H]
and let J = 7C[G]. Suppose that I affords 9 (viewing I as a C[H]-module).
Show that J affords 9G.
(5.11) Let N <i G and /elrr(G). Suppose that [/N, 1N] # 0. Show that
N £ leer/.
Hint Use Lemma 5.11.
(5.12) Let cpG = / e Irr(G) with <p e Irr(H). Show that Z(x) £ H.
(5.13) (L. Solomon) Show that each row sum in a character table is a
nonnegative rational integer.
Hint Let G act on G by conjugation. Consider the permutation
character.
Note The column sums are also integers. They can be negative.
(5.14) Let G be nonabelian and let/ = min{x(l)|xeIrr(G), x(l) > l}.Show
(a) If|G'| < /thenG's Z(G),
(b) if | [G, G'] | <, f, then G' is abelian,
(c) ifHs Gand|G:H| </,thenG'£H.
Hint For (a). If x e G, then | Cl(x) | :£ | G' |.
(5.15) Let HsG and suppose <p is a character of H with det cp = 1H. Let
X = (pG and show (det x)2 = 1G.
(5.16) Let H be a maximal subgroup of G and let / = (1H)G. Let \p be a
nonprincipal irreducible constituent of /. Show ker \ji = ker /.
(5.17) Let H <=, G and let / = (1H)G. Fix a positive integer n. For g e G, let
m(#) = [/<»>> !<»>] and define 9(g) = nm<9). Show that 9 is a character of G.
Hint Let Q be the set of right cosets of H in G. Fix a set S with | S | = n
and let A be the set of all functions Q -> S. Then G acts on A.
Note This result provides another necessary condition to add to the
list in Theorem 5.18.

76
Chapter 5
(5.18) Let G be a group. Write a(n) = |{geG\o(g) = n}\. Then the
polynomial
/(*) = ( 1/| G|)£a(m)x'Gi/m
m
takes on integer values whenever xeZ.
Hint Use Problem 5.17.
(5.19) Let G be doubly transitive on fi and let H £ G with \G: H\ < |fi|.
Show that H is transitive on Q.
(5.20) Assume the notation and hypotheses of Problem 4.11 and show that
\G\ = q(q - l)(q + 1).
Hints Let N = N(S) and show \N\ = q(q- 1). Let Xelrr(N) with
1 ^ In and S £ ker 1. Let xeS, x 76 1. Show lG(x) = 1(1). Write 1G =
E^irr(o axx and conclude
4D = E «,(Z(1) - 1) = IG: N|l(l) - q £ flr
Finish the problem by showing that if ax =£ 0, then x(l) > q + 1 so that
Y.ax ^ W)- To show /(l) > <j + 1, compute [/s, Is] and use the fact that
since fe, 1] > 0 and / is real, then [/s, ls] ^ 2.
Note Using the result of this problem, it is not very hard to prove
that G = SL(2, q), a theorem originally due to Brauer, Suzuki, and Wall.
(5.21) Let P(G) denote the set of all Z-linear combinations of characters of
of G of the form 1HG for H £ Q. Show that P(G) is a ring.
(5.22) Let / be a rational valued character of G and let nG(x) be the least
positive integer such that nG(x)x £ P(G). Let N = ker / so that / may be
viewed as a character of G/N. Show that
nG(x) = nGIN(x).
Note The existence of nG(x) is guaranteed by Corollary 5.23.
Hint If nx = £%(1h)g, then nx = £ Ah(1\h)g- Show this by writing
(1H)G = (1,vh)g + £<H), where £m is a character of G with the property that N
is not contained in the kernel of any of its irreducible constituents.
(5.23) Let x be any character of G. Show that
\G\x = la,XG,
where 1 runs over the set of linear characters of cyclic subgroups and ax e Z.
Hint Use Problem 5.3.

Problems
77
(5.24) Let G be a doubly transitive permutation group on Q and let a, /? e Q,
with a#^. Let q> e Irr(Ga) and assume that (pG f e Irr(Ga^). Show that
[<pG, <pG] <; 2.
Hint Use Problem 5.6.
(5.25) Assume that every Sylow subgroup of G has a faithful irreducible
character. Show that G has one also.
Hint Let S be the product of all of the minimal normal subgroups of G.
(This is called the socle of G.) Then S is a direct product of simple groups.
Find 9 e Irr(S) such that ker 9 contains no nontrivial normal subgroup of G.
(5.26) (Passman) Let X £ G - {1} be a subset and let n = \X\. Assume
for all primes p < n, that a Sylow p-subgroup of G is cyclic. Show that there
exists i e Irr(G) such that X n ker(/) = 0.
Hints Reduce to the case that G is abelian by observing that it suffices
to assume that G = <AT> and that if [x, y] $ ker(x), then both x and y £ ker /.
In the abelian case, let N s G be maximal such that N r\ X = 0. Show that
G/N is cyclic.
Note If n = 2, the hypothesis on Sylow subgroups is vacuously satisfied.

Normal subgroups
Let x e Irr(G) and H s G. In general, very little can be said about the
restriction in • The situation is quite different if H is normal in G.
Let H <i G. If 9 is a class function of H and g e G, we define 99: H -> C
by 9('(/i) = $(ghg~l). We say that 9" is conjugate to 9 in G.
(6.1) lemma Let H o G and let <p, 9 be class functions of H and x, y e G.
Then
(a) (p* is a class function;
(b) {ffY = qf>\
(c) [>*,9*] = [>,9];
(d) [Zh, <P*] = [in, <p] for class functions % of G;
(e) <?)•* is a character if <p is.
Proof (a) If /i and k are conjugate in H, then so are /i9 and kg for all
geG. Take g = x_1.
(b) Compute. [Note that had we defined Bg(h) to be &(hg), then (b) would
fail.]
(c) The two sums are identical since xhx~l runs over H as h does.
(d) We have (i„f = i„. Apply (c).
(e) Let X afford cp and define 3E* by Xx(h) = X(xhx~l). Check that Xx
is a representation which affords cpx. |
It follows from Lemma 6.1 that G permutes Irr(H) by g: 9i-> 99. Note
that H acts trivially and thus G/H permutes Irr(H).

Normal subgroups
79
(6.2) theorem (Clifford) Let H <i G and let / e Irr(G). Let 9 be an
irreducible constituent of Xh and SUppOSC *? — i?i, i?2 > < • • •> **t
are the distinct
conjugates of 9 in G. Then
Xh = e£9i,
i=l
where e = [/„, 9].
Proo/ We compute (9G)H. For h e H, we have
I«I*eG IWI*eG
since xhx~1eH for all x e G. Thus |H|(9G)H = ^£fi 9* and hence if
<p e Irr(H) and <p i {9;}, we have 0 = [£ 9*, <p] and therefore [(9G)H, <p] = 0.
Since x is a constituent of 9G by Frobenius reciprocity, it follows that
[.Xh' 9] — 0- Thus all irreducible constituents of Xh are among the 9f and
Xh = p= i [Zh. 9i]9.- However, [*„, 9f] = [&,, 9] by Lemma 6.1(d) and the
proof is complete. |
Theorem 6.2 is so important that we digress to consider another proof in a
more general, module theoretic setting. If X is an F-representation of H <i G
for an arbitrary field F, we define the conjugate representation 3E9 for g e G
by X9(h) = X(ghg~l). Note that 3E9 is an F-representation which is irreducible
iff X is. Also, if 3Et and 3E2 are F-representations of H, then Xtg and 3E2* are
similar iff Xt and 3E2 are.
(6.3) definition Let H <i G and let Wt and W2 be F[H]-modules. Then
Wt is conjugate to W2 if there exist bases for the Wf such that the corresponding
F-representations of H are conjugate.
(6.4) lemma Let Vbe an F[G]-module and let H <G. Suppose W £ V
is an F[/TJ-submodule.
(a) If g e G, then Wg is an F[H]-submodule of V and is conjugate to W.
(b) If M is an F[H]-module conjugate to W, then M ^ Wg for some
#eG.
(c) If U s F is an F[H]-submodule isomorphic to W, then Ug = Wg
as F[H]-modules.
Proof We have (Wg)/i = W(ghg~ v)g =■ Wg since ghg~1 e H. Thus Wg
is an F[H]-submodule. Let wl5 w2, ..., w„ be a basis for W. Then wvg,
..., w„g is a basis for Wg. Let 3E and 9) be the F-representations of H
corresponding to W and Wg respectively, with respect to these bases. We claim that
9) = 3E».

80
Chapter 6
If X(ghg l) = (ai}), we have
w{ghg-1 =Z<*ijWj
}
and thus
(wtg)h = £ aij(Wj9l
j
Therefore, 9)(/i) = (ay) = Xe(h), establishing the claim. The proof of (a) is
now complete.
If M is conjugate to W, then M corresponds to the representation X9 for
some geG. Since Wg corresponds to the same representation, we have
Wg s M, proving (b). Finally, if U = W then with respect to a suitable basis,
U corresponds to X. Thus Ug and Wg both correspond to X° and (c) follows. |
(6.5) theorem {Clifford) Let H -a G and let 7 be an irreducible F[G]-
module. Let W be any irreducible F[/T]-submodule of V. Then
(a) 7 = Y,' Wf where the Wt are irreducible F[H]-submodules of V.
(b) Each Wj is of the form Wgx for some gteG and thus is conjugate to W.
(c) In the notation of Lemma 1.13, nw(V) = nM(V) for every F[H]-
module M conjugate to W.
Note Another way of saying (c) is that each isomorphism class of F[H]-
modules conjugate to W is represented equally often among the Wt.
Proof of Theorem 6.5 Clearly, £9eG Wg is G-invariant and hence
V= Yw3
geG
by the irreducibility of V. The F[H]-modules Wg are conjugate to W and
hence are irreducible. By Lemma 1.11, it follows that V is the direct sum of
some of the Wg. This proves (a) and (b).
Now let M be an F[H]-module conjugate to W. Since dim(M) = dim( W),
it follows from Lemma 1.13(b) that it suffices to show that dim(M(V)) =
dim(W(V)) in order to prove that nM(V) = nw(V).
Since M is conjugate to W, it follows by Lemma 6.4(b) that M = Wg
for some geG. Now 6.4(c) yields W(V)g £ M(V) and M(*0g-1 £ W(V).
Thus W(V)g = M(V) and the result follows. |
(6.6) corollary Let H -a G and let 7 be an irreducible F[G]-module for
some field F Then when viewed as an F[H]-module, V is completely
reducible.
Proof Use Theorems 1.10 and 6.5. |

Normal subgroups
81
Note that Corollary 6.6 is independent of Maschke's theorem and is
valid even if char(F) divides \H\.
Also note that Theorem 6.2 is a consequence of Theorem 6.5.
We now return to the study of C-characters and obtain some
consequences of Theorem 6.2.
(6.7) corollary Let H <i G and suppose that / e Irr(G) and [/H, 1H] ^ 0.
Then H £ ker /.
Proof We have Xh = e £ 9h where 9t = 1H and the 9f are conjugate.
Since (\Hf = 1H for all g e G, we have Xh = z(l)l« and H £ ker /. |
Corollary 6.7 can also be obtained as a consequence of Lemma 5.11 (see
Problem 5.11) since /is a constituent of (1H)G and ker(lH)G = f]HB = H.
An interesting class of problems in character theory arises from
considering how the structure of a group G and the set {/(1) | / e Irr(G)} are related. We
give one result of this type as an application of Theorem 6.2. Many more
results like this will be found in Chapter 12.
(6.8) lemma Let H <i G and suppose x £ Irr(G) and 9 e Irr(H) with
[XH,9]*0.Then9(l)|x(l).
Proof Since 99(1) = 9(1), we conclude that /(l) = et9(l) where Xh =
e Y!i= i fy as in Theorem 6.2 with 3 = 3^ I
(6.9) theorem Suppose x(l) is a power of the prime p for every / e Irr(G).
Then G has a normal abelian p-complement.
Proof If N <i G, then by Lemma 6.8,9( 1) is a power of p for all 9 e Irr(N).
Working by induction on | G |, we see that it suffices to find a normal
subgroup N of index p, since the normal abelian p-complement for N will be one
forG.
We have
\G\ = \G:G'\+ I /(I)2.
xeIrr(G):;(<l)>l
If G is abelian, the result is trivial. Otherwise, some / e Irr(G) is nonlinear and
thus p 11G |. Since the last sum is divisible by p, we see that p 11G: G' |. It
follows that G has a normal subgroup of index p and the proof is complete. |
The converse of Theorem 6.9 is true and will be proved later in this
chapter.

82
Chapter 6
(6.10) definition Let H <i G and let 9 e Irr(H). Then
7G(9)={0eG|9» = 9}
is the inertia group of 9 in G.
Since 7G(9) is the stabilizer of 9 in the action of G on Irr(H), it follows that
it is a subgroup and that/G(9) 2 77. Also | G : 7G(9) | is the size of the orbit of 9
and so in the formula
i=l
of Theorem 6.2, we have t = \G: /G(9)|. In particular, t divides \G:H\. It
turns out that e divides \G:H\ also, but that is much more difficult to show.
We shall prove some special cases in this chapter and the general result will be
proved in Chapter 11, using projective representations.
The following result is of fundamental importance in the character theory
of normal subgroups.
(6.11) theorem Let H -a G, 9 e Irr(H), and T = 7G(9). Let
s/ = Welrrmirjfo, 9] * 0}, £ = {/e Irr(G)|[/H, 9] * 0}.
Then
(a) If \jiestf, then i/fG is irreducible;
(b) The map \j/\-npG is a. bijection of si onto 3$\
(c) If t/fG = /, with xjiestf, then i/f is the unique irreducible constituent of
Xt which lies in si;
(d) If >AG = x, with tf, e.«/, then [>„, 9] = [zfl, 9].
Proof Let \fiesi and choose any irreducible constituent / of \jiG. Then
\ji is a constituent of /T and since 9 is a constituent of \pH, we conclude / e 34.
Let 9 = 9t, 92 9, be the distinct G-conjugates of 9. Then t = | G: T|
and xH = e £',_ i fy f°r some integer e. Since 9 is invariant in T, we conclude
from Theorem 6.2 that \pH = /9 for some/. Since \p is a constituent of xT,
we have/ < e.
Therefore
et9(l) = z(l) < tfrG(l) = ^(1) = /t9(l) < et9(l)
and hence equality holds throughout. In particular, x(l) = \jiG{\) and we
conclude that / = i/fG so that (a) follows. Also
[Zh,9] = <> = / = [>„, 9]

Normal subgroups
83
and (d) is proved. Statement (c) follows from the last equality since if i/f t e j/,
\pi 76 \p, and \pt is a constituent of xT, then
a contradiction.
The map in (b) is well defined by (a) and its image lies in Si by (d). It is one-
to-one by (c). To prove that it maps onto SH, let / e 38. Since 9 is a constituent
of Xh , there must be some irreducible constituent i/f of %T with [i/fH, 9] ^ 0.
Thus i(/6j/ and x is a constituent of \\iG. Therefore / = \\iG and the proof is
complete. |
(6.12) corollary Let /elrr(G) be primitive. Then for every Af<G,
XN is a multiple of an irreducible character of N.
Proof We have %N = e£j=1 9j, where the #,- are distinct and t =
|G:/G($i)|. By Theorem 6.11, i = i>° for some i/felnt/o^)). Since / is
primitive, it follows that /G(#i) = Gandr = l.Thusx,v = e9t and the proof is
complete. |
The converse of Corollary 6.12 is false. For instance the irreducible
character of A5 of degree 5 is imprimitive and yet the conclusion of the
corollary is (trivially) valid for this character. Characters which are multiples of an
irreducible are called homogeneous characters. Irreducible characters whose
restriction to every normal subgroup is homogeneous are often called
quasi-primitive characters.
(6.13) corollary Suppose G has a faithful primitive character and let
A <i G be abelian. Then A £ Z(G).
Proof Let / e Irr(G) be primitive and faithful. Then Xa = ek for some
linear character k. Thus A £ Z(/) = Z(G). |
(6.14) corollary Every nilpotent group is an M-group.
Proof Let G be nilpotent and let / e Irr(G). Choose H £ G minimal, such
that there exists \p e Irr(H), with / = \pa. By transitivity of induction (Problem
5.1), ij/ is a primitive character of H. It follows that \p is a faithful primitive
character of H/ker(i/f) = H and thus every normal abelian subgroup of H is
central. Since every nilpotent group contains a normal self-centralizing
subgroup, we conclude that H is abelian. Then \p is linear and the proof is
complete. I
We shall soon give a much more general sufficient condition for a group to
be an M-group. The preceding result is included here simply to illustrate the
usefulness of Theorem 6.11.

84
Chapter 6
(6.15) theorem (ltd) Let A o G be abelian. Then x(l) dividies | G: A \
for all x e Irr(G).
Proof Let A e Irr(A), with [ja, ^] # 0, and let T = 7G(1). Then for some
t^elrr(T), we have x = i>G and \jiA = el. Thus A £ Z(i/f) and hence t/f(l)
divides | T: A | by Theorem 3.12. Since /(1) = | G: T | i/f( 1), we conclude that
X( 1) divides | G: A | and the proof is complete. |
Note that the converse of Theorem 6.9 is immediate from Theorem 6.15.
Therefore, the purely group theoretic condition that G has an abelian normal
p-complement is precisely equivalent to the condition that /(1) is a power of
p for all x £ Irr(G).
Suppose N <i G and 9e Irr(N). Write 9G = £ e^,- for x,e Irr(G) and
e( > 0. Then (x^ = et £}=1 9Jf where 9 = 9t 9, are the distinct
conjugates of 9 in G. Let T = 7G(9). By Theorem 6.11 it follows that 9T =
£ e,i/fj with t/f j e Irr(T) and i/^6 = /;. It follows that for the purpose of
investigating the nature of the integers eh it suffices to assume T = G, that is, that
9 is invariant in G.
For the remainder of this discussion we assume that 9 is invariant. We
have
9G = £ elXl and (Z|)K = Cj9.
Therefore, Xj(1) = e,-9(l) and
\G: N\8(l) = 9G(1) = 5>jZj(l) = J e,29(l),
so that
$>.2 = |G:/V|.
It was remarked earlier (without proof) that ef divides \G: N\. (Note that
if N is abelian then ef = x,-(l) and in this case we know that et\\G: N\ by
Theorem 6.15.)
We see that in some respects, the integers et behave like character degrees
for G/N. If they were really character degrees, then one of them, say elt would
equal 1. That would mean that (Xi)N = 9. In other words, that 9 is extendible
to G. (Note that if 9 is extendible then it is automatically invariant.)
A consequence of the next result is that if some ex = 1, then the eK are
exactly the degrees of the irreducible characters of G/N.
(6.16) theorem Let N <i G and let <p, 9 e Irr(N) be invariant in G.
Assume cpS is irreducible and that 9 extends to / e Irr(G). Let Sf =
{P e Irr(G) | [<pG, ff] * 0} and 3T = {if, e Irr(G) | [(<p9)G, £| * 0}. Then 01-» fo
defines a bijection of y onto ^".

Normal subgroups
85
Proof We have (cpG)N is a multiple of (p and hence ((pG)N = \ G: N \ cp by
comparing degrees. Thus \_<pG, <pG] = \_q>, ((pG)N~\ = \G:N\\_<p,<p\ = \G:N\.
Similarly, [(<p9)G, (<p9)G] = | G: A/1. Also, by Problem 5.3, (<p9)G = (pGx-
Now write <pG = £,eJ, e,/J. We have |>G, <pG] = |G: N| = |>Gz, <pG/]
and hence
Since [fo, y/] ^ 0 and [fix, Pxi ^ 1 and all e0 > 0, this forces [fix, YX] = 0
if fi t6 y and [/?/, /?/] = 1. Thus the fix are distinct irreducible characters for
distinct fi. These are all of the irreducible constituents of q>Gx = (<p9)G. The
proof is complete. |
What is probably the most important special case of Theorem 6.16 is
when (p = In .
(6.17) corollary (Gallagher) Let N o G and let / e Irr(G) be such that
Xn = 9 e Irr(N). Then the characters fix for /? e lrr(G/N) are irreducible,
distinct for distinct fi and are all of the irreducible constituents of 9G.
Proof This is exactly Theorem 6.16 in the case (p = lN. |
Note that in Corollary 6.17 we have identified Irr(G/A0 with a subset of
Irr(G). In this situation, we see that the ex are exactly the fi(\) for fi e lrr(G/N).
There are other situations where we have control of the ef's. The following
"going down" theorem is useful in studying the characters of solvable groups.
(6.18) theorem Let X/L be an abelian chief factor of G. (That is, X,L<i G
and noM<G exists with L< M < K.) Suppose 9 e Irr(X) is invariant in G.
Then one of the following holds:
(a) 9LeIrr(L);
(b) 9L = ecp for some (p e Irr(L) and e2 = |X: L\;
(c) 9L = Yj= i <Pi where the <pte Irr(L) are distinct and t = \K\L\.
Proof Let cp be an irreducible constituent of 9L and let T = IG((p)-
Since 9 is invariant in G, every G-conjugate of q> is a constituent of 9L and
hence is X-conjugate to <p. It follows that \G:T\ = \K:K r\T\ and hence
KT = G. Since X/L is abelian, K r\ T o KT = G and thus either K r\ T =
K or K r\T = L.
If X n T = L, then 9L = e £j=! <?;, where t = |X: L|, <p! = <p, and the
<p{ are distinct. Thus 9(1) = e\K: L|<p(l). Since 9 is a constituent of <pK, we
have 9(1) < |X: L|<p(l) and therefore e = 1. This is situation (c).

86
Chapter 6
Now assume K r\ T = K so that <p is invariant in K and 9L = e<p for
some e. Let 1 e Irr(X/L). Since 1 is linear, 19 e Irr(X). Also (19)L = 9L = e<p.
Suppose that all of the characters 19 are distinct as 1 runs over Irr(X/L). Each
of these \K: L\ characters is an irreducible constituent of cpK with
multiplicity e and we have
e|K:L|9(l)<«/(l) = |K:L|<p(l).
Therefore,
e>(l) = eS(l) <: <p(l)
and e = 1. This is situation (a).
In the remaining case, 19 = /z9 for some X, fieIrr(X/L) with 1 j= fi. Let
1/ = ker(l/I). Then L ^ U < K and 9 vanishes on K — U. Since 9 is
invariant in G, it follows that 9 vanishes on K — Ua for all g e G. Since
D»eG Ue = L, we conclude that 9 vanishes on X — L. By Lemma 2.29 we
have
|X:L| = |X:L|[9,9] = [9L)9L] = e2
and the proof is complete. |
(6.19) corollary Let N<iG with \G:N\=p, a prime. Suppose
1 e Irr(G). Then either
(a) Xn is irreducible or
(b) Xn = E?= i &i> where the 9f are distinct and irreducible.
Proof TakeK = GandL = Nand apply Theorem 6.18. Case (b) of that
theorem cannot occur since p is not a square. |
(6.20) corollary Let N<iG and suppose \G:N\=p, a prime. Let
9 e Irr(iV) be invariant in G. Then 9 is extendible to G.
Proof Let / be an irreducible constituent of 9G. Then Z\ = e9 for some e.
Comparison with Corollary 6.19 yields e = 1. |
Actually, Corollary 6.20 would still be true if the hypothesis that | G: N \ is
prime were weakened to G/N cyclic. This stronger result will be proved in
Chapter 11.
(6.21) definition Let N o G and let / e Irr(G). Then / is a relative M-
character with respect to N if there exists H with N s H s G and t/f e Irr(H)
such that iiG = x and iAjv e Irr(iV). If every / e Irr(G) is a relative M-character
with respect to N, then G is a relative M-group with respect to N.
Note that / e Irr(G) is a relative M-character with respect to 1 iff it is a
monomial character, and G is a relative M-group with respect to 1 iff it is an

Normal subgroups
87
M-group. Also, it is clear that if G is a relative M-group with respect to JV,
then G/N is an M-group.
The converse of the last statement is false. If G = SL(2,3) and JV = Z(G),
then G/N = A 4, which is an M-group. However, G is not a relative M-group
with respect to JV since G has an irreducible character of degree 2 and has no
subgroup of index 2 as would be required by the definition.
(6.22) theorem Suppose JV <a G and G/N is solvable. Suppose,
furthermore, that every chief factor of every subgroup of G/N has nonsquare order.
Then G is a relative M-group with respect to JV.
Note The hypotheses on G/N, above, are automatically satisfied if G/N
is nilpotent or supersolvable since then all chief factors of subgroups have
prime order.
Proof of Theorem 6.22 Let / e Irr(G). We must show that / is a relative
M-character with respect to JV. If %N is irreducible, this is clearly the case and
hence we assume Xn reduces. Let K -o G be minimal such that K 2 JV and
Xk is irreducible. Then K > JV and we may choose L <i G, L 2 JV such that
K/L is a chief factor of G. In particular, K/L is abelian of nonsquare order.
We apply Theorem 6.18 to the chief factor K/L and the G-invariant
character xKeIrr(X). Since xl is reducible and \K:L\ is nonsquare, we
conclude that Xl = S=i •?■> wnere trie <?;eIrr(L) are distinct and t =
\K:L\ > 1.
Let T = IG((pi) 2 L 2 JV. By Theorem 6.11, we conclude that x = 4>G
for some \ji e Irr(T). Since T/N < G/N, we may apply induction on | G: JV|
to conclude that T is a relative M-group with respect to JV and that \ji = 9-T
for some 9 e Irr(H) where JV s H s T and SN e Irr(JV). Now x = ^° = (8Tf
= 9G by Problem 5.1 and the proof is complete. |
Note that Corollary 6.14 follows immediately from this result as does the
somewhat more general fact that all supersolvable groups are M-groups.
We prove a still more general sufficient condition.
(6.23) theorem Let JV o G and suppose that all Sylow subgroups of JV
are abelian. Assume that G is solvable and is a relative M-group with respect
to JV. Then G is an M-group.
Proof Let / e Irr(G). Since / is a relative M-character with respect to JV,
choose HsG with JV s H, \\i e Irr(H), \j/N e Irr(JV) and \pa = x- Now choose
U s H, minimal such that there exists 9 e lrr(U), with 9H = \p. By Problem
5.1, 9G = (9G)H = \jiG = x and it suffices to prove that 9 is linear.
Let M = U n,N. Since (&NV)tl = \p and \jiNV is irreducible, it follows
that \jiNV = $NV and thus (Sw% = ^ e Irr(JV). By Problem 5.2, ($M)N =
(&NV)N e Irr(JV) and hence 9M e Irr(M).

88
Chapter 6
By the minimality of U and Problem 5.1, 9 is a primitive character of U.
Let K = ker_9, V =_U/K and M = MK/K. Then 9 is a faithful primitive
character of U and M has all of its Sylow subgroups abelian. Now let Z =
Z(M) <a L7. If Z < M, let AjZ be a chief factor of {J with A <= M. Then X/Z
is a p-group for some prime p. Let P e Sylp(y4). Then P is abelian and A = PZ,
so that A is abelian. Since A <i I/, we conclude from Corollary 6.13 that
A s Z(E7). This contradicts A > Z.
We conclude that Z = M, and hence M is abelian. Since 9M is irreducible,
so is 9MK. Thus 9MK e Irr(M) and hence 9(1) = 1. The proof is complete. |
We introduce some notation. If / is a character of G, let det / = 1 be the
uniquely defined linear character of Problem 2.3. Now write o(%) = o(X), the
order of X as an element of the group of linear characters of G. (We call o(%)
the determinantal order of /.) Since \G: ker X\ = o(X), it follows that o(%)
divides \G\.
Now let JV o G and suppose 9 e Irr(iV) is invariant in G. We wish to find
sufficient conditions that 9 be extendible to G. If (| G: N |, 9(1)) = 1, then 9 is
extendible iff det 9 is extendible. We shall prove this now under the additional
hypothesis that G/N is solvable and defer the general proof to Chapter 8. As
will be seen, there is a gain in replacing the problem of extending 9 by that of
extending the linear character, det 9.
(6.24) lemma Let N <i G and suppose 9 e Irr(iV) is extendible to G and
that (| G: N \, 9( 1)) = 1. Let X = det 9 and let fi be an extension of X to G. Then
there exists a unique extension / of 9 to G such that det / = fi.
Proof Let >/ be an extension of 9 to G and let v = det r\ so that vN = X.
Let fiv = a. so that txN = lN and hence a|G:iV| = 1G. Write/ = 9(1) so that
det(afc>/) = abf det rj = a.bfv for beZ. Since (/, \G: N\) = 1, we can choose
b e Z such that ft/ = 1 mod|G:iV| and thus a?f = a. Let / = a.brj. Then
det i = <xbfv = av = y.. Since <xN = \N, we have /^ = >/w = 9 as desired.
To prove uniqueness, suppose %0 is an extension of 9 with det(/0) = fi.
By Corollary 6.17, we have %0 = //? for some fi e lrr(G/N). Since /0(1) =
9(1) = /(l), we have 0(1)= land
fi = det(z0) = p det x = PfH
and/?' = lG.Since(/, |G:iV|) = 1, this forces /? = lGand/0 = /.The proof
is now complete. |
(6.25) theorem Let N <a G and suppose G/N is solvable. Let 9 e Irr(iV)
be invariant in G and suppose (9(1), | G: N\) = 1. Then 9 is extendible to G
iff det 9 is extendible to G.

Normal subgroups
89
Proof If 9 is extendible, then obviously so is det 9. We prove the
converse. Let y. be an extension of 1 = det 9 to G. We work by induction on
| G: JV|. (We may assume G > JV.) Let G0 be a maximal normal subgroup of
G with G0 ^ JV. Since X extends to /zGo, the inductive hypothesis guarantees
that 9 is extendible to G0. By Lemma 6.24, there is a unique extension,
Xo e Irr(G0) with det(x„) = Hg0-
We claim Xo is invariant in G. For g e G, we have fo0)% = 99 = 9 and
det((x0)») = det(x0)9 = 0zG/ = /zGo. The uniqueness of /„ now yields (xoY =
Xo and establishes the claim.
Since G/N is solvable, | G: G01 is prime and by Corollary 6.20, Xo is
extendible to G. The proof is now complete. |
We now discuss some sufficient conditions for extending linear
characters. A version of the following theorem is true without the assumption
of linearity. The general result will be derived from this special case in
Chapter 11.
(6.26) theorem Let JV<G and suppose X is a linear character of JV
which is invariant in G. For each prime p | o(l), choose Hp £ G with
Hp/N e Sylp(G/JV), and assume that X is extendible to Hp. Then X is extendible
toG.
Note If p Jf | G: JV |, then Hp = N and 1 is automatically extendible to
Hp. It is only necessary, therefore, to check the hypothesis for primes dividing
|G:JV|.
Proof Let m = o(X). For each p | m, we may choose lp, a power of 1, such
that X = Y\ Xp and o(Xp) is a power of p. We shall show that Xp is extendible to
fip e Irr(G). Then fi = J~[ fip is an extension of 1.
Since lp is a power of 1, which is extendible to Hpi it follows that Xp is
extendible to Hp. Also lp is invariant in G. We now see that it is no loss to
assume that m is a power of p.
Let v be an extension of X to Hp. Since p Jf\G: Hp\ and v(l) = 1, it follows
that p/fv^l) and hence there exists an irreducible constituent / of vG with
p^(l).Letz(l)=/.
We have \_xH , v] ^ 0 and hence \_xN, X~\ ^ 0. Since X is invariant in G,
we conclude that Xn = /^ and thus (det x)w = J/. Let (5 = det /. Since p Jf f
and m is a power of p, we may choose b e Z with/ft = 1 mod m. Then ((5% =
Xfb = X and the proof is complete. |
(6.27) corollary Let JV <i G and suppose X is a linear character of JV
which is invariant in G. Assume (|G: JV|, o(l)) = 1. Then X has a unique
extension /z to G with the property that (| G: JV |, o(fi)) = 1. In fact, o{y) = o(l).

90
Chapter 6
Proof The existence of an extension is immediate from Theorem 6.26
since the hypotheses are trivially satisfied. Let v be an extension of X and
choose beZ, with b\G:N\ = 1 mod (o(l)). Let fi = vfc|G:JV|. Then fxN =
Xb\G-.N\ = x In particuiar) 0^ > 0(X). On the other hand, (voW)N = lN,
so that v0<A) e Irr(G/JV) and
n«W _ ivoWbyG:N\ _ j
Therefore, o(y) = o(X).
For uniqueness, suppose t is an extension of X with (o(t), \G: N\) = 1.
Then (o(/zf), \ G: N |) = 1 and (fit )N = lN so that /zf e Irr( G/JV). It follows that
\ix = 1G and n = t as desired. |
(6.28) corollary Let JV <i G with G/N solvable and suppose 9 e Irr(iV)
is invariant in G. Assume (|G: JV|, o(9)9(1)) = 1. Then 9 has a unique
extension, x g Irr(G) with (| G: JV|, o(z)) = 1. In fact o(%) = o(9).
Proof By Corollary 6.27, let y. be the unique extension of X = det 9 to
G with (|G: JV|, o(/z)) = 1. By Theorem 6.25 and Lemma 6.24, let x be the
unique extension of 9 to G with det x = V- Then o(x) = o[/j.) = o(X) = o(9).
If Xo extends 9 and (| G: JV |, o(x0)) = 1, let /z0 = det(z0). Thus
(|G:JV|,o(/z0))=l
and hence /z0 = fi by the uniqueness of /z. Then z0 = z by the uniqueness
of /. This completes the proof. |
Note that since o(9) and 9(1) divide |JV|, the condition that
(|G:JV|, o(9)9(l))= 1
in Corollary 6.28 will be automatic if (|G: JV|, |JV|) = 1. The hypothesis of
solvability in Corollary 6.28 will be removed in Chapter 8.
We shall give one further extendibility criterion now. It can be used when
G/N is a p-group for p^2. Unlike the previous results, this condition is
independent of character degree.
(6.29) definition Let x be a character of G and let p be a prime. Then x is
p-rational if there exists an integer r with pjfr, such that x(g) £ Q[e2"'/r] for
every g e G.
We use the notation Qk = Q[e2"i/k] for 1 < keZ. As is well known,
Qk n Q, = Qd, where d = (k, I). By Lemma 2.15, it follows that /(#) e Q|G|
for every character x of G and every g e G. Write \G\ = n = p"m, with p Jfm.
It follows that x is p-rational iff all of its values lie in Qm.
Let yp(G) denote the Galois group 9(QJQm). If / is a character of G,
let x" be defined by x"ig) = lig)" for a e 9(QJQ). Then / is p-rational iff/" =
X for all a e ^p(g) [Note that by Problem 2.2, /" is necessarily a character. In
particular, @P(G) permutes Irr(G).]

Normal subgroups
91
From Galois theory we have 9(QJQm) = 9(Qpa/Q), where as above
n = mp" and p Jf m. This isomorphism is the restriction map. It follows if
p 76 2 and a > 0, that e2nilp is not invariant under 9(QJQm) and hence if
p#2 and X is a linear character of G with p | o(l), then 1 is not p-rational. Also,
when p 7^ 2, we have <S^G) s 0(OP„/O) is cyclic.
We have one more general remark. If H £ G, then Q|H| £ Q|G| and so x"
is defined for characters / of H with a e ^P(G). It follows that / is p-rational
ifiFz" = z for all (xe^G).
(6.30) theorem Let JV <G with G/JV a p-group, p#2. Suppose 9 e Irr(JV)
is invariant in G and p-rational. Then 9G has a unique p-rational irreducible
constituent /. Furthermore, Xn = &
Proof Use induction on \G:N\. We suppose |G:JV| > 1. Let K<G
with JV £ X and |K : JV| = p. By Corollary 6.20, 9 can be extended to
rj e Irr(X). Since p ¥= 2, ^P(K) is cyclic and we choose a generator er. Now
>/" e lrr(X) and (>/% = 9" = 9 since 9 is p-rational. By Corollary 6.17, we
have rf = rjX for some (linear) X e lrr(K/N).
If X = lK, then rf = r\ and >/ is p-rational. Suppose that this is not the
case. Then p = o(X) and since p ¥= 2v/e have X" ^ X and so 1" = lm for some
m e Z, with m # 1 mod p. Choose b e Z with (1 — m)ft = 1 mod p and let
^ = X\
Since AN = ljv, we have \pN = rjN = 9. Also, mb + 1 = b mod p and thus
ji' = (A)" = Xmb+lti = Xbrj = \j/
and \j/ is p-rational. Thus in any case, 9 has a p-rational extension, \p e Irr(X).
If <p e Irr(X) is any p-rational extension of 9, then cp = ip/j. for some unique
fi e lrr(K/N) by Corollary 6.17. We have cp = cp" =,(^n)a = ^"n" = #" and
thus n" = fi. Since p # 2, it follows that p Jf o(n) and thus fi = lK. Therefore
<p = \p.
Now let g e G. Then W = (\p"Y = \pe and 9» = 9 so that \pe is a p-
rational extension of 9 to K and by the preceding paragraph, we have
\jig = ip and \ji is invariant in G.
Since \G:K\ < \G: JV|, the inductive hypothesis guarantees that ipG has
a unique p-rational irreducible constituent / and that Xk = "A s° that Xn = &
Let x0 be any p-rational irreducible constituent of 9G. We show Xo = X-
Let cp be an irreducible constituent of (x0)k- Since (x0)N is a multiple of 9, it
follows that cpN = 9and<p = t/n> for some v e Irr(X/JV). Since K/N £ Z(G/JV),
we have v° = vforgeG and thus cpe = \peve = \pv = cp and cp is invariant in G.
It follows that (xo)k = e<P for some integer e and thus cp is p-rational since
cp(k) = (l/e)x0(k) for keK. Therefore cp = \ji and Xo is a constituent of i/fG. It
follows that Xo = X and the proof is complete. |

92
Chapter 6
We can use some of our results on character extendibility to prove Tate's
theorem. This result, which was originally proved in an entirely different way,
serves as a "booster" for transfer theory, as will be explained.
We define Op(G) to be the unique minimal normal subgroup of G such
that G/Op(G) is a p-group (for the fixed prime p). Similarly, let AP(G) be the
unique minimal normal subgroup of G such that G/AP(G) is an abelian p-
group.[Note that AP(G) = G'Op(G).]
Let P e Sylp(G) and let JV 2 P. Then JV is said to control p-transfer if
JV/AP(JV) s G/AP(G), or equivalently, JV n AP(G) = AP(JV). Several of the
standard transfer theorems assert that under suitable hypotheses, certain
subgroups JV control p-transfer. [Usually, JV = N(W) for some subgroup,
W <a P.] Tate's theorem guarantees that whenever JV/AP(JV) ^ G/AP(G),
then also JV/Op(JV) z G/Op(G).
(6.31) theorem (Tate) Let PeSylp(G) and JV 2 P. Suppose that
JV n AP(G) = Ap(JV).ThenJV n Op(G) = Op(JV).
Proof (Thompson) Since JV 2 P, we have JVOp(G) = G. Let U =
N r\ Op(G) so that U -a JV and N/U S G/Op(G) is a p-group. Thus U 2
Op(JV). Assume U > Op(N) and choose V <a JV, 7 2 Op(JV) such that t//7
is a chief factor of JV. Let 1 be a nonprincipal linear character of U/V. Since
U/V is a chief factor of the p-group JV/Op(JV), we conclude that [//Ks Z(JV/F)
and hence 1 is invariant in JV.
Let 9 = kOP(G). If/ is an irreducible constituent of 9 and x e JV, then since
A* = A, we have
[X, 9] = [Xv. *] = l(Xx)u, X*] - [_(xx)v, X] = If, 8].
Since P £ JV, it follows that we may write
A
as A runs over the sums of the orbits of Irr(Op(G)) under the action of P.
Now 9(1) = \Op(G):U\ = | G: JV | is prime to p. It follows that for some A,
we have p Jfa^ and p Jf A(l). Write A = £^0 / where 0 is an orbit. Since /(l)

Normal subgroups
93
is constant for / e 6, we conclude that pjf\&\. Since 0 is an orbit under the
p-group P, we have 10 \ = 1 and A = /, where / e Irr(Op(G)), / is invariant
underP,p^x(l)andp^[9,x].
Since POp(G) = G and / is invariant under P, it follows that / is
invariant in G. We claim that / is extendible to G. By Corollary 6.28, it suffices to
show that p X Z(l)o(z)- Now Op(G) has a normal subgroup, ker(det /) = K,
such that |Op(G):K| = o(x) and Op(G)/K is abelian. Since Op(Op(G)) =
Op(G), it follows that Op(G) has no nontrivial p-factor group and thus p Jf o(x).
Since p X Z(l)» Corollary 6.28 does apply.
Let i/f be an extension of / to G and write
<pelrr(N)
We have
E K&v, W = D/V i] = Lxv, i~] = Ez, 9],
which is prime to p. We may therefore choose q> e lrr(N) with p ^ [pj,, 1]. In
particular, \_(pv, l~] =£ 0 and since 1 e Irr(t/) is invariant in JV, we conclude
that <pD = el and thus <p(\) = e = \_(pv, 1] is prime to p. Since V £ ker I
and <pj, = el, we conclude that 7 £ ker <p and <p e lrr(N/V). Since JV/7 is a
p-group, <p(l) is a power of p and hence <p(l) = 1.
Now JV/ker <p is abelian and thus AP(JV) £ ker q>. However, U =
N r\ Op(G) £ JV r\ AP(G) = AP(JV) and therefore 1/ £ ker (p. Since <?„ = el, it
follows that I = \v. This contradicts the choice of I and completes the proof
of the theorem. |
We discuss one further topic in this chapter. Let JV <i G so that G
permutes Irr(JV). It is also true that G permutes the set of conjugacy classes of JV
and it is natural to consider the relationship between these two actions. It is
not the case that they are necessarily permutation isomorphic although they
are closely related.
(6.32) theorem (Brauer) Let A be a group which acts on Irr(G) and on
the set of conjugacy classes of G. Assume that %(g) = x'ig") for all / e Irr(G),
aeA and geG; where g" is an element of Cl(g)a. Then for each ae A, the
number of fixed irreducible characters of G is equal to the number of fixed
classes.
Proof Let /, and Jf; be the irreducible characters and conjugacy classes
of G for 1 < i, j < k. Choose g} e Jf 3 and write g" = gs if Jf," = Jts. Let
X = (Xiigj)), the character table of G, viewed as a matrix.

94
Chapter 6
For as A, let P(a) = (py), where py = 0 unless Xi" = Xj> in which case
ptJ = 1. Similarly, define Q(a) = (<jy), where qtJ = 1 if Jf," = Jf j and is zero
otherwise.
The (w, v) entry of the matrix P(a)X is
since only when /,■ = Xu is pui j= 0. Similarly, the (u, v) entry of the matrix
XQ(a) is
Yxuidj)Qjv = Xuidav~')
j
since only when gj = g"„~l is <?;„ ,£ 0.
The hypothesis of the theorem now implies that P(a)X = XQ(a). Thus
Q(a) = X~lP(a)X since the orthogonality relations guarantee that X is
nonsingular. We conclude that tr P(a) = tr Q(a). Since tr P(a) is the number
of / e Irr(G) which a fixes and tr Q(a) is the number of a-fixed conjugacy
classes, the proof is complete. |
(6.33) corollary Under the hypotheses of Theorem 6.32 the numbers of
orbits in the actions of A on the irreducible characters and conjugacy classes
of G are equal.
Proof Apply Corollary 5.15 to the result of Theorem 6.32. |
We may apply Theorem 6.32 to obtain information about the characters
of "Frobenius groups."
(6.34) theorem Let N <i G and assume that CG(x) £ N for every
1 t6 x e N. Then
(a) For cp e Irr(iV), with cp ^ \N, we have IG((p) = N and <pG e Irr(G).
(b) For x e Irr(G) with N £ ker /, we have x = 9° f°r some q> e Irr(iV).
Proof Let <peIrr(iV), ^ 1^. To show that <pGeIrr(G), it suffices by
Theorem 6.11 to show that IG((p) = N. In order to prove (a), therefore, it
suffices to show that no element g e G — N can normalize any nontrivial
conjugacy class of N and then apply Theorem 6.32.
Suppose then, geG — N normalizes Jf, a class of N. Let x e Jf. Then
xaeJf and thus xa = x" for some neN. Therefore gn~leC(x). Since
gn~l $N and xeN, the hypothesis yields x = 1 and thus Jf = {1}. This
proves (a).
Now let x £ Irr(G) with N £ ker /. Choose an irreducible constituent cp
ofxN with cp =£ lN. Then x is a constituent of <pN which is irreducible and so
cpN = x- The proof is now complete. |

Problems
95
It turns out that the groups G satisfying the hypotheses of Theorem 6.34
and for which N < G are exactly the "Frobenius groups" which are
discussed at some length in the next chapter.
Problems
(6.1) Let JV -a G and 9 e Irr(JV). Show that 9G e Irr(G) iff 7G(9) = JV.
(6.2) Let N <i G and suppose G/N is abelian. Let C be the group of linear
characters of G/N so that C acts on Irr(G) by multiplication. (See Problem 2.6.)
Let 9 e Irr(JV). Show that
where/is an integer and the & e Irr(G) constitute an orbit under C.
Hint Let / e Irr(G). Then (%N)G = p%, where p is the regular character
of G/N.
(6.3) Let N -a G and let / e Irr(G) and 9 e Irr( JV) with \_%N, 9] ± 0. Show that
the following are equivalent:
(a) Z„ = e9, withe2 = \G:N\;
(b) x vanishes on G — JV and 9 is invariant in G;
(c) x is the unique irreducible constituent of 9G and 9 is invariant in G.
JVote In the situation of this problem, we say that / and 9 are fully
ramified with respect to G/N.
(6.4) Define Jf(G) = {H £ G | cpa e Irr(G) for some cp e Irr(H)}. Let H(G) =
C\Jf(G). Show that if G is an M-group, then H(G) is abelian.
JVote By Problem 5.12, Z(G) £ H(G) for all groups, G. Of course, H(G)
is characteristic in G.
(6.5) Let H(G) be as in the preceding problem. Show that H(G) centralizes
N/N' for every JV <i G and conclude that if G is solvable, then H(G) is nil-
potent.
Note In fact we may conclude that if G is solvable, then H(G)' £
Z(H(G)). This is so because a group which is nilpotent of class > 2 necessarily
has a characteristic abelian noncentral subgroup; namely, the next to the last
nontrivial term in its lower central series.
Hint Ucpe Irr(JV) with N o G, then H(G) £ IG(cp). Use this to show that
[JV, H(G)] £ JV'.

96
Chapter 6
(6.6) Let G be solvable and assume that every / e Irr(G) is quasi-primitive.
Show that G is abelian.
(6.7) Let JV <G and assume that G/N is solvable. Let / e Irr(G) and
9 e Irr(JV), with [*„, 8] * 0. Show that x( 1 )M 1) divides \G:N\.
Note The conclusion of this problem is valid even if G/N is not solvable.
(6.8) Suppose that G has exactly one nonlinear irreducible character. Show
that G' is an elementary abelian p-group.
(6.9) (Dornhoff) Let G be an M-group and suppose JV -o G with
(|JV|,|G:JV|) = 1. Show that JV is an M-group.
Hint Let 9eIrr(JV). Find H £ G and lelrr(H), 1(1) = 1 such that
[(!*%, 9] * 0 and XG e Irr(G). Use Problem 6.7 to show that lra(l) = 9(1)
and then use Problem 5.8.
(6.10) Let JV«a G with (|G: JV|, |JV|) = 1. Suppose that every subgroup of
G/N is an M-group. Show that G is a relative M-group with respect to JV.
(6.11) Let A o G with A abelian. Suppose / e Irr(G) is a monomial
character. Show that / is a relative M-character with respect to A.
Hint Write / = 1G, with k a linear character of H <=, G. Then every
irreducible constituent of (kAnll)A has multiplicity 1.
Note Since Problem 6.11 is false if A is not abelian, it does not follow
(and is not true) that an M-group is necessarily a relative M-group with
respect to every normal subgroup.
(6.12) Let K/L be an abelian chief factor of G. Let <p e Irr(L) and T = IG((p).
Assume that KT = G. Show that one of the following occurs:
(a) <pKelrr(K);
(b) <pK = e9 for some 9 eIrr(X), where e2 = \K:L\.
(c) cpK = Y!t=i &h where the 9;eIrr(K) are distinct and t = \K:L\.
Hint Use Problem 6.2 and the ideas in the proof of Theorem 6.18.
(6.13) Show that the number of real classes of a group G is equal to the
number of real valued / e Irr(G).
Note It is easy to see that if | G | is odd, then 1 is the only real element. It
follows that Problem 6.13 generalizes Problem 3.16.
(6.14) Let F be a field with Q £ F £ C. Say that g e G is an F-element if
X(g) e F for every / e Irr(G). Let G be a p-group with p^2 and show that the

Problems
97
number of classes of F-elements of G is equal to the number of / e Irr(G) with
values in F.
Hint Let e be a primitive | G | th root of unity. Let ^ be the Galois group
of F[e] over F. Define actions of <& on Irr(G) and on the set of classes of G.
Note Problem 6.14 is false without some assumption on G.
Counterexamples with G a 2-group and G of odd order have been found by Thompson
and Dade respectively.
(6.15) Let F be an arbitrary field and let N-a G. View F[iV] £ f[G]. Show
that J(FrjV]) £ J(F[G]). (See Problem 1.4.)
(6.16) Let P be a p-group and Q a <j-group, where p and q are distinct odd
primes and P £ Aut(Q). Show that \P\ < \\ Q| by carrying out the following
steps.
(a) It suffices to assume that Q is elementary abelian, that is, is an F\_P]-
module with F = GF(q).
(b) It suffices to assume that Q is an irreducible F[P]-module.
(c) We may assume that P is not cyclic, so that by theorems on p-groups,
it follows that there exists A <a P with A elementary abelian of order p2.
(d) Q is imprimitive as an F[P]-module and we may choose H £ P,
\P:H\ = p, with Q = £ • f= i Wf, each W{an F[H]-module with all | Wt\ equal.
(e) Let N{ be the kernel of the action of Hon Wt. Then
iPi<Pnitf--^-i-
(f) Complete the proof by induction.
Hint Clifford's theorem (6.5) is used in (d). Show that F[A] cannot have
a faithful irreducible module.
Notes The result of Problem 6.16 may be stated as a theorem in
arithmetic. If Q is elementary abelian of order q", then | Aut(Q) | = f] "r/ (qn — q')
and hence this number is not divisible by p" if p" > \q".
The induction in the previous step (f) would not go through to prove the
weaker theorem that | P | < \Q\.
(6.17) Let N o G with G/N cyclic. Let 9 e Irr(iV) be invariant in G and
assume that (9{l),\H:N\)= 1. Show that 9 is extendible to G.
Hint Show that G/ker(det 9) is abelian.
Note The result of Problem 6.17 is true without the assumption that
(9(l),|G:iV|)= 1.

98
Chapter 6
(6.18) Let N <a G and suppose G = NH with N n H = 1. Let 9 e Irr(JV) be
invariant in G and assume (9( 1), | G: JV |) = 1. If H is solvable, show that 9 is
extendible to G."
Note The result of Problem 6.18 is true without the assumption that H
is solvable. However, even if H is abelian, the condition (9(1), | G: N\) = 1
cannot be removed.
(6.19) Let N o G with G/N a p-group and let 9 e Irr(JV) be invariant and p-
rational and assume p Jf o(9) and p ^ 9(1). Let / be the extension of 9 to G
with p X o(x) (which exists by Corollary 6.28.) Show that / is p-rational.
(6.20) (Thompson) Let EP(G) denote the minimal normal subgroup of G
such that G/EP(G) is an elementary abelian p-group. Suppose N £ G with
p Jf | G: N\ and EP(JV) = EP(G) n JV. Show that Op(N) = Op(G) n JV.
Hints Sharpen the proof of Theorem 6.31 as follows. Choose 1 so that
1" = lv. Let <$ = 0(O|G|/OP) and let A e Sylp(0). Now A permutes Irr(H) for
all H £ G and X is invariant under A. Now proceed with the proof of
Theorem 6.31 but arrange matters so that x, i> and q> are invariant under A.
(6.21) Let 1 = H0 < H[ < • • ■ < H„ = G. Assume that Hj/H,-_ t is non-
abelian. Show that there exists / elrr(G) with /(l) > 2".
Hint Use Corollary 6.17.

T.I. sets and exceptional characters
Suppose we know that G has a subgroup H with certain specified
properties and assume that we have some information about how H is embedded
in G. How can we draw conclusions about G? We might try inducing the
irreducible characters of H to G. Usually this gives little information since
if | H | is much smaller than | G |, the characters 9G tend to have large numbers
of irreducible constituents with large multiplicities, even if 9 e Irr(H).
In certain situations, however, one can find a difference of two characters
9t — 92 of H where ($t — 92)G = Xi — Xi and Xi afld Xi are under control.
In these situations, information about Irr(G) can be obtained.
The earliest example of the use of this technique is in the proof of
Frobenius' theorem. We shall give this before discussing any of the later
refinements of these ideas.
In Theorem 6.34 we considered groups G having a normal subgroup JV
such that CG(x) £ JV for all 1 j= x e JV. It follows immediately using Sylow's
theorem and the fact that nontrivial p-groups have nontrivial centers that
(|JV|, |G:JV|) = 1. By the Schur-Zassenhaus theorem we conclude that
there exists H s G with NH = G and JV r\H = 1.
In this situation, let g e G — H. Write g = xn with x e H and 1 j= n e JV.
If y e H r\ Ha, then y e H" and y = h" for some he H. Since y e H, we have
[h, n] = h~ 1h" = h~lye H. Since JV <a G, we have [h, n] e H r\ N = 1 and
h e C(n) £ JV. Thus h = 1 and y = 1. We conclude that H r\ HB = 1.
(7.1) definition LetH £ G, with 1 < H < G. Assume that H r\ Hg = 1
whenever g e G — H. Then H is a Frobenius complement in G. A group which
contains a Frobenius complement is called a Frobenius group.
99

100
Chapter 7
We have proved in the foregoing that groups satisfying the hypotheses
ofTheorem 6.34 with 1 < JV < G are Frobenius groups. Frobenius' theorem
is the converse of this.
(7.2) theorem (Frobenius) Let G be a Frobenius group with
complement H. Then there exists JV*«a G with HN = G and H r\ JV = 1.
The fact that the group JV of Theorem 7.2 satisfies the condition that
CG(x) s JV for all 1 t6 x e JV is not difficult to prove and is left to the reader.
Before beginning the proof of Theorem 7.2, we mention the curious fact
that it is trivial to find JV. What is hard is to prove that JV is a subgroup.
(7.3) lemma Let H be a Frobenius complement in G. Let
n = (g- uh*)u{1}.
Then|JV| = \G:H\. If M<c G with M n H = 1, then M £ JV.
Proof Since H = NG(H), there are \G: H\ distinct subgroups of the
form Hx. These contain exactly |G: H|(|H| — 1) nonidentity elements. The
remaining elements of G constitute the set JV. We have
\N\ = \G\ - |G:H|(|H| - 1) = \G\ - \G\ + \G:H\ = \G:H\.
IfM<GandMnH=l, then M n Hx = 1 for all xeG and thus
M ^ N. The proof is complete. |
We mention that except for some special cases, no proof of Theorem 7.2 is
known which does not use characters.
(7.4) lemma Let H be a Frobenius complement in G. Let 9 be a class
function of H which satisfies 9(1) = 0. Then (9G)H = 9.
Proof Let 1 ¥= h e H. Then
9G(/I) = (l/|tf|)E9V*"1)-
xeG
lf$°(xhx-1) ^ 0,then I * xhx~l eH n Hx~l andxeH.Then 9°(x^_l) =
9(h). We have
9G(/I) = (l/|tf|)£9(/I) = 9(/I).
xeH
Since 9G( 1) = | G: H \ 9( 1) = 0, the proof is complete. |
The proof of Theorem 7.2 may be motivated as follows. Assuming the
theorem is true, let / e Irr(G) with JV £ ker /. Then Xh e Irr(H). Now given
Xh = (p e Irr(H) we try to find / e Irr(G). We do this for each <p e lrr(H) and
check that (~)ker x is the desired normal subgroup.

T. I. sets and exceptional characters
101
Proof of Theorem 7.2 Let \H j= (pelrr(H) and write 9 = cp — <p(l)lH
so that 9(1) = 0. Now [9G, 9G] = [9, (9G)H] = [9, 9] by Lemma 7.4. Thus
[9G 9G] = 1 + <p(l)2. Now [9G, 1G] = [9, 1H] = -<p(l). We may therefore
write 9G = q>* — <p(l)lG, where <p* is a class function of G, \_q>*, 1G] = 0, and
1 + <p(l)2 = \_(p*,(p*l + <p(l)2,sothat \_q>*, <p*] = 1. Since 9 is a difference of
characters, so is 9G and hence <p* is a difference of characters also. Since
[cp*, cp*2 = 1, it follows that ± q>* e Irr(G). Furthermore, if he H, then
<p*(/j) = 9G(/z) + <p(l) = 9(/j) + <p(l) = <p(h).
In particular, <p*(l) > 0 and thus cp* e Irr(G).
For every nonprincipal cp e Irr(H), we have now chosen an extension,
<p*eIrr(G).LetM = f)„ ker <p*.Ifx eM n H,then<p(x) = <p*(x) = <p*(l) =
<p(l) for all (p elrr(H) and thus x = 1. By Lemma 7.3, M £ N.
Conversely, if g e G lies in no conjugate of H, then
<?*(#)-<p(l)=9G(g) = 0
and g e ker cp*. It follows that M = N and hence the normal subgroup M
satisfies |M| = |G:H|. We have \MH\ = \M\\H\ = |G:H||H| = \G\ and
the result follows. |
The normal subgroup whose existence is guaranteed by Theorem 7.2 is
called the Frobenius kernel of G. By Lemma 7.3, it is uniquely determined by
H. We mention without proof that in fact a Frobenius group has a unique
conjugacy class of complements and a unique kernel.
An entirely equivalent version of Frobenius' theorem may be stated as
follows.
(7.5) corollary Let G be a transitive permutation group on Q with
character /. Assume /(g) < 1 for all g eG with g j= 1. Then the set
{g e G | xig) = 0} u {1} is a transitive normal subgroup.
Proof Let a e Q and assume | Q| > 2. If there exists any geG,g j= 1 with
X(g) t6 0, then Ga is a Frobenius complement. By Theorem 7.2. and Lemma
7.3, {geG|x(g) = 0} u {1} = JV is the Frobenius kernel. It is transitive
since NGX = G. |
We shall now discuss some of the ways that the ideas in Frobenius' proof
have been extended and used in other contexts. We need to introduce some
terminology.
(7.6) definition Let XcGbea subset. Then X is a T.I. set (trivial
intersection set) if for every g eG, either I9 = IorIsnIs {1}.

102
Chapter 7
(7.7) lemma Let X be a T.I. set in G and let <p and 9 be class functions on
JV = NG(X). Assume that cp and 9 vanish on JV - X and that 9(1) = 0. Then
9G(x) = 9(x) for all x eX and [9G, <pG] = [9, <p].
JVoqf Let x e X. Then 9G(x) = (1/1 JV|) £yeG 9P(yxy-l). If 9°(yxy-') *
O.then^'eXnl'"' srndyxy'1 # 1. It follows that yeJV and ^(yxy'1)
= 9(x). The first statement now follows.
Wehave[9G,<pG] = [(3%, <p]. Since <p vanishes on JV - Xand(9% - 9
vanishes on X, we conclude that [(9% - 9, <p] = 0 and hence [9G, 9G] =
[9, <p] as claimed. |
Because of the requirement that 9(1) = 0, the preceding lemma cannot be
applied if 9 is a character. We usually take 9 to be a difference of two
characters. Such a difference is called a generalized character. Note that 9 is a
generalized character of G iff [9, /] e Z for all / e Irr(G). Also, the set of
generalized characters of G is a ring.
We are now ready to consider groups whose Sylow 2-subgroup P is
generalized quaternion. If | P \ > 16, we shall prove the theorem of Brauer and
Suzuki which asserts (among other things) that such a group cannot be
simple. This theorem is also true if | P \ = 8 but it is more difficult to prove in
that case.
We shall use the following facts about a generalized quaternion group P:
(a) P has a cyclic subgroup of index 2;
(b) \P:P'\ = 4;
(c) |Z(P)| = 2;
(d) noncyclic subgroups of P are themselves generalized quaternion;
(e) P contains a unique involution.
We shall need to use the result of Problem 3.9 in the proof.
(7.8) theorem (Brauer-Suzuki) Let P e Syl2(G) be generalized
quaternion with \P\ > 16. Then there exists JV<i G with |JV| odd and such that
G/N has a normal subgroup of order 2.
Note that possibly JV = 1 in 7.8. We first prove the following weaker
version. The full result will then follow easily.
(7.9) theorem Let PeSyl2(G) be generalized quaternion with \P\ > 16.
Then there exists M <i G such that | M | is even and G/M is nonabelian.
Proof Let H s P be cyclic with \P: H\ = 2. We have FsH and
|P'| = |P|/4 > 4, so that F > Z(P) and hence Cp(P') = H. Let C = CG(P')
and JV = NG(P'). Now P e Syl2(JV) and C <i JV so that H = PnCe Syl2(C).
Since a Sylow 2-subgroup of C is cyclic, it follows (for instance from Burn-
side's transfer theorem) that C has a normal 2-complement K and K <i JV.

T. I. sets and exceptional characters
103
Now N/C is isomorphic to a subgroup of Aut(P'). Since P' is a cyclic 2-group,
it follows that N/C is a 2-group. We therefore conclude that JV = KP. Since
C = KHwehave|JV: C\ = 2.
Let U £ F with \F: U\ = 2. Let X = C- UK. We claim that X is a
T.I. set and N = N(X). Now C/K s H is cyclic and UK/K is its unique
subgroup of order equal to | U \. It follows for y e C that y e X iff o{yK) in C/K
exceeds | U \. We conclude that y e X iff (211/1) | o(y).
Now |P'| = 2|t/| and P'-a C. Since P' contains all elements of C of
order 21 U \, it follows that F £ <x> for all x e X. If x e X n X9, then P' £
<x> and (P'f £ <x>. Since |P'| = |(P7I, we have F = (F)B and geN.
Since clearly N £ N(X), it now follows, as claimed, that X is a T.I. set and
N = N(X). Since | P' | > 4, we also have 41 o(x) for every x e X.
Now C/l/K is cyclic of order 4. Let X be a linear character of C with
ker 1 = UK. Let 9 = 1N - (lcf. Since ker 1N = UK £ ker(lcf, we
conclude that 9 vanishes on t/X and in particular 9(1) = 0. Clearly, 9 vanishes
on N — C and hence 9 vanishes on N — X. We may therefore apply Lemma
7.7 with cp = 9 to conclude that [9G, 9G] = [9, 9].
To compute [9, 9], observe that (lc)^ = In + fi where ker y. = C. We
claim that AN e Irr(iV). Otherwise, XN is a sum of linear characters and F £
N' £ ker 1" = UK, which is not the case. Thus 9 = 1N - fi - lN and
[9, 9] = 3.
Now [9G, 9G] = 3 and [9G, 1G] = [9, 1„] = -1. It follows that
9G= ±Zi±Z2- 1g>
where %u ite Irr(G) are not principal. Since 9G(1) = 9(1) = 0, we conclude
that the signs above are opposite and without loss we may write
9G = Zi - X2 ~ 1G-
Since 9G(g) = 0 unless g is conjugate to an element of X, we conclude
(1) Xi(Q) - 12(9) = 1 if 4Jfo(g).
Since P has a unique involution, it follows that G has a unique conjugacy
class of involutions. Let Jf t be this class. Define the class function <p on G by
(p(9) = \{(x,y)\x,yeJfu xy = g}\.
lf(p(g) ¥= 0, then g = xy for involutions x and y and hence g* = yx = g~l.
\io(g) is even, let a be the involution in <g>. Then <x, er> is an abelian, non-
cyclic group of order 4. Since P has no such subgroup, neither does G
by Sylow's theorem. We conclude that q>(g) = 0 if o(g) is even. Thus
<p(9)(Xi(g) - Xiia) - 1) = 0 for all geG. Therefore
(2) [cp, Iv-li- IgJ = 0.

104
Chapter 7
In C[G], let Kv be the class sum corresponding to the conjugacy class Jf v.
Then in the notation of Problem 3.9,
^1^1 = Lflllv^v>
andifge Jfv, then (pig) = allv.
By Problem 3.9 we have for g e Jf v and x e Jf t that
«»(ff) = fliiv = (l-^il2/|G|) £ )(W2i/z(l).
Xelrr(G)
Since /(x) and <p(#) are real, we may rewrite this equation as
(\G\/\Jt1\2)(p= £ (ZW2/Z(D)Z
Xelrr(G)
and
(|G|/|jf1|2)[<p,z] = zW2/z(l)
for all / e Irr(G).
We conclude from Equation (2) that
ZiW2/Zi(i) - Z2W2/z2(i) = i-
From Equation (1) we have /2M = ZiW — 1 since 4 Jf o(x) and also /2(1) =
Xi(l) — 1. Substitution into the preceding equation yields
ZiW2/Zi(l) - (ZiW - l)2/(Zi(l) - 1) = I-
Simplifying this, we obtain
(ZiW - Zi(l))2 = 0
and we conclude that x e ker Xi-
Since o(x) = 2, we have |ker Xi\ is even. Now, ^(l) = 1 + /2(1) ^ 2
and thus G/ker /t is nonabelian. The proof is now complete with M =
ker/!. |
Proof of Theorem 7.8 Let U be the (normal) subgroup generated by all of
the involutions in G. If U has a cyclic Sylow 2-subgroup, then U has a normal
2-complement JV and JV <i G. In this case, U/N <i G/JV and U/N is a cyclic
2-group. The result follows.
Assume then that the Sylow 2-subgroups of U are noncyclic. We derive a
contradiction. Since 8| | U\, we may choose V, with U £ V £ UP such that
| 7: U | < 2 and 1611 V \. Now Theorem 7.9 applies to V and we may choose
M<F with 7/M nonabelian and | M | even.
Since a Sylow 2-subgroup of V contains a unique involution, it follows
that all involutions of V are conjugate. Since M<F contains an involution,
it follows that M contains all of the involutions of V and hence U £ M. Thus
17: M | < | V: C/1 <2 and this contradicts V/M being nonabelian and
completes the proof. |

T. I. sets and exceptional characters
105
By the Brauer-Fowler theorem (Corollary 4.12), there are at most
finitely many nonisomorphic simple groups which contain an involution
whose centralizer is isomorphic to some given group C. Much work has been
done in recent years to find all of the simple groups corresponding to various
specific C. A key step is to find all possible orders of these simple groups and
this often involves character theoretic techniques related to those in this
chapter. As an illustration of this we prove the following.
(7.10) theorem Let G = G' and suppose reG is an involution with
Cg(t) dihedral of order 8. Then \G\ = 168 or 360.
We need a lemma.
(7.11) lemma (Thompson) Let SeSyl2(G) and suppose M^S with
| S: M | = 2. Let t e S be an involution which is not conjugate in G to any
element of M. Then t £ G'.
Proof Let G act by right multiplication on Q = {Mg\geG}. Then
|Q| = 21G: S|. Now if Mgx = Mg, then gxg~l e M, which is not the case.
Thus t has no fixed points on Q and since \G:S\ is odd, it follows that t
induces an odd permutation on Q. Therefore, there exists A <i G with
| G: A | = 2 and i^A. The result follows. |
Proof of Theorem 7.10 Let D = Cg(t) and D £ SeSyl2(G). Then
Z(S) £ Z(D) = <t> and hence S £ C(t) = D. Thus D e Syl2(G). Let M £ D
be cyclic of order 4 so that t is the unique involution in M. Since G = G',
Lemma 7.11 guarantees that every involution in G is conjugate to t.
Now M is a T.I. set in G since if M n M" # 1, then teM* and thus
t = t* and x e D. Since M<Dwe have M = Mx. This argument also shows
that D = NG(M).
Let 1 be a faithful linear character of M and let 9 = (1M - Xf. Since kD is
irreducible, it follows that [9, 9] = 3. Also, 9(1) = 0 and 9 vanishes on
D - M. Thus by Lemma 7.7 we have [9G, 9G] = 3 and also (9% = 9.
Since 9G(1) = 9(1) = 0, it follows that we may write 9G = 1G + / - i//,
where /, \ji e Irr(G). Calculation in D yields 4 = 9(t) = 9g(t) and we have
(1) 0=1 + z(l) - Ml), 4=1+ jrft) - ^t).
Let Jf denote the (unique) conjugacy class of involutions in G and define
the class function cp by
9(g) = \{(x,y)\x,yejf,xy = g}\.

