Modular Representations of Finite Groups (Pure and Applied Mathematics (Academic Press), Volume 73)

Modular Representations of Fiiite Groups Pure and Applied Mathematics A Series of Monographs and Textbooks Editors Sa...

Author: B. M. Puttaswamaiah | John D. Dixon

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Modular Representations of Fiiite Groups

Pure and Applied Mathematics A Series of Monographs and Textbooks Editors Samuel Ellenberg and Hymen B e r m Columbia University, New York

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Modular Representations of Finite Groups B. 11. Puttaswamaiah John D. Dixon Deparrment of Mathematics Carleton Unicersity Ottawa, Canada

ACADEMIC PRESS

New York

San Francisco London

A Subsidiary of Harcourt Brace Jovanorich, Publishers

1977

COPYRIQHT 0 1977, BY ACADEMIC PRESS, INC. ALL RIQHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN A N Y FORM OR BY A N Y MEANS, ELECTRONIC OR MECHANICAL, INCLUDINQ PHOTOCOPY, RECORDINQ, OR A#Y INFORMATION STORAGE A N D RKTRIEVAL SYSTEM, WITHOUT PERMISSION IN WRlTINQ FROM THE PUBLISHER.

ACADEMIC PRESS, INC.

111 Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 14/18 Oval Road. London

NWI

Libtary of Congress Cataloging in Publication Data Puttaswamaiah,B.M. Modular representationsof finite groups. (Pure and applied mathematics, a series of monographs and textbooks ; ) Bibliography: p. Includes index. 1. Modular representations of groups. 2. Finite groups. I. Dixon, John D., joint author. 11. Title. 111. Series. QA3.P8 [QA1711 510'.8s [ S 12'.22] ISBN 0-12-568650-1 AMS (MOS) 1970 Subject Classifications: 2OC05,2OC15, 2oc20 PRINTED IN THE UNITED STATES OF AMERICA

76-21038

Contents

ix

Preface Note

to

xi

the Reader

xiii

Notar ion

Chapter I 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Group algebras and modules Reducible and irreducible modules Semisimple rings and the Wedderburn structure theorem Tensor products The number of irreducible KG-modules Indecomposable modules Absolutely indecomposable and absolutely irreducible modules Principal indecomposable modules Composition factors and intertwining numbers Notes and comments

Chapter I1 2.1 2.2 2.3 2.4

Representation Modules

Induced Modules and Characters

1 1 5 8 13 15 20 24 26 29 32

33 33 37 39 43

Induced modules Clifford's theorem Group characters The theory of ordinary characters V

CONTENTS

Vi

2.5 2.6 2.7 2.8

Induced characters Brauer's theorem on induced characters Splitting fields Notes and comments

Chapter I11 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

The padic integers padic algebras Ordinary and modular representations Lifting idempotents The case where p does not divide I G I Modular characters Cartan invariants, decomposition numbers, and orthogonality relations Modular characters of psolvable groups Notes and comments

Chapter IV 4.1 4.2 4.3 4.4 4.5 4.6 4.7

The Theory of Indecomposable Modules

Relatively projective modules Vertices and sources Green's theorem The degrees of indecomposable modules Vertices and defect groups Restriction of indecomposable modules Jordan's theorem in characteristic p Notes and comments

Chapter VI 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

Blocks of Group Algebras

Blocks Classifying modules, characters, and idempotents into blocks Defect groups Further analysis of the Cartan matrix and decomposition matrix The characters in a block of given defect Blocks of small defect Notes and comments

Chapter V 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Modular Representations and Characters

The Main Theorems of Brauer

The Brauer homomorphism Blocks with normal psubgroups The Brauer correspondence: The First Main Theorem Extension of the First Main Theorem Generalized decomposition numbers: The Second Main Theorem Principal blocks: The Third Main Theorem The characters in the principal block Notes and comments

48 52 56 60 62 62 65 69 73 75 77 79 84 88 89 89 93 100 104 107 112 114 116 116 120 123 127 129 130 133 140 142 142 144

149 152 157 162 164 168

vii

CONTENTS

Chapter VII

Fusion of 2-Groups

169

Further results on generalized decomposition numbers Some technical lemmas Groups with Sylow 2-subgroups of type (2"'. 2"') Groups with quaternion Sylow 2-subgroups Glauberman's 2.-theorem Notes and comments

169 174 178 184 191 196

Chapter VIII Blocks with Cyclic Defect Groups

197

7.1 7.2 7.3 7.4 7.5 7.6

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

Extending characters from normal subgroups Blocks with normal cyclic defect groups Groups with cyclic Sylow psubgroups Some technical lemmas Groups of order g = pg, with p $8, Groups with a faithful representation of degree d < f ( p - 1) Criteria for normal Sylow p-groups Notes and comments

197 199 205

206 208 215 220 226

References

229

Index

239

This Page Intentionally Left Blank

Preface

Our purpose in this book is to give a direct and readable account of the basic concepts and techniques of the theory of modular representations of finite groups and some significant applications. If we have succeeded in our aim, then the book will be useful both as a textbook for students and as an introductory guide to the rather extensive literature in the subject. The theory of modular representations has a reputation of being much more difficult than the theory of ordinary representations, and to some extent this reputation is justified. At least at the beginning, the modular theory requires a more sophisticated background in algebra, and it takes a little longer to reach the stage where we can apply the theory. On the other hand, a knowledge of elementary properties of rings and modules up to, say, the Artin-Wedderburn structure theorem for Artinian rings together with a reasonable background in the theory of finite groups will be adequate prerequisite for almost all that we shall do in this book. At the same time we shall see that there are serious applications of the theory at quite an early stage, and before the end of the book we shall be using the theory to prove some of the deep and important results on finite groups that have been obtained only in this way. In presenting this material we have chosen to work throughout with representations over splitting fields; indeed, much of the work is carried out over a specially constructed field described in Chapter 111. This seems appropriate in an introduction such as this since the great majority of ix

X

PREFACE

applications of the theory so far can be given in this context. In fact, at all points where we have had to make a choice we have preferred to treat a particularly important case as clearly as possible rather than give the most general result. To balance this we have included notes and comments at the end of each chapter; in these sections we offer guides to pertinent literature. It is our hope that anyone using this book will thus be able to go much further into the subject than we have been able to do within the confines of this book. Throughout the volume we have included examples, and at the ends of most sections there are exercises for the reader to test his understanding. Many of the latter are essentially corollaries and extensions of the main results, and some are counterexamples to show how our theorems are limited. The basic foundations of modular representation theory were laid by Richard Brauer, and his monumental contribution to the subject spreads over the past five decades, through his numerous papers and the work of his students and collaborators. His influence is evident in every chapter of this book. In writing this book we are indebted to Dr. Roderick Gow who read a draft of the manuscript and whose suggestions led to a number of important corrections and clarifications. We also wish to thank Gillian Murray who typed the final version of the manuscript.

Note to the Reader

The first chapter is essentially introductory, detailing the basic algebraic concepts and setting the notation. Much of this material will be quite familiar and the reader should pass over this chapter quite quickly. To a lesser extent the same may be true of the second chapter; parts of it will be familiar to someone who has studied the theory of ordinary representations, although some will not. The real differences between the ordinary and modular theories become apparent in Chapter 111. Each chapter ends with a section of comments and references. These should be read to put our results into context and also for suggestions for further reading. Our use of notation follows current mathematical usage (such as, for example, Lang [l]), and a list of special notation has been included. One symbol which may not be familiar is the sign “:=” which may be read as “is defined by.” The reader should be warned that certain conventions are introduced and used throughout the book. In particular, group always means finite group, and ring means a ring with unity so the unity is mapped onto unity under ring homomorphisms. Each “module” or “algebra” is finitely generated and unital. The letter p always denotes a prime. Following 83.3 further general conventions are made.

xi


Notation

Relations and operations 4

.-

normal subgroup of defined to be equal to p divides m p does not divide m U is isomorphic to a direct summand of V the subgroup generated by the subset S tensor product direct sum direct product restriction off to S set difference

Alphabetic matrix with (i, j)th entry ail decomposition matrix reduced decomposition matrix decomposition matrix of H Cartan matrix of k H induced character xiii

xiv

NOTATION

automorphism group of G automorphism group of R-module V alternating group on n symbols ring of polynomials over A principal pblock ring of ordinary generalized characters ring of modular generalized characters algebra of class functions over k centralizer of S in G dimension of V over k ring of R-endomorphisms of V inner product index of H in G p‘-elements of G commutator subgroup of G commutator subgroup of G’ general linear group of degree n over A group of homomorphisms from U to V intertwining number of the pair U,V if and only if image of the mapping f kernel of the homomorphism f kernel of the module V kernel of the character C ring of d x d matrices over A normalizer of S in G largest normal psubgroup of G largest normal p’mbgroup of G orbit of s under G projective special linear group field of padic numbers Jacobson radical of R rank of V over A stabilizer of s symmetric group on R trivial module inner tensor product dual of U restriction of V to H conjugate of V associated with x induced module ppart of x

xv

NOTATION

Z(kG)

p’-part of x center of the group G center of the group algebra kG ring of p-adic integers center of G/O,.(G)


CHAPTER

I

Representation Modules

The object of this chapter is to give a short introduction to material of a general algebraic nature which will be used in the later chapters. The topics include group algebras, reducibility and decomposability of a module over a ring, tensor products, the structure theory of semisimple algebras (the Wedderburn structure theorem), and the relation that holds between composition factors ofa module and intertwining numbers. Some standard theorems are stated without proof; proofs are easily available and can be found, for example, in Lang [l]. 1.1 Group algebras and modules By convention all groups which we consider in this book will be finite unless otherwise specified. Let G be a group and S an arbitrary set. An action of G on S is a homomorphism 8 of G into the symmetric group Sym(S) of all permutations of S . We shall also say that G acts on S (via 8). Thus if G acts on S via 8, then for each x E G, 8(x) is a bijection of S onto itself satisfying the conditions: 8(l) = 1 (the identity on S), 8(x- ') = 8(x)- and O ( x y ) = 8(x)O(y) for all x, y in G. The action is calledfaithful if the kernel of the action Ker 0 equals 1, the trivial subgroup; in this case 8 is an injection of G into Sym(S). When the action is understood we shall suppress the symbol 8 and write instead SI+ sx or SH sx for 8(x) (x E G, s E S ) .

',

EXAMPLE 1 Let H be a subgroup of G. Then H acts on the set G by right multiplication; namely sx :=sx (product) for s, x E G and by left multiplication by sx := x-ls (product) for s, x E G . 1

2

I EXAMPLE 2

REPRESENTATION MODULES

G acts on the set I:of all its subgroups by conjugation; namely E C and x E G ) .

H" :=x- 'Hx (H,x- 'Hx

For any action of a group G on a set S we define the stabilizer of an element s E S by Stab(s) I= {x E G I sx = s} and the orbit of an element s E S by Orbit(s):={sXI x E G } . G acts transitiuely if Orbit(s) = S for each s E S. Note It is readily seen that Stab(s) is always a subgroup of G and that the orbits in S form a partition of S. Since sx = sy if and only if Stab(s)x = Stab(s)y, the elements of Orbit(s) are in one-to-one correspondence with the right cosets of Stab(s) in G ; hence I Orbit(s) I = I G : Stab(s)I. This important relation shows in particular that the length (Orbit(s)(of each orbit divides the order I G I of G .

Let A be a commutative ring. By convention all rings that we shall consider will be associative rings with unity 1 # 0. For any group G we define the group algebra AG of G over A to be the set of all functionsf: G + A with the operations + and defined by

( f +m + = f ( x ) + d x )

and

(fg)(x):=

c f(ylg(y-'x)

YEG

for all x E G . It is routine to verify that AG is a ring under these operations. We can embed G into AG by the mapping x ~ f , wheref, , is defined by

1 fx(y) := (0

ify=x otherwise.

Then, if we identify x andf,, eachfc AG can be written uniquely in the form f= axx, where a, E A ; indeed a, =f(x) for each x E G . Addition and multiplication in AG, then, correspond to

cxEG

( c axx)( c Bxx) c c axBYxY. XE

c

= XEG

XE

yeC

G

We shall use this latter notation almost exclusively. We next consider modules over a ring R. By convention (unless explicitly stated otherwise), we shall use the term R-module to refer to a finitely generated unital right module V over R. Thus as well as general module properties, V is assumed to have a finite subset {ul, u q , ..., u,} such that V = u , R + u2 R + ... + u, R, and for each u E V we have u 1 = u (1 is the unity of R). In the case R is commutative, a (right) R-module V is also a left R-module, and it is sometimes convenient to write the elements from R on the left as well as the right; so ru = ur for all u E V, r E R, in this case. We I

1.1

3

G R O U P ALGEBRAS A N D M O D U L E S

define the annihilator of the R-module V to be Ann( V) = {r E R I Vr = 0). Clearly Ann( V) is an ideal of R and V can be considered as an R/Ann( V)module. Obviously the annihilator of the R/Ann( V)-module V is (0). An R-module V is free (or more precisely R-free) if it has a (finite) set ul, u 2 , . . ., u, of elements such that each u E V can be written in exactly one way in r,,~ t h e f o r m u = u , r l + u 2 r 2 + ~ ~ ~ + u , r , w i t h r 1 , r 2 , . . . ,R;inthiscaseu,, u 2 , . . ., u, form an R-basis of V . EXAMPLE

3 If K is a field, then every K-module is K-free.

EXAMPLE 4 If A is a (commutative) principal ideal domain, then a basic theorem on the structure of 4-modules states that each A-module V can be written as a direct sum V = u , A 0 u2 A 0 * * * 0 u, A for suitable ui in V (see Lang [I], Chapter 15, for a proof). It follows that V is A-free if and only if ua # 0 for each u # 0 in V and a # 0 in A. (If this latter condition holds then we say that V is rorsion-free over A). The structure theorem also shows that for a given torsion-free A-module, there is a number r called the rank of V such that all A-bases of V have r elements. This example will be important later.

EXAMPLE 5 For each ring R we have the regular R-module consisting of the same set of elements as R and with the natural multiplication by elements from the ring R. The regular R-module R is R-free and has {I} as a basis. Let A be a commutative ring. Then an A-algebra is a ring R which is at the same time an A-module such that the two kinds of multiplication are compatible in the sense

a(xy) = (ax)y = x(ay)

for all a E A and x, y

E

R.

EXAMPLE 6 If A is a commutative ring and G is a group, then the group algebra AG is an A-algebra. The set of elements of G forms an A-basis for AG and so AG is A-free. Let A be a commutative ring and AG the group algebra of a group G over A. Then for any AG-module V we have the action of G on V defined by ux = ux (module product) for all u E V and x E G. Conversely, suppose that we are given an A-module V and an action of G on V which is " linear" in the sense that

(

for all ai in A, u iin V, and x operation

a i u i ) x= i= 1

E

aiu; i= 1

G. Then V becomes an AG-module under the

4

I


( u E V and all a, E A). Thus for an A-module V there is a one-to-one correspondence between the possible definitions of an AG-module structure on V and the possible linear actions of G on the A-module V. It is often useful to define the structure of an AG-module in this way.

If U and V are modules over a given ring R, then we shall write Hom,(U, V) to denote the set of R-homomorphisms of U into V. Then Horn,( U,V) is a module over the ring of rational integers under the addition defined by u(f+ g ) I= uf+ ug for allf, g E Hom,(U, V), and u E U.In particular, End,( U ):= Horn,( U,U )is a ring with multiplication defined by u ( f g ) = (uf)g for allf, g E End,(U) and u E U.Finally, Aut,( V )is the group of units of End,( U),consisting of all R-automorphisms of U. If A is a commutative ring, and if U and V are A-modules, then Horn,( U,V) is an A-module with module operation given by u(fa) := (uf)a for allfE Hom,(U, V), u E U,and a E A. Note Suppose AG is a group algebra of a group G over a commutative ring A and R is a subring of AG. Then for any AG-modules U and V we have Hom,,(U, V) C Hom,(U, V). If A E' R then g E Hom,(U, V) lies in Horn,,( U,V) if and only if (ug)x = (ux)g for all u E U and x E G.In particular, HornAG( U,V) c Horn,( U, V).

Consider the case where U is a torsion-free module over a commutative ring A. Assume that U has a basis {ul, u 2 , . .., ud}over A and that each basis of U over A has d elements. Then the usual argument from linear algebra shows that there is a corresponding algebra isomorphism End,(U) z Mat(d, A), the ring of all d-by-d matrices over A. Under this mapping Aut,(U) = Gqd, A), the general linear group of degree d over A. Here G 4 d , A) is the group of units of Mat(d, A) and consists of all matrices whose determinants are units in A. Let A be a commutative ring and G a group. A (linear)representation of G on a nonzero A-module V is a homomorphism T: G Aut,( V); the representation is faithful when the kernel Ker T equals 1. Sometimes Ker T will also be denoted by Ker K A representation of G is just a linear action of G on V in the sense referred to above. Thus to each representation T of G on V there corresponds an AG-module structure on V given by

for u E V and all a, E A, and conversely. Hence there is a one-to-one correspondence between the class of all nonzero AG-modules and the class of all linear representations of G on A-modules. Moreover, if V is A-free, then Aut,( V) z Gqd, A), where d is the number of elements in an A-basis of V.

1.2

5

REDUCIBLE A N D IRREDUCIBLE MODULES

(Again we assume here that any two bases of V have the same number of elements.) Thus to each representation of G on the A-free A-module V there is a matrix representation T*: G + G q d , .4). In particular, if A = K is a field, then all K-modules are K-free. In this case the study of KG-modules, linear representations of G on K-modules, and matrix representations of G over K are essentially equivalent. The theory of representations of groups is concerned with the study of these objects. Every homomorphism T from G into Aut,( V) can be extended to a nonzero homomorphism of AG into End,(V) by T

a,x (XEG

)

=

1 a,T(x)

with a, E A.

XEG

Conversely, every nonzero homomorphism T from AG into End,(V) determines a unique representation of G on V. Thus the study of representations of G and the study of nonzero representations of AG are equivalent. EXERCISES

1. Suppose that a group G acts on a set S and that for each s E S and each subgroup H of G we put Orbit,(s):={s” 1 x E H}.Let P be a Sylow psubgroup of G. Show that for each subgroup H of G, 1 Orbit&) 1 divides )Orbit(s)l for all s E S. In particular, show that )Orbit,(s)) is equal t o the largest power of p dividing I Orbit(s) I. 2. Let G act transitively on S. For each s in S, show that the normalizer N,(Stab(s)) of Stab(s) ( s E S ) is transitive on the points fixed by Stab(s). Also show that the normalizer of a Sylow p-subgroup P of G acts transitively on the points fixed by P. 3. Let G be a group of order 108. By considering the action by conjugation of G on the Sylow 3-subgroups of G, show that G has a normal subgroup of order 9 or 27.

1.2 Reducible and irreducible modules Let R be a ring and V an R-module. Then V is called Noetherian (or said to satisfy the maximal condition on R-submodules ”) if every nonempty set R of R-submodules of V contains a submodule W, which is maximal in R in the sense that W ER and W 2 W, implies W = W,. Dually, V is Artinian (or satisfies the “minimal condition on R-submodules ”) if every nonempty set R of submodules ‘of V contains a submodule W, which is minimal in R in the sense that W E i2 and W, z W implies W = W,. The ring R itself is called Noetherian or Artinian, respectively, if the regular R-module R has “

’

1

6


these properties; in this case '' submodule " is equivalent to "right ideal." EXAMPLE 1 If R is an algebra over a field K, then R is a finitely generated K-module by definition. More generally, since (by convention) each Rmodule V is finitely generated over R, V is also finitely generated as a K-module. In other words, every R-module V is finite dimensional over K. Since every R-submodule of V is a K-submodule, it follows (by reasons of dimension) that V is both Noetherian and Artinian. In particular, R itself is both Noetherian and Artinian.

Lemma 1.2 (Fitting's lemma) Let U be an R-module and letfe End,(U). (i) If U is Artinian andfis injective, thenfis an R-automorphism. (ii) If U is Noetherian andfis surjective, thenfis an R-automorphism. (iii) If U is both Artinian and Noetherian, then we can write U = Uf" @ Kerf" as a direct sum of submodules for some integer n 2 1.

Proof (i) Since U is Artinian there exists an integer n such that U 3 Uf 3 3 Uf" = Uf"+l. Then for each u E U there exists u E U such that uf" = of"". Sincefis injective, u = of€ Uf.Hencefis surjective, and so an R-automorDhism. (ii) Since U is Noetherian there exists an integer n such that 0 c K e r f c * * c Kerf" = Kerf"' Since f (and hence f")is surjective, each u E Kerf may be written in the form u = vf" for some u E U.Since uf= 0, this shows that u E Kerf"" = Kerf", and so u = 0. Thus Kerf= 0, and sofis injective and hence an R-automorphism. (iii) Since U is Artinian, Uf"= Vf"" for some integer n 2 1. Then the restriction offto the submodule Uf" is surjective, and so by (ii) this restriction is also injective. Hence Uf " n Kerf" = 0. On the other hand, since Uf " = Uf'", for each u E U there exists u E U such that (u - ufn)fn = 0. But the latter shows that u E uf" + Kerf" E Uf" Kerf". Hence U = Uf"@ Kerf" as asserted.

'.

+

Let V # 0 be an R-module. Then V is reducible if it has a proper Rsubmodule U # 0; otherwise V is irreducible. (The zero module 0 is neither reducible nor irreducible.) A composition series for an R-module V is a series of submodules of the form (1)

v = V, v, 3 v, ... v, = 0 3

3

3

such that for each i 2 1, the factor module 6- l / q is irreducible. The JorduwHiilder theorem states that if a module V has a composition series (l), then every series (2)

V=Uo3U,~*.*~U,=0

1.2

REDUCIBLE A N D IRREDUCIBLE MODULES

7

of submodules can be refined by inserting additional submodules where necessary to make a composition series and, moreover, if (2) is also a composition series, then m = 1 and there is a permutation i~ n, of { 1,2, .. ., I } such that 2: Urn,- /Un, for i = 1, 2, ..., 1 - 1, 1. An irreducible R-module U is called a composition factor or an irreducible constituent of V if LI z y- ,/Y for some i; and it has multiplicity j if U z y - ,/V for exactly j values of i.

v- /v

EXAMPLE 2 Not all R-modules have composition series. For example, the ring Z of integers as a Z-module does not. In the case when R is an algebra over a field K , however, every R-module V does have a composition series. Indeed we saw in Example 1 that V is both Noetherian and Artinian. By the Noetherian condition we can find a sequence V = V,, V,, V,, ... of RBy the Artinian condition submodules of V such that is maximal in the sequence terminates after a finite number of steps; hence for some 1 2 0, V, = 0. The resulting series (1) is clearly a composition series.

v

v-

Note Suppose I is a maximal right ideal of the ring R. Then R/I is an irreducible R-module. Conversely, suppose V is an irreducible R-module, and choose any u # 0 in V. Since u = u . 1 E uR, and uR is an R-submodule of V, the irreducibility of V implies that uR = V. The mapping r w or is an R-homomorphism of R onto V with kernel I, say. Then I is the annihilator of the element u, and R/I N c! Since V is irreducible, I is a maximal right ideal.

Finally, a basic property of irreducible modules is given by Schur’s lemma. If U and V are irreducible modules over a ring R, then Horn,( U , V) = 0 if U is not isomorphic to V. If U is isomorphic to V, then each nonzero mapping in Hom,(U, V ) is bijective; in particular, End,(U) is a division ring.

[Proof If f f 0 in Hom,(U, V), then K e r f # U . Since U is irreducible, K e r f = 0, and so the image I m f # 0. Hence I m f = V because V is irreducible.] EXERCISES

1. Let V be a submodule of an R-module U . Then prove that U is Noetherian (Artinian) iff V and U / V are Noetherian (Artinian). 2. Prove that a finite direct sum of Artinian (Noetherian) R-modules is Artinian (Noetherian). 3. Let G be the cyclic group of order 3 generated by a, Q the rational field, and V the vector space of 2-tuples of Q. If V is a QG-module such that (a, 8). = ( - 8 , a - /I) show , that V is irreducible.