106
Chapter 7
If xy = g for involutions x and y, then gx = g~l and conversely, if x e Jf,
x 76 g, and g* = g"l, then y = xg is an involution. It follows that cp(g) =
\{xejf\x ¥= g, gx = g~l}\- If 1 ¥= geM and gx = g~\ then ix = 1 and
xeD. We conclude that cp(g) = 4 for 1 =£ geM.
Reasoning as in the proof of Theorem 7.9 and using Problem 3.9, we have
9 \g\ s£(G)m)q'
Since | Jf | = | G | /8 this yields
rQG -, |G|f. yix)2 m2'
Also, [9G, cp] = [(1M - X), <Pm\- Since <p has the constant value 4 on
M - {1}, this yields [9G,<p] = (1 + i) + (2) + (1 - i) = 4. We conclude that
XW2 -AW2
(2) 2s = \G
1 +
z(i) 'A(i)
Write a = x(l) and ft = /(t) so that t/^(l) = a + 1 and i/^(t) = ft - 3 by
Equations (1).
By the second orthogonality relation we have
8 = I C(t) I > 1 + X(x2) + ^(t)2 = 1 + ft2 + (ft - 3)2.
Since ft e Z, we conclude that ft = 1 or 2.
Assume ft = 1. Equation (2) yields 28 = |G|[1 + (1/a) - (4/(a + 1))] and
\G\ = 28a(a + l)/(a - l)2.
Now 2\a(a + 1) but 24/|'|G| and we conclude that 23\(a — 1). Therefore
22 X a(a + 1) and since 2311G | we have 24 X (« — !)• No odd prime divisor of
a — 1 can divide 28a(a + 1) and we conclude that a — 1 is a power of 2 and
thus a = 9. This yields |G| = 22 x 9 x 10 = 360.
Now assume ft = 2. Equation (2) yields 28 = |G|[1 + (4/a) - (l/(a + 1))]
and
\G\ = 28a(a+ l)/(a + 2)2.
Reasoning exactly as above, we conclude that a + 2 = 8 and \G\ =
22 x 6 x 7 = 168. The proof is complete. |
We mention that GL(3, 2) s PSL(2, 7) is the unique group G of order
168 with G = G' and A6 s PSL(2, 9) is the unique one of order 360.
We go back to the problem of obtaining information about the
irreducible characters of a group from information about a subgroup. Let

T. I. sets and exceptional characters
107
y £ Irr(JV). We introduce the notation Z[5^] for the set of Z-linear
combinations of elements of Sf. (Thus Z[Irr(JV)] is the set of generalized characters
ofJV.)
Suppose JV £ G and that we can find a map *: Z[y~] -► Z[Irr(G)] such
that * is Z-linear and that |>*, 9*] = |>, 9] for all cp, 9 e Z\_y]. (Such a map
is called a //near isometry.) In that case we have [/*, /*] = 1 for / e y and
thus + x* e Irr(G). Write e(x) = + 1 so that e(/)x* e Irr(G). The characters
e(/)X* are called the exceptional characters associated with y and*.They are
in one-to-one correspondence with y.
How might such a map * be constructed? An easy example of a linear map
1\_y\ -» Z[Irr(G)] is the induction map 9 -» 9G. This map, however, is rarely
an isometry on Z[5^]. By Lemma 7.7, we see that there are situations when
induction is an isometry on Z[^]° = {9 e ~l\_y~\ 19(1) = 0}. This occurs, for
instance, if JV = N(X), where X is a T.I. set and y = {/e Irr(iV) | / vanishes
on JV - X}. The problem then is to extend a linear isometry from Z[^]°
to all of Z[^].
(7.12) definition Let JV £ G and 9> £ Irr(JV) with \y\>2. Suppose
t:Z[^]° -» Z[Irr(G)]° is a linear isometry. We say that (y, t) is coherent if
t can be extended to a linear isometry * defined on Z[5^].
If t is the induction map and (5^, t) is coherent, we simply say y is
coherent. It should be emphasized that even in this case, the map * usually is
not induction. The prototypical example of coherence is where JV is a
Frobenius complement in G. There, Irr(JV) is coherent; and the proof of that
fact is the essence of the proof of Frobenius' theorem.
If (y, t) is coherent, the map * is not always uniquely defined. Nevertheless
the set of exceptional characters e(x)x* is uniquely determined by (y, t).
(7.13) lemma Let (y, t) be coherent and let * be a linear isometry which
extends t. Then there exists e = + 1 such that ex* e Irr(G) for all / e y. The
function f:Sf-> Irr(G) defined by /(/) = ex* is one-to-one. The image off
is {ip e Irr(G)| [9T, t/0 * 0 for some 9 e Z[^]°}.
Proof For / e y, [/*, /*] = 1, and we may choose e(/) = + 1 so that
e(x)X*e'lrr(G). We claim e(x) = e(£) for all ZeSf. We have 9 = *(1)£ -
£(l)xeZ[,ST and thus
Z(l)^-^(l)Z*=9IeZ[Irr(G)]°.
Evaluation at 1 yields 0 = x(l)£*(l) - £(l)z*(l) a"d hence £*(1) and z*(l)
have the same sign. Thus e(x) = £(£) as claimed.
The foregoing also shows that [9T, /(/)] ^ 0 for some 9 e Z[^]°.
Conversely, suppose ^elrr(G) and [9T, ^] # 0, where 9eZ[^]°. Then

108
Chapter 7
9 = Yxesr ax* and & = X ax**- Therefore 0 # [/*, t/f] = e[/(z), i/f] for
some xey and hence/(z) = \p.
To show that/is one-to-one, it suffices to show that * has trivial kernel.
This follows since if 9* = 0, then 0 = [9*, 9*] = [9, 9] and hence 9 = 0. |
Suppose N £ G, if ^ Irr(iV), |^| > 1 and t: Z[^]° -»Z[Irr(G)]° is a
linear isometry. We seek conditions on if sufficient to guarantee that (if, t)
is coherent. One such condition is that all z e if have equal degrees. Although
this is not hard to prove directly, we shall derive it as a corollary of the
following more general result of Feit.
The main point of Feit's theorem may be summarized as follows. List the
degrees z(l) for z e if in increasing order. Assume that the smallest of these
divides all of the others. Then if is coherent if the degrees do not increase too
rapidly.
(7.14) theorem (Feit) Let N £ G, if £ Irr(JV) and let v. I\_ifJ ->
Z[Irr(G)]° be a linear isometry. Suppose if = if0 u {/}, where (if$, t) is
coherent. Assume that there exists \p e if$ such that i/f(l)|z(l) and
zd) < Y^r, X £(D2-
2^(1) st^o
Then (if, t) is coherent.
Proof Let *: Z[^0] -» Z[Irr(G)] be a linear isometry which extends t on
Z[y0]°. We shall define z* so that the map * can be extended to all of Z[^]
by linearity.
Let /(l) = dip(l) so that z - dip e l\_ifj. Define A by
(i) (z - #r = a - #* + x &<{*,
where [A, <!;*] = 0 for all £,eif0. Of course, A is a (possibly 0) generalized
character of G and all b^ e Z. We shall show that [A, A] = 1 and all b4 = 0.
Now
i + d2 = to - #), (Z - #)] = [(Z - #r, (z - dm
and thus
1 + d2 = [A, A] + £ V + (ft, - rf)2-
Therefore
(2) [A,A] + XV = 1 + 2<tfV
Furthermore, since t maps into Z[Irr(G)]°, we have (/ - #)T(1) = 0 and thus
(3) 0 = 5>^*(1)-#*(1) + A(1).
i

T. I. sets and exceptional characters
109
For £ e &0 let S = t/,(l)£ - £(l)t/f e Z[^0]°. Then
[e\ (x - dm = [s, a - #)] = dm.
Therefore
«(D = [HT, E M*] - rf[Sr, ^*] + [E\ A]
= 0(l)be - £(1)6, + #(1) + 0
since ST = ^(1)^* — £(l)t/'*. We conclude that
Ml)fce = £(1)V
Now define /z = ^/^(l). We have then
(4) fc« = tf(D
for all <* e 5V
Also, since ST(1) = 0, we have
W)TO = fll)lTO
and we may write
(5) TO = ofll)
for all £ e ^0 and fixed a#0,
By hypothesis we have
(6) 2#(1)2 < £ £(1)2.
Suppose /j^O. Then by (4) we obtain
2<v = M2(2#(i)2) < ti2 E £(i)2 = E V
and therefore
i + 2rfV^EV-
Now (2) yields
[A, A] +2db/ <2dbr
Since b^, is an integer j= 0 (since ^#0) and [A, A] > 0, we obtain A = 0 and
*V = 1- Thus A(l) = 0 and (3), (5), and (4) yield
#(l) = X^(l) = Ml£(l)2-
Since b^, = l,n= 1/^(1) and thus
#(1)2 = ££(1)2-
This contradicts (6) and proves /z = 0.

no
Chapter 7
It now follows that all bi = 0 and (1) yields
(X - #)T = A - #*.
From (2) we have [A, A] = 1 and we extend * to Z[5^] linearly by defining
Z* = A.
We now show that * is a linear isometry on Z[5^] and that it extends t. If
9 e Z[^]°, with [9, /] = a, we can write
9 = a(x - #) + <p,
where <p e Z[^0]°. Then
9* = a(z - #)* + <p* = a(x - #)T + <px = F
and thus * extends t. To show that * is an isometry, it suffices to show [/z*, v*]
= [/i, v] for fi, v e y. If fi, v e Sf0, we already know this. Since [A, £*] = 0 for
^ey0 and [A, A] = 1, the result now follows. |
(7.15) corollary Let N c G and ¥ £ Irr(iV) with |^| > 2. Suppose
t:Z^]° -► Z[Irr(G)] is a linear isometry. Suppose that all xe ^ have equal
degrees. Then (Sf, t) is coherent.
Proof Use induction on n = \Hf\. Let Xu Xi e -^ be distinct. Then
[(Zi - Xi)\ (Zi ~ lif] = E(Zi - lil (Zi ~ XzTi = 2
and since (xi — Z2)T(1) = 0, we may write (xi — XiY = a — P where a,
/? elrr(G) are distinct. If n = 2, define xt* = a and /2* = /? and extend by
linearity to Z[^]. In this case Z[^]° = {a(xt - z2)|aeZ} and thus *
agrees with t on Z[^]°.
If n = 3, let x3 e <f - {xu Xi)- Reasoning as above, (xi - XsY = fi - v
where fi, v e Irr(G) and ^v, Also
[(Zi - to)', (Zi - XsYl = 1
and we conclude that either y. = a or v = /? but not both. If /z = a, define
Zi* = ^ Xi* = A and Xz* = v. If v = /3, define *!* = -0, *2* = -a, and
X3* = — /z. In either case, extend * linearly to Z[5^] and check that * agrees
with t on Xi — Xi and on Zi — Xz and that * is an isometry. Since
ZDST = {a(Xl - Xi) + HXi - X3)\a,beZ},
the result follows in this case.
Suppose n > 3. Write & = £f0 u {/}, where 3 < | ^01 = " — 1. By the
inductive hypothesis, (£f0, t) is coherent. Choose ipeSf0 and observe that

T. I. sets and exceptional characters
111
^(1)|Z(D since ^(l) = z(l). Now
and thus (Sf, t) is coherent by Theorem 7.14. |
The most common applications of coherence are in the situation of a
"tamely imbedded " subgroup as defined in the paper of Feit and Thompson
on the solvability of groups of odd order. For the purposes of this book, we
give a considerably more restrictive definition which, nevertheless, is
important in applications.
(7.16) definition Let K £ G. Assume
(a) K is a T.I. set in G;
(b) K<NG(K);
(c) CG(fc) £ K for all 1 ¥= k e K.
Then K is a T.I.F.N. subgroup of G.
The "F.N." in the foregoing definition stands for "Frobenius normalizer."
Note that if N = NG(JC), where K is T.I.F.N. in G, then N is a Frobenius
group and K is the Frobenius kernel. By Theorem 6.34, the irreducible
characters of N are thus of two types, those with kernel containing K and
those induced from nonprincipal q> e hr(K). If y = {/ e Irr(JV) | K £ ker /},
then the object of what follows is to prove that y is coherent. Results of Feit
and Sibley do, in fact, prove this in most cases. Note that if K £ G satisfies
condition (a) of Definition 7.16, then (c) is equivalent to CN(k) £ K for all
l^keK, where N = NG(K).
An important theorem of Thompson (not involving characters) asserts
that Frobenius kernels are necessarily nilpotent. The nilpotence of T.I.F.N.
subgroups will be assumed in what follows. (The reader may simply consider
this to be an additional hypothesis in Definition 7.16.)
A more elementary fact about a T.I.F.N. subgroup K of G is that
(\G:K\, \K\) = 1, that is, K is a Hall subgroup. To see this, let p||X|, let
PeSylp(X) and suppose P £ SeSylp(G). Then Z(S) £ C(P) £ K by part
(c) of 7.16. Now Z(S) ^ 1 and so S £ C(Z(S)) £ K, again by 7.16(c). Thus
PX\G:K\.
We mention some situations where T.I.F.N. subgroups arise. If
PeSylp(G),|P| = pandP = CG(P) < NG(P), then Pis T.I.F.N. This happens,
for instance in permutation groups of degree p and in the groups PSL(2, p).
A doubly transitive permutation group is a Zassenhaus group if some
nonidentity element fixes two points but none fixes three. Let G be a
Zassenhaus group on a set Q and let a, /? e Q with a j= /?. Then Ga is a Frobenius

112
Chapter 7
group with complement G^. (We are assuming | Q | > 2 to avoid triviality.)
Let K be the Frobenius kernel of Ga. The nonidentity elements of K are
exactly those elements of G which fix a and no other point. It follows easily
that K is T.I.F.N. in G and Ga = NG(K).
(7.17) lemma Let K be T.I.F.N. in G and let JV = NG(K) and <f =
{ZeIrr(JV)|K £ ker /}. Then induction defines a linear isometry
Z[^]° -+ Z[Irr(G)]°.
Also, if 9 e Z[^]°, then (9G)N = 9.
Proof UieHf, then x = £N for some £ e Irr(X) by Theorem 6.34. Thus %
vanishes on JV - K. If cp, 9 e Z[^]°, it follows that <p, 9 vanish on N - K
and Lemma 7.7 yields \jpG, 3G] = [<p, 9] and the first assertion is proved.
Also, (9G)K = 9K by Lemma 7.7. Since 9 vanishes JV - K, it suffices to
show that 9G vanishes on JV — K in order to prove that (9G)N = 9. However,
no element of JV — K can be G-conjugate to an element of K since if
xeN - K and o(x)||K|, then K < <X, x> and <X, x> = X<x> so that
|<X, x>: K| is not relatively prime to \K\. This contradicts the fact that
(| G: K |, | K |) = 1. It now follows from the definition of 9G and the fact that
9 vanishes on JV — K that 9G also vanishes on JV — K and the proof is
complete. |
(7.18) corollary (Brauer-Suzuki) Let K be an abelian T.I.F.N.
subgroup of G. Let JV = NG(X)ande = | JV: K |. Let ^ = {/eIrr(JV)|K £ ker/}.
Then either
(a) \Sf\ = \,\K\ = \ + e and K is an elementary abelian p-group or
(b) y is coherent and there exists a one-to-one function/: y -»Irr(G)
and e = ± 1 such that (z - if = e(f(x) - /(0) for all z, £ e ^.
Proof All of the orbits of nonprincipal irreducible characters of K under
the action of JV have size e. Thus /(l) = e for all / e ^ and \Sf\ = (\K\ — l)/e.
U\Sf\>2, then Corollary 7.15 yields that Sf is coherent. In this case, (b)
follows from Lemma 7.13.
If \y\ = 1, then \K\ — 1 = e and the nonidentity elements of K are all
conjugate in JV. It follows that they all have the same prime order and (a)
follows. |
The theorems of Feit and Sibley generalize Corollary 7.18 by dropping the
assumption that K is abelian. If | K | - 1 ^ e, they prove that either Sf is
coherent or K is a 2-group. Before proceeding with the proofs, we discuss
some implications of coherence in the T.I.F.N. situation.
(7.19) lemma Let X£G be T.I.F.N. and let JV = NG(X). Suppose
3C £ {/eIrr(JV)|X £ ker /} and that 9C is coherent. Let * denote an iso-

T. I. sets and exceptional characters
113
metry Z\_3T\ -> Z[Irr(G)] such that * extends induction on Z[#"]°. Let
« = Ez6arZ(l)Z-Then
(a) If \pelrr(G) and \p$ {±y*\is3C}, then i//N = aa. + 9, where a is
rational and [9, Z] = 0 for all Z e 9C.
(t>) ((x*)n ~ X)/x( 1) is independent of Z e 3C.
(c) (x*)n - X = a<* + # for x g #\ where a is rational and [9, £] = 0 for
alUeST.
Proof (a) Let Z, £ e SC. Then
[>K,Z(1K; - £(1)Z] = DMzdK; - £(Dz)G]
= [>, Z(D£* - flDzM = 0.
Thus x(1)|>n, £] = ^(1)[^, z] and hence |>N, z]/z(l) = a is independent
of Z e #\ Therefore, \j/N = aa. + 9, where [9, Z] = 0 for all Z e #\ Thus (a) is
proved.
(b) Let xA e X. Then A = z(l)£ - £(l)z e Z[^° and so (AG)W = A by
Lemma 7.17. However, AG = A* = Z(l)£* — £(l)z* and thus
Z(l)£ - £(1)Z = z(l)«*)K - {(l)(z*)K
and
«D((Z*)W - Z) = zU)(tf*)w - 0.
Thus (b) follows.
(c) Let £,»/ e 3C and write A = £{\)r\ - >/(l)£. Then
[(Z*)w, A] = [z*. AG] = [/*, A*] = [z, A]
so that [(/*)n - /, A] = 0. It now follows as in (a) that (x*)N - Z = «a + #,
where a = [(z*)K - z, £]/£(l) and K, 9] = 0 for £ e ST. |
(7.20) theorem Let X c G be T.I.F.N. and let N = NG(X). Suppose
y={^e Irr(iV)|X £ ker Z} is coherent and let S £ Irr(G) be the
corresponding set of exceptional characters. Let * be an isometry Z[5^] -» Z[<f]
which extends induction on Z\_Sf]° and let e = +1 so that ez* e $ for Z e y.
Then
(a) S={i//e Irr(G) | ^ is not constant on K - {1}}.
(b) If \ji = ex* e S, then i/fK - eZx is constant on K — {1}. This constant
has the form mZ(l)/|./V:K| for some m e Z which is independent of the choice
ofZe^.
(c) If g e G is not conjugate to an element of K — {1}, then t/K^Vi/Kl)
is independent of the choice ofipeS.
(d) If z e ^, then Z*( l)/z( 1) is independent of the choice of Z.

114
Chapter 7
Proof We apply Lemma 7.19 with 3C - £f. Then a = pN - pNjK
where pN and pNjK are regular characters. In particular, a. is constant on
X - {1}. Also, if 9 is a generalized character of JV and [9, /] = 0, for all
X e S?, then X is in the kernel of every irreducible constituent of 9 and hence
9K is constant.
Now, if \p $ 8, then 7.19(a) yields \pN = aa. +9 and i/f is constant on
X - {1}. If \p e S, then 7.19(c) yields eipN = x + <*<* + & where &// = X*-
However, / e if cannot be constant on X - {1}, otherwise, by Problem 3.8,
every nonprincipal irreducible character of X is a constituent of Xk- This
would force \9'\ = 1 which is not the case. Thus \pN is not constant on
X — {1} and (a) is proved.
The first part of (b) is immediate from 7.19(c). To evaluate the constant,
choose ZeSP with £(1) = |JV:X|, that is, £, = XN for some nonprincipal
linear 1 e Irr(X). Now (£*)K — £K is a generalized character of X with constant
value m on X - {1}. It follows that m e Z. If / e Sf is arbitrary, 7.19(b)
yields that m'/x(l) = m/£(l) where m' is the constant value taken on by
(X*)k ~ Xk on X - {1}. Since {(1) = \N;K\, we have tri = mx(l)/|JV:X|
and (b) follows.
Let x, £ e &> so that £(l)z* - z(l)£* = (£(l)z - X(1)£)G, which vanishes
on elements g e G not conjugate to elements of X - {1}. Thus £(l)x*(g)
- xO)£*(ff) = 0 and (c) follows.
Finally, (d) is immediate from 7.19(b). The proof is complete. |
We now begin work toward the theorems of Feit and Sibley.
(7.21) lemma Let JV be a Frobenius group with Frobenius kernel K.
Suppose | JV:X | is even. Then X is abelian.
Proof Let t e JV be an involution. Then t $ X since (|X|,JV:X|)=1
and thus x' ¥= x for x e X - {1}. Now map X -» X by x i-> x~ 'x'. This map
is one-to-one since if x_1x' = y~V, thenyx-1 = (yx~ly and thus yx~l = 1.
Therefore the map is onto and every xeX has the form u~lu'. Thus
x' = (u~vux)x = (u')~1u = x"1. We now have (xy)' = (xyy1 = y~ix~l =
y'x' = (yxf. Thus xy = yx for x, y e X and the result follows. |
(7.22) lemma Let P e Sylp(G) with p ^ 2 and suppose P is T.I.F.N. in G
and is nonabelian. Let JV = NG(P) and let 1 < Z £ Z(P) with Z <\ JV. Let
3C = {x e Irr(JV)|Z £ ker /}. Then SC is coherent.
JVote By Corollary 7.15, the lemma is also true if P is abelian except
in the degenerate case that \3C\ = l.InthatcaseZ = Pand|P| = |JV:P| + 1.
Proof of Lemma 7.22 Let e = | JV: P |. By Lemma 7.21, e is odd and thus
|JV| is odd. Since Z > 1, 3C ¥=0. Let £ e 3C have minimum possible degree

T. I. sets and exceptional characters
115
andlet^0 = {xeSC\x(\) = {(1)}. Since \N\ is odd, $± £ and thus | ^01 > 2.
By Corollary 7.15, st0 is coherent. Now define sets #", for i > 0 by
Sft = ^"i-! u {/,•}, where the x, are chosen from <Sf— ^oSothat/.O) < z(+i(l).
Thus f0cf1c...cft = | We use induction on i to show that SC{ is
coherent.
Now the degrees of the / e Sf are all of the form ep" and so £(l)|x(l) for
all xe3C.lt follows by Theorem 7.14 that in order to show that Sfi is coherent,
it suffices to show for i > 1 that
2£(l)Zj(l)< X X(D2-
We have
\N\= Z z(l)2 = |N:Z|+ XZ(D2
xelrr(JV) XEiT
and so |P:Z| divides £Z6ir Z(l)2- Thus
e2|P:Z| divides £ *(1)2.
If/ e #, then / = S" for some 9 e Irr(P) and 9(1)2 < | P: Z | by Corollary 2.30.
Thus x(l)2 = e29(l)2 divides e2|P:Z| and hence /((I)2 divides ^.y x(l)2.
Also,Zi(l)2IZj<l)2forj > j and we conclude that/((I)2 divides ^ZEa.(_1 *(1)2.
In particular, since 2£(1) < p£(l) < #,(1), we have
2«l)Zi(l) < Z,(D2 < I Z(D2
and the result follows. |
(7.23) lemma Let N £ G and ^ c Irr(N). Let a: Z[^]° -» Z[Irr(G)]°
be a linear isometry and let S£, <W £ ^ with #n <^ = 0 and (3C, a) and
(<^, a) coherent. Let a and t be isometries on Z[^] and Z[<^], respectively,
where er and x are as implied by coherence. Then
(a) if, rf] = b for all / e SC and i/e?'.
(b) Suppose x0 e#" and >/0 e <& with
(ZoOfoo - »7o(l)Zo)a = ZoOfoo1 - r,0(l)x0a.
Then 3C u <^ is coherent.
Proof For /z, v e ST, write A(/z, v) = /z(l)v - v(l)/z e Z[^]°. Let 5, e = ± 1
so that (5/" and erjz e Irr(G) for ](ef and t] e<W. To prove (a), suppose
If, 1*] * 0. Then btf = 9 = eif for some 9 e Irr(G). Pick Xi e 3T - {/}
and (|[6f- {>/} and write t/f = «5xiCT and ^ = £>//. Then ^S^ and

116
Chapter 7
we have
0 = [A(z, Xl\ A(i,, ijj] = [A(z, XiT, Afo, i^f]
= rf[(zdWf-Zi(l)%(f(l)«-fi(l)0)]
= e«5(Zi(D'7i(l) + Zd¥l)[^a).
Since [i/f, <!;] ^ 0, this is impossible, and (a) is proved.
Now fix Xo e #" and >/0 e <^ and assume A(/0, rj0f = /0(l)'7ot - fo(l W-
Define * on 3C u 'W by/* = /"for/e^andf/* = if for >/ e "^ and extend * by
linearity to Q\_3C u <¥]. It follows from (a) and the fact that a and t are
isometries, that * is an isometry and *: 1\_3C u <¥] -» Z[Irr(G)]. It thus
suffices to show that * agrees with the linear extension of a to Q[_SC u W]°.
However, * agrees with a on Q[^"]°, Q[^]° and A(/0, rj0). A dimension count
shows that these span Q\_3C u W]° over Q and the result follows. |
(7.24) theorem (Sibley) Let P e Sylp(G) be T.I.F.N. in G with p # 2.
Let N = NG(P) and 9> = {/eIrr(JV)|P £ ker/}. Let e = \N:P\. Then
one of the following occurs.
(a) \Sf\ = 1, P is elementary abelian, and \P\ = e + 1.
(b) ^ is coherent.
Proof If P is abelian, this is included in Corollary 7.18. Assume that P is
nonabelian so that e is odd by Lemma 7.21, and thus |JV| is odd. Let Z =
P'nZ(P)> landlet<^= {i\ s^\Z g ker^.Let^- = {/e^F = ker/}.
Then <^ is coherent by Lemma 7.22 and every ^ef satisfies /(l) = e.
Since / ,£ / for / e #", we have #" coherent by Corollary 7.15. Since Z £ P',
we have SC r\ <W = 0. Most of the proof is devoted to showing that SC u <W
is coherent.
Let *: l\dC] -» Z[Irr(G)] be a linear isometry which extends induction
on J.\_9T]°. Similarly, let t be an isometry on I\W] which extends induction.
Since SC r\ <W = 0, Lemma 7.23 yields [/*, if] = 0 for / e SC and >/ e <^.
Let a = (1/e) £ ,6^ >/(l)>/. Since e|>/(l) for >/ e ^ a is a character of JV.
For %e3£,x* ¥= ±i]z for any ij e<W and Lemma 7.19(a) yields
(X*)n = #* + aza,
where 9Z is a generalized character of JV such that \_9X, if] = 0 for r\ e 'W
and az e Q.
Since all x(l) are equal for / e 3C, Lemma 7.19(b) yields that
(X*)n ~ X = $x ~ X + ax<*
is independent of / e SC. It follows that 9X — x = A and ax = a are
independent of x £ 3C and
(1) (X*)n = X + A + ace.
The next several paragraphs are devoted to proving that a e Z.

T. I. sets and exceptional characters
117
Since Z is contained in the kernel of / e 3C and of every irreducible
constituent of A, we have for z e Z that x(z) = x(l) and A(z) = A(l) and hence
Equation (1) yields
Z*(l) - z*(z) = a(a(l) - a(z)).
We also have pN = pNjZ + ea. where pN and pNjZ are regular characters.
For z eZ — {1} we thus have
\N\ = pN(l)- pN(z) = e(a(l) - a(z))
and hence
(2) x*(l)-X*(z) = a\N\/e = a\P\.
We now compare j(*(l) and x*(z) in a different way. Let Jf 0, jf 1;... be the
conjugacy classes of G, numbered so that
(a) Jf0 = {1};
(b) Xtr\Z ±0 iff i<r;
(c) Jf;nP^0ifn<s.
Note that since P is a T.I. set with N = N(P), we have Jf, n P is a conjugacy
class of N for i < s and Jf; n P c z <i N for i < r.
Write X; e C[G] for the class sum corresponding to Jf, and KtKj =
X fly/i^/i. where ay„ e Z. Now fix i, _/ < r. Let t/f e Irr(G) and let co(KJ =
t/f(x) | X ^ | /t/f( 1) for x e Jf^ as in Chapter 3. Let R <=, C be the ring of algebraic
integers so that (o(K^ e R. Write t/^(l) = wp' with pjfm, and let <j = \P\/p'.
We have
eo(KjMK/) = Zflf^K/J-
For /z > s, we claim that <y(K,J e <?/?. From the T.I.F.N. property and
the fact that Jflir\P=0, it follows that \P\ divides |jfj and thus
m(a>(K^lq) e R. Since also q(a>(K^lq) e /? and (w, <j) = 1, it follows that
(x^K^/q e R as claimed. Thus
eo(K,)eo(Ky) = £ ai)MKd mod «*■
/i=0
Now let r < p. < s and x e Jf^ n P. Let C = Cp(x) and let Q =
{(w, v)\ueJth veJfj, uv = x}. Then a,^ = |I1| and C acts on Q by
(m, t>)c = (if, vc). If c e C - {1} and (m, vf = (u, v), then u, v e CG(c) £ P
and so w e Jf, n P c z and similarly ceZ. However wt> = x e Jf^ and
Jf^ nZ=0. This contradiction shows that all orbits of C on fi have
size | C | and thus | C| divides alJtl.

118
Chapter 7
We have CG(x) ^ p and so CG(x) = C and \P:C\ divides | jfj. It now
follows that miau^aiK^/q) e /? and since q(ai}lloi(K^lq) e /?, we have
a^miK^) e 4/? and
r
oiKiMKj) = X ai}MKd mod ^/?-
/i=0
Now suppose t/f e Irr(G) is such that i/f is constant on Z — {1}. Then all
ai(KJ are equal (to a, say) for 1 < fi < r and ca(K0) = 1. Also write
aH = S=i fly,/, so that
co2 = ayo + aya) mod qR.
Since | JV | is odd, the nonidentity elements of Z are not conjugate in JV
(and hence not in G either) to their inverses. Thus all0 = 0. Assume Jf 2 is the
class of inverses of Jf t so that a12o = I Jfil = |G|/|P|. We thus have
(3) ana> = a2 = \G\/\P\ + al2oi mod qR.
In the special case that i> = 1G, we have co = \G\/\P\ and q = \P\ so that
flii|G|/|P| = |G|/|P| + fl12|G|/|P|mod|P|
and atl = I + a12 mod |P|. Now (3) yields
(1 + al2)co = \G\/\P\ + al2co mod qR
and
(4) (o = \G\/\P\ mod qR.
We apply this to the character i/f = e/* e Irr(G) with e = + 1 and ^ef.
Note that \\i is constant on Z - {1} by (2) and thus (4) applies.
We have for z e Z - {l}that
Z*(z)|G:P|/z*(l) = co = \G:P\ mod qR.
Since \P\ divides <j/*(l), this yields
Z*(z)|G:P| = z*(l)|G:P|mod|P|K.
Now (2) yields
\G:P\-\P\a = \G:P\(x*(l) - x*(z))e\P\R
and hence \G:P\aeR. Since also \P\a = x*(l) - z*(z)e/? and (|G:P|, |P|)
= 1, we have ae/?n<Q = Zas desired.
Now let x, Xi e % and >/ e ^. Note that x(l) = e divides >/(l) and we write
c = r](l)/e. Let <p = c* - r\ e Z[^]°. We have
[<PG, Zi*] = l>, (Zi*)w] = |>, Zi + A + fla].

T. I. sets and exceptional characters
119
Since [/, a] = 0 = [r\, xt + A], this yields
[<PG. Zi*] = c[x, /i + A] - fl[ij, a].
However, [>/, a] = c by definition of a and since a e Z, we have c| [<pG, /t*].
This calculation also shows that
[>G, Z*] = c + WG, Zi*]
for xt t6 /. This yields
q>G = c(\ + b)X* + cb X Z* ~ r>
where ft e Z and [T, £*] = 0 for all £ e X.
We also have
[>G, «pG] = [<p, <p] - 1 + c2
and thus there are two possibilities:
(i) b = 0 and [T, T] = 1; or
(ii) fc = -l,|aT| = 2and[r,r| = 1.
In situation (ii), we can replace* by** where /** = — Z2*and/** = — Zi*
for 9C = {xu Xi)- The result of this change is to put us into situation (i). We
thus assume that (ex — rjf = ex* — T.
Now let i]l e 'W — {r\\ so that
fid) = 19, f(l>h - fi(Df] = l>G. '/(D'/it - 71(1)71
= [r, fftdjff1 - frtitoii.
Thus again we have two possibilities:
(i) T = if; or
(ii) T = -riSm<\ri(\) = ril(\).
If T ¥= rf, then rjtz = -T for every i^e*- {>/} and thus \<&\ = 2
and <^ = {>/, >/!} with >/(l) = ^(l). In this situation, we can redefine t and
thus we may assume that (i) occurs. Thus
(Q - if = Q* - 1Z
and hence #T u <W is coherent by Lemma 7.23(b).
To complete the proof, we observe that £,6» l(l)2 = \N\ - \N\Z\
and so is divisible by \P\Z\ and hence by e2\P:Z\. Let i// e 9> - (9C u <^).
Then as in Lemma 7.22, we have \p = 9N, with 9 e Irr(P) and 9( l)2 < | P: Z |.
Thus
iP(l)2 = e2B(l)2<e2\P:Z\< ££(1)2-
4e»

120
Chapter 7
Let ](el so that x(l) = e and \p(l) > px(l) > 2/(1) since \ji £ J and so
9(1) ^ p. Thus
2jrflMl) < ^(l)2 < X £(1)2.
Since #" u <& is coherent and x(l)|t/^(l) for all ii sif, repeated application
of Theorem 7.14 yields that if is coherent and the proof is complete. |
(7.25) theorem (Feit-Sibley) Let K £ G be T.I.F.N. with JV = NG(K),
e=\N:K\ and ^ = {/ e Irr(JV)|K £ ker /}. Then one of the following
occurs.
(a) \Sf\ = 1,\K\ = e + \ and K is an elementary abelian p-group.
(b) K e Syl2(G) and | K: K' \ < 4e2.
(c) if is coherent.
Proof If K is abelian, either (a) or (c) occurs by Corollary 7.18. Assume
then, that K is nonabelian and so e is odd. If \K:K'\ = e + 1, then K/K'
is an elementary abelian p-group and thus K is a p-group. Since e + 1 is even
p = 2 and (b) follows. Assume therefore, that \{xeif\K' c ker /}| > 2.
This set is thus coherent by Corollary 7.15. Let L<i N with L £ K' minimal
such that X = {/ e ^ | L £ ker /} is coherent and assume that if is not
coherent so that L > 1.
Let M o N be such that L/M is a chief factor of N. Since
{ze^|M£ker/}
is not coherent, repeated application of Theorem 7.14 yields \ji e Irr(iV)
such that
2^(1) > XX(D2 = \N:L\ - \N:K\ = e[\K:L\ - 1)
and M £ ker i/f. Let Z/M = Z(K/M). Then Z n L > M and hence LqZ.
Also ^ = 9N for some 9eIrr(X) and 5(1)2<|K:Z| so that iA(l)2 <
e2|X:Z|.Thus
4e4\K:Z\ > 4eV(D2 ^e\\K:L\ - l)2.
Now write a = |K:L|andfc = \Z: L|.We have b\K : Z\ = \K:L\ = a and
4e2a = 4>2fc|K:Z| > b(\K:L\ - l)2 = fc(a - l)2.
If 4e2 < b(a - 2), this would yield ba(a — 2) > ft(a - l)2, which is not the
case. Thus b(a — 2) < 4e2.
If Z = L, then Z(K/M) = L/M is a p-group for some prime and thus
K/M is a p-group. Since M £ X', it follows that K is a p-group. By Theorem
7.24, we have p = 2. Also, a — 2 < 4e2 so that
\K:K'\ < \K:L\ = a < 4e2 + 1.

Problems
121
Since | K : K' | is a power of 2 and e > 1 is odd, we have \K:K'\ < 4e2 and
(b) holds.
If Z > L, then we must have e\(\Z: L\ - 1) and b > e + 1. Also, K/M
is nonabelian and thus Z < K and e|(|X : Z\ — 1). Thus
a > (e + l)b > (e + l)2 > e2 + 2.
Thus 4e2 > b(a — 2) > (e + l)e2 > 4e2 since e > 3. This contradiction
proves the theorem. |
We remark that situation (b) can actually occur with y not coherent. The
inequality | K : K' \ < 4e2 can be sharpened. In fact, the groups in which (b)
occurs have been classified.
Problems
(7.1) Let N <i G, H £ G, with NH = G and N n H = 1. Show that the
following are equivalent:
(a) CG(n) £ iV for all 1 *neN;
(b) C„(n)= 1 for all 1 *neN;
(c) CG(h) £tf for all 1 *h eH;
(d) Every x e G — N is conjugate to an element of H;
(e) If 1 ¥= h e H, then h is conjugate to every element of Nh.
(f) H is a Frobenius complement in G.
iVofe Problem 7.1 does not involve characters. It is included to acquaint
the reader with some elementary properties of Frobenius groups. Much
deeper information is known.
(7.2) (Wielandt) Let H £ G with M <i H and suppose that H n Hx £ M
whenever x £ H. Show that there exists N<i G with NH = G and N n H
= M.
Hint Note that H - M is a T.I. set. Mimic the proof of Frobenius'
theorem.
(7.3) Let H £ G and £ e Irr(tf). Suppose (£ - £(1)1H)G = 9 and [9, 9] =
1 + £(1)2. Show that there exists No G with N n, H = ker £, and every
xeG — N conjugate to some element of H.
(7.4) Let H < G and suppose induction to G is an isometry on Z[Irr(#)]°.
Show that H is a Frobenius complement in G.
(7.5) Let N £ G and y s Irr(iV). Assume that induction to G is an isometry
on Z[^]° and \<f\ > 2. Suppose ^0 = {£ e ^ is extendible to G} ^0.

122
Chapter 7
Let d = g.c.d.(K(l)|^e^0}) and assume d|£(l) for all ZeSf. Show that
y0 = y, y is coherent, and that every £,e y has a unique extension to G.
(7.6) Let Pe Sylp(G) with \P\ = p, CG(P) = P, and |NG(P): P| = e. Show
that G has at most e + (p — l)/e irreducible characters with degree not
divisible by p.
Hints Note that P is T.I.F.N. in G if e > 1. If e < p - 1, let S be the
corresponding set of exceptional characters of G. Show Y^xes IzWI2 > p — e,
where 1 ^ x e P. If *(x) = 0, then p | *(1).
Note It is a consequence of some of R. Brauer's deep results in block
theory that in the situation of Problem 7.6, the upper bound is always
attained.
(7.7) In the situation of the preceding problem, assume that G has exactly
e + (p - l)/e irreducible characters of p'-degree. Let P = <x>. If / e Irr(G)
is not exceptional (in particular if e = p — 1), show that /(x) = — 1, 0, or 1
and that /(l) = /(x) mod p. If e < p — 1 and $ is the corresponding set of
exceptional characters, let a = Y.xssl{x)- Show that a = ±1 and x(l) =
— eer mod p for / e <f.
Note The results of Problem 7.7 together with the equation
Z z(DzW = o
Xelrr(G)
and the facts that x(l)||G| and X^1)2 = \G\ provide a powerful tool
for computing the degrees of the irreducible characters of a group which
satisfies the hypotheses of Problem 7.6.
(7.8) In the situation of Lemma 7.13, assume that \Sf\>2. Prove that *, e,
and/are uniquely determined by (Sf, t).
(7.9) Let K c G be T.I.F.N. Let / be a (possibly reducible) character of G
which is constant on K - {1}. Assume that y = {\p e Irr(NG(X)) | K £ ker \j/}
is coherent. Show that the multiplicity with which each exceptional
character appears in / is proportional to its degree.
(7.10) Let K be T.I.F.N. in G and assume that
y = {xlie Irr(N(X))|X £ ker xfi}
is coherent. Let &" be the corresponding set of exceptional characters
of G. Let M = f)XEr ker /. Show that M n K = 1.
(7.11) In the situation of Problem 7.10, show that either M = ker / for all
X e F or else MK <a G.

Problems
123
Hint Suppose / e 2T and U = ker x > M. Then U/M is a rc-group
where n is the set of prime divisors of | K | and so U £ MK. Use the Frattini
argument.
(7.12) Let K be T.I.F.N. in G and let e = |N(X):K\. Suppose that G has a
faithful irreducible character of degree < 2e. Show that K is abelian.
(7.13) In the situation of Problem 7.12, assume that K -?6 G. Show that K is
an elementary abelian p-group.
Hint Let k be a linear character of K with | K : ker 11 = p. Show that
there exists faithful /e Irr(G) with /(l) < 2e and [/K, 1] ^ 0. Use Problem
7.11.
(7.14) Let G be a simple Zassenhaus group in which the stabilizer of two
points has even order e. Let the degree of G (the number of points permuted)
be k + 1. Show that k is a prime power.
Hint Count involutions to prove that G has a nonprincipal irreducible
character of degree < 2e.
Note Problem 7.14 is true without the assumption that e is even. This is a
theorem of Feit and the case where e is odd is proved in his book.
(7.15) Let N £ G and if £ Irr(JV). Suppose that v. I\_ifJ - Z[Irr(G)]° is
a linear isometry. Assume that | if \ > 3 and that for every /, \p e if, we have
(X(l)t/f - ii{\)tf = a9 - b<p for some <p, 9 e Irr(G) with a, be Z. Show that
{if, t) is coherent.
Hint Suppose /, \j/, r\ e if are distinct. Write
(*( W - 4>{\)xy = a9- b<p, (n(\)xl> - ^W = Cfi - dv
with a, b,c,d> 0. Then exactly one of 9 = y. or cp = v holds.
(7.16) Let JV £ G, ^ £ lrr(JV) and suppose t: Z[^]° - Z[Irr(G)]° is a
linear isometry. Let ifu if2^if such that if^ c\if2 ^0. Assume either
that \SfY\ > 3 or that ifY = {\ji, /} with \p(l) = *(1). Show that
(^! u if 2, t) is coherent.
Hint Let/J: y, -► Irr(G) and e, = ± 1 be such that £;/; defines an
appropriate extension of x on Z[^,]. If 1^1 > 3 for i = 1, 2, use the hint for
Problem 7.15 to show that et = e2 and/^x) = /2(z) f°r Ke^i c\if2.
(7.17) Let X £ G be T.I.F.N. with JV = NG(X) and
^={ZeIrr(N)|KgkerZ}.
Suppose that \\i is an exceptional character of G corresponding to /. Show
that the values of \ji and / generate the same field over Q.

124
Chapter 7
(7.18) Let Ku K2 c G be T.I.F.N. with Nt = NG(Kf) and
^{zelrrMIK.gkerz}
and assume that the ^,- are coherent with St the corresponding sets of
exceptional characters. Suppose SY n 82 *t 0- Show that St = S2 and
Xt is conjugate to K2.
Hint To show that K2 = K^, it suffices to show that x° e K2 for some
xeKux¥= 1. Use the fact that the K( are nilpotent to prove this.
(7.19) Suppose \G\ is odd and that C^x) is abelian for every 1 ^ xeG.
Assume that G is neither abelian nor a Frobenius group and show that every
nonprincipal / e Irr(G) is exceptional for some T.I.F.N. subgroup.
Note A group with all C(x) abelian for 1 =£ x e G is called a CA-group.
Problem 7.19 is a step in M. Suzuki's determination of all simple CA-groups.
(7.20) Let K c G be T.I.F.N. with |X| even. Let N = NG(X) < G and
y = {i/f e Irr(iV)|X £ ker i/f}. Assume that y is coherent. By carrying out
the following steps, show that G has a nontrivial normal subgroup of odd
order.
(a) Show that G has a unique conjugacy class of involutions and that JV
contains exactly e = | N: K | involutions.
(b) Let il/eSf and let \ji* be the corresponding exceptional character of
G. For 1 t6 x e K we have tA*(x) = et/^x) + a^, where a^eZis independent of
x. Show that
V2OA*) = ev2(W + (tfr*(l) - #(1) - <i,)/|K|
where v2 is as in Theorem 4.5.
(c) Show that (ip*)K = ipK + a^,lK.
(d) Show that there exists \ji e & with ker(i/f*) t6 1.
(e) Complete the proof.
Hints (a) See the hint for Problem 4.11.
(b) Let s/t = {xeG|x # 1, x2 = 1};
si2 = {x e GIx2 t6 1, x is conjugate to an element of K};
s/3 = G - (rfi u si2\
Compute £t/f*(x2) separately on each of s/u si2, and s/3. Note that
(c) Let»v = (^*(l)-#(l)-fl#)/|K|.
Use (b) to study the behavior of m^ as i/f varies over £f and conclude that
m^ = 0. Use the fact that m^ is proportional to \ji(\).
(d) Note that K is not elementary abelian.

Problems
125
Notes A corollary of Problem 7.20 is that a simple Zassenhaus group
of odd degree has degree 1 + 2". This is another special case of Feit's theorem.
There do exist simple groups with T.I.F.N. subgroups of even order. If
G = SL(2, 2"), with n > 2, then the Sylow 2-subgroup K of G is T.I.F.N.
It is elementary abelian of order 2". In this case, coherence fails because
|N(X):X| = 2" - 1 and the set Sf contains only one character.
If G = Sz(2") with odd n > 1 [the Suzuki simple group of order
(2" - 1)(22")(22" + 1)],
then again the Sylow 2-subgroup is T.I.F.N. Here, case (b) of Theorem 7.25
holds.
(7.21) (Sibley) In the situation of Theorem 7.20, let e = \N:K\ and
k = |K\. If m is as in 7.20 (b), show that
-eKsfk + 1) < m < el(yfk - 1).
In particular, if K is not an elementary abelian p-group, show that m = 0.
Hints Let 1 =£ x e K and use the inequality
|CG(x)|> 5>(x)iAW,
ill bS
where S is the set of exceptional characters. Derive that
2m + e > m2(k - \)/e.
(7.22) Let K £ G be T.I.F.N. with N = N(K) and
<?={xeln(N)\K<£kerx}
coherent. Let \\i e Irr(G) with K £ ker \ji.
(a) If \ji is nonexceptional, show that |X| < \ji(\) + 1 - [»/>*, 1K]-
(b) If \ji is exceptional, show that | K \ < t/f(l)2.
Hints (b) Reduce to the case that \jiN = / + A, where / e y and either
A = 0 or K c ker A. If A ^ 0, appeal to Problem 7.21. If A = 0, pick an
irreducible constituent, r\ =£ lGof\pij/ and apply part (a) to n.

Brauer's theorem
There are many ways in which class functions of a group can be
constructed. For instance, we could define 9(g) = |NG«g»| or 9(g) = 1 if g2
is conjugate to some fixed x e G, and 9(g) = 0 otherwise. It is perhaps too
much to expect that such arbitrarily defined functions will turn out to be
characters. (For instance, in the second example given, 9(1) = 0 if x # 1.)
One might hope, however, that a well chosen 9 will be a generalized
character, that is, 9 e Z[Irr(G)], or at least that it is an R-linear combination of
irreducible characters for some specified ring R with Z £ R £ C. (We
denote the set of these R-generalized characters by /?[Irr(G)].)
How can one decide if a given class function 9 is an R-generalized
character? Of course, if one knows Irr(G), the answer is easy: simply check that
[9, %] e R for all / e Irr(G). A more typical situation is that one has enough
information about some family 3f of subgroups of G so that it can be shown
that 9H e /?[Irr(H)] for every H eJf. The main point of Brauer's theorem is
that for suitable families 3f the last relationship is sufficient to guarantee that
9e/?[Irr(G)]. For instance, this will be true if Jf is the collection of all
nilpotent subgroups of G.
(8.1) definition Let R be a ring with Z £ R £ C and let ^f be a family of
subgroups of G.
(a) 3>tR(G, tf) is the set of class functions 9 of G such that 9H e R [Irr(H)]
forallHe^f.
(b) JR(G, Jtf) is the set of /^-linear combinations of characters t/^G for
tAelrr(H), He#.
126

Brauer's theorem
127
If R = Z, we delete the subscripts and write 9t(G, M) and J(G, tf).
(8.2) lemma Let W be a collection of subgroups of G and let R be a ring
with Z £ /? £ C. Then
(a) ./(G, JT) £ A(G, ■#") £ K[Irr(G)] £ ^«(G, Jf).
(b) 3tR(G, Jtf) is a ring in which JR(G, Jf) is an ideal.
(Addition and multiplication are pointwise.)
Proof The containments in (a) are all obvious from the definitions. To
prove (b) observe that Z[Irr(H)] is a ring for every He# Since Z £ R, it
follows that K[Irr(H)] is a ring. If <p, 9 e ^R(G, ^f), then
(<p% = <pH3He/?[Irr(tf)]
and hence q>9 e 3%R(G, #?), which is, therefore, a ring.
To prove that JR(G, df) is an ideal, we use the fact that if a is a class
function of H £ G and /? is a class function of G, then (a/?H)G = aG/J. This is
immediate from the definition of induction (and appears as Problem 5.3).
Let cpeJR(G, Jf) and 9e@R(G, Jf). Then cp = YMefWanf with
tA(H)e/?[Irr(H)]. Now
<p3 = £ 0/WG9 = I (•/'(hA)6
Since \pm9HeR[ln(H)], it follows that (p9eJR(G,J^) and the proof is
complete. |
It follows from Lemma 8.2 that in order to prove that JR(G, W) =
/?[Irr(G)], it suffices to show that 1G e J(G, Jff). Furthermore, if this can be
done for some family Jf, it follows that
JR(G, Jf) = /?[Irr(G)] = St AG, Jf)
for all R with Z £ R £ C.
(8.3) definition (Brauer) A group E is p-elementary (where p is a prime)
if E is the direct product of a cyclic group and a p-group. We say that E is
elementary if it is p-elementary for some prime.
(8.4) theorem (Brauer) Let Z £ R £ C, where R is a ring.
(a) A class function 9 of G is an R-generalized character iff BE e R[lrr(EJ]
for every elementary E £ G.
(b) Every / e Irr(G) is a Z-linear combination of characters of the form
1G for linear characters 1 of elementary subgroups of G.
Let S be the set of elementary subgroups of G. Statement (a) of Brauer's
theorem is exactly the assertion that /?[Irr(G)] = 3fcR(G, $). This part of the
result is often called the "characterization of characters."

128
Chapter 8
Every E e $ is nilpotent and hence is an M-group by Corollary 6.14. If
cp e Irr(£), then q> = XE, where X is a linear character of some E0 £ E. Since
E0 e $ and cpG = 1G, it follows that J(G, S) is exactly the set of Z-linear
combinations of XG for linear X e Irr(£) with E e i. Therefore, statement (b) of
Brauer's theorem amounts to the assertion that Z[Irr(G)] = J(G, S). This
is often called the "theorem on induced characters."
By Lemma 8.2, both parts of Brauer's Theorem 8.4 will be proved when
we show that 1G e J(G, i\ We work toward this goal. The proof given here is
based on an idea of B. Banaschewski. It is suggested that the reader also
consult the Brauer-Tate paper for another approach.
(8.5) lemma (Banaschewski) Let S be a nonempty finite set and let R be
a ring of Z-valued functions defined on S (with pointwise addition and
multiplication). If the function ls with constant value 1 does not lie in /?, then there
exists x e S and a prime p, such that p divides/(x) for every/ e R.
Proof For each xeS, let Ix = {f(x)\f eR}. Then Ix is an additive
subgroup of Z. If for some x e S, we have Ix < Z, then Ix £ (p) for some prime
and the result follows. Assume then, thatIx = Z for every xeS. For each x,
we may therefore choose/* e R with/*(x) = 1. Thus/* — ls vanishes at x and
Ekes (/* — *s) = 0- Expanding this product yields an expression for ls as a
linear combination of products of the functions fx. Thus ls e R. |
To obtain a ring to which we can apply Lemma 8.5, we consider
permutation characters of G, that is, characters of the form (1H)G for subgroups
H£G.
(8.6) lemma Let H, K £ G. Then (1H)G(1K)G = X av(\v)G for subgroups
U £ H and integers av > 0.
Proof Write 9 = (1K)G so that (1H)G(1K)G = 9(1 H)G = (9„f by
Problem 5.3. By Lemma 5.14, 9 is the permutation character of G acting on the
set of right cosets of K. It follows that 9H is a permutation character of H and
hence 9a = Xfltf(li/)H for subgroups U £ H and integers av > 0. Since
((l vff = (l Dft the result follows. |
(8.7) corollary The set of Z-linear combinations of characters of G of
the form (1H)G is a ring P(G). Let ^f be a collection of subgroups of G with
the property that if K £ H e JV, then Ke/. Let P(G, Jt?) denote the set of
Z-linear combinations of characters of the form (1H)G with H eJf. Then
P(G, tf) is an ideal of P(G).
To apply Corollary 8.7, we define a class of groups more general than
elementary groups.

Brauer's theorem
129
(8.8) definition A group H is p-quasi-elementary if H has a cyclic normal
p-complement for the prime p. We say that H is quasi-elementary if it is p-
quasi-elementary for some p.
Note that subgroups of p-quasi-elementary groups are themselves p-
quasi-elementary. Thus in the notation of Corollary 8.7, P(G, Jf) is closed
under multiplication, where Jf is either the set of all quasi-elementary
subgroups of G or is the set of all p-quasi-elementary subgroups of G for some
fixed prime p.
(8.9) lemma Let x e G and let p be a prime. Then there exists a p-quasi-
elementary subgroup, H c G, such that (lH)G(x) is not divisible by p.
Proof Let C be the p-complement in the group <x>. (Possibly C = 1 or
C = <x>.) Let JV = NG(C), so that x e JV. Since <x>/C is a p-group, we may
choose H/C e Sylp( JV/C) with <x> ^ h. Then C is a normal p-complement for
H and H is p-quasi-elementary.
Now (1H)G is the permutation character of G acting on the right cosets of
H, and so (l„f(x) = \{Hy\Hyx = Hy,yeG}\. If Hyx = Hy, we have
xyl e H and hence Cy~' c h. However, C is the unique p-complement in
H and hence Cy~' = C and y e JV. We therefore need to count the number of
fixed points in the action of <x> on the cosets of H in JV.
Since C <i JV, and C £ H, we see that C is in the kernel of the action of JV
on the cosets of H in JV. Since <x>/C is a p-group, it follows that the number of
nonfixed cosets is divisible by p and hence (lH)G(x) = \N:H\ mod p. By the
choice of H, pX\ N: H | and the proof is complete. |
(8.10) theorem (L. Solomon) Let ^f be the set of quasi-elementary
subgroups of G and Wp the set of p-quasi-elementary subgroups for some
prime p. Then
(a) lGeP(G,#).
(b) mlG e P(GT ^f p) for some m e Z with pjfm.
Proof By Corollary 8.7, P(G, Jf) is a ring of Z-valued functions on G.
If 1G £ P(G, ^f), then by Lemma 8.5, there exists x e G and a prime p with
p | <p(x) for all (p e P(G, JF). This contradicts Lemma 8.9 and so (a) is proved.
For (b) let R = {<p + np\G\<p e P(G, Jt?p), n e Z}. Then R is a ring. If there
exists x e G and a prime q, with <j | <p(x) for all (peR, then since plG e R, we
have q = p and thus we have a contradiction to Lemma 8.9. By Lemma 8.5,
we conclude that lGe/? and hence (1 — np)lGeP(G, ^fp) for some neZ.
This completes the proof. |
Only part (a) of Theorem 8.10 is needed to prove Brauer's theorem;
however, part (b) is useful for certain refinements of the result. Recall that we

130
Chapter 8
need \GsJ(G, $). The method of proof is essentially to use 8.10 to reduce
the problem to quasi-elementary groups. We need a lemma.
(8.11) lemma Suppose G = CP where C o G, P is a p-group and p/HC|.
Let 1 be a linear character of C which is invariant in G and suppose
Q(P)£ ker 1. Then 1 = lc.
Proof Let X = ker k. Since 1 is linear, it takes on distinct constant values
on the distinct cosets of K in C. Since k is invariant under the action of P, it
follows that P normalizes each coset Kx for xeC.
In the conjugation action of P on Kx, the number of moved points is
divisible by p. Since pJf\K\, we have pJf\Kx\ and thus Kx r\ CC(P) ^ 0.
Since CC(P) £ X by hypothesis, we conclude that Kx = K and thus K = C
and 1 = lc as claimed. |
Proof of Theorem 8.4 As was remarked following the statement of the
theorem, it suffices to prove that 1G e J(G, $). We use induction on | G |. We
may therefore suppose that lHe<f(H, SH) whenever H < G, where SH is the
set of elementary subgroups of H. Thus Z[Irr(H)] = J(H, SH) for these
H by Lemma 8.2. By transitivity of induction (Problem 5.1), it follows for
<peJ{H,SH), that <pG e J(G, SH) £ J(G, $). Thus for all H<G and
<p e Irr(H), we have <pG e J{G, S) and it will suffice to show that 1G is a Z-
linear combination of characters induced from proper subgroups.
By Solomon's Theorem 8.10(a) we are done if G is not quasi-elementary
and so we may assume that G has the cyclic normal p-complement C. Let
P e Sylp(G) and Z = CC(P). Since we may clearly assume that G is not
elementary, we have Z < C and E = PZ < G.
Write (1£)G = 1G + E, where S is a possibly reducible character of G.
We shall show that every irreducible constituent of S is induced from a proper
subgroup and thus 1G = (1£)G - S e J(G, S) as desired.
Let i be an irreducible constituent of S. Now CE = G and C n E = Z
so that
lc + Sc = ((1£)G)C = (lz)c
by Problem 5.2. Thus
1 = E(lz)C, lc] = [lc + Sc, lc]
and hence [Sc, lc] = 0. Thus [jc, lc] = 0.
Let k be an irreducible constituent of %c. Then k ^ lc. However, Z <i G
and so Z £ ker((l£)G) and hence Z £ ker /. We therefore have Z £ ker k
and thus by Lemma 8.11, k is not invariant in G. Let T = IG(k) < G. By
Theorem 6.11, / = \pG for some \ji e Irr(T). The result now follows. |
The following is a useful special case of part (a) of Brauer's theorem.

Brauer's theorem
131
(8.12) corollary Let x be a class function of G. Suppose
(a) Xe is a generalized character for every elementary E £ G;
(b) [/,/] = 1;
(c) Z(D>0.
Then / e Irr(G).
Proof Using (a) and Theorem 8.4(a), we have l = Y,az^ for <!; e Irr(G)
and ai e Z. From (b) we conclude that at most one ai^=Q and / = + ^ for
some £, e Irr(G). Then (c) yields the result. |
Before going on to reap some of the harvest of applications of Brauer's
theorem, we digress briefly to derive the "local" version of the result. In the
Brauer-Tate proof, this is obtained as an intermediate result.
(8.13) corollary Let Sp be the set of p-elementary subgroups of G for
some fixed prime p. Then
m\GeJ(G,Sp)
for some m e Z with pjfm.
Proof By Theorem 8.10(b), there exists m e Z with pjfm such that m 1G is
a Z-linear combination of characters of the form (1H)G for p-quasi-elementary
H £ G. Let SH denote the set of elementary subgroups of H. If H is p-quasi-
elementary, then SH £ Sp and by transitivity of induction, we have
(lH)Ge^(G,^H)£^(G,^p).
The result now follows. |
We shall use Brauer's theorem to remove the solvability hypothesis from
Theorem 6.25 on extending characters. To do this we need part (c) of the
following result. [Part (b) will be used in Chapter 13.]
(8.14) lemma Let N<iG and let zelrr(G) with Xn = # e Irr(iV). Let
\ji e Irr(G). For each coset Ng of JV in G compute
r\(Ng) = (\/\N\)YJ -AWzW-
xsNg
We have
(a) If [\pN, 9] # 0, then i\ e Irr(G/N).
(b) If !>„, 9] = 0, then tj(Ng) = 0 for all g.
(c) If x = iA, then rj(Ng) = 1 for all g.

132
Chapter 8
Proof Conjugates of Ng in G/N all have the form Ngh = h~l(Ng)h for
h e G. It follows that r\ is a class function on G/N. We may thus write r\ =
£ b^, where £ runs over Irr(G/JV) and ft,, e C. Now
*>< = [*. fl = TTt^t I *NgWg)
|G.JV| NS6G/jv
l"l lefi
where in the last expression, we view t, e Irr(G) and use the fact that as such, ^
is constant on cosets of JV.
We thus have b4 = \j//, /<!;]. Now by Corollary 6.17, x£ e Irr(G) and the
characters x£ are distinct for distinct <!; e Irr(G/JV). Therefore r\ = Y.'b^
satisfies rj = 0 unless \\i = x£ for some £, e Irr(G/JV) in which case r\ = £.
By Corollary 6.17, \ji = x£ for some £ iff [t/^, 9] t6 0. The result now
follows. |
(8.15) theorem (Gallagher) Let JV o G with ,9 e Irr(JV) invariant in G.
Suppose (9(1), |G : N\) = 1. Then 9 is extendible to G iff det .9 is extendible.
Proof The "only if" part of the assertion is trivial. Assume X = det 9
is extendible to \i e Irr(G). If JV s H £ G with H/JV solvable, then 9 is
extendible to H by Theorem 6.25. By Lemma 6.24, there is a unique extension
X(H), such that det(/(H)) = fiH. Define the function / on G as follows. If
g e G, let H = <JV, g} so that H/JV is cyclic. Set x(g) = X(H)(q)- We shall use
Corollary 8.12 to show that / e Irr(G).
First we establish that / is a class function. If g, x e G, let H = <JV, g},
so that Hx = <JV, g*>. Define \p on H* by i/f(/i*) = x(H)(/i) for /i e H. Since the
map /i h» A* is an isomorphism and /(H) e Irr(H), we have \p e ln(Hx). Also,
(deti/f)(/i*) = (det(/(H)))(/i) = /z(/i) = fi(hx), so that det i// = /zH*. Furthermore,
if neJV, we have \ji(n) = Xml"* ) = #(«*') = 9x(n) = 9(n). It now follows
from the uniqueness of z(H*,, that x(H*) = \\i and hence
*(»*) = Xw*)(9x) = ^(9X) = Xwis) = X(g)
and / is a class function.
Next, let E s G be elementary and set H = JV£ so that H/JV s £/(JV n E)
is nilpotent and x(H) is defined. We shall show that the restriction Xh = Xan
and thus Xe = (X{H))e which is a character of E. If g e H, then <JV, g> =
K £ H. Clearly, (x(H))K is an extension of 9 to K and det((/(H))K) = fiK so
that (x(H))K = X(Ky Therefore, xio) = fojfo) = X<hM hence la = Xw as
claimed, and Xe is a character.
Next, we compute [/, /]. For each coset Ng of JV in G, we have
I IzWI2 = I l/(H,WI2,

Brauer's theorem
133
where H = <JV, g} = <JV, x> for all xeNg. Since (x(H))n = 9eIrr(iV),
Lemma 8.14(c) yields £„„, lx(H)(x)|2 = |JV| and it follows that X«g IzWI2
= |G:JV|.|JV| = |G|andrjf,Z] = l.
Finally, /(l) = 9(1) > 1 and hence / e Irr(G) by Corollary 8.12. Clearly,
%N = 9 and the proof is complete. |
(8.16) corollary Let JV <a G and 9 e Irr(iV) with 9 invariant in G.
Suppose (|G: JV|, o(9)9(l)) = 1. Then 9 has a unique extension, /e Irr(G) with
(| G: N |, oft)) = 1. In fact, oft) = o(9). In particular, this holds if (| G: N |, | N |)
= 1.
Proof This is immediate from Corollary 6.27 and Theorem 8.15. (See
the proof of Corollary 6.28.) |
We have already seen several conditions sufficient to guarantee that a
character value is zero. The following is a powerful one.
(8.17) theorem Let /elrr(G) and suppose pjf( | G|/z(l)) for some prime p.
Then %(g) = 0 whenever p | o{g).
Proof Define 9 on G by 9(g) = x(g) if pJfo(g) and 9(g) = 0 if p | ofcr). We
shall use Brauer's theorem to show that 9 is a generalized character.
Let £cGbe elementary. Since E is nilpotent, we may write E = P x Q,
where P is a p-group and pjf\ Q \. If x e E and p^x), we have x e Q and so 9
vanishes on E — Q and 9Q = %Q.
Let t/f e Irr(£). We have
IE| [9£, tfr] = £ z(x)iAW = |Q| [ft, ^].
xbQ
Since | £ | = | P11Q |, we conclude that | P | [9£, i//~\ e Z.
Now let co = <yz be the algebra homomorphism Z(C[G]) -» C associated
with / so that co(X) = x(g)| Jf |//(1), where Jf is the conjugacy class
containing g and K = EJf e C[G]. We shall write co(gr) for the algebraic integer
co(K\ so that
X(g) = z(l)co(g)/|X| = x(l)co(g)|CG(g)|/|G|
Now
|£| [S*. -A] = Z zWiAW = t^t Z co(x)lA(x")|CG(x)|.
Since P Q CG(x) for x e Q, we have
^~ [S*. *] = -J^jjfl [9£, £| = £ co(x)i/Kxl|CG(x): P|,

134
Chapter 8
which is an algebraic integer. Since \P\ [9E, \p~\ e Z, we have [S£, \p~\eQ and
thus (| G11Q | /z( 1)) [9E, tA] e Z. Since | G11Q | //( 1) e Z and is relatively prime
to |P|, we conclude that [$£, i/f] e Z and hence $£ is a generalized character
of£.
Brauer's theorem now yields that 9 is a generalized character of G. In
particular [9, z]eZ. Now [9, z] = (1/|G|) £ {lz(0)l2lpM0)}, and so
0 < [3, *] < [/, /] = 1. We conclude that [9, /] = [/, /] and
0 = [(z-.9),z]=^X{lz(9)l2lp|o(g)}.
The result now follows. |
A character i e Irr(G) is said to have p-defect zero if p does not divide
|G|//(1).
In Chapter 2 we discussed the question of what information can be
obtained about a group from a knowledge of its character table. Since D8 and
Q8 have the same table, one cannot determine the orders of the elements in
the various classes. Nevertheless, it is possible to determine the sets of prime
divisors of these orders. To do this, we shall apply Theorem 8.4(a) in a
situation where R =£ Z.
If geG, we shall use the notation n(g) to denote the set of prime divisors
ofofef).
(8.18) lemma Let geG and let n be a set of primes. Then there exist
uniquex,yeG with
(a) g = xy = yx;
(b) n(x) ^ n and n(y) n % = 0.
Furthermore, x, y e <g>.
Proof Write o(g) = mn such that every prime divisor of m is in n and
no prime divisor of n is in n. Then (m, n) = 1 and we have km + In = 1 for
some k, leZ. Let x = gin and y = gkm so that g = xy = yx and x, ye<g>.
Since xm = 1 = /, (b) follows.
Now suppose g = uv = vu with n(u) ^ n and n(v) r\n = 0. Then
u c C(g) c C(x) and hence n(xu~l) c n(x) u 7t(w) c n. Similarly,
n(y~vv) c 7t(t>) u n(y),
so that 7r()>~ lv) n 7t(xu~') = 0. Since ^_1t> = xw_1, we have xw_1 =
1 = y~ 'u and uniqueness follows. |
If g, n, x, and y are as in Lemma 8.18, we write x = g„ and y = g„,. If
n = {p}, we write x = gp and y = gp,.