I

8


4. If in Exercise 3 the field Q is replaced by the complex field, show that V is reducible.

Note This example shows that the property of irreducibility depends on the underlying field. 5. Let G be the group generated by a, b where a2 = b2 = ( ~ b=)1,~Q the rational field, and V a QG-module of dimension 3 over Q. If {ul, u 2 , u3} is a basis of V such that (a1u1 a2 u2 a3u3)a = a2 u1 a1 u2 a3 u3 and (al u1 a2 u2 a3u3)b= a1 u1 a3 v2 a2 u 3 , show that V is reducible. Write out a composition series of V. Determine the irreducible constituents of V. 6. Let R be a ring and let U and V be Artinian and Noetherian R-modules. If Horn,( U,V) # 0, prove that U and V have at least one common composition factor.

+

+

+ +

+ +

+

+

1.3 Semisimple rings and the Wedderburn structure theorem Let R be a ring and R the set of all maximal right ideals of R. For any I E Q R/I is an irreducible R-module and Ann(R/I) is an ideal of R. The Jacobson radical of R is defined to be rad R := I E n Ann(R/I). It is readily verified that rad R = (7 I E n I and rad(R/rad R) = 0. Since rad R c Ann ( V ) for every irreducible R-module, V can be considered as an irreducible R/rad R-module, and conversely. Thus R and RJrad R have the “same ” irreducible modules and irreducible representations. Remark Because R is assumed to have a unity, a simple application of Zorn’s lemma shows that R # 0.Note also that whenever U and V are isomorphic R-modules, then Ann( U )= Ann( V). In particular, since each irreducible R-module V is isomorphic to R/I for some maximal right ideal I, rad R is the intersection of Ann( V) taken over all irreducible R-modules K

K,and let V be an irreducible R-module. Put D:=End,(V) (so D is a division ring by Schur’s lemma). Consider the representation T: R + EndK(V) afforded by K and put S I = Im T The ring End,( V) has an R-module structure given by ua := uT(a) for all u E End,( V) and a E R, and S is both a subring and submodule of EndK(V ) . Lemma 2.3 Let R be an algebra over a field

(i) All R-composition factors of S are isomorphic to K (ii) S is a simple ring. (iii) S = End,( V ) .

1.3

SEMISIMPLE RINGS AND THE WEDDERBURNSTRUCTURE THEOREM

9

Proof (i) Put d :=dim,( V). Then End,( V) N Mat(d, K) and it is readily seen that End,( V) = V, 0 @ V,, where the 5 are R-submodules that are isomorphic to V (and correspond to the rows of the matrices). Thus all composition factors of End,( V) are isomorphic to C: and hence from the Jordan-Holder theorem the same is true of the submodule S. (ii) Let J be a proper ideal of R containing Ann(V). Choose I as a maximal right ideal of R containing J. Since R/Ann( V) N S, (i) shows that R/I = V as R-modules, and so Ann(R/I) = Ann( V). But R J E J G I, and so J c Ann(R/I). Hence J = Ann(V). This shows that R has no ideal lying properly between Rand Ann( V); and so S, which is isomorphic to R/Ann( V), is simple. (iii) Let U be a minimal right ideal in S. Then U is an R-composition factor of S, and so U = V as R-modules by (i). Thus, putting D o:= End,( V) and E := EndDo(V), we have Do N D and E N End,( V) as R-modules. We shall first prove that S = E. Consider U as an S-module, and let To denote the representation of S afforded by U . By the definition of D o , Im To c E. We claim that To(U) is a right ideal in E. To see this, consider the mappings lu E End,( U ) for u E U defined by ul, := uu (left multiplication). Clearly I,, E Doand so l , f = f l , for allfE E and u E U . Thus for all u, u E U a n d f c E we have

To(uf) E To(U)for all u E U a n d f c E. Since To(U)is closed Thus To(u)f= under subtraction, this shows that To(U)is a right ideal of E as claimed. Now SU = S by (ii) because SU is a nonzero ideal in S. Therefore Im To= To(S)To(U ) , and so Im Tois also a right ideal in E. Since Im To contains 1, Im To= E. Finally, since S is simple, Ker To= 0, and so S = Im To= E as asserted. Finally, it is clear that S E End,( V) by the definition of D.Since S 1: E N End,( V) as R-modules, comparison of the R-composition lengths shows that S = End,( V). This completes the proof of the lemma. We shall now prove the basic structure theorem for an algebra over a field. Recall that a right ideal is nil if each of its elements is nilpotent. Theorem 1.3A (Wedderburn structure theorem) Let R be an algebra over a field K. Put d := R/rad R. Then

(i) rad R is nilpotent and contains every nil right ideal of R. (ii) R = I, Q * - .@ I,, where 11, ..., I, are the minimal ideals of d.

10

I


(iii) For each j there is an irreducible d-module V, such that all Rcomposition factors of I, are isomorphic to 5, 0,*=End,( 5)is a division ring containing K in its center, and I, 2: End,JT) as K-algebras. Proof (i) Since R is an algebra it has an R-composition series, say, 3 R, = 0. For each i, R,- JR, is an irreducible R-module, R = R, 3 R, =I and so Ann(R,- JR,) =I rad R. Thus R,- ,(rad R) E R, for each i, and hence (rad R)’ G R(rad R)’ = 0. This shows that rad R is nilpotent. Conversely, suppose that x is an element of some nil right ideal J of R; we have to show that x E rad R. Let V be any irreducible R-module. We shall show that Vx = 0. Suppose on the contrary that Vx # 0 and choose u E V so that ux # 0. Since uxR is a nonzero submodule of and V is irreducible, uxR = K In particular, for some a E R, uxa = u ; and so by induction u(xa)” = u for all integers n 2 1. Since xu E J is nil, the latter implies that u = O contrary to the choice of u. Thus V x = O for each irreducible R-module K and so x E rad R as required. (ii) As we saw in the last section, each irreducible R-module is isomorphic to an R-composition factor of R. Therefore, by the Jordan-Holder theorem, there is a finite set V,, . . ., V, of nonisomorphic irreducible R-modules such that each irreducible R-module is isomorphic to one of them; and by the Ann(VJ. Let T, be the representation of R remark above, rad R = afforded by the module V,, and define the K-algebra homomorphism

n;-,

S

T: R + @ End,(

V,)

by

T ( a ):= (T,(a),. . . , T,(a)).

i=l

Since Ker 17; = Ann(V,), Ker T = rad R, and so R/rad R 2: Im T E @ S,, where S, I = Im T,. Since S, 2: R/Ann(v), it follows from S, Lemma 1.3(i) that the multiplicity of V, as a composition factor in S, S, is no greater than its multiplicity as a composition factor in R/rad R. Thus comparing R-composition lengths we conclude that Im T = S, @ @ S,, and so d 2: S , @ @ S, as K-algebras. Define I, as the inverse image ofS, under this latter isomorphism. From the direct decomposition it is clear that each I, is an ideal in R, and since S, is simple by Lemma 1.3(ii),I, is a minimal ideal. To complete the proof of (ii) it remains to show that d has no other minimal ideals. However, if I is a minimal ideal of 8, then I = Id c 11, + + 11,, and so I I , # 0 for some j. But 11, is an ideal contained in both I and I,; therefore the minimality of the latter ideals shows that I = II, = I,. Thus (ii) is proved. (iii) As we noted at the beginning of this section, the structure of the ‘.; as R-modules and as d-modules is the same. Thus (iii) follows a t once from the proof above together with Lemma 1.7. This completes the proof of the theorem. @ . a * @

1.3

SEMISIMPLE RINGS AND THE WEDDERBURN STRUCTURE THEOREM

11

Two R-modules U and V are said to be distinct if U 34 V. A set { V,, V, , . . ., V,} of irreducible R-modules is called a full set of irreducible modules if and 5 are distinct for i # j and if for any irreducible R-module V there is a

5 . If U and I/ are distinct, the associated representations are said to be inequivalent; if U ‘v r! then the associated representations are equivalent. Note I Under the hypothesis ofTheorem 1.3A each irreducible R-module V (considered as an R-module) is isomorphic to one and only one 5 ( j = 1, 2, . . ., s). [See the proof of part (ii).]

j (1 I j Is) such that V z

Definition Let K be a field and R an algebra over K. Then K is called a splirtingjeld for R if for each irreducible R-module End,( V) = K 1 [the reason for the term “ splitting ” field will become clearer in $2.71. In particular, a splitting field K of the group algebra KG is called a splitting field for G. Note 2 If V is an irreducible R-module, then by Schur’s lemma the K-algebra End,( V) is a division ring. Note 3 If, in Theorem 1.3A, K is a splitting field for R, then D, = K,and so I, ‘v End,( 5).Conversely, since each irreducible R-module is isomorphic to some V, (Note I), the conditions D, = K for all j imply that K is a splitting field. Note 4 If K is algebraically closed, then it is a splitting field for every K-algebra R. For suppose c E End,( V) for some irreducible R-module V. Since K is algebraically closed, the K-linear mapping c has an eigenvalue y E K. Then c - y * 1 is a singular K-linear mapping in End,( V). Since the latter is a division ring by Schur’s lemma, c - y . 1 = 0; hence c = y * 1. This proves that End,(V) = K . 1. (Burnside) Under the hypothesis of Theorem 1.3A suppose that K is a splitting field and V is an irreducible R-module. Then the image of R under the representation of R associated with V is all of End,( V). Corollary

Proof Immediate from Lemma 1.3(iii) since D = K in this case. A ring R is called semisimple if rad R = 0. If R is an algebra over a field K, it follows from Theorem 1.3A that R is semisimple iff R is a (finite) direct sum of simple rings. An R-module V is called completely reducible if it can be written as a direct sum of irreducible submodules.

+ +

Note 5 If an R-module V can be written as a sum V = V, V, ... + V, of irreducible R-submodules, then it can be written as a direct sum of some subset of { V,, V, , . . ., V,} and so is completely reducible. In fact, after suitable reordering we may suppose that U = V, @ V, 0 0 V, is a

12

I


+

direct sum for some n I m, but U is not a direct sum for any j > n. This means that U n V;. # 0 for eachj, and hence, since Vj is irreducible, U 2 5 for eachj. Thus U = V and we have V = V, @ V, @ . . . @ V,. Note 6 If R is a semisimple algebra over a field K, then R is completely reducible; in other words, R = @ =; I i is a direct sum of minimal (that is, irreducible) right ideals I i . By Theorem 1.3A, R = @ =; S i , where Si is a simple ideal of R. Since R is Artinian, Si contains a minimal right ideal I, say, of R. For any r in R, rl is either 0 or a nonzero right ideal of R which is isomorphic to I and hence irreducible. Since Si is an ideal, rl G S i . Moreover, because R is Noetherian there exist r l r . . ., I,E,, R such that J := rl I + * + rmI contains the right ideal rI for all r E R. Clearly J is a nonzero two-sided ideal of R contained in S i ; hence J = Si because Si is simple. So Siis a sum of irreducible R-submodules and thus is completely reducible by Note 5. Therefore R is completely reducible as an R-module.

Theorem 1 3 B Let R be an algebra over a field K. Then (i) R is semisimple iff every R-module is completely reducible; and (ii) (Maschke's theorem) the group algebra R = K G of a group G over K is semisimple iff the characteristic of K does not divide I G I.

Proof (i) Let R be semisimple and V an R-module. Then R = @ y= I i , where each I i is a minimal right ideal and V = u1 R + u2 R + * + un R for some u l , u 2 , . . ., on in V . Then n

V = CviR = i= 1

n

m

1 C viIj.

i=l j=1

But either ui I j = 0 or ui I, is isomorphic to I j . Hence V is a sum of irreducible R-modules. By Note 5, V is completely reducible. Conversely, if every R-module is completely reducible, then in particular R is completely reducible; let R = @ =; I i , where each I i is irreducible. If J i is the sum of all I j except I i , then J i is a maximal right ideal of R and = ; J i= 0. Hence rad R = 0, and so R is semisimple. (ii) Suppose that R = K G is semisimple. Then the element r := CxE x belongs to the center of K G and r z = r I G I. Thus if the characteristic of K divided I G 1, then r would be a nilpotent element in the center of KG, and hence r R would be a nonzero nilpotent ideal. Since rad R = 0 this is impossible by Theorem 1.3A. Hence the characteristic of K does not divide I G 1. Conversely, suppose K G is not semisimple and choose EX. a, x # 0 in rad R. Multiplying by a suitable element of KG, we may assume that c = 1 + Ex+y,x belongs to rad R. Consider the K-linear transformation t E End,(KG) defined by ur := uc (right multiplication by c). Since

n

1.4

13

TENSOR PRODUCTS

rad(KG), cn = 0 for some integer n 2 1 by part (i) of Theorem 1.3A. This implies that t" = 0, and so all eigenvalues o f t are 0. Therefore tr t = 0. But the matrix of r relative to the basis x (x E G) clearly has all its diagonal entries equal to 1, so tr t = I G I . 1. Thus 1 G I . 1 = 0 in K, which implies that the characteristic of K is a factor of I G I. This completes the proof of the theorem.

cE

EXERCISES

U 10 U 2 0 ... 0 U,, where each U iis a sum of isomorphic irreducible R-modules and U iand U j have no isomorphic irreducible constituent in common for i # j . Prove that End,( U ) 2 End,( U , ) 0 End,( U z )0 ... 0 End,( U , ) as rings. 2. Let G be a group generated by an element a of order 3, K a field of characteristic 3, and V a KG-module of dimension 2 over K. If {ul, u 2 } is a basis of V such that ( a l u l a2 u,)a = - a 2 u1 + (al - a 2 ) u z ,show that V is reducible, but not completely reducible. 1. Let U be a completely reducible R-module and let U

=

+

1.4 Tensor products

We shall assume a basic knowledge of tensor products of modules and algebras and their most elementary properties (references to this material are given in 41.10). In all cases in which we are interested, the modules and algebras with which we shall deal will be free over some underlying commutative ring or field, and in this case the tensor product construction is especially simple. The following examples are essentially the cases that we shall need. EXAMPLE 1 Let AG be the group algebra of a group G over a commutative ring A, and let U and I/ be AG-modules. Then V may also be considered as a left A-module, and so we have the tensor product U 0 A V as an Amodule. If U and V are free A-modules with A-bases ulr uz , . . .,u, and uI, u , , . . , , u,, respectively, then U 0 A V is a free A-module with an A-basis u i0 uj ( i = 1, 2, . . ., in, j = 1, 2, . . ., n) and an action of G on U 0 A V is defined in terms of its action on U and V by (Ui0 Uj)X := uix 0 u j x

for all i, j and x E G, and then extended to U @ A V and K G by linearity. The AG-module U 0 A V is called the (inner) tensor product of U and V. EXAMPLE 2 Let K be a subfield of the field L, and let V be a K-module (that is, a finite dimensional vector space over K). Then I/@, L is an L module. If u,, u 2 , . . ., u, form a K-basis of V, then I/@, L has o1 0 1, u z 0 1, . . ., u, 0 1 as an L-basis; in particular, dim, V = dim,( V 0 , L). If K G is the

1

14


group algebra of a group G over K, and V is a KG-module, then V @ , L is an U-module. The action of G on V @, L is given by (ui @ 1). := ui x @ 1 for all i and all x in G. EXAMPLE3 Let K be a field and A a subring of K such that K is the field of quotients of A. Let AG be the group algebra of a group G over A, and let V be an AG-module. Then' I @ K is a KG-module. If V is a free A-module with an A-basis u l , u 2 , . . ., u,, then V @ K has the K-basis u1 @ 1, u2 0 1, . . ., u, @ 1, and the action of G on V @ K is defined by (ui @ 1). := ui x @ 1 for all i and all x E G. In particular, rank, V = dim,( V @ K) in the case V is A-free.

,

,

,

,

Note In these last two examples L need not be finite dimensional over K, nor K finitely generated over A, so they need not be modules by our usual convention. EXAMPLE

4 Let A be a commutative ring and G a group with subgroup

H. Let {xl, x,, ..., x,,,} be a right transversal for H in G, so H x l , H x , , . .., H x , are the distinct right cosets of H in G. Since we may consider AH as a subring of AG in a natural way, AG can be considered as a (free) left AH-module. Therefore for any AH-module V we can define V @ AH AG, and the latter is an AG-module. If V is free as an A-module with an A-basis ul, u, , ..., u,, then V @ AH AG is free as an A-module with A-basis ui @ x j (i = 1, 2, .. ., n;j = 1,2, . . ., m). In this case the action of G on V 8 AG is given by (ui @ x j ) x = ui z @ x,, where x, is the coset representative defined by xjx E Hx, and z E H is defined by zx, I = x, x (i = 1, . . . n ; j = 1, 2, .. ., m ; and all x E G). The AG-module V @ AH AG is called the induced module. This example will be very important later when we deal with induced modules.

,!

EXAMPLE 5 Let A be a commutative ring and I an ideal of A. Then A / I is a two-sided A-module. Let G be a group and V an AG-module. Then V @ A ( A / ] )is defined and is an AG-module. The annihilator of this module contains the ideal I G consisting of all linear combinations of elements of G with coefficients in I . Since AG/IG z (A/I)G,the group ring over A/I, we can consider V @ ( A / I )as an (A/I)G-module. In particular, if V is A-free with an A-basis olr u,, ..., u,, then V @ , ( A / I ) is (A/I)-free with (A/I)-basis 01 @ 1, 0 2 8 1, ..., u n @ 1.

,

EXAMPLE 6 Let G be a group and K a field. Let U and V be KG-modules. The dual U* of U is defined as follows. As a K-module U * t= Horn,( U , K), the K-space of linear functionah on U ; in particular dim, U s = dim, U . The action of G on U* is defined by u(fx) I= (ux-')ffor all u E U , ~ U*, E and x E G. This is clearly a linear action and so defines a KG-module structure on U*. called the dual of U . Next we describe a K-isomorphism of U* @, V onto Hom,(U, V) as follows. Iff,,f, ,...,f, is a basis of U* and ulr

1.5

KG-MODULES

THE NUMBER OF IRREDUCIBLE

15

0 2 , . . . , u, is a basis of V over K , thenf, 0 uj (i = 1,2, . . ., m ;j = 1,2, . . . , n ) is a basis of U* 0 V over K . Define h i j : U + V by uhij := (uf;:)ujfor all i, j and u E U (note that uf;:E K ) . Then hij E Hom,(U, V ) , and the hij are linearly independent since if uij E K then uij hij = 0 implies

xi,

for all u

E

U , and so rn

m

0 = c u i j u f ; := u x u i j f , i=1

i =1

for all u

E

U and all j . But the latter implies m

Uijfj =0 i=1

for all j , and so all uij = 0. This shows that the hij are linearly independent and hence form a K-basis of Horn,( U , V ) since dim, Horn,( U , V ) = mn. Thus the mapping f;: 0 ujw h, ( i = 1, 2, .. ., m ; j = 1, 2, . . ., n) defines a K-linear mapping which is a K-isomorphism of U* 0 V onto Hom,(U, V ) . Finally, if we define the action of G on Horn,( U , V )by hc :=x- hij x, then it can be seen that the mapping above gives a KG-isomorphism of U* 0 V onto Horn,( U , V ) with this KG-module structure. EXERCISES

1. Let H be a subgroup of a group G, K a field, and V a KH-module. If Ker V = 1, prove that Ker(V@,,, K G ) = 1. 2. Let GI,Gz be finite groups and let K be a field. Then prove that K ( G , x G 2 ) ‘v K G , 0, K G z as K-algebras. 3. Let R be an algebra over a field K , and let Mat(& R ) be the algebra of n x n matrices over K . Prove that Mat(n, R ) 2: Mat(n, K ) O KR as K-algebras.

1.5 The number of irreducible KG-modules In one sense the theory of representations of a group over a field K is the classification of KG-modules to within isomorphism. In the case that K G is semisimple, every KG-module is a direct sum of irreducible KG-modules, and then the problem reduces to a classification of the irreducible KGmodules (see $1.3). If K G is not semisimple (and this is really the case in which we shall be interested) the situation is more complicated, but it is still

16

1


of basic interest to classify the irreducible KG-modules even though this is only the beginning of a solution of the general problem. In the present section we shall consider the irreducible KG-modules when K is a splitting field for KG. Lemma 15A (Binomial theorem) Let K be a field of characteristic p > 0 and let R be an algebra over K. Define S as the K-subspace of R spanned by the set of all ring commutators ab - ba (a,b E R) and write u = u (mod S) for u, v E R when u - u E S. Then

(i) ( x + y)P = x p + y p (mod S ) for all x , y E R; and (ii) ( x + y)P = x p + y p whenever x, y E R and x y = y x .

Proof Suppose 1 Ii Ip - 1. Then any two products consisting of i factors equal to x and p - i factors equal to y are congruent (mod S).There are (p) I=p ! / i !(p - i ) ! of these products, and since (7) = 0 (mod p), we conclude that the sum of these products (for fixed i ) is in s. Since this is true for each i, 1 I i I p - 1, it follows that ( x Y ) =~ x p y p (mod S). This proves (i), and (ii) then follows if we replace R by the commutative subalgebra generated by 1, x, and y , since then S becomes 0.

+

+

Lemma 15B Let

G be a group and K a splitting field for KG. Let S be the K-subspace of KG spanned by all ab - ba (a, b E KG).

(i) If K has characteristic p > 0, then T : ={c E K G I cPmE S for some integer m 2 I} is a K-subspace of KG, and the number of minimal ideals of KG/rad KG [see Theorem 1.3A(ii)] equals dim,(KG/T). (ii) If K has characteristic 0, then the number of minimal ideals of KG equals dim,(KG/S). Proof (i) First note that c E S implies cp" E S for all m 2 1. Indeed by Lemma lSA, (ab - ba)P = (ab)P - (ba)P= a{(ba)P-' b } - {(ba)P-'b}a = 0 (mod S). Hence an inductive argument using Lemma 1.5A shows that cp E S for all c E S. In particular, S E T. Next we see from Lemma 1.5A that ( a + b)P"= ( u p + bp)P"-' = = up" + bP" (mod S) for all m 2 1. Thus if a, b E T, then for sufficiently large m, ( a + b)p" = u p m + bP" E S; hence a + b E T. Clearly aa E T for all a E K and a E T ; therefore we conclude that T is a K-subspace of KG. Since rad(KG) is nilpotent [Theorem 1.3A(i)], it follows from the definition of T that rad(KG) G T. Now write R := KG and put I? := R/rad R and ?:= T/rad R; note that dim,(KG/T) = dim,(I?/p). In accordance with Theorem 1.3A we write k? = I, @ Z2 CD @ I , , where I , are minimal ideals of R. Since K is splitting for R, Zj % End,( Y), where 5 is a minimal right ideal of Zj (see Note 3 of $1.3). Hence Zj % Mat(dj, K ) for some integer d j dim, 5 2 1. 9 . .