Brauer's theorem
13S
(8.19) lemma Let | G | = mn with (m, n) = 1 and let n be the set of prime
divisors of m. Let g e G, with g = #„-, and define 9 on G by
,9(x) = {n 'f *"' iS conJu8ate to &
(0 if xK> is not conjugate to g.
Let /? be the ring of algebraic integers in Q[e], where e is a primitive nth
root of 1. Then 9 is an /^-generalized character.
Proof Let E £ G be elementary. It suffices to show that 9E e /?[Irr(£)].
Since E is nilpotent, we may write E = H x K, where H is a rc-group and K
is a rc'-group. Then for xe E, we have x = uv,ue H,veK and thus w = x„
and v = x„..
If g is not conjugate to any element of X, then 9£ = 0 and there is nothing
to prove. Suppose then, that gug2,...,graKths distinct elements of K
which are conjugate to g. Then the elements xeE with x„. conjugate to g are
exactly the elements of the cosets Hgh 1 < i <r.
Let i/f e Irr(£). By Theorem 4.21, we have \ji = £ x « for some £ e Irr(H)
and w e Irr(K). We have
IE | [3£, *] = £ £ 9(ugtWUg~d = n £ ^) £ W\
i- 1 ueH i= 1 usH
Since £„eH £(m) = 0 unless £ = 1H, we have either [9E, i/f] = 0 or
|£|[9£,.A] = n|H|£^7).
i= 1
Now |£| = |X||H|and |K ||n and thus [S£,i/f] is a multiple of £ «(#() which
lies in R. The result now follows. |
(8.20) theorem Let R be the ring of algebraic integers in Q[e], where e
is a primitive |G|th root of 1. Let pe Z be a prime and let / be a maximal
ideal of R with pel. Let x, yeG. Then xp. and yp. are conjugate in G iff
X(x) = X(y) mod J for every / e Irr(G).
Proof Write \G\ = p"n, with p^n, and define 9 on G by 9(g) = n if gP' is
conjugate to xp. and 9(g) = 0, otherwise. By Lemma 8.19, 9 is an
R-generalized character of G.
Suppose /(x) = %(y) mod / for every / e Irr(G). It follows that 9(x)
= 9(y) mod /. Now 9(x) = n. If 9(y) = 0, then n = 0 mod / and n e I. Since
pel and p^n, it follows that 1 = (p,n)el and this is a contradiction. Thus
9(y) = n and hence yp* is conjugate to xp. as desired.
Conversely, we may suppose xp. = g = yp- and we show that /(x) = %(g)
= x(y) mod /. Write x<x> = £ 1;, where the X{ are linear characters. Since
g e <x>, it suffices to show that X(g) = Mx) mod / for linear 1.

136
Chapter 8
Now x = gh, where h = xp e <x> and l(x) = l(g)l(/i). Let l(/i) = 8 so
that (5pm = 1 for some m. For a e R, let a* denote its image in R/I which is a
field of characteristic p. We have 0* = (8*ym - 1* = (8* - l*)pm and thus
,5* - 1* = 0* and 8 = 1 mod /. It follows that l(x) = Mg) mod / and the
result follows. |
Note that by Theorem 8.20, it follows that if xp- and yp. are conjugate and
X e Irr(G), then /(x) — %(y) lies in every maximal ideal of R which contains p.
(8.21) theorem (G. Higman) Let X Y, Jf2, ..., Jft be the conjugacy
classes of G. Then the character table of G determines the sets of primes
n^ = n(g^ lor g^X^.
Proof Let Irr(G) = {/; 11 < i < k}. We are given the complex numbers
ain = XiiSii) f°r 0/ie -*V ^ 1 e •%"*>tnen v is tne unique integer 1 < v < k,
such that aiv e Z and a,v > \aifl\ for all i, /z and hence we know nv = 0. For
notational simplicity, we now assume 1 e jf t.
Next we compute £f (aa)2 = |G| and let /? be the ring of algebraic
integers in Q[e], where e is a primitive |G | th root of 1. For each prime, p11G |,
we choose a maximal ideal of R, Ip 2 pR. Construct the equivalence
relation ~p on {/i\ 1 < fi < k} by setting fi ~p v if a;/i = aiv mod 7P for all
i, 1 < i < k. By Theorem 8.20, /z ~p v iff the elements of Jf ^ and Jfv have
conjugate p'-parts. In particular, if (i~pv, then n^ u {p} = 7tv u {p}.
Furthermore, given fi, there exists v, with p$nv, such that /z ~ p v. To see this,
take g e JT^ and choose v so that gp, e Jf v.
Let n be a set of prime divisors of G. Using induction on | n \, we show how
to construct the set y, = {/zl^ = ?c}. We have ^0 = {1}. If 7t j= 0, write
7t = 7t0 u {p}, where p$n0. By the preceding remarks, it follows that
■5^= {\i\\ii^no and /z~pv for some vg^J.
The proof is now complete. |
Transfer theory is a tool commonly used for producing normal
subgroups of a group, especially normal p-complements or normal 7t-comple-
ments. Many of these transfer theorems can also be proved using characters,
and in particular using Brauer's theorem. An example of a typical transfer
theorem is the following: Let H be a Hall subgroup of G (that is, (| H \, \ G: H \)
= 1) and suppose that H is nilpotent and that if x, y e H are any two
elements which are conjugate in G, then x and y are already conjugate in H.
Then H has a normal complement in G.
In the above situation, let n be the set of prime divisors of | H \. Using the
assumption that H is nilpotent, it is not too hard to prove that if U £ G is a

Brauer's theorem
137
nilpotent rc-subgroup, then U is conjugate to a subgroup of H. Because of
this, the following result is a generalization of the above transfer theorem.
(8.22) theorem (Brauer-Suzuki) Let H be a Hall subgroup of G and
suppose that whenever two elements of H are conjugate in G, then they are
already conjugate in H. Assume for every elementary subgroup E £ G, that
if | E111H |, then E is conjugate to a subgroup of H. Then there exists K<G
with HK = G and H n K = 1.
Proof Let n be the set of prime divisors of |H|, so that by hypothesis,
every elementary rc-subgroup of G is conjugate to a subgroup of H. Let 9
be a class function of H. We define a class function 5 on G as follows. For
g e G, <#„> is an elementary rc-group and hence g„ is conjugate to some
element x e H. Set §(g) = 9(x) and observe that this is well defined since if g„
is also conjugate to y e H, then x and y are conjugate in H by hypothesis and
so 9(x) = %).
For x e H, let /x(x) denote the number of elements g e G with g„ conjugate
to x. Let xu x2 x, be representatives for the conjugacy classes of H. If 9
and cp are class functions of H, we have
(*) & 9\ = yir t SixMxdiAxd.
\G\ (=i
Now let R be a ring with ZsKcC and suppose that 9 is an /?-gener-
alized character of H. We claim that S is an /^-generalized character of G. To
prove this, let E £ G be elementary. We show that (5)£ is an /^-generalized
character of E. Write E = U x V, where U is a 7t-group and V is a 7t'-group.
We may assume that U £ H.
If w e 1/ and veV, then (w)„ = w and hence §(uv) = 9(u). Since U £ H,
we may write 9^ = £ a^ for t/f e Irr(t/) and a^ e /?. It follows that (9)£ =
5> fli/rC/' x V)> which is an R-generalized character of E. Thus 5 is an R-
generalized character of G by Brauer's theorem as claimed.
Now let R be the ring of algebraic integers in Q[e], where e is a primitive
|H|th root of 1. For 1 < i < t, define 9, = £V€irr(H) q>(x^q>, so that 9;(xj) = 0
if i t6 j and 9;(xj) = | CH(x.) | by the second orthogonality relation. Also, 9,
is an /^-generalized character of H and hence [5;, 1G] e R and
[S«. lG] = ^ i W/4*j) = Mx'),|Cf(Xi)l.
luU=i |G|
We conclude that /x(x,)|CH(x,)|/|G| is a positive rational number which
is an algebraic integer. It follows that it is a positive rational integer and thus
/Axd\C^x,)\/\G\ > 1

138
Chapter 8
and fi(x{) > |G|/|CH(x,)| for all i. We now have
Therefore, equality holds throughout and /z(xj) = |G|/|CH(x,)|. Thus
Equation (*) yields
Now suppose / e Irr(H). By taking R = Z, we conclude that % is a
generalized character of G. Since [J?, %~] = \j, /] = 1 and %(\) = x(\) > 0, it
follows that % e Irr(G).
Let K = fifkerOOIZ e Irr(H)}. Then X -c G. If u e H n X, then z(w) =
jf(M) = jj(l) = /(I) for all /elrr(H) and thus u = 1. On the other hand, if
P^tc, PeSylp(G), and veP, then t>„ = 1 and %(v) = z(l) = x(l) for all
/elrr(H), so that t>eX. Thus P s K and hence |X| > \G:H\. It follows
that HK = G. |
To demonstrate the power of Theorem 8.22, we show how easily Burn-
side's transfer theorem follows from it.
(8.23) corollary Let P e Sylp(G) and suppose P £ Z(N(P)). Then there
exists K~a G with KP = G and K r\ P = 1.
Proof If E £ G is a p-group, then E is conjugate to some subgroup of P
by Sylow's theorems. It therefore suffices to assume x, y e P are conjugate in
G and show that x and y are conjugate in P.
We have then, y = xa for some g e G. Since P is abelian, P £ Cg(j>) and
Pg £ CG(x9) = CG(y). Thus P»c = P for some c e CG(x9) and hence
gc e N(P) £ C(x) by hypothesis. Now y = x° = (xe)c = x°c = x and the
proof is complete. |
We also remark that Theorem 8.22 provides an alternate proof of
Frobenius' Theorem 7.2. It is routine to check that a Frobenius complement
necessarily satisfies the hypotheses of Theorem 8.22.
Our next results depend on Corollary 8.13 where we restrict attention to
p-elementary subgroups for some fixed p.
(8.24) theorem (Dade) Let N <a G with G/N a p-group. Let 9 e Irr(iV)
be invariant in G. Then there exists a p-elementary subgroup E £ G, and

Brauer's theorem
139
<p e Irr(£ n JV), such that
(a) NE = G;
(b) <p is invariant in E;
(c) [9£nJV,<p]isprimetop.
Proof By Corollary 8.13, we have
(*) m\G = £ a^G,
where \p runs over irreducible characters of p-elementary subgroups, a^ e Z
and p^m. If £ is p-elementary and i/f e Irr(£), we compute [/?, (i/^n] for
G-invariant class functions p of JV. Now (/?<% = \G:N\P and (j3£JV)n =
\EN:N\p. Since 0G and 0£JV both vanish on £JV - N, we have (0G)£JV
= |G:£JV|0£Jv.Now
where the latter equality follows since \jiG = (\liEN)G. This yields
[fi, W%] = \G:EN\IPEN, rNl = \G:EN\[fi, (*"%]
= \G:EN\[p,WEnN)Nl
where we have used Problem 5.2 to obtain the last equality.
We apply this with /? = 95. For each \ji in Equation (*), write \ji e Irr(£^)
for p-elementary £^ s G. We obtain
m = [95, my = £ a^[95, (,/,%]
= 5>*|G:JVEJ[8S,WrE,nN)w]-
Since p^m, there exists ^ such that p does not divide | G : NE | [95, (i/f£ n A,)iV],
where we have written E = £^. Since G/N is a p-group and pjf\ G: JVE |, we
conclude that NE = G, and (a) follows.
Now write L = E r\ N so that
[55, (^f] = [$L§L, tfrj = [9L, ^L9J
is prime to p. Since 9 is invariant in G we have 9L is invariant in £ and we may
write 9L = £A eA A, where A runs over sums of orbits of the action of E on
Irr(L) and eA e Z. Write y = \\iL 9L so that y is invariant in £. We have
[A. Ml] = 5>aIA 7]
A
is prime to p. Choose A such that p/feA[A, }>] and write A = <pt + • • • + <p„
where the <pt e Irr(L) and are conjugate under £.

140
Chapter 8
Since y is invariant under E, all [<p;, y] are equal and [A, y] = t\_<pu y].
However t is a divisor of|£:L| = |G:JV| and so is a power of p. Since
pjf\_\ y], we have t = 1 and <pt = q> is invariant in £. This is (b). Finally,
[#l> <p] = eA is prime to p and the proof is complete. |
An interesting special case is when N = G.
(8.25) corollary Let x £ Irr(G) and let p be a fixed prime. Then there
exists p-elementary E £ G and <p e Irr(£) with [jE, <p] ^ 0 mod p.
Examples exist which show that in Corollary 8.25, E cannot always be
taken to be a p-group.
Dade's Theorem 8.24 has a nice application to the question of character
extendibility.
(8.26) theorem Let JV -a G with G/N a p-group. Let P e Sylp(G) and
assume that P n JV £ Z(P) and that every linear character of P r\ JV is
extendible to P. Then every G-invariant irreducible character of JV is
extendible to G.
Proof Let 9eIrr(JV) be invariant in G. By Theorem 8.24, choose a
p-elementary subgroup £ £ G, with £JV = G, and q> e Irr(£ n JV) such
thatp^rA^.p],
We have E = Q x C, where Q is a p-group and C is cyclic. We may
assume that C is a p'-group so that EnN = (QnN)xC and we write
(p = a. x X where a eIrr(Q n JV) and 1 elrr(C). We may assume Q £ P so
that Q n, N ^ P n N, which is abelian. It follows that a is extendible to
P n JV and thence to P by hypothesis. Therefore a has an extension a e Irr(Q)
and <p = a x A is an extension of <p to £.
Now since EN = G, we have (<pG)N = (q>EnN)N = 9N and hence
which is prime to p. Therefore, there exists an irreducible constituent / of
cpG with pJf\_XN, #]• Since 9 is invariant in G, it follows that xn = e& with
pjfe. Now e divides | G: JV | by Problem 6.7 and since G/N is a p-group it
follows that e = 1 and / extends 9. (Note that Problem 6.7 follows
immediately from Corollary 6.19 by induction.) |
We remark that in the notation of Theorem 8.26, the hypothesis on P will
automatically be satisfied if P is abelian. More generally, it suffices for
F n JV = 1.
It has been mentioned that if JV -a G and 9 e Irr(JV) is invariant in G, then
it suffices to show that 9 is extendible to the inverse images in G of the Sylow
subgroups of G/N in order to prove that 9 is extendible to G. It is for this
reason that it is useful to obtain results like Theorem 8.26 which give
sufficient conditions for extendibility when the factor group is a p-group.

Problems
141
Problems
(8.1) Let Z £ R £ C, where R is a ring and the additive group (Z, +) is a
direct summand of (R, +). Let W be a family of subgroups of G. Show that
A(G, JT) n Z[Irr(G)] = ./(G, jf).
Hint Write /? = Z 4- S. If 9 = £ a^G, where all a^, e S, then [,9, *] e S
for all / e Irr(G).
Note The Brauer-Tate proof of Brauer's theorem uses this fact applied
to R = Z[e], where e is a root of unity.
(8.2) Let s# be the collection of abelian subgroups of the group G. Show
that the following statements are equivalent:
(a) \GeJ(G,sf)\
(b) «(G,j/) = Z[Irr(G)];
(c) every Sylow subgroup of G is abelian.
Hint If a Sylow p-subgroup of G is nonabelian, consider the function
9 = (l/p)p, where p is the regular character of G.
(8.3) Let i be a character of G. For g e G and prime p, write g = gpgP' in
the notation introduced following Lemma 8.18. Define the class function xP
by Ipig) = X(9P\ Show that (xp)N is a character for every nilpotent JV £ G.
Show that if G is not nilpotent then xP 1S not a character of G for some
X e Irr(G) and some prime p.
Note In fact, xP is a character for every / e Irr(G) iff G has a normal
Sylow p-subgroup. Of course, by Brauer's theorem, xP is always a generalized
character.
(8.4) Let x and \ji be characters of G and suppose \G\ = mn, where (m, n) = 1.
Let
where the sum is taken over those g e G such that o(g) \ m. Show that a/m e Z.
(8.5) Let /elrr(G) and suppose |G| = mn with (m, n) = 1. Assume that
X(x) = 0 for all 1 # x e G such that x" = 1. Suppose y e G and /" ^ 1. Show
that x(y) = 0.
(8.6) Let G and H be groups with classes Jf, and if, respectively and
irreducible characters /, and t/^, respectively. Assume that whenever gejft
and /i e if, we have x,<0) = \jifti) for all i and j. (In short, G and H have
identical character tables.) Let x e Z(P) where P e Sylp(G). Suppose x e Jf,.
Let y e if (. Show o(x) = o(y).

142
Chapter 8
Hint If o(y) > o(x) and z e H, with <y> = <z> and yp = zp, show that
z e if, and conclude that p 11 if,-1. See the hint for Problem 2.12.
(8.7) Suppose G and H have identical character tables and that G is solvable
and has an abelian Sylow p-subgroup. Show that G and H have isomorphic
Sylow p-subgroups.
Hint If G has no nontrivial normal p'-subgroup, then its Sylow p-
subgroup is normal. Use Problem 8.6.
(8.8) (Dade) Let N <a H £ G. Assume the following:
(a) (|tf:JV|,|G:tf|)=l.
(b) If x, x9eH for some g eG, then there exists heH and neN with
x9 = xhn.
(c) If £ £ G is elementary and (|£|, |G:H|) = 1, then EB £ H for
some 3 e G.
Show that there exists M <i G such that HM = G and H r\M = N.
Hint Mimic the proof of Theorem 8.22. Consider only those class
functions 9 of H which are constant on cosets of JV.
Note Many of the important "transfer" theorems follow from Problem
8.8, for instance, the Focal Subgroup Theorem:
LetPeSylp(G)andletJV = {x~lxg\xeP,xBeP,g eG>.Then AP(G) n P
= N.
Here, AP(G) denotes the normal subgroup of G minimal such that the
factor group is an abelian p-group. Problem 8.8 yields M <i G with MP = G
and M r\ P = N. It follows that M 2 AP(G). The reverse inclusion follows
from N £ G' £ AP(G).
(8.9) Show that a group G is quasi-elementary iff 1G cannot be written in the
form £ %(1h)g for proper subgroups H, with aH e Z.
Hint For "only if": Assume 1G = £ aH(lH)G, where the H run over
representatives for the conjugacy classes of proper subgroups. Let C be a
cyclic normal p-complement for G. Choose H0 minimal, such that
pJf(aHo\G:H0\), and let 1 be a faithful linear character of C/(C r\ H0).
Consider the numbers [(aH(lH)G)c, 1].
Note Problem 8.9 essentially says that Theorem 8.10(a) is the best
possible.
(8.10) Show that the integers m in Theorem 8.10(b) and Corollary 8.13 can
be taken to be the p'-part of | G |.

Problems
143
(8.11) Let x £ Irr(G) be a character of defect zero with respect to the prime p.
Let Q^G, with pJf\Q\ and let PeSylp(CG(Q)). Show that (l/|P|)xQ is a
character of Q.
(8.12) Let P oG, where P is a p-group, and let / e Irr(G/P) be of defect zero
in G/P with respect to p. Let 9 e Irr(P) be invariant in G. Define the function
\ji on G by
#7) = 0 if g„{P
*l>(g) = HgP)x(gP) if gP£P-
(a) Show that \ji is a generalized character of G.
(b) If G = PCG(P), show that ip e Irr(G).
(8.13) Let N o G be a Hall 7t-subgroup. Define the function v on N by v(x) =
| {y e CG(x)\y is of 7t'-order} |. Show that v is a generalized character of N.
Hint If 9 is a G-invariant character of N (which is not necessarily
irreducible), define 5 on G by
kg) = %*)•
Show that 9 is a generalized character of G and compute [9, 1G].

9 Changing the field
So far we have restricted our attention almost exclusively to characters,
representations, and modules over the complex numbers. In this chapter we
digress from our study of the properties of Irr(G) to consider irreducible
group representations over arbitrary fields. (In prime characteristic, this falls
far short of the general case, since not all representations are completely
reducible.) In particular, if F £ E is a field extension we explore the
connections between the irreducible ^-representations and F-representations of a
group. In prime characteristic, we shall see that this situation is (surprisingly)
under somewhat better control than it is in characteristic zero, which will be
more fully considered in Chapter 10.
Let F c E and let 3E be an F-representation of a group G. Then 3E maps
G into a group of nonsingular matrices over F that, of course, are also non-
singular over E. We may, therefore, view 3E as an F-representation of G. As
such we denote it by 3E£. (The superscript thus merely indicates a change in
point of view.) If 3Et and 3£2 are similar F-representations, then 3Et£ and 3E2£
are similar and it follows that if 3E corresponds to the F[G]-module V, then
there is a uniquely defined (up to isomorphism) £[G]-module VE that
corresponds to 3E£. (For those readers familiar with tensor products, we remark
that VE = V ® FE.) We shall not, however, need to refer to VE again, since
it is usually easier to work with the representation 3E£.
The F-representation 3E may be extended by linearity to obtain a
representation of F\_G~\, which we shall continue to call 3E. Under this
convention, the £[G]-representation 3E£ is an extension of the F[G]-represen-
tation 3E.
144

Changing the field
145
0
1
0
0
1
0
0
0
0
0
0
1
°\
o\
-7
0/
X(b) =
0
' 0
1-1
^ o
0
0
0
-1
1
0
0
0
0^
1
0
0
If 3E£ is irreducible, then clearly so is 3E. However, 3E£ may well reduce, even
if 3E is irreducible. To illustrate what can happen, we consider two examples,
both for the field extension RcC. Let G = <g> be cyclic of order 3 and let 3E
be the R-representation defined by
Then 3EC affords the character k + I, where 1 is a faithful linear character of
G and so 3EC is reducible. Since k is not real valued, it is not afforded by any
R-representation and it follows that 3E is irreducible over U.
Now let QB = <a, b} be the quaternion group of order 8, where o(a)
= 4 = o(b). Define
X(a) =
It is not hard to check that this defines a representation of Q8. (In fact this is
the representation whose module is the quaternion algebra H = IR + Ui +
Uj + Uk with respect to the basis {1, i,j, k}, where a = i and b = j act by
right multiplication.) Now 3EC affords the character 2/, where / e Irr(Q8) and
X(l) = 2. By Problem 2.5(b), / is not afforded by any R-representation of QB
and thus 3E is irreducible over U. (This can also be deduced from the fact that
H is a division algebra.)
(9.1) definition Let 3E be an F-representation of G. Then 3E is absolutely
irreducible if 3E£ is irreducible for every field £2F.
(9.2) theorem Let 3E be an irreducible F-representation of G with degree n.
The following are equivalent.
(a) 3E is absolutely irreducible.
(b) 3E£ is irreducible for every finite degree extension E 2 F.
(c) The centralizer of 3E(G) in the matrix ring Mn(F) consists of scalar
matrices.
(d) X(FlG-]) = Mn(F).
Proof That (a) implies (b) is trivial. Now assume (b) and let M e M„(F),
with MX(g) = H(g)M for all geG. Let £ be a finite degree extension of F,
chosen so that M has an eigenvalue, keE. Since 3E£ defines an irreducible
representation of £[G] and M - kl is a singular matrix centralizing its
image, it follows from Schur's Lemma 1.5 that M - kl = 0. Thus (c) follows.
That (c) implies (d) is immediate from the Double Centralizer Theorem
1.16. Finally, assume (d) and let L s F. Since every M e Mn(L) is an L-linear

146
Chapter 9
combination of matrices in Mn(F), it follows that 3E(L[G]) = Mn(L). Thus
for every L representation 9) similar to 3E, we have 9)(L[G]) = M„(L) and so
9) cannot be in reduced form. Thus 3E is irreducible over L and the proof is
complete. |
(9.3) definition The field F is a splitting field for G if every irreducible
F-representation of G is absolutely irreducible.
(9.4) corollary If F is algebraically closed, then F is a splitting field for
every group.
Proof Since F has no proper finite degree extension, condition (b) of
Theorem 9.2 holds for every irreducible F-representation of G. |
Suppose F is a splitting field for G and E 2 F. Then every irreducible
F-representation 3E determines an irreducible F-representation 3E£. If {£,} is
a set of representatives for the similarity classes of irreducible F-represen-
tations of G, then {3E,£} is a complete set of representatives for the irreducible
£-representations of G. In order to prove this, we need to discuss the "
irreducible constituents " of a possibly reducible representation.
If V is an F[G]-module, then a composition series for V is a chain of
submodules
V=V0>Vl> — >Vn = Q
such that each Vt_ JVj is an irreducible module. The modules V{_ JVi are the
factors of the series. The Jordan-Holder theorem asserts that the factors of
any two composition series for V are the same up to isomorphism (and
counting multiplicities). An irreducible module W isomorphic to a factor of
some (and hence all) composition series for V is called an irreducible
constituent of V. If W is isomorphic to an irreducible submodule of V, then W is
a constituent of V. We call it a bottom constituent in that case. Similarly any
irreducible homomorphic image of V is a constituent. These are the top
constituents of V. If V is completely reducible [e.g., if char(F)/|'|G|], then
every irreducible constituent of V is both a top and a bottom constituent.
(Caution: the converse is false.)
All of the preceding remarks may be translated into the language of
representations. If 3E is an F-representation of G corresponding to the F[G~\-
module V, then 3E is similar to a representation 3 m triangular block form
/3ite) * \
\ o ' 3„(0)/
where the irreducible representations 3; correspond to the factors Vi_1/Vi.

Changing the field
147
Representations similar to the 3; are the irreducible constituents of 3E; those
similar to 3i being top constituents and those similar to 3„ being bottom
constituents. (Of course some of the other irreducible constituents of 3E may
also be top or bottom constituents since 3 is not necessarily the only
representation in triangular block form to which 3E is similar.) Note that the
character afforded by a representation is the sum of the characters of all of
the irreducible constituents (counting multiplicities). We emphasize that
by the Jordan-Holder theorem, the representation 3E has only finitely many
irreducible constituents (up to similarity), namely those which appear in any
single triangular block representation similar to 3E.
(9.5) corollary Let F be a field and G a group.
(a) Every irreducible F-representation of G is a top constituent of the
regular F-representation 5R.
(b) There exist only finitely many similarity classes of irreducible
F-representations of G.
(c) If E 2 F and 9) is an irreducible F-representation of G, then 9) is a
constituent of 3E£ for some irreducible F-representation 3E.
Proof Statement (a) is immediate from Lemma 1.14 and (b) follows from
(a) via the Jordan-Holder theorem. Let 3 be any F-representation of G. The
irreducible constituents of 3£ rnay be found by taking the irreducible
constituents 3E; of 3 and then finding the irreducible constituents of the 3E,£. Now
(c) follows by applying this remark to 3 = 5R. |
The following result is a useful tool for establishing the similarity of two
F-representations of G and for other purposes.
(9.6) theorem Let 3E be an irreducible representation of F\_G] and let
a e F[G]. Then there exists b e F[G1 such that X(b) = 3E(a) and 9)(b) = 0 for
every irreducible F[G]-representation 9) which is not similar to 3E.
Proof Let {3E,} be a set of representatives for the similarity classes of
irreducible F[G]-representations. Let It= {xe F[G] |3E,(x) = 0} so that J{ is
an ideal of F[G] and in the notation of Problem 1.4, J(F[G]) = f)/{. Let
A = F\_G]/(f]Ii), so that each 3E( may be viewed as a representation of the
algebra A. As such, the 3E,- are irreducible and pairwise nonsimilar. In
particular, J(A) = 0 and by Problem 1.5, A is semisimple. By Theorem 1.15, A
has minimal ideals M{, such that X/M,) = 0 if j j= i and 3E,(Mj) = 3E,(y4) =
3Ej(F[G]). Now suppose 3E = 3Et. Choose b in the inverse image in F\_G] of Mt
with X(b) = 3E(a). The result follows. |
(9.7) corollary Let 3E and 9) be irreducible F-representations of G.
Suppose F £ E and that 3E£ and 9)£ have a common irreducible constituent.
Then 3E is similar to ?).

148
Chapter 9
Proof Let 3 be an irreducible constituent of 3EE and of 9)£. If 3E and 9) are
not similar, view them as representations of F[G] and choose b e F[G] with
X(b) = 3E(1) and 9)(b) = 0. It follows that if U is any £[G]-representation
similar to 3E£, then \X(b) is the identity matrix and hence 3(b) is the identity
matrix. A similar argument, working with 9)£, shows that 3(b) = 0 and this
contradiction proves the result. |
Note that Corollary 9.7 asserts that the representation 3E in Corollary
9.5(c) is unique up to similarity.
(9.8) corollary Let F be a splitting field for G and let {3E,} be a set of
representatives for the similarity classes of irreducible F-representations.
Suppose £2F, Then £ is a splitting field and {3E,£} is a set of representatives
for the irreducible ^-representations of G.
Proof Since the 3E; are absolutely irreducible, the 3E£ are absolutely
irreducible. They are pairwise nonsimilar by Corollary 9.7. Finally, suppose
9) is any irreducible £-representation. By Corollary 9.5(c), 9) is a constituent
of 3E,£ for some i. Since X,£ is irreducible, it is similar to 9) and the proof is
complete. |
(9.9) theorem Let £ be a splitting field for G and let F £ £. Then F is a
splitting field iff every irreducible ^-representation of G is similar to 9)£ for
some F-representation 9).
Proof Suppose F is a splitting field. Then by Corollary 9.8, every
irreducible £-representation is as desired. Conversely, let 3 be any irreducible
F-representation and let 3E be an irreducible constituent of 3£. By hypothesis,
there exists an irreducible F-representation 9), such that 9)£ is similar to 3E
and hence 9)£ and 3£ have an irreducible constituent in common. By
Corollary 9.7,9) is similar to 3, and thus 3£ is absolutely irreducible. It follows that
3 is absolutely irreducible since the only F-matrices which centralize all 3(9)
are scalar. The proof is complete. |
(9.10) corollary Let F be any field and G a group. Then some finite
degree extension of F is a splitting field for G.
Proof Let F be the algebraic closure of F, so that F is a splitting field.
Let {3Ej} be a set of representatives for the similarity classes of irreducible
F-representations. By Corollary 9.5(b), | {3E,} | < oo and hence only finitely
many elements of F occur as entries in any of the matrices 3tt(g) for g e G.
Adjoin all of these elements to F so as to obtain the field E. Since F is
algebraic over F, it follows that | E: F\ < oo. Since each 3E, may be viewed as an
F-representation of G, it follows from Theorem 9.9 that £ is a splitting field. |

Changing the field
149
(9.11) corollary Let Q £ F s C. Then F is a splitting field for G iff
every / e Irr(G) is afforded by some F-representation.
Proof If F is a splitting field and / e Irr(G), choose a C-representation
3E that affords /. By Theorem 9.9,3E is similar to 9)c for some F-representation
% Then 9) affords /.
Conversely, suppose that every / e Irr(G) is afforded by some
F-representation. Let 3E be an irreducible C-representation and let 9) be an F-
representation which affords the same character. Then by the linear
independence of Irr(G) and using the fact that C has characteristic zero, it follows
that 3E is the unique irreducible constituent of 9)c and occurs with
multiplicity 1. Thus 9)c is similar to 3E. The result now follows from Theorem 9.9. |
We shall now discuss the character theory of a group over an arbitrary
splitting field. The following result is actually true without the assumption
that £ is a splitting field. We shall prove the more general fact later.
(9.12) lemma Let £ be a splitting field for G. Then the characters of non-
similar irreducible ^-representations of G are nonzero, distinct, and linearly
independent over E.
Proof Let {3E,} be a set of representatives for the similarity classes of
irreducible ^representations, and let /, be the character afforded by 3E,.
View Xi as being defined on all of £[G]. Since 3Ef(£[G]) is a full matrix ring
over E, we may choose aieE\_G] with x,(a,) = 1. By Theorem 9.6, we may
assume that //a,) = 0 if i =£ j. The result is now immediate. |
If £ is a splitting field for G, we shall use the notation Irr^G) to denote the
set of characters of the (absolutely) irreducible ^representations of G. The
point of the next result is that in some sense Irr^G) is not as dependent on the
particular field £ as the notation would indicate.
(9.13) lemma Let £ be a splitting field for G and let / e Irr^G). Suppose
K £ £ is a subfield which contains %(g) for all g e G and let F 2 K be another
splitting field for G. Then / e IrrF(G).
Proof We may replace F by a K-isomorphic copy and assume that £,
F £ L for some field L. By Corollary 9.8, L is a splitting field for G and
IrrF(G) = IrrL(G) = Irr^G). The result follows. |
The next result is of great importance in studying representations in
prime characteristic. Its proof depends on Wedderburn's theorem which
asserts that finite division rings are commutative. As the quaternion group
of order 8 shows, Theorem 9.14 would be false in characteristic zero.

ISO
Chapter 9
(9.14) theorem Let 3E be an absolutely irreducible ^representation of G,
where £ is of prime characteristic. Suppose that 3E affords the character / and
that %(g) e F for all g e G, where F is some subfield of E. Then X is similar to
9)£ for some absolutely irreducible F-representation 9).
Proof First we consider the crucial special case that | E \ < oo. Working
by induction on | E \, it is no loss to assume that F is a maximal subfield of E.
Let A be the set of all F-linear combinations of the matrices X(g) for g e G.
Then A is an F-subalgebra of the matrix ring M„(E), where n is the degree of 3E.
If a e Z(A), then a is a matrix over E which commutes with all X(g). Since 3E is
absolutely irreducible, a is a scalar matrix and thus F • 1 £ Z{A) £ £ • 1.
It follows that Z(A) is a finite integral domain, and hence is a field. Since F is
a maximal sufield of E, we conclude that either Z(A) = F • 1 or Z(A) = E • 1.
If Z(A) = £• 1, then A = £• A = 3E(£[G]). Since 3E is absolutely
irreducible, we conclude that A = M„(£) and thus if a e £ — F, we can choose
a e A with tr(a) = a. However, a is an F-linear combination of matrices of the
form X(g), each of which has trace in F. This contradiction shows that
Z(A) = F-l.
Next, we claim that the F-algebra A is algebra isomorphic to the full
matrix algebra Mk(F) for some integer k. To show this, we first establish that
A is semisimple. By Problem 1.5, it suffices to show that A has no nonzero
nilpotent ideals. If / is a nilpotent ideal of A, then £ • / is a nilpotent ideal in
£ ■ A = 3E(£[G]) = M„(£). It follows that 1 = 0. (Any nonzero matrix has a
multiple which is not nilpotent.)
By Theorem 1.15, the semisimple F-algebra A is a direct sum of minimal
ideals. Since dimF(Z(y4)) = 1, it follows that there is only one direct summand
and hence in the notation of 1.15, A = M(A) s AM for some irreducible
A-module M. Let D = EA(M), so that D is a division ring by Schur's Lemma
1.5. Since dimF(M) < oo and \F\ < oo, we have |D| < oo and thus D is
commutative by Wedderburn's theorem.
By the Double Centralizer Theorem 1.16, we have AM = ED(M) and
thus D s Z(AM) s F. Since D contains the scalar multiplications, 1 • F, of F
on M, we conclude that D = IF. Thus if k = dimF(M), then A s AM =
Ed(M) = EF(M) ^ Mk(F) and we have an F-algebra isomorphism
9: A -► Mk(F). Thus ?) = 3E9: F[_G^\ ->• Mk(F) is a representation of F[G].
It is absolutely irreducible since it maps onto Mk(F).
We claim that 9)£ is similar to 3E. To see this, let 3 be an irreducible F-
representation such that 3E is a constituent of 3£- If 31S not similar to ?), use
Theorem 9.6 to choose b eF[G] such that 3(b) = 0 and ?)(£>) ^ 0. Since
3E is a constituent of 3£, we have X(b) = 0 and thus ty(b) = (3E(ft))9 = 0. This
contradiction shows that 9) is similar to 3 and thus 3E is a constituent of 9)£.
Since 9) is absolutely irreducible, the claim follows and the theorem is proved
when £ is finite.

Changing the field
151
We now consider the general case where £ may be infinite. Since we may
replace £ by a larger field, it is no loss to assume that E is algebraically closed.
Let K £ E be the prime field and let L 2 K be a splitting field with
\L:K\ < oo (by Corollary 9.10). Since E is algebraically closed, we may
assume L <=, E. Since L is a splitting field, 3E is similar to 3£ for some L-
representation 3 and 3 affords / which takes values in L n F.
Since \L\ < oo, the first part of the proof yields an absolutely irreducible
(L n F)-representation 9), such that 9)L is similar to 3 and hence 9)£ is similar
to 3E. Now 9)F is the desired F-representation and the proof is complete. |
Theorem 9.14 remains true if X is irreducible but not absolutely
irreducible. This will be proved in Corollary 9.23.
By the exponent of G we mean the least positive integer n, such that
g" = lforallgeG.Clearlyn||G|.
(9.15) corollary Let G have exponent n and assume the polynomial
x" — 1 splits into linear factors in the field F. If F has prime characteristic,
then it is a splitting field for G.
Proof Let E 2 F be a splitting field and let / e Irr^G). Then %(g) is a
sum of nth roots of unity and hence %(g) e F. The result follows by Theorems
9.14 and 9.9. |
An entirely different proof shows that Corollary 9.15 also holds in
characteristic zero. This result of R. Brauer depends on his theorem on induced
characters and will be proved in Chapter 10.
Let E be any field, a an automorphism of E, and 3E an £-representation of
G. We can apply a to every entry in the matrix H(g) for every g e G. What
results is a new representation, denoted 3E", which may or may not be similar
to 3E. Similarly, if 9: G -» E is a function, we write 9-"(g) = {S(g))a. Suppose 3E
affords the £-character / and that / takes values in F £ E. If t e Aut(F), it is
not obvious that /T is an F-character of G.
(9.16) lemma Let £ be a splitting field for G and let / e Irr^G). Suppose
1(g) e F £ £ for all g e G and let t e Aut(F). Then /* e Irr£(G).
Proof Let E be an algebraic closure for £ and let F £ £ be an algebraic
closure for F. Then Irr£(G) = Irr£(G) = Irrf(G) by Corollary 9.8. Since t is
extendible to an automorphism of F, the result follows. |
Let K £ £ be a field extension and suppose / is an £-character of G. We
write X(j) to denote the subfield of £ generated by K and the character values
X(g) for g e G. Note that K(%) is contained in a splitting field for a polynomial

152
Chapter 9
of the form x" — 1 over K. Since this polynomial yields a Galois extension
with an abelian Galois group, it follows that K(x) is a finite degree Galois
extension of K and the Galois group ^(K(x)/K) is abelian.
Assume that £ is a splitting field for G and that F £ E. If /, \\i e Irr£(G), we
say that / and i/f are Galois conjugate over F if F(x) = F(\p) and there exists
a e g(F(x)/F) such that x" = \ji. It is clear that this defines an equivalence
relation on Irr£(G).
(9.17) lemma Let £ be a splitting field for G and let / e Irr£(G). Let Sf be
the equivalence class of / with respect to Galois conjugacy over F where
F £ E. We have
(a) If K £ E, K(x) = K, and a e &(K/K n F), then x" e if.
(b) If F0 £ F and \ji e if, then \ji is Galois conjugate to / over F0.
(c) \y\ = \F(x):F\.
Proof (a) By 9.16, x" e Irr£(G). Now (K r\ F)(x) 2 K r\ F is a normal
extension and so (X n F)(x) £ X is invariant under a. It is thus no loss to
assume that K = (K n F)(/) and so X £ F(/) and no proper subfield of F(x)
contains both F and K. It follows by Galois theory that restiction maps
g(F(x)/F) onto <S{KjK r\ F) and so x" = f for some t e 9(F(x)/F). Now
f(Zt) = fa)andsoz" = zte^.
(b) This is immediate by applying (a) to F0 and taking K = F(x).
(c) We have that if is the orbit of / under &(F(x)/F). By definition of
F(x), the stabilizer of / in this group is trivial and so \!f\ = |^(F(/)/F)|
= \F(X):F\. |
Now let F be any field and let 3E be an irreducible F-representation of G.
Let E 2 F be a splitting field for G. What does 3E£ look like? We shall prove
that it is completely reducible; that all irreducible constituents occur with
equal multiplicity; that the characters of these constituents constitute a
Galois conjugacy class over F and that the common multiplicity is 1 except
possibly when F has characteristic zero.
(9.18) lemma Let F £ E with \E:F\ = n< oo and let V be an irreducible
£[G]-module corresponding to the £-representation 3E. Then V may be
viewed as an F[G]-module and as such let it correspond to the
F-representation 3- Then
(a) deg 3 = n deg 3E.
(b) 3 has a unique (up to similarity) irreducible constituent. It is the
F-representation 9) such that 3E is a constituent of ?)£.
(c) If 3E affords the £-character / and F(x) = F, then 3 affords nx-
Proof We may certainly view V as an F-space. Let v1,v2,---,vm be an
£-basis for V and let eu e2,. ■ ■, e„ be an F-basis for E. It is routine to check

Changing the field
1S3
that {vte}} is an F-basis for V and so V is finite dimensional over F, 3 is
defined, and (a) follows.
Let 9) be a (unique up to similarity) irreducible F-representation of G
such that 3E is a constituent of 9)£. By Theorem 9.6, choose b e F[G~\ c £[G],
such that ?)(£>) = 9)(1) and 9)0(fe) = 0 for every irreducible F-representation
9)0 not similar to 9). Since 3E is a constituent of 9)£ and ?)(£>) is an identity
matrix, then so is X(b) an identity matrix. Thus b acts as an identity on V and
hence 3(^) is an identity matrix and ?),(£>) ,£ 0 for every irreducible
constituent 9)f of 3- Thus every 9), is similar to 9) and (b) is proved.
For (c), let g e G and write v$ = £; Vjoty, where a(j- e £ and x(g) = £, ««■
Now write
v
for 1 ^ /z, v ^ n and /?y/iV e F. Let 3 afford the character \ji. Then
(v.ejg = £ fjev^(J> and i/%) = £ &,„„.
We have
and since we are assuming that £; a„ = /(#) e F, we conclude that
I a« = I &;„„
for each /z, 1 < fi < n. The result now follows. |
(9.19) corollary Let F c E with |£:F| = n < oo. Let 3E be an
irreducible F-representation of G and let 9) be an irreducible F-representation
such that 3E is a constituent of 9)£. Then deg 9) divides n(deg 3E).
Proof Let 3 be an F-representation obtained by viewing an F[G]-
module corresponding to 3E as an F[G]-module. By Corollary 9.7 and Lemma
9.18(b), we conclude that 9) is the unique irreducible constituent of 3 and so
deg 9) divides deg 3- The result now follows from 9.18(a). |
The following corollary is what survives of Theorem 9.14 when the
hypothesis of prime characteristic is dropped.
(9.20) corollary Let 3E be an absolutely irreducible F-representation of
G which affords the character /. Let F £ £ be such that F(#) = F. Then there
exists an irreducible F-representation 9), such that 3E is the unique (up to
similarity) irreducible constituent of 9)£. In particular, 9) affords m/ for some
integer m.

154
Chapter 9
Proof If F has prime charactersitic, then by Theorem 9.14, there exists 9)
such that 9)£ is similar to 3E and there is nothing to prove. Assume then that
char(F) = 0.
It will suffice to show for some positive integer n that the character nx is
afforded by some F-representation 3- This is sufficient since by the linear
independence of IrrL(G) for a splitting field L 2 E, it follows that every
irreducible constituent of 3L affords / and so is similar to 3EL. We may thus
take 9) to be any irreducible constituent of 3 and quote Corollary 9.7 to
complete the proof.
To produce 3> let K 2 F be a splitting field with \K :F\ = n < oo. By
Lemma 9.13, /eIrrK(G). Let 3 be the F-representation which results by
taking a K[G]-module which affords / and viewing it as an F[G]-module.
Then 3 affords nx by Lemma 9.18(c) and the result follows. |
Of course it follows in the above situation that if 9)0 is any irreducible
F-representation such that 3E is a constituent of (9)0)£, then 9)0 is similar to 9)
and hence 3E is the unique irreducible constituent of (9)0)£-
(9.21) theorem Let F s E, where £ is a splitting field for G. Let 9) be an
irreducible F-representation of G. Then
(a) The irreducible constituents of 9)£ all occur with equal
multiplicity m.
(b) Ifchar(F)^ 0, then m = 1.
(c) The characters /, e Irr£(G) afforded by the irreducible constituents
of 9)£ constitute a Galois conjugacy class over F and so the fields F(xi) are all
equal.
(d) Let L = F(x). The irreducible constituents of 9)L occur with
multiplicity 1.
(e) If 3 is any irreducible constituent of 9)L then 3£ has a unique
irreducible constituent. Its multiplicity is m.
(f) 9)L and 9)£ are completely reducible.
Proof Let 3E be an irreducible constituent of 9)£ and suppose 3E affords
X e Irr£(G). Let L = F(x) and let 3 be an irreducible constituent of ?)L such
that X is a constituent of 3£. By Corollary 9.20, 3E is the unique irreducible
constituent of 3£. Let m be its multiplicity, so that 3 affords the character m/.
If char(£) =£ 0, then Theorem 9.14 yields m = 1 and 3 is absolutely
irreducible.
Let x = Xu li > ■ ■ ■ > In be the distinct Galois conjugates of / over F so that
n = | L: F | by Lemma 9.17(c). For a e @(L/F), form the L-representation 3"
defined by 3<r(sf) = 3idf- As a runs over @(L/F) we obtain n representations
3" affording the characters mx„ 1 <i < n. Since m = 1 when char(£) ^ 0,
the mxi are distinct in all cases and thus the 3" are pairwise nonsimilar. Also,

Changing the field
155
each (3")£ has a unique irreducible constituent and it occurs with multiplicity
m. This follows from the characters if char(£) = 0 and it holds in prime
characteristic since then all of the 3" are absolutely irreducible.
We claim that the n = \ L: F | representations 3" are exactly the
irreducible constituents of 9)L and that each occurs with multiplicity 1. Since the
irreducible constituents of (3<r)£ and (3I)£ are nonsimilar if a j= t, statements
(a)-(e) will follow when the claim is established.
Since 3 is a constituent of 9)L and C^L)a = 9)L for a e <&(L/F), it follows
that 3" is a constituent of 9)L for every a. Therefore n(deg 3) ^ deg 9) since
n = \&(L/F)\. By Corollary 9.19, deg 9) divides n(deg 3) and we thus have
equality. Therefore the 3" are the only irreducible constituents of 9)L and
each has multiplicity 1 as claimed.
All that remains is to show that 9)L and 9)£ are completely reducible. This
follows from Maschke's Theorem 1.9ifchar(£) = 0 so we assume char(£) j= 0.
We may assume without loss that 3 is a bottom constituent of 9)L (since some
constituent certainly is). Since (9)L)or = 9)L, it follows that 3" is a bottom
constituent for every ae&(L/F). Now let V be an L[G]-module
corresponding to 9)L and let W be the sum of all of the irreducible submodules of V.
By Theorem 1.10, W is completely reducible and it suffices to show W = V.
Since every irreducible constituent of 9)L is a bottom constituent, every
composition factor of V occurs as a composition factor of W. Since the
composition factors of V all have multiplicity 1, V/W is necessarily trivial. The
proof for ?)£ is similar since 3£ is irreducible in this case. |
We obtain some consequences now. The first generalizes Theorem 9.14.
(9.22) corollary Let F be any field. Then the characters of nonsimilar
irreducible F-representations of G are nonzero, distinct and linearly
independent over F.
Proof Let E 2 F be a splitting field for G. By Theorem 9.21, the
characters of nonsimilar irreducible F-representations of G are nonzero multiples
of sums of disjoint subsets of Irr£(G). The result follows from 9.12 |
(9.23) corollary Let F £ E be fields of prime characteristic. Let 3E be an
irreducible F-representation of G which affords the character /. Let ?) be an
irreducible F-representation such that 3E is a constituent of ?)£. Then deg 9)
= \F(x) :F\ deg 3E. In particular, if F(x) = F, then 3E is similar to ?)£.
Proof Let L 2 E be a splitting field for G and let C e IrrL(G) be the
character of an irreducible constituent of HL. Let & and !T be the Galois
conjugacy classes of £ over E and F respectively. Since E has prime
characteristic, it suffices by Theorem 9.21 to show that \2T\ = \F(x): F\ \y\.

156
Chapter 9
By Lemma 9.17(b), we have ^cf and thus z = £ if takes on values in
F(Q and so F(x) £ F(Q. Since |.T | = \F(Q: F| by 9.17(c), we must show that
1^1 = 1 F(0: F(Z) |. Let <S = 0(F(O/F(z))- If * e 0, then
I if = Z = f = I ff
tieSf tisSf
and it follows from the linear independence of IrrL(G) and Lemma 9.16 that
0 permutes if. Since only the identity of 0 can fix £, this yields \if\ > \<S\ =
| F(0: F(z) |. However, F(x) ££n F(Q and thus
\if\ = |£(Q:£| = |F(Q:£nF(0| < |F(0:F(z)l
and the proof is complete. |
Problems
(9.1) Let fc£be fields and let Fbea finite dimensional £-space. Let
W £ V be an F-subspace. We write V = W*F E provided V = WE and
dim£(F) = dimF(W).
(a) If V = W *F E and U £ W is an F-subspace, show that UE
= U*FE.
(b) If 3E is an F-representation of G, show that dimF(3E(F[G])) =
dim£(X£(£[G])).
Note The symbol *F denotes an internal tensor product.
(9.2) Let 3E be an F-representation of G and let E 2 F. Let V be an £[G]-
module corresponding to 3E£.
(a) Show that V = W *F E for some G-invariant, F-subspace W £ V
such that the F[G]-module W corresponds to X.
(b) Let U £ W be an F[G]-submodule. Let U and W/I/ correspond to
the F-representations ^) and 3> respectively. Show that l/F, is an F[G]-
submodule of V which corresponds to ^)£ and that V/UE corresponds to 3£-
(9.3) Let W be an F[G]-module and let F £ E with |£ : F\ = n < oo. Let
W£ be an £[G]-module corresponding to X£ where 3E is an F-representation
corresponding to W. Now view WE as an F-space. Show that the resulting
F[G]-module is isomorphic to the direct sum of n copies of W.
Note It follows from the Krull-Schmidt theorem that if V and W are
F[G]-modules and V®V@---@V=*W®W®---®W, where each
direct sum has n terms, then V = W. We conclude via Problem 9.3 that if 3E
and ^) are F-representations of G and 3E£ is similar to ^)£, then 3E is similar to
% provided \E:F\ < oo.

Problems
1S7
(9.4) Let F £ E be a field extension of possibly infinite degree. Let 3E and ^
be F-representations of G and assume 3E£ and tyE are similar.
(a) Show that there exist matrices Ml,M2,...,Mk over F and elements
eu e2, ■ ■ ■, ek e E such that X(g)Mt = M^g) for all g e G and 1 < i < k and
such that eiMt + • • • + ekMk is nonsingular over E.
(b) If | F | > deg 3E, show that 3E is similar to $).
Hint [For (b)] Letf(xu x2, ..., xk) be a nonzero polynomial over F
and assume that the degree off in x, is < \F\ for each i. Then there exists
aua2,..., ak eF such that/(a,, a2,..., ak) =£ 0.
(9.5) Under the hypotheses of Problem 9.4, show that 3E and ^ are similar
without assuming anything about | F | or | E: F |.
Hint Combine the results of Problem 9.4(b) and the note following
Problem 9.3.
(9.6) Let F £ E and let V be an irreducible £[G]-module which affords the
character /. Assume that E = F(%). Show that V is irreducible when viewed
as an F[G]-module.
(9.7) Let F have prime characteristic and let 3E be an irreducible F-repre-
sentation of G. Let D be the centralizer of 3E(G) in the matrix algebra M„(F)
where n = deg 3E. Show that D is a field.
Hint Let E 2 F be a splitting field and consider the centralizer of 3E£
in Mn(E). Use Theorems 9.21 and 9.2.
(9.8) In the situation of Problem 9.7, let / be the character of an irreducible
constituent of XE where E 2 F is a splitting field for G. Show that the natural
isomorphism between F and F • 1, the scalar matrices in M„(F), extends to an
isomorphism F(%) = D.
Hint Let L = F(%) and let V be an L[G]-module affording /. View V as
an F[G]-module. Then D =* EF[G](V). Use Problem 9.6.
(9.9) Let 3E be an F-representation of G and assume 3E£ is completely
reducible for some E 2 F. Show that 3E is completely reducible.
Hint Consider 3E(J(F[G])). Use Problem 1.10.
(9.10) Let 3E be a completely reducible F-representation of G and let E 2 F.
Show that 3E£ is completely reducible.
Note The analogous statement for algebras other than group algebras
does not hold, in general.
(9.11) Let F££ and view F[_G^\ <= £[G]. Show that J(E[GJ) =
J(F[G]) *F E (in the notation of Problem 9.1).

158
Chapter 9
Hint To obtain J(E[G]) £ (J(F[GJ)E, use Problems 9.2(b), 9.10, and
1.10 to argue that £[G]°/(J(F[G]))£ is a completely reducible £[G]-module.
(9.12) If 3E is an F-representation of G let s(3E) denote dimF(3E(F[G])). Let
^Du ..., H)t be the distinct irreducible constituents of 3E. Show that
I^,)<S(X)
with equality if 3E is completely reducible.
(9.13) Let 3E be an irreducible F-representation of G. Let ^) be an
irreducible constituent of 3E£, where E 2 F is a splitting field. Show that ms(3E) =
deg(3E) degC?)), where s(3E) is as in Problem 9.12 and m is the multiplicity of ^)
as a constituent of 3E£.
(9.14) Let 3E be an irreducible F-representation of G and let D be the cen-
tralizer of 3E(F[G]) in the matrix ring M„(F), where n = deg(3E). Let d =
dimF(D). Show that s(3E) = n2/d.
Hint Let V be an F[G]-module corresponding to 3E so that V becomes a
vector space over the division ring D. Express s(3E) = dimF(3E(F[G])) in terms
of dim^K) and d. Use Theorem 1.16.
Note Applying Problem 9.14 in the situation of Problem 9.13 we obtain
mn2/d = ms(3E) = n degC^)and mn = d degC?)). Since n = wdegC?))|F(x): F|,
where / e Irr£(G) is afforded by ^), we obtain-rf = m2 \F(%) :F\. Compare this
with Problem 9.8. This formula can be used to generalize Problem 9.8 to the
characteristic zero case.
(9.15) Let 3E be an irreducible F-representation of G and let / be the
character afforded by an irreducible constituent ^) of 3E£, where E 2 F is a
splitting field for G. Let D be the centralizer of 3E(F[G]) as in Problem 9.14.
Show that the isomorphism FsF'lcfl extends to an isomorphism
F(X) = Z(D).
Hints Let L = F(%) and let 3 be the irreducible constituent of 3EL such
that ^) is a constituent of 3£. Let V be an L[G]-module corresponding to 3-
Let D0 = EL[G](V) and £>! = EF[G](V). Check that D0^DV^D. Use the
note following Problem 9.14 to show that D0 = Dt.
Let Z = Z(D0) 2 L • 1. Observe that D0 = EZ[G](V). Let 2B be the Z-
representation corresponding to V viewed as a Z[G]-module, and compute
the multiplicity of ^) as a constituent of 2B£. Now use the note following
Problem 9.14 again to show Z = L • 1.
(9.16) Let 3E be an irreducible F-representation of G and let E 2 F. Show
that g e G lies in ker(3E) iff g e kerC^) for every irreducible constituent ^) of 3E£.

Problems
159
(9.17) Let F have prime characteristic p. Show that Op(G) (the largest
normal p-subgroup of G) is the intersection of the kernels of the irreducible
F-representations of G.
(9.18) Let k = char(F).
(a) If k t6 2, show that Qe has a unique (up to similarity) nonlinear
irreducible F-representation.
(b) If k ¥= 0, show that F is a splitting field for Q8.
(c) If k = 0, show that F is a splitting field for Q8 iff — 1 is a sum of two
squares in F.
(9.19) Let F c E, where £ is a splitting field for G. Does there necessarily
exist a splitting field K such that F £ X £ E and |X: F| < oo? Consider
zero and nonzero characteristic separately.
Hint Consider F = Q and E = Q(a, /?) £ C, where a, /? are suitable
transcendentals. Take G = QB.
(9.20) Let G be cyclic of order n and let F have characteristic not dividing n.
Let £ = F(e), where e is a primitive nth root of unity. Let m = | E: F \. Show
that every faithful irreducible F-representation of G has degree m and that
there are exactly q>{n)/m similarity classes of such representations.
Note If \F\ = q < oo, then m can be characterized as the least positive
integer such that n \(qm — 1).
(9.21) Let G = GL(n, pe) be the full group of nonsingular n x n matrices
over GF(pe). Show that G has an irreducible representation of degree ne over
GF(p).
(9.22) Let F £ £ be fields of prime characteristic and let 3E be an
^representation of G. Assume that 3E affords a character / such that F(%) = F.
Do not assume 3E is irreducible.
(a) Show that / is afforded by some F-representation.
(b) Find an example where 3E is not similar to ^)£ for any
F-representation ^.

The Schur index
The main question considered in this chapter is the following. If x e Irr(G),
then for which fields F c c is % afforded by an F-representation? If F c c is
not one of these fields, we wish to measure the extent to which x fails to be
afforded over F. This and the results of Chapter 9 (and, in particular, Theorem
9.21) suggest the following definition.
(10.1) definition Let F £ E, where E is any splitting field for G. Let
X e Irr£(G). Choose an irreducible ^-representation 3E which affords / and an
irreducible F-representation ^ such that 3E is a constituent of tyE. Then the
multiplicity of 3£ as a constituent of tyE is the Schur index of / over F. It is
denoted by mF(/).
Note that given / as above, representations 3E and ^) do exist and are
unique up to similarity and so mF(x) is well defined. Actually, mF(x) does not
really depend on E. If L 2 F(x) is another splitting field, then / e IrrL(G) by
Lemma 9.13. A routine argument shows that mF(x) is the same when
computed in E or in L.
By Theorem 9.21(b), Schur indices are always trivial (that is, equal to 1)
in prime characteristic. For this reason we now restrict our attention to the
characteristic zero case. In fact, it is really no loss to confine ourselves to
subfields of the complex numbers, C.
Many important results about Schur indices appear to depend on deep
facts about division algebras and number theory. Nevertheless, as this
chapter will demonstrate, much can be done by elementary means. In
particular, some of the results of Chapter 8 will prove invaluable.
160

The Schur index
161
We collect some facts.
(10.2) corollary Let x e Irr(G) and F £ C. We have
(a) mF(x) = mF(x)(x).
(b) Let y be the Galois conjugacy class of / over F. Then mF(x)(£ if)
is the character of an irreducible F-representation of G.
(c) If S is the character of any F-representation, then mF(x) divides
[S.Z].
(d) mF(x) is the smallest integer m such that mx is afforded by an F(x)-
representation.
(e) mF(x) is the unique integer m such that mx is afforded by an
irreducible F(/)-representation.
(f) If F £ E £ C, then m£(/) divides mF(/).
(g) If F £ £ £ C and | £: F\ = n < oo, then mF(x) divides nm^x)-
(h) mF(x) divides/(I).
Proof Part (a) follows from Theorem 9.21(e) and part (b) is a
consequence of 9.2 l(a, c). Let ^) be the F-representation in (b). Then ^ is the unique
(up to similarity) irreducible F-representation whose character \ji satisfies
[>> X] ^ 0- Since mF(/) = \_\ji, /], part (c) follows.
To prove (d) and (e), we may assume (by (a)) that F = F(x\ Now y = {/}
and the representation ^ of the preceding paragraph affords m/. Parts (d)
and (e) are now immediate.
Now let F £ E £ C. Any character afforded by an F-representation is
also afforded by an F-representation. Now (f) follows from (b) and (c).
Assume \E:F\ = n < oo. Then n0 = \E(x):F(x)\ = \E: E n F(x)\ which
divides n. Since m£(/)x is afforded by an £(x)-representation, we conclude
from Lemma 9.18(c) that n0 m£(/)x is afforded by an F(/)-representation. Now
(c) and (a) yield (g).
Finally, let p be the character of the regular F-representation of G. Then
[p> Xl = Z(l) and (h) follows from (c). The proof is complete. |
A useful method for obtaining information about Schur indices is to use
induced representations. Let F £ C and H £ G. If 9 is a character of H
which is afforded by an F-representation of H, then 9G is afforded by an
F-representation of G. (See Theorems 5.8 and 5.9.) This idea underlies much
of the remainder of the chapter. We use it to prove the following celebrated
result.
(10.3) theorem (Brauer) Let G have exponent n and let F = Q(e2*iln).
Then F is a splitting field for G and every / e Irr(G) is afforded by an F-
representation.

162
Chapter 10
Proof The two conclusions are equivalent by Corollary 9.11. Let
/elrr(G). Since F(%) = F, it suffices by 10.2(d) to show that mF(x) = 1. By
Brauer's Theorem 8.4 we may write
z = ZMG,
where 1 runs over linear characters of subgroups of G and ax e Z. Now a
linear k e Irr(H) is obviously afforded by an F-representation of H and hence
kG is afforded by an F-representation of G. By 10.2(c), we have mF(/)| [1G, /]
and since
i = Dr. z] = 2>i[*G, z],
X
it follows that mF(x) = 1 and the proof is complete. |
We wish to exploit the ideas in the above proof in order to obtain some
more delicate information about the Schur index.
(10.4) lemma Let H s G and F s C. Suppose \}i e Irr(H) and / e Irr(G).
Then mF(x) divides mF(t/0|F(x, t/0: F(x)\ \jpa, *].
Proof We may replace F by E = F(%). This is so since m£(/) = mF(x)
by 10.2(a) and m£(t/^) | mF(i^) by 10.2(f) (and of course F(%, ijj) = E(\li)). Thus
we assume F = F(x).
Let & be the Galois conjugacy class of \ji over F so that mF(i/f) £ ^ is
afforded by an F-representation of G. Therefore mF(%) divides
by 10.2(c).
Now if ne£f, then \ji = n" for some er e ^(F(t/^)/F). We have X = X" and
[fG,z] = [(fGr.zn = [^G.z]
and thus mF(/) divides mF(i/>) | ^ | [\pa, /]. Since | V \ = \ F(\p): F |, the proof is
complete. |
The next theorem is a variation on a result of Brauer and Witt. It allows
us to analyze mF(x) one prime at a time in terms of certain sections of a group
with sharply defined properties. (A section of G is a factor group, H/K,
where K -a H £ G.)
(10.5) definition Let F £ C. Then (H, X, 9) is an F-triple provided
(a) H is a group, X<H,X = CH(X);
(b) 9elrr(H) is faithful;
(c) the irreducible (linear) constituents of Sx are Galois conjugate
over F(9).