I=

1.5 THE NUMBER OF IRREDUCIBLE KG-MODULES

17

s

Put = (S + rad R)/rad R. Then clearly 9 is the K-subspace of Z? spanned by all a6 - 6a (a, 6 E a) and = {c E a 1 cpmE for some integer m 2 l}. Let S,, q be the analogous K-subspaces defined for the K-algebra Zi . Since Z? is a direct sum of the Z i , it is easily seen that S, = I , n and q = I , n for each i, and as we saw above, S , E q. We claim that dim,&i/q) = 1. Indeed, since I, z Mat(di, K) we can carry out the calculations in the latter ring. Let e,, E Ii correspond to the d, x d, matrix with its (s, t)th entry equal to 1 and all the other entries 0. Then for any s # t, e,, = esjej, - ej,eSjE Si and e,, - e,, = e,, e,, - e,, e,, E S , . From this it follows that Sicontains the d? - 1 linearly independent elements e,, (s, t = 1, 2, ..., d, with s # t ) and ess - e l , ( s = 2, ..., d,). Since Si G Ti, this shows that dirnK(Zi/q)= dim&,) - dim,(l7;,) I d: - (df - 1) = 1. On the other hand, the elements of S , are linear combinations of elements of the form a6 - 6a and so correspond in Mat(d, , K )to elements of trace 0. Since e l , does not correspond to an element of trace 0, e l , # Si. However eP”11 - e l l for all m 2 1, so e l l # q . Therefore IT; # I , , and so from dirn,(Z,/TJ I 1 we conclude dim,(Zi/TJ = 1 as asserted. dimK(Zi/q)= r, and part (i) Finally, dim,(KG/T) = dim,(a/T) = of the lemma is proved. (ii) The proof is analogous to the last part of (i) taking Si in place of q .

c!=

Let x be an element of a group G and let p be a prime. Then x is called a p-element (or “p-singular”) if its order is a power of p , and x is called a p’-element (or “p-regular”) if its order is relatively prime to p (so 1 is the unique element which is both a p-element and a p’element). In general, suppose that x has order p“n ( p X n). Since p“ and n are relatively prime there exist integers r and s such that rp” + sn = 1. If we put xp:=XS” and x p ,:= X’P” , then we have (1)

x = x p x p ,= x p , x p,

where x p is a p-element and x p , is a p‘element. We call x p and x p ,the p-part and p’-part of x, respectively. Note The elements x p and xp, are completely determined by the conditions given in (1). Indeed, if y is a p-element and z is a p’-element such that x = yz = zy, then y and z commute with x and so commute with x p and x p , . Therefore xp ly is a p-element and xp,z - is a p’element. However from (1) these are equal and so must both equal 1. Hence y = x p and z = x p . .

Theorem 2.5 Let G be a group and K a splitting field for KG. Let r be the number of distinct irreducible KG-modules (up to isomorphism). Then

I

18


(i) r equals the number of conjugate classes of G if K has characteristic 0; (ii) r equals the number of conjugate classes of p'elements of G if K has characteristic p > 0. Proof Let S be the K-subspace of K G spanned by all ab - b a (a, b E KG). Then it is readily verified that S is also spanned by all xy - yx (x, y E G),and hence that an element u = CXEa,x (axE K )of KG lies in S if and only if ExEY a, = 0 for each conjugate class V of G. It follows from Note 1 of $1.3 that the number of irreducible KG-modules (up to isomorphism) is equal to the number of minimal ideals of KG/rad KG. Therefore with the notation of Lemma lSB, r = dim,(KG/S) when K has characteristic 0, and r = dim,(KG/T) when K has characteristic p > 0. (i) ( K has characteristic 0). Let Vl, V 2 , .. ., V, be the conjugate classes of G and let K' denote the vector space ofall t-tuples over K.Then there is a K-linear mapping 4 from K G into K'defined by

C axxw(Bli B 2 ,

P,)

where

C ax.

Pi!=

xsc

XSYi

Clearly 4 is surjective and its kernel is S by the observation at the beginning of this proof. Hence r = dim,(KG/S) = dim, K' = t as asserted. (ii) ( K has characteristic p > 0). Let a,, a,, . . ., a, be a K-basis of T. Choose the integer m so large that a;" E S for each i and p" is at least as large as the order of a Sylow p-subgroup of G. By Lemma 1.5A we have

(

P

ilyiui)

I

= C yf'"a?" = 0

(mod S)

i= 1

for all yi E K . Therefore ap"E S for all a E T. Now let Ex a, x by Lemma 1.5A we have

(C

(4

P"

a,.)

XOC

= 1 a!fxP"

E

KG. Then

(mod S).

XEC

Let V,, U 2 , ..., V, denote the conjugate classes of p'elements of G,and put V::={XE GIxP"E U,} ( i = 1, 2, ..., s). Then V:, V:, ..., V: is a partition of G and it follows from (2) that

c

xsc

11

\ D"

1

a,xE T

iff

a,).

x E Yi+

=

1 ar =O

XEY{+

zxoG

..., s (using Lemma 1.5A). Thus a x x E T iff for each i. Finally, define $ from KG into Ks as the K-linear mapping givenby C x E ca x X + + ( B 1 , B 2..., , PS)rwhere~i~=~xsYiZax.Then for i = 1, 2,

~ x 6 r ra,

=0

1.5

19

THE NUMBER OF IRREDUCIBLE KG-MODULES

t,h is clearly surjective and T is the kernel of t,h. Thus r

= dim,(KG/T) = dim,

K"

and so s = I as asserted. Corollary

Under the hypothesis of the theorem,

(i) if G is a p-group and K has characteristic p, then the only irreducible KG-module is the trivial module; (ii) if G is abelian, then each irreducible KG-module has dimension 1; (iii) if the characteristic of K does not divide I G I, then there are (up to isomorphism) exactly 1 G : G' I irreducible KG-modules of dimension 1 ( G denotes the commutator group of G). Proof (i) By the theorem I = 1. (ii) Let V be an irreducible KG-module. The image of G under the representation associated with V maps G into End,,( V )because G is commutative. Since K is splitting for KG, End,,( V) = K 1, and so each x E G acts as a scalar on V .Thus End,( V) 2: End,,( V) 'Y K 1,and so dim, V = 1. (iii) In the case where G is abelian (G' = l), G has I G I conjugate classes. Hence by the theorem there are IG I irreducible KG-modules, and then each has dimension 1 by (ii). In general, if V is a KG-module of dimension 1, then each x E G acts as a scalar on K Thus G/Ker V is an abelian group, and hence Ker V 2 G'. Conversely, since G/G' is an abelian group we have just seen that there are IG/G'l irreducible K(G/G')-modules of dimension 1. On each of these modules V we can define the action of G by ux := u(Gx) ( u E V , x E G, and G'x E G / G ) . This gives exactly J G: G I KG-modules of dimension 1. Since each KG-module of dimension 1 has G' in its kernel, it must be isomorphic to one of these. a

9

EXAMPLE Let K be a field of characteristic p > 0, and G a group. Let V be a KG-module such that each composition factor of V is trivial. Then G/Ker V is a p-group. Indeed, let V = V, 3 V, 3 3 V, = 0 be a composition series of V. By hypothesis 4-1 / < is trivial for i = 1, 2, ..., m ; that i s , u x + ~ = ~ + < f o r a l l u E V a, n- d~ x E G . Hence V , - l ( x - l ) ~ & f o r each i, and so V(x - 1)" C Vl(x - l)"'-' c c V, = 0 for each x E G.

Choose n so that p" 2 m. Since K has characteristic p , the binomial theorem (Lemma 1.5A) shows that (x - 1)p"= xp" - 1 for all x E G, and so V(xP" - 1) = V(x - 1)P" = 0. Hence for all u E V and all x E G, u(xP" - 1) = 0; that is xp" E Ker V . Thus G/Ker V is a p-group as asserted. EXERCISES

1. Let G be the group of order 10 generated by a, b satisfying the relations a2 = bS = ( ~ b=)1.~ Determine the number of distinct irreducible KG-

20

I


modules over a splitting field K of (i) characteristic 0, (ii) characteristic 2, (iii) characteristic 3, and (iv) characteristic 5. Also determine the onedimensional irreducible KG-modules, where K is a splitting field of characteristic 0. 2. Let G be the symmetric group on { 1, 2, 3, 4) and let Q be the rational field. Set a = (12), b = (23), and c = (34). Let U and V be QG-modules which afford the irreducible matrix representations T and S, respectively, with respect to some bases, where

T(a)=[

i :], 0 -1

T(b)=E

-; -I], 4

and 1

3

0

0

8

-2

-3

-8

S(a) = -1

0 -1

-1

-1

-4

Show that T and S are inequivalent. Determine other distinct irreducible matrix representations of G over Q. (It can be shown that Q is a splitting field for G.) 1.6 Indecomposable modules

Let R be any ring. Then an R-module U # 0 is called decomposable if it can be written as a direct sum U = I/ @ W of nonzero R-submodules V and W; otherwise U is called indecomposable.(The zero module is not classified as being either decomposable or indecomposable.) EXAMPLE Any irreducible R-module is indecomposable, but as we shall see later, the converse is not usually true. However, if R is a semisimple algebra over a field, then every R-module is completely reducible (Theorem 1.3B); therefore in this case an R-module is irreducible iff it is indecomposable.

A direct sum decomposition of an R-module U into a sum U = U , @

U, @ .-.@ U, of R-submodules may be described by n projection homomorphismsf;. from U to U iand n injection homomorphisms g ifrom U ito U

21

1.6 INDECOMPOSABLE MODULES

by requiring gifj = 0 for i # j, g i J = 1 on U ifor all i, and U. We observe thatff = J for all i andJ E End,(U).

C:=,

J g , = 1 on

Lemma 1.6A Let U , , U,, V,, and V, be R-modules, a be an isomorphism from U , to V,, and /?be an isomorphism from U , 0 U , to V, 0 V2such that u p E ua + V, for all u E U , . Then U , 2: V,. Proof Let f,and f, be the canonical projections from V, 0 V2 onto V, and V,, respectively. Define the R-homomorphism p: U , 0 U 2-,V, @ V, by p :=p( 1 -J, a- Ifif,). We first show that p is an isomorphism. Indeed, if u E U , then ubf, = ua by the condition on p. Therefore up = up upf, = up',, and hence U , p = V,. O n the other hand, since V,fl = 0, ( V 2 ~ - 1 ) p = V 2 ( 1 - f , a - ' ~ f 2 ) = VThus 2. V , @ V , ~ 1 m p and so p is surjective. On the other hand, if w E Ker p then w p = ( w p ) f ,a-'pf2. This shows first that wp E V , , and then that w p = 0 because V,f, = 0. Since p is invertible, Ker p = 0. This proves that p is also injective, and hence is an isomorphism as asserted. Finally, since U1p = V, from above, we have U2

(U,0 U,)/U,

2

( U , 0 U,)p(IU,p 'v ( 4 0 V,)IV,

V2

*

This proves the lemma. An element e in a ring R is called an idempotent if e # 0 and e2 = e. A ring R is called local if the sum of any two nonunits in R is a nonunit. Lemma 1.6B Let R be a ring. (i) R is local iff rad R is the unique maximal right ideal of R. (ii) If R is local, then 1 is the unique idempotent in R. Proof (i) First suppose that R is local and let 1 be a maximal right ideal in R. Let J be any proper right ideal in R. Since I and J consist of nonunits of R, 1 + J # R; and since 1 is maximal, this shows that J c I . Thus 1 is the only maximal right ideal of R. Since the intersection of the maximal right ideals of any ring is equal to the radical, rad R = 1 is the unique maximal right ideal. Conversely, suppose that rad R is the unique maximal right ideal of R. Since every nonunit of R generates a proper right ideal, rad R contains all the nonunits of R and therefore is precisely the set of nonunits. Thus R is local. (ii) It follows immediately from the definition that (in any ring) 1 is the only idempotent that is also a unit. Now suppose that e is a n idempotent in a local ring R. From the relation e + (1 - e) = 1 we deduce that either e or 1 - e is a unit. Since both e and 1 - e are idempotents, the relation e( 1 - e) = 0 shows that either e = 0 or 1 - e = 0. Since e is an idempotent, e # 0, and so e = 1. This completes the proof of the lemma.

22

I


We now have the following basic theorem on direct sum decomposition of R-modules. Let R be a ring and V be a

Theorem 2.6A (Krull-Schmidt-Azumaya) nonzero R-module such that m

V= @ i= 1

n

ui=@ q j= 1

for some submodules Ui and 5. Suppose that for each i and j the rings EndR(Ui) and EndR(6) are local. Then m = n and for some permutation i-n, of {I, 2, ..., n} we have U ,2: K, for each i. Proof We may assume that m 2 n, and use induction on m. For m = 1 the result is trivial, so suppose m > 1. Let& be the canonical projection from V onto Ui,f; the projection from V onto I(, gi the injection of Ui into V, and g: the injection of & into V. Then gifj = 0 for i # j , .gifi= 1 on U i , xy=lfig, = 1 on r! g;f; = O for i P j , g;f; = 1 on y , and C1=l f;g; = 1 on I/: Put ai :=gif; E Horn,( Ui, Vl) and Pi :=g;& E Horn,( V,, Vi). Then each Pi ai E End,( V1) and

Since EndR(V1) is local, at least one of the piai is a unit, and hence an automorphism of V,. Without loss in generality we may suppose that P1a1 is and g I = a l . Thenfg = 1 on V, and gf an automorphism; putfi= (PI al)- '/I1 is an idempotent Of h d R ( U l )since g(fg)f= gf# 0.Hence by Lemma 1.6B, gf= 1 on U,, and so g is an isomorphism from Ul onto V,. Now it follows from Lemma 1.6A (taking fl as the identity mapping on V and g for a ) that U,@*.*cD U r n = V , @ * . * @v,.

Finally, by the induction hypothesis, m - 1 = n - 1, and there is a permutation it+ n, of (2, 3, .. ., n} such that Ui 2 V,,for i = 2, ..., n. Since U , E V,, the theorem is proved. Corollary (Krull-Schmidt theorem) Let R be an algebra over a field K, and let V be a nonzero R-module. Then V can be written as a direct sum V = 437' CI,, where the U,are indecomposable submodules. Moreover, if V = $;=, V' is another decomposition of this kind, then n = m and (after possibly reordering the 5)we have I( E Ui for each i. Proof Since V is a finite dimensional vector space, it is clear that at least one such decomposition exists. Thus the corollary will follow from the

1.6

23

INDECOMPOSABLE MODULES

theorem once we have shown that End,(U) is a local ring for each indecomposable R-module U . Put E := End,(U). Let I be a maximal right ideal of E and letfe I. Sincef is a nonunit, Kerf # 0 by Lemma 1.2, and so Kerf" # 0 for all n 2 1. Since U is indecomposable, Lemma 1.2 now shows that Uf" = 0 for some n 2 1, and sofis nilpotent. On the other hand, suppose that g E E and g q! I. Then, since I is maximal, gE I = E, and so g h + f = 1 for some h E E. Since f" = 0 for some n 2 1,

+

gh{l + f + ... +f"-'}= (1 -f){l + f + *.. +f"-'} = 1 -f" = 1, and so g is a unit. Thus I is precisely the set of all nonunits of E, and so E is local. This proves the corollary. A set {el, e2 , . . ., en} of idempotents in a ring R is called orthogonal if O whenever i # j . An orthogonal set of idempotents such that 1 = el + e2 + ... en is called a complete orthogonal set. In this case it is easily seen that R = el R 0 e , R 0 . @ enR. Conversely, suppose R = R, 0 R, 0 @ R,, where each Ri is a nonzero right ideal. There are unique elements el, e , , . . ., en such that 1 = el e , ... e, and ei E R,. Since eiej =

+

+ + +

ei - e: = elei E

+ ... + e i - , e i + e i + l e i+ + e,ei * * *

Ri n ( R l O . . . O Ri - l @ R i + l @ . . . @ R , ) = ( 0 ) ,

it follows that e; = ei and eiej = 0 for i # j . Since Ri # 0, e, # 0 and {el, e 2 , . . ., en} is a set of orthogonal idempotents whose sum is 1. Thus there is a one-to-one correspondence between the decompositions of R as a direct sum of nonzero right ideals and the decompositions of 1 as a sum of orthogonal idempotents. If R = R , @ R, @ ... 0 R,, then R i = e i R for some idempotent e,. A summand Ri is indecomposable if and only if eicannot be written as a sum of orthogonal idempotents. An idempotent e is called primitive if it cannot be expressed as a sum of two orthogonal idempotents. Note If R is any ring and U is an R-module, then U is decomposable iff End,(U) possesses an idempotent e # 1. Indeed, if such an e exists then U = Ue @ U ( 1 - e) is a decomposition of U into a direct sum of nonzero R-submodules. Conversely, if U = U , 0 U 2 ,where U , and U 2 are nonzero R-submodules, then we can define e to be the canonical projection of U onto U,.It is readily seen that e is an R-endomorphism of U such that e2 = e # 0 or 1.

We shall find the following result on idempotents useful in the following section.

24

I


Theorem 2.6B Let K be a field of characteristic p > 0 and let R be an algebra over K. Let J be a nilpotent ideal in R. If e + J is an idempotent in R/J, then for some integer m > 0, e, :=ep" is an idempotent in R such that e + J = eo J .

+

+

Proof By hypothesis e" + J = (e J)" = e + J # 0 for all integers n 2 1. Therefore e2 - e E J and en # J for any n 2 1. Since K has characteristic p, the binomial theorem shows that (ep"')'- ep"= (e2 - e)P" = 0 for sufficiently large m because J is nilpotent. This shows that e,:=eP" is an idempotent in R, and from above, e, + J = e + J . EXERCISES

1. Let G be a group generated by an element a of order 2, K a field of characteristic 2, and Y a KG-module such that (a1u1 + ct2 u2)a= a l u1 (al+ a2)u2for a fixed basis { u l , u,} of V , Prove that Vis indecomposable but not irreducible. 2. Let R be a ring and U an Artinian and Noetherian R-module. Suppose U = U , @ U,QI ..-@ U,,where each U,is an indecomposable submodule. Then prove that any direct summand V # 0 of U is isomorphic to a direct sum of a subset of the modules U,,U,,..., U,. 3. Prove that an Artinian module U over a ring is a finite direct sum of indecomposable submodules. Is the converse true?

+

1.7 Absolutely indecomposable and absolutely irreducible modules Let R be an algebra over a field K.An R-module V is absolutely indecomposuble if for each field extension L of K, V B KL is indecomposable. Any absolutely indecomposable module is indecomposable, but the converse is not true. However we shall see below that if K is algebraically closed, then each indecomposable R-module is absolutely indecomposable.

Theorem 2.7A Let R be an algebra over a field K, and let V be an Rmodule. Put E I=End,( V). Then we have the following criteria. (i) V is indecomposable iff E/rad E is a division ring. (ii) V is absolutely indecomposable iff E/rad E = K. (iii) If K is algebraically closed and V is indecomposable, then V is absolutely indecomposable.

Remark It is easily seen that a ring is a division ring iff 0 is its only proper right ideal. Thus E/rad E is a division ring iff rad E is a maximal right ideal of E. Since rad E is the intersection of all maximal right ideals, Lemma 1.6B shows that E/rad E is a division ring iff E is local.

1.7

ABSOLUTELY INDECOMPOSABLE AND IRREDUCIBLE MODULES

25

Proof (i) If V is indecomposable, then the proof of the corollary to Theorem 1.6A shows that E is local, and so E/rad E is a division ring by the remark above. Conversely, if E/rad E is a division ring, then E is local, and so by Lemma 1.6B, 1 is the unique idempotent of E. In particular, 1 is a primitive idempotent, and so by $1.6, V is indecomposable. (ii) For each field extension L of K we define E,:=End,,(VBKL), where RL I= R B KL. Elementary properties of the tensor product show that EL = E B KL, rad EL = (rad E) @, L, and E,/rad E L 1: (Elrad E) 8, L (for a proof, work with fixed bases for the K-spaces). Thus, if E/rad E = K, then EJrad EL = K 6,L z L for each extension field L of K.Hence V is absolutely indecomposable by (i). Conversely, suppose that V is absolutely indecomposable; then (i) shows that D = E/rad E is a division ring with the property that D B , L is a division ring for each field extension L of K. We have to prove D = K. Suppose D # K, and choose a E D\K. Since D is a K-algebra, it is finite dimensional over K, and the powers 1, a, a2, ...are not linearly independent over K. Thus there exists a nonzero polynomialf(x) in K [ x ] with leading coefficient 1 such that f ( a ) = 0. Let L be a splitting field off@) over K and write f ( x ) = (x - a,)(x - a 2 ) (x - a,,,). Then in D B K L we have ( a @ 1 - 1 6a , ) ( a B 1 - 1 0 a 2 ) ( a 6 1 - 1 a,,,) = 0 with e . 1

a B 1 # 1 0 ai

for any i because a $ K. But this shows that D O KL contains nonzero divisors of 0 and so is not a division ring. Since this is contrary to the hypothesis on V we conclude that D = K, and (ii) is proved. (iii) Suppose V is not absolutely indecomposable. Then it follows from what we proved in (ii) that there exists an algebraic extension L of K for which V @ , L is decomposable. But K is algebraically closed, so L = K and V itself is decomposable. Let R be an algebra over a field K. An R-module V is called absolutely irreducible if for each field extension L of K, V B , L is irreducible. Any absolutely irreducible R-module is obviously irreducible, but the converse is not true. The following theorem clarifies the concept of splitting field introduced in 41.3. Theorem 1.7B Let G be a group and K a field. Then K is a splitting field for KG iff each irreducible KG-module is absolutely irreducible.

Proof First suppose that each irreducible KG-module is absolutely irreducible. Then for each irreducible KG-module V and each extension L of K, V O KL is irreducible. In particular, taking L as an algebraic closure of K,

26

I


L is a splitting field for LG, so End,,( V @ , L) = L * 1. On the other hand, if End,,(V), then c commutes with the action of G, and so c @ 1 E End,,( V @, L) = L * 1. This shows that c E K . 1. Since K . 1 E End,,( V), we conclude therefore that End,,( V) = K . 1. Hence K is a splitting field for G. Conversely, suppose that K is a splitting field for KG, and let V be an irreducible KG-module of dimension n. Let T be the representation of K G associated with V. Then by the corollary of Theorem 1.3A we know that T ( K G ) = End,(V) as a K-space. For any extension field L of K, let T, denote the representation of LG associated with V @ , L. Then TL(LG) = T ( K G ) @ , L , and so dim, T,(KG) = dim, T(KG) = n 2 ; hence T,(LG) equals all of End,( V @, L). Since Aut,( V @, L) acts transitively on V @, L, this shows that V @, L is an irreducible LG-module. Hence V is absolutely irreducible. CE

EXERCISES

1. Let G be the cyclic group of order 3, x a generator of G, Q the rational field, and V a QG-module of dimension 2. If { u l , u 2 } is a basis of V over Q such that (a1u1 + a2 u2)a = -(al + a2)ul + a l u 2 , show that V is irreducible but not absolutely irreducible. 2. Let G be the group generated by a, b that satisfy a3 = b3 = (ab)2= 1, let K be a field of 3 elements, and let V be a KG-module of dimension 3. If {ul, u 2 , u3} is a basis of V over K such that

(al u1

+ u2u2 + a 3 4 a = a1 ul + (a1 + a2)u2 + (a1 + a3)u3

(alu1

+ a2u2+ a3u3)b= (a1 + a2)ul + u2 + (al - a & ,

and show that V is absolutely indecomposable. 1.8 Principal indecomposable modules Let R be an algebra over a field K . Then the corollary to Theorem 1.6A shows that we can write R = Ul U, (1) as a direct sum of indecomposable R-submodules U , where the Ui are uniquely determined up to isomorphism and the order in which they appear. We shall call the summands in (1) the principal indecomposable R-modules. From 51.6 we know that there exists a complete set {el, . . ., en}of primitive @ * a * $

1.8

27

PRINCIPAL INDECOMPOSABLE MODULES

orthogonal idempotents such that U i= e, R for i = 1, . . ., n. Except when R is semisimple, the principal indecomposable R-modules form only a small subclass ofthe class ofall indecomposable R-modules. However, it is the class which plays a very important role in the theory of modular representations.

Theorem 1.8 Let R be an algebra over a field K. Then

(i) each principal indecomposable R-module U = eR (e an idempotent) has a unique maximal submodule U' = e rad R = U n rad R, and (ii) if U and V are two principal indecomposable R-modules with maximal submodules U' and V', respectively, then U = V iff U/U' = V / V ' . Proof (i) Since R is a K-algebra, U certainly contains at least one maximal submodule. Moreover it is easily verified that the submodules e rad R and U n rad R are equal. Thus to prove (i) it is enough to show that each proper submodule of U is contained in e rad R. Suppose that V is a submodule of U such that V $ e rad R. Then we shall show that U = V, from which it follows that e rad R is the unique maximal submodule of U. Assume that V # U . Then there is a maximal submodule W of U with V G W Since V G W $ e r a d R and W is maximal, U = W e rad R, and so e E Ue = We + e(rad R)e. Thus e = w h for some w E We and h E e(rad R)e. Note that eh = he = h and that h is nilpotent by Theorem 1.3A. Then for some n 2 1,

+

w(e

+ h + + h"-') = (e - h)(e + h + ... + A"-')

+

=e

- h"

= e.

This implies e E wR G W and so eR = W , a contradiction. This proves the first part. (ii) Clearly U 'v V implies U/U' z V / V .O n the other hand, suppose g is an R-isomorphism of U / U onto V / V ' . Then U H (w U')g is an Rhomomorphism of U onto V / V . Let e be the idempotent of R such that U = eR, and choose uo E V such that (e + U')g = uo + V.Since uo R s V, we can define an R-homomorphism f: U + V by uf:= oo u. Because g is an R-homomorphism, uo e + V' = uo + L", and since g is surjective, the latter is not 0. Thus the image offcontains uo e and the latter is not contained in V. Hence by (i), I m f = V . In particular, this shows that dim, U 2 dim, V . Now interchanging the roles of U and V we can show similarly that dim, V 2 dim, U . Hence U and V have the same dimension, and the surjectivity offthen implies thatfis bijective. Thusfis an R-isomorphism from U onto V. This completes the proof of the theorem.

+

Corollary Suppose that the decomposition (1) holds for R. Then there is a unique integer r such that (possibly after reordering the Ui) { U , , U , ,. . ., U,} is a full set of nonisomorphic principal indecomposable R-modules and { U 1/ U i , U 2/ U ; , .. ., U , / U : } is a full set of irreducible R-modules.

I

28


Proof From part (ii) of the theorem it is enough to show that each irreducible R-module W is isomorphic to U/U'for some principal indecomposable R-module U.Since W = WR, (1) shows that WU, # 0 for some i. Then choose w E W such that w U , # 0; the submodule wU, of W equals W because W is irreducible. Thus the mapping U H wu is an R-homomorphism of U ionto W, and since W is irreducible the kernel must be a maximal submodule of U , . Since U;is the unique maximal submodule of U,,this shows that U , / U ; W. EXERCISES

1. Let

G be the group of order 4 generated by a and b that satisfy

a2 = b2 = (ab)' = 1, K a field of characteristic 2, and V a KG-module of dimension 2. If {ul, u2} is a basis of V over K such that

(a1ul

+ a2u2)a= (a1+ a2)u1+ a2u2

and

(a1u1 + a2u2)b= (al

+ za2)u1+ a 2 u 2 ,

where z is a fixed element of K, then show that V is an indecomposable KG-module, which is not isomorphic to a principal indecomposable KG-module. 2. With the notation of Problem 1, show that (1 a)KG is an indecomposable but not an irreducible KG-module. 3. An R-module U is called projective if given R-modules V and W and a surjective homomorphism g E Horn,( V, W), then for any h E Horn,( U , W ) there is an f~ Hom,(U, V) such that h =fg. Assume that R is Artinian. Prove that a projective R-module is isomorphic to a direct sum of principal indecomposable R-modules. Hence a projective indecomposable R-module is isomorphic to a principal idecomposable R-module. 4. Let R = K G be the group algebra of a p-group G over a field K of characteristic p > 0. Prove that the regular R-module R is indecomposable and hence that R is the unique principal indecomposable R-module. 5. If R = KG is the group algebra of a group G over a field K,prove that every direct summand of R is projective and hence is a direct sum of principal indecomposable R-modules. 6. Let U be a principal indecomposable KG-module and U* the dual of U (see Example 6 of 01.4). Show that U* is isomorphic to a principal indecomposable KG-module. If U' # 0, show that the set of all elementsfof U*such thatf(U') = 0 is the unique minimal submodule of U*.Deduce that U has a unique minimal submodule isomorphic to UIU'.

+

1.9

COMPOSITION FACTORS AND INTERTWINING NUMBERS

29

7. If U is a principal indecomposable KG-module, prove that U z U* iff U/U'z ( U / U ) * .

1.9 Composition factors and intertwining numbers Let K G be the group algebra of a group G over a field K. Let V be a KG-module and suppose that V = Vo 3 V, 3 * * * 3 v, = 0 (1) is a composition series of V . Then by the Jordan-Holder theorem we know that the number of the irreducible constituents ( i = 1,2, .. ., I ) that are isomorphic to a given irreducible KG-module W is independent of the series (1). The theorems of the present section show how to calculate the multiplicities of the composition factors.

v- ,/v

Theorem 1.9A Let K be a splitting field for the group G, and let V be a KG-module. Let U be a principal indecomposable KG-module with maximal submodule U'. Then the multiplicity of U/U' as an irreducible constituent of V is equal to dim, HomKG(U,V). Proof By the corollary of Theorem 1.8 each irreducible constituent of V is isomorphic to U/U' for some principal indecomposable KG-module U . Suppose that (1) is a composition series for V ;we shall proceed by induction on the length 1. Since the result is trivial when 1 = 0, suppose that 12 1. To simplify notation we write Hom for HornKG. Let h: V + V / V , denote the canonical homomorphism and define +: Hom(U, V)+Hom(U, V/V,) by r$(f):=fh. Then 4 is a K-linear mapping such that f~ Ker iff U f c V, ; hence Ker 4 z Hom( U , V,) as vector spaces. On the other hand, 4 is surjective. Indeed, since U is a principal indecomposable module we can write U = e ( K G ) for some idempotent e. Suppose fo E Hom(U, V/V,) and efo = o + V,, say. Then since fo is a K G homomorphism, afo = (ea)fo = (efo)a = oa + V, for all a E e ( K G ) = U . If we definefE Hom(U, V ) by af:= oa; then +(f)=fo. This proves that is surjective. We next show that Hom(U, V / V , ) 'Y Hom(U/U', V / V , ) (as vector spaces). Since V/V, is irreducible, each g # 0 in Hom( U , V / V , )has Im g = V/V, and hence Ker g = U',since U' is the unique maximal submodule of U . Thus for all g E Hom( U , V/V,), Ker g 2 U',and so there is a natural isomorphism Hom(U, V / V , ) z Hom(U/U', V/V,). Finally, since K is a splitting field for G we have, using Schur's lemma,

+

+

I

30


Therefore, counting, we obtain dim, Hom( U , V) = dim, Hom( U , Vl)

=I

+ dim,

Hom(U, VIV,)

dim, Hom(U, Vl) + 1 dim, Hom(U, Vl)

if u/U'N V/V1 otherwise.

Induction now completes the proof of the theorem. Let U and V be any KG-modules. Then the intertwining number for U and V is defined to be i( U , V) := dim, Horn,,( U , V).

Definition

Note 1 s h e Horn,,( U @ V , W ) Horn,,( U , w)@ Horn,,( V, W),We have i( U @ V, W ) = i( U , W ) + i( V, W) for any KG-modules U , V, and W. Similarly i( U , V @ W ) = i( U , V ) + i( U , W ) . Note 2 Theorem 1.9A shows that when K is a splitting field and U is a principal indecomposable KG-module, then i( U , V) equals the multiplicity of U/U' as an irreducible constituent of V. Theorem 2.9B

Let K be a splitting field for the group C and let

KG = U , @ U , $ * * . @ U , (3) be a decomposition of K G into a direct sum of principal indecomposable KG-modules. Let U be any principal indecomposable KG-module. Then the number of U i isomorphic to U [in (3)] equals dim,(U/U').

Proof Suppose that U is isomorphic to m of the U i . Since Hom,,(KG, V) N K dim,(U/U') = i(KG, U / V ) = i ( U , @ U,, U/U') = i ( U l , U/U') + i ( U 2 , U / U ' ) + + i ( U , , U/U').By Theorem 1.8(ii) U j / U ; 'v U / U iff U., N U . Therefore by Theorem 1.9A, @ . . a

i ( U j , U / U ' )=

if U j l : U otherwise.

1 0

Thus dim,(U/U') = m as required. Let U , , ..., CJ, be a full set of nonisomorphic principal indecomposable KG-modules and let K be a splitting field for G. Put d., :=dim,( UjU'.,).Then r

Corollary

di dim, U i

[GI = i= 1

In particular, if K G is semisimple (see Theorem 1.3B), then each U i is irreducible. Hence in this case, U ; = 0 and r

i= 1

1.9

31

COMPOSITION FACTORS AND INTERTWINING NUMBERS

Later on we shall need two other facts about the intertwining numbers. We record these in the following result. Theorem 1.9C Let K be a splitting field for G.

(i) If K G is semisimple, then i( U , V) = i( V , U ) for any KG-modules U and V. (ii) For any idempotent e of KG, and any KG-module V, i(e(KG), V) = dim, Ve. Remark In general Ve need not be a KG-module.

Proof (i) Since K G is semisimple, by Theorem 1.3B every KG-module is completely reducible, and so we can write U = U , 0 U , 0 - . .0 U , and V = V, 0 V, 0 ... 0 V, as sums of irreducible modules. By Note 1 above m

n

and there is a similar expression for i( V, U ) . However, since K is a splitting field for G, Schur's lemma shows that i(U,, 5)= 0 if U , 6 and i( U , , 5 )= 1 if U , N 6. Therefore i( U , V) = i( V , U ) . (ii) Define 4: Hom,,(e(KG), V ) + V by 4(f):= e$ Then clearly 4 is a K-linear mapping; and Ker 4 = 0 since ef= 0 implies (ef)a = (ea)f= 0 for all a E KG. O n the other hand, Im 4 = Ve. Indeed, Im 4 c Ve because for allJ 4(f)= e2f= (ef)e E Ve, whilst Im 4 2 Ve because for all u E V we can definefe Hom,,(e(KG), V) by af:= ua and then $(f)= ef = ue.Hence 4 is injective with image Ve, and so

+

i(e(KG), V ) = dim, Hom,,(e(KG), V ) = dim, Ve. EXAMPLE Suppose that G is a p-group and that K is a splitting field of characteristic p for G; then the regular KG-module K G is indecomposable. (It follows from Theorem 2.7B that any field of characteristic p is a splitting field for every p-group.) Indeed, by the corollary of Theorem 1.5 the trivial KG-module is the only irreducible KG-module; so the corollary of Theorem 1.8 shows that (up to isomorphism) there is a unique principal indecomposable KG-module, say U , and U/U' is the trivial KG-module of dimension 1. Hence Theorem 1.9B shows that K G = U , and so K G is indecomposable.

EXERCISES

1. Let U be a KG-module. An invariant of U is an element u E U such that E G. Show that the set of all invariants of U is a KG-submodule.

ux = u for all x

I

32


2. Let U and V be KG-modules. Then show that i(U, V) is equal to the dimension of the space of invariants of U* B KV, where U*denotes the dual of u. 3. Let G be a group and K a splitting field for G. Show that a completely reducible KG-module is irreducible iff i(U, V )= 1. 4. Let G be a group, K a field, and U = e ( K G )a principal indecomposable KG-module generated by the idempotent e. Show that a KG-module V has a composition factor isomorphic to U/U'iff Ve # 0. 5. Let K be a field, U1,U 2, . .., U, a full set of principal indecomposable KG-modules, and U;the unique maximal submodule of U,.If ).

IG I = I =

[dimK(Ui/U:)]2, 1

prove that K G is semisimple. 1.10

Notes and comments

The concept of a group acting on a set is a generalization of the concept of permutation, and so a permutation group G on a set R acts naturally on R. The definition of a group algebra AG given in $1.1 can be extended to the group algebra RG of any (not necessarily finite) group G and any ring R.For ring theoretic properties ofRG see Passman [ 11. Other proofs of Wedderburn structure theorem (Theorem 1.3A) may be found in Lang [l], pp. 443-445, or Curtis and Reiner [l], &25-26. These two books also give detailed treatments of the tensor product construction. The proof of Theorem 1.5 is due to Brauer [8]. Theorem 1.5 has been generalized to the case where K is not necessarily a splitting field; the case where the characteristic of the field does not divide the order of the group is dealt with in Berman [l] and the case of modular representations over an arbitrary field in Reiner [l]. The principal indecomposable KG-modules form only a small subclass of the class of indecomposable KG-modules (when the characteristic of K divides I G I ). The determination of all indecomposable KG-modules is a difficult step, except when the Sylow p-subgroups of G are cyclic. For the proof of this result see $8.3, and for further theory of indecomposable modules see Green [l, 2, 31. Most of the results of $1.9 hold for a finite dimensional algebra over a field. For further results on intertwining numbers, see Curtis and Reiner [l].

CHAPTER

1

Induced Modules and Characters

This chapter describes the connection between the representations of a group and the representations of its subgroups. In the first two sections we shall prove two crucial results which have important roles in our work. The first of these results, due to Mackey, describes the behavior of the restriction of an induced module, while the second result, due to Clifford [13, describes the behavior of an irreducible module on restriction to a normal subgroup. As an immediate application of Clifford’s result, we shall prove a theorem of Blichfeldt on nilpotent groups. The third section is an introduction to the theory of group characters and the fourth describes the orthogonal property of ordinary characters. The next two sections deal with induced characters and Brauer’s induction theorem. In the seventh section we use Brauer’s induction theorem to show that every group has “small ” splitting fields. 2.1 Induced modules

Let A be a commutative ring. Let G be a group and H a subgroup. Then we can identify the group ring AH as a subring of AG. If V is an AG-module then we shall denote by VHthe AH-module obtained by the restriction of the ring; thus as an A-module VHequals V, but only action of H (and not all of G) is defined on ’V . This process will be called restriction and it permits us to go from any AG-module V to a (uniquely determined) AH-module VH. There is a dual process of induction. Let W be any AH-module. Then we 33

11

34

INDUCED MODULES AND CHARACTERS

saw in Example 4 of 81.4 that we can define an AG-module structure on the tensor product W @ AH AG. This is the induced module and we denote it by

w.

Note 1 If V and Ware A-free modules (in particular, if A is a field), then rank, V' = rank, V , but rank, w = J G: Hlrank, W (see 51.4). Note 2 If V = V, @ V, and W = W, @ W,, then it is easily seen that VH= ( V1),+ @ ( V2)" and = ( W,)" @ ( W,)'. Note 3 If two modules are isomorphic then restriction (respectively, induction) gives modules which are also isomorphic. Moreover induction and restriction are transitive in the following sense. If L is a subgroup of G with H c L c G, then

( VL)'

= V,

and

(W')" z W .

The latter relation follows from (W")" = (W@AHAL)@,LAG 'V W@,H ( A L @ , L A G ) z W@,,AG

= W".

The process of induction is related to the concept of imprimitivity of a module.

Definition Let A be a commutative ring and G a group. Then an AGmodule V is called imprimitiue if V can be written as a direct sum V = V, @ V, @ @ V, of A-submodules J( with m > 1 such that G acts on the set { V,, V,, ..., V,} by right multiplication. The set { V,, V,, . . ., V,} is called a system ofimprimitioity for V. If V cannot be written in this form, then V is primitiue. Note 4 If R = { V,, V,, . . ., V,} is a system of imprimitivity for K then the fact that G acts on this set means that for each x E G, &I+J ( x is a permutation of R. If V is irreducible, it is easily seen that fl forms a single orbit under G; so G acts transitively on R. EXAMPLE Let H be a proper subgroup of G and let W be an AH-module for some commutative ring A. Then WG cannot be primitive. For suppose = ( W @ y,) @ that y,, y,, . . ., yh is a right transversal of H in G. Then ( W @ y,) @ . @ ( W @ y h ) (as A-modules). Moreover, it is clear that W @ y,, ..., W @ y, is a system of imprimitivity for WC and the action is transitive. Indeed h = I G : HI > 1, and for each i and each x E G we have (W @I yi)x = W @ y j , where y j is defined by H y j = H y , x.

This example gives a method of constructing imprimitive AG-modules. The following lemma shows that essentially all imprimitive AG-modules can be constructed in this way.

2.1

35

INDUCED MODULES

Lemma 2.1 Let A be a commutative ring and G a group. Let V be an imprimitive AG-module with { V,, V,, . . ., V}, as a system of imprimitivity and suppose that G acts transitively on {V,, V', ..., V,}. Define H *={x E G 1 V, x = V,} as the stabilizer of V, under the action of G. Then V, is an AH-module and V 'v V y . Proof Since G acts transitively on { V,, V,, . . ., V,} and H = Stab( V,), it follows from $1.1 that rn = G : H I. Thus we can choose a right transversal of H in G, say y, = 1, y,, . . ., y, such that = V, yi ( i = 1 , 2 , . . ., m).Since G = (V1 o Y , ) o (V1o ~ 2 O) * . * O (V1 O Y,) and

I

V = V,y,O V,.Y~O"'O V1Ym

as A-modules, we have an A-isomorphism 8 of m

onto V given by

m

where u,, u , , . . ., u, E V,. However, for each x E G and each i there exist uniquely determined z E H and j such that yi x = z y j ; and (ui @I y i ) x = ui z @ y j and ( ui yi)x = (ui z ) y j for ui E V,. Thus 8 is an AG-isomorphism and the lemma is proved. A basic relation between the process of induction and restriction is given by the following theorem.

Theorem 2.IA (Mackey) Let H and L be subgroups of the group G and let {x,, x, ,. . .,x,,} be a set of double coset representatives for (H, L) in G. Let A be a commutative ring and let V be an AH-module. For each x E G, define V, to be the A ( x - ' H x n L)-module V @ x with the action ( u @ x ) y : = U X ~ X - ~ @ for X all U E Vand y e x - ' H x n L. Then (v")L 'v @;=, (Vxi)".

Proof Let y,, y,, ..., y, be a right transversal of H in G. Then V C = ( V O y , ) O ( V O y , ) ~ . . . O ( V O y , ) a sA-modules.Put

Then G, and in particular L, acts on R. Moreover V @ y i and V B y , lie in the same orbit under L iff yi and y, lie in the same double (H, L)-coset. Let W;. denote the sum of the V @ y j for which y j E H x , L. Then each W;. is an AL-module and (v")" = W, 0 W, 0 O W,. It remains to show that each W;. = (VJ". Consider W,, say, and suppose W, = ( V O y,) 0 u H y , = H x , L. 8 ( V @ yl), where by definition of W,, Hy, u H y , u Without loss of generality we may assume that the y j are chosen so that one of them, say y,, equals x,. Now L acts transitively on the set { V O y,,

11

36 V 8 y,, is


..., V @ y l } and under this action the stabilizer of V @ y ,

I

{z E L V @ x , z = V @ x l }= (z E

LI

= V@ x1

x 1 z x ; ' E H}= x ; ' H x , n L.

Hence by the definition of VXi and Lemma 2.1 we conclude that = ( Vx,)" for each i, and the theorem is proved.

W,z (V',)". Similarly M(

Our next result also relates the induction and restriction processes. Theorem 2.2B Let A be a commutative ring, G a group, and H a subgroup of G. Let U be an AH-module and Van AG-module, and suppose that U and V are both free as A-modules (in particular, the latter always holds if A is a field). Then (U@ VH)% U G @ V .

Proof Let u l , u 2 , ..., u, and u,, u,, ..., u, be A-bases of U and V, respectively. Let x l , x 2 , ..., xh be a right transversal of H in G. Then ( u i @ u,)@ xl ( i = 1, 2, ..., m ; j = 1, 2, ..., n;and I = 1,2, ..., h) isan A-basis of ( U 8V,)" and ( u i @ xi)@ u j x I ( i= 1,2, ..., m ; j = 1,2, ..., n;and I = 42, . . ., h) is an A-basis of UG8 V. Hence (1) (ui 8 0,) 8 X I - (ui 6 X I ) ujxl defines an A-isomorphism of (U@ V,)' onto UG€3 V; it remains to check that this is an AG-isomorphism. To do this we must show that it respects the action of G.If y E G, then for each 1 we have uniquely defined z E H and k such that x , y = zxk, since x , , x 2 , .. ., x,, is a right transversal of H.But then ((ui 8 V j ) @ xl)y = (ui @ u,)z 8 x k = ( u i z

8 V j z ) @ xk

*

On the other hand ((ui 8 X I ) 8 ujxI)y = (ui @ xl)y @ ujxIy = ( u i z

8xk) @ (ujz)xk

*

This shows that the A-isomorphism defined by (1) is in fact an AGisomorphism, and so the result is proved. EXERCISES

1. Prove that in Theorem 2.1A, ( VX)"N ( V,)" when x and y lie in the same double (H,L)coset. 2. Suppose that H and L are subgroups of the group G such that G = HL and H n L = 1. Let K be a field and let V be a KL-module. Put n :=dim, !I Show that (v"), is isomorphic to the direct sum of n copies of KH (as a KH-module). 3. Let G be a group and K a splitting field for G. If K has characteristic 0, deduce from Theorem 2.1A that the number of distinct irreducible representations of G is equal to the number of conjugate classes of G.

2.2

CLIFFORD'S THEOREM

37

4. Let A be a commutative ring, H a subgroup of the group G, and V an AH-module. Define an action of G on Hom,,(AG, V) by af := (ay- ')ffor all a E A G , ~ EHom,,(AG, V), and y E G. Show that under this action the A-module Hom,,(AG, V) becomes an AG-module which is isomorphic to

VG.

2.2

Clifford's theorem

Let K be a field and let G be a group with H as a normal subgroup. In this situation we can say more about the restriction-induction process. The key result is Clifford's theorem. In the case where H 4G we say that two KH-modules V, and V, are conjugate (under G) with respect to some y E G if there is a K-isomorphism V H uY of V, onto V, such that for all x E H and u E V we have ( U X ) ~= uYy-'xy (note that y- ' x y E H because H Q G). Clearly if V, and V, are KH-modules that are conjugate under G, then V, is irreducible (respectively, indecomposable) iff v, has this property. EXAMPLE Let H be a normal subgroup of G and suppose that V is a KG-module. Let U be a KH-submodule of V,. Then for each y E G, U y is also a KH-module and U is conjugate to U y with respect to y.

Theorem 2 2 4 (Cliflord) Let H be a normal subgroup of the group G and let K be a field. Let V be an irreducible KG-module. Then

(i) V, is completely reducible and its irreducible components are all conjugate (under G); (ii) there exists a subgroup S of G such that H E S c G and for some irreducible KS-module W we have V N W ; (iii) each irreducible component of V, occurs with the same multiplicity, and there are exactly I G : S I nonisomorphic irreducible constituents. Proof (i) Let U be an irreducible submodule of V,. Put R :={ U y I y E G}. Then by the example above each element of R is a KHmodule which is conjugate under G to U and hence also irreducible. The sum of the modules in R is clearly a KG-module and so equals V by the irreducibility of V. Thus V, is a sum of irreducible KH-modules U y (y E G); hence by Note 5 of $1.3, V, is completely reducible and can be written (1)

V, = uy, a3 u y 2 a 3 * * * 0U y ,

for some yi E G with y , = 1. This proves (i). (ii) Let U1, U,, ..., U , denote the distinct (up to isomorphism) irreducible constituents of V, , and define (i = 1, 2, .. ., m) as the sum of all U y

38

11 INDUCED MODULES AND CHARACTERS

(y E G) such that U y 'Y U,.Then W; is a KH-submodule of V, which is a direct sum of certain Uy ; in particular, every irreducible constituent of is isomorphic to U,. Thus by the Jordan-Holder theorem, n W - 0 for each j since 4 and y. have no isomorphic jT irreducible constituent in common. Therefore from (1) we conclude that

w)

xiti

V, = w,(33 W2(33**.@ w,. (2) On the other hand it is clear that G acts on the set { W,, W,, . . ., Wm}. This action is transitive because V is irreducible. If we define S to be the stabilizer of W, under the latter action, then I G : S = m, and it follows from Lemma 2.1 that W, is a KS-module and V z ( W,)". Finally W, is irreducible because V is (see Note 2 of 52.1). Hence taking W = W,, (ii) follows. each (iii) Let d = dim, V. Since G acts transitively on { W,, W2,. .., Wm}, W; has the same dimension, and so by (ii) we have dim, Pi( = d/m. On the other hand, by (i) each irreducible constituent of V has dimension equal to dim, U.Therefore by (ii) the multiplicity of U ,as an irreducible constituent in V, is dim w./dim U i = d/m dim U , which is independent of i. As we noted above, m = I G : S 1, and so (iii) is proved.