The Schur index
163
(10.6) corollary Let (H, X, 9) be an F-triple and let 1 be a linear
constituent of &x- Then
(a) lis faithful and X is cyclic;
(b) I„(k) = X, k" = 9 and F(S) £ F(l);
(c) Sx is afforded by an irreducible F(9)-representation;
(d) H/X s 0(F(1)/F(9)).
Proof By part (c) of the definition, all linear constituents of $x have the
same kernel. Thus ker k c ker 9 = 1 and (a) follows. Since k takes on distinct
values at distinct elements of X, it follows that if kh = k, then h e C(X) = X.
Thus X = Ij^k) and kG is irreducible by Theorem 6.11(a). Now (b) follows.
Also Sx is the sum of 9(1) = \H:X\ distinct linear characters. These must
constitute the full Galois conjugacy class of k over F(9) since if a e ^(F(l)/F(9)),
then \_k", 9*] = \_k, 9*] ^ 0. Since mm)(k) = 1, Corollary 10.2(b) asserts that
9jf is afforded by an irreducible F(9)-representation and (c) is proved.
Finally, for each heH, there exists a unique ah e ^(F(k)/F(9)) such that
I""' = X"\ Now l""'k"' = (k'-f1 = (kk")ah = kaka\ Therefore, akh = akah
and we have a homomorphism, a: H -> ^(F(l)/F(9)), defined by a(h) = ah.
Now ker a = IH{k) = X. Since it was shown above that the H-conjugates of
k constitute a full Galois conjugacy class over F(9), it follows that a maps onto
<g(F(k)/F(S)) and the proof is complete. |
(10.7) theorem Let /elrr(G) and F £ C. Assume p" divides mF(x) for
some prime p. Then there exists an F-triple (H, X, 9) such that
(a) H is a section of G;
(b) pa|mF(9);
(c) H/X is a p-group;
(d) pJ|F(Z,8):F(z)|.
Proo/" By Lemma 9.17(b), it follows that if (H, X, 9) is an F-triple for
some E 2 F, then it is an F-triple. By 10.2(f, a) we may therefore replace
F by F(x) and so we assume F = F(x).
Use induction on \G\. If there exists K < G and t/f eIrr(X) such that
pJf bl>a, X] and pJf\F(\l/):F\, then by Lemma 10.4 it follows that p" \ mF(i/0 and
we may apply the inductive hypothesis to K with respect to ip and obtain
an F-triple (H, X, 9) which satisfies (a), (b), and (c) and such that
pjf\ F(\p, 9): F(t/01. Since also pJf\F(\ji): F |, we conclude that pJf\F(9):F\ and
the proof is complete in this case. We therefore assume that no such pair
(K, \ji) exists.
By Solomon's Theorem 8.7 we can write 1G = Z ao.(^o)G' where aQeZ
and Q runs over ^f, the set of quasi-elementary subgroups of G. It follows that
1 = ZlG = Z aQx(lQf = X aQ(xQf

164
Chapter 10
and thus
1 = [*,*] = Z «q[(Xq)g. *]■
Choose QeJP such that p does not divide \_(Xq)g, x\ = [Jq, Xq]- If "A is any
irreducible constituent of xQ, then since F(x) = F, every element of the
Galois conjugacy class of \ji over F has equal multiplicity as a constituent of
XQ and we may write Xq = X ^aA. where 6AeZ and A runs over sums of
Galois conjugacy classes over F in Irr(Q). Thus p does not divide £ 2>a2[A, A]
and we choose A such that pJfbA2\_A, A]. Write A = £ !?, where y is a
Galois conjugacy class over F and let i/^ey. Then ftA = [/Q, i/f] = [/, i/fG]
is not divisible by p and [A, A] = |!? \ = \ F(\ji): F | is not divisible by p. By
the result of the second paragraph of the proof, we conclude that G = Q is
quasi-elementary.
Thus G has a cyclic normal ^-complement C for some prime q. By
Theorem 6.15 we have x(l) is a power of q. Now if a = 0, the result is trivial,
taking H = X = 1, and so we assume a > 0 and p|mF(x). Since mF(x)|/(l)
by 10.2(h), it follows that q = p and G/C is a p-group. Let X o G, with
X 2 C be chosen maximal such that X is abelian. By the inductive hypothesis
applied to G/ker /, we may assume that / is faithful. We shall show that
(G, X, x) is an F-triple to complete the proof.
We have X £ CG(X) -a G. If X < CG(X), then since G/X is a p-group,
there exists U^G with X £ U £ CG(X) and | U : X \ = p. Thus U is abelian
and this contradicts the choice of X. Thus part (a) of Definition 10.5 is
established.
Now let X be a linear constituent of Xx and let S = {g e G | X9 is Galois
conjugate to X over F}. If st, s2 e S, there exist e^, er2 e ^(F(l)/F) with Is' =
1"'. It follows that Is'52 = X"2"' and thus sis2 e S and S is a subgroup of G.
Let T = 7G(1) £ S. By Theorem 6.11(b), there exists a unique >/ e Irr(T) such
that rja = x and [r\x, 1] ^ 0. Let \p = rjs so that t/^G = / and \p e Irr(S). We
claim that F(\p) = F.
Let ae^(F(iii, X)/F). Then /" = x and so [1", Xx] = E^, X*] * 0. Thus
X" = X? for some geG and since X" is Galois conjugate to X over F, we have
g g S. Since Galois conjugate characters have the same inertia groups, we
have T~a S and so fa")9"' e Irr(T). Now
((>7T",)G = (>7'r)G = X'r = X
and
[iwr'h, i] = [(*")*, ^] = Kriix, n * o.
By the uniqueness of r\ we conclude that (rja)B"' = >/ and so
•A" = (ri")S = {r\B? = ris = ^

The Schur index
165
since g e S. Since F(i/f, X) is a Galois extension of F, it follows that \ji takes
values in F and thus F(i/f) = F as claimed. By the result of the second
paragraph of the proof applied to (S, \ji), it follows that S = G. Thus condition (c)
of Definition 10.5 is satisfied and the proof is complete. |
Before proceeding to our major applications of Theorem 10.7 we derive
two easier consequences which demonstrate its power. These are included
in Theorem 10.9.
(10.8) lemma Let H c g and suppose H has a complement in G. Let
(pelrr(H) and suppose q>G = /elrr(G). Then mF(x) divides <p(l) for every
F£ C.
Proof Let U c G with UH = G and U n H = 1. Since (1[,)G is afforded
by an F-representation, we have mF(x) | [(1 vf, /]. Now [(1 vf, /] = [(1 vf, <pG]
= [((1i/)g)h> <?]■ However, ((1[,)g)h = (1i/m«)h = P«, the regular character
of H. Thus [(1[;)G, xl = lPh> <p] = <P(1) and the result follows. |
(10.9) theorem Let /elrr(G) and assume p|mF(x) for some F £ C and
prime p. Then the Sylow p-subgroups of G are not elementary abelian and
prrifix) divides \G\.
Proof Let p" be the p-part of mF(x) so that a > 0. Since mF(x)|/(l), the
second statement will follow if pfl+1 divides | G |. Let (H, X, 9) be the F-triple
whose existence is guaranteed by Theorem 10.7. Thus 9 = 1H for some linear
lelrr(X) by Corollary 10.6(b). Since mF(9) > 1, we conclude from Lemma
10.8 that X is not complemented in H.
Since H/X is a p-group, it follows that a Sylow p-subgroup of H is not
elementary abelian. Also, if PeSylp(H), then P n X ^ 1 and so 9(1) =
|H:X| < |P|. Since pfl<mF(9) < 9(1), we have f < |P| and so pfl+1||H|.
The result now follows. |
There is an important case when we can decide whether or not mpiS) = 1
for an F-triple (H, X, 9). If H/X is cyclic, the problem "reduces" to a question
in field theory. If E 2 F is a Galois field extension and a e E, we define
^£/F(a) = riffeiW) a"- ^ne image of this norm map is clearly contained in F.
(10.10) theorem Let (H, X, 9) be an F-triple for some F e c and assume
H = XC, where C is a cyclic group. Let 1 be a linear constituent of 9* and
write E = F(l) and K = F(S). Let m = \ X n C | and let e be a primitive mth
root of unity in C. Then e e K. Also, mf{9) = 1 iff e lies in the image of NEjK.
Proof Since X r\ C ^ Z(H) and 9 e Irr(H) is faithful, we can choose a
generator y of X n C, such that 9(y) = e9(l) and thus eeF(9) = K. Write
| X: X n C | = s and \H: X\ = \C: X r\ C\ = t and choose generators x and
c for X and C such that xs = y = c\ Note that X(xf = X(y) = e. By Corollary

166
Chapter 10
10.6 we have
t = 9(l) = \9(E/K)\ = \E:K\.
Write xc = xr for some integer r.
By Definition 10.5 there exists a e <g(E/K) such that Xc~' = X". It follows
that the orbits of X under C and under <er> are identical. Since C is transitive
on the t distinct linear constituents of Sx, we conclude that |<<x>| > t =
\9(E/K)\ and thus <g(E/K) = <<x>. Therefore,
NEIK(z)= art"2-/"1
for ae E. Also, X(x)r = X(xr) = Xc~\x) = X(x)°.
Let V be the (unique up to isomorphism) irreducible K[X]-module
which affords Sx- (See Corollary 10.6(c).) Now mF(9) = 1 iff 9 is afforded by
a X[H]-module (by Corollary 10.2(d)) and this happens iff there exists a
X-linear action of C on V such that for v e V we have
(a) v ■ c' = ve;
(b) ((vc~l)-x)-c = vxr.
(The sufficiency of this condition involves an easy generators and relations
argument on H.)
View E as an £[X] module affording X so that /? ■ u = fiX(u) for u e X.
With this same action of X on E, we now view £ as a X[X]-module of
dimension | E : K | = t. By Lemma 9.18(b) it follows that V is an irreducible
constituent of the X[X]-module E. (This is because X is a constituent of tyE
where ^ is a K-representation corresponding to V.) Since dimK(F) = 9(1)
= t = dimK(E), it follows that V = E as X[X] modules. We conclude that
mF(9) = 1 iff there exists a X-linear transformation c of £ such that for
all 0 e £
(a') B •£' = /?£•
(b') ((B-c-l)X(x))-c = B(X(x)f.
Now suppose that mF(9) = 1 and let c be as above. Write d = X(x),
a. = 1 ■ c and let n > 0 be an integer. We show that ((5") ■ c = ad"' by induction
on n. This is clear for n = 0. For n > 0, the inductive hypothesis and (b')
yield
(dn) • c = (((a(5(n-1)r) ■ £-1)8) ■ c = a(5(n- i)rdr = tx8"r
as claimed. Since d" = 5" we have (dn) ■ c = a((5")or. Since E = K(X) and d =
X(x), the d" span E over K. The map /? -» olB" for Be E is X-linear and agrees
with c on the 8". We conclude that B-c = a.B" for all 0 e £. Now (a') yields
e = 1 • (c)' = aa'a"2 ■ ■ ■ a"'"' = JV£/K(a)
as desired.

The Schur index
167
Conversely, suppose NEjK(a) = e for some a e E. Define c on E by /? ■ c
= a/?" and observe that c is X-linear. Check that (a') and (b') are satisfied.
Thus mF(9) = 1 and the proof is complete. |
(10.11) lemma Let p be a prime and let 1 ^hZ. Assume that p\(k — 1)
and if p = 2, assume 4|(/c — 1). Write
f(e) = (kT - l)/(/c - 1)
for 0 < e e Z. Then pe is the exact power of p dividing/(e).
Proof Since/(0) = 1, we assume e > 1 and use induction on e. Since
k""' = 1 + (k - \)f(e - 1), we have
kT=l+ p(k - \)f(e - 1)
+ ^-^ (k - l)2f(e ~ l)m + (k- Iffie - l)n
for suitable integers m, n. Thus
f(e) = f(e - 1)
p+P(pl)(/c_ 1)m + (/c_ 1}:
-]
and it suffices to show that the second and third terms in the brackets are
divisible by p2. Since p\(k — 1), this is clear if p j= 2. If p = 2, the hypothesis
that 4\(k - 1) yields the result. |
We next give our principal result about the Schur index. Note that it
generalizes Brauer's Theorem 10.3.
(10.12) theorem (Goldschmidt-Isaacs) Let / e Irr(G), where G has
exponent n and let e be a primitive nth root of unity in C. Let F ^ C and assume
that ^(F(e)/F(x)) has a cyclic Sylow p-subgroup P. Then pjfmpix) except
possibly when p = 2, P > 1 and ^/--l $ F(%).
Proof We may replace F by F(%) and assume F(%) = F. Suppose
p | mF(x) and let (H, X, 9) be the F-triple whose existence is given by Theorem
10.7 so that p|mF(9). Let 1 be a linear constituent of Sx so that F(%) =
F £ F(9) c F(l) <= F(e). By 10.6(d), H/X S 0(F(1)/F(9)) which is a section
of 0(F(e)/F(x)). Since H/X is a p-group, it must be cyclic. Also, since 9(1) > 1,
we have H > X and we conclude that P > 1.
There exists a cyclic p-group C with XC = H. Let |C| = pc and \C r\ X\
= p" and write E = F(l) and X = F(3). Since mF(9) # 1, Theorem 10.10
yields that no primitive pfcth root of unity lies in the image of NE/K, but that
K does contain such a root. Since NEjK(l) = 1, we have b > 0 and K contains
a primitive pth root of unity.

168
Chapter 10
Let y be a primitive pcth root of unity in C. Then | K(y): K | divides pc b
and K(y) £ F(e). Also | E: K | = | H: X \ = pc~b and E £ F(e). Since 0(F(e)/K)
is abelian and has a cyclic Sylow p-subgroup, its subgroups of p-power
index are linearly ordered by inclusion. The fundamental theorem of Galois
theory now yields that K(y) £ E (since \K(y) :K\ < \ E : K |) and so y e E.
We compute NE/K(y).
Write t = pc~b. If y e X, then NE/K(y) = y\ which is a primitive pfcth root
of unity, a contradiction. Thus y £ K. Write ^(E/K) = <er>, where o(<r) = t
and y" = yk for some integer k ^ 1. Then y"' = yk' and
W£/K(}') = }7V2---}''lt"1 = y",
where q = (k' — l)/(/c — 1). If pc_fc is the exact p-power divisor of q, then yq
is a primitive pfcth root of unity which is not the case. By Lemma 10.11 then,
pjf(k- l)ifp^2and4^(fc- 1) if p = 2.
Let 8 e K be a primitive p"th root of unity with a chosen as large as
possible. Then 0 < b < a. Since y $ K, we have a < c and 8 e <y> so that 8 =
8" = 8k and p"\(k — 1). Since a > 0, we have p = 2 and a = 1 so that
■v/—1 £ X. The proof is complete. |
(10.13) corollary (Fong) Let G have exponent n = mp", where p is
prime and pjfm. Suppose F £ C and F contains a primitive mth root of unity.
Let i e Irr(G). Then mF(x) = 1 unless p = 2 and v*^^ £ -F> m which case
mF(x) < 2.
Proof Ifp = 2andy^T^F, let £ = F[V^] so that \E:F\ = 2 and
rrifix) | (2mE(x)) by Corollary 10.2(g). Thus it suffices to assume that sf--l e F
if p = 2 and to prove mF(x) = 1.
Let e be a primitive nth root of unity in C. The hypotheses on F now
guarantee that ^(F(e)/F) is cyclic. If q is a prime divisor of mF(/), then Theorem
10.12 yields q = 2 and ,/-^l $ F. Thus p ¥= 2,4Jfm, and mF(/) is a power of 2.
We have 4Jfn and thus a Sylow 2-subgroup of G is elementary abelian and
2J(mF{x) by Theorem 10.9. The result follows. |
(10.14) corollary (Roquette) Let G be a p-group and /elrr(G).
Let F £ C. Then mF(x) = 1 unless p = 2 and s/-\ $F in which case
mF(x) < 2.
Proof Immediate from 10.13. |
Note that taking G to be the quaternion group of order 8 shows that the
exceptional cases in the three preceding results can, in fact, occur.
(10.15) corollary (L. Solomon) Let k be the product of the distinct
prime divisors of \G\ and let F s C. Assume that F contains a primitive
(2k)th root of unity. Then mF(/) = 1 for all / e Irr(G).

The Schur index
169
Proof Note that if 211G |, then ^f — 1 e F. Let e be a primitive nth root of
unity, where n is the exponent of G. Since every prime divisor of n divides k,
the hypothesis on F guarantees that ^(F(e)/F) is cyclic. If / e Irr(G) and p is a
prime divisor of mF(x), then Theorem 10.12 yields p = 2 and y/--l $ F. Since
mF(x) | x(1) we conclude that 211G | and we have a contradiction. |
Most of the above results are directed to showing that Schur indices are
small. We have not yet seen an example to show that mF(x) can be greater than
2. In fact every positive integer can occur as a Schur index. Here, we shall
settle for an indication of how to prove that every prime can occur. We need
two facts from number theory
(a) Let F = Q(e), where e is a root of unity. Let R = Z[e] and let / be
a proper ideal of the ring R. Let a e F. Then a = u/v for some u, v e R with
not both u,vel.
(b) (Dirichlet) Let a,beZ with (a, b) = 1. Then there exists a prime of
the form ak + b for some k e Z.
Let p be a prime. By (b) above, choose a prime q = p2k + (p + 1) for
some k. Then p\(q — 1) but p2Jf(q — 1). We now construct the semidirect
product G = QP, where Q is cyclic of order q and P is cyclic of order p2 and
acts nontrivially (but not faithfully) on Q. There exists faithful / e Irr(G) with
X(l) = p. We claim that mQ(x) = p.
(10.16) theorem Let G = XP, where X -o G is cyclic of order pq,P *&G
is cyclic of order p2 and |G| = p2<? for primes p and <j such that p2Jf(q — 1).
Then there exists faithful / e Irr(G) such that mQ(x) = p.
Proof Note that X = CG(X). Let 1 be a faithful linear character of X
and let / = 1G. Then /e Irr(G), /(l) = p, and (G, X, /) is a Q-triple. Since
™Q(x)\p by 10.2(h), it suffices to show that mQ(x) ¥= 1. Let /C = Q(/), £ =
Q(l), and co e K be a primitive pth root of unity. By Theorem 10.10, it suffices
to show that co is not in the image of NEjK.
Let v be a primitive qth root of unity in E and write e = cov so that e is a
primitive (p<j)th root and £ = Q(e). Let R = Z[e] and let I 3 qRbea.
maximal ideal of R. Since /?// is a field of characteristic q, the only qth root of
unity in R/I is 1. Thus if a e ^(E/K), then v = 1 = v" mod / and since coeK,
we have e" = cov" = cov = e mod /. Since R = Z[e], it follows that r" =
r mod / for all reR. Since | E : X | = p, we conclude that NEjK(r) = rp mod /
for all re/?.
Suppose, by way of contradiction, that co = NEjK(a) for some a e E. By
Fact (a), write a = ft/c, where b,ceR but not both b,cel. We conclude that

170
Chapter 10
bp = NEIK(b) = NE/K{c)NEIK{a) = cvm mod /. Since not both b, eel, we
conclude that co* = (b*/c*f, where *: R -» R/I is the natural homomorphism.
Now 1 + x + x2 + ■ ■ ■ + xp~l = l\?=i (x - ^ and setting x = 1, we
see that 1 — co divides p in R. Since p* j= 0 in R/I, we conclude that co* ^ 1.
Thus b*/c* is a primitive p2 root of unity in R/I.
Since p|(<? — 1), there exists a e Z such that a* is a primitive pth root of
unity in R/I. Since (e*)p = 1, we have e* = (a*f for some k and hence
eeZ + I. Thus /? = Z[e] = Z + / and /?// S Z/(Z n I) = Z/qZ. Since
P2^(<? — 1)» 2/qZ does not contain a primitive p2 root of unity and this
contradiction completes the proof. |
Suppose F c E c c, with | E: F\ = n < oo and let / e Irr(G) with F(%)
= F. By 10.2(g), rrifix) divides nm^x). In particular, if / is afforded by an E-
representation, then m^x) = 1 and so mF(x) divides \E:F\. This suggests the
question of finding minimal fields £2F over which / is afforded.
(10.17) theorem Let x e hr(G) and F £ C with F(x) = F. Let m = mF(x).
Then there exists E 2 F such that / is afforded by an ^-representation and
\E:F\ = m.
Proof Let V be an irreducible F[G]-module which affords mx (by
Corollary 10.2(b)). Let D = EF[G](V), the centralizer ring of V, so that D is
a division ring by Schur's lemma. Let E be any maximal subfield of D. Then
E acts on F and V may be viewed as an £-space. Since E commutes with the
action of G, we may view V as an £[G]-module. Now EE[G](V) is the
centralizer of E in D. The maximality of E thus yields EE[G](V) = E and thus the
£[G]-module V corresponds to an absolutely irreducible ^-representation X
by Theorem 9.2(a, c). Let 3 be the F-representation corresponding to the
F[G]-module V. By Lemma 9.18, X is a constituent of 3£- Since 3 affords
mx and 3E is absolutely irreducible, we conclude that X affords /. Finally, by
9.18(a) we conclude that m = \E:F\ and the proof is complete. |
We can, of course, find an F-isomorphism of E into C in the above
situation, and so the result remains true if we add the condition that E ^ C.
However, it is not always true that if F c L £ C, where L is a splitting field
for G and / e Irr(G) with F(x) = F, then there exists a field E with F £ E £ L,
| E: F\ = mF(x) and such that / is afforded by an £-representation.
We close this chapter by stating a few more facts about the Schur index.
The proofs of some of these seem to require a fairly deep knowledge of number
theory and division algebras.
(a) (Fein) If /e Irr(G), then mu(x) divides n\_x", 1g] for every positive
integer n.

Problems
171
(b) (Brauer-Speiser) (Corollary of (a)) If / e Irr(G) is real valued, then
mQ(x) < 2.
(c) (Fein-Yamada) If /e Irr(G), then mu(x) divides the exponent of G
and mQ(x)2 divides \G\.
Problems
(10.1) Let H £ G, 9 e Irr(H) and / e Irr(G). Suppose F £ C.
(a) If Xh = $> show that mF(x)|w,(9) and mF(9) < |G: H|mF(x).
(b) If 9G = x, show that mF(S)\mF(x) and mF(x) < \G:H\mF(S).
(10.2) Let N o G and <p e Irr(JV). Let F £ C and put H = IG((p). Let 9 be an
irreducible constituent of <pH and let / = 9G. Show that mF(x) divides
|NG(H):H|mF(9).
Hint Consider {g e G | cpg and <p are Galois conjugate over F(/)}.
(10.3) Let /elrr(G) and F £ C. Define lF(x) to be the greatest common
divisor of the integers nx = \ F(x, ^) ■' F(x) | [iG, xl as X runs over linear
characters of subgroups of G. Show that mF(/) | /F(/) but that examples exist
where mF(/) < /F(/).
Hint For the example, take G = Q8.
(10.4) Let /e Irr(G) and F £ C. We say that / is F-semiprimitive if there
does not exist H < G and \ji e Irr(H) such that iiG = x and F(\j/) = F(x).
(a) If x is F-semiprimitive and N <i G, show that the irreducible
constituents of Xn are Galois conjugate over F(x).
(b) Let G be nilpotent and assume s/~—l e F if | G | is even. Show that
/e Irr(G) is F-semiprimitive iff/(l) = 1.
(c) In the situation of (b) and the notation of Problem 10.3, show that
1F(X)= lforallxGlrr(G).
Hint A noncyclic p-group in which every normal abelian subgroup is
cyclic is necessarily a 2-group and contains a cyclic maximal subgroup.
Note Since mF(x) \ lF(x), this problem provides an alternate (and more
elementary) proof of Corollary 10.14. Since lF(x) can, in general, exceed
mF(x), this result strengthens Corollary 10.14 slightly.
(10.5) Let F £ C be a field in which — 1 is a sum of two squares. Let G be a
2-group and /e Irr(G). Show mF(x) = 1.
Hint Reduce to the case where x(l) = 2 and Q8 £ G.

172
Chapter 10
Note This strengthens Corollary 10.14 in a different direction and
suggests a strengthening of Theorem 10.12. In fact, B. Fein has proved that the
exceptional case in Theorem 10.12 can only occur if — 1 is not a sum of two
squares in F.
(10.6) Let ze Irr(G) and F s C. Let H = G x G x ■ ■ • x G be a direct
product of copies of G and \et\p = xxXx'"xXe Irr(H). Show that
mF(i/f) divides mF(x).
Note In fact if H is the product of mF(/) copies of G, then mF(\ji) = 1.
Conversely, if mF(\p) = 1 and H is the product of n copies of G, then mF(x) \ n
provided \F: Q\ < oo.
(10.7) (Fein) Let H be the product of n copies of G and let / and \ji be as in
Problem 10.6. Show that mQ(i/0 divides \jn, 1G].
Note Problem 10.7 and the note preceding it prove that mu(x) divides
n\_f, 1G] for every n > 0 and / e Irr(G).
(10.8) Let H = G x G and let /elrr(G). Let i> = x x /elrr(H). Show
that mQ(\p) = 1.
(10.9) Let G = Q8 x Z3 and let x e Irr(G) be faithful. Show that mQ(x) = 1.
(10.10) Let G = H x X and \ji e Irr(H) and 9 e Irr(X). Let * = \p x 9. Let
F£ C.
(a) Show that mF(/) divides mF(t/^)mF(9).
(b) Show that equality occurs in (a) provided (mF(i/f), 9(1)|F(9): F\) = 1
and(mF(9),^(l)|F(^):F|)=l.
(c) Let p, q be primes such that pjf(q — 1) and qjf(p — 1). Show that pq
occurs as a Schur index.
(10.11) Let x e Irr(G) and p | mu(x) for some odd prime p. Show that G
contains an element of order pq, where <j is some prime such that p\(q — 1).
(10.12) Let p be an odd prime and let P be a nonabelian p-group of order p3
and exponent p. It is possible to find H s Aut(P) such that H s Q8 and
CP(H) = Z(P) = Cp(t), where t is the involution in H. Let G = PH, the semi-
direct product. Let 9 e Irr(P) with 9(1) = p.
(a) Show that 9 can be extended to 9e Irr(G) such that %) = ± 1 and
§(r) = p mod 4.
(b) Show that [£„, /] = (p - 9»)/4, where * e Irr(tf), *(1) = 2.
(c) Conclude that — 1 is a sum of two squares in Q(e2"ilp) if p = 3
or 5 mod 8.

Problems
173
Hints Obtain § via Corollary 6.28. Compute %) by working in JV =
<P, t>- Note that |CN(f)| = 2p and that the p - 1 Galois conjugates of &N
are all equal at t. If p = 3 or 5 mod 8, show that mF(x) = 1, where F =
0(e2*ilp).
Note Conversely, if — 1 is a sum of two squares in 0(e2*ilp), then
p ^ 7 mod 8. Some primes = 1 mod 8 can occur, however.
(10.13) Let PeSylp(G) be abelian of exponent p". Show that paJfmQ(x)
for x g Irr(G).
Hint If X -o G is cyclic and G/X is a p-group, then there exists F £ G
such that V n X = 1 and \G: VX\ < p".
(10.14) Show that the following two statements can be added to the
conclusions of Theorem 10.7 provided p =£ 2.
(e) If Y is the p-complement in X, then CH(Y) = X.
(f) If U e Sylp(X) and P e SylP(H), then U £ <D(P), the Frattini
subgroup.
Hints If CP(Y) > U, show that H has a noncyclic normal abelian p-
subgroup. If M £ P and MU = P, then 9My is irreducible.
(10.15) Suppose mF(x) = z(l) for some / e Irr(G) and F £ C.
(a) If H £ G, show that all irreducible constituents of Xh are Galois
conjugate over F(x) and that mF(9) = 9(1) for each such constituent 9.
(b) If / t6 1G, show that G is not simple.
(c) If 2^/(1) or >/^T g F, show that G/ker / is solvable.
Hint Use Lemma 10.4.
(10.16) Suppose that mF(x) = z(l) for all / e Irr(G) with F £ C. Show that
every subgroup of G is normal.

Projective representations
Let JV -o G and suppose 9 e Irr(iV) is invariant in G. For each irreducible
constituent % of 9G we have that Xn = e(/)$» where e(/), the ramification, is
a positive integer. In Chapter 6 we obtained some information about these
mysterious integers. If 9 is extendible to G [which is equivalent to saying that
some e(x) = 1], then the e(x)'s are exactly the degrees of the irreducible
characters of G/N. (See Corollary 6.17.) We shall see that, in general, the
e(x)'s are the degrees of irreducible "projective representations" of G/N.
(11.1) definition Let G be a group and F a field. Let X: G -»GL(n, F) be
such that for every g,heG, there exists a scalar a.(g, h)eF such that
X(g)X(h) = X(gh)z(g, h).
Then X is a projective F-representation of G. Its degree is n and the function
a: G x G -» F is the associated factor set of 3E.
Note that the "factor set" a has nonzero values and is uniquely
determined by X. Both of these observations follow from the fact that the matrices
X(g) are nonsingular.
Let Z(n, F) s GL(n, F) be the group of nonzero scalar matrices. [Note
that Z(n, F) = Z(GL(n, F)).] By definition, PGL(n, F) = GL(n, F)/Z(n, F) is
the projective general linear group. If X is a projective F-representation of G of
degree n, then the composition of X with the canonical homomorphism
GL(n, F) -» PGL(n, F) is a homomorphism G -» PGL(n, F). Conversely, if
7T.- G -»PGL(n, F) is any homomorphism, we can define a projective
representation X of G by setting 3E(gf) equal to any element of the coset n(g) of
Z(n, F) in GL(n, F).
174

Projective representations
17S
Before proceeding with the study of projective representations, we digress
to give an instance of how they can arise in the study of ordinary
representations.
(11.2) theorem Let JV<G and suppose ^) is an irreducible C-represen-
tation of JV whose character is invariant in G. Then there exists a projective
C-representation X of G such that for all n e JV and g e G we have
(a) X(n)=y(n);
(b) X(ng) = X(n)X(g);
(c) X(gn) = X(g)X(n).
Furthermore, if X0 is another projective representation satisfying (a), (b),
and (c), then X0(g) = X(g)/i(g) for some function fi: G -* C *, which is
constant on cosets of JV.
Proof For geG and neN, write V(gng~ l) = tye(ri). Since ^ affords a
G-invariant character, we conclude that ^) and tya are similar representations
ofJV.
Now choose a transversal T for JV in G (that is, a set of coset
representatives). Take 1 e T. For each t e T, choose a nonsingular matrix P, such that
P,tyPt~l = ^)'. Take P± = I. Since every element of G is uniquely of the
form nt for n e JV and t e T we can define 3E on G by 3E(nt) = '^(n)P,. Properties
(a) and (b) are immediate and (c) follows since
X(nt)X(m) = V(»)PtV{m) = WW(m)Pt = ^(nrmr1)?,
= X(ntmt~v ■ t) = X(nt • m).
Properties (a), (b), and (c) yield
mVin) = *(gn) = X{gng-l-g) = Wsng-1)^)
and
X{gmny%grl = V{gng-1)
for all g e G and n e JV. If A is any nonsingular matrix such that
A^(ri)A-' = ^(gng-1)
for all neN, then A ~ lX(g) commutes with all ^(n) for n e JV and thus /I" lX(g)
is a scalar matrix by Corollary 1.6. If X0 also satisfies (a), (b), and (c), we may
take A = X0(g) and conclude that X0(g) = X(g)y.(g) for some y.(g) e C *. Also
X(g)X(hW(n)X(hy1X(g)-1 = X(gmhnh~')X(gYl = ^(ghnh-'g-1).
Comparing this with
X(ghmn)X(ghy' = y(ghnh-'g-1)

176
Chapter 11
yields X(g)X(h) = X(gh)a.(g, h) for some a(g, /i)eC and thus X is a projective
representation.
All that remains now is to check that fi is constant on cosets of N. We have
X(n)X(g)f4g) = X0(n)X0(g) = X0(ng) = X(ng)y.{ng).
Since X(ri)X(g) = X(ng) is nonsingular, the result follows. |
Of course, the above argument does not really require that ^ be a
complex representation. Using Theorem 9.2 and Lemma 9.12, any absolutely
irreducible representation will do.
We begin our study of projective representations by considering the
associated factor sets.
(11.3) lemma Let a: G x G -» F* be the associated factor set of a
projective F-representation of G. Then
u(xy, z)a(x, y) = <x.(x, yz)cc(y, z)
for all x,y,ze G.
Proof Let X be the projective representation. We have
X(x)X(y)X(z) = X(xy)X(z)a.(x, y) = X(xyz)a.(xy, z)a.(x, y).
Also
X(x)X(y)X(z) = X(x)X(yz)a(y, z) = X(xyz)a(x, yz)a(y, z).
The result now follows because all of the matrices are nonsingular. |
(11.4) definition Let A be a possibly infinite abelian group and let G be
any group. Then an A-factor set of G is a function a: G x G -» A such that
a(xy, z)cc(x, y) = a(x, yz)a(y, z)
for all x, y, z e G.
Thus the factor set of a projective F-representation is an F * -factor set
where F * denotes the multiplicative group of F. Conversely, we shall show
that every F * -factor set is associated with a projective F-representation. To
do this we introduce the "twisted group algebra."
Let G be a finite group and F a field. Let a be an F* -factor set of G. Let
FX\_G] be the F-vectorspace with basis {g\g e G}. (That is, there is a specified
basis of Fa[G] which is in one-to-one correspondence with G.) Define
multiplication in F"[_G] hyg-h = g~ha{g, h) and extend via the distributive law. To
establish that the multiplication thus defined is associative, it suffices to check
it on the basis elements g for g e G. That it holds there is immediate from the

Projective representations
177
definition of a factor set. The finite dimensional algebra F*[G~\ is the twisted
group algebra with respect to a. Note that if a is the trivial F * -factor set, that is,
a{g, h) = 1 for all g, h, we can identify F*[G] with F[G].
In order to see that F*\_G] has a 1, we need the following.
(11.5) lemma Let a be an /1-factor set of G. Then a(l, x) = a(l, 1) = <x.(x, 1)
for all x g G.
Proof We have
a(l-l,x)a(l, 1)= a(l, l-x)a(l,x).
Canceling a(l, x) yields a(l, 1) = a(l,x) for xeG. That a(l, 1) = a(x, 1)
follows symmetrically. |
Let a be an F*-factor set of G and let v = a(l, 1)"l e F. Now (vl)g =
vga(l, g) = g and similarly g{v\) = g. Thus vT is the unit element in Fa[G].
It is now immediate that the elements g e F*[_G~i for g e G all have inverses.
Now let a be an F * -factor set of G and let ^ be any representation of the
algebra F"\_G]. Define X(g) = ty(g). Then X(g) is nonsingular and
x^xw = mW(Ii) = W • K) = y(gK*(g, h)) = *(gh)*(g, h)
so that X is a projective representation of G with factor set a.
Conversely, if X is a projective F-representation of G with factor set a, we
can define a representation ^ of Fa[G] by setting ty(g) = X(g) and extending
by linearity. In other words, the projective F-representations of G having
factor set a are in a natural one-to-one correspondence with the
representations of the twisted group algebra F*[G]. The situation is analogous to the
connection between ordinary representations and the ordinary group algebra.
Exactly as in the case with ordinary representations, we define two
projective representations X and ^) to be similar if ^) = P~ lXP for some non-
singular matrix P. Also X is irreducible if it is not similar to a projective
representation in the form
Note that similar projective representations have equal factor sets and
correspond to similar representations of the appropriate twisted group
algebra. Also, irreducible projective representations correspond to
irreducible representations of the algebra. Since every finite dimensional algebra
has irreducible representations, we have proved the following.
(11.6) corollary Let a be an F * -factor set of G. Then G has irreducible
projective F-representations with factor set a.

178
Chapter 11
The set of /1-factor sets of G forms a group under pointwise multiplication.
In the language of group cohomology, this group is denoted Z2(G, A), the
group of "2-cocycles." If fi: G -» A is an arbitrary function, we can define
8(fi): G x G -» A by
d(fi)(g,h) = fi(g)fi(h)fi(gh)-\
It is routine to check that 8(fi) is a factor set.
Note that 8 is a homomorphism from the group of /1-valued functions on
G (with pointwise multiplication) into Z2(G, A). The image of 8 is the
subgroup B2(G, A) c Z2(G, A) which is called the group of "2-coboundaries."
The factor group Z2(G, A)/B2(G, A) can be identified with the second
cohomology group H2(G, A). (More generally in cohomology theory, one
assumes that G acts on A. Here we are considering only the "trivial action"
case.)
The superscript 2 above refers to the fact that we are discussing functions
of two variables on G. Since that will be the only situation considered here, we
write Z(G, A) for the group of A -factor sets B(G, A) for the image of d and we
define H(G, A) = Z(G, A)/B(G, A).
We say that two /1-factor sets of G are equivalent if they are congruent
mod B(G, A). Thus H(G, A) is the set of equivalence classes of /1-factor sets
onG.
If X is a projective F-representation on G with factor set a, and y.:G -» F *
is any function, define ^ = Xfi by ty(g) = X(g)(j(g). It is trivial to check that
^ is a projective representation of G with factor set /? = o«5(/z). Thus X and 9)
have equivalent factor sets.
If X and ^ are projective F-representations of G, we say that X and ^) are
equivalent if ^ is similar to X/i for some function fi: G -> F*. This is easily
seen to define an equivalence relation which preserves irreduciblility.
We can now give a necessary and sufficient condition for an invariant
irreducible character of a normal subgroup to be extendible.
(11.7) theorem Let JV -a G and let 9 e Irr(iV) be invariant in G. Let ^ be a
representation affording 9 and let X be a projective representation of G
satisfying conditions (a), (b), and (c) of Theorem 11.2. Let a be the factor set of
X. Define p e Z(G/N, C *) by 0(gJV, WV) = ofo, h). Then /? is well-defined and
its image /? e H(G/N, C *) depends only on 9. Also, 9 is extendible to G
iffjg= 1.
Proof For m,neN and g.heG, we have
a(gfn, hm)X(gnhm) = X(gn)X(hm) = X(g)X(nhm)

Projective representations
179
using 11.2(b) and (c). Furthermore,
X(g)X(nhm) = X(g)X(hnhm) = X(g)X(h)X(nhm)
= a(0, h)X(gh)X(nhm) = a(g, h)X(ghnhm)
by 11.2(c). Since X(gnhm) = X(ghnhm) is nonsingular, we have a.(gn, hm) =
a{g, h) and /? is well defined. That /? is a factor set is clear.
If X0 were chosen in place of X, we have X0 = X/j. where fi: G -* C * is
constant on cosets of JV so that we can define v(gN) = fi{g). If X0 has factor
set a0, then
tx0(g, h) = x(g, h)n{g)n(h)n(ghyi
= p(gN, hN)v(gN)v(hN)v(ghN)-1
and hence /? is independent of the choice of X. If PtyP~l were chosen in place
of ty, we can replace X by PXP~l which leaves a, /? and ft unchanged. Thus /?
is uniquely determined once 9 is given.
If 9 is extendible to G, we can choose X and ^) so that 3E is a representation.
Thus a = 1 and hence /? = 1.
Conversely, if /? = 1, there exists v: G/N -> C * such that
/JfoJV, MV) = vigNXhNXghNy1.
Define /z: G -» Cx by fi(g) = v(gN). Now define X0(g) = 3E(g)/z(g)_1 so that
X0 is a projective representation of G with factor set
cc0(g, h) = cc(g, h)y.{gY ^(h)' lf4gh) = 1
and thus X0 is a representation of G. Furthermore, since X(l) = ^)(1) = /,
we have
l = a(l,l) = /z(l)/z(l)/z(l)-1 = /z(l).
Thus /z(n) = v(l) = /z(l) = 1 for all neN and X0(n) = 3E(n)/z(n)_1 = ^(n).
Thus X0 is an extension of ^) and the proof is complete. |
A word of caution about Theorem 11.7 is appropriate. In the notation of
the theorem, if a e B(G, C *) so that a = 1 in H(G, C *), it does not follow that
9 is extendible to G. For instance, take G = QB, N = Z(G), and 9 the non-
principal linear character of JV. Then 9 is not extendible to G and yet
H(G, Cx) = 1. (See Problem 11.18.)
The theory of projective representations is closely related to that of
central extensions.
(11.8) definition A central extension of a group G is a (possibly infinite)
group T together with a homomorphism n of T onto G such that
ker 7t £ Z(r).

180
Chapter 11
(11.9) lemma Let (r, 7t) be a central extension of G with A = ker n. Let X
be a set of coset representatives for A in T and write X = {xg\g e G}, where
n(xg) = g. Define a:GxG->/l by xgxh = a(g, h)xgh. Then aeZ(G,A).
Furthermore, the equivalence class of a is independent of the choice of X.
Proof That a is an /1-factor set follows by computing xgxhxk two ways,
using the associative law in r. If Y = {yg} is another set of coset
representatives, then y9 = fi(g)xg for some fi(g) e A. Now
yBy>, = KdMh)xgxh = fi(g)n(h)tx(g, h)xgh
= V(g)v(hMghylcc(g,h)ygh
and the result follows. |
If A and U are abelian groups and 1 e Hom(A, U), then for each
ae Z(G, A), we define 1(a) by 1(a) (g, h) = l(a(g, h)). It is routine to check
that 1(a) eZ(G,t/).
(11.10) corollary Let (r, 7t) be a finite central extension of G and let
A, X = {xg} and a be as in Lemma 11.9. Let 9) be an (ordinary) F-represen-
tation of T such that the restriction tyA is the scalar representation XI for some
leHom(A,F*). Define X(g) = ^(x9) for geG. Then X is a projective F-
representation of G with factor set 1(a). Furthermore,
V(y) = my)My)
for all jieT, where /z: T -» F * is the function defined by fi(y) = Myx^). Also
X is irreducible iff ^) is and the equivalence class of X is independent of the
choice of coset representatives X.
Proof We have
X(g)X(h) = WxM*J = ^W0> hK*) = M9, hWgh)
and X is a projective representation with factor set 1(a). If y e r, we have
y = axg, where g = n(y) and ae A. Thus
SK)0 = SfoWfl) = X(n(y))l(yx«tt-
In particular, 3E(G) and ^)(r) span the same vectorspace of matrices over F and
the assertion about irreducibility follows.
Finally, if Xt is the projective representation determined by an alternate
choice of coset representatives, we have
*i(nO0) = V(y)^iy)'1 = sWjOMOmiOO" '
and X and 3Et are equivalent. |

Projective representations
181
Note that if ^) is an irreducible C-representation (or any absolutely
irreducible F-representation), then the condition that tyA be a scalar
representation in Corollary 11.10 is automatically satisfied.
Henceforth we shall consider only C-representations.
(11.11) definition Let (r, n) be a finite central extension of G. Let X be a
projective C-representation of G. We say that X can be lifted to T if there
exists an ordinary representation ^ of T and a function /z: T -» C * such that
$(*) = X(tcW)/zW
for all x g T. Furthermore, (r, 7t) has the projective lifting property for G if
every projective C-representation of G can be lifted to T.
Note that if X is lifted to the representation ^ on T, then ^ is necessarily
a scalar representation and by Corollary 11.10 we can construct a projective
representation 3Et of G by choosing coset representatives for ker n in T. Then
XMx))^) = V(x) = X(n(xMx)
and hence Xt is equivalent to X. Thus if (r, rc) has the projective lifting
property for G, then all projective C-representations of G are equivalent to ones
obtained via the construction in Corollary 11.10.
We shall prove a theorem of Schur which asserts that every finite group
has a finite central extension with the projective lifting property. The point
of Schur's theorem is that it allows us to apply what we know about ordinary
representations to the study of projective representations. For instance, in
the situation of Corollary 11.10, if ^) is irreducible and F = C, then we have
deg X = deg ^ divides | T: A | = | G | by Theorem 3.12. Thus a consequence of
Schur's theorem is that the degree of every irreducible projective
C-representation of G divides \G\.
(11.12) definition The Schur multiplier of G is the group H(G, C"). It is
denoted M(G).
For a finite abelian group A we use the notation A to denote the group
\xx{A). If (r, n) is a central extension of G with A = kern finite, we construct a
homomorphism n: A -» M(G) as follows. Choose a set X of coset
representatives for A in T and write X = {xg\geG}, where n(xg) = g. Let
ocgZ(G, A) be defined by xgxh = <x.(g, h)xgh as in Lemma 11.9. For XeA
define rj(l) = i(a), where 1(a) g Z(G, C) is defined by 1(a) (g, h) = l(a(g, h))
and the bar denotes the canonical map Z(G, £*)-> H(G, C*) = M(G). Note
that r\ is a homomorphism.
The map >/: /?->M(G) is independent of the choice of coset
representatives X. This follows since another choice would yield a factor set
/? g Z(G, A), which is equivalent to a by Lemma 11.9. Since iz^'e B(G, A), it

182
Chapter 11
follows that l(a)l(/3)- l=Jia.fT^) e B(G, C *). Thus 1(a) and l(/3) are
equivalent in Z(G, C") and 1(a) = Wl
We shall call the unique homomorphism n: A-> M(G) constructed above
the standard map.
(11.13) theorem Let (r, n) be a finite central extension of G and let rj be
the associated standard map. Let X be a projective C-representation of G
with factor set y. Then X can be lifted to r iff y lies in the image of n. In
particular, (r, n) has the projective lifting property iff n maps onto M(G).
Proof. Let A = kern and let X = {xg\g e G} be a set of coset
representatives for /I in T with n{xg) = g. Write xgxh = a(g, /i)xg)l so that a e Z(G, /I).
Now suppose y = n{X) for some Xe A. Then 1(a) is equivalent to y and we have
X(oig,h))=y(g,h)t4g)fi(h)fi(ghrl
for some function fi: G -* C *.
Define ^ on T by ^(ax9) = X(a)X(g)fi(g) for a e 4 and ge G. We have
mxM*-) = X(g)*(hMgMh) = X(gh)y(9, %(<?)/#)
= X(a(g, h))X(ghMgh) = ^(oi(g, h)xgh) = V(xgxh).
Since ty(axg) = X(a)ty(xg), it follows that ^ is a representation. Thus ^) lifts
Xtor.
Conversely, if X can be lifted, we have ty(x) = X(n(x))n(x) for some
representation ^ of T and function fi: Y -> C. Thus X(l) = /z(l)~ M|)(l) and for
every aeA we have ^(a) = X(l)fi(a) = /z(a)/z(l)-1,?)(l) and so 1(a) =
H(a)/i(l)~i is a linear character of A. Now write v(g) = fi{xg). We have
l(a(g, ^))^)(x9,) = ^)(x9)^)(x,) = X(g)X(h)v(g)v(h).
Thus
l(a(#, h)Wgh)X{gh) = y(g, h)v(g)v(h)X(gh)
and 1(a) is equivalent to y. Therefore
tfl) = m = y
and the proof is complete. |
Given an arbitrary finite group G we shall show that M(G) is finite and
construct a central extension (r, ri) of G with A = ker rc such that the standard
map n:A-> M(G) is an isomorphism. This will thus be a group with the
projective lifting property for G with smallest possible order, namely | G11 M(G) |.
Such a group T is called a Schur representation group for G.
An abelian group Q is divisible if for every x e Q and positive integer n
there exists yeQ, with/ = x. For instance, F * is divisible if F is algebraically
closed.

Projective representations
183
(11.14) lemma Let A be an (infinite) abelian group and let Q c A with Q
divisible. Assume | A : Q | < oo. Then Q is complemented in A.
Proof Use induction on | A : Q\. We may assume A > Q and choose
ae A - Q.Letn = o(aQ) in A/Q and let u = a" e Q. Let ve Q with t>" = u by
divisibility and let ft = av~l so that ft" = 1. Since aQ = bQ, it follows that
n = o(bQ) in A/Q and thus <b_> n Q = 1.
Now let ,4 = 4/<fc> so 0 = Q(b}/(by satisfies 0 s Q and \A: 0| =
|/1: 8<ft>| < |/I: Q\. By the inductive hypothesis, Q is complemented in A
and thus there exists B ^ A with B n Q<£>> = <£>> and QB = A. Now
gnB = Qn Q<£>> nB=2n <£>> = 1 and the proof is complete. |
The above result remains true without the assumption that | A: Q | is
finite. We will not need that more general fact, however.
(11.15) theorem Let F be an algebraically closed field and G a finite
group. Then H(G, F *) is finite and each of its elements has order dividing \G\.
Furthermore, B(G, F*) is complemented in Z(G, F").
Proof First, we argue that B(G, F *) is divisible. If 0 e B(G, F *) and n is a
positive integer, write /? = d(/i) for some function fi: G -» F *. For each g e G,
choose v(g)e F* such that v(g)n = fi(g). Then c5(v)n = (5(^) = p. We can thus
apply Lemma 11.14 when we show that \H(G, F*)| < oo.
Let a e Z(G, F *) and define
tAo) = EI a(0>;,(:)-
xeG
For fixed g, /i e G, we have a(g, hx)a.(h, x) = o{gh, x)a.(g, h). Taking the product
over all x e G yields
fi(g)ti(h) = ii{gh)z(g, h)M
and thus a|G| e B(G, F *). This shows that H(G,F*) has exponent dividing | G |.
Now let t/ = {aeZ(G,Fx)|a|G| = 1}. For aeZ(G,Fx), let /4 =
<B(G, Fx), a>. By the result of the previous paragraph \A:B(G, F*)\
divides \G\. Thus by Lemma 11.14, B(G, F*) is complemented in A and the
complement is a subgroup of U. Thus a e B(G, Fx )t/ and hence B(G,F*)U
= Z(G,F*).
Now every element of U is a function from G x G into {ye F|>>|G| = 1}.
Since this is a finite set, it follows that | U | < oo and thus
|tf(G,Fx)| = |B(G,Fx)t/:B(G,Fx)| < \U\ < oo
and the proof is complete. |
Next we need a general method for constructing central extensions.

184
Chapter 11
(11.16) lemma Let A be an arbitrary abelian group and let ae Z(G, A).
Then there exists a central extension (r, n) of G with kern = A and such that
a set of coset representatives {xg\g e G} exists with n(xg) = g and xgxh =
Proof Let T = G x A as a set and define (g, a)(h, b) = (gh, a.(g, h)ab).
That this multiplication is associative follows from the fact that a is an A-
factor set. Let a(l, l)~l = zeA. Then (1, z)(g, a) = (g, a(l,g)za) = (g, a) by
Lemma 11.5. Also
(g-\a-ia(g-1,gr1z)(g,a) = (l,z)
and hence T is a group with 1 = (1, z).
Clearly n: F -> G defined by n(g, a) = g is a homomorphism with ker 7t =
/?"= {(1, fl)|ae i4}. That A £ Z(G) follows since a(l, g) = a(g, 1) by Lemma
11.5.
Since (1, za)(l,zb) = (1, zab), it follows that a<->(l,za) defines an
isomorphism A ^ A. We identify /I and A via this isomorphism.
Let X = {(g, l)\geG}. Then n maps X one-to-one and onto G and so X
is a set of coset representatives for A in T. Now
(g, 1)0, 1) = (gh, a(g, h)) = (1, zafo, fc))(gfc, 1)
by Lemma 11.5. Since (1, za(g, h)) = a(g,h) under our identification, the
proof is complete. |
(11.17) theorem (Schur) Given G, there exists a finite central extension
(r, n) which has the projective lifting property for G. Furthermore, (r, n) can
be chosen such that ker n = A = M(G) and the standard map A -> M(G) is
an isomorphism.
Proof Let M be a complement for B(G, C *) in Z(G, C *) by Theorem
11.15. Let A = M. Define <x(g, h)e A by <x(g, h)(y) = y(g, h) for y e M. It is
clear that in fact a(gr, h)eM = A. Next
(a(g/i, /c)a(g, /i))(y) = y(gh, k)y(g, h)
using the definition of multiplication in M. Similarly,
(a(#, ^/c)a(^, /c))(y) = y(g, hk)y(h, k)
and since y runs over a set of factor sets, it follows that a e Z(G, A).
Now let (r, 7t) and X = {xg\g e G} be as in Lemma 11.16 so that xgxh =
a.(g, h)xgh. Let >/: A -» M(G) be the standard map. We show that rj maps onto.
For y eM(G) = Z(G, C*)/B(G, Cx), there exists yeM r\y since M
complements B(G, Cx) in Z(G, Cx). Now define X on A = M to be the
evaluation map at y. Note that Xe A. Now
!(%, h)) = a(g, ^)(y) = y(g, h)

Projective representations
185
so that 1(a) = y. Therefore
as desired.
By Theorem 11.13, it follows that T has the projective lifting property for
G. Also,
\A\ = \A\>\r,(A)\ = \M(G)\ = \M\ = \A\
and so rj must be one-to-one and A = M(G). However A = A by Problem
2.7(c) and the proof is complete. |
(11.18) corollary Let X be an irreducible projective C-representation
of G. Then deg(3E) divides \G\.
Proof See the discussion preceding Definition 11.12. |
Before going on to exploit Schur's theorem, we give another result which
is useful in the computation of M(G). We have defined a Schur representation
group of G to be a minimal central extension with the projective lifting
property. The next theorem will yield another characterization.
(11.19) theorem Let (r, n) be a finite central extension of G with A =
ker n and let r\: A -> M(G) be the standard map. Let A0 = A n F. Then
ker rj = {AeA\A0^ ker 1}.
In particular, r\ is one-to-one iff A <=, F.
Proof Let X = {xg\g e G} be as usual and write xgxh = a.(g, h)xgh with
a e Z(G, A). Suppose 1 e ker rj. Then 1 = rj(X) = 1(a) and so 1(a) e B(G, C *).
Thus
Mg,h)) = fi(gMh)fi(ghyl
for some function fi: G -* C *. Now define 1 on T by \axg) = l(a)/*(g) and
check that
l(xg)l(xh) = Uxgxh).
(This is essentially the same calculation as in the proof of Theorem 11.13.)
It follows that X is a linear character of T which extends 1. Now F ^ ker 1
and so A0 c ker 1.
Conversely, let A0 c ker 1. Then 1 is extendible to l0e Irr(/IF/F) by
l0(a:>c) = 1(a) for x e F. Since T/F is abelian, 10 is extendible to lt e Irr(r/F).
Now
h(xg)^i(xh) = Xi(xgxh) = l(cc(g, /j))1i(x9A).

186
Chapter 11
Define ^g) = li(x9). Then
and the result follows. |
(11.20) corollary Let (r, 7t) be a finite central extension of G with
A = ker n.
(a) If A £ F, then A is isomorphic to a subgroup of M(G).
(b) Assume |/1| = |M(G)|. Then /I £ F iff r has the projective lifting
property for G. In this case M(G) = A.
Proof (a) follows since A = A and (b) is immediate from Theorems 11.19
and 11.13. |
Thus we see that r is a Schur representation group for G iff T/A = G for
some A <= Z(F) such that \A \ = \M(G)\ and A <= r.
(11.21) corollary Let p be a prime divisor of |M(G)|. Then a Sylow
p-subgroup of G is noncyclic.
Proof Let T be a Schur representation group for G (which exists by
Theorem 11.17). We may assume that T/A = G with A c Z(r)- Let
P//1 e Sylp(G) and assume P/A is cyclic. Since /I is central, it follows that P is
abelian and thus r has an abelian Sylow p-subgroup. By Theorem 5.6,
pJC\r nZ(r)|. Since A <= F by Corollary 11.20, we have pJf\A\. Thus
pi|M(G)|. |
The next corollary is quite important. It can, of course, be proved without
all of the complex machinery which we have developed.
(11.22) corollary Let Af<G with G/N cyclic and let 9eIrr(N) be
invariant in G. Then 9 is extendible to G.
Proof By Corollary 11.21, M(G/N) = H(G/N,C) is trivial. The
result now follows from Theorem 11.7. |
The above theory, together with some rather technical computations
yields a tool which is very useful when studying the characters of groups with
normal subgroups. Let N <i G and let 9 e lrr(N) be invariant in G. Under
these hypotheses we say that (G, N, 9) is a character triple. The analysis of this
situation is much easier if 9 is linear. We shall use a Schur representation
group for G/N to replace (G, N, 9) by another character triple (I\ A, X) in
which r/A = G/N and X is linear. Of course, this would not be of much value
unless we knew that the character theory of r was somehow closely tied to

Projective representations
187
the character theory of G. For instance, we would want lr and 9G to have
the same numbers of irreducible constituents with the same ramifications.
We define below the notion of "isomorphism" of character triples. This
rather complicated definition makes precise some of the ways in which the
character theory of two character triples can be related. We first introduce
some notation. If (G, N, 9) is a character triple, let Ch(G 19) denote the set of
(possibly reducible) characters / of G such that Xn is a multiple of 9. Let
Irr(G|9) be the irreducible characters among these, so that Irr(G|9) is the
set of irreducible constituents of 9°. Note that if N c H c g, then (H, N, 9)
is a character triple and Xh e Ch(H 19) whenever / e Ch( G19). If r. U -» V is an
isomorphism of groups and (pelrr(U), let (pTelrr(V) denote the
corresponding character, so that q>x(ux) = q>(u).
(11.23) definition Let (G, N, 9) and (r, M, cp) be character triples and
let v. G/N -» T/M be an isomorphism. For N £ H c G, let Hr denote the
inverse image in T of i{HjN). For every such H, suppose there exists a map
aH: Ch(H|9) -> Ch(HT|<p) such that the following conditions hold for H, K
withN c K c He Gand/, t/^eCh(H|9).
(a) o„(x + ip) = a„(x) + (TH(^);
(b) [x,./0 = [>h(xW'/')];
(c) ^(Zk) = (erH(x)V;
(d) <rH(Z/?) = <7H(z)/T for /? e Irr(tf/JV).
Let a denote the union of the maps erH. Then (t, a) is an isomorphism from
(G, N, 9) to (r, M, <p).
Note that if (t, a) is an isomorphism from (G, N, 9) to (r, M, <p), then erH
is determined by its restriction to Irr(H 19). (This follows from (a).) By (b) it
follows that aH maps Irr(H|9) one-to-one into \xx{Hx\q>). Therefore, to
construct an isomorphism (t, a) it suffices to define erH on Irr(H 19) to be one-
to-one, then extend the definition by (a) and check that (c) and (d) hold for
Xelrr(tf|9).
(11.24) lemma Let (t, a): (G, N, 9) -* (r, M, cp) be an isomorphism of
character triples. Then aH is a bijection of Ch(H 19) onto Ch(HT | <p) for all H
withN c H c G. Furthermore, x(l)/9(l) = (7H(x)(l)/<p(l)forallxeCh(H|9).
Proof If OalXi) = oH(x2) for /(eCh(H|9), we have [/;, i/^] =
["ateX <th('/')] is independent of i for all \ji e Irr(H 19). It follows that Xi = li
and hence erH is one-to-one.
For /eCh(H|9) write e(x) = z(l)/9(l) and similarly set e(rj) = rj(l)/(p(l)
for rieCh(Hr\(p). Note that aN(9)elrr(M\(p) and so cr^(9) = <p. We have

188
Chapter 11
Xn = e(z)9 and rjM = e(r])(p and thus
e(<rH(x))<P = (oh(X))m = On(Xn) = ^(e(z)9) = e(x)<p
and thus e(aH(x)) = e(x) as claimed.
By Frobenius reciprocity, we have
9H= Z e(X)x
zelrr(H|8)
and comparing degrees yields £e(x)29(l) = |H:JV|S(1) so that £e(x)2 =
\H:N\ where / runs over Irr(H|9). Similarly £ e(rf)2 = \Hr: M\ = \H:N\
for rj e Irr(HT | cp). Since aH maps Irr(H 19) one-to-one into Irr(HT | cp\ we have
\H:N\ = Ze(Xf = £e(o„(X))2 < I<W2 = \H:N\.
It follows that every r\ e Irr(HT | <p) is of the form oH{x) for some / e Irr(H 19).
The result now follows. |
(11.25) corollary Isomorphism is an equivalence relation on
character triples.
Proof. The reflexive and transitive properties are obvious. If
(T,(7):(G,JV,9)^(r,M,<p)
is an isomorphism, then aH: Ch(H|9)-> Ch(K|<p) is one-to-one and onto
where K = HX. We can, therefore, define a"1 by (<x~')K = {oH)~l for
M £ K £ T where H = KT~'. It is routine to check that
(T-',(j-1):(r,M,<p)^(G,iV,9)
is an isomorphism. |
A nearly trivial example of an isomorphism of character triples is given
by the following result.
(11.26) lemma Let (G, N, 9) be a character triple and let /z: G -» T be an
onto homomorphism with ker y. £ ker 9. Let M = /z(iV) and let q> e Irr(M) be
the character corresponding to 9 e Irr(iV/ker /z). Then (G, JV, 9) and (r, M, <p)
are isomorphic character triples.
Proof We have r. G/N -» T/M defined naturally from /z. For N £ H £ G
and / e Ch(H 19) we obtain ker fi £ ker / and we may view / as a character
of H/ker fi. Let oH{x) be the corresponding character of (i(H) = H/ker fi.
Check that (t, er) is the desired isomorphism. |
Another example of an isomorphism of character triples is provided by
the following.

Projective representations
189
(11.27) lemma Let (G, JV, cp) be a character triple and let r\ e Irr(G) be such
that r]N(p = 9e Irr(JV). For JV £ H £ G, define aB: Ch(H\cp) -+ Ch(H|9) by
atM>) = </"7h ■ Let i: G/N -> G/N be the identity map. Then
(i, (J): (G, JV, p) -»(G, JV, 9)
is an isomorphism of character triples.
Proof It is clear that aH does map Ch(H | q>) to Ch(H 19) and that
properties (a), (c), and (d) of Definition 11.23 hold. It thus suffices to show that aH
maps Irr(H | <p) one-to-one into Irr(H 19). This follows from Theorem 6.16. |
Given invariant X, 9eIrr(JV) for JV-a G, with /(l) = 1, it follows that
(G, JV, X) and (G, JV, 9) are isomorphic if 91"i is extendible to G.
(11.28) theorem Let (G,N, 9) be a character triple and let (r, n) be a
finite central extension of G/N having the projective lifting property. Let
A = ker n. Then (G, JV, 9) is isomorphic to (r, A, X) for some Xe A.
Proof By Theorem 11.7, the triple (G,N,9-) determines an element
peH(G/N, C*) = M(G/N). Since T has the projective lifting property for
G/N, we can find X e A such that rj(X) = fi~l where r\ is the standard map
(Theorem 11.13).
Now let G* £ G x T be defined by G* = {(g, x)\g = n(x)}, where g =
gN, the image of g in G/N. Note that G* is a subgroup of G x T. Let L =
N x A and observe that L -a G*. Define 9* and 1* on L by &*(n, a) = &(n)
and X*(n, a) = 1(a). Note that 9*, 1* e Irr(L) are invariant in G*.
We have projection homomorphisms fiG: G* -> G and /zr: G* -> T. These
maps are onto and ker (iG = 1 x /I s ker 9* and ker /zr = JV x Is ker 1*.
It follows by Lemma 11.26 that (G*, L, 9*) is isomorphic to (G, JV, 9) and
(G*, L, 1*) is isomorphic to (r, A, X). By Corollary 11.25, it suffices to show
that (G*, L, X*) and (G*, L, 9*) are isomorphic and by Theorem 11.27 we
will be done when we show that 9*(1*)~l is extendible to G*.
Let 9) be a representation of JV affording 9 and let 3E be a projective
representation of G as in Theorem 11.2. Let a be the factor set of G belonging
to 3E and let /? be the corresponding factor set of G/N as in Theorem 11.7. Let
{xg\g e G/N} be a set of coset representatives for A in T with n(xg) = g. Take
xi = 1. Write xsxh = y(g, h)xsh so that y e Z(G/N, A). Since >/(!) = /T \ we
have
X(y)peB(G/N,C*)
and hence
^(7(ff,«))ate,*) = v(g)-1v(«)"1v(ff«)
for some function v: G/N -» C *.

190
Chapter 11
Every element of G* is uniquely of the form {g, axs) for g e G and a e A.
Define 3 on G* by
3(ff,axJ) = X(g)l(a)"1v(g).
We compute
3(0, axs)3(h, bxh) = HgKMpbY'"fo. MaMfi)
and
3tefc, aby(g, h)xsh) = l£{gh)X(ab)- lMy{g, R))" 1v(^).
Since these two expressions are equal, we conclude that 3 is a representation
ofG*.
Now y(l, 1) = 1 and a(l, 1) = 1 and it follows that v(n) = v(l) = 1 for
neN. Therefore 3(n, a) = 9)(n)l(a)"l and 3L affords S*(A*)~K The proof is
now complete. |
We now consider some applications.
(11.29) corollary Let JV -a G and /elrr(G). Let 9eIrr(JV) be a
constituent of Xn . Then /(l)/9( 1) divides |G:JV|.
Proof Let T = 7G(9), the inertia group, and let \jie\xx(T) such that
\pG = x and i/^ = e9 (Theorem 6.11). Since x(l) = |G: T|i/<1), it suffices to
show that i/f(l)/9(l) divides | T: JV|. Let (r, A, X) be a character triple
isomorphic to (T, JV, 9) with 1 linear. Let C e Irr(r|l) correspond to \p e Irr(T|9).
Then t/<l)/9(l) = C(l)/l(l) = C(l) by Lemma 11.24. Since A £ Z(Q we have
C( 1) divides \r-.A\ = \T:N\by Theorem 3.12. The result follows. |
Note that by repeated application of this result we can weaken the
hypothesis that JV is normal in G and assume only that JV is subnormal, that is,
that there exist subgroups JV, such that
JV -a JV, -a ci JVt <a G.
In particular, if JV is subnormal and abelian then since 9(1) = 1 we obtain
the following generalization of Ito's Theorem 6.15,
(11.30) corollary Let N S: G be subnormal and abelian. Then x(l)
divides | G: JV | for every / e Irr(G).
(11.31) corollary Let JV-a G and let 9eIrr(JV) be invariant in G.
Suppose for every Sylow subgroup P/N of G/N that 9 is extendible to P. Then
9 is extendible to G.
Proof Let (r, A, X) be a character triple isomorphic to (G, JV, 9) and with
1 linear. If JV £ H s G and /I £ X £ r are such that H and X correspond,

Projective representations
191
then 9 is extendible to H iff X is extendible to K. This follows from Lemma
11.24 since extendibility of 9 is equivalent to existence of / e Irr(H|9) with
X(l)/9(1) = 1 and similarly for extendibility of X. Thus X is extendible to the
inverse image in T of every Sylow subgroup of Y/A. By Theorem 6.26, we
conclude that X is extendible to r and hence 9 is extendible to G, as desired. |
We can use Corollary 11.31 to obtain an alternate proof of Gallagher's
Theorem 8.15, a proof independent of the results of Chapter 8. We are
assuming that (G, N, 9) is a character triple in which (9( 1), | G: N |) = 1 and that
det(9) is extendible to G. By Theorem 6.25, 9 is extendible to H for every H
with JVcHsG and H/N solvable. The result now follows from
Corollary 11.31.
The following combines several of our extendibility criteria.
(11.32) theorem Let N o G and 9eIrr(iV), with 9 invariant in G.
Assume for every prime divisor p of (9( 1 )o(9), | G : N |) that a Sylow p-subgroup
of G is abelian or if p ^ 2, that 9 is p-rational. Then 9 is extendible to G.
Proof By Corollary 11.31, it suffices to assume that G/N is a p-group for
some prime p. If a Sylow p-subgroup of G is abelian, then 9 is extendible to
G by Theorem 8.26. If p ^ 2 and 9 is p-rational, then 9 is extendible by
Theorem 6.30. In the remaining case, (|G:JV|, o(9)9(l)) = 1 and
Corollary 6.28 yields the result. |
As another application of the theory of projective representations, we
prove a theorem of T. Berger about the characters of solvable groups. Recall
that x £ Irr(G) is said to be quasi-primitive if %N is homogeneous (that is, a
multiple of an irreducible) for every JV o G. Primitive characters are
necessarily quasi-primitive and, in general, the converse is false.
(11.33) theorem (Berger) Let G be solvable and suppose /elrr(G) is
quasi-primitive. Then / is primitive.
In order to prove Berger's theorem, we shall exploit the fact that there
exists a central extension of G with the projective lifting property. We break
the proof into several intermediate results. The first of these is independent
of projective representations.
(11.34) theorem Let L^H<G with LoG,H maximal and G/L
solvable. Let 9 e Irr(H) such that 9G = / and 9L are irreducible. Then there
exists M <i G such that Xm is not homogeneous and M ^ L.
Proof We may assume without loss that L = coreG(H), the largest
normal subgroup of G contained in H. Let K/L be a chief factor of G so that

192
Chapter 11
K/L is an abelian p-group and X £ H.ThusG = XHbythemaximalityofH.
Since K/L is abelian, we have H n Ko K and since HnK<Hwe
conclude that H nK-^G and thus H r\ K = L.
Let C = CG(K/L). Then C<aG and so CnH-^H. However,
K £ N(C n H) and hence C n H ~^G. The maximality of L forces L =
CnH.We have C = K(CnH) = X.
If X = G then H = L <a G and since 9G e Irr(G) we have 7G(y) = L by
Problem 6.1 and thus Xl is not homogeneous. We assume, therefore, that
K < G and let M/K be a chief factor of G so that M/K is an abelian <j-group
for some prime q. If q = p, then (K/L) n Z(M/L) ,£ 1 and the minimality of
K/L yields X/L £ Z(M/L). This contradicts CG(K/L) = K and we conclude
We suppose that Xl and /M are homogeneous and write Xm = &t\ for some
if e Irr(M). Now /(1)/5(1) = | G: H | = | K : L | is a power of p. Since 5L e Irr(L),
it is a constituent of r\L and hence >/( l)/y(l) is an integer which divides /(l)/5( 1).
Therefore >/(l)/y(l) is a power of p. Since M/X is a <j-group for <j j= p, we
conclude that >/KeIrr(X) by Theorem 6.18 or Corollary 11.29.
Since Xl is homogeneous, it is a multiple of 5L and thus >/L is also. Let
R = M r\H and let t be any irreducible constituent of >/R. Then [tl, 5 J ^ 0
and hence t = yR/? for some /? e Irr(/?/L) by Corollary 6.17. Since M = KR
and X n /? = L, we can find y e Irr(M/X) such that yR = /?. Now
*M = (W)" = (SRyRf = (»Rfy.
However, since MH = G and M n H = R, we have
(V = (y°)M = Zm = erj
and thus tm = er\y. Since >/ is a constituent of iM by Frobenius reciprocity, we
have [rj, rjy~\ ^ 0. Since >/KeIrr(X), Corollary 6.17 yields y = 1M and hence
/? = 1R and t = yR. Thus >/R is homogeneous and since (SR)M = er\ we have
rjR = e3R and ij(1) = ey(l). This yields y(l)|M:/?| = ((9R)M)(1) = e2y(l)
and [tjr^r] = e2 = \M:R\. It follows that r\ vanishes on M - R by
Lemma 2.29. As r\ is G-invariant, it vanishes on M — Ra for all g e G and
hence vanishes on M — f]Ra = M — L. In particular, >/ vanishes on M — X
and so \M : X | = [>/K, >/K] by Lemma 2.29. This contradicts rjK e Irr(X) and
completes the proof. |
(11.35) lemma Let (t, a): (G, N, y) -> (T, M, cp) be an isomorphism of
character triples and let JV £ H £ G. Suppose i/f is a character of H. Then
ipeCh(H\&) iff tAGeCh(G|y). Also, if ipeCh(H\&l we have crG(^G) =

Projective representations
193
Proof The first assertion is immediate from Frobenius reciprocity. To
prove that <rG('/'G) = (<JH('/'))r when \ji e Ch(H 19), it suffices to check that
for all /e Irr(r|<p). Since aG maps Irr(G|9) onto Irr(r|<p) by Lemma 11.24,
we have i = crG(^) for some £, e Irr(G 19). Now
i°Gw% °Gm = bi>G, n = W, «fl]
and the result follows. |
(11.36) lemma Let Z £ Z(r) be such that T has the projective lifting
property for G = T/Z with respect to the natural homomorphism n: T -» G.
Suppose K< r, H g r, M = T and H n K 2 Z. Let 9e Irr(H/Z) and
assume HnKs Z(9). Then there exists / e Irr(r) such that K £ Z(/) and
XH = 19 for some linear 1 e Irr(H).
Proof Let 9) be a representation of H which affords 9. Let L =
H n X <i H. Since L £ Z(9), 9)L is a scalar representation and we can apply
Corollary 11.10 to H/(L n ker 9) and construct a projective representation
3E of H/L such that 9)(/i) = X(hL)/i(h) for some function /z: H -> C\
Now let t: G -» H/L be the composite of the natural maps
G = T/Z ^ T/K = HK/K s H/L
so that x(hkZ) = hL for heH and fceX, Let 3E* be the projective
representation of G obtained by composing t with 3E so that X*(hkZ) = 3E(/iL) =
Now lift 3E* to the representation 3 of T so that 3M = 3E*(xZ)v(x) for
x e r, where v: T -» C *. This yields
3(hk) = y(h)fi(hriv(hk)
for he H and keK. Let 3 afford /.
Restricting the above to H, we have 3W = (9(h)i4hy1v(h) and hence
l(/i) = /z(/i)_1v(/i) defines a linear character lelrr(H) and Xh = ^1. Thus
Zelrr(r).
For /c e K we have
3(/c) = 9)(l)/z(ir1v(/c))
which is a scalar matrix. Thus K £ Z(#) and the proof is complete. |
The next result is a generalization of Theorem 11.34 which includes
Berger's Theorem 11.33.

194
Chapter 11
(11.37) theorem Let L^H <G with L~^G and G/L solvable. Let
9 e Irr(H), with 9G = x e Irr(G). Then there exists M oG such that M 2 L
and xm is not homogeneous.
Proof We may assume that there do not exist subgroups L0 £ H0 < G
and 90 e Irr(H0) such that L < L0 -a G and (90)G = /. We may also assume
that H is maximal in G. We have L = coreG(H). We assume that %L is
homogeneous and write %L = acp with <p e Irr(L) so that (G, L, <p) is a character
triple and 9 e Irr(H | <p).
It follows from Lemma 11.35 that the hypotheses of the theorem remain
invariant if (G, L, <p) is replaced by an isomorphic triple, and similarly for the
conclusion. Therefore, by Theorem 11.28, we may assume that q>{\) = 1.
Thus L £ Z(x) and in particular, L £ Z(9).
Let (r, 7t) be a finite central extension of G having the projective lifting
property. Let Z = kern and identify G with T/Z via n. For subgroups 1/ £ G,
let I/* denote the inverse image of 1/ in T so that t/*/Z = U and G* = T.
Let X/L be a chief factor of G so that X/L is abelian and K £ H. Thus
KH = GandKnH<KHso that X n H = L. We have 9 e Irr(H*/Z)
and L* £ Z(9). Also X*H* = T and X* n H* = L* 2 Z. Lemma 11.36
thus applies and yields r\ e Irr(r) such that h\H* = 9 for some linear k e Irr(H*)
andX* £ Z(rj).
We have
Z = 9r = (l^H.)r = Fri
and hence lr e Irr(r). Since 1L. e Irr(L*), Theorem 11.34 applies and we can
find M ~3. L* with M<T and (Ar)M not homogeneous.
If M = L*, let vt and v2 be distinct irreducible constituents of (lr)M-
Let y. be an irreducible constituent of rjL*. Since L* £ X* £ Z(>/) we have
/z(l) = 1 and thus /zvt and fiv2 are distinct irreducible constituents of (rjXr)L*
= Xl*- This is a contradiction since /L* is homogeneous.
Therefore, M > L*. Since (lr)M is not homogeneous, Theorem 6.11
yields a subgroup T S M with T < T and \}i e Irr(T) such that i/fr = lr. Thus
z = r\lT = #r = (riT\l/)r.
Since Z £ ker i we have Z £ Ysx{y\t$) and thus in G, % is induced from
the proper subgroup T/Z < G. Also, T/Z 2 M/Z > L and this contradicts
the first sentence of the proof. It follows that %L is not homogeneous and the
proof is complete. |
Problems
(11.1) Let a, /? e Z(G, F *), where F is a field. If a and /? are equivalent, show
that F'lG] s f*[G].

Problems
195
(11.2) Let aeZ(G, A) and let geG. Show that a(x, y) = a(y, x) for x,
Note In general, xy = yx does not imply that a(x, y) = <x(y, x).
(11.3) Let a e Z(G, A) and let x,yeG commute. Assume that A is divisible.
Show that a(x, y) = a(y, x) iff the restriction of a to <x, y) is equivalent to the
trivial factor set.
Hint Use Lemma 11.16.
(11.4) Let a e Z(G, A). Say that geG is a-special if a(g, c) = a(c, g) for every
c e CG{g). Show that if g is a-special, then so is every conjugate of g in G.
Hint Let (r, 7t) be as in Lemma 11.16 with respect to G, A, and a. Then
7t(x) is a-special iff 7t(Cr(x)) = CG(n(x)\
Note If a and /? are equivalent, then geG is a-special iff it is /^-special.
Thus one can speak of a-special elements for a e H(G, A).
(11.5) Let a e Z(G, A) and let X be a conjugacy class of G. Show that X
consists of a-special elements (see Problem 11.4) iff there exists a function
\i: jf -»/I such that
l4gWg, h) = /i(ghMh, gh)
for all g e X and h e G.
Hint See the hint for Problem 11.4. Consider the conjugacy class of xg
in T, where xg is the coset representative of A in r corresponding to g.
(11.6) Let (r, 7t) be a finite central extension of G with ker n = A. Choose a
set of coset representatives for A in r and let a e Z(G, A) be as in Lemma 11.9.
Let X e A. For each / e Irr(r 11), choose a representation 9) affording / and
construct the projective C-representation X as in Lemma 11.10. Show that
this defines a bijection of Irr(r | X) onto the set of similarity classes of
irreducible projective C-representations of G with factor set 1(a).
(11.7) Let a e Z(G, C *). Show that Ca [G] is semisimple.
Hint Show that £ (dim M)2 = | G |, where M runs over a representative
set of irreducible Ca[G]-modules. Use Problem 11.6.
(11.8) Let a e Z(G, F *). Show that dim Z(Fa[G]) is equal to the number of
conjugacy classes of a-special elements in G.
Hint Use Problem 11.5.
(11.9) Let (G, N, 9) be a character triple. If g e G/N, say that g is ^-special if 9
is extendible to <iV, g, c> for all c e G with [g, c] e N. Check that this is well

196
Chapter 11
defined. Let peH(G/N,C*) be as in Theorem 11.7. Show that geG/N is
9-special iff it is /^-special. (See the note following Problem 11.4.)
Note It is clear (without appeal to Problem 11.4) that conjugates of
9-special elements in G/N are 9-special.
(11.10) (Gallagher) Let(G, N, 9) be a character triple. Show that | Irr(G 19) |
is the number of 9-special conjugacy classes of G/N.
Hint Several of the previous problems are relevant.
(11.11) The character triple (G,N,9) is fully ramified if there exists
X eIrr(G|9) such that (x(l)/9(l))2 = \G:N\. Suppose (G, JV, 9) is fully
ramified and G > N. Show that no cyclic subgroup of G/N can be its own cen-
tralizer in G/N.
Hint Use Problem 11.10.
(11.12) Let (G,N,9-) be a character triple. For x,yeG with \_x,y~\eN,
define the complex number <x, y^ as follows: Let \ji be an extension of 9 to
H = (N, y). Write ipx = ipl for X e ln(H/N) and put «x, y» = X(y). Show
that C> is uniquely defined and that
(a) «x, y» = «xn, >>m» for n, m e N\
(b) (xlX2, y» = «*!, y»«x2>>» whenever [xu yl [x2, y'] e N;
(c) «*, y» = 9([x, y]) if 9 is linear.
Note By (a) we can view <,> as being defined on commuting pairs of
elements of G/N.
(11.13) Let (G, N, 9) be a character triple. If N £ H £ G, i/f eCh(H|9)
and g = gN e G/N we define \pB e Ch(HB | 9) by \pB(hB) = \p(h). Check that this
is well defined. We say that the isomorphism.
(T,(7):(G,JV,9)^(r,M,<p)
is strong provided (o„(\li))m = aH9(\jiB) for all g e G/N, all H with N £ H £ G
and all i/f e Ch(H 19). Show that the isomorphism constructed in Theorem
11.28 is strong.
(11.14) In the notation of Problem 11.12, show that <x,.y> = (.y, x»-1
and «x, yty2^ = «x, y^ix, y2» if [x, y{\, [x, y2] e N.
Hint Viewing (,)as defined on commuting pairs of elements of G/N,
it is invariant under strong isomorphisms of character triples. (See Problem
11.13.)

Problems
197
(11.15) Let (G, N, 9) be a character triple.
(a) If g e G and <g, x> j= 1 for some x e G, show that /(#) = 0 for all
ZeIrr(G|9).
(b) If g e G/N, show that g is 9-special iff <g, x^ = 1 for all x e G such
that <g, x> is defined.
See Problems 11.12 and 11.9 for definitions.
(11.16) Let £ be elementary abelian of order p". Show that M(E) is
elementary abelian of order pn{n~1)/2.
Hint Consider a Schur representation group for E to get \M(E)\
< pn<n-1'/2. For equality, let {x,|l < i < n} be a generating set for E.
Define ay: E x E -* C by a.i$\ x/*, Y[ */*) = e"'bj> where e is a primitive
pth root of unity.
(11.17) Let G = A5. Show that |M(G)| = 2.
Note In fact, | M(A„) \ = 2 for all n > 4 except for n = 6, 7, where the
Schur multipliers have order 6.
(11.18) Let G = CH, where H and C are cyclic, C<G,andHnC = Z(G).
Show that M(G) = 1.
Hint If T is a Schur representation group for G, show that \G'\ = |F|.
(11.19) Let (r\, 7^) and (r2, n2) be Schur representation groups for G.
Define r £ r\ x T2 by r = {(*, y) | tc^x) = n2(y)} and show that Ft s F s F2.
Hint Let /lj = ker nj and define 7t: r -» G by 7t(x, y) = 7ti(x), so that
ker n = /lt x 42 and (F n) is a central extension of G. Use Theorem 11.19
to show that |F| < \r\ |. Note that |F: F n (/4t x A2)\ = |G'|.
iVote One cannot conclude that r\ s T2 as the two nonabelian groups
of order p3 show. If G = G' then rt = Tj and r\ = F2 in this case.

Character degrees
Let c.d.(G) denote* the set {x(l)|xeIrr(G)}, where G is a group. In this
chapter we consider what can be said about G when c.d.(G) is known. We have
already seen in Theorem 6.9 that if every/ 6 c.d.(G) is a power of the prime p
then G has an abelian normal p-complement. The first results in this chapter
consider a weaker hypothesis, namely that p divides / for every / e c.d.(G)
with/ > 1.
(12.1) theorem Fix a prime p. Write ^(G) = {xeIrr(G)|p/Cx(l) and
PMX)}- Let s(G) = £y(G) X(l)2- Then
|Op(G)| = s(G)modp.
Proof Let N = Op(G). Since Op(JV) = N, N can have no irreducible
character with determinantal order divisible by p. Thus Irr(iV) =
<?(N) u {tAeIrr(JV)|p|t/<l)} and hence \N\ = £,6lrr(K)M)2 = s(N) mod p.
It will therefore suffice to show that s(N) = s(G) mod p.
Now G acts on Sf(N) and the resulting orbits are of p-power size. Let
Sf0 = {\jieSf(N)\\j/ is G-invariant}. Since all characters in an orbit have
equal degrees, it follows that s(N) = ^£j,0 t/^(l)2 mod p. We show that
restriction defines a one-to-one map of ^(G) onto Sf0 and this will complete the
proof.
If /e^(G) then (z(l), \G:N\) = 1 and so xweIrr(iV) by Problem 6.7.
Thus xNeSf0. Conversely, if \\i e ^0 then by Corollary 6.28, \ji is extendible to
G and a unique extension of \ji lies in ^(G). The result now follows. |
* The initials c.d. stand for "character degrees."
198

Character degrees
199
(12.2) corollary (Thompson) Suppose p|x(l) for every nonlinear
X e Irr(G), where p is a prime. Then G has a normal p-complement.
Proof In the notation of Theorem 12,1, all / e £f(G) are linear and have
kernels which contain Op'(G). It follows that ^(G) = Irr(G/G'Op'(G)) and
s(G)= |G:G'Op'(G)|. Thus PJfs(G) and hence by 12.1, p^|Op(G)|. Thus
Op(G) is a normal p-complement for G. |
The following lemma is very useful for inductive proofs of theorems
giving information about G when c.d.(G) is known.
(12.3) lemma Let G be solvable and assume that G'is the unique minimal
normal subgroup of G. Then all nonlinear irreducible characters of G have
equal degree/and one of the following situations obtains:
(a) G is a p-group, Z(G) is cyclic and G/Z(G) is elementary abelian of
order/2.
(b) G is a Frobenius group with an abelian Frobenius complement of
order/ Also, G' is the Frobenius kernel and is an elementary abelian p-group.
Proof If Z(G) t^ 1, we must have that Z(G) is a cyclic p-group and
G' £ Z(G) with | G' | = p. Every / e Irr(G) with #(1) > 1 is faithful and satisfies
X(l)2 = | G: Z(G)| by Theorem 2.31. If x, y e G, then since [x, y] e Z(G), we
have [xp, y] = [x, y~]p = 1 since | G' | = p. Thus x" e Z(G) for all x e G. This
completes the proof in situation (a).
Now suppose that Z(G) = 1. Certainly, G' is an elementary abelian
p-group for some prime p. Choose q \ \ G |, a prime different from p and let
QeSyl,(G). Then G'QoG and it follows by the Frattini argument that
G = G'N where N = NG(Q). Now N n, G' -a N since G'^aG and
N r\G' <G' since G' is abelian. Thus N n G' <a G. Now Q ^ G and so
N < G and N ^ G'. By the minimality of G', it follows that N n G' = 1 and
thus N is abelian.
If 1 t^ x e N, then N normalizes CG.(x) since N centralizes x. Since
CG.(x) o G' we have CG(x)<a NG' = G. If CG(x) ¥= 1, then x centralizes G'
and it follows that x eZ(G), a contradiction. It follows that CG(x) = 1 and
thus G is a Frobenius group with kernel G' and complement N by Problem
7.1. We conclude that all nonlinear / e Irr(G) are of the form XG for linear
X e Irr(G'). Thus | G: G' | = | N | is the common degree of all nonlinear
irreducible characters of G. The proof is complete. |
The way Lemma 12.3 is applied in practice is the following. Given non-
abelian G, let K <i G be maximal such that G/K is nonabelian. Then (G/K)'

200
Chapter 12
is the unique minimal normal subgroup of G/K. Thus either G/K is non-
solvable or else satisfies the hypotheses of Lemma 12.3.
If/is a character ofG, we introduce the notation V(x) = (geG\x(g) ^ 0>.
Then \(x), the vanishing-off subgroup, is the smallest subgroup, KsG such
that x vanishes on G — V. Of course, V(/) <a G.
The relevance of \(x) to character degrees is this. Suppose / e Irr(G) and
V(z) £ JV«a G. Then by Lemma 2.29, we have [xn, %N] = \G:N\. Now
write Xn = e £j= i 9( where the 9( e Irr(JV) are distinct and of equal degree
(Theorem 6.2). We have |G:JV| = [xn,Xn1 = e2t and /(l) = et9(l), where
9 is one of the 9,. It follows that |G: JV|9(1)2 divides /(l)2. In particular,
IG: JV | < x( I)2 and /(1) is divisible by every prime divisor of | G: JV |.
The following result is useful when K -a G and G/K satisfies the
conditions of case (b) of Lemma 12.3.
(12.4) theorem Let K o G be such that G/K is a Frobenius group with
kernel N/K, an elementary abelian p-group. Let \ji e lrr(N). Then one of the
following holds.
(a) |G:JV|i/f(l)ec.d.(G).
(b) V(i/0 s K and thus \N:K\ divides \}i(l)2.
Proof For 1 e Irr(JV/K), let T(l) denote 7G(i/fl), so that T(l) 2 N. If
T(l) = JV for some 1, then (li/0G e Irr(G) and hence \G:N\ip(l)e c.d.(G). We
suppose then, that T(l) > JV for all k
Let W = \(ip)K so that K £ W £ JV and let S = NG(W) 2 JV. Since
IG(\I/) normalizes V(t^), we have 7G(i/f) £ S. However, V(i/f) = \(X\ji) for all
1 e Irr(JV/X) since the 1 are linear, and it follows that T(l) £ S for all X.
Now view JV/X as an F[S/JV]-module, where F is the field with p
elements. Since G/K is a Fro benius group, we have | G/N | divides | N/K \ — 1
and thus pJf\G/N\. In particular, pJf\S/N\ and hence N/K is completely
reducible as an F[S/JV]-module by Maschke's Theorem 1.9. Since W/K is
a submodule, we can write N/K = (W/K) x (U/K), where S normalizes U.
Now suppose W > K so that 1/ < JV. Let leIrr(JV/t/) with 1 ^ 1^.
Since G/K is a Fro benius group, we have 7G(1) =(JV and thus 7S(1) = JV.
Therefore, |Irr(JV/t/)| > 1 + |S/JV|. We claim that there exist distinct
l,fie Irr(JV/t/) with T(l) n T(/z) > JV. If not, then since JV < T(l) £ S for
all 1 e Irr(JV/t/) we have
\S/N\ - 1 £ X(|T(1)/JV| - 1) :> |Irr(JV/t/)| > |S/JV|,
where the sum runs over X e Irr(JV/t/). This contradiction shows that 1,
/z e Irr(JV/t/) exist with T(l) n T(/z) > JV and 1 ^ /z, as claimed.
Now let x e (T(l) n T(/z)) - JV and write v = 1/7. Then A* = /z*v* and
ipX = (\pX)x = \pxXx = (\px/ix)vx = (\pfiYvx = ip/ivx

Character degrees
201
and thus \jiv = \jivx and we have \(\p) £ ker(vv*). Also U £ ker(vv*) and thus
JV = V(t/^)t/ £ ker(vv*) so that vv* = 1^ and v* = v. Since X ± fi, we have
v ^ \N and 7G(v) = JV. This contradicts x £ JV and thus proves that W = K
and \(\p) £ K as desired. That this implies |JV: X||i^(l)2 follows from the
remarks preceding the statement of the theorem. |
In particular, in the situation of Theorem 12.4, we have for \ji e Irr(JV)
that either | G: JV|i/f(l) e c.d.(G) or plt/^l). This most important consequence
of 12.4 could have been obtained with somewhat less work than was needed
to prove the full strength of the theorem.
(12.5) theorem Let c.d.(G) = {1, m}. Then at least one of the following
occurs.
(a) G has an abelian normal subgroup of index m.
(b) m = pe for a prime p and G is the direct product of a p-group and an
abelian group.
Proof Suppose there exists K<G such that G/K satisfies conclusion
(b) of Lemma 12.3. Let N/K = (G/K)' so that \G: JV| = m and N/K is a
p-group. Since G/K is a Frobenius group, m\(\N/K\ — 1) and so pjfm. We
claim that JV is abelian.
Let \ji e Irr(JV) and let / be an irreducible constituent of \pG. Then x(l) = 1
or m and i/f(l)|x(l). In particular, p^^l). By Theorem 12.4, mi/f(l)ec.d.(G)
and thus i/f(l) = 1. This establishes the claim and situation (a) of the theorem
holds in this case.
We now suppose that no K < G as above exists. Let n be the set of prime
divisors of m. By Corollary 12.2, G has a normal p-complement for every
p e n. If | n \ > 1, then G has no irreducible character of p-power degree and
hence G/Op(G) is abelian for all p. It follows that G' is a 7t'-group. Since the
elements of c.d.(G') divide elements of c.d.(G), we conclude that c.d.(G') = {1}
and G' is abelian. In particular, G is solvable. Now let K -a G be maximal such
that G/K is nonabelian. Thus G/K satisfies the hypotheses of Lemma 12.3. By
assumption, we are in case (a) of the lemma and this contradicts m not being
a prime power.
We may now assume that n = {p}. Let A be the normal p-complement of
G so that A is abelian. Let k e \xx(A) and let T = IG(ty- If i> is any irreducible
constituent of 1G, then by Clifford's theorem, \G:T\ divides \p(l), and thus
| G: T | <, m. Now (1T)G is not irreducible and has degree <m. It follows that
all of its irreducible constituents are linear and thus G' £ ker((lT)G) £ T.
(Note that we have just done Problem 5.14(c).)
Now if D = f|;uirru) IM, then A £ Z(£) for all £eIrr(D) and thus
A £ Z(£>). Thus D = A x P for P e Syl^D). Since G' £ D, we have D -a G
and hence P -a G. If G' £ P then [G, /4] £ A r\ P = 1 so that A £ Z(G)

202
Chapter 12
the result follows. Otherwise, let K 2 P, K <i G with K maximal such^ that
G/K is nonabelian. Since G is solvable, the hypotheses of Lemma 12.3 are
satisfied. By assumption we are in case (a) of the lemma and G/K is a p-group.
Thus K 2 Op(G) = A. We have K 2 AP = D 2 G' and this contradicts
G/K being nonabelian and completes the proof. |
(12.6) corollary If | c.d.(G) | = 2, then G' is abelian.
Proof Let c.d.(G) = {1, m). If A -a G with |G : /4| = m then G/4 can
have no nonlinear irreducible characters and so is abelian. Thus if case (a) of
Theorem 12.5 holds, we are done.
In case (b) of 12.5, G is nilpotent and hence is an M-group by Corollary
6.14. The result follows by Theorem 5.12. |
In the case that m is a prime we can sharpen Theorem 12.5 to give a
necessary and sufficient condition on G that c.d.(G) = {1, m). The following
results prove more than is needed for this and will be used again. Theorem
12.7 is actually a generalization of Theorem 6.16.
(12.7) theorem Let JV o G and suppose 9t, 92 e Irr(JV) are invariant in G
and 9^2 e Irr(JV). Let /, be an irreducible constituent of (9j)G for i = 1, 2
and let \ji be an irreducible constituent of X\li- Then
W)Zi(l) St /2(l)9i(l)2-
Proof We have 0 =£ [Zifo, <A] = Lfo, "AZi] and thus Xi is a constituent
of tftfi. Also (za)w = to(l)/S2(l))S2 and it follows that
Z2W2O) =S L^OMtfi)*] = [92(Zi)iv, >M.
However, (x\)n = (Zi(l)/#i(l))#i and this yields
Z2O) < zi(i) r9 9 , -,
Now 9^2 is the unique irreducible constituent of 1/^ and thus [9i$2, *Pn]
= t/^(l)/91(l)92(l)- Substitute this in the above and simplify to obtain the
result. I
(12.8) corollary Let JV o G and suppose /? e Irr(G) with JV £ Z(/J). Let
Selrr(JV). Then there exists an integer b such that £>9(l)ec.d.(G) and
b2 > /?(l)t, where t = |G:/G(S)|.
Proof Let T = 7^9) and let y be an irreducible constituent of fiT. Then
/? is a constituent of yG and so /?(1) ^ y(l)r. Also, JV £ Z(y) and we write
yN = y(lU. Let ^ be an irreducible constituent of 9-T and let r\ be an irreducible

Character degrees
203
constituent of £y. Since 91 e Irr(iV), Theorem 12.7, applies and so
£(1>7(1) > y(l)9(l)2.
Thus
ZG(l)rjG(l) > y(l)t29(l)2 > j8(l)t9(l)2.
Now T = 7G(9) = 7G(91) and hence £G, rjG e Irr(G) by Theorem 6.11. Let / be
whichever of £G, r\G has larger degree. Then /(1) = b&( 1) for some integer b and
ft29(l)2 = Z(l)2 > ^(1)^(1) > Al)t9(l)2
and the result follows. |
(12.9) corollary Let c.d.(G) = {1, p}, where p is a prime. Then there
exists abelian A <i G with | G : /I | = p or p2.
Proof By Theorem 12.5, we may assume that G is a p-group. Let K<G
be maximal such that G/K is nonabelian and let Z/K = Z(G/X). By Lemma
12.3 it follows that |G : Z\ = p2.
Let 0eIrr(G/K) with 0(1) = p and let 9eIrr(Z). Then since Z <= Z(0),
Corollary 12.8 yields an integer b such that b2 > 0(1) = p and 69(1) e c.d.(G).
This forces 9(1) = 1. Thus Z is abelian and the proof is complete. |
(12.10) lemma Let A ^ Gbe abelian and let b = max(c.d.(G)). Then
(l/|^|)XlCG(a)|>|G|/ft.
aeA
Proof Since | CG(a) I = £* I *(fl) 12 for X e Irr(GX we have
(l/l^l)ZlCG(a)| = (l/|^|)XZlz(a)l2 = Z[^,zJ-
" X a x
However, iA is the sum of x(l) linear characters and hence [%A, %A~] > /(l).
Thus
(l/l^l)Z|CG(a)|^Xx(l)-
a X
Furthermore, \G\ = £z Z(l)2 ^ b £z Z(l) and thus £z /(l) > \G\/b and the
result follows. |
(12.11) theorem Let G be nonabelian and let p be a prime. Then cd.(G)
= {1, p} iff one of the following holds.
(a) There exists abelian A <i G with \G: A\ = p.
(b) |G:Z(G)| = p3.
Proof If (a) holds, then /(1) | p for every / e Irr(G) by Ito's Theorem 6.15.
Since G is nonabelian, c.d.(G) = {1, p}.If |G :Z(G)| = p3, then x(l)2 < p3 by

204
Chapter 12
Corollary 2.30 and x(l)|p3 by 6.15 (or 3.12). Again it follows that c.d.(G)
= {l,p}.
Conversely, suppose c.d.(G) = {1, p} and assume that (a) is false. We may
assume that G is a p-group. By Corollary 12.9, there exists abelian A <i G
with | G: A \ = p2. By Lemma 12.10, we have
(l/|,4|)X|CG(a)|>|G|/p.
aeA
Now A acts by conjugation on G — A and by Corollary 5.15, the number
of orbits of this action is
(i/|/H)E(|Cri(fl)|-|/i|)£(|G|/p)-M|.
asA
It follows that the average size a of these orbits satisfies
*<(\G\-\A\)/((\G\/p)-\A\) = p+l<p2
and so A has an orbit of size 1 or p on G — A, Since G does not have an abelian
subgroup of index p, A = CG(A) and thus there exists x e G — A in an orbit
of size p. Let K = (A, x>. Then Z(X) = CA(x) has index p in A and index p3
in G. We shall show that Z(G) = Z(X) to complete the proof.
Let Z = Z(X). If / e Irr(G) and %K is irreducible, then Z £ Z(/) and
[G, Z] c ker /. On the other hand, if %K reduces, then all irreducible
constituents are linear and K' c ker /. Suppose [G, Z] > 1. We also have
K' > 1 but K' n[G,Z]^ ker / for every / e Irr(G) so that K' r\[G,Z] = 1.
Now X'[G, Z] is the direct product of two nontrivial groups and so has an
irreducible character 9 with K' £ ker 9 and [G, Z] £ ker 9. Let / be an
irreducible constituent of 9G. Then K' £ ker / and [G, Z] £ ker /. This
contradiction shows that [G, Z] = 1 and completes the proof. |
We remark that the last several sentences of the proof could be replaced by
an appeal to Problem 5,26.
Next we refine Theorem 12.5 in a somewhat different direction. Namely,
if c.d,(G) = {1, m} and G has no abelian normal subgroup of index m, then
the nilpotence class of G is < 3.
(12.12) lemma Let A<i G with A abelian and G/A cyclic. Then \A\ =
\G'\\AnZ(G)\.
Proof Let G/A = (Ag} and let a: A -> A be defined by o(a) = a~la".
Then a is a homomorphism and ker a = CA(g) = A n Z(G). Let / be the
image of a. Then g e N(7) and so I <i G. Since G = (A, g) and g centralizes
A mod 7, it follows that G/I is abelian and G' s /. Clearly, I £ G' and hence
|/l| = |ker(T||/| = |/lnZ(G)||G'|
and the proof is complete. |

Character degrees
205
(12.13) lemma Let c.d.(G) = {1, m} and suppose A, B £ G are abelian of
index m with A ± B. Then \G'\ < m and G' £ Z(G).
Proof If X £ G with | G: X | < m, then G'sK and hence K<G
by Problem 5.14(c). In particular, A <aG. Let A < K £ G, then K<G
and thus K'<G. If X' < G', let 1G. # /z e Irr(G'/K') and let / be an
irreducible constituent of fiG. Now K' £ ker / and so %K has a linear constituent
1 and /(I) < 1G(1) = |G: X| < m. Thus *(1) = 1 and G' £ ker /. It follows
that G' £ ker fi, contradicting the choice of fi. Therefore K' = G'.
Since A ^ B, let b e B - A and let X = </l, b). Thus K//1 is cyclic and
\G'\ = \K'\ = \A:A nZ(X)| by Lemma 12.12. However,
A n Z(X) 2 /4 n B
and so |-4;/ln Z(X)| <|/lMnB|<|G:B| = m and the first assertion is
proved.
For g e G, the conjugacy class of g is contained in the coset gG' and so has
size < \G'\ < m. Thus \G:C(g)\ < m and G' £ C(g) by the first sentence of
the proof. Since g is arbitrary, we have G' £ Z(G) and the proof is complete. |
(12.14) theorem Let c.d.(G) = {1, m} and suppose that G has no abelian
normal subgroup of index m. Then [G, G'] £ Z(G), that is, G is nilpotent of
class ^3.
Proof By Theorem 12.5 we may assume that G is a p-group. If G has a
faithful irreducible character / then / = 1G for linear 1 e Irr(X) and K £ G
since G is an M-group. We have | G: K | = /(l) = m and so X o G by
Problem 5.14(c). Thus all irreducible constituents of Xk are linear and K' £ ker /
= 1, a contradiction.
Thus G has no faithful irreducible character and hence Z(G) is not cyclic.
Let Zt, Z2, Z3 £ Z(G) be distinct subgroups of order p. If G/Zt is of nil-
potence class <3, then [G, G, G, G] ^ Zt. If this happens for two distinct
Z;, we conclude that [G, G, G, G] = 1 and we are done.
Assume then that G/Zt and G/Z2 (say) do not have class < 3. Working by
induction on \G\, it follows that there exist A, B<i G with |G: A\ = m =
| G: B| and /4' £ Zt and B' £ Z2. Let Z = Zt Z2 so that AZ/Z and BZ/Zare
abelian. Since G/Z is nonabelian, we have m e c.d.(G/Z) and so | G: AZ\ >m
and | G: BZ| > m. It follows that Z £ /I n B.
UA/Z = B/Z, then A' £ Zt n Z2 = 1 and /I is abelian, a contradiction.
Thus /4/Z ^ B/Z and Lemma 12.13 applies to yield G'Z/Z £ Z(G/Z). Thus
[G', G] £ Z £ Z(G) and the proof is complete. |

206
Chapter 12
We now consider groups G for which |c.d.(G)| = 3. Although the
information obtained is not as detailed as when |c.d.(G)| = 2, we do prove a
result analogous to Corollary 12.6.
(12.15) theorem Let |c.d.(G)| = 3. Then G" = 1, that is, G is solvable of
derived length < 3.
Before proceeding with the proof of this result we make some general
remarks. As the group A5 shows, we cannot conclude from |c.d.(G)| = 4
that G is solvable. It has been conjectured by G. Seitz that for solvable
G, d.l.(G) < |c.d.(G)|. (Here, d.l.(G) is the derived length of G; the smallest
integer k for which G(k\ the kih commutator subgroup of G, is trivial.) By
Theorem 5.12, this conjecture holds for M-groups. It has also been proved
when | cd.(G) | = 4 (S. Garrison) and when | G | is odd (T. Berger). It is known
that for any solvable group, d.l.(G) < 3|c.d.(G)|.
(12.16) lemma Let JV-aGwith(|JV|,|G:iV|)= 1. Then
|c.d.(JV)|<|c.d.(G)|.
Also if G/N is supersolvable and N is solvable with d.l.(iV) < | c.d.(iV) |, then
d.l.(G)< |cd.(G)|.
Proof Let n be the set of prime divisors of | JV |. If / e Irr(G), let 9 be an
irreducible constituent of %N. By Corollary 11.29, /(1)/9(1) divides |G:JV|.
Since 9(1) divides |JV|, it follows that 9(1) is exactly the rc-part of x(l). Since
every 9 e Irr(iV) arises this way, we see that c.d.(JV) is exactly the set of rc-parts
of the elements of c.d.(G). The first assertion follows.
Now assume that N is solvable and G/N is supersolvable. It follows that
G/N' is an M-group by Theorems 6.22 and 6.23 and thus d.l(G/N') <
\c.d.(G/N')\ by Theorem 5.12. We may assume that N' > 1 and observe that
d.l.(G) <; d.l.(G/JV') + d.l.(JV') < |c.d.(G/JV')| + d.l.(iV) - 1
< |c.d.(G/JV')| + |c.d.(JV)| - 1,
where the last inequality follows from the assumption that d.l.(N) < \ c.d.(N) \.
Now every/ ec.d.(G/iV') divides |G:iV| by Ito's Theorem 6.15 and thus
the 7t-part off is trivial. We conclude from the first part of the proof that
I c.d.(JV) | < | c.d.(G) | - | c.d.(G/iV') | + 1 and the result follows. |
Proof of Theorem 12.15 Let c.d.(G) = {1, m, n}. If (m, n) ^ 1, then G has
a proper normal p-complement N for some prime p by Corollary 12.2. By
Lemma 12.16, |c.d.(N)| < 3 and so N is solvable and d.l.(N) <, | c.d.(JV) | by
induction if | cd.(iV) | = 3 and by Corollary 12.6 if | cd.(iV) | < 3. In this case
we are done by Lemma 12.16.

Character degrees
207
Assume now that (n, m) = 1. Suppose there exists K<G with G/K
solvable and nonabelian. We may assume that G/K satisfies the hypotheses
of Lemma 12.3, If G/K is a p-group then (say) n is a power of p. Let z e Irr(G)
with %(1) = m. Since pjfm, we have %K e Irr(X) and thus by Gallagher's
theorem (Corollary 6.17), //? e Irr(G) for /? e lrr(G/K), This is a contradiction
since mn $ cd.(G).
Therefore G/K is a Frobenius group with kernel N/K = (G/K)', where
N/K is a p-group and \G : N\ = n (say). Suppose \ji e lrr(N) with i/f(l) > 1.
Since ni/f( 1) £ c.d.(G), we conclude from Theorem 12.4 that p | \j/{ 1). Let z be an
irreducible constituent of \jiG. Then plz(l) and since pjfn, we have z(l) = m.
Thus Xn is irreducible since (|G:JV|, x(l)) = 1. We conclude that %N = \ji
and \p(l) = m. Thus c.d.(JV) = {1, m} and dl(JV) < 2 by Corollary 12.6.
Since G/JV is abelian, we have d.l.(G) < 3 as desired.
Suppose now that no such K exists. If/ elrr(G) with z(l) ^ l,thenG/kerz
is nonabelian and thus is nonsolvable. However | c.d.(G/ker /) | < 3 and
working by induction on \G\ we must have ker / = 1. Therefore, every
nonlinear irreducible character of G is faithful.
Now let n < m and let %eIrr(G) with z(l) = «• By Theorem 4.3, each
irreducible character of G is a constituent of z' for suitable integers t. Let t be
minimal such that z' has an irreducible constituent \p of degree m. Thus
Ez£, <A] t6 0 for some irreducible constituent £ of z'~ '• Now <j;(l) j= 1 or else
z£ is irreducible forcing z£ = tA which is not the case. Also ^(1) # m by the
minimality of t. Thus £(1) = n.
Since [zi, t/f] = 0 for linear 1, we conclude that [1, i/^] = 0 and \p% has
no linear constituents. Thus
a b
•AZ = E & + E fj'
i= 1 j= 1
where £t, >/; e Irr(G), £,-(1) = n, >/j{1) = m and a > 1 since £ is one of the <!;,.
Comparing degrees yields mn = an + bm and since (m, n) = 1, we have m\a.
Since ft > 0 and a > 0, this yields a = m and ft = 0 and t/^z = E?= i &•
We claim that each ^ is of the form 1;Z for some linear character, Xt. It
suffices to find a linear constituent of xii ■ Suppose (for some i) that x£t has no
linear constituent. Then
r s
j= 1 k=l
where Zj, iK £Irr(G), z/1) = » and t/^(l) = w. Thus n2 = rn + sm and n\s.
However, 0 j= [t/^z, ^] = [i/f, z£/] and \ji is one of the i/ft. Thus s > 1 and
hence s > n. Since r > 0, this yields n2 > nm and n > m, a contradiction.
We now have & = l,z for linear 1( and thus f^ = E £i = X E ^i an(*
hence
('/'g')(Zg') = wzG'-

208
Chapter 12
Since ker \j/ = 1, we have il/(x) ¥= m for x ¥= 1 and thus x(x) = 0 for all
x e G' - {1}. It follows that [/G-, 1G.] ^ 0 and thus G' £ ker x- Thus
contradiction completes the proof. |
Another technique for studying character degrees is based on the
following lemma.
(12.17) lemma (Garrison) Let Hs G and let 9e\xx(H). Suppose for
every irreducible constituent / of 9G that Xn = $• Then \(9) <i G.
Proof We have (9G)„ is a multiple of 9 and so (9G)H = |G: H|9. Let
heH with 9(/j) ^ 0 and let S = {x eG|/i*eH}. By definition of 9G we have
(l/\H\)ZHhx) = SG(h)=\G:H\$(h)-
xeS
However, since 9G(/i*) = 9G(h), it follows for x e S that 9(h*) = 9(/j). This
yields \S\9(h) = \G\9(h) and since 9(h) ± 0, we have \S\ = \G\ and S = G.
Thus ft'eH for all g eG and 9(hg) = 9(h) ± 0. Thus G leaves the generating
set for \(9) invariant under conjugation and the result follows. |
(12.18) lemma Let x g Irr(G) and let ker / < JV o G. Then
ker x < N n \(x).
Proof We may assume ker / = 1. If V(/) n JV = 1 then / vanishes
on N — {1} and so [/^, 1N] ^0. This forces JV £ ker /, a contradiction. |
In the following, F(G) denotes the Fitting subgroup of G, the (unique)
largest normal nilpotent subgroup.
(12.19) theorem (Broline-Garrison) Let/elrr(G)andX = ker/.Either
of the following conditions guarantees the existence of t^elrr(G) with
t/f(l)>X(l)andkert/f <K:
(a) K£F(G);
(b) K = F(G), G/K is solvable and K < G.
Proof Suppose H is a maximal subgroup of G and that KH = G. Then
Xh = 9 e lrr(H). If \\i is an irreducible constituent of 9G such that \jiH ^ 9,
then i/'h reduces and t/^(l) > 9(1) = /(l). Also, we cannot have (ker \j/)H = G
or else \jiH would be irreducible. The maximality of H thus yields ker \ji £ H
and since 9 is a constituent of i/fH, we have ker \ji £ ker 9 = H r\K < K.
Thus the result follows in this situation and we may suppose that whenever
KH = G for a maximal subgroup HsG, the character 9 = Xh satisfies the
hypotheses of Lemma 12.17. In particular, \(9) -o G.

Character degrees
209
Now we choose L <a G as follows. If K £ F(G), take L = K. If K £ F(G),
then we are in situation (b) and F(G) = K < G. Here, G/K is solvable and we
take L > K with L/K an elementary abelian chief factor of G. Thus L
is not nilpotent and we choose a nonnormal Sylow subgroup P of L. Note
that in the case K = F(G) and L > K, we must have L = KP. By the Frattini
argument, G = LNG(P) = KNG(P) and NG(P) < G. Let H 2 NG(P) be a
maximal subgroup of G. Thus G = HX and we let 9 = /H. Then V(9) -a G
and thus V(3) nL<G.
Now K r\H = ker 9 £ V(9) and so in the case K = L, we have
P ^ K r\H ^ L r\ V(9). In the case that L/K is a chief factor of G, we have
(L n H)/(ker 9)isachieffactorofHandthusL n H £ V(3) by Lemma 12.18.
Thus in any case, P s L n V(S)<GandP is Sylow in L n V(S). The
Frattini argument now yields G = (L n V(9))NG(P) £ H. This is a
contradiction and proves the theorem. |
(12.20) corollary Let /elrr(G). If either x(l) = max(c.d.(G)) or ker/
is minimal among kernels of irreducible characters of G then ker / is nil-
potent.
(12.21) corollary (Garrison) Let G be solvable and let | c.d.(G) | = n.
Then there exist JV( -a G with
1 = N0 £ Ni £ • • • £ N„ = G
such that Ni+ JNi is nilpotent for 0 < i < n.
Proof Let Nl = F(G), the Fitting subgroup. Then c.d.(G/JV!) £ c.d.(G)
and if JVt < G, then cd^G/iVJ does not contain the largest element of c.d.(G)
by Theorem 12.19. In this case, \c.d,(G/Nl)\ < n and the result follows by
induction on | G |. |
Suppose A £ G is abelian. By Problem 5.4 (or 2.9(b)), | G : A \ is an upper
bound for c.d.(G). Conversely, suppose we know max(c.d.(G)). Can we
conclude that there exists an abelian subgroup with bounded index in G? We
can, although it is certainly not true that there necessarily exists abelian
A £ G with \G:A\ = max(c.d.(G)).
We use the notation b(G) = max(c.d.(G)). Note that if H £ G and
\ji e Irr(H), then \ji is a constituent of 1h f°r some / e Irr(G), and thus
^(l)<Z(l)<ft(G)
and hence b(H) < b(G).
(12.22) lemma Let b(G) = b. Then there exists xeG- {1} such that
|G:CG(x)|<fc2.

210
Chapter 12
Proof Let k = | Irr(G) | - 1. Then
|G|-l = £z(l)2</cft2,
where the sum runs over nonprincipal / e Irr(G). Let
m = min{|G:CG(x)||xeG,x ¥= 1}.
Then | G | — 1 > mk since each of the k nonidentity conjugacy classes has
size > m. We now have kb2 > mk and the result follows. |
(12.23) theorem Let b(G) = b. Then G has an abelian subgroup of index
<(ft!)2.
Proof Use induction on b. If b = 1, the result is trivial and if b = 2,
then c.d.(G) = {1, 2} and we are done by Corollary 12.9. Assume then,
that b > 3.
Let K <i G be maximal such that G/K is nonabelian. It follows that
b(K) <, b/2 since if \ji e Irr(X) with \ji(\) > b/2, then necessarily \ji = %K for
some zelrr(G). By Corollary 6.17, we have 0xeIrr(G) for 0eIrr(G/K).
Since G/K is nonabelian, we may choose /? with 0(1) > 2 and this yields
0(1)2(1) > ft, a contradiction.
By Lemma 12.3, we have three main cases to consider namely: G/K is non-
solvable, G/K is a Frobenius group and G/K is a p-group.
Case I G/K is nonsolvable. Since b[G/K) < ft, Lemma 12.22 yields
x e G/K such that x ¥= 1 and
|(G/K):CG/K(x)|<ft2.
Let C/K = CG/K(x). If b(C) < b - 1, then there exists abelian A s C with
\C:A\ <((b- l)!)2andso|G:/4| = \G:C\\C:A\ < (ft!)2 and we are done.
Assume that b(C) = b and let \ji e Irr(C) with t/^(l) = b. Then every
irreducible constituent / of \jiG satisfies /(l) = b and thus %c = \p. It follows by
Lemma 12.17 that V(t/0 <G and thus XV(tA) <iG. If KV(t/0 > K, then
G/KV(t/0 is abelian and soC<G since KV(t/0 s C. If Z/K = Z(C/K)
then Z <i G. Also, x e Z/K and so Z > X. It follows that G/Z is abelian
and G/K is solvable, a contradiction. We conclude that V(t^) c x and hence
\C:K\ < tA(l)2 < b2 by the remarks preceding Theorem 12.4. Thus
\G:K\ <b\
By the inductive hypothesis, K has an abelian subgroup of index
<(rj>/2]!)2. For b > 5, we have ([ft/2]!)2ft4 < (ft!)2 and the result follows.
Assume then, that ft < 4. If 2 £ c.d.(G/X), then c.d.(G/X) s {1, 3,4} which
contradicts the nonsolvability of G/K by Theorem 12.15. Thus we may
choose 9 e Irr(G/X) with 9(1) = 2. We have G/ker 9 is nonabelian and hence
ker 9 = K. Let M/K = (G/K)'. Then M' £ X = ker 9 and hence 9M is

Character degrees
211
irreducible. Pick y e M/K of order 2. Since y e ker(det(9)), it follows that
yeZ(9) = Z(G/K) = 1, a contradiction. This completes case 1.
Case 2 G/K is a Frobenius group. Let N/K = (G/K)', the Frobenius
kernel. By Lemma 12.3, \G:N\e cd.(G) and so |G :N\ < ft. If b(N) < ft/2,
then application of the inductive hypothesis to N yields an abelian subgroup
A £ N with \N :A\ < ([ft/2]!)2. Since ft([ft/2]!)2 < (ft!)2 we are done.
Suppose then, that ft(N) > ft/2 and let \j/e lrr(N) with t/^(l) > ft/2. Then
ip(l)\G:N\^c.d.(G) and by Theorem 12.4 it follows that t^(l)2 > \N:K\
and hence | G: K \ < ft3. Since ft > 3, we have ([ft/2] !)2ft3 < (ft!)2 and we are
done in case 2 by applying the inductive hypothesis to K.
Case 3 G/K is a p-group. Let Z/K = Z(G/K). By Lemma 12.3, \G:Z\ =
0( l)2, where 0eIrr(G) and Z £ Z(0). In particular, | G: Z | < ft2.Ifft(Z) > ft/2,
pick i// e Irr(Z) with \p( 1) > ft/2. Then \p = Xz f°r some / e Irr(G). It follows
from Theorem 12.7 or 6.16 that /?(l)t/f(l) ec.d.(G) and this is a contradiction
since /?(1)^(1) > ft. Thus ft(Z) < ft/2 and the result follows as in previous
cases since ([ft/2]!)2ft2 < (ft!)2. The proof is now complete. |
It is apparent in the above proof that in cases 2 and 3, the inequalities
obtained are far from being best possible. The limiting factor in this proof is
the nonsolvable case 1. If one assumes that G is solvable it is possible to
obtain a better bound. It is not known what the best possible bound is either
in the general situation or for solvable groups.
The following result provides a tool which can be used to find an abelian
subgroup in G of index <ft(G)4 when G has an abelian normal subgroup with
nilpotent factor group. (Compare Theorem 12.24 with part (a) of Theorem
12.19.)
(12.24) theorem (Broline) Let /e Irr(G) and let K = ker / and F/K =
F(G/K). Suppose F is not nilpotent. Then there exists \ji e Irr(G) with
ker \ji < K.
Proof Let Q e Syl,(F) with Q -^ F and let H 2 NG(Q) be a maximal
subgroup. Now QK <i G since F/K is nilpotent and normal in G/K and
hence HK = G by the Frattini argument. Thus 9 = in e Irr(H).
Since ker(9G) <=, H, we have ker(9G) £ K. Let \ji be an irreducible
constituent of 9G with K £ ker \p and let L = ker \ji. If L £ H, then L £ ker 9 =
H n K < K and we are done. Suppose then, that L £ H so that LH = G
and \jiH is irreducible. Then \jiH = 9 and L r\ H = ker 9 = X n H. Write
JV = ker 9.
Now KnL<GandK>KnL2KnH.It follows that
(K n,L)H < KH = G
and thus K n, L s H. Therefore K n L = K n, H = N

212
Chapter 12
We have
\F:FnH\ = \G:H\ = \L:N\ = \KL:K\.
Since Q £ F r\ H and QeSyl,(F), it follows that qJC\KL/K\. Since
XQ/X -a G/X, we conclude that the commutator [XL/X, XQ/X] = 1 and
thus [L, Q] £ X. Since L -a G, we have [L, Q] £ X n L = N and hence
L £ N(JVQ). However,
NQ = (H n, K)Q = H r\KQ~aH
since XQ -a G. Therefore, G = LH <= N(JVQ) and JVQ -a G. The Frattini
argument yields G = JVN(Q) £ H, a contradiction. |
(12.25) lemma Let G" = 1 and let m = |G:F(G)|. Then m < b(G) and
b(F(G)) <L b(G)/m.
Proof Let / e Irr(G) be such that ker / is minimal. Since G is a relative
M-group with respect to G' which is abelian, there exists H 2 G' and
linear X e Irr(H) such that kG = %. Also, H 2 ker /. Since H -a G, all
irreducible constituents of Xh are linear and thus H/ker / is abelian. Thus
H/ker / £ F(G/ker /) and hence H is nilpotent by Theorem 12.24. In
particular,// £ F(G)andm < \G:H\ = /(l) < b(G).
Now let 9eIrr(F(G)) and let \\i be an irreducible constituent of 9G.
Choose x £ Irr(G) with minimal ker / £ ker \ji and let H be as above with
| G: H\ <L b(G), H/ker / abelian and H £ F(G). We have W £ ker / £ ker \p
and thus t/^H has linear constituents. Thus 9H has linear constituents and
hence 9(1) < | F(G): H \ = | G: H \ jm < ft(G)/w and the proof is complete. |
(12.26) theorem Suppose U -a G is abelian and G/U is nilpotent. Then
there exists abelian A £ G such that | G: A | < fe(G)4.
Proof Use induction on b = b(G). We may assume b > 1 and so G is
not abelian. First, suppose that every nilpotent factor group of G is abelian.
Then G' £ U, G" = 1 and Lemma 12.25 applies. Since G is not nilpotent,
F(G) < G and b(F(G)) <, b/m < b, where m = \G: F(G)| <; ft. By the
inductive hypothesis, there exists abelian A £ F(G) with |F(G):/1| < (b/m)4
and hence \G:A\< b*/m3 < b4.
Now suppose that G does have a nonabelian nilpotent factor group.
Choose X <a G, maximal such that G/X is nonabelian and nilpotent. By
Lemma 12.3, G/X is a p-group and |G: Z\ = /2, where Z/X = Z(G/X) and
Z = Z(0) for some 0 elrr(G) with 0(1) = /. By Corollary 12.8, we conclude
that b(Z) ^ b/f1'2 < b. By the inductive hypothesis, there exists abelian
A sZ, with |Z:/4|^ft(Z)4^ft4//2- Thus |G:/4| = \G:Z\\Z:A\ £ b4
and the proof is complete. |

Character degrees
213
Next we discuss the "p-structure" of G, where p is a prime which is in
some sense large when compared to b(G). The objective here is to show that
| G: Op(G) | is not divisible by too large a power of p. We prove that ifft(G) < p,
then pjf\G: Op(G)\ and that if b(G) < p3/2, then p2X\ G: Op(G)|.
(12.27) lemma Let G act transitively on a set Q with |Q| > 1. Let H = Ga
for some a e Q. Then the average size of the orbits of H on Q — {a} is < b(G).
Proof Let 9 = (1H)G, the permutation character of G on Q. Write
$ = 1g + X flx& where the sum runs over the nonprincipal irreducible
constituents of 9. The number of orbits t of H on Q — {a} is given by t =
Now |fi - {a}| = 9(1) - 1 = 2Xx(l) < b Xaz, where ft = fc(G). Thus
the average orbit size s is given by
s = (I «*x(i))/(I ax2) < (b x <g/(E«,) = *>• I
(12.28) corollary If G has more than one Sylow p-subgroup, then
there exist distinct P, Q e Sylp(G) such that
b(G) > |NG(P): NG(P) n NG(Q)| > |P:P n Q|.
Proof Apply Lemma 12.27 to the conjugation action of G on Sylp(G).
Let P e Sylp(G) and H = NG(P). Then some orbit of H on Sylp(G) - {P} must
have size <b(G). Let Q be in such an orbit. Then b(G) >\H: NH(Q)| and we
have the first inequality. Also
\H:N„(Q)\>\P:NP(Q)\
and since NP(Q) = P n Q, the result follows. |
(12.29) theorem Let p be a prime and let b(G) < p. Then G has a normal
abelian Sylow p-subgroup.
Proof Let P e Sylp(G). If P ^ G, then by Corollary 12.28, we can find
eeSylp(G) such that Q ^ P and |P:P n Q\ < b(G) < p. This is a
contradiction and shows P <i G.
That P is abelian follows since every/ e c.d.(P) is a power of p satisfying
/ < p. The proof is complete. |
(12.30) lemma (Burnside) Let P e Sylp(G) and let X, Y £ P be normal
subsets of P which are conjugate in G. Then X and Y are conjugate in NG(P).
Proof Suppose 7 = X9 so that P £ N(Y) and P9 £ N(X)9 = N(X9)
= N(Y). By Sylow's theorem in N(Y), we have P9" = P for some weN(Y)
and gu e N(P). However, X9" = Y" = Y and the proof is complete. |
(12.31) lemma LetHs Gwith|G:H| = p, a prime. Then Op'(H) <i G.

214
Chapter 12
Proof Let K be the kernel of the action of G on the right cosets of H
(acting by right multiplication). Then K ^ H and K<G. Also \G:K\ divides p!
and so \H: K | divides (p - 1)! and is prime to p. Thus Op'(H) £K<H
and so Op'(H) = Op'(K) -a G since K<G. |
(12.32) theorem Suppose b(G) < p3/2 for some prime p. Then
p2^|G:Op(G)|.
Proof We may assume that Op(G) = 1. Let PeSylp(G) and assume
\P\ > p2.LetJV = NG(P). By Corollary 12.28, choose Q ± P, with Q e Sylp(G)
such that
p3/2>ft(G)>|iV:Nw(0|>|P:PnQ|
and set D = P n Q. Then |P: D\ = p. Let M = NG(D) 2 P. Now P £ N(Q)
and so
N„(Q) < PN„(Q) £ JV n M.
It follows that
p3/2 > IJViNjrfQ)! >p\N:Nr\M\
and |JV: N n M| < p1/2. Note that M < G.
We consider the action (by right multiplication) of M on fi = {Mx | x e G}.
Suppose r of the orbits of M on fi — {M} have size < p2 and s have size > p2.
Let fi0 be the union of the r smaller orbits.
We claim
(*) If Mx e fi0, then DDX = DXD and x e MNM.
Assuming (*), let us complete the proof. We have |fi — {M} | > p2s so that
Lemma 12.27 yields
p3/2 > b(G) > \Q - {M}\/(r + s)> p2s/(r + s).
If s = 0, then MxeQ0 for every xeG — M and hence DDX = DXD for all
x e G by (*). Thus <£>* | x e G> is a p-group that is normal in G, a contradiction.
Thus s > 0 and p1/2 < (r + s)/s < 1 + r. Now if Mx eO0, then
x e MNM because of (*) and hence the orbit of Mx under M contains an
element of the form Mn for n e N. The number of distinct Mn for neN is
| iV: N n M | < p1/2 and thus at most p1/2 M-orbits of Q contain an element
of the form Mn. These include the trivial orbit {M} and so we have
1 + r < p1/2. This contradicts a previous inequality.
We now work to establish (*). Let Mx e Q0 so that x $ M and D ^ Dx.
If either of D or Dx normalizes the other, then DDX = DXD is a p-group
properly containing D. Since \P:D\ = p it follows that DDX e Sylp(G) and
|D£>*: D\ = p so that £> <i £>£>* and DDX ^ M. Since also P c m we have
P = (DDT for some m e M by Sylow's theorem. Now Dm, Dxm <i P and

Character degrees
21S
hence Lemma 12.30 yields Dm = Dxmn for some n e JV. Thus xmnm l e N(D)
= M and x e MNM as desired.
The remaining case is where D £ Mx and Dx £ M. Let W = M n Mx so
that WD and WDX are groups. Since Mx e Q0, we have \M :M n Mx\ < p2
and hence \DW:W\ < \M:W\ < p2 and |DW:W| = p. Similarly
|D*W: w\ = p. By Lemma 12.31, Op'(W) is normal in both WD and WD*.
Write X = Op'(W) and L = NG(X).
Since |M:W| < p2 and |DW: W| = p it follows that |M:DW| < p and
DW contains a full Sylow p-subgroup of M and hence of G. It follows that
DK/K and DXK/K are Sylow subgroups of L/K. Also, D*X ^ DX since
DK^M and D* £ M. Thus |Sylp(L/X)| > 1 and b(L/K) > p by
Theorem 12.29.
By Corollary 12.8, b(L) > ab(K) for some a with a2 > b(LIK) > p.
Thus p3/2 > b(L) > pll2b(K) and b(K) < p. Thus X has a normal Sylow p-
subgroup U by Theorem 12.29. Now UD e Sylp(M) and UD <i WD since 1/
is characteristic in X <i WD. Thus |M:NM(t/D)| < |M :DW\ < p and it
follows that UD <i M. Thus | Sylp(M) | = 1 which is a contradiction since
P,Qe Sylp(M). This completes the proof. |
We close this chapter by considering the opposite of the stituation with
which we began. Suppose no/ e cd.(G) is divisible by the prime p. A sufficient
condition for this to happen is that G has a normal abelian Sylow p-subgroup.
(This follows by Ito's Theorem 6.15.) It is conjectured that this condition is
also necessary. The next result shows that to prove the conjecture, it would
suffice to check simple groups.
(12.33) theorem Suppose G does not have a normal abelian Sylow p-
subgroup and that no element of c.d.(G) is divisible by p. Then G has a non-
abelian simple composition factor S of order divisible by p such that no
element of c.d.(S) is divisible by p.
Proof If N <i G and t/^eIrr(iV), choose /elrr(G) with [zN,^] ^ 0.
Then t/^(l)|x(l) and so p)(\ji(\). In particular, if JV e Sylp(G), then JV is
necessarily abelian and thus G does not have a normal Sylow p-subgroup. Also,
if G is simple, the result is trivial and we assume G is not simple.
Let JV be a maximal normal subgroup of G. Working by induction on | G |,
we may assume that JV has a normal Sylow p-subgroup P. Then G/P does not
have a normal Sylow p-subgroup and if P > 1 we complete the proof by
applying the inductive hypothesis to G/P. Suppose then, that P = 1.
We must have p 11G: JV | and since G/N is simple we can take S = G/N
unless G/N is abelian, that is, | G/N \ = p. We suppose this is the case. Let
QeSylp(G)sothat|Q| = p.

216
Chapter 12
If \j/ e lrr(N), then i/fG cannot be irreducible since p | \jiG(\). Thus /G(i/0 > JV
and hence 7G('/') = G and \ji is invariant in G. Thus Q acts trivially on Irr(G).
It follows by Brauer's Theorem 6.32 that Q acts trivially on the set of con-
jugacy classes of JV.
If jf is a class of JV then Q permutes jf. Since pjf\ N\, we have pX\tf\
and thus Q fixes an element of Jf. It follows that C = CN(Q) meets every
conjugacy class of JV. Thus JV = [jxeN Cx and this yields
|JV| - 1 < |JV:N^(C)|(|C| - 1) < \N:C\(\C\ - 1) = \N\ - \N:C\.
Thus | JV: C | < 1 and C = JV. We conclude that Q o G, a contradiction. |
(12.34) corollary {ho) Let G be solvable. Then G has a normal
abelian Sylow p-subgroup iff every element of c.d.(G) is relatively prime to p.
Problems
(12.1) The normal subgroups JV,- o G are a S_y/ow tower if
1 = JV0 c JVt c ... c Nk = G
and JV,+ i/JV; is a Sylow subgroup of G/N; for each i, 0 < i < k. Suppose for
every m,ne c.d.(G), either m \ n orn | m. Show that G has a Sylow tower.
(12.2) Suppose that every/ ec.d.(G) is a power of the integer m. Assume
that m is not a prime power. Show that there exists abelian A<=,G with | G: A \
= b(G) and that such an A is necessarily normal in G.
Hint This generalizes part of Theorem 12.5. Mimic the proof of that
theorem.
(12.3) Suppose that G is solvable and that for every m, nec.d.(G) with
m t6 n, we have (m, n) = 1. Show that |c.d.(G)| < 3.
(12.4) Let G be solvable with b(G) = b > 1 and suppose that G has no factor
group which is a nonabelian p-group. Show that there exists L ^ G and an
integer r with 2 <r < b such that b(L) < b/r and | G: L | < br.
(12.5) Let G be solvable with b(G) = b. Show that G has an abelian
subgroup of index <kbloi2{b) for a suitable constant k independent of b.
(12.6) Let c.d.(G) = {1, pe}, where p is a prime and e > 1. If a Sylow p-
subgroup of G is nonabelian, show that G is nilpotent.
Hint Use the fact that abelian Frobenius complements are cyclic to
show that if G is not nilpotent, then there exists abelian H <i G with \G:H\
= pe and G/H cyclic. Now let G/K be as in Lemma 12.3(a) and consider HK.
(12.7) Let G be solvable with b = b(G). Let p be a prime.