I

w

Remark The KH-modules defined in part (ii) of the proof above are called the homogeneous components of V, . Corollary

Under the hypothesis of the theorem

(i) if V, has at least two nonisomorphic irreducible components, then V is imprimitive; (ii) if V, has at least one trivial irreducible component, then H c Ker V. Proof (i) It follows from (iii) that S # G. Hence G is imprimitive by the example of 52.1. (ii) If U is a trivial KH-submodule of V,, then its conjugates under G are all trivial. Hence by part (i) of the theorem, all irreducible components of the completely reducible KH-module V, are trivial. This means H acts trivially on V, and so H E Ker V.

If a KG-module V has dimension 1, then its structure is very simple (see for example, the corollary of Theorem 1.5). The situation is only a little more complicated when V can be written as an induced module V z V , where W is a KH-module of dimension 1 for some subgroup H (see Exercise 3). In the latter case we say that V is monomial; in particular, every KG-module of dimension 1 is monomial. The following classical theorem of Blichfeldt shows that for a nilpotent group and a splitting field, the irreducible KGmodules are all monomial.

2.3

39

GROUP CHARACTERS

Theorem 2.2B (Blichfeldt) Let G be a nilpotent group and K a field such that K is a splitting field for each subgroup of G.Then every irreducible KG-module is monomial. Proof We shall proceed by induction on I G 1 ; the result is certainly true if G = 1. Let V be an irreducible KG-module. If dim, V = 1, then V is monomial, so we may assume that dim, V > 1. Put G = G/Ker V . Then V is also an irreducible KG-module, and so by the corollary of Theorem 1.5, G is not abelian because dimk V > 1. This means that the center 2 of G is a proper subgroup; choose j j E G\Zso that zjj lies in the center of G/Z.(Note that the center of G/Zis nontrivial because G is nilpotent.) Let fi be the subgroup of G generated by and j ; then I?/Z -a G/Zand It? is abelian because 2 centralizes j . Take H as the full inverse image of I? under the canonical homomorphism G + G. The subgroup H is normal in G since f i / Z -a G/Z and HIKer V 2: fi is abelian. Choose y E G mapping onto J . Since j # 2, the action of y on V is not scalar. Now by the corollary of Theorem 1.5, the irreducible constituents of V, all have dimension 1. Since the action of y is not scalar, and y E H,therefore not all irreducible constituents of V, are isomorphic. Hence by Theorem 2.2A, there exists a proper subgroup S of G and an irreducible KS-module W such that V = w .By the induction assumption, W is monomial (since I S I < 1 G I ). Hence V is monomial by Note 3 of $2.1. EXERCISES

1. Let G be the dihedral group of order 8 and let C be the complex field. Determine a full set of irreducible CG-modules (use Theorem 2.2B). 2. Let G be a group with a faithful KG-module V of dimension n over any field K. Suppose that V is monomial. Show that G has a normal abelian subgroup H such that G/H is isomorphic to a subgroup of the symmetric group Sym(n) on n symbols. Moreover, if V is irreducible, then show that the' latter group is transitive. 3. Let K be a field, G a group, and V a completely reducible KG-module. Use Theorem 2.2A to show that V, is a completely reducible KH-module whenever H is a subnormal subgroup of G. (Recall that H is subnormal in G if it appears in some composition series for G.)

2.3 Group characters Let G be a group and K a field. Then to each KG-module V we have the associated representation T: G + Aut,( V) defined by uT(x):= ux for all u E V and x E G. Since T ( x )is a linear transformation of the vector space V , we can define its trace tr T(x). The function (: G + K given by ((x) := tr T ( x )is called the (Frobenius) character of G afforded by V.

40

11

INDUCED MODULES A N D CHARACTERS

EXAMPLE The trivial KG-module V is a one-dimensional vector space such that ux = u for all u E V and x E G.The character afforded by V is denoted by 1, and is called the principal character; we have I&) = 1 for all x E G.

Note 1 A character C is a classfunction on G;that is, its value is constant on any conjugate class W of G.Indeed, for all x , y E G we have ( ( y - ' x y ) = tr T ( y - l x y )= tr{T(y-')T(x)T(y)} = tr T ( x )= ( ( x ) by standard properties of the trace. Note 2 If U and V are isomorphic KG-modules and afford the characters cu and Cv, respectively, then C,, = Cv. For suppose that f: U V is a KG-isomorphism, and let Tu and Tv be the representations of G associated with U and V, respectively. Then for all u E U,x E G we have (uf)x = ( u x ) ~ ; which gives (uf)T,(x)= (uTu(x))f.Hence fT,(x)f-' = Tu(x)for all x E G. This shows that c,(x) = tr Tu(x)= tr{fTv(x)f- I} = tr Tv(x)= c,(x) for all x E G;hence cu = Cv. Note 3 Let V be a KG-module and U a proper nonzero submodule; so V 3 U 3 0. If Cl and Cz are characters afforded by U and V/U,respectively, then V affords the character C2. [Proof Take a K-basis u,, u z , ..., u, of V such that u,, u z , ..., urnis a basis for U and urn+ U,.. ., u, U is a basis for V / V . Calculate the traces relative to these bases.]

cl +

+

+

Note 4 If U and V are KG-modules which afford the characters Cv and

cv,respectively; then U 8 V (see 51.4) affords the character laCv (the product), [Proof Let ul, u,, ..., urnand ul, u,, ..., u, be K-bases of U and V, respectively; then ui@ u, ( i = 1, 2, ..., m ; j = I, 2, ..., n) is a K-basis of U @ V. Calculate the traces relative to these bases.]

Let G be a group and K a field. Then every representation T : G Aut,( V ) of G can be extended to a unique representation T : K G -+ End,( V) of the K-algebra K G (we denote the latter representation also by T ) .The annihilator Ann V of V is then equal to {a E KG I T(a)= 0). Given any representation T of G (or KG)associated with V (also said to be aflorded by V) and any fixed K-basis u,, u z , ..., u, of V, there is a corresponding matrix representation T*: G -+ Gyn,K),where n I=dim, V. The relation between T and T* is given by T*(x)I=[ar,(x)],where ui T(x)= ui x = ai,(x)u, for i = 1, 2, . . ., n. The following classical theorem is due to Burnside, Frobenius, and Schur.

E=,

Theorem 2.3

Let G be a group and K a splitting field for KG. Let V,, V,,

..., V , be a full set of nonisomorphic irreducible KG-modules, and let T:, Tf , ..., T,* be matrix representations of G afforded by these modules. Suppose

v(x)[ai,(x)] for each x E G and put dl s=

s=

dim, V,. Then the set of

2.3

41

GROUP CHARACTERS

functions az: G + K (i, j = 1, ..., d,; I = 1, 2, over K. Remark The functions a t (i, j = 1, 2, functions of the matrix representation T:.

..., s) is linearly independent

. . ,, d,)

are called the coordinate

Proof We shall use the notation of Theorem 1.3A with R = KG. The note following that theorem shows that we may take our 5 to be the irreducible KG-modules referred to in the theorem. Furthermore, it was also shown there that Ann V, contains each a E K G such that a + rad K G E Ii for some i # I; in other words, KG/rad K G = (I, + Ann &)/rad KG. On the other hand it follows from the corollary of Theorem 1.3A that the image of K G under the representation afforded by V, is the whole of End,( because K is a splitting field. Interpreting this in terms of the matrix representation shows that the image of KG under T: is Mat(d,, K). Thus for any rn, n = 1, 2, ..., d, we can choose a E KG such that af,,,(a) = 1 and aij(a)= 0 when (i, j) # (m,n). Moreover, we may suppose that a + rad K G E I, since as we saw above, KG/rad K G = (I, + Ann y)/rad KG. But when a + rad KG E I , , then from above a E Ann for each t # I, and so T:(a) = [a:-i(a)] = 0. Thus we have shown that for each function a:,, there exists a E KG such that a f ( a ) = 1 but a;j(a)= 0 for all ( t , i, j) # (I, m, n). This clearly implies that a,1 (m,n = 1, 2, ..., d,; I = 1, 2, ..., s) are linearly independent, and the theorem is proved.

v)

el,

With the hypothesis and notation of the theorem, let c2, . . ., [, be the characters afforded by the irreducible KG-modules V,, V, , . . ., V , , Then these characters are linearly independent (as functions from G into K). In particular, two irreducible KG-modules are isomorphic iff they afford the same character. Corollary I

Proof

Since dr

:=

C ali i= 1

(I = 1, 2, . .., s),

the first assertion follows at once from the theorem. Since every irreducible KG-module is isomorphic to one (and only one) of the 5, the second assertion now follows using Note 2. Corollary 2 Let G be a group and K a field of characteristic 0 such that K is a splitting field for G. Then two KG-modules are isomorphic iff they afford the same character.

Proof Let U and V be two KG-modules and suppose they afford the characters lo and respectively. Then using the notation of the theorem, let mj (respectively, nj) be the multiplicity of vj as an irreducible constituent

cv,

11

42


of U (respectively, V ) for i = 1, 2, ..., s. Since K has characteristic 0, Maschke's theorem (Theorem 1.3B) shows that both U and V are completely reducible. Therefore U 'Y V iff mj = nj for j = 1,2,..., s. On the other hand, if cj is the character afforded by 6, then using Note 3 it follows that

c a

cv =

c I

mjcj

j= 1

and

cv =

njcj.

J= 1

cV

Thus by Corollary 1, ca = iff mj = nj for j = 1,2, . . ., s (here we are using the fact that K has characteristic 0). Hence we conclude U 5 V iff = iv.

cv

Let G be a group and K a field. A classfunction from G into K is defined to be any function 8: G + K with the property that 8 is constant on each conjugate class V of G. [Equivalently, 8(y- 'xy) = 8(x) for all x, y E G.]We shall denote the set of all class functions of G into K by Class,(G). It is readily verified that Class,(G) is a K-algebra under the operations defined by (el 8,)(x) I=8,(x) + e,(x), (8, e,)(x) := el(X)e,(x) and (ae,)(x) := a(O,(x)) for all x E G,a E K, and el, 8, E Class,(G). In particular, if G has t conjugate classes, then it is evident that dim, Class,(G) = t ; for example, we could choose as a K-basis the t functions el, 8,, . . .,O,, where Oi is 1 on the ith conjugate class and 0 on all other conjugate classes. The conjugate classes of G also come up in what at first appears to be a different context. Consider the group algebra KG, and let Z ( K G )denote the a x x (a, E K) lies in Z ( K G ) iff it commutes center of KG. Then a = with all elements of KG; and since {y I y E G} is a K-basis of KG, this is equivalent to ay = ya for all y E G. Comparing coefficients, this gives CXpG a,x E Z ( K G ) iff ax = ay-lxyfor all x, y E G.Let Wl, W,, ...,V, be the conjugate classes of G and define the ith class sum in K G as ci I= Ex %, x for i = 1, 2, ..., t. Then cl, c , , . . ., c, are linearly independent over K (since the elements of G are); and from what we have just shown, cl, c,, ..., c, is a a,x lies in Z ( K G ) iff the K-basis of Z ( K G ) . It is now easy to see that function XH a, lies in Class,(G). Since cl, c,, ..., c, is a K-basis for Z ( K G ) , there exists yijl (i, j, 1 = 1, 2, . . ., t) in K such that

+

xxsG

cxeG

I

cicj =

C yijlcl I= 1

for

i, j = 1, 2, ..., t.

Rewriting these equations in terms of the basis {x I x E G} of KG, we see that the yijl are nonnegative integer multiples of the unity 1 of K. Moreover, if wj = {x-' otherwise.

Ix E Wi}

2.4

THE THEORY OF ORDINARY CHARACTERS

43

EXERCISES

1. Let G be a finite group, K an arbitrary field of characteristic 0, T',

T2,

. .., T" a set of pairwise inequivalent irreducible matrix representations of G over K, di the degree of Ti,and atl, the coordinate functions of T'.Prove that {a:" I u, u = 1, 2, . . ., d,; i = 1, 2, . . ., n} is a linearly independent set. 2. Let G be a group gnd K an arbitrary field of characteristic 0. Prove that the irreducible characters of G over K are linearly independent.

2.4 The theory of ordinary characters Throughout this section we shall consider a group G of order g and a field

K where K is a splitting field of characteristic 0 for KG. Under these hypotheses we shall call the characters and representations of G over K the ordinary characters and representations. There is an extensive theory of ordinary characters, but here we shall just touch on a few basic results which we shall be using later. Since K has characteristic 0, all KG-modules are completely reducible (Theorem 1.3B); and since K is also a splitting field, the number of nonisomorphic irreducible KG-modules is equal to the number of conjugate classes of G (Theorem 1.5). By Corollary 1 ofTheorem 2.3 this latter number is also the number of distinct characters afforded by irreducible KG-modules; we shall denote this common number by s. It follows from Corollary 2 of Theorem 2.3 that two KG-modules are isomorphic iff they have the same character. Therefore (in this situation of ordinary characters) we can refer unambiguously to a character [ of G as being reducible or irreducible depending whether it is afforded by a reducible or irreducible KG-module, and define the irreducible constituents of [ as the characters of the irreducible constituents of a KG-module that affords [. Similarly, the kernel of [ (written Ker [) is defined as the kernel of any KG-module that affords [, and the degree of [ (written deg [) is the dimension of this module. Let V be a KG-module that affords [ and let T be the representation of G afforded by V. Each element x E G has order dividing g := 1 G I, and so T ( x )=~1. Let o be a primitive gth root of unity over K. Then X e - 1 splits into linear factors in the field extension K ( w )of K, and its roots are precisely 1, o,wz, . . . , d - Each eigenvalue of T ( x )is then one of the roots of unity. Without loss of generality we may assume that the field Q of rationals is contained in K (since K has characteristic 0) and identify Q ( w ) with the corresponding subfield of the complex field. In particular, the values of 5 all

'.

11 INDUCED MODULES A N D CHARACTERS

44

.

lie in Q(w). Moreover, if T ( x ) has the eigenvalues w", d = deg c :=dim, V), then we have (1)

Il(x)l = Jwil+ w'2+ *.. +wid( I la'' I

. wl2,

..., wid(where

+ Iw'q + + *-.

)wid1

=d

with equality iff all mi#are equal. Since K has characteristic 0, it follows from T ( x Y = 1 that the minimal polynomial for T ( x )has distinct roots. Thus if all the eigenvalues 0'1 of T ( x )are equal, then the minimal polynomial has the single root wit and T ( x ) is scalar. Thus from (1) we conclude (2)

lC(x)l I c(l) for all x E G and

Ic(x)l = ((1)

iff T ( x ) is a scalar.

In particular, x E Ker ( iff T ( x ) = 1, so from (2) we have (3)

Ker [ = { x E G I c ( x ) = c( 1)).

Next we define an inner product ( , ) on the K-algebra Class,(G) of class functions by

Clearly ( , ) is bilinear and symmetric. (Note that this definition does not require K to be a splitting field for KG.) A set of functions O,, O,, . .., O n in Class,(G) is called orthogonal if (O,, O j ) = 0 for all i # j, and orthonormal if it is orthogonal and (Oi, ei) = 1 for each i. Since G has s conjugate classes, dim, Class,(G) = s (see 82.3). On the other hand, we also know that G has s distinct irreducible characters and each of these lies in Class,(G). Since these characters are linearly independent by Corollary 1 of Theorem 2.3, they must form a K-basis for Class,(G). In the next theorem we shall prove an even stronger result. To standardize the notation let Wl = (11, V,, .. ., Ws be the conjugate classes of G and let ci I = y p i x be the corresponding class sums in KG. For each class Wi, the set {x- 1 x E Wi) is also a conjugate class which we denote by Wi.. Put h := 1%" I and for each character [ of G define Ci as the value c takes on the class Wi ( i = 1, 2, . . ., s). We shall denote the s irreducible characters of G over K by c', c2, . . ., r" (using upper indices). As usual 6, denotes the Kronecker delta (equal to 1 if i = j and 0 otherwise).

qxe

EXAMPLE Consider the KG-module K G and let p be the character that it affords ( p is called the regular character of G).Since K G is semisimple by Maschke's theorem (Theorem 1.3B), it follows from Theorem 1.3A that K G = I, Q3 I , CD ... CD I,, where the irreducible constituents of I, are all isomorphic to a single irreducible KG-module 5. By Note 3 of $1.3, I, 1: End,( 5)as K-algebras because K is a splitting field. Hence, counting dimensions, we see that dim, I, = d j , where d,:=dim, VJ, and so 5 occurs with multiplicity dj as an irreducible constituent of I,. By the note following

2.4

45


Theorem 1.3A, the characters afforded by the C; are precisely the distinct irreducible characters of G, so we may suppose that C; affords c j . Thus we conclude p

=

S

S

j= 1

j= 1

C d j c j = C cj(l)[j.

On the other hand, using the basis {y I y E G} for KG to calculate traces, we find at once that p ( 1) = g and p ( x ) = 0 for all x # 1 in G. Theorem 2.4A

(Orthogonality relations) With the above notation

and (2) for i, j = 1, 2, . . ., s. In particular, the characters normal basis of Class,(G).

c', cz, .. .,cS form an ortho-

Proof For each character 5' choose a KG-module U i which affords ci and let be the representation afforded by U i . Since cj lies in the center Z ( K G ) of KG, T(cj)E End,,( Ui). But K is a splitting field and so the latter equals K 1. Hence for each i and j there exist wi E K such that T ( c j )= wj * 1. Taking traces then shows that = %, ['(x) = tr T ( c j )= wi . ti(l), and

-

cx.

so (3)

oj = h j c j / c i ( l )

for all i, j .

On the other hand it follows from (1) of $2.3 that for each i and j and each I we have 5

T,(ci)T,(cj)=

1

Yijm

T,(cm)-

m= 1

Therefore using (3) we obtain S

(4)

Now consider the regular character p of G (see the example above). Since C;= c'( l)(', (4) shows that

p =

46


because p1 = g and pi = 0 for i # 1. Finally from (2) of 42.3 we conclude yijl = h, a,, . Replacing j by j* (and j* by j) this proves (1). Now the equations in (1) can be written as a matrix identity in the form Y Z = I, where Y and Z are s-by-s matrices over K whose (i, j)th entries are h i ( { and (&/g, respectively. Hence Y = Z - I , and so ZY = I. Using the fact that (I*)* = 1 and h, = h l z , the latter matrix identity is equivalent to Eqs. (2). Finally (2) can be written in the form

1 9

c

c

l S h1(;(;*= aij XEG 9 1=1 and this shows that (', (', ..., rsare orthonormal. Since Class,(C) has dimension s, they form an orthonormal basis for this space. ((i,

(')= -

( i ( x ) ( j ( x - 1)

=-

I;=

Corollary If ( is any character of G, then we can write ( = rn, ( I , where the nonnegative integer rn, is the multiplicity of in (. Since ( I , lz,. . ., ts are m:. orthonormal, this shows that m, = ((, (') for each I, and ((, () = In particular, ( is irreducible iff ((, () = 1.

('

I;=

Since K has characteristic 0 its prime subfield is isomorphic to Q, so without loss of generality we may suppose that K contains the field Q of rational numbers and hence the ring Z of integers. Consider the ring Z[X] of polynomials over Z; a polynomialf(X) E Z[X] is called monic if its leading coefficient is 1. An element a E K is called an algebraic integer if it is a root of some monic polynomial f(X) E Z[X]. Note 1 It is well known (and easily proved), that a E K is an algebraic integer iff the subring Z[a] is a (finitely generated) Z-module. It follows immediately that if a and fi are algebraic integers of K,then a k fi and as are also algebraic integers of K.Thus the set Int(K) of all algebraic integers of K is a subring of K containing Z. Moreover, if m and n are nonzero relatively prime integers, f(X) I= Xd + a t Xd-' + + ad E Z[X] and f(m/n) = 0, then md/n = -al m d - l - a 2md-2n - - adnd- E Z and so n = k 1. This shows that Q n Int(K) = Z. [Compare with $3.2.1 Theorem 2RB With the notation above and h, (f/t;'( 1) are algebraic integers in K for all i, 1 = 1, 2, .. .,s; and (i) (f (ii) ('(1) divides g.

Proof (i) As we saw earlier, (fis a sum of 8th roots of unity. Since each E Int(K). 8th root of unity is clearly an algebraic integer, this shows that (f Now fix i and let C be the s x s matrix whose (j, m)th entry is yi,, . As we

2.4

47


noted in 82.3 the yijm are nonnegative integers in K. The equation (4) above can be written

which shows that hi[:/['(l) is an eigenvalue of the matrix C. Since C has integer entries, the characteristic polynomial det(X . 1 - C) is a monic polynomial in Z[X], and so (i) is proved. (ii) Theorem 2.4A shows that for any 1 we have g/", = i:*(hil:/['(l)). But and hic:/['(l) both lie in Int(K) by (i), and so g/", E Q n Int(K) = Z. Hence "(1) divides g.

x;=,

rfz

EXERCISES

1. Let G and H be two groups and let K be a splitting field of characteristic 0 for both K G and KH. Suppose 5 is a character of G over K and let x be a character of H over K. Show that @ E Class,(G x H) defined by t,b((x,y)) := c ( x ) x ( y ) for all (x,y) E G x H is a character of G x H over K and prove that it is irreducible iff and x are both irreducible. 2. Let G be a group and K a field of characteristic 0 which is splitting for G. If Z ( G ) is the center of G, prove that the degree of an irreducible representaon the set tion of G over K divides I G: Z ( G )1. [Hinr: Define the relation of conjugate classes of G by V i Vj iff V i= Vjz for some z E Z ( G ) . Show that if xi # 0, then the --class containing Vi has I Z ( G ) I members.] 3. Let G be a finite group, K a splitting field for G of characteristic 0, a character of G, n the number of conjugate classes of G, and iithe value of C at the ith conjugate class. Prove that Cl. C2, . . ., inare the eigenvalues of a matrix with nonnegative integer coefficients. [Hint: Consider aij[j, where i', i2, .. ., (" are the irreducible characters of G.] CiC = 4. In Exercise 3, prove that il,c2, . . ., i nare the eigenvalues of a permutation matrix iff il = 1. 5. Let 8 be an ordinary character of G over a subfield of the complex numbers. Show that e(x- ') equals the complex conjugate of O(x). If 8 is irreducible and x is an involution, show that O(x) is an integer and O(x) = O( l)(mod 2). 6. Let x be a faithful ordinary character of a group G. If x takes only rational values show the following: (i) the values ofx are all rational integers ) n ( n : = x ( l ) ) ; and (ii) g:= [ G I divides (2n)!. [Hint: Put with - n I ~ ( x I 0= !.,, (x - j ) and calculate ( 0 , Q . ]

-

nj.;

-

48

11


2.5 Induced characters

Let G be a group and let K be any field. Suppose that H is a subgroup of G and V is a KH-module. Consider the induced KG-module P. If ul, u 2 , ..., u,, is a K-basis of V,and x l , x 2 , ..., xh is a right transversal of H in G,then u1@I x, ( l = 1, 2, ..., d ;j = 1, 2, . .., h) is a K-basis for P. By definition of P we have (ul @I xl)yr= u l ( x i y x i’) @I x, for each y E G, where x, is the coset respectively, representative such that x,yx; E H.Suppose that V and P, afford the matrix representations S and T with respect to the given bases. Then for each y E G, T ( y )can be written as the “block” matrix

’

T(Y)=

S(x1yx;’)

S(x,yx;’)

-*.