Problems
217
(a) Show that there exists proper K<G and integer e > 0 such that
pel4b(K) <b and pe+1Jf\G:K\.
(b) Show that if pf 11G : Op(G) |, then pf < b4.
(12.8) Let G" = 1 and assume that all Sylow subgroups of G are abelian.
Show that b(G) = \ G : A | for some abelian A o G.
(12.9) Let G be solvable and suppose that all Sylow subgroups of G are
abelian. Show that | G: F(G)| < b(G)2.
Hints F(G) = CG(F(G)). Let leIrr(F(G)) be such that |G:/G(1)| is
maximal. Let T = IG(1) and s = \G: T\. Show that b(T/F(G)) < b/s and
let F/F(G) = F(T/F(G)). Show b(F) < s. Use Corollary 11.32.
(12.10) Suppose that every/ e c.d.(G) divides pe where p is a prime. Show
that there exists abelian A ^ G with index dividing p4e.
(12.11) Suppose 0eIrr(G) with 0(1) = / and that Z = Z(0) satisfies
|G:Z| = f2. LetZ <= H £ G.
(a) If | G: H | > / show that fc(tf) < fc(G).
(b) If | G: H| = /and / e Irr(H) with /(l) = ft(G), show that \(x) <= Z.
(12.12) (a) Suppose A £ G is abelian and |G: CG(A)| > b(G). Show that
there exists A0 ^ A such that | A : A0 \ < b(G)2 and CG(A0) > CG(A).
(b) If G is a p-group, improve part (a) to read \A : A01 < b(G),
Hint Use Lemma 12.10.
(12.13) Show that there exists a function/defined on positive integers such
that for any group G if b(G) = b, then there exists H £ G with | G: H\ <b
and |tf:Z(tf)| </(&).
Hint Use repeated applications of Problem 12.12.
(12.14) Let G be solvable and let p be a prime. Suppose p2Jff for all
/ e c.d.(G). Show that either Op(G) > 1 or a Sylow p-subgroup of G is abelian.
Hint In a minimal counterexample, let M be a minimal normal
subgroup. Show that | Op(G/M)| = p. Now show that Op(G/M) is a direct factor
of a Sylow subgroup of G/M. Produce the other factor by considering 7G(1)
for suitable X e Irr(M).
(12.15) Let N<i G with G/N a p-group and p ^ 2. Let Selrr(iV) be
invariant in G. Suppose that every irreducible constituent of 9G has degree
5Sp9(l). Show that b(G/N) < p.
Hints Extend 9 to §eIrr(H) with JVcJf and \G:H\ = p. For
<p e lrr(H/N) with <p(l) = p, consider (<p9)G. Conclude that there exists linear

218
Chapter 12
fi e lrr(H/N) such that (px = cp/i for x e G, with /z independent of the choice of
cp. Use this and the hypothesis p ,£ 2 to show that cp is invariant in G.
Note If p = 2, then Problem 12.15 is actually false.
(12.16) Let G be a p-group and suppose b(G) = p2. If p =£ 2, show that
n{kerz|z(i) = p2}=l.
Hint Use Problem 12.15.
Note Passman has conjectured that if G is any p-group with b(G) = pe
and e < p, then P){ker x|/(l) = pe} = 1. He has proved this when G has
class 2.

Character correspondence
We have already seen several examples of the following situation (for
instance, Theorems 6.11 and 6.16). We are given H, a subgroup or factor
group of G. Subsets if £ Irr(H) and 3" c Irr(G) are specified and it is proved
that there exists a "natural" one-to-one correspondence between if and 3".
Here, the word "natural" is intended to mean that the correspondence is
uniquely described by some general rule and thus more is being said than
merely that |if \ = \3~\. We shall not attempt to give a precise definition of
naturalness. Most of this chapter is devoted to the study of a particular
character correspondence which was discovered by G. Glauberman.
We introduce some notation. Let S and G be groups such that S acts on G.
[That is, we are given a homomorphism S -» Aut(G).] In this situation, we
can construct the semidirect product r of G by S so that G <i T, S £ r,
GS = T, G n S = 1 and the given action of S on G is the action by
conjugation in T. (In fact these properties characterize the semidirect product.)
If x is a character of G and seS, then as usual we define the character
Xs of G by xs(gs) = 1(g)- Then S permutes Irr(G). We write
Irrs(G) = {x e Irr(G) \f = X for all s e S}.
(13.1) theorem (Glauberman) For every pair of groups (G, S) such that
S is solvable and acts on G and (|G|,|S|)= 1, there exists a uniquely
defined one-to-one map n(G, S): Irrs(G) -»Irr(CG(S)). These maps satisfy the
following properties:
(a) If T <i S and B = CG(T), then n(G, T) maps Irrs(G) onto Irrs(B).
(b) In the situation of (a), n(G, S) = n(G, T)n(B, S/T).
219

220
Chapter 13
(c) Suppose S is a p-group and C = CG(S). Let / e Irrs(G) and i/f =
(x)7t(G, S). Then i/f is the unique irreducible constituent of Xc such that
In the situation of (a) in the theorem, note that B is S-invariant since
T <i S. Thus S acts on B and in fact S/T acts on B. Therefore n(B, S/T) is
defined on Irrs/T(B) = Irrs(B) which is the image of Irrs(G) under n(G, T).
Also, n(B, S/T) maps to the irreducible characters of CB(S/T) = CB(S) =
CG(S). Thus the equation in (b) makes sense.
Also note that if T = S in (a), then B = CG(S) and Irrs(B) = Irr(B). Thus
this special case of (a) asserts that n(G, S) always maps Irrs(G) onto Irr(CG(S)).
There is enough information in the statement of Theorem 13.1 to
determine n uniquely. Thus suppose that n0(G, S) is defined whenever (G, S)
satisfies the hypotheses of 13.1 and that n0 satisfies (a), (b), and (c). We claim
that n0(G, S) = n(G, S). By 13.1(c), this is certainly the case if S is a p-group so
we may work by induction on | S | and assume that S has composite order.
Let T <i S have prime index and let B = CG(T). Then n(B, S/T) = n0(B, S/T)
and n(G, T) = n0(G, T) and 13.1(b) yields n(G, S) = n0(G, S).
The preceding argument suggests how to construct the map n(G, S);
namely, prove that if S is a cyclic p-group and / e Irrs(G), then %c does have a
unique irreducible constituent /? such that [/c, /?] $ 0 mod p, where C =
CG(S). Define n(G, S) in this case by d)n(G, S) = (i. For general solvable S,
define n(G, S) by working along a composition series for S, There are
numerous technical difficulties with this approach, not the least of which is to show
that the map constructed is independent of the composition series. The key
to overcoming these difficulties is to find a uniform definition for n(G, S) for
all cyclic S. Following Glauberman, this is what we shall do.
We establish some notation which will be used repeatedly.
(13.2) hypothesis Let S act on G and suppose (|G|, \S\) = 1. Let C =
Cc(S) and let T be the semidirect product T = GS.
(13.3) lemma Assume the situation in 13.2 and let / e Irrs(G). Then there
exists a unique extension % of / to T such that (o{%), \S\) = 1. Also, % is the
unique extension such that S £ ker(det #).
Proof The first statement is just Corollary 8.16. Since o(#s) divides both
\S\ and o(%), we have o(#s) = 1 and S c ker(det #). If i> is an extension of/
with S £ ker(det $), then 0(1^) = o(det \ji) and divides | Y: ker(det \ji) \ which
is prime to | S \. Thus \p = % and the proof is complete. |
We call the character % of Lemma 13.3 the canonical extension of /.
For positive integers n, we write Q„ = 0(e), where e is a primitive nth
root of unity. If (m, ri) = 1, it is well known from Galois theory that Q„ n Qm

Character correspondence
221
= Q. Thus if a e Q„ is invariant under the Galois group ^(Qmn/Qm), then
aeO.
If M<i H with \M\ = m and \H :M\ = n such that (m, n) = 1, we shall
use the notation &(H/M) to denote 0(Onm/OJ. Note that &(H/M) permutes
Irr(H) (by Problem 2.2 or Lemma 9.16). Suppose 9 e Irr(M) is extendible to
H. Since ^(H/M) fixes 9, it permutes the set of extensions of 9 to H.
In the situation of 13.2, if / e Irrs(G), and % is the canonical extension of
X to T, then for t e^(T/G), (xY is an extension of / and o(xz) = o(x). Thus
f = % and hence f has values in Q|G| and %s has values in Q|G| n Q|S| = Q.
We have thus proved the following corollary.
(13.4) corollary In the notation of Lemma 13.3, %s is rational valued.
The following strengthens this.
(13.5) lemma Assume Hypothesis 13.2. Let /eIrrs(G) and let % be the
canonical extension of / to T. Note that CS = C x S. For each irreducible
constituent /? of Xc there exists a (possibly reducible) character \jip of S such
that #cs = Y,t (P x •M- This equation uniquely determines the \jip. Also,
the \pp are rational valued.
Proof We have
Irr( CS) = {^x <p\pelrr(C), <peIrr(S)}.
Write %cs = Yji.q, apJiP * <?)• Set \\it = £„ a$v9 and observe that \pp = 0
unless \jc, PI ¥=■ 0. Thus #cs = Yj (P x ^f\ where the sum runs over those
/?eIrr(C), which are constituents of Xc- Also, this equation uniquely
determines the \jips.
If t e 0(r/G) then (%)z = X and pz = p. Thus
xcs = ((mcs = KP*(hn
Thus \jip is invariant under 0(r/G) and hence has values in Q(G|. Since its
values also lie in Q|S|, the result follows. |
(13.6) theorem Assume Hypothesis 13.2 and that S is cyclic. Then for
each x £ Irrs(G), there exist unique /? e Irr(C) and e = +1 such that %(cs) =
eP(c) for all c e C and all generators s of S, where % is the canonical extension
of x to T. Also, p is a constituent of Xc and the map / h-» $ is one-to-one.
Proof Write %cs = Y P x ^fi as m Lemma 13.5 and fix a generator s of S.
Write 9(c) = #(cs) for c e C. Then 9 = Jj, »A^(s)jS. In particular, 9 is a class
function on C. We claim that [9, 9] = 1.
Let T be a set of representatives for the right cosets of C in G. If t1; t2 e T
and x e (Cs)'1 n (Cs)'2, then there exist c1; c2e C with
qV' = (c^)'1 = x = (c2s)H = c2t2st2.

222
Chapter 13
We thus have two factorizations of x into products of commuting elements of
orders dividing \G\ and \S\. By Lemma 8.18, s'1 = s'2 and thus t^-1 eC(s)
= C since S = <s>. Thus rt = t2. Hence the sets (Cs)' are disjoint for distinct
teT and \\J,(Cs)'\ = \T\\C\ = \G\ = \Gs\. Since s's'^G, we have
(Cs)' £ Gs and hence Gs = (J, (Cs)' is a disjoint union.
We have
|C|[M]= II2WI2= II2(*')I2
for all teT.lt follows that
|T||C|[M]= II2WI2-
*eGs
By Lemma 8.14(c), the latter sum equals | G\ = | T11 C\ and thus [9, 9] = 1
as claimed.
We now have
l = [9,9] = £l'/'^)|2.
However, i/^(s) is a rational algebraic integer by Lemma 13.5 and so lies
in Z. Thus i/^(s) is nonzero for some unique /? and for that /?, i/^(s) = e = +1.
This yields %(cs) = e/?(c) for ceC. This equation clearly determines e and /?
uniquely.
We now show that /? is independent of the choice of the generator s of S,
If S = <s0>, then s0 = sm for some m with (m, \S\) = 1. Thus there exists an
automorphism a of the field Q|S| such that X"(s) = X(sf = l(s0) for all
1 e Irr(S). Therefore 0 j= i/^s)" =» ^p(s0) and hence replacing s by s0 yields
the same /?.
Finally, suppose /l5 /2 e Irrs(G) determine the same character /? so that
j?i(cs) = e;f}(c) for c e C and i = 1,2. Since Gs = (J(Cs)', it follows that ft^gs)
= £1^2 Xi{9s) f°r all g e G and thus by Lemma 8.14(b), it follows that
Xl = (Xi)g = (l2)G = #2
and the map /i-> /? is one-to-one. The proof is complete. |
(13.7) definition Assume Hypothesis 13.2 and that S is cyclic. Construct
maps y(G, S): Irrs(G) -+ Irr(C) and e(G, S): Irrs(G) -> {-1, 1} by (z)y(G, S) = fi
and (x)e(G, S) = e = +1 where #(cs) = e0(c) for ceC, <s> = S, and % is the
canonical extension of / to Y.
It will turn out that the map n(G, S) of Theorem 13.1 equals y(G, S) when
y(G, S) is defined, that is, for cyclic S.
The map y(G, S): Irrs(G) -»Irr(C) of Definition 13.7 is one-to-one by 13.6.
It is also onto. One way to prove this is to show that |Irrs(G)| = |Irr(C)|.

Character correspondence
223
Since S is cyclic, it follows from Brauer's Theorem 6.32 that | Irrs(G) | is equal
to the number of S-invariant conjugacy classes of G. It is true that each S-
invariant class of G intersects C nontrivially and, in fact, the interesction is a
single conjugacy class of C. It follows that the number of S-invariant classes of
G is equal to the total number of conjugacy classes of C, and hence equals
|Irr(C)|. Thus y(G. S) maps onto.
The assertion about S-invariant classes of G in the preceding paragraph
follows from the Schur-Zassenhaus theorem. We digress from the
discussion of characters in order to give a proof.
(13.8) lemma (Glauberman) Let S act on G with (|S|, |G|) = 1. Assume
that one of S or G is solvable. Let S and G both act on a set Q such that
(a) (a • g) ■ s = (a • s) • gs for all a e Q, g e G, and seS.
(b) G is transitive on Q.
Then S fixes a point of Q.
Proof Let T = GS, the semidirect product. For gs e T and a e Q,
define a • (gs) = (a • g) ■ s. Condition (a) above guarantees that this is an
action. Pick a efi and let H = Ta. Since |fi| = \G:G r\H\ = |r:H|by(b),
it follows that \H: G n H\ = \S\. By the existence part of the Schur-
Zassenhaus theorem, let T be a complement for G n H in H.
Then \T\ = \S\ and T is a complement for G in T. Now the conjugacy
part of the Schur-Zassenhaus theorem yields S = Tx for some x e T. Thus
S £ H* and S fixes a • x e Q. The proof is complete. |
(13.9) corollary In the situation of Lemma 13.8, the set of S-fixed
points of Q is an orbit under the action of CG(S).
Proof If a e Q is fixed by S and c e CG(S), then (a • c) • s = (a • s) • cs =
a • c and a • c is S-fixed. Now suppose a, /? e Q are S-fixed. Let X =
{g e G|a • g = /?}. Then X is a left coset of Gfi and is S-invariant. Let Gp act
on X by right multiplication. Note that G^ is S-invariant and is transitive
on X. For xeX,geGfi and s e S, we have (x ■ g) ■ s = (xgf = xsgs = (x • s) • gs
and Lemma 13.8 applies to the actions of S on G0 and S and Gp on X. Thus S
fixes a point xeX. Then x e CG(S) and a • x = /?. The proof is complete. |
(13.10) corollary Assume Hypothesis 13.2 and that at least one of G or S
is solvable. Then Jf i-^ Jf n C defines a bijection from the set of S-invariant
conjugacy classes of G onto the set of conjugacy classes of C.
Proof Let Jf be an S-invariant class of G. The conjugation action of G
on Jf is transitive. For keJt,geG, and s e S, we have
(k-g)-s = (g- lkgf = (g' lyk°gs = (k-s). g\

224
Chapter 13
By Lemma 13.8 we conclude that Jf n C ¥= 0 and by 13.9, Jf n C is a
class of C.
Since the classes of G are disjoint, the map Jf i-> Jf n C is one-to-one.
If ceC, then the G-class of c is S-invariant. It follows that the map
jf i-> Jf n C is onto. |
(13.11) corollary The map y(G, S) of Definition 13.7 maps Irrs(G) onto
Irr(C).
Assume Hypothesis 13.2 and that S is solvable. Let if be a composition
series for S with
y:l = So<S!< xSk = S.
Let Q = CG(S,) so that
G= C0 2 Ct 2 ••■ 2 Ct= C.
Also, Si+1 £ Nr(C,-) for 0 < i < k and we view Si+l/S; as acting on Cf.
Since S;+ l/Si is cyclic,
y» = y(C„s,+1/S|)
is defined for each i, 0 ^ j < k. We have
y,:IrrSl + 1(C,)-In(Q + i).
Each yt is one-to-one and maps onto Irr(Ci+1) and so we can define
yrl:hr(Ci+i)^lrr(Cd.
(13.12) definition Assume Hypothesis 13.2 with S solvable. Let if be a
composition series for S and use the above notation. Put 9C{G, if) =
(\xx(Q)y;}vy-k}2 ■ ■ ■ 7l~ ly0~l £ Irr(G) and let n(G, if): SC(G, if) -»Irr(C) be
given by n(G, if) = y0yi • • • Ik-1-
Thus 3C(G, if) is the largest set on which the composite function yoTi • • •
yt_! is defined. Since each yt is one-to-one, n(G, if) is one-to-one and by
construction of 3C(G, if), we see that n(G, if) maps SC(G, if) onto Irr(C). Our
objective now is to show that 3C(G, if) = Irrs(G), that n(G, if) is independent
of the choice of if and that for cyclic S, n(G, if) = y(G, if). We obtain these
results by considering the case that S is a p-group.
(13.13) lemma Let Z s R s C, where R is a ring containing the values of
all ^ e Irr(G). Let 9 be a generalized character of G with values in an ideal / of
R. Assume / n Z £ pZ for some prime p not dividing \G\. Then p divides
[9, <J] for all £ 6 Irr(G).

Character correspondence
225
Proof We have | G | [8, £| = £ %)£(#) e / n Z. Thus p 11G | [8, £] and
the result follows since pjf \ G \. |
(13.14) theorem Assume Hypothesis 13.2 and that S is a p-group. Let
/eIrrs(G). Then there exists a unique /?eIrr(C), with \jc, P] ^ 0 mod p.
Furthermore
(a) [XoP] = ±lmodp;
(b) if S is cyclic, then 0 = (x)y(G, S) and (x)e(G, S) = [&, j8] mod p.
(c) If ^ is a composition series for S, then 3C{G, if) = Irr^G) and
(x)7t(G, if) = /?. In particular, n(G, if) is independent of the choice of if.
Proof First assume that S is cyclic and let /? = (x)y(G, S) and e =
(x)e(G, S) so that %(cs) = e/?(c) for c e S and S = <s>, where # is the canonical
extension of / to T.
Let R be the ring of algebraic integers in Q|r| and let / be a maximal ideal
of R with p e /. In the notation of Theorem 8.20, we have (cs)p, = c and thus
that theorem yields /(c) = %{c) = %{cs) mod /.
We therefore have /(c) = e/?(c) mod / for all c e C and thus Xc — e/? is a
generalized character of C with values in /. Since 1 £ / and p e /, we have
I n I = pZ and since p^|C|, Lemma 13.13 yields [/c - e/?, £] = 0 mod p
for all £ e Irr(C). It follows that [_Xc, £1=0 mod p for £ ^ 0 and [/c, 0] =
e mod p. When S is cyclic, this proves everything but (c).
If \S\ = p, then ^: 1 <a S and ^(G, if) = Irrs(G) and tc(G, S) = y(G, S).
In particular, part (c) of the theorem holds when \S\ = p.
Now assume \S\ > p and drop the assumption that S is cyclic. Work by
induction on \S\. Let
if: 1 = S0 <i • • • <i Sk = S
and write T = St_ t and
5": 1 = S0 <i • ■ • <i St_! = T.
Let B = CG(T). Then tc(G, if) = n(G, S~)y(B, S/T). Also, 3C(G, if) is the
inverse image in 3C{G, 3~) of Irrs/T(B) = Irrs(B) under the map
n(G, 3T); %(G, 3T) -+ Irr(B).
By the inductive hypothesis applied to T, we have / e Irrs(G) £ IrrT(G)
= 3C(G, 9~) and Xb = P® ± £, where 9 is a character of B or is zero and
Z = (x)n(G, 3-). UseS, we have [/B, £] = [_(f)B, £] = [/B, « and thus
[.Xb, F\ ^ 0 mod p. Thus £ = <f and £ e Irrs(5). In particular, x e %(G, if).
If 0 e Irr(C), we have
\lc, PI = E(P8 ± Oc « = ± Kc, « mod p

226
Chapter 13
and since £, e Irrs/T(B) and S/T is cyclic, it follows from the first part of
the proof that there is a unique /?eIrr(C) with \_£c, /?] ^ 0 mod p,
namely 0 = (£)y(B, S/T). Thus fi = (z)tc(G, if) is unique in Irr(C) such that
[Xc> 0] ^ 0 mod p and in fact [/c, 0] = ± [<^c, 0] = ± 1 mod p.
We already have Irrs(G) £ 3C(G, if). Now suppose
tA e %(G, if) £ #"(G, 3T) = IrrT(G).
Let >/ = (\p)n(G, 3~) so that >/eIrrs(B). Let seS. Then 0 ^ |>B, >/] =
[(^b, yf\ = [(i/Ob. >/] mod p. Since T <i S and \p e IrrT(G), it follows that
\lis e IrrT(G) = 3C(G, 9~) and thus r\ = (^s)n(G, £T) since W)B, rf\ £ 0 mod p.
Since 7t(G, 5") is one-to-one, we have \p = \j/s and thus \p e Irrs(G).
We have now shown that 3C(G, if) = Irrs(G) and have given a description
of the map n(G, if) which is independent of if. This completes the proof. |
Assume Hypothesis 13.2 and that S is cyclic so that
y(G, S): Irrs(G) -> Irr(C)
is defined. Suppose errr^r-! is an isomorphism and that er(G) = Gi
and ct(S) = Si. Then a(C) = C} = C^St) and y(Gu St): IrrSl(Gi) -»Irr(d)
is defined. Because y(G, S) is uniquely defined, independently of any arbitrary
choices, it is clear that if
(x)y(G,S) = (S, .
then
(XiMGi,Sl) = p1,
where Xi and /?! correspond to / and /? via the isomorphism, a. That is,
Xi(9a) = X(g) and jS^c") = P(c) for g e G and c e C. (Recall that the
computation of (x)y(G, S) requires choosing a generator s of S, but that the result is
independent of this choice.)
An important special case of this invariance under isomorphism of
y(G, S) is when a e Aut(r) and G" = G and S" = S. In that case we have
(Xa)y(G, S) = ((xMG, S))°.
(13.15) lemma Assume Hypothesis 13.2 and let To S with T cyclic and
B = CG(T). Then y(G, T) maps Irrs(G) onto Irrs(B).
Proof Since Irrs(G) £ IrrT(G), y(G, T) is defined on Irrs(G). Let H =
GT o r. If s e S, then s defines an automorphism of H with Gs = G and
Ts = T. Therefore, by the above discussion, we have
(f)y(G, T) = ((z)y(G, T)f
for all x e Irr7(G). It is immediate that y(G, T) maps Irrs(G) into Irrs(B). Since
y(G, T) maps onto Irr(B) and is one-to-one, the result follows. |

Character correspondence
227
(13.16) theorem Assume Hypothesis 13.2 and that S is cyclic. Let p be a
prime and let T be the p-complement in S. Let B = CG(T). Then y(G, S) =
y(G,TMB,S/T) on Irrs(G).
Proof If pJf\S\, then T = S, B = C, and y(C, 1) is the identity map on
Irr(C) (as is clear from Definition 13.7). The result is thus trivial in this case
and we assume p 11S |. In particular, pjf \ G \.
Let x e Irrs(G) £ IrrT(G) and let £ = (z)y(G, T) e Irr(B). Let 0 =(z)y(G, S).
By Lemma 13.15, £ e Irrs(B) = Irrs/T(B). We must show that fi = (£)y(B, S/T).
Let p0 = (£)y(B,S/T). Since S/T is a p-group, Theorem 13.14 yields
£c = P<P + £o^o> where <p is a character (or is zero) and e0 = +1. Let R be
the ring of algebraic integers in Q|r| and let / be a maximal ideal containing p.
We have then £(c) = e0fi0(c) mod / for c e C.
By definition of y(G, T), we have %(bt) = e£(b), where T = <t>, 6eB,
e = +1 and 2 is the canonical extension of % to r. Applying this to c e C £ B
yields
#(ct) = e£(c) = ee0)So(c) mod 1
for any generator, t of T.
Now let S = <s> and let t = (s)p, in the notation of Theorem 8.20. Then
<t> = T and (cs)p, = ct for c e C. Thus by 8.20, we have
%(cs) = %(ci) mod /
and thus
t(cs) = eeopo(c)modl.
By definition of y(G, S), we have %(cs) = 5f](c) for c e C with d = ± 1. Thus
)S(c) = 8ee0(i0(c) mod I for all ceC. By Lemma 13.13, [)S - (5ee0/?<,,/?] is
divisible by p and thus /? = )S0 as desired. |
(13.17) corollary Assume Hypothesis 13.2 and that S is cyclic with
\S\ = pq, where p and <j are primes. Let Sf be a composition series for S.
Then 3C(G, if) = Irrs(G) and n(G, if) = y(G, S).
Proof If p' = q, the result is immediate from Theorem 13.14(b) and (c).
Assume then that p ^ q. Write if: 1 <i T <i S. We may assume | T | = q.
Now Lemma 13.15 yields that SC(G, if) = Irrs(G). Theorem 13.16 asserts that
n(G, if) = y(G, T)y(B, S/T) = y(G, S)
where B = CG(T). The proof is complete. |
(13.18) theorem Assume Hypothesis 13.2 with S solvable. Let if and 9~
be composition series for S. Then SC(G, if) = 3C(G, 3~) and n(G, if) =
n(G, $~).

228
Chapter 13
Proof Use induction on the composition length k of S, If' k = 1 then
y = 3~ and there is nothing to prove so assume k > 1. Write
y:i = s0<--<st = s,
Let £f* and ^"* be the composition series for Sk _ t and Tt _ 1 respectively,
obtained by deleting S from y and ^".
Consider the case that St_ t = Tt_ t. Let B = CG(St_ t). Then Irrs(B) =
(In(QMB.S/Slk_1r1andso
%(G,<?) = (lrrs(B))n(G,y*rl
and
SC(G,3-) = (lxxs(B))n(G,3-*rx.
By the inductive hypothesis, n(G, £f*) = n(G, 5"*) and hence 3C(G, Sf) =
SC(G, F). Also
n(G, ST) = n(G, <?*)y(B, S/Sk-1) = n(G, <T*)y(B, S/Tk_ t) = n(G, T).
Assume now that Sk-t =£ Tt_t and let M = St_! n Tk_v Let ^ be a
composition series for M and extend ^ to composition series y° and 3~° for
S which run through St_ t and Tt_ 1; respectively. By the preceding paragraph,
3C{G, y°) = 3C{G, <f) and n(G, Sf°) = n(G, <f\
We may thus replace Zf by y° and similarly replace 9~ by 9~°. We may now
assume that S; = Tt for i < k — 2 and that
Jt: 1 = S0<i aSt_2 = M.
Let
F: 1 -a St_ t/M -a S/A#, F: 1 -a Tt_ t/M <a S/A#
and let D = CG(M). It follows that
%(G, <f) = (%(D, &)MG, JtY'
and
3C(G, $~) = (3C(D, F))n(G, Jt)' \
Also
n(G, <f) = n(G, Jt)n(D, 3s) and n(G, $~) = n(G, Jt)n(D, W\
Therefore, it suffices to prove that
3C{D, &) = 3C{D, F) and n(D, F) = n(D, F),
We may therefore assume that M = 1 and£> = G.ThusS = Sk-t x Tk_t
is abelian of order pq for primes p and q. If p = q, then
3C(G, <f) = Irrs(G) = 3C(G, F) and n(G, <f) = n(G, 9~)

Character correspondence
229
by Theorem 13.14. If p ^ q, then G is cyclic and Corollary 13.17 yields the
result. The proof is complete. |
Assume Hypothesis 13.2 with solvable S and choose a composition series
y for S. By Theorem 13.18, 3C(G, Zf) is independent of the particular
composition series and so we can write 3C(G, S) = 9C{G, Sf). Similarly, we can
write n(G, S) = n(G, y\ Then 3C(G, S) and n(G, S) are unambiguously
defined. We have SC(G, S) £ Irr(G) and n(G, S) is a bijection of 3C(G, S) onto
Irr(C).
As in the discussion preceding Lemma 13.15, suppose a: r -» r\ is an
isomorphism and let Gt = a(G), St = a(S\ and C\ = a{C) = Cq^S^). Then
a induces bijections Irr(G) -» \rr(G{) and Irr(C) -»Ir^CJ. Since #"(G, S) is
uniquely defined, we have 3C(GU St) is the image of #"(G, S) in Ir^GJ and
0 = (z)tc(G, S) implies that ft = (xMGu Si), where Zl(g') = fa) and
ft^") = 0(c). In particular, if a e Aut(r) and G" = G and S" = S, then er
leaves SC(G, S) setwise invariant and (x")n(G, S) = ((z)tc(G, S))" for / e SC(G, S).
(13.19) corollary Assume Hypothesis 13.2, with S solvable, and let
SC(G, S) and n(G, S) be as above. Then SC(G, S) = Irrs(G). Also, if T <i S
and B = CG(T), then
(a) n(G, T) maps Irrs(G) onto Irrs(B);
(b) n(G, S) = n(G, T)n(B, S/T).
Proof If T <i S and B = CG(T), use a composition series for S, which
runs through T in order to construct n(G, S). Then (b) is immediate and
%(G, S) = (3C(B, S/TMG, T)~l.
Thus (a) will follow once we prove the first statement.
We show that 3C(G, S) = Irrs(G) by induction on the composition length
k of S. If k = 1 then SC(G, S) = (Irr(C))y(G, S)~' = Irr^G). Suppose then,
that k > 1 and let T <i S with 1 < T < S. Let H = GT <i T. For s e S, we
have Gs = G and Ts = T and hence by the discussion preceding the
statement of the corollary, we see that
(f)n(G, T) = ((x)n(G, T)f
for x g SC(G, T) = IrrT(G). Since n(G, T) is a bijection from IrrT(G) onto
Irr(B), where B = CG(T), we see that n(G, T) carries the S-invariant
characters in IrrT(G) onto Irrs(B). Since Irrs(G) £ IrrT(G), it follows that
Irrs(G) = (Irrs(B))7t(G, 7T1 = 3C(B, S/T)n(G, T)~l = 3C{G, S\
where the second equality is by the inductive hypothesis applied to S/T. |
(13.20) definition Assume Hypothesis 13.2 with S solvable. Then the
Glauberman map is the map n(G, S): Irr^G) -»Irr(C) constructed above.

230
Chapter 13
We have now completed the proof of Theorem 13.1. The Glauberman
map satisfies conditions (a) and (b) of 13.1 by Corollary 13.19. It satisfies
condition (c) by Theorem 13.14(c). Also, by Theorem 13.14(a), we have the
following.
(13.21) corollary Let n(G, S) be the Glauberman map with S a p-group.
Let C = CG(S) and let xeIrrs(G) and 0 = (x)n(G, S). Then [zc, 0] = ±1
mod p.
There is one further loose end.
(13.22) corollary Assume Hypothesis 13.2 with S cyclic. Then n(G, S)
= y(G, S).
Proof Use induction on | S \. Let p 11S \. If S is a p-group, the result follows
from Theorem 13.14(b). Assume that S is not a p-group and let T be the p-
complement in S and B = CG(T). Then n(G, T) = y(G, T) and n(B, S/T) =
y(B, S/T) by the inductive hypothesis. The result now follows from
Corollary 13.19(b) (or Theorem 13.1(b)) and Theorem 13.16. |
Let S act on G. Then S permutes Irr(G) and S permutes the set C1(G)
of conjugacy classes of G. By Brauer's Theorem 6.32, the permutation
characters of S on Irr(G) and C1(G) are equal and it is natural to ask if these
actions are permutation isomorphic. That is, does there exist a bijection
a: Irr(G) -> C1(G) such that a(/s) = a(/)s for all / e Irr(G) and s e S? In general,
the answer is no. However, if (| G |, | S |) = 1 and S is solvable, it follows via
Glauberman's Theorem 13.1 that the actions of S on Irr(G) and C1(G) are
permutation isomorphic.
(13.23) lemma Let the group S permute two sets Q and A. Suppose that
for every T £ S, the number of fixed points of T on Q equals the number on A.
Then Q and A are permutation isomorphic.
Proof We prove the existence of a bijection a: Q -> A such that a(eo • s) =
a(eo)• s for all coeQ and seS by induction on |Q|. (Note that taking T = 1
yields |fi| = |A|.)
Let T s S be maximal such that T has a fixed point on Q. (Possibly
T = S.) Let T fix co e Q and 1 e A. By the maximality of T, we have S^ =
T = Sx. Let 6a be the orbit of co and Ox the orbit of 1 under S. Write Q =
ff.uQ, and A = &x u At where the unions are disjoint. Map a0: 6a -> 6X
by a0(co • s) = Is and check that a0 is well-defined, one-to-one, onto and
that a0(v• s) = a0(v)• s for all ve(9m and seS,
Since every H s S has equal numbers of fixed points on Om and 6X,
it follows that H has equal numbers of fixed points on fit and At. By the

Character correspondence
231
inductive hypothesis, there exists a permutation isomorphism at: fit -» At.
Now define a on Q by combining a0 and al. |
(13.24) theorem LetSactonGwithSsolvableand(|G|,|S|)= 1. Then
(a) S fixes the same numbers of irreducible characters and conjugacy
classes of G.
(b) The actions of S on Irr(G) and C1(G) are permutation isomorphic.
Proof It suffices to prove (a) since (b) follows via Lemma 13.23 by
application of (a) to all subgroups of S.
Since n(G, S) maps Irrs(G) one-to-one and onto Irr(C), it follows that
|Irrs(G)| = |Irr(C)| = |C1(C)|.
By Corollary 13.10, intersection defines a bijection from the set of S-fixed
conjugacy classes of G onto C1(C). The result now follows. |
Theorem 13.24 becomes false if the hypothesis that (|G|, \S\) = 1 is
dropped. (See Problem 13.16 for an example.)
Suppose we continue to assume that (|G|,|S|)= 1, but drop the
assumption that S is solvable. If S is nonsolvable, then 2||S| by the Feit-
Thompson theorem and thus 2xK IGI - Thus G is solvable, again by Feit-
Thompson. In this situation, where Hypothesis 13.2 is satisfied with G
solvable of odd order, it is possible to construct a natural character
correspondence from Irrs(G) onto Irr(C) by a method entirely different from Glau-
berman's. (And thus Theorem 13.24 remains valid without the hypothesis
that S is solvable.) We shall describe the map Irrs(G) -» Irr(C) in this case but
without giving the proof since the only known proofs are too long to include.
The key step is the following.
(13.25) theorem Assume Hypothesis 13.2 and that G is solvable of odd
order. Let C £ H £ G. Suppose that there exist S-invariant normal
subgroups, K and L of G such that
(a) L s K and K/L is abelian;
(b) G = KC;
(c) H = LC.
Then for each x £ Irrs(G), there exists a unique \p e lrrs(H) such that [jH, i/f]
is odd. The map / h-» \ji is a bijection from Irrs(G) onto Irrs(H).
Proof Omitted. |

232
Chapter 13
To construct the correspondence between Irrs(G) and Irr(C) we construct
a chain of subgroups
G = C0 > Ct > ■ ■ ■ > Ck = C
and apply Theorem 13.25 to obtain maps Irrs(Cj)-> lrr^Cl+l). The
composition of these maps is the desired correspondence.
Assume Hypothesis 13.2 and that G is solvable of odd order. We consider
the set Jf of S-invariant subgroups H with C s H £ G. For H ejf, define
H* = [H, SJC. Since [H, S] -a HS it follows that [H, S]' -a HS and thus
H* e JT. If H > C, then [H, S] > 1 and [H, S] > [H, S]' by solvability.
Since [H*, S] £ [H, S]', it follows that H* < H. We now define the
subgroups C,- e jf by C0 = G and C»+1 = (C,)* and we have the desired chain of
subgroups.
We must now check that the hypotheses of Theorem 13.25 are satisfied by
taking H = G*. Let K = [G, S] and L = [G, S]'. Then K and L are normal
S-invariant subgroups of G and K/L is abelian. That G = KC follows fairly
easily from Glauberman's Lemma 13.8. Condition (c) of Theorem 13.25 is
automatic from the definition of G*.
It has been conjectured for every group G and prime p that if JV = NG(.P)
for P e Sylp(G), then the numbers of irreducible characters of p'-degree of G
and of JV are equal. (For simple groups G, this conjecture is due to McKay.)
Using Glauberman's Theorem 13.1, we prove a result which includes the
special case of this conjecture when G has a normal p-complement.
(13.26) theorem Let G = KH with X<i G, H solvable, and (|H|, |X|)
= l.LetJV = NG(H)andput
<r={zeIrr(G)|(|tf|,/(l))=l}
and
^=faeIrr(iV)|(|H|,irtl))=l}.
Then there exists a uniquely defined bijection of SC onto <&.
Proof We have N = (N n K)H and the commutator
[JVnX, H]sHnX = l
so that JV n K = C = CK(H) and JV = C x H. It follows that
<W={px l|/JeIrr(C),leIrr(tf), 1(1)= 1}.
Now if x g 3C, let 9 be an irreducible constituent of %K. Then | G: 7G(^)I
divides both *(1) and |G:X| = |H|. Thus JG(9) = G and 9eIrrH(X). Let

Character correspondence
233
9 be the canonical extension of 9 to G. Then x = H for some unique
£eIrr(G/K) by Gallagher's theorem (Corollary 6.17). Since (x(l), |H|) = 1,
we must have £(1) = 1. Also, / uniquely determines 9 by Xk = & and thus
determines 9 and £. We can now map 3C to <& by /1-» {(9)n(K, H)) x £H. Since
restriction defines a bijection of Irr(G/X) onto Irr(H) and since &; e S£ for
every 9 e IrrH(X) and linear £ e Irr(G/K), it follows that we have mapped
SC onto $/. The map is one-to-one since the r\ e <W are uniquely of the form
Suppose S acts on G and that JV <i G is S-invariant. Also, assume that
(|G:JV|,|S|) = 1. We consider a pair of " dual" questions:
(a) Let x e Irrs(G). Does there exist 9 e Irrs(iV) with [/N, 9] ^ 0?
(b) Let 9 e Irr^N). Does there exist / e Irrs(G) with [9G, /] # 0?
Both questions can be answered in the affirmative. By the Feit-Thompson
theorem, at least one of S and G/N is solvable and we assume this. Also,
note that if S is a p-group, both facts can be proved relatively easily by
counting arguments.
(13.27) theorem Let S act on G and leave JV <i G invariant. Assume that
(| S |, | G: JV |) = 1 and that one of S or G/N is solvable. Let / e Irrs(G). Then
XN has an S-invariant irreducible constituent.
Proof Let fi = {9eIrr(N)|[^, 9] ^ 0}. Then G/N permutes fi
transivively and since / is S-invariant, S permutes Q. Also, the action of S on
G induces an action on G/N. Let 9 e Q, s e S, and g e G. For x e JV, we have
(9T(xs)=9"(x) = 9(9*0-1)
and
(9TV) = SVxV)"1) = %^-1)-
The hypotheses of Glauberman's Lemma 13.8 are thus satisfied and the
result follows. |
The situation of question (b) is more difficult and interesting. First we
consider the case where G/N is solvable.
(13.28) theorem Let S act on G and leave JV <i G invariant. Assume that
(|S|, |G: JV|) = 1 and that G/N is solvable. Let 9eIrrs(JV). Then 9G has an
S-invariant irreducible constituent.
Proof First assume that G/N is abelian, and let A be the group of linear
characters of G/N. Let fi = {/ e Irr(G) | [9G, /] ¥= 0}. If / e fi and k e A, then

234
Chapter 13
#1 e fi since (%1)N = Xn. This defines an action of A on Q. We claim that this
action is transitive.
Let /, \ji efi. Then 0# \_xN, \pN~] = [(xN)G, t/f] and \ji is an irreducible
constituent of (xNf = fel^)G = x(\Nf. However, (lNf = ^a^ and thus
(Xn)g = Z Z^- Since #1 e Irr(G) for each XeA,it follows that \ji = %X for some
k and A is transitive on Q as claimed.
Since 9 is S-invariant, it follows that S permutes Q and also S acts on the
group A since (l/z)s = ls/zs. If / e Q, 1 e A, and s e S, we clearly have
(ztf = W
and the hypotheses of Glauberman's Lemma 13.8 are satisfied since \A\ =
\G:N\ and thus (| X |, | S |) = 1. The result follows in this case.
We now assume that G/N is nonabelian and work by induction on
| G: N\. Let M/N = (G/N)'. Then M <i G is S-invariant and M < G since
G/JV is solvable. By the inductive hypothesis, 9M has an S-invariant
irreducible constituent \ji and by the first part of the proof, \jiG has an S-invariant
irreducible constituent. The result now follows since 9G = (9M)G. |
To handle the case that G/N is not solvable, we appeal to the Glauberman
correspondence, Theorem 13.1. We first restrict attention to the situation
where(|G|,|S|)= 1.
(13.29) theorem Assume Hypothesis 13.2 and that S is solvable. Let
JVo T with N £ G. Let / e Irrs(G) and 9 e Irrs(JV). Write £ = (x)n(G, S) and
q> = (S)n(N, S). Then [9G, Z] * 0 iff [>c, £] * 0.
Proof We first consider the case that S is a p-group. Then [/c, £] ^
0 mod p and every irreducible constituent of Xc other than £, occurs with
multiplicity divisible by p. Since XcnN = (Xc)cnN> it follows that
(1) [XCnN'fpl = DfoflKcnN.^modp.
Now write Xn = X ^a A, where A runs over sums of orbits of the action
of S on Irr(iV). If A = £ 0 where (9 is such an orbit, then [ACnJV, cp] =
ICIfocnw.p] forijeffi If |0|> 1, then p||0| and [ACnJV, cp] = 0 mod p.
Thus
[Zcniv.<p]= Z [*iv>'7][>7cniv><p]mocip.
i,eIrrs(JV)
However, for >/ e lrrs(N), we have \_rjc n N, <p] = 0 mod p unless cp = (rj)n(N, S),
that is, unless r\ = 9. Thus
(2) [XcnN, <l>] = [.Xn, 9][9CnN, <p] mod p.
Since [jc, Q p 0 p [9CnJV, <p], comparison of Equations (1) and (2)
yields that \_Xn> 9] = 0 mod p iff \_iCr,N> <P\ = 0 mod p. Since JV <i G and

Character correspondence
23S
pjf | G | it follows that [Zw, 9] = 0 mod p iff [zK, 9] = 0. Similarly, [£c n N, <p]
= 0 mod p iff [iJCfllv, <p] = 0. The result thus follows if S is a p-group.
To complete the proof, we use induction on | S \. We may choose T <i S
with S/T a nontrivial p-group. Let B = CG(T). By Theorem 13.1, we have
n(G, S) = n(G, T)n(B, S/T) and similarly, n(N, S) = n(N, T)n(B n N, S/T).
Now [9G, jf] * 0 iff [((9)7t(iV, T))B, (x)n(G, T)] ^ 0 by the inductive
hypothesis. The result now follows by the first part of the proof. |
(13.30) corollary Let S act on G with S solvable and (| S \, \ G \) = 1. Let
N <i G be S-invariant. Let 9 e lrrs(N). Then 9G has an S-invariant irreducible
constituent.
Proof Let C = CG(S) and cp = ($)n(N, S) so that cp e Irr(C n N). Let £
be an irreducible constituent of <pc and let / = (<p)n(G, S)~1. Then / e Irrs(G)
and [9G, /] j= 0 by Theorem 13.29. The proof is complete. |
To complete our analysis of question (b), we need to consider the case
where S is solvable and (| S \, | G |) ¥= 1. Of course, we continue to assume that
(\S\, \G:N\) = 1. First, we observe that it is no loss to assume that 9 is
invariant in G since if U = IG(9), then S leaves U invariant. Now, if we can
find an S-invariant irreducible constituent \\i of 9V, then t/^GeIrrs(G) as
desired. We shall finish the proof by an appeal to the theory of projective
representations of Chapter 11. The following proves slightly more than we
need.
(13.31) theorem LetJVs G«a r with JV«a Tand (|r:G|, |G:JV|) = 1.
Assume that one of Y/G or G/N is solvable. Let 9 e Irr(iV) be invariant in T.
Then 9G has some T-invariant irreducible constituent.
Proof By Theorem 11.28, we can find a character triple (Tl,Nl,9l) and
an isomorphism (t, a): (T, N, 9) -> (Tu Nu 9t) with 9t(l) = 1. Let Gi = G1
so that the isomorphism v. F/N -> r^/iV! carries G/N to G^iV*!. Suppose we
can find i/^ elntG!^) with i/^ invariant in rt. Let t^eIrr(G|9) with
<tg(1/') = "Ai- We claim that \ji is invariant in r. To see this, let / be an
irreducible constituent of t/^r so that /elrr(r|9). Let Xi = °r(x)- Then (/i)Gl
= eipi for some integer e and it follows that /G = ei/f. Thus ip is invariant as
desired.
The argument of the preceding paragraph shows that it is no loss to
assume that 9 is linear. We may thus factor 9 = X/j. where (o(X), \G:N\)= 1
and (o(/z), | T: G |) = 1 and X and y. are powers of 9. Thus X and y. are invariant
in T. By Corollary 6.27, X has a unique extension, X e Irr(G) such that o(X)
= o(X). Because of the uniqueness, it follows that X is invariant in T. Suppose
we can find a T-invariant irreducible constituent \ji of fiG. Then \jil e Irr(G)
is T-invariant and 9 = fiX is a constituent of (^1)^,.

236
Chapter 13
We may therefore assume that 9 = fi, that is, that 9 is linear and
(o(9), |T: G|) = 1. Also, we may replace T by T/ker 9 and thus assume
that 9 is faithful. Thus |iV| = o(9) is relatively prime to |T: G| and hence
(|G|, |T: G|) = 1. By the Schur-Zassenhaus theorem, we can find S £ r
with SG = T and S n G = 1. Thus S acts on G and leaves N and 9 invariant.
Also, one of S or G/iV is solvable. Now Corollary 13.30 or Theorem 13.28
yields an S-invariant irreducible constituent of 9G. The result follows. |
To close the chapter, we obtain some further information in a special
case of Theorem 13.6.
(13.32) theorem Assume Hypothesis 13.2 and that S is cyclic. Also
suppose that C = CG(x) for all x e S - {1}. Let / e Irrs(G) and 0 = (x)n(G, S).
Let I be the canonical extension of / to r. Then there exists e = +1 and
fi e Irr(S) with fi2 = ls such that
(a) %{cx) = £P(c)(i(x) for all x e S - {1};
(b) Zc = IS19 + e/?, where 9 is a character of C or is zero;
(c) Crfl)-e0(l))/|S| = *eZ.
(d) Xs = kps + e.fi(V)fi, where ps is the regular character.
(e) fi ¥= ls iff IS| is even and k is odd.
'Proof If 1 < T s S then n(G, S) = n(G, T) = y(G, T) and hence %(cx)
= + Ac) for all x e S - {1} where the sign is independent of c. It follows that
in the notation of Lemma 13.5, we have t/^(x) = ± 1 for all x j= 1 and iji^x)
= 0 for x ^ 1 and 0 ^ 9 e irr(C).
Write i/f = i//0. We work to express \ji in terms of Irr(S). Let lu X2 e Irr(S)
and compute
[(*i--U£l = (i/ls|) Z ±(i1W-i2w).
JceS; jc?s 1
Since Ut(x) — A2(x)| < 2 for xeS, this yields
IQ^i - ^2X«A:i < 2(|S| - 1)/|S| < 2.
Therefore, the multiplicities with which lt and X2 occur in \ji differ by at most
1. It follows that \p = aps ± fi where a e Z and fi is a sum of distinct linear
characters of S with /z(l) < |S|/2. Also, /z(x) = ±tA(x) = ±1 for 1 ^ x eS
and hence
/z(l) = [>,/z] = (l/|S|)Ml)2 + |S|-l)
and (i(l)2 - |S|/*(l) + \S\ - 1 = 0. It follows that /z(l) = 1 and /zelrr(S).
Define e = +1 by the equation \p = aps + e/z. (Note that if \S\ = 2, then

Problems
237
neither e nor fi is uniquely defined.) Now xicx) = 0(<#(x) = e/^M*) f°r
l#x£S and (a) is proved. Also fi(x) = ± 1 for x e S so that fi2 = ls.
Since all i//v for <p ^ /? vanish on S — {1}, each is a multiple ofps and hence
Up x </v)c is a multiple of | S\ <p. Since (/? x i/^)c = a| S| /? + e/?, statement (b)
follows and (c) is immediate from (b). Also (d) now follows.
We now prove (e). Since \i2 = ls, it follows that fi = ls if | S | is odd.
Suppose then, that \S\ is even and let xeIrr(S) with x2 = ls but x ^ ls. (This
uniquely defines x.) Now det(ps) = n^irns) ^ = t. By (d) it follows that
ls = det(zs) = tV (1).
Since 2| \S\, we have 2J((J(1) and therefore fi = xk. Statement (e) now follows
and the proof is complete. |
(13.33) corollary Let (r, N, 9) be a character triple and let N s G <i T.
Suppose T/G is cyclic and that r/N is a Frobenius group with kernel G/N.
Assume that 9G = ex for zelrr(G) (and thus e2 = \G:N\). Then |T:G|
divides e — e for some e = ± 1.
Proof We may replace (r, N, 9) by an isomorphic character triple and
assume 9(1) = 1. Write 9 = pk, where (o(l), \G:N\) = 1 and (o(/z), |T: G|)
= 1 and 1 and fi are powers of 9 and thus invariant in T and 1 is extendible
to G by Corollary 6.27. Let v be an extension of I. Then
^ = (l9)« = (Vjv9)G = v9G = e(vz).
Since v/eIrr(G) we may replace 9 by fi and assume (o(9), |T: G|) = 1. We
may also assume that ker 9 = 1 so that N c Z(H and (|G|, |r:G|) = 1.
Let S be a complement for G in T. If 1 ^ x e S, then CG/A,(x) = 1 and it
follows that N = CG(x) and we are in the situation of Theorem 13.32. (Note
that x £ Irrs(G) since / is the unique irreducible constituent of 9G.) The result
now follows from 13.32(c). |
Problems
(13.1) Assume Hypothesis 13.2 with S solvable. Prove the following facts by
using the statement of Theorem 13.1, but do not appeal to any of the results
used in constructing the Glauberman map.
(a) If x e Irrs(G) then (x)7t(G, S) is a constituent of Xc-
(b) If p = (x)n(G, S), then Q(X) = QW).
(c) If C = G, then n(G, S): Irr(G) -> Irr(G) is the identity map.
(13.2) In the situation of Theorem 13.1, let 0 = (x)n(G, S) for / e Irrs(G).
Show that x( 1) divides |G: C|j8(l).
Hint In the case that S is cyclic, consider a = eoj as in Chapter 3.

238
Chapter 13
(13.3) In the situation of Theorem 13.1, let JVo G be S-invariant with
NC = G. Let 9 e Irrs(JV). Show that
IM&MN, S)) = U») n C.
(13.4) Let S act on G and leave JV <i G invariant. Assume that (| S |, | G: JV |)
= 1 and that one of S or G/N is solvable. Suppose CGIN(S) = 1. If / e Irrs(G),
show that there exists a unique i/f e Irrs(JV) such that [x^, i/f] ^ 0. Write
(X)5 = \ji. Show that 5 maps Irrs(G) onto Irrs(JV).
(13.5) Assume Hypothesis 13.2 and that S is solvable. Let C £ JV £ G with
jv<ir.
(a) Show that CG/JV(S) = 1.
(b) Let 5: Irrs(G) -»Irrs(iV) be as in Problem 13.4. Show that 5 is one-
to-one.
(c) Show that dn(N, S) = n(G, S).
Hint (For (a)) Use Glauberman's Lemma 13.8.
(13.6) In the situation of Theorem 13.1, assume that G is solvable. Let
X e Irrs(G) and /} = (i)k(G, S). Show that 0(1) divides *(1).
Hints Let JV < G be normal, S-invariant and maximal such. Then
either NC = G or N 2 C. Use Problem 13.3 or 13.5(c) and induction on | G |.
(13.7) In Problem 13.4, assume that G/N is solvable. Show that 8 is one-to-
one.
Hint Use induction on | G: JV |.
(13.8) Assume Hypothesis 13.2 and that G is nilpotent. Show that |Irrs(G)|
= |Irr(C)|. Do not assume Theorem 13.25.
Hint Use Problem 13.7.
(13.9) Assume Hypothesis 13.2 and that G is solvable. Show that C > 1
iff | Irrs(G)| > 1. Do not assume Theorem 13.25.
(13.10) Let JVo T with N ^G-oT and (|T: G|, |G: JV|) = 1. Assume
that one of Y/G or G/N is solvable. Let K/N be a complement for G/N in
T/JV and assume CGjN(K/N) = 1. Let 9 e Irr(JV) be invariant in K. Show that
there exists a unique T-invariant / e Irr(G) with [9G, /] ^ 0.
Hints Use the argument of Theorem 13.31 to reduce to the case
(| G |, | T: G |) = 1. Use Problem 13.5(b) if T/G is solvable.
(13.11) Let S be solvable and let H be a group with (\S\, \H\) = 1. Let
G = HxHx-xH where there are \S\ factors and let S act on G by

Problems
239
permuting the factors regularly. Let C = CG(S) and note that
C = {(h,h,...,h)\heH} = H.
Also, if / e Irrs(G), then / = 9x9x---x9 for some 9 e Irr(H). If
(x)n(G, S) = p, show that p((h, h,...,h)) = 9(/j|s|).
(13.12) Let JV <i G and suppose 9 e Irr(JV) is invariant in G. Suppose 9G = ex
for some / e Irr(G). (Thus (G, JV, 9) is a fully ramified character triple and
e2 = | G: JV|.) Assume that S is solvable, acts on G, leaves JV and 9 invariant
and that (|G: JV|, |S|) = 1. Let C/N = CG/N(S). Show that Xc = ft with
£ e Irr(C) and/2 = |G:C|.
Hwf Use the technique of the proof of Theorem 13.31 to reduce to the
case that (| G |, | S\) = 1 and 9(1) = 1 with 9 faithful. In this case, C = CG(S).
Use Theorem 13.1.
(13.13) Let S act on G and leave JV -a G invariant. Assume that (| S \, \ G: N \)
= 1 and that S acts trivially on G/N. Let 9eIrr(JV) and /elrr(G) with
[Xiv> #] t6 0- Show that 9 is S-invariant iff / is S-invariant. Do not assume
the Feit-Thompson theorem.
Hint In showing that 9 e Irrs(JV) implies / e Irrs(G), it suffices to assume
that G/N is cyclic.
(13.14) In the situation of Theorem 13.1, let JV<i G be S-invariant with
NC = G. Let 9 e Irrs(JV) and <p = (9)tc(JV, S). Let / = 7G(9) (so that I n C =
Ic(<P) by Problem 13.3). If \ji is an irreducible constituent of 9', show that
(ii>G)n(G, S) = (W)n(l, S)f.
(13.15) Let E be elementary abelian of order p2 for p j= 2 and let S be
dihedral of order 2p.
(a) Define an action of S on E such that Irrs(£) = {1E} but C^S) > 1.
(b) Define an action of S on E such that CE(S) = 1 but |Irrs(£)| > 1.

Linear groups
Suppose x is a faithful character of G. In this chapter we are concerned
with drawing conclusions about G when given information about /. For
instance, we already know that if / is irreducible, then Z(G) is cyclic and that
if all irreducible constituents of / are linear, then G is abelian. Another, less
trivial example which we have seen is Theorem 3.13.
A faithful F-representation of G with degree n is an isomorphism of G
with a linear group of degree n over F; in other words, a subgroup of GL(n, F).
For our purposes, we will restrict attention to finite linear groups. (It should
be pointed out, however, that stating that an infinite group is a linear group
imposes a type of finiteness condition on it, that is, it guarantees that the
group is not "too badly" infinite.)
We shall also restrict attention to complex linear groups. Thus from now
on, a "linear group" is a finite subgroup of GL(n, C) for some n. A group is
thus isomorphic to a linear group of degree n iff it has a faithful character of
degree n. We say that a linear group is irreducible if the identity map is an
irreducible representation.
(14.1) theorem (BUchfeldt) Let G be a linear group of degree n and let
7t = {p|p is prime, p > n + 1}. Then G has an abelian Hall rc-subgroup.
We need a lemma. Recall that a character / is p-rational (where p is prime)
if its values lie in Qr for some r with pjfr. (This is Definition 6.29.)
(14.2) lemma Let p 76 q be primes such that G has no element of order pq.
Then each / e Irr(G) is either p-rational or ^-rational.
240

Linear groups
241
Proof Let /elrr(G) and suppose that z is neither p-rational nor q-
rational. Then in the notation of the discussion following Definition 6.29,
there exists a e 0P(G) and t e &q(G) with f # X ^ t- Also, by Problem 2.2(b),
Z",ZreIrr(G).
If g eG and pJfo(g), then %(g)e Qm, where \G\ - pam and pjfm. Thus by
definition of 0p(G), we have /(g)" = x{g). Similarly, if qJfo(g), then z(#)
= x(g)-
Since G has no element of order pq, it follows for every g e G that either
P^o(9)or 4^o(g) and we conclude that
l(x-xa)Ax-xl']=o.
Since [/, z1] = 0 = [z°, /], we obtain
0 = [z, z] + Df, ZT = 1 + [Z*. Z1] > 1
and this contradiction proves the result. I
The following easy lemma is a special case of a more general result due to
Schur which we will prove later.
(14.3) lemma Let x be a faithful p-rational character of G and assume that
p divides \G\. Then z(l) > p - 1.
Proof Since z is p-rational, it has values in Qr for some r with pjfr. Let
P^ G with |P| = p. Then zPhas values in Qr n Qp = Q. Let <S = <S(QP/Q).
Then ^ fixes zP and thus permutes the linear constituents of zP. Since z is
faithful, zP has a nonprincipal linear constituent X and l(x) t6 1, where
1 t6 x e P. Since ^ is transitive on the p — 1 primitive pth roots of 1, it follows
that the images of X under ^ take on p — 1 different values at x and thus there
are at least p — 1 different characters in the orbit of X under <§. Since each is a
constituent of xP, the result follows. |
An observation that is often useful when working with linear groups is
that if {Xj} is a family of normal subgroups of G with f]Kt = l,thenGcanbe
isomorphically embedded in the direct product
U(G/Kd via g^(...,gKt,...).
Proof of Theorem 14.1 Use induction on n and for groups with degree n,
induct on | G |. Let z be a faithful character of G with z( 1) = n. Suppose that z
is reducible and write z = Zi + Xi- Let Kt = ker Xt so that G/Kt is
isomorphic to a linear group of degree z,(l) < n. By the inductive hypothesis,
G/Kt has an abelian Hall 7t,-subgroup where nt = {p\p > 1 + z.-(l)}- Since
n c m, it follows that G/X,- has an abelian Hall 7t-subgroup H/K;. If H,- < G,
then the inductive hypothesis yields an abelian Hall 7t-subgroup H of Ht.
Since no prime in n divides | (G/X,): (H,/X,) | = | G: H; |, it follows that H is a

242
Chapter 14
Hall 7t-subgroup of G and we are done if Ht < G. We may thus assume that
H, = G and G/Kt is an abelian rc-group for i = 1, 2. However, Kt n K2 =
ker xi ^ ker /2 = ^er / = 1 and thus G is isomorphic to a subgroup of the
abelian rc-group, (G/K]) x (G/K2)- The result follows in this case. We now
assume that / is irreducible.
If H £ G is a rc-subgroup, let 9 be an irreducible constituent of %H. Then
8(1) divides |H| and so a prime divisor p of 9(1) satisfies n < p <. 9(1) < n
and this contradiction shows that 9(1) = 1. Thus Xh is a sum of linear
characters and since / is faithful, it follows that H is abelian.
Suppose there exists M <i G with \G:M\ = perc. Let H be a Hall n-
subgroup of M, which exists by the inductive hypothesis. If H <i G, let
P e Sylp(G). Then //P is a Hall rc-subgroup of G and we are done. Suppose
then, H~fi G and choose a Sylow subgroup Q of H with Q~& G. Then Q is
Sylow in M and thus G = MNG(Q) by the Frattini argument. Now H
is abelian and so H £ NG(Q). We have |G :NG(Q)| = |M :NM(Q)|
which divides \M\H\. Thus |G: NG(Q)| involves no primes from n. Since
NG(Q) < G, it has a Hall 7t-subgroup which is one for G. We may thus assume
that no such M exists.
Next,supposeZ £ Z(G)with \Z\ = pen.ThenXz = X( 1Mwitho(^) = P-
Thus (det(/))z = 1*(1) # lz since /(l) < p. Therefore, p|o(x) and it follows
that p 11G : ker(det /) |. This yields a normal subgroup of index p, a
contradiction. (Note that we have reproved part of Theorem 5.6 here.) Thus no
such Z exists.
Now let H £ G be a 7t-subgroup of maximum possible order. Suppose H
is not a Hall 7t-subgroup. Then there exists qen with q \ \ G : H \. In particular,
H ¥= 1 or else H < QeSyl,(G) which violates the maximality of H. Let
x e H have prime order pen and let C = CG(x). Then C < G by the previous
paragraph and thus C has a Hall 7t-subgroup K by the inductive hypothesis.
Since H is abelian, we have H £ C and thus | H | < | K \. The maximality of
|H | yields \H\ = \ K \ and H is a Hall 7t-subgroup of C. In particular, qJ(\C:H\
and it follows that q \ \ G : C \.
LetxePe Sylp(G). Then P is abelian and so P £ C(x) = C and pj( \ G: C \.
Thus p t6 q. Since p can be any prime divisor of \H \, we have q^|H \.
Therefore, q)(\C\ since H is a Hall 7t-subgroup of C. We conclude that x centralizes
no element of order q in G.
We claim that G contains no element of order pq. Otherwise, there exist
commuting y, zeG with o(y) = p and o(z) = q. However, H contains a full
Sylow p-subgroup of G since C does and H is a Hall 7t-subgroup of C. Thus
H contains a conjugate of y which we may suppose to be x. Since x centralizes
no element of order q, this is a contradiction and proves the claim.
Lemma 14.2 now yields that/ is r-rational for r = por<j.Thusx(l) > r — 1
by Lemma 14.3. This is a contradiction since ren and the proof is complete. |

Linear groups
243
(14.4) corollary Suppose x(l) is prime for every / e Irr(G) with x(l) > 1.
Then G is solvable.
Proof The hypothesis is inherited by factor groups and by normal
subgroups and so we may assume that G is simple. Let /elrr(G) with
minimal /(l) > 1. By Problem 3.3, /(l) = p > 2. Let n = {q\q is prime,
q > p + 1}. Since G is simple, / is faithful and Theorem 14.1 yields an abelian
Hall rc-subgroup A £ G.
Let 1 ¥= a e A. Then A £ CG(a) and the only prime divisors of | G: CG(a) |
are < p + 1 and hence <p. If \ji e Irr(G) and t/^(l) £ {1, p}, then
(.A(l),|G:CG(a)|)= 1
and by Burnside's Theorem 3.8, we have \p(a) = 0 or a e Z{\p). Since G is
simple, Z(t/f) = 1 and hence i/f(a) = 0. Now
0 = £ flDfti) = 1 + p £ £(a)
«€lrr(G) <SeIrr(G);4(l) = p
and hence 1/p is an algebraic integer. This contradiction completes the
proof. |
In the situation of Corollary 14.4, Problem 12.3 applies since G is solvable.
It follows that |c.d.(G)| < 3 and thus G" = 1 by Theorem 12.15.
Our next results concern finding normal subgroups of a linear group of
degree n whose order is divisible by a prime which is large when compared
with n.
(14.5) theorem (D. L. Winter) Let G be a solvable irreducible linear
group of degree n. Suppose that a Sylow p-subgroup of G is not normal. Then
n is divisible by a prime power q > 1 such that q = — 1, 0, or 1 mod p.
Proof Let / e Irr(G) be faithful with /(l) = n. Suppose G is a
counterexample to the theorem with minimum possible order. We argue first that
every proper normal subgroup of G has index divisible by p and has a normal
Sylow p-subgroup.
Let M <i G be proper and let 9t,..., 9r be the distinct irreducible
constituents of Xm ■ Letm be the common degree of the fy. Then m \ n and hence no
prime power q > 1 with q = -1, 0, or 1 can divide m. Since |M\ < \G\, it
follows that M/ker fy has a normal Sylow p-subgroup. Now (~)ker 9, £ ker /
= 1 and thus M is isomorphic to a subgroup of
(M/ker St) x ■ • ■ x (M/ker 9r).

244
Chapter 14
It follows that M has a normal Sylow p-subgroup which is necessarily normal
in G. Since G does not have a normal Sylow p-subgroup, we conclude that p
divides | G: M | as claimed.
Now let K be a maximal normal subgroup of G and let P e Sylp(K). Then
P <iG and \G:K\ = p. Since G is not a p-group, we have P < K and we
choose L^P such that K/L is a chief factor of G. Write | K : L \ = a and let
SeSylp(G).
Now LS < G and pJc \ G: LS \. Thus LS ^G G. Since K(LS) = G, it follows
that CK/L(S) < K/L. However, CK/L(S) -a G/L and we conclude that CK/L(S)
= 1. Since S is a p-group, it follows that a = 1 mod p.
We claim that Xl reduces. Otherwise, Xls is irreducible and the minimality
of G yields S <i LS. Since p/fa(l) it follows that Xs is a sum of linear
constituents and hence S is abelian. Thus S £ CG(P) -a G and hence pjf\ G : CG(P)|.
By the first part of the proof, we conclude that CG(P) = G and P £ Z(G).
Therefore, L = P x N where JV = Op.(L) <i G. Since S <i LS and S n JV = 1,
it follows that S £ CG(JV) <a G and thus pjf \ G : CG(N) |. Therefore, CG(JV) = G
and JV c Z(G). Thus L = NP ^ Z(G) and since /L e Irr(L), we have n = 1
and G is abelian. This is a contradiction and proves that Xl reduces, as claimed.
Since pJfx(l\ we have Xk e Irr(K). Since /L ^ Irr(L) there are two
possibilities by Theorem 6..18. Either Xl is a sum of |K : L| = a distinct irreducible
constituents or else Xl = ei>, with "A e Irr(L), and e2 = a. In the first case,
a > 1 is a prime power dividing n and a = 1 mod p, a contradiction. In the
second case, e > 1 is a prime power dividing n and e2 = a = 1 mod p. Thus
e = ± 1 mod p and the proof is complete. |
(14.6) corollary (ho) Let G be a solvable linear group of degree n and
let p > n + 1 be a prime. Suppose that a Sylow p-subgroup of G is not
normal. Then G is irreducible, p = n + 1, and n is a power of 2.
Proof If G is irreducible, then by Winter's Theorem 14.5, there exists a
prime power q > 1 that divides n such that q = — 1, 0, or 1 mod p. Thus
p— l>n><j>p— 1 and hence n = p — 1 is a prime power. We
conclude that n is a power of 2 and the proof is complete in this case.
If G is reducible, write / = Xi + li - where / is a faithful character of G of
degree n. Let Kt = ker Xi so that K± c\K2 = \ and G is isomorphic to a
subgroup of (G/Ki) x (G/K2). Each G/Kt is isomorphic to a solvable linear
group of degree x,(l) < n. Working by induction on n, it follows that each
G/Kt has a normal Sylow p-subgroup and hence so does their direct product.
Since G does not have a normal Sylow subgroup we have a contradiction and
the proof is complete. |
The hypothesis that G is solvable in Corollary 14.6 can easily be relaxed
to the assumption that G is "p-solvable." We say that G is p-solvable if there

Linear groups
245
exist subgroups Nt <i G with
1 = N0 c JVt c . •. c Nk = G
and such that each factor Ni+ l/Ni is either a p-group or of order prime to p.
(14.7) corollary Let G be a p-solvable linear group of degree n < p — 1
and suppose that a Sylow p-subgroup of G is not normal. Then n = p — 1 is
a power of 2 and G is irreducible.
Proof Assume either that G is reducible, n < p — 1 or that n is not a
power of 2. Use induction on \G\. Let M be the next to last term in a p-
solvable series for G so that M < G, G/M is either a p-group or a p'-group and
M is p-solvable. By the inductive hypothesis, M has a normal Sylow p-
subgroup P and thus we may assume that G/M is a p-group.
Let Q/P e Sylq(M/P) for some prime q ^ p. The Frattini argument yields
MNG(Q) = G and thus |G:NG(Q)| = |M:NM(Q)|. Since P £ Q <= NM(Q)
and P e Sylp(M), we conclude that p/f | G : NG(Q) |. Let S e Sylp(NG(Q)). Thus
SeSylp(G).
Now SQ is solvable and hence S <i SQ by Corollary 14.6. Since | G : Q |
is not divisible by <j, we conclude that qJf\G: NG(S)|. Since |G:NG(S)| is
independent of the choice of S e Sylp(G) and q =£ p is arbitrary, the result
follows. |
It is conjectured that Winter's Theorem 14.5 holds for all p-solvable
groups. The conjecture is known to hold for n < 2p + 1. Unlike the ease with
which Corollary 14.7 follows from Corollary 14.6, the facts for n > p seem
quite deep. Even the special case where G has a normal p-complement is open
if n is significantly larger than 2p. The results which have been obtained all
seem to depend heavily on the Glauberman correspondence, and in
particular on Theorem 13.14. The proofs are too complicated to give here.
Now we drop all hypotheses of solvability or p-solvability. Suppose G is a
linear group of degree n and a Sylow p-subgroup of G is not normal. How big
can p be? The first bound was established by Blichfeldt and the best possible
bound, p <, 2n + 1, was proved by Feit and Thompson. The Feit-Thompson
proof depends on a very deep result of Brauer which gives the bound under
the additional assumption that p^ld. Here we shall prove p < (n + l)2
and also show how the Feit-Thompson reduction to the case p2Jf\ G\ works.
We need some preliminary results.
(14.8) lemma Let H £ G be abelian and let y be the set of nonprincipal
irreducible constituents of (1H)G. Compute a = min{/(l)/[x, (lH)G]lze^}-
Then t/^(l) > a — 1 for every nonlinear ijj elrr(G).

246
Chapter 14
Proof Let \ji e Irr(G) with \ji(X) > 1. Write i/^ = \G + £ axx, where the
sum runs over nonprincipal x £ Irr(G). Since H is abelian, we have
•A(l) < [>«, -Ah] = E-Ah-Ah, lfl] = [-/"A, (1h)°] = 1+1 afo, (l„)Gj\
However, [/, (1H)G] <, Z(l)/a for / e S? and this yields
Ml) < 1 + (l/a)£az/(l) < 1 + (l/a)(^(l)2 - 1).
Thus o#(l) - 1) < i/<l)2 - 1 and since i/f(l) > 1 we obtain a < i/f(l) + 1
as desired. |
(14.9) lemma Let N <i G and suppose g e G with JV n C(gr) = 1. Let / be a
character of G such that \jN, ljv] = 0. Then %(g) = 0.
Proof Let g = gN e G/JV. Then CG/N(g) 3 CG(g)N/N and hence
|CG/JV(0)| > |CG(9)JV/JV| = \CG(g)\. Thus
I l«%)|2> I K(9)l2-
i^eIrr(G/N) ^elrr(G)
Viewing Irr(G/JV) £ Irr(G), we conclude that £(g) = 0 for £ e Irr(G) with
JV ^ ker <!;. The result follows. |
Note that the above proof is essentially a repetition of the proof of
Corollary 2.24. Also, if g and JV are as in the Lemma 14.9, then all elements of
the coset Ng are conjugate in G and thus the result follows from Problem
2.1(b).
(14.10) theorem (Feit-Thompson) Let H £ G be abelian with H n Z(G)
= 1. Assume for every geG — Z(G) that CG(g) has nontrivial intersection
with at most one conjugate of H in G. Let / e Irr(G) with H £ ker /. Then
(a) Z(1)2>|H|[ZH, 1H]2.
Also, if z(l) > 1 and H is contained in no proper normal subgroup of G, then
(b) (l + z(l))2>|H|.
Proof Let JV = NG(H) and
X = {xeG- Z(G)|C(x)nH > 1}.
Note that H - {1} £ X and that JV £ N(X). We claim that X is a T.I. set
and that JV = N(X). In particular, X £ JV.
Let g e G be such that X r\ X9 ^ 0. Choose x e X with x9 e X. Then
C(x) n H > 1 and so C(x9) n HB > 1. Also, C(x9) nH>l and hence
H = Hg by hypothesis. Thus g e JV and X = Xg and the claim is established.
We have
\G\ = \G\[x, Z] = I l*(0)l2 > Z(l)2 + IG: JV| £ |Z(x)|2
geG JceX

Linear groups
247
and thus
(i) IlxWI2<|N|.
xeX
Now write Xn = a + P, where [aH, 1H] = 0 and fiH = 0(1)1H. (This is
possible since H <i JV.) Since H £ ker /, we have a ^ 0 but we allow the
possibility that /? = 0.
Write Z = Z(G). Then £Z€Z |/(z)|2 = |Z|/(1)2 and since Z r\X =0,
we conclude from (1) that
(2) |Z|/(1)2 + |JV|> £ lz(x)|2= £ |a(x)+ /?(*) |2.
By Lemma 14.9, we have a(y) = 0 for y e JV — (X u Z) and thus
(3) X |a(x) + 0(x)|2 = |JV|[a,a]+2|JV|[a,/?]+ £ |0(x)|2.
Since [a, /3] = 0 and [a, a] > 1, (2) and (3) yield
\Z\X(l)2 + \N\ > \N\ + £ |/?(x)|2.
xeX u Z
Because HZ ^ X u Z, we obtain
(4) |Z|Z(1)2 > £ |/?(x)|2.
*eHZ
Write Xz = Z(1M- Then /?z = /?(1)1 and since /?H = /?(1)1H, we have
/?(*«) = 0(l)A(z) for he H and z e Z. Thus | 0(x) |2 = 0(1)2 for x e HZ and (4)
yields
/?(l)2|tf||Z|<|Z|x(l)2.
Thusx(l)2 > 0(l)2\H\ and since 0(1) = [/H, lH], the proof of (a) is complete.
Now assume that no proper normal subgroup of G contains H. Let Sf
be the set of nonprincipal irreducible constituents of (1H)G. If ip eSf, then
H £ ker ip and (a) yields i/<l)2 > \Jia, \„~\2\H\ and
Itfl1'2 < a = min{t/<l)/[tMlH)G]|</>e.n
Lemma 14.8 then gives for /(l) > 1 that
Z(l) + 1 >a> l#|1/2
and (b) follows. |
(14.11) theorem (Feit-Thompson-Blichfeldt). Let G be a linear group of
degree n and let p be a prime. Assume one of the following.
(a) Every subgroup of G of order not divisible by p2 has a normal
Sylow p-subgroup and p > n + 1.
(b) p>(n + l)2.
Then G has a normal Sylow subgroup.

24S
Chapter 14
Proof Use induction on \G\. We may thus suppose that every proper
subgroup of G has a normal Sylow p-subgroup. Let P e Sylp(G) and assume
P -56 G. Since P has a faithful character of degree n < p and this character can
have no nonlinear irreducible constituents, we conclude that P is abelian.
If P n Z(G) ^ 1, then by Theorem 5.6, we cannot have P n Z(G) £ G'
and thus p||G: G'|. It follows that there exists N <i G with |G:JV| = p.
Since JV has a normal Sylow p-subgroup, we conclude that G is p-solvable.
Then G violates Corollary 14.7 since n < p — 1 and P ~fi G. We conclude
that P n Z(G) = 1.
Now if 1 # y e P n P9, then P, P9 e Sylp(C(y)) and this forces P = PB
since C(y) < G. Thus P is a T.I. set. Now suppose that xeG — Z(G). Then
C(x) has a unique Sylow p-subgroup S and if S > 1, then S is contained in a
unique conjugate of P. It follows that C(x) has nontrivial intersection with
at most one conjugate of P.
IfP £ M -a GwithM < G,thenP <M and thus P -a G, a contradiction.
We have now shown that P satisfies the hypotheses of Theorem 14.10(b) and
thus \P\ < (i/^l) + l)2 for every nonlinear ifr e Irr(G).
Since G is nonabelian, the given faithful character must have some
nonlinear irreducible constituent and we conclude that \P\ < (n + l)2.
Thus p <(n + l)2, contradicting hypothesis (b). We are therefore in the
situation of hypothesis (a) and in particular, p2||G|. Thus p2 < \P\ <
(n + l)2 and p < n + 1 which is our final contradiction. |
As was mentioned before, Brauer proved that if G is a linear group of
degree n and p2Jf\G\ where p > 2n + 1, then a Sylow p-subgroup of G is
normal. It then follows from Theorem 14.11(a) that the hypothesis p2X IG | is
unnecessary.
There is a great deal of information known about linear groups of degree
n in which a Sylow p-subgroup is not normal and n < p <2n + 1. This all
requires deep results from Brauer's "modular character" theory and we will
not discuss it further here.
There are many ways in which the structure of a group is limited in terms
of its degree as a linear group. For example, there exist integer valued
functions/, such that for linear groups G of degree n we have:
(a) If G is solvable, then d.l.(G) < /t(n).
(b) If G is p-solvable and pe divides | G: Op(G)|, then e < f2(n).
(c) If G is a p-group and | G: 3>(G) | = pe, then e <; /3(n).
A much more general result of this type is Jordan's theorem.

Linear groups
249
(14.12) theorem (C.Jordan) Let G be a linear group of degree n. Then
there exists abelian A <i G such that
|GM| ^(n!)12""t"'+1)+1)
where 7t(/c) denotes the number of primes <, k.
In fact, the existence of the functions/,,/2, and /3 mentioned above
follows easily from Jordan's theorem, although the best known bounds for
/i,/2, and/3 are very much better than those which can be derived from the
inequality in Theorem 14.12 (or from any other known bound for Jordan's
theorem). There is no reason to suppose that the index of an abelian normal
subgroup of maximum possible order in a linear group of degree n can be
anywhere nearly as large as (n!)12n('t(''+1)+1). (For instance, if n = 2, the
"correct" bound is 60 rather than 2 ■ 126.) The good bounds for the functions
j\ are proved by methods independent of Jordan's theorem.
We begin work now on a proof of Theorem 14.12 which is due to Fro-
benius. As the reader will see, this proof has an unusual geometric flavor.
Recall that a square complex matrix U is said to be "unitary" if UJ =
U"1. Note that the unitary n x n matrices form a subgroup of GL(n, C). Also,
if U is unitary, then there exists unitary V such that V~lUV is diagonal.
(More generally, this holds for all normal matrices U, that is, those which
satisfy UU* = U*U, where U* = UJ.) Also, the diagonal matrix VlUV
is unitary since both U and V are. Write V1UV = diag(lt,..., kn) so that
the Xj are the eigenvalues of U. Since V~lUV is unitary, it follows that
kfl = X] and this proves that the eigenvalues of a unitary matrix lie on the
unit circle.
(14.13) lemma Let A and B be n x n complex unitary matrices. Assume
that the eigenvalues of B lie in the interior of some arc of length n on the unit
circle. Suppose that A commutes with A~ iB~ iAB. Then A commutes with B.
Proof We may conjugate A and B by a unitary matrix so as to diagon-
alize B and hence we may assume that B = diag(ft1,..., b„). Since a
permutation matrix is unitary, we may rearrange the bj in any desired order by
conjugating both A and B by an appropriate permuation matrix. We may
thus write b} = emj where &t < 92 < ■ ■ ■ < 9„ < 9t + n.
Write A = (a^). We shall show that a^ = Qiib^^ bv. This immediately
yields that AB = BA. Now write C = A~lB~lAB. We have A"1 = A1
and B~1 = B so that C = A1 BAB. Also, since AC = CAv/e have
A1 BAB = C = AC A-1 = B-lABA~v = BAB A1
Now evaluate the diagonal entry, cmm of C. We have
li v

250
Chapter 14
and thus
We compare imaginary parts and obtain
£ |a„J2 sin(9m - 9,) = I |amv|2 sin(9v - 9J.
/i V
Thus for all m, 1 < m < n, we have
(*) £(|amv|2 + |avJ2)sin(9v-9J = 0.
V
We now use (*) to prove that if 9; j= 9k, then aJk = 0. Suppose this is false
and choosey minimal such that there exists k with 9,- < 9k and either aJk ^ 0
or akj ¥= 0. Take m = j in (*). If v < j, then either 9V = 9; in which case
sin(9v — 9;) = 0 or 9V < 9j in which case avj = 0 = aJV by the minimality of
j. Thus (*) yields
Z(|flJ-vl2 + |flvjl2)sin(9v-9j) = 0.
Now for v > j we have 9^ < 9V < 9; + n and thus sin(9v — 9^) > 0. Also,
Ifl/vl2 + l«vjl2 ^ 0- We conclude that for each value v > j, either
Ifljvl2 + |flVJ-l2 = 0
or sin(9v — 9j) = 0. Now 0 < 9k — 9j- < n and hence sin(9k — 9;) > 0.
Therefore |a;k|2 + |akJ|2 = 0 and thus ajk = 0 = akj. This is a contradiction
and completes the proof. |
(14.14) lemma Let A and B ben x n complex unitary matrices and
suppose that the eigenvalues of A lie in an arc of length a < n on the unit circle.
Then the eigenvalues of A" iB~ iAB lie on the unit circle between — a and a.
Proof Certainly A" iB~ iAB is unitary and so its eigenvalues lie on the
unit circle. Let a be an arc of length a which contains the eigenvalues of A
and write C = B~ lAB. We need to find the eigenvalues of A" lC.
Let 1 be an eigenvalue of A" iC and let the column vector x be a
corresponding eigenvector so that A" 1Cx = he and thus Cx = A Ax and 3cTCx =
M^x1 Ax). Thus it suffices to show that arg(3cTy4x)ea and arg(3cTOc) ea,
where arg(a) = a/|a| for 0 j= aeC. Since both A and C are unitary with
eigenvalues in a it suffices to show that
arg(yJUy)ea
whenever y#0 and U is unitary with eigenvalues in a. Let U and y be such
and let V be unitary with V~lUV = D, a diagonal matrix. Let z = V~ly so
that
yJUy = zJVJUVz = zJDz

Linear groups
251
sinceFT= V '.NowifD = diag(rflf..., d„)andz = co\(zu ..., z„),wehave
zJDz = 1\zt\2dt.
Since the dt lie in a and \zi\1 > 0, it follows that £ |Zj|2rfj lies in the
infinite wedge which is the union of all rays from the origin through points of a.
Since a < n and some |z,|2 is nonzero, it follows that £ |z(|2dj ^ 0 and
arg(£ I Zj |2rfj) e a as desired. |
(14.15) theorem (Frobenius) Let G be a linear group, let A, B e G and
suppose that the eigenvalues of A and £ lie in arcs of length a and /?
respectively, on the unit circle. Then
(a) If /? < n and A commutes with the commutator [A, B], then
LA, B] = 1.
(b) If a < 7t/3 and ft < n, then [A, B] = 1.
Proof By Theorem 4.17 we can conjugate all of the elements of G by a
matrix such that the resulting matrices are all unitary. We may thus suppose
that A and B are unitary matrices. Now (a) is just a restatement of
Lemma 14.13.
To prove (b), conjugate by a unitary matrix which diagonalizes A. We may
thus assume that A is diagonal. Construct elements Bt e G by setting B0 = B
and B, = \_A, B^J for i > 1.
By Lemma 14.14, the eigenvalues of Bt lie between — n/3 and 7t/3 on the
unit circle for i > 1 and thus for each i > 0, they lie in the interior of some
arc of length n. Thus by part (a) we see that if Bi+1 = 1 for any i > 0, then
Bj = 1. It thus suffices to prove that Bt = 1 for some i.
If M is any complex matrix, we define 9(M) = tr(MTM) so that if M =
(m^), we have 9(M) = £,„ |m,J2. Thus 9(M) > 0 and 9(M) = 0 iff M is a
zero matrix. If U is unitary, we have
9(UM) = tr(MTUTUM) = tr(MTM) = 9(M).
Now we compute
(*) 9(Bi+1 - 1) = 9(B,A(Bl+1 - 1)) = BiB^A-'Br'AB, - 1))
= 9(/4Bi - BtA) = 9iA(Bt - 1) - (B, - 1)A).
Now write B, — 1 = (ft^v). Since A is diagonal, we can write A =
diag(alf a2,..., a„), where all of the a} lie inside some arc of length n/3. The
(fi, v) entry of A(B{ — 1) - (B; - 1)A is a^fc^v - t^a, = ^(a,* - «v)- Note
that \all — av| < 1. Equation (*) now yields that
9(Bi+1 - 1) = 11VI2K - «vl2 ^ 11VI2 = S(B, - 1),
t*t V P> V
where in fact the inequality is strict unless all & = 0.

252
Chapter 14
If no Bt = 1, we therefore have
9(B0 - 1) > 9{Bi - 1) > 9(B2 - 1) > ■ ■ ■.
Since G is a finite group, there are only finitely many Bt and this is a
contradiction. We conclude that Bt = 1 for some i and the proof is complete. |
Observe that we have used our standing assumption that the groups we
consider are finite twice in the above proof. The first time was in the appeal to
Theorem 4.17 which is false for infinite groups. This use of finiteness could
have been avoided by assuming that G was contained in the group of unitary
matrices. The second application of finiteness in the last paragraph of the
proof is more fundamental. For instance, part (b) of the theorem is not valid
for the full unitary group.
(14.16) theorem Let G be a linear group of degree n. Then there exists
abelian A <i G such that
\H:H nA\ < 12"
for every abelian subgroup H ^ G.
Proof Since G is a finite group, the eigenvalues of every element of G
lie on the unit circle. If g e G, let a.(g) be the length of the shortest closed arc
which contains the eigenvalues of g. Note that a is a class function on G.
Put A = <a e G| a(a) < 7t/3>. Then /1<G and A is abelian since the
generators of A commute by Theorem 14.15(b).
Now let H c G with H abelian. By Maschke's Theorem 1.9 we may
conjugate all of the matrices in G so as to assume that all h e H are diagonal
matrices.
Partition the unit circle into twelve half-open arcs, abQ2 o12, each
of length n/6. Each element heH determines a function fh from {i\ 1 < i < n}
into {j\ 1 < j < 12} by writing h = diag(alf..., a„) and setting fh(i) = j if
a.ea,-. Note that there are at most 12" different functions fh that can be
obtained this way.
Suppose x,yeH with fx = fy- Write x = diag(alf..., a„) and y =
diag(/?lf..., /?„). Then for each i, a; and /?,- lie in the same a,- and hence a,/?,
lies strictly between — n/6 and n/6 on the unit circle. Thus xy~l eA and it
follows that elements in distinct cosets of A n H in H determine different
functions. Thus \H :A n H\ < 12" and the proof is complete. |
(14.17) lemma Let P be a linear p-group of degree n. Then there exists
abelian A <i P such that | P : A \ divides n!
Proof By Corollary 6.14, P is an M-group and hence if \ji e Irr(P), then
there exists H c p with \P: H\ = t/^(l) and such that t/^H has a linear
constituent. Choose such a subgroup H^ for each \p e Irr(P).

Linear groups
253
Let x be a faithful character of P of degree n and let Q be the set of all
right cosets of all H^ for irreducible constituents \p of /. Then | Q | < n and G
acts on Q by right multiplication. Let A be the kernel of this action. Thus
| P: A | divides n!
To show that A is abelian, it suffices to show that Xa is a sum of linear
characters. However, for each irreducible constituent \p of / we have A £ H^,
and thus i//A has a linear constituent. Since A <i G, it follows that all
irreducible constituents of i/^ are linear and the result follows. |
Proof of Theorem 14.12 Let A -a G be as in Theorem 14.16 so that A is
abelian and \H :H n A\ < 12" for every abelian subgroup H £ G. We claim
that |G:X| < (n!)12n('t(''+1)+1). Factor |G:X| into prime powers. For each
prime p let rp = \G:A\P.
By Blichfeldt's Theorem 14.1, there exists an abelian Hall subgroup H0 of
G for the set of primes >n + 1. We have \G : A\ = \G :H0A\\H0A inland it
follows that
I! rp = | H0 A :A\ = \H0:H0nA\< 12".
p>n+l
For p < n + 1, let P eSylp(G) and let Hp = P be abelian with |P: Hp|
dividing n! (using Lemma 14.17). Then \P : Hp\ < (n!)p and
rp = IPAMI = \PA:HPA\\HPA:A\ = |P:PnHpA||Hp :Hpn A|
<\P:Hp\\2n<{n%W.
Thus
EI rP ^ (n (" Op )12""("+ ° = (n !)12"'t"'+ !>.
p<n+l \p /
The result now follows. |
We can also use Frobenius' Theorem 14.15 in a different way.
(14.18) theorem Let p > 7 and let G be an irreducible linear group of
degree n <2p with n^p. If the Sylow p-subgroups of G are nonabelian, then
p2Jf\ G : Op(G)|. Thus in any case, G/Op(G) has abelian Sylow p-subgroups.
Proof Let PeSylp(G) be nonabelian so that P'nZ(P)^l. Let
U S F nZ(P)with|l/| = p.
Let x £ Irr(G) be the character of the given faithful representation so that
/(l) = n. Since P is not abelian, zPhas some nonlinear irreducible constituent
\ji. We must have t/^(l) = p. Since /(l) < 2p, we conclude that xP = i> + A,
where A is a sum of linear characters.
Since U £ P', we have U £ ker A and since U £ Z(P), we also have
\jiv = pfi for some linear fi e lrr(U). Thus Xv = W + (n — P)lu- Since / is
faithful, we have n ¥= lv

254
Chapter 14
Now pick ueU, with fi(u) = e2lti/p, and note that 2n/p < n/3 since p > 7.
Therefore, u has only two distinct eigenvalues, namely 1 and e2*'/p, and these
lie in an arc of length < 7t/3 on the unit circle. By Frobenius' Theorem 14.15(b)
it follows that u commutes with all of its conjugates in G and thus A =
(ua\g e G> <i G is abelian.
Write Xa = e Yj= i ^i>wnere trie i; are distinct linear characters of A and
et = n. Since 1/ = <u> £ A, we conclude that e divides \%v, /z] = p. Since
also e | n, we have e = 1 and t = n. Now let K be the kernel of the permutation
action of G on {!,}. Then IG: K | divides n! and hence p2Jf\ G: K |. If 9 is any
irreducible constituent of xK, then 9^ is the sum of an orbit of the action of
K on {1J. By definition of K, it follows that 9(1) = 1. We conclude that K is
abelian and so the Sylow p-subgroup of K is contained in Op(G). The result
follows. |
We now present the promised generalization of Lemma 14.3.
(14.19) theorem (Schur) Let G have a faithful p-rational character of
degree n and let P e Sylp(G) with \P\ = f. Then
a < t ln/{p - l)p'] < np/{p - l)2.
i = 0
The brackets above denote the greatest integer function and thus the
"infinite" sum has only finitely many nonzero terms. If n < p — 1 then all
terms in the sum are zero and the result that a = 0 is Lemma 143.
Note that G is really irrelevant in Theorem 14.19; the result is really about
P. Also, a p-rational character of a p-group is necessarily rational valued
since the values lie in Qpa r\ Qr where pjfr.
Suppose x £ Irr(P) is faithful and we wish to bound \P\. Obviously,
knowledge of /(l) is not sufficient since even if z(l) = 1, P could be an unboundedly
large cyclic group. However, | P \ is bounded in terms of /(1) and the degree of
the field extension Q(/) 2 Q. If z eZ(P) with o(z) = p, then x(z) = /(l)e,
where e is a primitive pth root of 1. Thus e e <Q>(/) and hence p — 1 divides
| Q(x): QI ■ Since |Qp«:Q| = (p- l)pfl_1, it follows that if P # 1, then
| Q(x): Q | is of the form (p — l)p' for some integer t > 0.
(14.20) theorem Let IPI = p", where p is a prime and a > 0. Let / e Irr(P)
be faithful with *(1) = / and | Q(x) :Q\ = {p- l)p'. Then
a < (f - l)/(p - 1) + (t + l)f.
Proof We use induction on/. Iff = 1, then P is cyclic and since / is
faithful, it follows that Q(/) = Qp« and thus t = a — 1. In this case, the
inequality is actually equality.

Linear groups
255
Now suppose / > 1. Since P is an M-group, there exists a maximal
subgroup H and 9 e Irr(H) with 9P = x- Thus \P : H\ = p and xH = £f= i 9;,
where the 9, are the distinct conjugates of 9 under P. Note that the fields
0(9,) are all equal and that since 8P = /, we have Q(x) £ 0(9).
Suppose Q(x) = 0(9). Let K,- = ker 9;. Thus f^i = 1 and H is iso-
morphically contained in (H/KJ x ■ ■■ x (H/Kp). The |H:K,| are all
equal, say to pb. Thus a < pft + 1. Since 9(1) = f/p, the inductive hypothesis
yields
b < ((//p) - l)/(p - 1) + (t + I)f/p
and hence
a < pb + 1 <; (/ - p)/(p - 1) + (t + l)f + 1 = (/- l)/(p - 1) + (t + l)f
as required.
The remaining case is where Q(/) < 0(9). Consider the Galois group
0 = 0(O(9)/Q(x)) so that |9| = |0(9): 0(/)|. In particular, <S ± 1 is a p-
group. Since ^ fixes /, it permutes the irreducible constituents 9; of /H.
Since the stabilizer of 9 in ^ is trivial, the orbit of 9 under ^ has size | # |. It
follows that \<S\ = p and |0(9): 0| = (p - l)p'+1. Also, the 9, are
^-conjugate and hence all ker 9, are equal. Since f)ker 9, = 1, we conclude that
ker 9 = 1. We can now apply the inductive hypothesis to 9 e Irr(H) and
obtain
a - 1 < ((f/p) - l)/(p - 1) + (t + 2)f/p = (f- l)/(p - 1) + (t + \)f/p
<(/-l)/(p-D + (t + D/-l
and the proof is complete. |
We note that if p ^ 2, it is always possible to choose H in the above proof
so that 0(x) = 0(9). The inequality in Theorem 14.20 is best possible.
Proof of Theorem 14.19 Define the function a on positive integers by
a(/c)= I[/c/(p-l)p'].
f = 0
Since \_k/(p — l)p'] < k/(p — l)p' and this inequality is strict for large i, we
have
a(/c)<(/c/(p-l))f>-' = /cp/(p-l)2
i = 0
and the second inequality in the statement of the theorem follows.
Since \_x + y] > |>] + [>], we have a.(k + I) > a(k) + a(/). We know
that Xp is rational valued, where / is the given character of G. If Xp = Xi + li >
where the /,- are rational valued, then working by induction on n we have
a; < a(/j(l)), where |P/ker Xi\ = p"'- By the usual argument, a ^ at + a2

256
Chapter 14
and so
a < a(Zl(l)) + afc2(l)) < a(Zl(l) + *2(1)) = a(n)
as desired.
We may now suppose that Xp is not the sum of two rational characters.
It follows that Xp = |j=i £,■> where the <!;,■ e Irr(P) constitute an orbit under
the Galois group 0 = 0(OP«/O). Let £ = £t. Then ker £,- = ker £ for all i
and thus 1 = f)ker £t = ker £ and ^ is faithful. The stabilizer in <§ of £ is
0(OP«/O(£)) and it follows that the orbit size r = |0(£): Q| = (p - l)p' for
some t > 0. Let/ = £(1) = £,(1) for all i so that n = (p - l)p'f. Now
a(n) = I + p + p2 + ■■■ + p'f
= (f- l)/(p - 1) + / + pf + ■ ■ ■ + p'f
>(f - 1)/(P - 1) + (t + 1)/ > a,
where the last inequality is by Theorem 14.20. |
As an application of Schur's Theorem 14.19, we prove the following.
(14.21) theorem Let G be a p-solvable linear group of degree n. Let
P e Sylp(G) and | P : Op(G) | = p". Then
a<, I[n/(p-l)p'].
i = 0
The proof of Theorem 14.21 depends on some standard facts about p-
solvable groups. The first is nearly trivial, namely that subgroups and factor
groups of p-solvable groups are p-solvable. The second fact is a special case
of what is often called "Lemma 1.2.3" since that was the designation in the
paper of P. Hall and G. Higman where it first appeared.
(14.22) lemma (Hall-Higman) Let G be p-solvable with Op(G) = 1.
Then CG(Op.(G)) £ Op.(G).
Proof Let H = Op.(G) and C = CG(Op.(G)). Then C<GandCis
p-solvable. Let D = Op.(C) so that D <i G and thus D = HUH ^ C, then
C/D is a non trivial p-solvable group and Op.(C/D) = 1. Thus Op( C/D) > land
we let E/D = Op(C/D) and P e Sylp(£) so that P > 1.
Now D = H = C(C) c N(P) and thus P <i P£> = £ <i G. It follows that
P <i G. Since P > 1 and Op(G) = 1, this is a contradiction and the proof is
complete. |
Proof of Theorem 14.21 Let U = Op(G) and H/U = Op.(G/U). By the
Schur-Zassenhaus Theorem, U is complemented in H and we may choose a
complement K. Let JV = NG(K). Since G permutes the set of complements for
U in H and H is transitive on this set, it follows that G = NH = NU. Let
S e Sylp(JV) and put G0 = KS. We have N n 1/ £ S £ G0 and
|G:l/|p = |JV:iVnl/|p=|G0:JVnt/|p.

Linear groups
2S7
We claim that JV n U = Op(G0) and thus it suffices to assume G = G0.
To prove the claim, note that JVn(/<JV and so JV n U <i G0 and thus
JV n t/ £ F = Op(G0). Since V -a G0, K -a G0 and FnK = l, we have
V £ C(X) and thus VU/U £ QKU/U) = C(H/t/). Since Op(G/t/) = 1,
Lemma 14.22 yields C(H/U) £ H/U and hence Ft//t/ £ H/U and Ft/ £ H.
Since Ft/ is a p-group and U e Sylp(H), we have VU = U and thus V £ t/.
Therefore, 7 £ JV n U and the claim is established.
We may now assume that G0 = G so that K is a normal p-complement for
G. Let / be a faithful character of G with /(l) = n. For each irreducible
constituent 9 of Xk, let 9 be the canonical extension of 9 to /G(9) and let 9* =
(9)G e Irr(G). Note that any field automorphism of Q|G| which fixes 9 also
fixes 9* and it follows that 9* is p-rational.
Choose a set y of representatives for the G-orbits of irreducible
constituents of Xk and define \p = £9€y 9*. Then
n = X{\)> X 9(1)|G:/G(9)|= £ 9*(1) = <«1).
Also /K and t/^K have the same sets of irreducible constituents and so ker \p r\ K
= ker x n K = 1. Thus ker »// is a p-group and ker \ji £ Op(G). Now
application of Schur's Theorem 14.19 to G/ker \ji yields the desired result. |
Write ap(/c) = Yj^o W(P ~ 1)p']- With further work one can replace
the inequality a < ap(n) in Theorem 14.21 by a < ap(/zn), where fi = f if
p = 2 and fi= (p — l)/p if p j= 2 is not of the form 2e + 1. With these
improvements, equality can be obtained for all n. If G is solvable, this
improvement is due to J. D. Dixon. The general p-solvable case was done by D. L.
Winter and depends on the results of Chapter 13.
Suppose G is a primitive linear group of degree n. In other words, the
identity map is a primitive representation in the sense of Chapter 5. By
Corollary 6.13, it follows that every abelian normal subgroup of G is central
and thus by Jordan's Theorem 14.12 there are only finitely many possiblities
for the group G/Z(G). These have been explicitly enumerated for certain
small values of n. We give a sample of this type of result although this proof
is not typical.
(14.23) theorem Let G be a primitive linear group of degree 2. Then
|G:Z(G)| = 12, 24, or 60.
Proof Write Z = Z(G) and let /elrr(G) be faithful with /(l) = 2. If
H £ G is nonabelian, then Xh e Irr(H) and it follows that Z(H) £ Z(/) = Z.
Thus if g e G — Z, then C(g) is abelian.

255
Chapter 14
Let Sf be the set of maximal abelian subgroups of G. Then clearly G =
\}S/> and Z = f)Sf. If A, Be & with Ar\B > Z, then C(A r\ B) is abelian
and contains both A and B. Therefore A = B and thus G is a disjoint union:
G = Z u \J(A- Z).
For A e 9> we have \j>A, %A~] = 2 and [*z, Xz] = 4. Thus
£ |Z(X)|2 = 2|A|-4|Z| = 2|A-Z|-2|Z|.
jce A — Z
This yields
|G|= IlzWi2= £lz(*)|2 + I I IzWI2
jceG xeZ AsSf xeA-Z
= 4\Z\ + 2%j\A-Z\-2\<?\\Z\
AeS?
= 4\Z\ + 2(\G\ - \Z\) - 2s\Z\
where s = \Sf\. We thus obtain | G : Z| = 2s - 2.
Now y is a union of conjugacy classes of subgroups. lfAeSf, we want to
compute JV = NG(y4). If A < JV then JV is nonabelian and %N e Irr(JV).
However, %A = X + fi with X =£ fi. It follows that | JV : IN(X) \ = 2 and the restriction
of i to IN(X) is reducible. Thus 7^(1) is abelian and hence A = IN(X). We
conclude that for all A eSf, we have |N(X) :A\ < 2. Also A -^ G.
We now consider the group G/Z = G of order 2s — 2. Let 9~ =
{A/Z\AeSf} so that 9" partitions G_and \2T\ = s. Also, 9~ is a union of
conjugacy classes of subgroups of G. Write ST = &t u 3~2 u ■ ■ ■ u 3~T,
where the 9~i are the distinct conjugacy classes. Let at be the common size of
the subgroups in ^"( and let t{= \&~t\.
IfBe^„ then |G:N(B)| = |G:B|/2or |G:N(B)| = \G:B\ and hence
(1) t( = (s - l)/fl( or t, = 2(s - l)/a,.
We also have
(2) £'< = *.
(3) ^ti(ai-l) = 2(s-l)-l,
and all t( > 1 and a,- > 1. We may assume that a^ < a2< ■■■ < ar and that
if flj = ai+1 then tt > ti+1.
If a! > 2 then £ tj(a; - 1) > 2 £ tt = 2s which contradicts (3). Thus
at = 2 and rt = s — 1 or (s — l)/2. However, if rt = s — 1, then (2) yields

Linear groups
259
t2 = 1, a contradiction. Thus rt = (s — l)/2. If a2 = 2, then t2 = (s — l)/2
forcing t3 = 1, which is impossible. Thus a2 S; 3. If a2 > 4, then
2(s - 1) - 1 = X ti(«i - 1) > (s - l)/2 + 3 £ t,
i=2
= (s - l)/2 + 3(s - (s - l)/2)
= 2(s - 1) + 3,
a contradiction. Thus a2 = 3 and f2 = (s — l)/3 or 2(s — l)/3.
Suppose t2 = 2(s — l)/3. Then
s>tl + t2 = (s- l)/2 + 2(s - l)/3 = 7(s - l)/6 > s - 1
and hence s = 7(s — l)/6 which yields s = 7 and | G : Z | = 12. We now
suppose that t2 = (s — l)/3. Since tt + t2 = 5(s — l)/6 < s, we have r > 3.
If a3 > 6, Equation (3) yields
2(s - 1) - 1 = £ trffl. - 1) = (s - l)/2 + 2(s - l)/3 + £ ^(a,. - 1)
i=3
> 7(s - l)/6 + 5(s - tt - t2)
= 7(s - l)/6 + 5(s - l)/6 + 5,
a contradiction. Thus a3 = 3, 4, or 5.
If a3 = 3 = a2 then t3 < t2 and so f3 = (s — l)/3. This yields
s>tl + t2 + t3 = (s- l)/2 + (s - l)/3 + (s - l)/3
and again s = 7(s — 1 )/6 and | G : Z | = 12. Suppose a3 = 4. If f 3 = (s — 1 )/2,
we have
s>(s- l)/2 + (s - l)/3 + (s - l)/2 = 4(s - l)/3 > s - 1
and thus s = 4. This yields | G | = 6, which is impossible since a3 must divide
| G\ by Lagrange's theorem. Thus t3 = (s — l)/4 which yields
s>(s- l)/2 + (s - l)/3 + (s - l)/4 = 13(s - 1)/12 > s - 1.
Thus s = 13(s - 1)/12 and s = 13 and | G : Z | = 24 in this case.
The remaining case is a3 = 5. If t3 = 2(s — l)/5, this yields
s>(s- l)/2 + (s - l)/3 + 2(s - l)/5 = 37(s - l)/30 > s - 1.
since this has no integer solution, we have t3 = (s — l)/5 and
s > 31(s - l)/30 > s - 1
and s = 31. This gives | G : Z| = 60 and the proof is complete. |

260
Chapter 14
We mention that all three cases of Theorem 14.23 can occur. If G =
SL(2, 3), we get G/Z =* A4 of order 12. If G = GL(2, 3), we get G/Z S Z4 of
order 24 and if G = SL(2, 5), we get G/Z S A5 of order 60. In fact, A4,
E4, and A5 are the only possibilities for G/Z.
Problems
(14.1) Let G be a linear group of degree n. Show that there exists another
linear group G* of degree n such that G/TXG) S G*/Z(G*) and det(g*) = 1
for g* e G*. Do this in such a way that G* is irreducible iff G is irreducible.
Hint Let / be the n x n identity matrix and let S = {a/|a e C, a" =
det(g) for some g e G}. Consider GS £ GL(n, C).
(14.2) Let x £ Irr(G) be p-rational and faithful and assume that
Z(l) < k(p - 1)
for some k < p. Show that the Sylow p-subgroups of G are elementary
abelian of order <pk.
(14.3) Let G be an irreducible p-solvable linear group of degree p" > 1.
Let N = Op.(G) and U/N = Op(G/N). Show that U is nonabelian.
(14.4) Let G be a linear group of degree n and suppose that n is not divisible
by any prime power q > 1 such that q = 1, 0, or — 1 mod p ,where p is some
prime. Assume that G has a solvable irreducible normal subgroup. Show that
a Sylow p-subgroup of G is normal.
Hint Let P0 e Sylp(S) where S is normal, irreducible, and solvable
and consider C = CG(P0) <i G.
(14.5) Let G have a normal p-complement N and assume that a Sylow p-
subgroup of G is not normal. Suppose that G is a linear group of degree p — 1.
Show that N/(N n Z(G)) is a 2-group and thus iV is nilpotent.
Hint Let P e Sylp(G). Show that CN(P) £ Z(N).
(14.6) Let G be a linear group of degree p — 1, where p is a prime. Suppose
there exists p-solvable M <i G such that M does not have a normal Sylow
p-subgroup. Show that G is solvable.
(14.7) Let G be an irreducible linear group of degree p + 1 where p is an
odd prime. Let P e Sylp(G) and suppose P^i G. Assume that G has a normal
p-complement, N. Show that p + 1 is a power of 2.
Hints Let G be a minimal counterexample and let q be an odd prime
divisor of | N: CN(P) \. Let Q e Syl,(JV) and show that NN(Q) is abelian.

Problems
261
(14.8) Replace the hypothesis that G has a normal p-complement in
Problem 14.7 by the weaker condition that G is p-solvable. Draw the same
conclusion.
Hints Let U = Op(G) and M/U = Op.(G/U). It is no loss to assume
that G/M is a p-group and that Op'(G) = G. If U £ Z(G), let T = IG(1),
where X is a linear constituent of %v and / is the given faithful character of G.
Show that Xt has an irreducible constituent of degree p and that M nT<M.
(14.9) Let G be a linear group which is generated by two elements, x and y.
Suppose that each of x and y have only two distinct eigenvalues. Show that
the irreducible constituents of G (that is, of the given faithful representation)
have degree at most 2.
Hint If V is a nonzero vectorspace over C and Vu V2, V3, and V± are
subspaces such that V = Vt + V} whenever i ^ j, then there exists a subspace
U = 7 such that dim 1/ = 2 and 1/ n V- ^ 0 for all i.
(14.10) Let G be a solvable linear group of degree n. Show that G has a
nilpotent normal subgroup of index < n\.
(14.11) Let p be a prime. Show that for every integer n > 0, there exists a
p-solvable linear group G of degree n with | P : Op(G) | > pa, where P e Sylp(G)
and

Changing the characteristic
Some of the deepest and most powerful results in group representation
theory involve "modular" representations, that is, representations over
fields of prime characteristic. These are important for at least two reasons.
First, if K, H o G, with K ^ H, and H/K is an elementary abelian p-group,
then H/K may be viewed as an F[G]-module, where F is the field of order p.
In this situation, the representation theory can give direct structural
information about G. Perhaps an even more important way in which the
characteristic p representations of G are relevant is that they can give new
information about the characteristic zero situation. This is especially the case
when | G | is divisible by p since then it is possible to obtain results which relate
the p-subgroups of G with the properties of Irr(G).
Our emphasis in this chapter will be on the relationship between the
absolutely irreducible characteristic p representations of G and Irr(G).
Following R. Brauer, who was the originator of this theory, we shall focus
our attention on characters rather than on modules or representations. (In
fact, we shall not even mention the indecomposable but not irreducible
modules which seem to be crucial for many of the deeper results.)
The objective of this chapter is to familiarize the reader with the principal
definitions and the most basic results of the theory. We do not attempt to give
a comprehensive treatment of the subject and we shall not prove every fact
that is mentioned.
We establish some notation which will remain fixed throughout this
chapter. Let R be the full ring of algebraic integers in C and let p be a prime.
We construct a particular field F of characteristic p by choosing a maximal
262
15

Changing the characteristic
263
ideal M 2 pR of R and setting F = R/M. (Note that there is a certain amount
of arbitrariness here since M is not uniquely determined.) Let * denote the
natural homomorphism /? -> F. Since pi* = p* = 0, it follows that char(F)
= p as claimed.
(15.1) lemma Let U = {eeC|em = 1 for some meZ with pjfm}. Let/?, F
and * be as above. Then
(a) 1/cR;
(b) * maps U isomorphically onto F *;
(c) F is algebraically closed and algebraic over its prime field.
Proof Clearly U £ Rand so * is defined on U. If ae U - {1}, then a is a
primitive nth root of 1 for some n > 1 with/^n and hence
n-l
1 + X + ■ ■ ■ + X"-1 = (x" - \)/(x - 1) = [] (* - **)■
i=l
Setting x = 1, we conclude that 1 — a divides n in R. If a* = 1, then (1 — a)*
= 0 and thus n* = 0. Since p* = 0 and (p, n) = 1, it follows that 1* = 0, a
contradiction. Thus * maps U isomorphically into F*.
If a e F, then a = a* for some a e R and there exists monic/ e Z[x], with
f(a) = 0. Let K £ F be the prime subfield. Then 0 # /* e K[x] and/*(a) =
/(a)* = 0. Thus F is algebraic over K.
To complete the proof, let E be an algebraic extension of F. Then
[/*cf c £* and it suffices to show that £* £ I/* in order to conclude
that I/* = F* and that F is algebraically closed. Let f} eE". Then 0 is
algebraic over F and hence over K and thus (}m — 1, where m = |K(/?)| — 1.
Now pjfm and so I/* contains m roots of xm — 1. Thus /Set/* and the proof is
complete. |
Continuing with the above notation, let 3E be an F-representation of a
group G. Let y be the set of p-regular elements of G, that is, elements of order
not divisible by p. We define a function q>: y -> C as follows. Let xe^ and
let eu£2,...,£feF* be the eigenvalues of 3E(x), counting multiplicities.
(Thus/ = deg 3E and £ et = t/^(x) where t/f is the F-character afforded by 3E.)
For each i, there exists a unique m, e 1/ such that (m,)* = e,-. Let cp(x) = £ m( .
The function <p: ^ -» R £ C is called the Brauer character of G afforded
by 3E. Note that similar F-representations afford equal Brauer characters and
that Brauer characters are constant on conjugacy classes. Both of these
statements follow from the fact that 3E(x)andP~ lX(x)P have the same
eigenvalues. Also, if x e y, then q>{x~l) = q>(x) since the eigenvalues of 3E(x~l) are
the reciprocals of those of H(x) and for u e U we have (u)* = (u ~l)* = («*) ~'.

264
Chapter 15
(15.2) lemma Let 3E be an F-representation of G which affords the Brauer
character cp and the F-character \ji. For g e G, let x = gp^ e if. Then cp(x)*
Proof We have g — xy, where x, y e <g>, pJfo(x) and o(y) is a power of p.
Replace 3E by a similar representation so as to assume that X(g) is in upper
triangular form. Since X(g) = X(x)X(y) it follows that the eigenvalues a, of
X(g) can be factored, a, = e,<5f, where e^ ..., ef are the eigenvalues of 3E(x)
and (dif^ = 1. Since o(y) is a power of p, we have 8{ = 1 and a, = e,. Thus
\p(g) = \p(x). That \p(x) = cp(x)* is immediate from the definition of cp. |
Lemma 15.2 provides one reason why we only bother to define Brauer
characters on p-regular elements: this is sufficient to reconstruct the full
F-character afforded by 3E.
Some words of caution are appropriate here. Given a group G and a
function, cp: if -> C, where if is the set of p-regular elements of G, it is not
always meaningful to ask if cp is a Brauer character unless the ideal M ^ Ris
specified or some other additional information is given. Examples exist where
cp is a Brauer character with respect to some choice of M and is not one when
some other maximal ideal is chosen. Also, if a is an automorphism of the
complex numbers and cp is a Brauer character of G, then cp" need not also be a
Brauer character where cp" is defined by cp"(x) = cp(x)a for xtif.
(15.3) lemma Let cp be a Brauer character of G. Then cp, the complex
conjugate function, is also a Brauer character.
Proof Let 3E be an F-representation affording cp, where F = R/M, as
usual. For g e G, define 9)(g) = 3E(gT 1)T and observe that 9) is an
F-representation of G. If £!,..., ef are the eigenvalues of H(g\ then g! ~\ ..., e^~'
are the eigenvalues of H(g)~1 = X(g~l) and hence of 9)(g). The result now
follows. |
Let 3Et,..., 3Er be a set of representatives for the similarity classes of
irreducible F-representations of G and let cp{ be the Brauer character afforded
by 3E,-. We say that the cpt are irreducible Brauer characters and we write
IBr(G) = {cpt}. When we use this notation, it is understood that a particular
prime p and maximal ideal M have been fixed.
It is routine to prove that sums of Brauer characters are Brauer
characters and that every Brauer character is of the form £ n,-<Pj, where the nx e Z
are nonnegative and not all zero. To prove that the cpt are linearly independent
we need the following fact from algebraic number theory. Although it is not
terribly deep, we omit the proof.

Changing the characteristic
265
(15.4) lemma Let <*!,..., ameC be algebraic over Q and let /be a proper
ideal of R, the ring of algebraic integers. Suppose that not all a.t = 0. Then
there exists /? e C such that /?a( e R for all i but not all /?a( e /.
Proof Omitted. |
(15.5) theorem The irreducible Brauer characters q>t are distinct and
linearly independent over C. Also, if £ u-i<p{x) = 0 mod M for all x e ^ with
a, e R, then all a, = 0 mod M.
Proof We prove the second statement first. Since the 3E, are absolutely
irreducible, we have X<(F[G]) = Mfi{F), where/J = deg 3E(. It follows that
we can choose b{ e F[G], with \jif]bi) = 1, where i/^ is the character afforded
by Xt. Also, by Theorem 9.6, we may suppose that il/£bj) = 0 if i =£ )■
We have £ u-i<p{x) = 0 mod M for all x e y and thus £ oi-*^,{g) = 0 for
all g e G by Lemma 15.2. It follows that £, a^ip^bj) = 0 for ally, and thus
a(* = 0 for all i. This proves the assertion.
Now let £ be the algebraic closure of Q in C and suppose that £ a(<p( = 0
with a(e£. If not all ol{ = 0, we apply Lemma 15.4 and choose /? with all
/fa, e R but not all /to, e M. Since £ (fiv-^<Pi = 0, this contradicts the first part
of the proof. Thus all a, = 0 and the cpt are linearly independent over E.
Since the q>t have values in E, it follows by elementary linear algebra that
they remain linearly independent over any extension field of E. |
The principal reason that Brauer characters are important is that they
provide a link which connects Irr(G) with the characteristic p representations
ofG.
(15.6) theorem Let x be an ordinary character of G and let % denote the
restriction of / to Zf, the set of p-regular elements of G. Then % is a Brauer
character of G (for any choice of M).
In order to prove Theorem 15.6, we need to consider a somewhat larger
ring than R. As always, we assume that we have fixed a particular maximal
ideal, M 2 pR. Let R = {a/0|a, $ eR, 0$ M} £ C and observe that R is
a ring and R 2 R. Let Si = {a//? | a e M, /? 6 R — M}. Then M is an ideal of
R and every element of R — M has an inverse in R. It follows that M is the
unique maximal ideal of R. We call R a ring of local integers for the prime p.
We extend the homomorphism *: R -> F to R by defining (a//?)* = a*//?*.
Note that if? is the kernel of this extension and M = At n /?.
(15.7) lemma (Nakayama) Let R and .A? be as in the preceding and let V
be a finitely generated R-module. Suppose V = VAi + U for some sub-
module U £ K Then 1/ = K

266
Chapter 15
Proof Since V is finitely generated, we can choose vuv2,- ■. ,vneV such
that V = Yj vt M + U.Do this with the minimal possible n. If n = 0, then
U = V and we are done. Suppose n > 1 and write
n
v „ = £ vimi + ">
i=l
where mteM and «e[/. Thus
n-l
Since m„ e M, we have I — m„eR —M and thus 1 — m„ is invertible in R and
t>„ e YjZl ViAl + U. It follows that V = Y]=l vtM+ U and this contradicts
the minimality of n and completes the proof. |
Suppose 3E is a C-representation of G with the property that all entries in
the matrices H(g) for g e G lie in R. We can construct an F-representation 3E*
of G by setting 3E*(gf) = 3E(g)*, that is, we apply * to every entry of X(g).
The following includes Theorem 15.6.
(15.8) theorem Let 3E be a C-representation of G. Then there exists a C-
representation 9) similar to 3E such that all entries in ?)(g) lie in R for all
g e G. If 9) is any such representation, then the F-representation 9)* affords
the Brauer character % where / is the ordinary character afforded by 3E (and
by$).
Proof Let E be the algebraic closure of Q in C. By the results of Chapter
9, every C-representation of G is similar to one of the form 3C for some £-
representation 3- It therefore suffices to assume that 3E is an £-representation
and to produce a similar £-representation ?) with entries in R.
Let V be an £[G]-module corresponding to 3E and let vu ...,v„ be an
£-basis for V. Let W be the £-span of the finite set {vtg\ 1 < i < n,geG} so
that W is a finitely generated R-module which is G-invariant. Let
t: W -> W/WM
be the natural homomorphism and view W/Whl as an F-vector space
via (w-r)a* = (woc)t for aeR. Let {vVj-r} be an F-basis for W/Wffi so that
(£ WjR)z = W/WM and W = WM + £ WjR~. By Nakayama's Lemma
15.7, we have W = £ w; R and thus the wj span V over £ since W contains
an £-basis for V.
We claim that the w} are linearly independent over £. Suppose that
£ Wj-a; = 0 with Zj e£ and not all a,- = 0. By Lemma 15.4, we can multiply
by a suitable /? e £ and assume that all a, e /? £ /? but that not all a;* = 0.
Now 0 = (£ Wj-0;)t = £ (wyT)a^* and this contradicts the linear
independence of the W;T and proves the claim.

Changing the characteristic
267
Now let 9) be the F-representation of G corresponding to V with respect
to the basis {w;} so that 9)(#) = (a;j), when w,g = £ WjOij. However, w,g e W
= £ wj^ and so all au e R. This completes the proof of the first assertion.
Now for p-regular g e G, we need to compute the eigenvalues of
V*(g) = Wq)*- Let / be the characteristic polynomial of 9)(gr) so that
f(x) = det(x/ - 9)(#)). Then /e/?[x] and /* eF|>] is the characteristic
polynomial of 9)(g)*. Write/(x) = Y[(x - 1{) and note that the eigenvalues 1;
lie in J? c r. Then/*(x) = Y\(x — A,*) and hence the eigenvalues of 9)*(g)
are the X;*. Thus %{g) = £ X-, is the value of the Brauer character afforded by
9)* at the element g. The proof is complete. |
In the foregoing proof, it is not the case that the character / uniquely
determines the F-representation 9)* up to similarity. It is possible that / is
afforded by another C-representation 3 with entries in R such that ?)* and 3*
are not similar over F. Of course, / does uniquely determine the Brauer
character % and we may write % = £ n9cp where cp runs over IBr(G) and
0 < n e Z. The coefficients n^ are uniquely determined because of the linear
independence of IBr(G) and since nv is the multiplicity of a particular
irreducible F-representation as a constituent of 9)*, it follows that 9)* and 3*
have the same irreducible constituents with the same multiplicities. Since
9)* and 3* need not be completely reducible, it does not follow that they are
similar.
(15.9) definition Let x e Irr(G) and let % be the restriction of / to the p-
regular elements of G. Write
«>ElBr(G)
The uniquely defined nonnegative integers dM are the decomposition numbers
of G for the prime p.
We view the decomposition numbers as forming a |Irr(G)| x |IBr(G)|
matrix, called the decomposition matrix. Although the decomposition numbers
are not even defined until the maximal ideal M is chosen, it is a fact (whose
proof we omit) that the decomposition matrix of G for the prime p is uniquely
determined up to permutations of the rows and columns.
(15.10) theorem The decomposition matrix (dxill) has linearly independent
columns. Also, IBr(G) is a basis for the space of C-valued functions defined on
p-regular elements of G and constant on conjugacy classes.
Proof Let V be the Space of p-regular class functions and let W c y be
the span of IBr(G). Let U be the span of the columns of (dX(/1) so that
dim U < |IBr(G)| = dim W < dim V,

268
Chapter 15
where the equality follows from Theorem 15.5. The theorem will follow when
we show that dim V < dim U. It therefore suffices to find a one-to-one linear
map from the dual space V of V into U. The elements of U are columns
indexed by / e Irr(G). For a e V, define u = col(a(#)). To show that ueU,
observe that a(#) = a(^ dx<p<p) = Y^ dx<pa.((p). Thus u is a linear combination
of the col(rf^) for <p e IBr(G). We map 9 -> U by a i-> col(a(#)).
If a(#) = 0 for all / e Irr(G), we claim that a = 0. To see this, let 9 e V
and extend 9 to a class function 9t of G. Write 9t = £ ax% so tnat ^ = X flx#
and a(9) = 0 as desired. The result now follows. |
(15.11) corollary The number of conjugacy classes of p-regular
elements of G is equal to | IBr(G) |. This is also the number of similarity classes of
irreducible ^-representations of G for every splitting field K of
characteristic p.
Proof The first assertion is immediate from Theorem 15.10. To prove
the second statement, we may replace K by its algebraic closure by
Corollary 9.8. Assuming that K is algebraically closed, it contains an isomorphic
copy of F which is an algebraic closure for Zp. By Corollary 9.8 again, we may
assume that F = K. The result now follows from the fact that nonsimilar
irreducible F-representations afford distinct Brauer characters. |
(15.12) corollary If q> e IBr( G), then there exists / e Irr(G) with dX(p =£ 0.
Note that for each / e Irr(G), it is trivial that there exists q> e IBr(G) with
d„± 0 since 0 * *(1) = £ dX9cp(l).
The interesting case of this theory is when p 11G |. The reason for this is
given by the following.
(15.13) theorem Suppose pJf\G\. Then IBr(G) = Irr(G).
Proof In this case, the group algebra F\_G] is completely reducible and
thus by Corollary 1.17 we have \G\ = dim F[G] = £ (deg 3E,)2, where the
3Ej are a set of representatives for the similarity classes of irreducible
F-representations. If 3E; affords (p; e IBr(G), then deg 3E, = <p;(l) and hence
x «ku' = igi= x x(i)2 = ife<w>(i)Y
«>eIBr(G) xelrr(G) X \d> J
= X <p(1M1)I^„.
If q> t6 (J-, then ^ dX(pdxll > 0 and if (p = fi, then ^ dx<pdxtl > 1 by Corollary
15.12. It follows that these inequalities are all equalities.
We now have ^ (dXll,)2 = 1 and hence for each q>, there is a unique /
such that dM ¥= 0, and in fact dxv = 1. Since ^ dXvdxll = 0 if cp ^ /z, it

Changing the characteristic
269
follows that for each / e Irr(G), there is a unique cp with dx<p ^ 0. We conclude
that % = % = q> and the result follows. |
A necessary condition that a function 9: £f -> C be a Brauer character is
that for each subgroup H £ G with pJf\H\, the restriction 9H is a Brauer
character and hence is an ordinary character of H. This is something which
can be checked.
It is trivial that every q> e IBr(G) is a C-linear combination of the Brauer
characters % for / e G. This may be seen by extending q> to a class function of G.
(15.14) theorem Let <peIBr(G). Then q> is a Z-linear combination of
ttlzelntG)}.
Proof Extend <p to a class function 9 of G by setting 9(g) = q>(gP'). It
suffices to show that 9 is a generalized character of G. We appeal to Brauer's
Theorem 8.4(a).
Let E £ G be elementary and write E = P x Q, where P is a p-group and
pX\Q|. If geE, write g = xy with xeP and y eQ. Then 9(g) = <p(y) and so
9£ = 1P x <pQ. By Theorem 15.13, q>Q is a character of Q and hence 9£ is a
character of E. It follows that 9 is a generalized character and the proof is
complete. |
We digress to show another way in which Theorem 15.13 can be used.
(15.15) lemma Let E be an algebraically closed field of characteristic not
dividing | N |, where N is a group. Let H act on N and suppose that CH(n) = 1
for all 1 ¥= neN. Let ?) be a nonprincipal irreducible £-representation of N
and write ty\nh) = 9)(n) for n e N and heH. Then the representations 9)*1
are pairwise nonsimilar for heH.
Proof Let K be the algebraic closure in E of the prime subfield of E.
Then $ = 3£ f°r some irreducible K-representation 3 of N. It suffices to
show that the 3*1 are pairwise nonsimilar by Corollary 9.7 and hence we may
assume that K = E.
If char(£) = 0, then (up to isomorphism) E £ C and the representations
(9)c)'1 are pairwise nonsimilar by Theorem 6.34 and Problem 7.1 which proves
this case.
Suppose char(£) = p. Choose a maximal ideal, M 2 pR of R and let
F = R/M as usual. Then F = E and we may assume F = E. Let ?) afford
q>eIBr(iV) = Irr(iV). By Theorem 6.34, the characters <ph for heH are all
distinct and the result follows. |
The following result is often quite useful.

270
Chapter 15
(15.16) theorem Let G be a Frobenius group with kernel N and let K be a
field with characteristic not dividing \N\. Let V be a K[G]-module and
suppose that CV(N) = 0. Let H be a Frobenius complement for G. Then V has a
basis which is permuted by H with orbits of size \H\. Also, if H0 £ H, then
dim CV(H0) =\H:H0\ dim CV(H).
Proof The second statement follows from the first since if b is a basis
permuted by H, then dim CV(H0) is equal to the number of orbits of the action
of H0 on b. Since each orbit of H0 on b has size \H0\, the assertion follows.
To prove the first statement we argue that it suffices to assume that K is
algebraically closed. Let 3E be a K-representation of G corresponding to V
and let E 2 K. The condition that CV(N) = 0 is equivalent to the restriction
3E,v having no principal constituent and this property is inherited by (XE)N
by Theorem 9.6. The conclusion of the theorem is equivalent to 3EH being
similar to a representation 9) in block diagonal form with each block being
the regular representation of H. If we can prove that (3EH)£ and 9)£ are similar,
then 3EH and 9) are similar by Problem 9.5. It follows that we may replace K
by any extension field and thus we assume that K is algebraically closed.
Now let Jt be a representative set of irreducible K[iV]-modules. Since H
acts on the set of similarity classes of ^-representations of JV, we can define a
corresponding action of H on Jt. Let Jt0 £ Jt be a set of representatives for
the H-orbits of nonprincipal K[iV]-modules in Jt. In the notation of
Definition 1.12, let
W = X M{V) £ V.
MeJ0
We claim that V = £ -hBll Wh. This will suffice to prove the result since
we obtain a basis for V by choosing any basis for W and taking all H-trans-
lates. To prove the claim, observe that Jt = [jhBH (Jt0)h u {I/}, where U is a
principal module. By Lemma 15.15, this union is disjoint. Since VN has no
principal constituent, Lemma 1.13 yields
V=I-hJ I M(V)\
We*)1 /
The result now follows from Lemma 6.4. |
We now resume consideration of the general case where p can divide \G\.
We introduce the concept of "blocks" which is at the heart of Brauer's
theory.
For each / e Irr(G), we have the algebra homomorphism cox: Z(C[G]) -» C
as in Chapter 3. The function cox is determined by its values on the conjugacy

Changing the characteristic
271
class sums Kt which form a basis for Z(C[G]). Furthermore, the values
cox(Ki) lie in R. For /, \ji e Irr(G), write / ~ \ji if co^K)* = co^Kf)* for all i.
This clearly establishes an equivalence relation on Irr(G).
(15.17) definition Ap-ft/oc/cofGisasubsetB £ Irr(G) u IBr(G) such that
(a) B r\ Irr(G) is an equivalence class under the relation ~ defined
above.
(b) B r\ IBr(G) = {q> e lBr(G)\dXll, ± 0 for some /efln Irr(G)}.
From its definition, it appears that the equivalence relation ~ on Irr(G)
depends on the choice of the maximal ideal M. In fact, this is not true.
(15.18) theorem Let x, i> £ Irr(G). Then / and \ji lie in the same p-block
iff cox(K) — co^(K) lies in every maximal ideal of R which contains pR for
every class sum K.
Proof The "if" statement is clear. Assume K~ i> and fix K. Let a =
cox(K) — oi^,(K). We shall show that a" e pR for some integer n and the result
will follow.
Let 0 = #(Q|G|/Q), the Galois group, and let a e<&. Let e be a primitive
| G|th root of unity so that e" = em for some m, with (m,\G\) = 1. Let g e G
be in the class with sum equal to K and let L be the sum of the class containing
gm. We have x(gf = X(gm). Also \C(g)\ = \C(gm)\ since <g> = <<f>. It
follows that cox(K)a = cox(L) and similarly co^(K)a = co^(L). It follows that
of = cox(L) — co^(L) e M since / ~ \ji.
Let f(x) = f\aEe (x - a") e(Q n /?)[*] = Zpc]. Since /eM for all
a e ^, all of the coefficients of/except for the leading one lie in M n Z = pZ.
Thus 0 = /(a) = a" mod pR, where n = | # |. The proof is now complete. |
Although it is clear from the definition that every / e Irr(G) lies in a unique
p-block, the analogous statement about IBr(G), though true, is not so obvious.
To prove it, we relate the p-blocks of G to the F-algebra, Z(F[G]).
We extend the map *: R -> F to a ring homomorphism *: £[G] -» F[G]
by setting g* = g for g e G. (Here, R[_G] is simply the R-span of the group
elements in C[G].) Since the class sums K{ form a basis for Z(C[G]) and their
images KfeFlG] form a basis for Z(F[G]), it follows that Z(R[G]) =
R[G] n Z(C[G]) maps onto Z(F[G]) via *.
Now suppose x e Irr(G). Then cox maps Z(^[G]) to R and we can define
cox*: Z(F[G]) -^ F by setting cox*(z*) = cox(z)* for z e Z(R[G]). This is well
defined since if zu z2 e Z(R[G]), with zt* = z2*, then zt - z2 has coefficients
in M and thus (eo^Zi - z2))* = 0. It is trivial to check that cox*: Z(F[G]) -» F
is an algebra homomorphism and that if /, \\i e Irr(G), then eo/ = co^* iff
X~ i>\ that is, iff / and \ji lie in the same p-block.