S(x1yxh1)

S(x,yx;’)

S(x,yx;’)

-..

S(x,yxhl)

S(xhyx; ’)

S(xhYx; ’)

’* *

S(x,yx, ’)

where for all z E G we define the d x d matrix S ( z ) by S(Z)

=

(,,,

if Z E H otherwise.

Thus we find that the characters x and c afforded by V and VG,respectively, are related by h

(1)

C(Y) =

c

h Y x ;

i= 1

where if Z E H if z E G\H.

i(.)

I=

The character c afforded by P and given in terms of y, by (1) is called the character induced from x, and we shall use the notation xG for C. It is clear from the definition of [ that the value of C given by (1) is independent of the choice of the right transversal x l , x,, ..., xh of H in G. We now specialize to the case where K has characteristic 0. In this case, since G can be written as a disjoint union of I H I right transversals of H in G, we have from (1) that 1 f(y) =-

1 i(x-lyx)

XEG

for all y

E

G.

2.5

49

INDUCED CHARACTERS

More generally, for any class function 8 E Class&) 8': G + K by

we shall define

where &z) :=e(z) if z E H and 0 otherwise. Clearly 8' E Class,(G), and from above 8' is a character of G whenever 8 is a character of H. The induction mapping Ow 8' is a function from Class,(H) into Class,(G). There is also the restriction mapping 4t+ 4Hfrom Class,(G) into Class,(H) (where 4H denotes the restriction of the function 4 to H). The basic properties of these mappings are given in the following theorem. Theorem 2 5 A Let K be a field of characteristic 0. Let G be a group and H and S subgroups such that H E S G G. Then for all 0 E Class,(H) and 4 E Class,(G) we have (i) (e")" = OC (ii) eG4= (64,)" (iii) (Frobenius reciprocity)

(e, 4 H ) H

=

(ec, 4)G .

Proof (i) As x ranges over S and z ranges over G, the product xz ranges over G taking each value I S I times. Therefore from (2) we have for all y E G 1 1 ( @ ) G ( y )= -- C ISIIHI xsG

c

xss

1

~(z-~x-~~ = x__ z)

C J(w-lyW) = eG(y).

WEC

(ii) (Compare with Theorem 2.1B) Put $ := &#+, E Class#). all x, y E G we have

Then for

ii(x-'yx)f$(y) = J ( x - ' y x ) ~ ( x - ' y x ) = J(x- ' y x )

because 4 is a class function on G. Therefore from (2) we get ( W ) ( Y ) = ec(Y)4(Y) = $"b)

and so 0'4 = $' = (&$,)", as asserted. (iii) Again we have for all x, y E G, & ( y - ' ) = 4(x-'y- 'x). On the other hand, x- 'yx ranges over G as y ranges over G.Therefore using (2) we have

50


Under the hypothesis of the theorem suppose that K is a splitting field for both H and G, and let x and [ be irreducible (ordinary)characters of H and G, respectively. Then the multiplicity of x as an irreducible constituent of CH is ( x , [,) (corollary of Theorem 2.4A) and the multiplicity of [ as an irreducible constituent of xG is (xG, [). These multiplicities are equal by part (iii) of the theorem. Corollary

EXAMPLE Let H be a subgroup of the group G and let 1, denote the principal character of H.Let xl, x2,. . ., x,,be a right transversal of H in G. Then

c i,(xiyx; h

lE(y) =

1).

i= 1

Since !,(x~Yx; ') = 1 if H x i y = H x i and 0 otherwise, this shows that l$(y) equals the number of cosets H x of H in G such that H x y = Hx. Equivalently, if we consider the action of G on the set R a= {dx 1 X E G} of right cosets of H in G by right multiplication, then lg(y) equals the number of points fixed by y. In particular, y E Ker :1 iffy E x- ' H x for all x E G ; therefore Ker :1 = x-'Hx, which is the largest normal subgroup of G contained in H.

nxaG

By Theorem 2.5A(iii), (1; , lc) = (I, ),1 = 1 and so l Gis an irreducible constituent of 1; with multiplicity 1. In general, [ is an irreducible constituent of 1; iff ,1 is an irreducible constituent of [". In the case where H 4 G, the corollary of Theorem 2.2B shows that (l, [,) # 0 iff H E Ker [. Hence in this case 1

1; =

1mi['

with m i= ti(l),

i= 1

where the sum is over all irreducible characters ci of G with H c Ker ['. (Compare with the example of $2.4, noting that p = lG,where 1 is the principal character of the subgroup 1). The use of the induction process is an important method of obtaining characters of a group G from the characters of a subgroup H.One problem which arises is that the induced character xG is not usually an irreducible character of G even when x is an irreducible character of H,and it is difficult to separate out the irreducible constituents. The following theorem due to Brauer and Suzuki [ 13 is therefore of importance. It shows how we can give a partial analysis of the induced characters under suitable restrictions. We shall only need this theorem in Chapter 8. To state the theorem we shall require the concept of a trivial intersection set. A subset S of a group G is called a trivial intersection set if (a) 1 $ S ; (b) S c N G ( S )and x-'Sx n S = 0 for all x E G\NG(S).

2.5

51

INDUCED CHARACTERS

Theorem 25B Let G be a group and let S be a trivial intersection set in G; put H := N G ( S ) .Let K be a field of characteristic 0 such that K is a splitting field for both H and G. (i) If 0 , 4 E Class,(H) that are 0 on H\S, then Oc(y) = e ( y ) and 4"(y) = 4 ( y ) for all y E S, e"(1) = 4G( 1) = 0, and (0, 4 ) H= (OG, 4G)G. (ii) Let x', x2, . . ., x" (n 2 2) be distinct irreducible characters of H over K such that for each i, f ( y ) = 0 for all y # 1 in H\S, and ~ ' ( 1 )= ~ ~ ( =1 ) . . = x"( 1). Then there exists E = 1 or - 1 and distinct irreducible characters (',C2, . . ., ("of G over K such that (xi - x j ) " = &(ci - ( j ) for i, j = 1, 2, .. ., n. Proof (i) Since S is a trivial intersection set and 8 is a class function, we have for all x E G and y E S

Therefore it follows from (2) that @ ( y ) = 8 ( y ) ; and similarly, @(y) = 4 ( y ) for all y E S. Again from the hypothesis on S, the number of distinct conjugates of S in G is I G : H 1 = I G: N G ( S )I, and any two distinct conjugates of S are disjoint. It follows from (2) and the hypothesis on 0 and 4 that e"(y) = 4'(y) = 0 for each y E G not conjugate to an element of S. Put So := { y E G I y E S and y - E S}. Then using the fact that 8' and 4' are class functions we obtain

'

(ii) First consider the case where n = 2, and put satisfies the hypothesis of (i), and so

(eG,eG)"= (e, e ) H

= <x'

e:= 2' - x2. Then 0

- x2, x' - x2)H = 2

by the orthogonality relations (corollary of Theorem 2.4A). By the same corollary it is clear that the relation (eG, d G ) G = 2 implies that for some k Since distinct irreducible characters (', of G we have 0' = eG(1) = 0 by (i), we have 8' = E((' - C2) with E = 1 or - 1. This proves (ii) in the case n = 2. In the case n 2 3 put O i : = x' - xi (i = 3, 4, ..., n). Then Oi satisfies the hypothesis of (i) in place of 4 and so we have

+[' c2.

c2

1 = (2' - x2,

x1 - xi)H = (e, e i ) H

= (eG,Of),

and

2 = (x' - xi, 2' - x i ) H = ( e i

1

6i)H =

(@,Of),

52

11


The second relation shows that 87 has exactly two irreducible constituents each with multiplicity f 1. Then the condition 8p(l) = 0 and the first relation show that either 8: = &(elor 87 = &(ci for some irreducible then character ci different from c' or c2. Now consider 8:. If8: = we interchange c' and in our notation; this also changes E to - 8 . As a result 8: becomes &(jl Thus we may choose the notation so that 8; = &(elThen from the relation

c2)

c2

c3).

1 = ( x ' - x3,

c2)

&(c3 c2),

c3).

x1 - xi)H = ( O , , 8 i ) H

=

(@,OF),

for all i 2 4, we see that 8y = &(el - ci) for all Oi (including 8, := 8 and dl :=O). Moreover, it follows from the relation 1 = (Op, 8;) for all i > j > 1 that 5' # c j when i # j . Finally, for all i, j = 1, 2, . . ., n we have W - x j) C - ( oj - ,gi)G = & ( e l - c j ) - & ( e l - ci) = & ( [ i - c j ) and the theorem is proved. EXERCISES

1. Let H and L be subgroups of the group G. Let V be a CG-module (C the complex field) which affords a character [. Put V, := { u E V 1 ux = v for all x E G}. Show that V, is a CG-module with dim, V, = (c, 1,). Deduce that if H and L together generate G, then

+ ( h 1),

( C H 1 ~ +) ( C L 1J. 2. Let H be a normal subgroup of G and x an irreducible ordinary charac. ter of G. Prove that there exists an irreducible character i,h of H such that x = i,h" iff x vanishes outside H and xH is a sum of distinct irreducible characters of H.

(CH

nL

3

1~ n

J

2

9

3

2.6 Brauer's theorem on induced characters Throughout the present section we shall be dealing with a group G and a field K of characteristic 0 such that K is a splitting field for KH for each . . .,cSdenote the set of all irreducible characters subgroup H of G. Let c', of G over K. Then a class function 8 E Class,(G) is called a generalized character of G if it can be written in the form 8 = m i l i for some integers mi. The set of all generalized characters is denoted by Char(G).

c2,

c;=

c2,

cs

Note 1 Since e l , ..., are linearly independent, Char(G) is a free Z-module of rank s. Moreover by Note 4 of 52.3 the product of two characters is again a character, and so it follows easily that Char(G) is a subring of Class,(G). It will be shown in Theorem 2.7A that the ordinary characters e l , . . ., are essentially independent of K.

c2,

cs

2.6

53

BRAUER'S THEOREM ON INDUCED CHARACTERS

If H is a subgroup of G, and 8 E Char(H), then clearly Oc E Char(G). The first of the important theorems of this section shows that Char(G) is generated as a Z-module by characters induced from an especially simple class of subgroups (the "elementary" subgroups). The second theorem is a dual; it characterizes the elements of Char(G) as those class functions whose restrictions to the elementary subgroups are always generalized characters. Definition A subgroup E of a group G is called elemenfury (or more explicitly q-elementary) if for some prime q it can be Written as a direct product E = Q x C, where Q is a q-subgroup and C a cyclic q'-group.

We shall call a subgroup H of G quasi-elementaryif for some prime q it can be written in the form H = QC, where Q is a q-subgroup and C a normal cyclic q'-subgroup of H ; briefly, H has a normal cyclic q-complement. Clearly subgroups of elementary subgroups are elementary and subgroups of quasi-elementary subgroups are quasi-elementary. Note 2 If R denotes the Z-module spanned by all characters of the form :1 with H quasi-elementary, then R is a subring of Char(G). Indeed it is clear that R is closed under addition. To see that it is also closed under multiplication, proceed as follows. If H and L are quasi-elementary and 8 :=:1 , then 1gO = (8,)' by Theorem 2.5A. Since 8 is a permutation character, OH = ( lf), = lii,where m is the number of orbits and H ithe stabilizer of a member of the ith orbit (compare with the example of 52.5). Since each H , E H , we have lglf = (8,)" = 1gi E R.

zy=

x'''l

Lemma 2.6 With the notation above the principal character '1

E

R.

Proof We first show that for each x E G and each prime q there exists a quasi-elementary subgroup H of G such that q X lg(x). Indeed, take C as the q-complement of (x) and put N : = N , ( C ) . We choose H/C as a Sylow q-subgroup of N / C ; clearly H is quasi-elementary. By the example of52.5 we know that lz(x) equals the number of right cosets H y in G such that H y x = H y . But H y x = H y implies (yxy- ') E H , which in turn implies y E N because C is the unique q-complement of H . Therefore lg(x) = 1;(x). Since C is normal in N and contained in H , C lies in the kernel of 1;. Since Cx is a q-element in N / C , the nontrivial orbits in the set of right cosets of H in N all have lengths which are multiples of q. Therefore the number of orbits of length 1 equals l;(x) = IN : HI f 0 (mod q). Now consider the set I, := {c$(x)1 c$ E R } for x E G . Clearly I , is an ideal in Z ; and from what we have just proved, I , $ qZ for any prime q. Therefore I , = Z. This means that for each x E G we can choose 4, E R such that c$x(x)= 1. Then the generalized character (9, - ),1 is 0.Expanding this product, we can write 1, as a Z-linear combination of products of 4,'s. Hence by Note 2 above, 1, E R.

nxeG


54

Theorem 2.6A (Brauer's theorem on induced characters) The ring Char(G) of generalized characters of G is generated as a Z-module by characters of the form A,, where 1 is a character of degree 1 of an elementary subgroup of G. Hence each character of G is a Z-linear combination of induced characters of this form.

Proof We first note that it is enough to prove that we can write m

for suitable integers mi,where the 4i are characters of some elementary subgroups Ei (i = 1, 2, . . ., m). Indeed, if 1, is of this form and x is any character of G, then m

x

=x

*

1, =

c

mi(XEi4JG

i= 1

by Theorem 2.5A(ii). Since each elementary subgroup is nilpotent, Theorem 2.2B shows that xE,4,is of the form 1;' for a character I, of degree 1 of some subgroup of E,. Since subgroups of elementary subgroups are elementary, x = xy= m, 1; has the required form. We shall proceed by induction on I G I. Since (1) is trivially true if G is elementary, we may suppose that G is not elementary. Case 1 G is quasi-elementary. Then for some prime 4, G = QC,where C is a normal cyclic 4'-subgroup and Q a q-group. Put H = N,(Q). Note that if M is a normal subgroup of G, then H E M implies M = G. Indeed, M contains all Sylow q-subgroups of G and so by the Sylow theorems, counting the number of Sylow q-subgroups of G gives I M : H I = I G : H I ; hence M = G. Also H # G because G is assumed to be not elementary. By the example of l2.5, lGis an irreducible constituent of 1: with multiplicity 1; so we can write 1: = 1, + mi xi,where xi are the irreducible constituents of 1; different from 1,. If we can show that each xi has the form where 4i is the character of some proper subgroup of G, then 1, = 1; - ~ , m , 4and ~, so (1) will follow from Theorem 2.5A(i) and the induction hypothesis. First note that C 4 G. It follows from the definition of 1; that

ci

Since G = CH, I G : H 1 = I C : C n HI,and so = (IH abelian and H n C 4C, and so by the example of $2.5, n

c)'.

Now C is

2.6

55

BRAUER’S THEOREM O N I N D U C E D CHARACTERS

where 11, I,, ..., In are the irreducible characters of C with H n C E Ker Aj . Moreover I,( 1) = 1 for each i, and n = I C : H n C 1 by the corollary of Theorem 1.5. Now let xi be an irreducible constituent of 1; different from 1,. The Frobenius reciprocity theorem (Theorem 2.5A) shows that ( l H , = (l;, x i ) 2 1. If xi(l)= 1, this would show that (xi), = l,, and SO Ker H. As we noted above, G is the only normal subgroup of G containing H; so Ker xi = G and xi = l,, contrary to the choice of xi. Hence xi( 1) > 1. On the other hand, since ( = ( 1, c)c = 1. I , , where the I , are distinct characters of degree 1, (xi)c is a sum of distinct characters of degree 1. Therefore by the corollary of Theorem 2.2A, any module affording xi is imprimitive. Hence xi = 4: for some character 4, of a proper subgroup of G. As we saw above, this proves the theorem in the case where G is quasi-elementary.

xi?

z=

Case 2 G is not quasi-elementary. In this case the theorem follows immediately from Lemma 2.6 and the induction hypothesis. This completes the proof of the theorem. Theorem 2.6B (Characterization of characters) Let 8 E Class(G) for the group G.Then (i) 8 E Char(G) if (and only if) 8, E Char(E) for every elementary subgroup E of G; (ii) 8 is an irreducible character of G if (and only if) 8 E Char(G), 8(l) > 0, and (8, 8) = 1.

Proof The proof of the “only i f ” statements is trivial so we consider only the direct statements. (i) Using Theorem 2.6A we write 1, = C; n

8 = 81, = C mi(84p) =

mi4:

as in (1). Then

n

C m,(8,ic$i)G i=1

1

by Theorem 2.5A. Since Char&) is a ring, it follows from the hypothesis that for each i, 8Ei4iE Char(E,); hence 8 E Char(G). n i x i , where the n, are (ii) Since 8 E Char(G), we can write 8 = integers and the xi are the distinct irreducible characters of G.The condition (8, 8) = 1 shows that cj= nz = 1 since ( x i , x j ) = 6,; hence for some j , nj = k 1 and n, = 0 for all i # j. Thus 8 = + x j and the condition 8(l) > 0 shows that 8 = x j .

cJI=l

EXERCISES

1. Let K = Q(w), where w is a primitive 1 G 1 th root of unity, and let Int(K) be the ring of integers of K. Assume that K is a splitting field for G. Show

11

56


that the set of all elements of Class,(G) that are linear combinations of irreducible characters of G with coefficients in Int(K) is a subring R of Class,( G). 2 With the notation of Exercise 1, prove that an element 8 E Class,(G) belongs to R iff (OH,A) E Int(K) whenever H is an elementary subgroup of G and A is a character of degree 1 of H. 27 Splitting fields

Let G be a group and K a field. Then recall that K is defined to be a splitting field for K G if EndKc(V) = K . 1 for every irreducible KG-module V . It was shown in $1.3 that an algebraically closed field is a splitting field for every group, and in $1.7 that K is a splitting field for K G iff each irreducible KG-module is absolutely irreducible. In the present section we give some further criteria for a field to be a splitting field. In particular, we shall show that if G is a group of order g and the polynomial X B- 1 splits into linear factors over K,then K is a splitting field for G. Lemma 2.7 Let G be a group, K a field, and L an extension field of K such that L is a splitting field for LG. Let U and V be completely reducible KG-modules. (i) If Hom,,(U @, L, V €3, L) # 0, then U and V have an irreducible KG-constituent in common. (ii) Suppose that V is irreducible and that each irreducible submodule of V L is isomorphic to an LG-module of the form U @, L where U is a KG-module. Then V 0, L is irreducible.

,

Proof (i) Write U = U , @ U , @ . . . @ U , a n d V = V,@V,CD...@V, as sums of irreducible KG-modules. Then as vector spaces m

Hom,G(U, V )

n

@ @ Hom,G(ui

i=l

j=,

9

6).

Suppose that no Ui is isomorphic to any 6.Then by Schur’s lemma, HomK,(U,, 5)= 0 for all i, j . Hence Horn,,( U , V) = 0, and so HOm,G(U @ K L, V@K L) ‘Y HOm,G(U, V )@ K L = 0. This proves (i). (ii) Choose U as a KG-module such that U O KL is isomorphic to an irreducible LG-submodule of V 0, L. Since U 0, L can be embedded into V 8, L, Hom,,(U B KL, V 8 , L) # 0. Moreover, U is irreducible because U 8, L is, and so (i) and the irreducibility of V show that U = K Thus V 8, L is isomorphic to U 0, L and hence is irreducible. Recall that the exponenr of a group G is the least integer d > 0 such that xd = 1 for all x E G. In particular, d always divides I G 1. Note also that if T is

2.7

SPLITTING FIELDS

57

a representation of G over some field K, then for each x E G, T ( x )=~ T ( 2 )= 1. Therefore the eigenvalues (in a suitable extension field of K ) are all dth roots of unity. It follows that the values of each character of G over K all lie in the field generated by the dth roots of unity. However it should be noted that often the character values may all lie in a much smaller field. Theorem 2.7A (Brauer) Let G be a group of exponent d, and let K be a field of characteristic 0 over which the polynomial xd - 1 splits into linear factors. Then

(i) every character of G over an extension field L of K is afforded by a KG-module ; (ii) K is a splitting field for G.

Proof (i) Let 5 be a character of G over L. Without loss in generality we may take L to be algebraically closed and hence assume that L is a splitting field for LG. Moreover, since it is enough to prove the result for each irreducible constituent of (, we may assume that ( is irreducible. By Theorem 2.6A we can write [ = mill: as a 2-linear combination of characters A: induced by characters Ai of degree 1 of certain subgroups Hi of G. Since K contains the dth roots of unity, each of the characters l iis afforded by some KH,-module 6,and so the character A? is afforded by the KG-module v[;. Hence separating out the terms with mi 2 0 from those with mi < 0, we can write 5 = 8 - $, where 0 and $ are both characters of G (possibly 4 = 0) that are afforded by KG-modules. Changing notation, we may suppose that 8 is a character of smallest possible degree such that 8 = [ + 4 and 8 and $ are characters (possibly $ = 0) afforded by KG-modules. It remains to prove 4 = 0. Suppose 4 # 0. Then there exist KG-modules U and V which afford 4 and 8, respectively, such that U B KL is isomorphic to a submodule of V B KL (all modules are completely reducible by Maschke’s theorem 1.3B). Then Hom,,(U B KL, V B KL) # 0 and so by Lemma 2.7(i) we conclude that U and V have a common irreducible KG-constituent. But this means that 8 and 4 have a common irreducible constituent (as characters over K ) contrary to the minimality of 8. This shows that $ = 0, and (i) is proved. (ii) Let V be an irreducible KG-module. It follows from (i) and Lemma 2.7(ii) that V B KL is irreducible for all extension fields L of K, and hence V is absolutely irreducible. Then (ii) follows by Theorem 1.7B. The situation for fields of characteristic p > 0 is similar but in some ways simpler; the result is stated in Theorem 2.7B. The proof is quite different and uses a classical result of Wedderburn, namely: every finite division ring is a field. (For a proof see Curtis and Reiner [l], $68.)

58


Theorem 2.7B Let G be a group, K a field of characteristic p > 0, L an extension field of K so that L is a splitting field for G,and I; an irreducible character of G over L. Then

(i) If I;(x) E K for all x in G,there is a KG-module of dimension deg I; which affords I;. (ii) If d is the exponent of G and the polynomial X d - 1 splits into linear factors over K, then K is a splitting field for G.

Proof (i) We first observe that K can be taken as a finite field. Since K has characteristic p > 0, the prime subfield K O of K has p elements. Each value I;(x) ( x E G)of I; is algebraic over K Osince it is a sum of roots of unity. Thus the subfield K 1 of K generated by the values of is algebraic over K O and hence finite. If we can prove that I; is afforded by a K1 G-module U,, then U := U 1 B K , K will be a KG-module which affords I;. Thus it is enough to prove (i) in the case K is finite. Let V be an U-module which affords I;, and let T be the representation afforded by V. By the corollary of Theorem 1.3A, T ( W )= End,( V); and so if n:=dim, V, then there exists an Lbasis T ( x , ) = 1, T ( x 2 ) , ..., T ( x n 2 )of End,( V) with each x i E G. Since K is a subfield of L, we can identify R := T ( K G ) as a K-algebra embedded in End,(V). We claim that T ( x , ) , T ( x , ) , . .., T(xn2)is a K-basis of R. Certainly those vectors are linearly independent over K since they are linearly independent over L. Thus it is only necessary to show that n2

~ ( x=)

1ai T ( x i )

i= 1

with ai E K for each x

E

G.This is equivalent to n2

1

~ ( x x=~ ) ai T ( x i x j ) i= 1

for all j = 1, 2, . . ., n2, and so taking the traces we have (1)

[(xx,) =

C ai[(xix,)

for j = 1, 2,

..., n2.

i= 1

We assert that the n2 x nz matrix [ I ; ( x i x j ) ] is nonsingular. For otherwise there are pi E L not all zero such that n2

c I=

1 p i x i $ Ann

V

and

tr{T(c)T(x,)}= 0

i= 1

forj = 1,2,. .., n 2 ;and since T ( x , ) , T ( x Z ).,. ., T ( x n 2 form ) a basis of T(LG), this would mean that tr(T(c)T(x))= 0 for all x E G. However since c 4 Ann V , this gives a nontrivial linear relation among the coordinate func-

2.7

59

SPLITTING FIELDS

tions of a matrix representation associated with T , contrary to Theorem 2.3. Therefore the equations (1) have a unique solution for a l r a z , .. ., an2which can be obtained using Cramer's rule. Since c ( y ) E K for all y E G, this shows that each a, E K as required. In particular, dim, R = nz. We next show that R is a simple ring. For if I is a proper ideal of R, then I is a K-subalgebra of dimension m, say, where rn < n2. This shows that Io:=I@,L is an Lspace of dimension m. But clearly I . is an ideal of T ( U )= End,( V) that is simple. Therefore I, = 0, m = 0, and hence I = 0. Since R is a simple K-algebra, Theorem 1.3A shows that R 1: End,( V) for some division ring D with K in its center and some R-module U.Since K is finite, R is finite, and so D is finite. By the theorem of Wedderburn quoted above, D is a field. Since L is a splitting field, End,,( V) = L 1. We claim that the center Z(R) of R is L * 1 n R = K * 1. Indeed, n2

c=

1a, T ( x i )E Z(R)

i=1

iff cT(x,) = T(x,)cfor all i = 1,2, ..., n2 and each a, E K. This is so if and only if c E L . 1 and a, E K. Thus c E Z(R) iff c = at T ( x , )= a1 1 E K 1. But the center of End,(U) is D . 1 because D is a field, and K c D. So we conclude that K = D.Since dim, R = n2,it follows that dim,( V) = n; hence R 1: End,( V) 1: Mat(& K). Let e,, E R (i, j = 1, 2, . . ., n) be elements of R corresponding to the matrices in Mat(& K) with an entry 1 in the (i, j)th position and 0 elsewhere. Then the isomorphism R z Mat(n, K) shows that leij RI = IKJ"for each i and j . Since VR # 0, there exists some u E V such that ue, # 0 for some i, j . Put W := ueij R. Then W is a KG-module with I W I II e , R I = I K I", so dim, W 5 n. Moreover LW is a nonzero LG-submodule of V. By the irreducibility of V, LW = V and dim,(LW) = dim, V = n. Therefore dim, W = n and V 2: W @, L. Hence W is a KG-module of dimension deg [ which affords [. This proves (i). (ii) Since X d- 1 splits into linear factors over K, K contains the values of all characters of G over L. If V is an irreducible KG-module, then (i) and Lemma 2.7(ii) show that V 0, L is irreducible for any field extension L of K, and so V is absolutely irreducible. The result follows from Theorem 1.7B, and (ii) is proved.

-

EXAMPLE Let G be a group of order g, K a field, and L the extension of the field K generated by all roots of X B- 1. Then by Theorems 2.7A and 2.7B, L is a splitting field for G.Moreover L is a separable normal extension of K. Let V be a LG-module which affords the character x, and let Q be an automorphism of L over K. Suppose u l , u z , ..., u, is a L-basis of V and ui x = aij(x)ujwith a,,(.) E L for all x E G. Then ~ ( x=) aii(x)for any x E G. Let V' be the LG-module whose underlying vector space is V and

60

11


E=

whose module action is given by u i x I= aij(x)"vjfor all x E G andj = 1, 2, ...,n, where aij(x)" is the image of aij(x)under Q. If xu is the character of G affordedby V', then ~ " ( x = ) ~ ( x ) "for all x E G. The character xu is called an algebraic conjugate of x. Thus the Galois group of L over K acts on the characters of G over L, and two characters of G over L are algebraically conjugate if and only if they are in the same orbit. EXERCISES

1. Let G be the group of order 8 generated by a and b with relations a4 = 1, az = bZ,and ab = b3a, Q the rational field, and K = Q(i), where iz = - 1. Show that the matrix representation T over K defined by

T(a)l=

[p i]

and

T(b)t=

0 -1 0]

is absolutely irreducible, and its character has its values in Q. Also show that T is not equivalent to a representation over Q. Finally verify that K is a splitting field for G, but Q is not. 2 Give an example to show that Theorem 2.7B(ii) cannot be sharpened in that a smaller field is not a splitting field in general. 3. Let k be a field of p (prime)elements and let K = k(t) be a transcendental extension of k. Let G be the group generated by a and b with relations aP = bP= 1 and ab = ba. Show that the matrix representation T of G over K defined by T(a)t=

1 0

and

T(b)l=

[: y]

is not equivalent to a representation over k. [This example shows that the conclusion of Theorem 2.7B(i) need not hold if the character is reducible.] 4. Let G be a group of order g, and let K be the extension of the rational field Q generated by the roots of X @- 1. Let x and 4 be characters of G over K and let 0 be an automorphism of K over Q. Show that ( x , 4) = (xu, 6). 5. Let G, g, K , and Q be as in Exercise 4, and let x be any character of G over K. Suppose Q is an automorphism of K over Q such that for some integer n relatively prime to g, w" = w" for all 8th roots o of 1. Show that f ( x ) = ~ ( x "for ) all x E G. 2.8

Notes and comments

There is an extensive theory of ordinary representations and characters. For example, see Burnside [I], Huppert [l], and Curtis and Reiner [l]. A comprehensive treatment of applications of ordinary characters is given in

2.8 NOTES AND COMMENTS

61

Feit [l] and Isaacs [3]. Also see Brauer [6] and Glauberman [2]. Brauer’s theorem on induced characters [Theorem 2.6AI was first proved in Brauer [7] and later a simpler ring theoretic proof was given by Brauer and Tate [l]. This result was generalized by Berman [3]. The proof we have given here is due to Goldschmidt and Isaacs [l]. A proof of Wedderburn’s theorem [which asserts the commutativity of a finite division ring] may be found in Herstein’s book “Topics in Algebra” [l]. Also see Curtis and Reiner [l]. The theorems of52.7 were first proved by Brauer [4] and later the proofs were simplified by Feit.

CHAPTER

Id

ModuIar Representations and Characters

When the characteristic of a field K is p > 0, the theory of representations of G over K (called “modular representation theory”) becomes more involved, since in this case the characters of G over K do not in general determine the representations (even up to equivalence). By defining the (modular) characters in a different way, the property that the characters determine the representations (Corollary 2 of Theorem 2.3) can be partly regained. In order to establish the relations between the ordinary and modular characters, it is simplest to work in a p-adic field. The necessary theory of p-adic fields is developed in the first sections and is then used to establish an important relation between the Cartan invariants and the decomposition numbers. This relation in turn implies the orthogonality relations for modular characters. In conclusion, we consider an application of the theory to prove a result of Fong and Swan on the representations of a p-solvable group.

3.1 The p-odic integers Let p be a prime and let Z denote the ring of integers. Consider the sequence of rings

3.1 THE P-ADIC

63

INTEGERS

together with the canonical homomorphisms. The ring of p-adic integers Z, is defined as the " inverse limit " of this sequence. In concrete terms it can be described as follows. Let S = (Z/p"Z)be the Cartesian product of the (finite) sets Z/pnZ. The elements of S may be written as sequences (x,), where x, E Z/pnZand S is a ring under componentwise addition and multiplication. Then

=:lf

Z,

:= {(x,,) E

S I x,v, = x,-

for n = 2, 3, . . .}.

Clearly Z, is a subring of S because the v, are ring homomorphisms. Moreover Z is embedded as subring of Z, by the mapping XI+

(x,)

where x,

I=

x

+ p"Z

for each n.

Theorem 3.IA The ring Z, has the following properties.

(i) (x,) E Z, is a unit in Z, iff x, # 0. (ii) pZ, is the unique maximal ideal of Z, ,and so rad Z, = pZ, ;moreover ZJpZ, N ZlpZ. (iii) The proper nonzero ideals of Z, are precisely p"Z, ( n = 1, 2, 3, . . .); and =: p"Z, = 0. In particular, Z, is a principal ideal domain. (iv) Z, is an integral domain of characteristic 0.

n ,

Remark The field of quotients of Z, denoted by Q, is the jield of p-adic numbers. Proof (i) If (x,) has an inverse, say (y,,), then in particular, x, y, = 1, so x1 # 0. Conversely, suppose (x,) E Z, and x, # 0. Since v, is surjective, it follows that x, E pZ/pnZ iff x,v, E pZ/p"-'Z. Since x1 # pZ/pZ, this shows that for each n, x, # pZ/pnZ. Thus x, is a unit in Z/p"Z, and hence has an inverse y,, say, for n = 1, 2, 3, ... . Then 1 = x n y , ; and 1 = (x, v,)(y, v,) = x,- l(y, v,) shows that y, v, = = y,for all n 2 2. Hence (y,) E Z, and (x,,)(y,,) = 1. Thus (x,) is a unit in Z,, and (i) is proved. (ii) It follows at once from (i) that Z, is a local ring and hence has a unique maximal ideal by Lemma 1.6B. (Note that there is no distinction between right and two-sided ideals because Z, is commutative.) On the other hand, it is readily verified that there is a ring homomorphism of Z, onto Z/pZ given by ( x , ) ~x i which has kernel pZ, .Hence Z,,/pZ, z Z/pZ and pZ, is the unique maximal ideal. Lemma 1.6B now shows that pZs.= rad Z,. (111) If (x,) E p"Z,, then x i = 0 for i = 1, 2, ..., m. Therefore ,"= pmZ, = 0. Now let I be any proper nonzero ideal in Z,. Then from what we have just shown, there exists rn 2 1 such that p"+'Z, 2 I; choose m as small as possible, so p'Z, 2 I. Let x E Ibm+'Z, . Then x = p"y for some y E Z,\pZ,. By (ii) y is a unit, and so I 2 p"yZ, = p'Z,. Hence I = p"Z,. This proves (iii).

,

0

64

111 MODULAR REPRESENTATIONS

A N D CHARACTERS

(iv) Since p"Z,p"Z, = p"+"Z, # 0, it follows from (iii) that the product of any two nonzero ideals in Z, is nonzero. Hence the product of any two nonzero elements is also nonzero. Thus Z, is an integral domain. As we noted at the beginning of this section, Z is embedded as a subring in Z,, so 2, has characteristic 0. This completes the proof of the theorem. The construction used for Z, can be applied in slightly more general situations. Let R be a free Z,-algebra and let ul, u 2 , . .., us be a Z,-basis for R. Then for each n 2 0 we have p"R = p"Z, u1 @ p"Z, u2 @ * * . @ p"Z, u,

(1)

Ti

as 2,-modules. Since p " 2 , = 0 by Theorem 3.1A, we see that p"R = 0. Let S, := :=1 (R/p"R). Then S, is a ring under componentwise addition and multiplication. We define a ring homomorphism I,$: R + S R by a$ = (a,,), where a,,i=a + p"R for each n. Since ;=o p"R = 0, Ker t j = 0, and so JI is injective. On the other hand it follows from (1) and the definition of 2, that (a,,)E S, lies in Im JI iff a,,p,,= a,,- for each n, where p,,: RJp"R + RIP"- 'R is the canonical homomorphism.

n=; n

EXAMPLE

Let R be a free Z,-algebra. Then every element of the form

+ + +

1 -pb ( b e R) is a unit in R. Indeed, put a,,:=l pb pn- 'b"p"R. Then a,,p,,= a,,- for each n 2 2, and so (a,,)E Im JI; hence there exists a E R such that a,, = a p"R for each n. Since (1 - pb)a, = 1 pnbn+ p"R = 1 p"R, this shows that (1 - pb)a - 1 E p"R for each n; and so (1 - pb)a - 1 E =; pnR = 0. Hence a = (1 - pb)- In particular, this shows that pR E rad R. Indeed if M is a maximal right ideal of R such that M 2 pR, then R = M pR, and so some element c E M has the form c = 1 - pb with b E R. Since c is a unit this contradicts the choice of M as a

'+

+

+

n

'.

+

proper right ideal of R. We now turn to a very important property of Z,-algebras. The following theorem may be compared with Theorem 1.6B. The final version appears as Theorem 3.4A. Theorem 3.1B (Idempotent refinement) Let R be a free Z,-algebra and suppose J is an ideal of R such that JJpR is a nilpotent ideal of RJpR. Letfbe an idempotent in RJJ. Then there exists an idempotent e in R such that e+J=J Remark Since pZ, annihilates R/pR, we may consider RJpR as a ZdpZ,-algebra. By Theorem 3.1A, ZJpZ, is a field with p elements, and so we can apply Theorem 1.3A to conclude that rad(RJpR) is nilpotent. By the example above rad R 1 pR, and so rad(R/pR) = rad RJpR. Thus the hypothesis of the theorem holds with J = rad R.

3.2

65

P-ADIC ALGEBRAS

Proof By Theorem 1.6B there exists an idempotent el of R/pR such that el maps onto f under the canonical mapping R/pR -+ R/J. We shall now define en (n 2 2 ) recursively such that for each n (2) en is an idempotent in R/pnR and e,p, = en- 1, where p,: R/pnR -+ R/p"-'R is the canonical homomorphism. Suppose enhas been defined and choose any a E R/p"R such that ap, = en- l . Then put b :=a2 - a; note that b E Ker pn = p"- R/pnR and so b2 = 0 because n 2 2. Define en:=a + (1 - 2a)6. Clearly ab = ba and so

+ (2a - 1)(1 - 2a)b + (1 - 2a)'b2

e,2 - e n = a 2 - a = (1 =

-4b2

1

+ 4a - 4a2}b

=o.

Since b E Ker p,, we also have e,p, = ap, = en- ; hence en satisfies (2). Thus in the notation above we can define an element (en)E SR such that enpn= en- for all n 2 1. Therefore from what we saw there, there is a unique e E R such that en = e + p"R for each n. Since (ef) = (en),therefore e2 = e. Finally, since el maps ontofunder the natural mapping R/pR -+ R/J, and el = e + pR, therefore e + J =f:In particular, e # 0, and so e is an idempotent with the required properties. EXERCISES

1. Show that the rational field Q is a subfield of the p-adic field Q, for any prime p. 2. The followingexercise is intended to provide an alternative approach for the study of the ring Z,. Let p be a fixed prime. For any integer n > 0, let x, be a primitive (complex)p"th root of unity, and let U be the group generated by the set {xl, x2, x j , . ..}. Show that every sequence of nonnegative integers

(1)

( k l , k2

9

kn,

.**)

such that 0 I k, -= p" and k,, = k, (mod p") determines a unique endomorphismfof U by x , f : = x ? for n = 1, 2, ... . Conversely, show that every endomorphismfof U determines a unique sequence of the form (1). The set E of all sequences of the form (1) forms a ring with respect to componentwise addition and multiplication. Prove that E N Z,. 3.2 p-adic algebras We continue with the notation of $3.1. Recall that Q, is the field of quotients of Z,. Let K be an extension field of finite degree over Q p ,and define A to be the set of all a E K that are roots of monic polynomials of

66

111 MODULAR REPRESENTATIONS A N D CHARACTERS

Z,[X]. Elements of A are said to be integral over Z,, and A is called the integral closure of Z, in K. Note 1 Suppose that a E K. Then a E A iff the ring Z,[a] is a finitely generated Z,-module. For suppose a E A. Then by definition there exists an integer m > 0 and a,- 1, a,- z , . . ., a, E Z, such that a" = a,- a". . * a,, and hence Z,[a] = Z, a"- + 2, a'"- + . - + Z, . Conversely, assume that Z,[a] is a finitely generated Z,-module. Suppose

+

+

Zp[al= Zp f1(a) + Zp fi(a) + * * * + Zp with eachf,(X) E Z,[X]. Choose N greater than the degrees of all theJ(X). Since aNE z,[a], we may write

+

+ +

aN= a , fl(a) u2 fz(a) a,, fn(a). This shows that a is a root of a monic polynomial of Z,[X], and hence a E A. Note 2 It follows at once from Note 1 that if a, a' E A, then a - a' and a d also lie in A. From the hypothesis Z,[a] and Z,[a'] are finitely generated Z,-modules, say Z,[a] = Zy= Z,ai and Z,[a'] = Cy= Z,Fi . Then n

Zp[a, a'] = (Zp[al)[a'l=

m

n

C1Zp[a]Bj = i C= l j C = 1 ZpaiFj

3

j=

and so every submodule of Z,[a, a'] is a finitely generated Z,-module (because Z, is a principal ideal domain). In particular, Z,[a a'] and Z,[aa'] are finitely generated Z,-modules. Hence A is a ring contained in K (and it contains Z,). We shall see below that A is finitely generated as a Z,-module and hence is a Z,-algebra.

+

Theorem 3.2 With the notation above: (i) (ii) (iii) (iv)

K is the field of quotients of A ; A n Q, = Z,; A is a Z,-algebra which is free as a Z,-module; A has a unique maximal ideal M (which equals rad A), and

nM ~ = o ; m

n= I

(v) the maximal ideal M = n A for some n E A, and the proper nonzero ideals of A are precisely n"A (n = 1,2, . . .) (in particular, A is a principal ideal domain); (vi) k I= A/nA is a finite field of characteristic p.

3.2

67

P-ADIC ALGEBRAS

Proof (i) First note that for each a E K there exists c # 0 in Z , such that ca E A. Indeed, since a is algebraic over Q,, there exists a polynomial

a,X"

+ a,-

I X"-

+ + a,

E

Z,[X]

with a,,, # 0

which has a as a root. Then u,a is a root of the monic polynomial X"

+ a,- xm- + a m - z a m x m - 2+ *..+ a am-' ~ z p [ X l 1

0 m

and so lies in A. In particular, K is the field of quotients of A. (ii) Each a # 0 in Q, can be written a = bp', where b is a unit in Z , and t E Z . Suppose a E A. We have to show that t 2 0, and then a E Z , . Indeed, let

X"

+ a,- xm- + + a, E Z , [ X ] 1

* * *

be a monic polynomial which has a as a root. If t < 0,then substituting a into this polynomial and multiplying through by pm1 ' I gives

b"=

-a,-,bm-'pIII

-...-a,p

"IrI E

PZ,.

Since b is a unit, this is a contradiction. Hence t 2 0 and a E Z , as asserted. (iii) Each a E K acts as a Q,-endomorphism h, of K by right multiplication. Temporarily, we shall write tr a:= tr ha, the trace of this linear transformation ha over Q,. (tr a is equal to the usual field trace introduced in algebraic number theory.) By definition tr a E Q, for all a E K.We also have tr a E Z , whenever a E A. Indeed, if a E A, then a is a root of some monic polynomial q ( X ) E Z , [ X ] . Let m ( X ) E Q , [ X ] be the minimal polynomial for the linear transformation ha. Then since q(h,) = 0, m ( X ) I q ( X ) in Q , [ X ] , and so each root of m ( X ) is a root of q ( X ) and hence is integral over Z , in some extension field of K. But all eigenvalues of ha are roots of m ( X ) , and the trace is the sum of the eigenvalues (with appropriate multiplicities). Since a sum of elements integral over Z , is again integral over Z , , this shows that tr a = tr ha is integral over Z , . But tr a E Q, from above, so by (ii), tr a E Z , whenever a E A. Now we show that for any Q,-basis ql, .. ., qn, the n x n matrix [tr(qi q j ) ] is invertible. Indeed otherwise the rows would be linearly dependent and (using the linearity property of the trace and the linear independence of the qi) that would mean that there exists a = aiqi # 0 with ai E Q,, such that tr(aqj) = 0 for all j . Since a # 0, we can write a- = C,?=bj q j for some b, E Q,, and then 0 = tr aa-l = tr 1, which is impossible because K has characteristic 0. Finally, using the observation at the beginning of the proof we can choose a Q,-basis qI, .. ., qn of K such that each q iE A. Since A is a ring, it then follows from above that tr(aqj) E Z , for every a E A. On the other hand each

68

111

MODULAR REPRESENTATIONS AND CHARACTERS

a E A can be written in the form a =

aiqi (ai E Qp).Then multiplying

by q, and taking traces we obtain n

( j = 1,

C a i tr(qiq,) = tr(aq,)

..., n).

i= 1

Since d I = det[tr(qiq,)] E Z, is not 0, we can use Cramer's rule to solve this system of equations and write each a, in the form bi/d, where bi E Z, because each tr(aq,) E Z,. Thus n

a = C a i q i E Zp(tll/d) i=1

+

+ Zp(qn/d)

for each a E A. Thus the ring A is contained in a Z,-module. Since Z, is a principal ideal domain, the structure theorem for (finitely generated) modules over principal ideal domains shows that A is also finitely generated as a Z,-module; and A is Z,-free because it is Z,-torsion-free (see 41.1). Because Z, is a subring of A, A is also a Z,-algebra. This proves (iii). (iv) Let M be a maximal ideal of A. Since A is commutative, AIM is a field, and so under the canonical homomorphism A + A / M , Z, maps onto an integral domain. Since pZ, is the unique maximal ideal of Z,, this shows that pZ, = Z, n M. In particular, p E M, and so M 2 PA. Hence we conclude that rad A 2 PA, and rad(A/pA) = (rad A)/pA. On the other hand, since pZ, c Ann(A/pA), A/pA may be considered as a Z,/pZ,-module. Since Z,/pZ, = Z/pZ is a finite field, it follows from the Wedderburn structure theorem (Theorem 1.3A) that rad(Alp.4) = rad A/pA is a nilpotent ideal, and A/rad A z (A/pA)/rad(A/pA) is a direct sum of simple rings. However, since A is an integral domain, it has only one idempotent 1, and so it follows from Theorem 3.1B that A/rad A has only one idempotent; therefore A/rad A is a simple, commutative ring (and hence a field). This shows that M := rad A is a maximal ideal in A, and so (from its definition) it is the unique maximal ideal. Finally, since M/pA = rad A/pA is nilpotent, M' E pA for some integer t 2 1. Since (7 ,"= pnZ, = 0 (Theorem 3.1A), and A is a free Z,-module, pnA = 0. Therefore (7=; M" = 0. (v) We saw in the proof of (iv) that M' E pA for some integer t 2 1. Choose t so that M' z pA but M'- $ pA; then choose /3 E M'- '\pA, and put n = p/B E K. Since B 4 PA, n- = B/p $ A; however

n:'

n-'M E ( p - l w - l ) M

E p-'(pA) = A.

We claim n- ' M $ M. Otherwise take any a # 0 in M.Then Z,[n-I]a c M , and so Z,[n- '1 C a- ' M E a- 'A. Since the last is a Z,-module, this shows that Z,[n-'] is a Z,-module, and so n-l E A, contrary to what we saw above. Thus we have shown that n- ' M is an ideal of A not contained in M, and so (iv) shows that x - ' M = A. Hence M = nA.

3.3

ORDINARY A N D MODULAR REPRESENTATIONS

69

The powers M" = 7c"A ( n = 1, 2, .. .) are certainly proper ideals of A. It remains to show that any nonzero proper ideal 1 of A equals one of these. Since I # 0, (iv) shows that for some n, M" $ I ; choose n so that M" 2 1 but M " + l $ I . Then n-"I E n-"M" = A, but ~ - " l $a-"M"+l ! = M . Thus a-"I is an ideal of A not contained in M which shows that a-"I = A ; and hence 1 = a"A as required. (vi) We saw in the proof of (iv) that a A = rad A 2 pA. Since A is a free Z,-algebra of rank I , say, I A/pA I = I Z,/pZ,, = p'. Since nA is maximal, k I= A / a A is a field, and I A / R AI divides I A/pA I = p'; so k is finite'of characteristic p. EXERCISES

1. Let K , be a finite extension of the rational field Q generated by a and let A , be &heintegral closure of Z in K,. Let K = Q p ( a )and A be the integral closure of Z, in K. Assume Q E Q,,. Show that A , c A. Let M , be a maximal ideal of A , such that p E M , and M , E aA, where a A is the unique maximal ideal of A. Then prove that A, / M , N A/aA = k. (Therefore a representation over k can also be considered as a representation over A, / M , . ) [Hint:One way to prove this result is to consider a p-adic completion of K,.] 2. With the notation of Exercise 1, let A, = { a / j E K , la, E A,, and j# M , } . Show that A , is an integral domain with a unique maximal ideal M , such that A , E A , E A, M , G M , E xA, and A , / M , N A , / M , N A/nA = k. Also prove that the nonzero ideals of A , are a"A2for some a E A , and n = 0 , 1,2,....