272
Chapter 15
Let B1(G) be the set of p-blocks of G. If BeBl(G), let kB = cax* for
X e B r\ Irr(G). Thus B >-* kB is a one-to-one map from B1(G) into the set of
algebra homomorphisms Z(F[G]) -» F.
(15.19) theorem Let cp e IBr(G). Then cp lies in a unique p-block B. If <i) is
an irreducible F-representation which affords cp, then ^(u) = AB(u)7 for all
ueZ(F[G]).
Proo/ By Corollary 15.12, dw ^ 0 for some /elrr(G). Let BeBl(G),
with / e B. Then <p e B. Let 9) afford <p. We will be done when we show that
sJ)(u) = kB(u)I for all u e Z(F[G]) since this equation uniquely determines kB
and thus uniquely determines B.
Let X be a C-representation which affords / and which has entries in R
(Theorem 15.8). Thus X* affords the Brauer character % and by the linear
independence of IBr(G), it follows that 9) has multiplicity dx<p > 0 as a
constituent of 3E*.Now letueZ(F[G])and write u = z*forsomezeZ(£[G]).
Then
X*(u) = X(z)* = (ojx(z)I)* = oi*(u)l = lB(u)I.
Since <i) is a constituent of X*, the result follows. |
We define a graph with Irr(G) as its vertex set by linking /, \\i e Irr(G) iff
there exists q> e IBr(G) such that dw and d^^ are both nonzero. This is called
the Brauer graph. If / and \ji are linked by cp, it follows that / and \\i lie in the
same p-block, namely the unique one which contains cp. This proves the
following.
(15.20) corollary Let B be a p-block of G. Then B r\ Irr(G) is a union of
connected components of the Brauer graph.
We shall see that in fact B n Irr(G) is a single connected component.
One of the principal benefits of considering blocks in various applications
of the theory is that in certain circumstances we can replace equations like
Xelrr(G)
(which arise from the second orthogonality relation) by equations like
Z x(x)xJy) = o,
Xelrr(G) nB
where B is a p-block. In particular, this holds whenever xp and yp are not
conjugate in G. We shall prove a weak form of this "block orthogonality."
For each cp e IBr(G) we define
%= 1 dx<pX.
Xelrr(G)

Changing the characteristic
273
The O's are called projective characters of G. (There is no connection with
projective representations in the sense of Chapter 11, but there is a connection
with projective modules in ring theory. The O's are also called "principal
indecomposable characters" in the literature.)
(15.21) lemma Let si ^ irr(G) be a union of connected components of
the Brauer graph and let ^ = {<peIBr(G)| dx<p ^ 0 for some /e si). Let
x, y e G with pjfo(x). Then
EzMxO0 = Z (plx)®^).
Proof For / e si, we have
X(x) = Z fiU<P(x)
since dxtl = 0 for all y. e IBr(G) — 28. Also, if <p e ^, then
since d^ = 0 for all £ e Irr(G) - j/.
Now
Z zMztF) = Z «U<pMx(F) = Z pM^OO
and the proof is complete. |
(15.22) corollary For each <peIBr(G) we have %(y) = 0 if p\o(y).
Furthermore, | P\ divides <5„(1) where P e Sylp(G).
Proof Let x e G be p-regular and let p | o(y). By the second orthogonality
relation we have
Z Z(x)zO0 = O.
Xelrr(G)
By Lemma 15.21 with si = Irr(G) and ^ = IBr(G), we conclude that
Z cp(xW& = 0.
«>ElBr(G)
Since this holds for all p-regular x e G, the linear independence of IBr(G)
yields Qjly) = 0 for all q>.
The second statement follows since \P\ [(Ov)P, 1P] = Ov(l). |
(15.23) corollary (Weak Block Orthogonality) Let x, y e G with pJfo(x)
and p | o(>>). Let B be a p-block of G. Then
Z X(*)ZOO = 0.
XeBnlrr(G)

274
Chapter 15
Proof Apply Lemma 15.21 with jsi = Irr(G) n B and Si = IBr(G) r\ B.
Since O^tv) = 0 for all q>, the result is immediate. |
Before going on to develop more of the theory of blocks we digress to give
an application of what we have already done.
(15.24) theorem (Brauer) Let G be a simple group of order p"qbr where
p, q, r are distinct primes. Let S e Sylr(G). Then S = CG(S).
The p-block containing 1G is called the principal block.
(15.25) lemma Let G be simple and let B be the principal p-block of G.
Suppose x £ B r\ Irr(G) and x(l) is a power of p. Then i = la-
Proof Let g e G and let K e C[G] be the sum of the elements in Cl(g),
the conjugacy class containing g. Then co^K) = \Cl(g)\ and since / and 1G
lie in the same p-block, we have
i \ x(g)ICl(g)|
(*) —-7JT— = ^(K)= |C%)|modM.
Now let P e Sylp(G). If P = 1, then since /(1) divides | G |, we have /(1) = 1
and by simplicity / = 1G. Assume then, that P > 1 and take g e Z(P) — {1}.
Then pJf\Cl{g)\ and so |C%)| #0modM since M n Z = pZ. Thus (*)
yields %(g) =£ 0. However (/(l), |C%)|)= 1 and Burnside's Theorem 3.8
yields that g e Z(/). Since g ¥= 1 and G is simple, it follows that / = 1G. |
Proof of Theorem 15.24 Suppose CG(S) > S. Then there exists xeG
of order pr or qr. Say o(x) = pr and let B be the principal p-block of G. By
Corollary 15.23, we have
-1= Z zd)zW+ I z(i)zW.
We claim that the second sum is zero. If /eB n Irr(G) and / j= 1G and
4^Z(1)> then r|/(l) by Lemma 15.25. Since r\o(x), we have z(x) = 0 by
Theorem 8.17 and the claim follows. We conclude that — l/q is an algebraic
integer and this contradiction completes the proof. |
For i e Irr(G), let ex e Z(C[G]) be the idempotent corresponding to /.
By Theorem 2.12 we have
7^7 IW)g.
|G| gBG

Changing the characteristic
27S
Note that in general, ex $ R\_G~\ (where R is a ring of local integers for p) and
so we cannot find idempotents in F[G] simply by applying * to ex. Since
ex e^ = 0 if x ¥= ip, sums of the various e^'s are also idempotents and we shall
consider when such a sum lies in R[G]. Note that if si <=, Irr(G), then the set
si can be recovered from s = Yxe* ex smce ^ = ixe^rr(G)\sex ¥= 6}.
(15.26) theorem (Osima) Let si be a connected component of the Brauer
graph and let/ = Yxe* ex. Write/ = Y agg. Then
(a) ag = (l/\G\)^x^X(DM;
(b) a9e£forallgeG;
(c) ag = 0 if p\o(g).
Proof Statement (a) is immediate from the formula for ex in Theorem
2.12. By Lemma 15.21, we have ag = (1/|G|) YjI> ^ ss VW^JsX where 3b =
{<p e \Br(G)\dx<p ¥= 0 for some / esi}. If p\o(g), then O^fer) = 0 by Corollary
15.22 and (c) follows.
Now assume that pjfoig). Lemma 15.21 then yields
ag = (l/\G\)£%(im).
However, \P\ divides <B„(1) where Pe Sylp(G) and thus <5„(1)/|G| ek~. Since
cp(g) e R, it follows that ageR and (b) is proved. |
(15.27) theorem The connected components of the Brauer graph are
exactly the sets Irr(G) n B for p-blocks B. Furthermore, every set si c Irr(G)
such that Yxe* exeR\_G] is a union of sets of the form Irr(G) n B.
Proof We prove the second statement first. If / e Irr(G) is afforded by
3E, then 5.(ex) is the identity matrix and 3E(e^) = 0 if \p =£ %. Thus oix(ex) = 1
and eo^) = Ofori/f ^ /.Write/ = Y* e^. It follows that/ej^ iff co^/) = 1
and otherwise cox(f) = 0. Thus / e si iff cox(f)* ¥= 0. Now if/ e R[G], then
all oix(f)* are equal as / runs over Irr(G) n B for a block B. The assertion
follows.
Now let si be a connected component of the Brauer graph. By Theorem
15.26, Y* exe ^[G] and tnus -^ is a union of sets of the form Irr(G) n B.
Since each Irr(G) n B is a union of connected components of the Brauer
graph, the result follows. |
We now have three different characterizations of the sets B n Irr(G).
In addition to their definition as equivalence classes under ~, they are also
the connected components of the Brauer graph and they are the minimal
nonempty subsets si £ Irr(G) such that Yxe* exe K[G].
The following is a strengthening of Theorem 15.14.

276
Chapter 15
(15.28) lemma Let B be a block of G and let <p e IBr(G) n B. Then q> is a
Z-linear combination of Brauer characters of the form # for / e Irr(G) n B.
Proof By Theorem 15.14, there exist bxeZ such that q> = Y.x^i"iO) bx%
Thus q> = 9B + 90, where
9B = X M and 9«= I bxt
*eIrr(G)nB *eIrr(G)-B
Now we express 9B and 90 in terms of IBr(G) using the decomposition
numbers dxll for fi e IBr(G). If / e B, then dxll = 0 when fi $ B and hence 9B
is a linear combination of fi e B. Similarly, if / £ B, then dxil = 0 when fieB
and hence 90 is a linear combination of /z £ B. The equation <p = 9B + 90 and
the linear independence of IBr(G) now yield q> = 9B and the proof is
complete. |
(15.29) theorem Let B be a p-block of G. Then
|BnIrr(G)| > |BnIBr(G)|.
Let / e B n Irr(G). Then the following are equivalent.
(a) |BnIrr(G)| = |BnIBr(G)|.
(b) p^(|G|/z(l)).
(c) B n Irr(G) = {/}.
Also in this case, B n IBr(G) = {#}.
Proof Let £> = (d^) be the decomposition matrix and let DB be the
submatrix corresponding to the rows and columns indexed by elements of B.
For each q> e B, the part of the corresponding column of D outside of DB
consists of zeros. Since the columns of D are linearly independent by Theorem
15.10, it follows that the columns of DB are linearly independent and thus
DB has at least as many rows as columns.
Now assume (a). Then DB is a square nonsingular matrix and we let
DB~l = (avX). For fixed / e Irr(G) n B, we have
and so / is a linear combination of the Ov and hence vanishes on elements of
order divisible by p (Corollary 15.22). If PeSylp(G), then \P\Ixp, lp] = Z(l)
and (b) follows.
Assuming (b) we have
^ = Ml)/|G|)X»ei?[G]
gsG
and so B n Irr(G) = {/} by Theorem 15.27.

Changing the characteristic
277
Finally assume (c). Then
0 < |IBr(G) n B\ < |Irr(G) n B\ = 1
and (a) follows. Also, if IBr(G) r\ B = {cp}, then cp = b% for some ft e Z by
Lemma 15.28. Thus % = dx<pcp = dx<pbf and dx<pb = 1. It follows that dX(p = 1
and the proof is complete. |
Note that Theorem 15.29 provides an alternate proof of Theorem 8.17.
Also observe that if pjf \ G |, then each p-block contains a unique irreducible
character.
We use Theorem 15.26 to obtain further connections between the p-blocks
of G and Z(F[G]). For each B eBl(G), writefB = XxeBnirriG) ex. This Osima
idempotent lies in Z(R[G]) and we let eB = fB* e Z(F[G]).
(15.30) theorem We have the following.
(a) lB(eB) = 1 and XB.(eB) = 0 for blocks B ± B'.
(b) The eB are idempotents and eB eB. = 0 for B =£ B'.
(c) eB is an F-linear combination of class sums of p-regular classes.
(d) ZeB=l.
(e) If kB(z) = 0 for all B e B1(G), then z is nilpotent.
(f) The kB are all of the algebra homomorphisms Z(F[G]) -» F.
(g) Every idempotent of Z(F[G]) is a sum of some of the eB.
Proof (a) For / e Irr(G), we have oix(fB) =1 if / e B and otherwise
oix(fB) = 0. If / e B', then a>x* = kB. and (a) follows.-
(b) Since/B/B. = dBB,fB we have eBeB, = 5BB,eB. Since X^eB) = 1, we
have eB j= 0.
(c) Immediate from Osima's Theorem 15.26(c).
(d) Z fB = lex =1 and thus 2>B=1* = 1.
(e) Let 9) be any irreducible F-representation of G and let ?) afford
cp e IBr(G) n B. Then 9)(z) = lB(z)7 = 0 by Theorem 15.19. Thus z is in the
Jacobson radical, J(F\_GJ) and hence is nilpotent (Problem 1.4).
(f) Let 1: Z(F[G]) -» F be an algebra homomorphism. Then ker k has
codimension 1 and so Z(F[G]) = ker 1 + F ■ 1 and k is determined by its
kernel. If k j= kB, then ker kB £ ker k and we can choose zB e ker 1B with
k(zB) ¥= 0. Assuming 1 £ {1B}, let z = f] zB. Then lB(z) = 0 for all B and
hence z is nilpotent by (e). However, l(z) = f] k(zB) =£ 0. This is a
contradiction.
(g) Let e e Z(F[G]) be an idempotent. Then e = e £ eB = £ eeB and
it suffices to show that either eeB = 0 or eeB = eB. Suppose eeB ^ 0. Then
since eeB is not nilpotent and kB,(eeB) = 0 for B' j= B, we have lB(eeB) ¥= 0
and thus lB(eeB) = 1 = lB(eB). Thus lB.(eB(l - e)) = 0 for all B' eBl(G) and
since eB(l — e) = (eB(l — e))2,wehaveeB(l — e) = 0and the result follows. |

278
Chapter 15
We now discuss some of the connections between block theory and the
collection of p-subgroups of G. We work in the situation of Theorem 15.30.
If Jf is a conjugacy class of G we consider the Sylow p-subgroups of CG(x)
for x e jf. These are called the p-defect groups for jf. They constitute a
conjugacy class of p-subgroups of G. The collection of p-defect groups for jf
is denoted 8( Jf).
We use the notation jf for the sum in F\_G] of the elements of the
class Jf so that the £ form a basis for Z(F[G]). If B e B1(G), write eB =
Y^a^yf)^ so that aB is a uniquely determined function from the set of
classes of G into F. In fact, aB(Jf) = ((1/|G|) X^imonB X(l)X(g))* for # e jf
by Theorem 15.26(a). By 15.30(c) we have aB(Jf) = 0 if Jf does not consist
of p-regular elements.
Since 1 = lB(eB) = £ aB(Jf)lBpf),it follows that for each B e B1(G), there
exists at least one class Jf such that aB(Jf) =£ 0 and 1B( jf) ^ 0. We call such a
class a defect class for B.
(15.31) theorem (Min-Max) Let Jf be a defect class for B e B1(G) and
let £> e <5(jf). Let if be any class of G.
(a) If aB(Jz?) t6 0, then D contains a defect group for if.
(b) If 1B(^) t6 0, then D is contained in a defect group for if.
To prove the min-max theorem, we define the Brauer homomorphism f}P.
Let P c G be a p-subgroup. Let N = NG(P) and C = CG(P). We map
/?P:Z(F[G])^Z(F[iV])by
jceX n C
and extend by linearity. Since Jf n C is a union of classes of JV, we do have
fr(jr)eZ(F[JV]).
(15.32) lemma The map 0P: Z(F[G]) -> Z(F[iV]) is an algebra
homomorphism.
Proof If suffices to check that 0P(i^) = PA30PA&) for classes Jf, if.
For ceC, let j/ = {(x, y)\xejf,ye£',xy = c} and
•^o = {(*> y)|xe jf nCyeif nC,xy = c}.
Then | sf |* is the coefficient of c in /?P(Jf J?) and | j/0 |* is the coefficient of c
in Pp{jf)Pp(^). It thus suffices to check that \sf\ = \sf0\ mod p.
Since P c C(c), it follows that P acts on d by (x, yf = (xu, y") for us P.
Then j/0 is exactly the set of fixed points of s# under the action of P. The
result follows. |

Changing the characteristic
279
(15.33) lemma Let P be a p-subgroup of G and let (tP be the corresponding
Brauer homomorphism. Let z = Yjax-X". Then /?P(z) j= 0 iff P is contained
in some D e 8( Jf) with a^ ,£ 0.
Proof The sets jf n C(P) are disjoint for distinct classes Jf and thus
the nonzero elements of the form /?p(jf) are linearly independent. It follows
that pP(z) ± 0 iff fiP(£) ± 0 for some Jf with ax * 0.
Now 0p(jO ^ 0 iff Jf n C(P) # 0 and this happens iff P £ C(x) for
some x e jf. Since P c C(x) iff P is contained in some Sylow p-subgroup of
C(x), the result follows. |
(15.34) lemma Let B e B1(G) and let if be a class of G with aB(S£) ^ 0. Let
P e (5(i?) and N = NG(P). Then there exists b e Bl(iV) such that
XB = pPXb. WIG]) -> Z(F[JV]) - F.
Proof By Lemma 15.33, /?P(eB) ¥= 0. Since /?P is a homomorphism,
e = /Sp(eB) is a nonzero idempotent inZ(F[JV]) and so is not nilpotent. Thus
there exists b eBl(JV) with Xb(e) # 0. Let fi = pPXb. Then /z(eB) = lfc(e) # 0
and /z is an algebra homomorphism Z(F[G]) -> F. Since fi(eB) ¥= 0, we have
H = XB and the proof is complete. |
Proof of Theorem 15.31 If aB(<£) # 0, let Pec5(if). Then 1B = 0Plfc
for some fceBl(JV(P)) by Lemma 15.34. Since Jf is a defect class, we have
0 ¥= Xg(Jf) = Xb(fiP(Jf)) and so $P(tf) ^ 0. Thus P is contained in some
defect group of jf by Lemma 15.33. Now (a) follows.
Now apply Lemma 15.34 to Jf. Since aB(Jf) #0we have XB = pDXb for
some b e B1(N(£>)). Suppose 1B(J^) # 0. Then 0D(Jzf) # 0 and hence D is
contained in a defect group of if.
(15.35) definition Let B be a p-block of G. Then the p-defect groups of the
defect classes of B are called defect groups of B. The set of these is denoted
8(B).
(15.36) corollary Let B e B1(G). Then 8(B) is a single conjugacy class of
subgroups.
Proof Let Jf t and Jf2 be defect classes for G. It suffices to show that
8(Jti) = 8(Jf2). Let £),-e (5(Jf,). By Theorem 15.31, each of the £>,• contains a
conjugate of the other. The result follows. |
Which p-subgroups of G can be defect groups for blocks? It is a fact (which
we shall not prove) that a defect group for a block is necessarily of the form
P r\ Q for some P, Q eSylp(G). We prove the weaker assertion that Op(G)
is contained in every defect group of a block.

280
Chapter 15
(15.37) lemma Let P = Op(G). Then Pgker 9) for every irreducible
F-representation 9) of G.
Proof Since P has a unique p-regular class, the principal representation
of P is its unique irreducible F-representation by Corollary 15.11. If 9) is an
irreducible F-representation of G, then the restriction 9)P is completely
reducible by Corollary 6.6. The result follows from these two facts. |
(15.38) theorem Let Jf be a class of G and assume that Jf n C(Op(G))
= 0. Then jf is nilpotent.
Proof Let P = Op(G) act on Jf by conjugation and let 6 be an orbit of
this action. Then \G\ > 1 and so p||0|. Let xe&. If ye&, then y = xu =
x[x, u] for some ueP. Since P o G, we have \_x, u]eP and so (9 £ xP.
Now let 9) be an irreducible F-representation of G so that P £ ker 9) by
Lemma 15.37 and thus 9) has the constant value 9)(x) on the coset xP.
Therefore Y,yse Viy) = WWW = 0 since p||0|. We thus have 9)(jf) = 0 for all
irreducible 9) and hence jf e J(F[G]), the Jacobson radical. It follows that
jf is nilpotent, |
(15.39) corollary Every defect group for a p-block of G contains Op(G).
Proof Let B e B1(G) and let Jf be a defect class for B. Then lB(jf) # 0
and hence jf is not nilpotent. Thus Jf n C(Op(G)) ¥= 0 and it follows that
Op(G) £ D for every D e 5(jf). I
(15.40) corollary Let G be p-solvable with Op.(G) = 1. Then G has a
unique p-block.
Proof Let B, B'eBl(G). We claim that lB(eB.) = aB-({l}). Since this is
independent of B, the result will follow. We have lB(eB) = XaB-(X')^Jt),
where the sum runs over classes Jf. Since 1B(1) = 1, it suffices to show that if
Jf ^ {1}, then either aB-(jf) = 0 or lB(jf) = 0.
Let P = Op(G). If lB(jf) ^ 0, then Jf n C(P) ^0. However, the Hall-
Higman "Lemma 1.2.3" yields C(P) £ Pand thus Jf c P. Since Jf 76 {1},
the elements of Jf are not p-regular and thus aB( Jf) = 0. The proof is
complete. I
If B e B1(G), let D be a defect group for B and write | D | = pd. We call d the
defect of B and write d = d(B). (This is well defined by Corollary 15.36.) We
show how to compute d(B) from a knowledge of B n Irr(G) or B n IBr(G).
We mention that if m, n e Z with pjfn, then m/n e R. Thus if p | m, we have
m/n epK Q: Ai.
(15.41) theorem Let IGI = p"m with /^m and let B e Bl( G) with d(B) = rf.
Then p"~d is the largest power of p which divides all x(l) for / e Irr(G) n B.

Changing the characteristic
281
Proof Let X be a defect class for B and let gsjf. Then |Jf| =
| G |/1 C(g)| and so f~d is the p-part of | JtT | since |D \ = pd for D e Sylp(C(g)).
Let / e B n Irr(G). We have
o * iB(jf) = (c»z)*(jf) = (zte)|jr|/z(i))*.
Since x(g)e7? and i{g)\ jf\/x(l)eR - U, we conclude that |jf|//(l)^M
and hence the p-part of | Jf | cannot exceed that of x(l). Thus pa~d divides /(l).
The coefficient of g in the Osima idempotent/B is given by
fl. = U/|G|) Z xWxW)
XeBnlrr(G)
and thus
fl. = d/|G|) Z %{\M9)
,j>EBnIBr(G)
by Lemma 15.21 (since g is p-regular).
We have 0 ^ a^Jf) = a* and ag$ id. However, p" I <!>,,( 1) by Corollary
15.22 and thus %(l)/\G\eR for all cpeIBr(G). We conclude that cp(g)4Al
for some cp e IBr(G) r\ B.
Now <p(g) is a Z-linear combination of x(g) for / e Irr(G) n B by Lemma
15.28_and thus xTgit® for some / eIrr(G) r\ B, Now j^)| jf |//(1) = a e/?
and xig) = a/(l)/| jf |. It follows that z(l)/| Jf | £j<? and so the p-part of *(1)
cannot exceed p"~d. The proof is complete. |
(15.42) corollary In the situation of Theorem 15.41, pa~d is the largest
power of p which divides all <p(l) for cp e IBr(G) r\ B.
Proof In fact {/(l)|x elrr(G) n B} and {cp(\)\cp e IBr(G) n B} have the
same greatest common divisor. This follows since each /(l) is a Z-linear
combination of the <p(l) using decomposition numbers and each <p(l) is a
Z-linear combination of the x(l) by Lemma 15.28. |
In connection with Corollary 15.42, we mention that <p(l) need not divide
|G|for<peIBr(G).
In the situation of Theorem 15.41, if / e B n Irr(G), then the p-part of/(l)
can be written in the form p"~d+h, where h > 0. The integer h is called the
height of/. Brauer has conjectured that all / e Irr(G) n B have height zero iff
a defect group of B is abelian. Note that if d(B) = 1 and / e B has positive
height, then p" divides /(l) and thus B n Irr(G) = {/} by Theorem 15.29.
This forces d(B) = 0, a contradiction. Thus if d(B) = 1, then all characters in
B have height zero.
Suppose H c G and b e B1(H). Let X = lb: Z(F[H]) -> F be the
corresponding central homomorphism. We construct a linear map 1G:Z(F[G]) -> F

282
Chapter 15
by setting
\xeX~nH J
It may be that kG is an algebra homomorphism, in which case XG = XB for
some unique B e B1(G). When this happens we say that ba is defined and we
write ba = B. The block ba is called the induced block.
(15.43) lemma Let b e B1(H) for H £ g and suppose bG is defined. Then
every defect group for b is contained in a defect group for ba.
Proof Let jf be a defect class for bG and let De8(b). We have
0 t6 (lb)G( jf) = lfc(X £\ where if runs over the classes of H contained in Jf.
In particular, Jf n H # 0 and there exists Jz? = Jf n H such that lfc(^) ^ 0.
By the Min-Max Theorem 15.31, there exists Pec5(if) with D c P. Now
P e Sylp(CH(x)) for some x e if £ jf and thus there exists S e Sylp(CG(x)),
with S^P. Thus DsSe8(Jf) = 8(bG). |
The Brauer homomorphism can be used to give a sufficient condition for
induced blocks to be defined.
(15.44) lemma Let P s G be a p-subgroup and let C(P) £ H c N(P).
Then ftG is denned for all ft eBl(H). If b eBl(H) and B e B1(G), then ftG = B
iff 1B = /Jpl;,, where /?P is the Brauer homomorphism.
Proof The image of 0P: Z(F[G]) -+ Z(F[N(P)]) actually lies in Z(F[H]).
Let 1: Z(F[H]) -> F be an algebra homomorphism and let
tx = pPk Z(F[G]) -> Z(F[H]) -> F
so that /z is an algebra homomorphism. We claim that y. = XG.
Let C = C(P) c H and let Jf be a class of G. Write £*
where w = X^^nc*-
Then lG(jf) = l(w + t>)and /z(jf) = Mfi^jt)) = l(u).
We must therefore show that X(v) = 0. Now v is a sum of elements of the
form &, where if is a class of H such that £? r\ C = 0. Since P<Hwe have
C(Op(H)) c C and it follows that g is nilpotent by Theorem 15.38. Thus
X(&) = 0 and hence l(t>) = 0 and kG = fi as claimed.
If 1 = lfc, then 1G = (ipl is an algebra homomorphism and ftG = B is
defined, where B is the unique block such that 1B = /?P1. |
Brauer's first and second "main theorems" concern induced blocks.
(15.45) theorem (First Main) Let D £ G be a p-subgroup and let
JV = N(D). Then ft i-> ftG is a bijection of
{fteBl(JV)|Z)ec5(ft)} onto {BeBl(G)\De8(B)}.

Changing the characteristic
283
(15.46) lemma Let P be a p-subgroup of G. Let C = C(P) and JV = N(P).
Then Jf h-» jf n C is a bijection of the set of classes of G with p-defect group
P onto the set of classes of JV with p-defect group P.
Proof Suppose P e 8(jT) and x, y e jf n C. Then P e Sylp(C(x)) and
P e Sylp(C(y)). Write y = x9. Then P, P9 e Sylp(C(y)) and hence P = P9C for
some c e C(y). Then gceN and y = xgc. It follows that if = jf n C is a class
of JV. Clearly P e (5(if) and the map Jf i-> jf n C is one-to-one.
If if is a class of N with P e (5(if), let jf be the unique class of G which
contains ^£. Now if £ C since if n C ^ 0 and C <i JV. If x e if, let
Pcje Sylp(CG(x)). If S > P, then P < NS(P) = S n JV £ C„(x). Since
S n JV is a p-group, this contradicts P e (5(if). We conclude that P = Se 8( Jf)
and jf i-> if. The proof is complete. |
Proof of Theorem 15.45 Write a = {b e Bl(JV) | D e 3(b)}. libeSi, then
bG is defined by Lemma 15.44. Let bG = BeB1(G). Since De8(b), we have
D £ P for some P e 8(B) by Lemma 15.43. We claim that D = P.
Let if be a defect class of b and let jf 2 if be a class of G. Then XnC
= ^£ and £> e 8(3f) by Lemma 15.46. By Lemma 15.44 we have
lB(jf) = kb(pD(£)) = h(&) * 0.
By the Min-Max Theorem 15.31, it follows that D contains some defect group
of B. Thus Pg £ D £ P for some g e G and hence D = P e 8(B) as desired.
Thus block induction maps SS into {B e B1(G)| £> e 8(B)}.
Now let B eBl(G) with De8(B). Let jf be a defect class for B. Then
aB(jf) ^ 0 and D e 8(Jf). Thus 1B = pDlb for some b e Bl(JV) by Lemma 15.34
and hence bG = B by Lemma 15.44. We must show that ft e Si. Let P e (5(b)
so that D £ P by Corollary 15.39 since £> -a JV. By Lemma 15.43, P is
contained in some defect group for B and we have D £ P £ D9 for some g e G.
ThusD = Pe«5(fc)andfce;#.
Finally, let bu b2e@ with bf = B = b2G. Let if be a class of JV with
D 2 P e (5(if). If £> > P, then by the min-max theorem, kbl(&) = 0 =
kbl(&). If D = P, then by Lemma 15.46, if = C n jf for some class Jf of G.
Then Xbl( <£) = Xbl(Pd#)) = ^bW for i = 1, 2 and hence the lfci agree on all
if for classes ^£ with defect group contained in D. By the min-max theorem,
it follows that Xb2(ebl) = Xb~(ebl) = 1 and thus b1 = b2. |
To state Brauer's "second main theorem" we need to introduce
"generalized decomposition numbers."
(15.47) lemma Let n e G with o(n) = pe and let C = CG(7t). For / e Irr(G)
and (p e IBr(C), there exist unique dnx<p e R r\ Qpe such that
X(xn)= X d"x<p(p(x)
«>ElBr(C)
for all p-regular xeC.

284
Chapter 15
Proof Write Xc = Z ^"A witn a<i>e z and "A e Irr(Q- We have \jiZ(C) =
t/Kl)/v for some linear ^ e Irr(Z(C)). Then i/^(x7c) = il/(x)(i^(n) and therefore
i^Elrr(C)i(j>ElBr(C)
and so we take dnM = ^^imo^fi^nld^. Uniqueness follows from the
linear independence of IBr(C). |
The algebraic integers d"x<p are called generalized decomposition numbers.
Note that if n = 1, then C = G and d\v = dXv. Also, if g e G is arbitrary, we
can take n = gp and x = gp. and thus express %(g) in terms of generalized
decomposition numbers and Brauer characters.
Note that if b is a block of C = C(n) for a p-element n then bG is defined
by Lemma 15.44.
(15.48) theorem (Second Main) Let /elrr(G) and <peIBr(C), where
C = CG(7t) for some p-element n e G. Let / eBeB1(G) and ipebeB1(C).
Thenrf^ = OiffcG^B.
Proof Omitted. |
Theorem 15.48 is extremely powerful and useful. We give a few
consequences.
(15.49) corollary Let x £ Irr(G) and g e G. Suppose gp is not contained
in any defect group for the p-block containing /. Then x(g) = 0.
Proof Let n = gp and write g = nx, where x e C(n) is p-regular. Then
X(g) = %,d*„cp(x). If d\„ ± 0 then cpebeB1(C(tc)) and x ebG, Let Q ed(b).
Then 7t e Op(C(7t)) c Q by Corollary 15.39. Also, Lemma 15.43 yields
P e d(bG) with Q1^ P and thus rc e P, a contradiction. Thus all d"x<p = 0 and
the result follows. |
Note that Corollary 15.49 generalizes Theorem 8.17 since if/ has p-defect
zero, then the subgroup 1 is the unique defect group for the p-block
containing x-
If 9 is a class function of G and B e B1(G), write 9B = XxeirrionB [#> l\l
so that 9 = Xbebko 9b.
(15.50) theorem Let 9 be a class function of G and let n e G be a p-element.
Suppose 9(7uc) = 0 for all p-regular x e C(7t). Then 9B(7uc) = 0 for all such x
andallBeBl(G).
Proof We have
0=9(tcx)= X [9,Z]Z(^)= X [9,X]^W
XElrr(G) ^eIrr(G);veIBr(C)

Problems
285
for all p-regular x e C(n) = C. The linear independence of IBr(C) yields
ZxEimG) [8, xW» = 0 for each cp e IBr(C). If cp e IBr(C) n ft with ft e Bl(C),
Theorem 15.48 yields d"„ = 0 for x i bG and thus £, e i„(C, n * W x¥% = 0.
Nowletj^= \J{b nIBr(C)|fceBl(C),fcG = B}. Then for each cp e j/ we
have Yxzb [9, x]</Jv = 0 and thus
X e B; ij> e jrf
(15.51) corollary (Block Orthogonality) Let g, h e G be such that gp
and /ip are not conjugate in G. Then
I X(<« = 0
*Elrr(G)nB
for every p-block B.
Proof Write 9 = Xxeirr(G) X(h)X and let n = gp. If x e C(n) is p-regular,
then n = (nx)p is not conjugate to hp in G and hence S(nx) = 0 by the second
orthgonality relation. Thus 0 = 9B(g) = ^eB xidixW) bv Theorem 15.50. |
Problems
(15.1) Let D be the decomposition matrix for G. The matrix DT D = C is the
Cartan matrix. (The rows and columns of C are indexed by IBr(G).) For cp,
9eIBr(G), let yvS = (1/|G|) Y.xey <P(x)Wx), where ¥ is the set of p-regular
elements of G. Define the matrix r = (y^). Show that T = C"l.
Hint Let X0 be the part of the character table corresponding to the
p-regular classes and let Y be the Brauer character table. Then X0 = DY,
(15.2) (a) Let cp, 9 e IBr(G) lie in different p-blocks. Show that
I <p(x)%) = 0.
xey
(b) Let /, t/' e Irr(G) lie in different p-blocks. Show that Y#ey XM^M
= 0.
(15.3) Let | G | = p"m with pjfm and let \ji e Irr(G) u IBr(G). Define the class
function, S^ by 5^(x) = pV(x) if pXo{x) and S^(x) = 0 if p | o(x). Show that
S^ is a generalized character of G. Conclude that det(C) is a power of p,
where C is the Cartan matrix as in Problem 15.1.
(15.4) (a) Show that {<I>J<p eIBr(G)} is a basis for the space of class
functions on G which vanish on G — ¥.
(b) If cp, fi e IBr(G), show that (1/1G |) £, E y <D,(x)£M = «5W.
(15.5) (a) Show that the product of two Brauer characters is a Brauer
character.

286
Chapter 15
(b) If cp, ji e IBr(G), define S on G by S(x) = Ov(x)fi(x) for p-regular x;
S(x) = 0 when p\o(x). Show that S is a nonnegative integer linear
combination of {Ov | v e IBr(G)}.
(c) If q>, /zelBr(G), show that O^O^ is a nonnegative integer
combination of {<5V| v e IBr(G)}.
(15.6) Let ^ be the collection of classes of G which are defect classes for
blocks of defect d. Show that | {B eBl(G)\d(B) = d) \ < \<€\.
Hint Let A c Z(F[G]) be the span of the Jf for Jf e<#. Then the
restrictions of the algebra homomorphisms 1B to A are linearly independent
for those B with defect d.
(15.7) Let N = Op.(G). Show that if /, \ji e Irr(G) lie in the same p-block,
then Xn and ^Pn have the same irreducible constituents. If n denotes the
number of classes of G contained in N, conclude that |B1(G)| > n.
(15.8) Let P e Sylp(G). Show that the number of p-blocks of G with defect
group P is equal to the number of p-regular classes of N(P) contained in C(P).
Hint Use Problems 15.6 and 15.7.
(15.9) Let N be a normal p-complement for G and let !T be the set of orbits
of the action of G on Irr(iV). For each BeBl(G), let 0(B) denote
{SGlrr(AT)|[xN, 9] ¥= 0 for some /eBn Irr(G)}. Show that B^(9(B) is a
bijection of B1(G) onto F.
(15.10) In the situation and notation of Problem 15.9, show that 3(B) =
U»6«(*>syM'G(%
Hints If /elrr(G), Selrr(iV), and [%N, 9] ^ 0, then co^K) = co^K),
where K is any conjugacy class sum of G for a class contained in N. If Jf is a
defect class for B and P e Sylp(/G(9)) for some 9 e (9(B), show that P fixes one
of the classes of N contained in Jf and conclude that P is contained in a
defect group for B.

Appendix
Some character tables
In the following tables, each conjugacy class is denoted by the order of its
elements. If there are more than one class of elements of a given order, they
will be distinguished by subscripts.
1. G = Z4
\G\ = 24 = 23 x 3
Class
|C(s)|
|Clto)|
Xi
Xi
X3
Xa
Xs
1
24
1
1
1
2
3
3
2i
4
6
1
-1
0
1
-1
22
8
3
1
1
2
-1
-1
3
3
8
1
1
-1
0
0
4
4
6
1
-1
0
-1
1
Note Class 2t is the class of transpositions.
287

2. G = SL(2, 3)
\G\ = 24 = 23 x 3
Class
\Qg)\
|C%)|
Xi
Xl
Xi
Xa
X5
X6
Xi
1
24
1
1
1
1
3
2
2
2
2
24
1
1
1
1
3
-2
-2
-2
4
4
6
1
1
1
-1
0
0
0
3i
6
4
1
CO
CO2
0
-1
-co
-co2
32
6
4
1
CO2
01
0
-1
-co2
— CO
6i
6
4
1
CO
CO2
0
1
CO
CO2
62
6
4
1
CO2
CO
0
1
CO2
CO
Irrational Entries co = e2*'13.
3. G = A5 S PSL(2, 5) S SL(2, 4)
|G| = 60 = 22 x 3 x 5
Class
\C(g)\
\Cl(g)\
Xi
Xi
Xi
Xa
Xs
1
60
1
1
4
5
3
3
2
4
15
1
0
1
-1
-1
3
3
20
1
1
-1
0
0
5i
5
12
1
-1
0
<*i
a2
52
5
12
1
-1
0
a2
<*i
Irrational Entries at = (1 + y/5)/2 = 1 + s + e4 and a2 = (1
1 + e2 + e3, where e = e2*"5.

Some character tables
289
4. G = PSL(2, 7) = GL(3, 2)
|G| = 168 = 23 x 3 x 7
Class
|C(3)|
|Cl(g)|
Zi
Xi
Xi
n
Xs
x<>
1
168
1
1
6
7
8
3
3
2
8
21
1
2
-1
0
-1
-1
4
4
42
1
0
-1
0
1
1
3
3
56
1
0
1
-1
0
0
7i
7
24
1
-1
0
1
a
a
72
7
24
1
-1
0
1
a
a
Irrational Entries a = (— 1 + iy/l)/2 = e + e2 + e4, where e = e2*'n.
5. G = A6 =£ PSL(2, 9)
|G| = 360 = 23 x 32 x 5
Class
|Cte)i
|Clto)|
Xi
Xi
Xi
X*
Xs
X6
Xi
1
360
1
1
5
5
9
10
8
8
2
8
45
1
1
1
1
-2
0
0
4
4
90
1
-1
-1
1
0
0
0
3i
9
40
1
2
-1
0
1
-1
-1
32
9
40
1
-1
2
0
1
-1
-1
5i
5
72
1
0
0
-1
0
«1
a2
52
5
72
1
0
0
-1
0
a2
ai
Irrational Entries at = (1 + y/S)/2 = 1 + e + e4 anda2 = (1 — y/S)/2
= 1 + e2 + e3, where e = e2*"5.

290
Appendix
6. G = SL(2, 8)
\G\ = 504 = 23 x 32 x 7
Class
|C(s)|
icito)i
X\
Xi
X3
X*
Xs
Xe
Xi
x%
X9
ational
*'n. P!
1
504
1
1
8
7
7
7
7
9
9
9
Entries
= -(8
2
8
63
1
0
— 1
-1
-1
-1
ai =
3
9
56
1
-1
-2
1
1
1
0
0
0
= e +
+ n p2 =
9i
9
56
1
-1
1
Pi
Pi
P3
0
0
0
e6, a2
-(«52
92
9
56
1
-1
1
Pi
Ps
Pi
0
0
0
93
9
56
1
-1
1
h
Pi
Pi
0
0
0
7i
7
72
1
1
0
0
0
0
1*1
a2
<*3
= e2 + e5 and a3
+ 81),
and /?3
= -
72
7
72
1
1
0
0
0
0
a2
<*3
ai
= e3 +
-(<54 +
73
7
72
1
1
0
0
0
0
<*3
«1
a2
e4, where
(55), where
7. G = PSL(2, 11)
\G\ = 660 = 22 x 3 x 5 x 11
Class
|C(0)|
|C%)|
Xi
Xi
Xi
X*
Xs
X6
Xi
x%
1
660
1
1
10
10
11
12
12
5
5
2
12
55
1
2
-2
-1
0
0
1
1
3
6
110
1
1
1
-1
0
0
-1
-1
6
6
110
1
-1
1
-1
0
0
1
1
5i
5
132
1
0
0
1
«i
a2
0
0
52
5
132
1
0
0
1
a2
«i
0
0
Hi
11
60
1
-1
-1
0
1
1
P
P
112
11
60
1
-1
-1
0
1
1
P
P
Irrational Entries at = (— 1 + ■y/S)/2 = e + e4, a2 = (— 1 — y/5)/2 =
e2 + e\ where e = e2*1'5. /? = (-1+ ujli)/! = 8 + 83 + 84 + 85 + 89,
where 8 = e2*'111.

Some character tables
291
. G =
\G\
Class
\C(g)\
icite)i
Xi
ii
Xi
n
Xs
Xf>
Xl
Xs
X9
Xn
Mn
= 7920 =
1
7920
1
1
10
11
10
10
16
16
44
45
,: 55
= 24
2
48
165
1
2
3
-2
-2
0
0
4
-3
-1
x 32 x 5 x
4
8
990
1
2
-1
0
0
0
0
0
1
-1
8i
8
990
1
0
-1
a
a
0
0
0
-1
1
11
82
8
990
1
0
-1
a
a
0
0
0
-1
1
3
18
440
1
1
2
1
1
-2
-2
-1
0
1
6
6
1320
1
-1
0
1
1
0
0
1
0
-1
5
5
1584
1
0
1
0
0
1
1
-1
0
0
Hi
11
720
1
-1
0
-1
-1
P
P~
0
1
0
112
11
720
1
-1
0
-1
-1
P
P
0
1
0
Irrational Entries a = ujl. 0 = (-1 + iy/ll)/2 = e + e3 + e4 + e5
+ e9, where e= e2"111.

Bibliographic notes
There exist several excellent bibliographies in group theory and character
theory (for instance in the books by Huppert [26], Dornhoff [15], and
Curtis and Reiner [12]) and so, instead of giving a comprehensive list of
publications, we shall only mention some of the items which are directly
relevant to the various chapters of this book.
General Some other books on characters and representations are
Dornhoff [15], Feit [16], and Curtis and Reiner [12]. Each of these has a
point of view somewhat different from the others and from this book. As a
reference on group theory we mention Huppert [26] which also has an
extensive chapter on characters. Finally, we come to Burnside [10]. This
classic, although somewhat difficult for the modern reader, contains a
wealth of material.
Chapter 1 Further relevant information on rings and algebras can be
found in Curtis and Reiner [12] and Herstein [25].
Chapter 2 Other methods exist for obtaining the basic results about
characters such as the orthogonality relations. For instance, instead of
using the central idempotents of C[G], Feit [16] and Dornhoff [15] use a
matrix approach which results in additional information, namely the Schur
relations which appear here as Problem 2.20.
Chapter 3 There is, of course, a large literature in algebraic number
theory. A reference for those parts of the subject most relevant to group
theory is the appropriate chapter in Curtis and Reiner [12]. The proof of
292

Bibliographic notes
293
Theorem 3.12 given here was discovered (independently) by G. Glauberman
and the author. For another proof, see (4.2) of [16]. Whitcomb's result on
isomorphism of integer group rings occurs in Whitcomb [37].
Chapter 4 The Brauer-Fowler proof [8] of Theorem 4.11(a) does not
depend on characters. They obtain a slightly better bound which is of the
same order of magnitude.
Chapter 5 There is a great deal more to be said about the relationship
between permutation groups and representation theory. For instance, see
Chapter V of Wielandt [38] and Sections V.20 and V.21 of Huppert [26].
Chapter 6 Clifford's results appear in Ref. [11]. This paper includes
significantly more than Theorem 6.5; it also has part of Theorem 6.11 and
bears heavily on Chapter 11. The " going down " Theorem 6.18 and its dual
Problem 6.12 appears in Isaacs [28], however Corollary 6.19 goes back to
Burnside's book [10]. A result on relative M-groups which is more general
than Theorem 6.22 occurs in Price [35]. Theorems 6.22 and 6.23 give
sufficient conditions for a group to be an M-group which generalize a result
of Huppert (Satz V.18.4 of Huppert [26]). The connections between the
extendibility of 9 and det(9) are due to Gallagher [19]. A proof of a version
of Tate's Theorem 6.31 without characters appears as Satz IV.4.7 of Huppert
[26] and Thompson's proof (including Problem 6.20) is found in Ref. [36].
Chapter 7 A proof of Theorem 7.8 for the case that |P| = 8 that does
not depend on "modular characters" has recently been discovered by
Glauberman [22]. For a proof of Theorem 7.10 without characters, see
Bender [2]. A large fraction of the known applications of character theory
to "pure" group theory are either directly or indirectly related to the
content of this chapter. We mention as examples Sections 28 and 32 of
Feit [16].
Chapter 8 Brauer's Theorem 8.4 occurs in Brauer and Tate [9] and in
some of Brauer's earlier papers. Banaschewski's Lemma 8.5 appears (in a
somewhat more complicated form) in Ref. [1]. Dade's Theorem 8.24 and its
consequence Theorem 8.26 appears in more general form in Ref. [13].
Chapter 9 An alternate source for much of this material is Curtis and
Reiner [12].
Chapter 10 The more standard version of Theorem 10.7 is due to Brauer
and Witt (Theorem 70.28 of [12] or Yamada's notes [41]). Theorem 10.12
appears in Goldschmidt and Isaacs [24]. What amounts to a special case of
Theorem 10.16 occurs in Burnside [10] as Exercise 8 on page 319. That every
integer can occur as a Schur index was proved by Brauer [5] using groups

294
Bibliographic notes
similar to those of Theorem 10.16. A great deal of further information can be
found in Yamada [41].
Chapter 11 Much of the theory relating projective representations with
the properties of a character triple was originated by Clifford [11]. Our
more character theoretic setting of Clifford's work occurs in Isaacs [31].
Berger's Theorem 11.33 appears in more general form in Ref. [3].
Chapter 12 Much of this material is the work of Passman and the
author and appears in Isaacs and Passman [33, 34], and Isaacs [29]. Also
relevant are Isaacs [32], Berger [4], and Garrison [20].
Chapter 13 This chapter is taken almost entirely from Glauberman's
paper [21]. Parts of Theorems 13.6 and 13.14 also appear in Isaacs [27].
A proof of Theorem 13.25 can be found in Isaacs [31].
Chapter 14 The literature on linear groups is very extensive and we
mention just a sample. Dixon's book [14] is a good reference. For
information on solvable and p-solvable linear groups, see Winter [39, 40] and
Isaacs [30]. We also mention the lecture notes by Feit and Sibley [18] for
results without solvability hypotheses.
Chapter 15 Brauer's papers [6] and [7] are good sources for further
reading on blocks and Brauer characters. There is also a chapter on the
subject in Curtis and Reiner [12]. The material is treated from a different
point of view in Part B of Dornhoff [15] and in Feit's notes [17].
Goldschmidt's notes [23] provide a development of the subject along lines
similar to those used here.

References
1. B. Banaschewski, On the character rings of finite groups, Canad. J. Math. 15 (1963),
605-612.
2. H. Bender, Finite groups with large subgroups, Illinois J. Math. 18 (1974), 223-228.
3. T. R. Berger, Primitive solvable groups, J. Algebra 33 (1975), 9-21.
4. T. R. Berger, Characters and derived length in groups of odd order, J. Algebra (to be
published).
5. R. Brauer, Gruppen linearer Substitutionen II, Math. Z. 31 (1930), 733-747.
6. R. Brauer, Zur Darstellungstheorie der Gruppen endlicher Ordnung I, II, Math. Z. 63
(1956) 406-444; 72 (1959), 25-46.
7. R. Brauer, Some applications of the theory of blocks of characters I-V, J. Algebra 1 (1964),
152-167; 1 (1964), 307-334; 3 (1966), 225-255; 17 (1971), 489-521; 28 (1974), 433-460.
8. R. Brauer and K. A. Fowler, Groups of even order, Ann. Math. (2) 62 (1955), 565-583.
9. R. Brauer and J. Tate, On the characters of finite groups, Ann. Math. (2) 62 (1955), 1-7.
10. W. Burnside, Theory of groups of finite order (2nd ed. 1911), reprinted by Dover, New York,
1955.
11. A.H.Clifford, Representations induced in an invariant subgroup, Ann. Math. (2)38(1937),
533-550.
12. C. W. Curtis and I. Reiner, Representation theory of finite groups and associative algebras,
Wiley (Interscience), New York, 1962.
13. E. C. Dade, Isomorphisms of Clifford extensions, Ann. Math. (2) 92 (1970), 375-433.
14. J. D. Dixon, The structure of linear groups, Van Nostrand-Reinhold, Princeton, New
Jersey, 1971.
15. L. Dornhoff, Group representation theory, Dekker, New York, 1971.
16. W. Feit, Characters of finite groups, Benjamin, New York, 1967.
17. W. Feit, Representations of finite groups, Mimeographed notes, Yale Univ., 1969.
18. W. Feit and D. A. Sibley, Finite linear groups of relatively small degree, Mimeographed
notes, Yale Univ., 1974.
19. P. X. Gallagher, Group characters and normal Hall subgroups, Nagoya Math. J. 21 (1962),
223-230.
29S

296
References
20. S. C. Garrison, On groups with a small number of character degrees, Ph.D. Thesis, Univ. of
Wisconsin, Madison, 1973.
21. G. Glauberman, Correspondences of characters for relatively prime operator groups,
Canad. J. Math. 20 (1968), 1465-1488.
22. G Glauberman, On groups with a quaternion Sylow 2-subgroup, Illinois J. Math. 18 (1974),
60-65.
23. D. M. Goldschmidt (to be published).
24. D. M. Goldschmidt and I. M. Isaacs, Schur indices in finite groups, J. Algebra 33 (1975),
191-199.
25. I. N. Herstein, Noncommutative rings, Cams monograph 15, Math. Assoc. Amer., 1968.
26. B. Huppert, Endliche Gruppen I, Springer-Verlag, Berlin, 1967.
27. I. M. Isaacs, Extensions of certain linear groups, J. Algebra 4 (1966), 3-12.
28. I. M. Isaacs, Fixed points and characters . .. , Canad. J. Math. 20 (1968), 1315-1320.
29. I. M. Isaacs, Groups having at most three irreducible character degrees, Proc. Amer. Math.
Soc. 21 (1969), 185-188.
30. I. M. Isaacs, Complex /(-solvable linear groups, J. Algebra 24 (1973), 513-530.
31. I. M. Isaacs, Characters of solvable and symplectic groups, Amer. J. Math. 95 (1973),
594-635.
32. I. M. Isaacs, Character degrees and derived length of a solvable group, Canad. J. Math. 27
(1975), 146-151.
33. I. M. Isaacs and D. S. Passman, A characterization of groups in terms of the degrees of their
characters I, II, Pacific J. Math. 15 (1965), 877-903; 24 (1968), 467-510.
34. I. M. Isaacs and D. S. Passman, Finite groups with small character degrees and large
prime divisors II, Pacific J. Math. 29 (1969), 311-324.
35. D. Price, A generalization of M-groups, Ph.D. Thesis, Univ. of Chicago, 1971.
36. J. G. Thompson, Normal/(-complements and irreducible characters, J. Algebra 14 (1970),
129-134.
37. A. Whitcomb, The group ring problem, Ph.D. Thesis, Univ. of Chicago, 1971.
38. H. Wielandt, Finite permutation groups, Academic Press, New York, 1964.
39. D. L. Winter,/^-solvable linear groups of finite order, Trans. Amer. Math. Soc. 157 (1971),
155-160.
40. D. L. Winter, On the structure of certain /(-solvable linear groups II, J. Algebra 33 (1975),
170-190.
41. T. Yamada, The Schur subgroup of the Brauer group, Springer-Verlag, Berlin, 1974.

Index
Abelian group, 16, 30, 182-183
extension of characters, 63
Abelian subgroup
and character degree, 30, 84, 190, 209-212,
216-217
of linear group, 240, 249
Absolutely irreducible representation, 145-
146
character of, 149-156
Action of group
on classes and characters, 93-94, 230-231
number of orbits, 68
/7-group on ^-group, 97
Algebra, 1
Algebra homomorphism, 3
Algebraic integer, 33-40, 44, 135, 265
Alperin, J., 52
Antisymmetric part of tensor square, 50, 56
Artin, E., 72
Augmentation map, 43
Automorphism of field, see Field
automorphism, Galois group,
Galois conjugate characters
Banashewski, B., 128
Berger, T., 191, 206
Blichfeldt, H. F., 240, 247
Block, 271-286
defect of, 280
defect group of, 279
number of with maximal defect, 286
numbers of characters in, 276
Block orthogonality, 273, 285
Bottom constituent, 146-147
Brauer character
and character over F, 264
definition, 263
linear independence, 265
from ordinary character, 266-267
Brauer-Fowler theorem, 54-55
Brauer graph, 272, 275
Brauer homomorphism, 278-279, 282
Brauer, R., 46, 49, 122
characterization of characters, 127
first main theorem, 282
second main theorem, 284
theorem on actions on classes and
characters, 93
theorem on groups of order p"qbr, 274
theorem on induced characters, 127
theorem on splitting fields, 161
theorem yielding proper subgroups, 70
Brauer-Speiser theorem, 171
Brauer-Suzuki theorem,
on coherence, 112
giving normal ^-complement, 137
on quaternion Sylow subgroups, 102
297

298
Index
Brauer-Suzuki-Wall theorem, 76
Brauer-Witt theorem, 162
Broline, D., 208, 211
Burnside-Brauer theorem, 49
Burnside, W., 36, 40, 46, 213
p'q" theorem, 37
transfer theorem, 138
C
Canonical extension, 220
Cartan matrix, 285
Center
of character, 26-28, 75
of group, 27, 29, 63
of group algebra, 9, 15, 45-46, 277, see
also Homomorphism, Multiplication
constants
Central extension, 179-182, 184-186, 195
Centralizer ring
of module, 3-4, 8
of representation, 26, 145, 157-158
Character, 14, see also Irreducible character
afforded over subfleld of C, 22, 29-30, 58,
74, see also Schur index
and field automorphism, see Galois group
Character correspondence, see
Correspondence of characters
Character degree, see Degree of character,
Set of irreducible character degrees
Character table, 17, 21, 45, 48, 287-291
construction for As, 64-65
of factor group, 24
fields generated by rows and columns,
96-97
information from, 22-25, 27, 36, 45,
134-136, 141-142
and isomorphism of groups, 24
row sum, 75
Character triple, 186-189, 195-197, 237
Character value, 20, 35, 44,46, 122
constant on G - {1}, 44
equals zero, 28-29, 36, 40, 133, 246, 284
equals zero off subgroup, 28, 31, 95, see
also Vanishing-off subgroup
exceptional character, 113, 123
rational, 31
real, 31
symmetric groups, 17-18,35
Characterization of characters, 127
Chief factor, 85, 87, 96
Class function, 16, 126
of direct product, 59
of form x"", 50, 60
induced, 62
Clifford, A. H., 79-80
Coboundary, 178
Cocycle, 178
Coherence, 107-125
Collineation group, 39
Commutator, 45
Commutator subgroup, 25, 63, 75, 96,
204-205
Completely reducible module, 4-7, 12, 80,
146
Completely reducible representation, 11,
154, 157-158
Complex linear group, 240-261, see also
Faithful character
/7-solvable, 245, 256, 260-261
solvable, 243-244, 260-261
Composition series, 146
Conjugacy class
in factor group, 24
number of, 16, 46
number of/7-regular, 268
prime power size, 37
in simple group, 37
in symmetric group, 17
Conjugacy class sum
in C[G], see Center of group algebra,
Homomorphism, Multiplication
constants
in Z[G], 41-44
Conjugate
class function, 78-79
module and representation, 79-80
Constituent
of character, 17
of representation, 146-147
Control of transfer, 92
Correspondence of characters
and coprime action, 219-239
and inertia group, 82
and products, 84-85
Cyclic center and faithful character, 29, 32
Cyclic subgroups and induced characters,
72,76
Cyclotomic polynomial, 72

Index
299
D
Dade, E. C, 97, 138, 142
Decomposition matrix, 267, 285
Decomposition number, 267
Defect class for block, 278
Defect group
of a block, 279
of a class, 278
Defect of a block, 280
Defect zero character, 134, 143, 276, 284
Degree of character, 14, 37-38, see also Set
of irreducible character degrees
and abelian subgroup, 30, 84, 190
and center, 28, 31,38, 44
and derived series, 67
in group of order 27, 25
in M-group, 67
in nilpotent group, 28
in/7-block, 274, 280-281
prime power, 39, 44, 274
in £3, 16
when/>2,f|G|, 122
Degree of projective representation, 174
Degree of representation, 9
Derived series, 67, 202, 206
Determinant of character, 29, 59, 75, 88
and extendibility, 88-90, 132-133
Determinantal order, 88-90, 133, 198
Difference of characters, 50, 60, 99, see also
Generalized character
Dihedral group, 30, 53, 105
Direct product, 59-60, 172
Dirichlet, P. G. L., 169
Divisible abelian group, 182
Division algebra, 4, 149
maximal subfleld of, 170
Dornhoff, L., 96
Double centralizer theorem, 8
Double transitivity, 69, 76-77, 111
E
Elementary group, 127
End(K), 1
Equivalent factor sets and projective
representations, 178
Even order group, 54-55
Exceptional character, 107, 113, 122-125
Exponent of group, 151, 161, 167
Extendible character, 63, 84-86, 88-91,
97-98, 132-133, 140, 178, 186, 190-191
F
F-triple, 162-163
Factor group
centralizer in, 26
characters of, 24, 51, 76, 84-85
representation of, 14
Factor set, 174, 176-180, 183, 194-195
Faithful character, 28-30, 32, 77, 83, 123,
see also Complex linear group
of direct product, 60
of /7-power degree, 39, 44
Fein, B., 170-172
Feit-Sibley theorem, 120
Feit-Thompson theorem
on abelian T.I. subgroup, 246
on odd order groups, 111, 231
Feit-Thompson-Blichfeldt theorem, 247
Feit, W., 52, 108, 123, 125
Field automorphism, 29, 151, see also
Galois group, Galois conjugate
characters
Field extension
and representations, 144-159
and Schur index, 161, 170
Field generated by character values, 151
and corresponding representation, 150,
153, 159
and order of /7-group, 254
Field restriction and modules, 152-154,
156-157
Fitting subgroup, 208-209, 211, 217
Focal subgroup theorem, 142
Fong, P., 168
Frattini subgroup, 60, 173
Frobenius complement, 99, 107, 121
Frobenius group, 99-101, 121, 200, 237
representations of, 94, 269-270
Frobenius kernel, 100-101, 111, 114
Frobenius reciprocity, 62
Frobenius-Schur theorem, 50-52, 58
Frobenius theorem
on existence of kernel, 100-101
on linear groups, 251
Fully ramified character, 95-96, 196, 237,
239

300
Index
G
Gallagher, P. X., 85, 132, 191, 196
Galois conjugate characters, 152, 154,
161-162, 173
Galois group, 31, 36, 46, 90-91, 165, 221,
see also Field automorphism
Garrison, S., 59, 206, 208-209
General linear group, 13
Generalized character, 102, 107, 141, 143,
see also Difference of characters
over a ring, 126, 135
Generalized decomposition numbers,
283-284
Generalized orthogonality relations, 19
Generalized quaternion group, 53, 102
GL(n,F), 13, 174
Glauberman, G.,41, 219
Glauberman map, 219, 229
Glauberman's lemma, 223
Going down theorem, 85
Goldschmidt-Isaacs theorem, 167
Group algebra, 2
Group ring over Z, 41, 43
H
Hall-Higman lemma, 256
Hall subgroup, 111,240
character degrees of, 206
and extendibility of characters, 90, 133
generalized character of, 143
and M-group, 96
normal complement for, 136-137
Height of a character, 281
Higman, D. G., 12
Higman, G., 136
Homogeneous character, 83
Homogeneous part of module, 6
Homomorphism
of algebras, 3
of center of group algebra, 35-36, 59,
271-272, 277
of modules, 3
of tensor product, 61
I
IBr(G), 264
Ideal, 2, 75, 127, 135, 224
Idempotents, 16, 19, 274-277
Imprimitivity decomposition, 65-66
Induced block, 282
Induced character, 63-77
center of, 75
and cyclic subgroups, 72, 76
explicit computation, 64
from inertia group, 82
kernel of, 67
module interpretation, 65
and right ideal of group algebra, 75
Induced class function, 62, 73-74
Induced module, 66, 74
Induction and restriction, 62
Inertia group, 82, 95
Inner product, 20-21
Integral group basis, 41-44
Intertwining matrix, 11
Invariant character, 84-85,91, 138-139, 175,
178, 233-236, see also Extendible
character, Character triple
Involution,
centralizer in simple group, 54, 105
as fixed point free automorphism, 114
number of, 51-54
Irr(G), 15
Irreducible Brauer character, 264
number of, 268
Irreducible character, 15
over arbitrary field, 149-156
number of, 16, 232
Irreducible constituent, see Constituent
Irreducible linear group, 240
Irreducible module, 4
Irreducible projective representation, 177
Irreducible representation, 11
of cyclic group, 159
of group, 13
structure over arbitrary field, 154
Isometry, 107-110, 113, 115, 121
Isomorphism of character triples, 187-189,
192, 196
Isomorphism of integer group rings, 41
Ito, N., 84, 190,216,244
J
Jacobson radical, 11-12, 97, 157
Jordan-H61der theorem, 146
Jordan's theorem, 249

Index
301
K
Kernel
of character, 23-24, 75, 77, 81, 208-209,
211
of induced character, 67
Krull-Schmidt theorem, 156
L
Lifting a projective representation, 181-182
Linear character, 14
and abelian group, 16, 30
action of group of, 95
and commutator subgroup, 25
extendibility, 89
Linear group, see Complex linear group,
General linear group
Linear isometry, see Isometry
Local integers, 265
M
M-group, 67, 86-87, 95-96
solvability and derived length, 67
sufficient condition for, 83, 87
McKay, J., 232
Mackey, G. W., 74
Maschke's theorem, 4
Mathieu group M22, 70
Maximal subgroup, 75, 191
Min-max theorem, 278
Module, 3
and corresponding representation, 10
finitely generated over Z, 34
and normal subgroup, 79-80
Monomial character, 67, 74, 86, 96
Multiplication constants, 2
of center of group algebra, 15,45
N
Nakayama's lemma, 265
Nilpotent group, 27, 83, 111, 171,212
character degree in, 28
Nilpotent ideal, 11
Norm map, 165
Normal matrix, 57, 249
Normal subgroup, 78-98
from character table, 23
in linear group, 243-245, 249, 260
prime index, 86
restriction of character to, 79
Normals-complement, 137-138
O
Odd order group, 46, 231
Order of character, see Detertninantal order
Orthogonality relations, 18-22, 273, 285
Orthonormality of Irr(G), 21
Osima idempotent, 277
Osima, M., 275
P
/7-block, see Block
/7-defect group, see Defect group
/7-defect zero, 134, 143, 276, 284
/7-elementary group, 127, 131, 138,140
/7-group, 29, 60, 91, 96-98, 168, 218
/7-quasi-elementary group, 129, 131
/(-rational character, 90, 98, 191, 240-241,
254, 260
/(-regular element, 263
/7-solvable group, 244-245, 256, 260, 280
Partitioned group, 30
Passman, D. S., 77, 218
Perfect group, 46
Permutation character, 68, 93, 101
necessary conditions for, 69, 75
ring of, 76, 128
Permutation group, 68, 76-77, 101,111
Permutation isomorphism, 230
Permutation module, 68
PGL(n,F), 174
Primitive character, 66, 83, 191
Primitive linear group, 257
Primitive module, 65
Principal block, 274
Principal character, 14
Principal indecomposable character, see
Projective character
Product of characters, 30, 47-49, 59-61,
84-85, 202, 285-286

302
Index
Projective character, 273, 285
Projective general linear group, 174
Projective lifting property, 181-182, 193
Projective representation, 174-197
Quasi-elementary group, 129-130, 142
Quasi-primitive character, 83, 96, 191
Quaternion algebra, 145
Quaternion group, 30, 145, 159, 168, 171,
see also Generalized quaternion group
Quotient group, see Factor group
R
/^-generalized character, 126, 135
Radical, see Jacobson radical
Ramification, 174
Rank of permutation group, 69
Rational character, 31, 72-73, 76, see also
/7-rational character
Rational integer, 34
Real character, 46, 50, 54-58, 96
Real element, 31,54, 96
Real representation, 30, 56-58
Regular character, 18
Regular module, 3
Regular representation, 18, 147
Relative M-character, 86, 96
Relative M-group, 86, 96
Representation, 9, see also Irreducible
representation
and corresponding module, 10
of group, 13
over R, 30, 56-58
Representation group, see Schur
representation group
Representative set of modules, 6
Restriction, 26
connection with induction, 62
to normal subgroup, 79
Rieffel, M., 8
Roots of unity, 20, 44, 46, 90-91, 220-221,
263
and splitting field, 161, 165-168
Roquette, P., 168
Schur, I.,
theorem on faithful /7-rational character,
254
theorem on lifting projectives, 181, 184
Schurindex, 160-173
and abelian Sylow subgroup, 165, 173
and direct products, 172
and exponent of group, 161, 167-168
and field extension, 161, 170
Schur multiplier, 181-186, 197
Schur relations, 32
Schur representation group, 182, 185-186,
197
Schur's lemma, 4
Schur-Zassenhaus theorem, 99, 223
Schwarz inequality, 53
Second orthogonality relation, 21
Section of group, 162
Seitz, G., 206
Semidihedral group, 53
Semidirect product, 219
Semiprimitive character, 171
Semisimple algebra, 4, 7-9, 12
Set of irreducible character degrees, 81,
198-218, 243
and abelian subgroup, 209-212, 216-217
cardinality three, 206
cardinality two, 201-205, 216
and derived length, 202, 206
divisible by p, 199
and Fitting length, 209
maximum, 74, 98, 203, 209-214, 216-218
minimum nonlinear, 75
and/7-structure, 213-217
power of/7, 81, 84
prime to/7, 215-216
Sibley, D., 116, 120, 125
Siegel, C. L., 46
Similar projective representations, 177
Similar representations, 9, 144, 147, 157
characters of, 14, 17
Simple algebra, 8
Simple group
and centralizer of involution, 54
characters of, 23,44
conjugacy classes of, 37
order pftfr, 274
order 360, 70

Index
303
SL(2, q), 76
Socle, 77
Solomon, L., 75, 129, 168
Solvable group, 24, 95, 142, 199, 209, 212,
216-217, see also M-group
character correspondence in, 231
complex linear, 243-244
and extendible characters, 88, 90, 98
quasi-primitive characters of, 96, 191, 194
sufficient conditions for, 37, 67, 206, 243
Special element, 195-196
Splitting field, 146, 148-149, 154, 159, 161
Standard map, 182
Strong isomorphism of character triples, 196
Subnormal subgroup, 190
Supersolvable group, 87, 206
Suzuki group, 125
Suzuki, M., 102, 112, 124
Sylow subgroup, 77, 114, 116, 122, 138, 142
all abelian, 87, 217
and character degrees, 213-215
and extendible character, 89, 190-191
of Galois group, 98, 167-168
in linear group, 39, 44, 243-245, 247, 253,
260
and Schur multiplier, 186
strongly self-centralizing, 61, 76
Sylow tower, 216
Symmetric group, 17-18, 35, 69
Symmetric part of tensor square, 50, 56
T
T.l. set, 101-102, 107, 111,246
T.I.F.N. subgroup, 111-125
Taketa, K., 67
Tamely imbedded subgroup, 111
Tate, J., 92, 128
Taussky, O., 53
Tensor power, 60
Tensor product, 47-48, 50, 55, 61, 156
and induced module, 66
Thompson, J. G., 46, 52, 92, 97-98, 105, 111,
199
Top constituent, 146-147
Trace of matrix, 14
Transfer, 63, 92, 136, 138, 142
Transitivity of induction, 73
Transversal, 62
Trivial intersection set, see T.I. set
Twisted group algebra, 176-177, 194-195
U
Unitary matrix, 57, 249-250
V
Values of characters, see Character value
Vandermonde determinant, 49
Vanishing-off subgroup, see also character
value, 200, 208
W
Weak block orthogonality, 273
Wedderburn, J. H. M.
theorem on finite division rings, 149
theorem on semisimple algebras, 7
Whitcomb, A., 41
Wielandt, H., 121
Winter, D. L., 243, 257
Y
Yamada, T., 171
Z
Zassenhausgroup, 111, 123, 125
Zeros of characters, see Character value,
Vanishing-off subgroup