3.3 Ordinary and modular representations We shall now apply these results on p-adic algebras ($3.2) to study the relation between representations of a group over a field K of characteristic 0 and the representations over a field of characteristic p > 0. At this point it is convenient to introduce notation for the various fields and rings; this notation will be used throughout the remainder of the book except in a few cases where the contrary is explicitly stated. NOTATION Let G be a group of order g and let p be a prime. Let K be a field extension of finite degree over the p-adic field Q, such that the polynomial X B- 1 splits into linear factors over K. Let A be the integral closure of Z,, in K and let a A be the unique maximal ideal of A. Put k I = A/aA. Then k is a finite field of characteristic p, and since the roots of X B- 1 in K lie in A (by the definition of A), X B- 1 splits into linear factors over k. Thus by $2.7, K and k are splitting fields for G, and all its subgroups and factor groups; in

70

111

MODULAR REPRESENTATIONS A N D CHARACTERS

particular, all irreducible KG-modules and all irreducible kG-modules are absolutely irreducible (Theorems 2.7A and 2.7B).

Remark A is a principal ideal domain (Theorem 3.2), and so by the structure theory for modules over a principal ideal domain, an A-module V is free iff V is “A-torsion-free” in the sense that a u = 0 (a E A, u E V) implies a = 0 or u = 0. In particular, an A-submodule U of a free A-module V is free and (again by the structure theorem) rank, U Irank, V. However, the homomorphic image of a free A-module need not be free. We use these observations frequently in what follows. Our first objective is to show how to go from a KG-module V to a related kG-module in a way which preserves at least part of the structure. We shall do this in two steps; first from KG-modules to AG-modules, and then from AG-modules to kG-modules. The first step is based on the following theorem. Theorem 3.3 (Burnside) Let V be a KG-module of dimension n. Then there exists an A-free AG-module W such that V z W @ , K, and rank, W = n.

Proof Let ul, u2, ...,u, be a K-basis of Y and put W I = u1 AG + u2 AG + + u, AG c V. The module W is finitely generated as an A-module (requiring at most ng generators), and is A-torsion-free since V is; therefore by the remark above, W is a free-A-module. Let wl, w2, ...,w, be an A-basis of W. Since no nontrivial linear combination of wl, w2 ,...,w, over A is 0, and K is * *.

the field of quotients of A (Theorem 3.2), the wl, w2, ..., w, are linearly independent over K.But from the definition of W,wl, w2, ...,w, span the vector space V and so form a K-basis. Hence rank, W = m = n. Finally W ~ H w, @ I gives a KG-isomorphism of V onto W @ K. This proves the theorem.

,

It is now clear how we can pass from a KG-module V to a kG-module.

First, using Theorem 3.3, we choose an A-free AG-module W such that V z W @ , K. Then W:=W@, k (where k = A/aA) is a kG-module. In terms of bases this second step can be described as follows. Let wl, w2, . .., w, be an A-basis of the free A-module W.Then W ,*=w, @ 1 (i = 1,2,. ..,n) is a k-basis for W,and the mapping from W + W is given by

where a, E A and 6, $=a,+ aA E k. We shall call the canonical mapping A + A/aA = k reduction modulo a and denote the image of a under this mapping by 6. More generally, if W is an A-free AG-module, then we refer to

3.3

71

ORDINARY A N D MODULAR REPRESENTATIONS

W I = W @ A k as the reduction of W modulo a ; note that if W is also an A-algebra, then W is a k-algebra. Suppose c is the character afforded by the KG-module V. Then W affords the same character [ (take an A-basis for W such that w1 0 1, w2 0 1, .. ., w, 8 1 is a K-basis of W @ A K N V). The character { afforded by the kGmodule W is then given by {(x)?

:r(x)

for each x E G;

this is clear if the traces are calculated with respect to the bases wl,w 2 , .. ., w, and w,,w,, . . ., w,, respectively. Note It follows that if W is an AG-module of rank n, say, and W is irreducible, then W 0 A K is also irreducible. Indeed, let c be the character afforded by W B AK. If [ were reducible, then c, and hence would be a sum of characters afforded by modules of dimensions less than n. But is afforded by the irreducible module W of dimension n; and so we have a contradiction by Corollary 1 of Theorem 2.3.

r,

r

Two obvious questions arise: (a) is the kG-module W determined (up to isomorphism) by the KG-module V ;and (b) is every kG-module isomorphic to a module W obtained in this way from some KG-module? In general, the answer to both these questions is no. (See Exercises 1 and 2.) However, as we shall see in the sections to follow, something can be said about the uniqueness of the W in (a), and the answer to both questions is yes whenever p does not divide 1 G 1. More importantly, the process always gives considerable information about the irreducible and principal indecomposable kG-modules. EXAMPLE 1 If W, and W2are two AG-modules such that Wlis irreducible and Wl C3 A K N W2B A K , then WlN W2.(So the answer to (a) is yes in this case.) Indeed, let c’ and c2 be the characters afforded by W, @ A K and W2C3,, K, respectively. Then by Corollary 2 of Theorem 2.3, W, 0 A K ‘v W20” K iff = c2. Similarly, the irreducibility of W ,and Corollary 1 of Theorem 2.3 show that & ‘v W2iff 7 = p.Since = L2 implies = p,the assertion follows. Note that in general this argument cannot be reversed to show that WlN W2 implies W, B A K N W 2 @ . , K because in general = 7 does not imply c’ = c2 (however, see 53.5).

[’

EXAMPLE 2 Suppose that W is an A-free AG-module. Then Ker W clearly contains Ker W. However it cannot be “too much” larger; the group Ker W/Ker W is always a p-group. To prove this it is enough to show that if x + Ker W has prime order q in Ker W/Ker W, then q = p. Since x E Ker W but x $ Ker W , it follows from Theorem 3.2 that there exists an integer n 2 1 such that W ( x - 1) E A”W but W(x - 1)$ n n + lW . On the

72

111 MODULAR REPRESENTATIONS AND CHARACTERS

+

+

+

other hand, XP - 1 = (1 x - 1)* - 1 = q ( x - 1) (j)(x - l), + (;)(x - 1)q by the binomial theorem. Therefore, since XP E Ker W, Wq(x - 1) c W(x4 - 1) W(X - 1)2 = W(X - 1)2 E n2"WE n"+lw because n 2 1. This implies q E nA; thus q equals 0 in AInA = k and so q = p by Theorem 3.2(vi).

+

EXERCISES

1. Let G be the alternating group of order 60 generated by a = (123) and b = (345). Let K be a field extension of finite degree over Q2 such that X60 - 1 splits into linear factors in K [ X ] , A the integral closure of Z, in K, nA the unique maximal ideal in A, k = A/nA, and o a root of X 2 X + 1 E k [ X ] .Prove that the kG-module Y of dimension 2 over k such that for a fixed basis {ul, u,} of V

+

+ a, u2)a = mal u1 + (1 + o)a2u2 (aclul + a,u2)b = a2ul + (a1 + a2)u2

(a1u1

is an irreducible kG-module. Show that there exists no AG-module U such that 0 z V . 2. Let K,Q,, Z,, A, nA, and k be as in Exercise 1, and let G be the group generated by a and b such that a3 = b2 = (ab), = 1. Suppose o is a root of the polynomial X 2 + X + 1 over Q, . Show that the matrix representations T and S of G over K defined by

and

are equivalent over K,but the representations 7 and S over k obtained from T and S by reducing modulo n are inequivalent. However, show that T and S have the same irreducible constituents. 3. Let K, Q 2 , Z,, A, nA, and k be as in Exercise 1, and let G be the permutation group generated by a = (12), b = (23), and c = (34). Show that the matrix representations T and S of G over K given by

T(a)i=[i

-3

0

-8

-:I,

-12

T(b)=[S

-: -fl. 4

3.4

and

73

LIFTING IDEMPOTENTS

s(a):=[s 8 -;:I, 3 -10 -1

S(+[

-1

-1

-2 S@):=[

-3

-8

0 2 -18 1 ,

41 -4

are inequivalent over K. Show that the representations i=and S of G over k obtained from T and S by reducing modulo n have the same irreducible constituents. 4. Let Q, be the field of p-adic numbers and K the extension of Q, generated by the roots of X B 1, where g is the order of a group G. Let A be the integral closure of Q p ,nA the unique maximal ideal of A, and k = A/nA. Prove that two kG-modules associated with the same KG-module have the same composition factors. 3.4 Lifting idempotents We continue with the notation of 53.3, and in this section examine the relationship between the idempotents of the group algebra AG and the group algebra kG. This will lead to a result which shows that if W is an AG-module and W is its reduction modulo n,then both Wand Wdecompose into direct sums in essentially the same way. (See 51.6 for elementary facts about idempotents.) Theorem 3AA (Lifting idempotents) Let R be a free A-algebra.

(i) If e l , ..., en is a complete set of orthogonal idempotents in R (so eie, = 0 for i # j and ei = l), then El, . . .,E,, (reduction modulo n) is a complete set of orthogonal idempotents of R = R/nR. (ii) Conversely, iffl, ...,f, is a complete set of orthogonal idempotents of R, then there exists a complete set of orthogonal idempotents e l , .. ., enof R such that zi =fi for each i.

c;=

Remarks We shall principally be applying the theorem in two situations. First where R = AG and R = kG, and second where R = Z(AG), the center of AG, and R = Z(kG), the center of kG. Note that (ii) shows that iff is an idempotent in R, then there is an (ii) to the set e, 1 - el. idempotent e in R such that E =![apply Proof (i) Since reduction modulo n is a ring homomorphism, E,E, = 0 for i # j , Z; = Zifor each i, and El + + P,, = 1. It remains to show that

74


Z, # 0 for each i. However Z, = 0 would imply that e, E nR. But then for each integer n 2 1, the idempotent e, = eg E n"R. Since R is a free A-module,

n n"R= OD

n= 1

m

(7 (nA)".R=O n= 1

by Theorem 3.2(iv).This implies e, = 0, contrary to the definition of idempotent. Hence Ci # 0, and (i) is proved. (ii) We prove (ii) by induction on n. For n = 1 the result is true (take el = 1). For n 2 2, induction shows that there exists a complete set of n - 1 orthogonal idempotents e,, . . ., en- 2 , e of R such that under reduction modulo n, 2, =fl, ..., Z,,-2 = f n - 2 , and z =f,- +f,. Now define R, I = eRe. Then Ro is a ring with unity e. Moreover, since A is finitely generated as a Z,-module, the same is true of R, and hence of Ro itself. Thus Ro is a Z,-algebra; and Ro is Z,-free because Ro c R and R is Z,-free. By Theorem 3.2 there exists an integer n 2 1 such that the ideal p A = n"A. Therefore (nR,)" c n"ARo = pRo E nRO,and so the hypotheses of Theorem 3.1B hold for R, with J = nR,. Since B =f,- +f, and the latter are orthogonal by hypothesis, it follows that f,- = Zf.- z and f, = Zf,z both lie in R, = Ro/nRo. Hence by Theorem 3.1B there exists an idempotent enof Ro such that z,,- =f,- ; and then enI= e - enis an idempotent in R, such that Z, =f.(note en # 0 since f, # 0). Clearly enen- = enen = 0. Since both en- and en lie in eRe and e = enen, this shows that el, , , .,en- enis a complete set of orthogonal idempotents of R with the required property.

+

As an immediate application of Theorem 3.4A we have the following result. Theorem 3RB (i) Let R be an A-free A-algebra. Then R is a local ring iff a is a local ring. (ii) Let U be an A-free AG-module. Then U is indecomposable iff 0 is an indecomposable kG-module. (iii) Let V be an A-free AG-module. Then the conclusion of the KrullSchmidt theorem (the corollary of Theorem 1.6A) holds for V.

Proof (i) Recall that a ring S is local iff SlradS is a division ring (see the remark following Theorem 1.7A). From the remark following the statement of Theorem 3.1B we know that rad R 2 pR, and rad R/pR = rad(R/pR) is the unique maximal nilpotent ideal in R/pR. By Theorem 3.2, p A = nnAfor some integer n 2 1, and so (nR)"E pR c nR. Thus nR/pR is a nilpotent ideal in R/pR, and therefore nR E rad R. This shows that rad R = rad(R/nR) = rad R/nR; and so Rlrad R = Rlrad R. Thus R is local iff R is local.

3.5

THE CASE WHERE

p DOES NOT

DIVIDE IG[

75

(ii) Put E I = End U )E End A( U).Since U is a free A-module, End A( V ) is also a free A-module; hence E is A-free. Moreover

(1) E = EndAG(U )@I A k N End,,( U €9A k ) = End,,( 0). By $1.6U (respectively, 0 )is indecomposable iff 1 is the only idempotent in E (respectively,E).By Theorem 3.4A,1 is the only idempotent of E iff 1 is the only idempotent of E (write 1 as a sum of primitive idempotents). Thus U is indecomposable iff 0 is indecomposable. (iii) Any direct summand of V is a free A-module of smaller rank, and so it is clear that V is a direct sum of a finite set of indecomposable AGmodules. The uniqueness of this decomposition will follow from Theorem 1.6A when we have shown that E:=End,,(U) is a local ring for each indecomposable A-free AG-module U.However, by (ii), 0 is an indecomposable kG-module, and so End,,( 0 ) is local by Theorem 1.7A.Then (1) shows that E is local and (i) shows that E is also local as required. 3.5 The case where p does not divide I G I We now consider the especially simple case where p Xg (where g I = Write the AG-module AG in the form (1)

I G I).

AG= Ul@-**$U,

where U1, U,, . .., U,are indecomposable AG-modules. Then reduction modulo II gives kG = 0, $ * * * @ 0, (2) and by Theorem 3.4B we know that the Oi are indecomposable kG-modules. However, in the case wherep X g,Theorem 1.3B (Maschke's theorem) shows that every kG-module is completely reducible, and so the indecomposability of the Oi implies that each Oi is irreducible. This leads to the following theorem. Theorem 35 Suppose that p X

I G I. Then the following hold.

(i) Each kG-module is isomorphic to a module of the form 0 for some A-free AG-module U. ' (ii) If U is an A-free AG-module, then the following three conditions are equivalent: 0 is an irreducible kG-module, U is an indecomposable AGmodule, and U @I A K is an irreducible KG-module. (iii) If c', ..., r' are the distinct irreducible characters of G over K,then p, ..,, r' are distinct and are the irreducible characters of G over k. In particular, for two irreducible characters C' of G over K,we have c' = iff

C' = p.

Y

ci,

['

111

76 (iv) If U and U @ AK Z V @ AK .

V


are

A-free

AG-modules, then

U

2

P iff

Proof (i) By Theorem 1.3B (Maschke's theorem), every kG-module is completely reducible; therefore it is enough to show that each irreducible kG-module has the form 0 for some A-free AG-module U . However each irreducible kG-module is an irreducible constituent of kG (see the note following Theorem 1.3A) and so is isomorphic to one of the summands in (2). This proves (i). (ii) If U = U'@ U",then U @ A K = (U' @ A K ) 0 (U" 8 A K). Hence, if U @ A K is irreducible, then U is indecomposable. Secondly, if U is indecomposable, then 0 is indecomposable by Theorem 3.4B and hence irreducible because 0 is completely reducible (Theorem 1.3B). Finally, if U is irreducible, then the note of 53.3 shows that U @ A K is also irreducible. Thus the three conditions are equivalent. (iii) From (1) we obtain K G 'Y A G O " K = ( U , @ A K ) @ . . . @ ( U , @ A K ) , (3) where the summands are all irreducible KG-modules by (ii). By the note following Theorem 1.3A every irreducible KG-module is isomorphic to one of the U i @ A K ; and we may assume that the U i are numbered so that each irreducible KG-module is isomorphic to exactly one of U i 8 A K ( i = 1, . . ., r) and that U i B AK affords the character c'. Similarly each irreducible kG-module is isomorphic to one of O,, . . . , 0,.Since irreducible kG-modules with the same character are isomorphic (Corollary 1 of Theorem 2.3), this implies that every irreducible kG-module is isomorphic to one of Oi (i = 1, ..., r). Since p $ I G I, Theorem 1.5 shows that the number of irreducible kG-modules (up to isomorphism) is the same as the number of irreducible KG-modules. Therefore the-modules Oi (i = 1, . . ., r) are nonisomorphic, and their characters p,. . ., are distinct and are all the irreducible characters of G over k. (iv) Write U = U 1 @ ... 0 U , and V = V, 0 . * * 0 V, as sums of indecomposable AG-modules. Then

c7=

q@..*@u,,v

=

V,@*-@V-,,

U @ , K = (U,O,K)@..~@(U,O,K),

and V @ , K = (V, @ , K ) @ . . . @ ( l / - , @ , K ) .

By (ii), each of the summands occurring in these latter modules is irreducible. Thus the proof of (iv) reduces easily to the case where U and V are indecomposable and 0, V, U @ A K, and V @ A K are irreducible. But in the latter case the result is immediate from (iii).

3.6

77

MODULAR CHARACTERS

Corollary If p X 1 G 1 and deg x divides 1 G I.

x

is an irreducible character of G over k, then

Proof By part (iii) of the theorem, deg x = deg for some irreducible character iof G over K. Then the result follows from Theorem 2.4B and $3.5. EXERCISE

Let G be the alternating group generated by a = (123) and b = (345), and let k be a finite field of characteristic 2 containing a root w of X 2 X 1. Let V be a kG-module of dimension 2 such that

+ +

+ a 2 u 2 ) a= ( 1 + w ) a l u , + wa2u (a1u1 + a2u2)b= a 2 u 1 + ( a l + a2)u2

(a1u1

where { u l , u2} is a fixed basis of V. Show that there is no AG-module U such that 0 = V. 3.6

Modular characters

Let W be a kG-module, let x be a p'-element of G, and put H := (x). Then the restriction W, is a kH-module. Since p $ I H 1, Theorem 3.5 shows that K is a there exists an A-free AH-module U such that V z W, and U KH-module uniquely determined up to isomorphism by W,. In particular, the character of U 0 A K is uniquely determined by W, and so we can define $(x) as the value of this character at x, where +(x) only depends on W. Defnifion We shall denote the set of all p'-elements of G by Go. Then the function 4 : Go + K defined above is called the modular character (or "Brauer character") afforded by W. The modular character C#J is called irreducible if it is afforded by an irreducible module (see the note below).

Note If x is the (Frobenius) character afforded by W , then it follows in the notation above that 0 affords the character xH of x restricted to H. ) each x E Go. Moreover, the modular character 4 Therefore qxj = ~ ( x for completely determines x. For suppose y is any element of G. Then for some integer m 2 1, yp'" E Go. In a representation of G corresponding to x, the linear transformation corresponding to y has eigenvalues wl, . . ., w,, say, in k, and ~ ( y=) w + . . . + 0,. Similarly x(ypm)= wt;" + * * . UP,". Since k has characteristic p, the Binomial theorem (Lemma 1.5A) shows that

+

(1)

x(y)P" = w?"

+ ... + 0;"= x(yP").

Now write y = y,y,,, where y, and y p , are the p-part and p'-part of y, respectively (see $1.5). Then yp'" = (y,.)Pm, and so corresponding to ( 1 ) we get x(yp,)Pm= ~ ( y p ' " ) .Hence the Binomial theorem shows that M Y ) - xbJP" = xbP7-

x(v"") = 0,

78

111


__

and so ~ ( y=) ~ ( y , . = ) 4(y,,,) because yp. E Go. Hence 4 completely determines x. In particular it follows from Corollary 1 of Theorem 2.3 that if 4 is affordedby an irreducible module W , then every kG-module V that affords Q, is isomorphic to W. Theorem 3.6 (Brauer and Nesbitt) Let U and V be kG-modules that afford the Frobenius characters D and T and the modular characters D , and T ~ respectively. Let S and T be the representations afforded by U and V . Then the following are equivalent.

(i) no = T , (on G O ) . (ii) For each x E Go, S(x) and T ( x ) have the same characteristic polynomial. (iii) U and V have the same kG-composition factors occurring with the same multiplicities. Remark The condition (ii) may be rephrased as saying that the eigenvalues of S(x) and T ( x )are the same (with the same multiplicities). Proof First assume that (i) holds. Let x E Go and put H * = ( x ) . By Theorem 3.5 there exist A-free AH-modules U o and V, such that 0, 2: U H and Po1: V,. By the definition of modular character, the modules U , @ A K and Vo K afford the ordinary characters (b0), and ( T ~ ) ! ,, respectively. Since these are equal by (i). Corollary 2 of Theorem 2.3 shows that U o @ A K 1: V, @I A K. Thus by Theorem 3.5, 0, z Po, and so U , = V,. Thus the restrictions of the representations S and T to H are equivalent, and (ii) follows. Next assume (ii) holds. Let xl, . . ., f be the irreducible characters of G over k. If mi (respectively, n,)is the multiplicity in U (respectively, V) of the irreducible constituent that affords the character xi (i = 1, ..., r), then D = mixi and T = nixi. Since both (ii) and (iii) only deal with the composition factors of U and V, there is no loss in generality in supposing that ) both U and V are completely reducible. By (ii) we know that a(x) = ~ ( xfor all x E Go,and so we conclude that D = T on all of G by the note above. Thus 0 = D - T = C(mi - ni)xi, and so by Corollary 1 ofTheorem 2.3, mi - n, = 0 in k for each i ; that is, p I (mi - n,) for i = 1, . . ., r. It remains to show that mi = n, for all i. We shall proceed by induction on the degree of D, and consider two cases. First, suppose that for some j, p X m j . Then p X nj, and so both mj and nj are at least 1. Since U and V are completely reducible, they have submodules U1 and V,, respectively, which afford the characters D - x j and T - x j . Then by the induction hypothesis it follows that mi = n, for each i. Second, suppose that p Imi and p [ n, for all i. Write in, = pm;. and n, = pni ( i = 1, . .., r). Then U and V have submodules U , and V,, respectively, which afford the

1

1

,

3.7

CARTAN INVARIANTS, NUMBERS, AND RELATIONS

79

characters c’:= Xmj xi and T‘ := En: xi. If S, and T2are the representations afforded by the modules U , and V,, then for each x E Go the characteristic polynomial for S(x) is the pth power of the characteristic polynomial of S 2 ( s ) ,and similarly, the characteristic polynomial for T ( x ) is the pth power of that of T,(.u). Hence the condition (ii) holds for S , and T,, and so by induction we conclude m:. = ni for all i. Hence mi = ni for i = 1, . . . , r and thus (ii) implies (iii). Finally, if (iii) holds, then (i) holds because the values of the characters depend only on the composition factors. EXAMPLE Let Q be a normal p-subgroup of G. Then each irreducible modular character 4 of G can be written in terms of some irreducible modular character I) of G/Q as follows: 4 ( x ) = I)(Qx)for all x E Go. For suppose W is an irreducible kG-module which affords 4. Then by Clifford’s theorem (Theorem 2.2A), W, is a completely reducible kQ-module because Q 4 G. Since Q is a p-group, the only irreducible kQ-module is the trivial module (corollary of Theorem 1.5). Therefore Q G Ker W,and W can be considered as an (irreducible) k(G/Q)-module.The modular character 4 afforded by the kG-module W is related to the modular character II/ afforded by the k(G/Q)module W by the equation +(x) = I)(Qx) ( x E Go).

3.7 Cartan invariants, decomposition numbers, and orthogonality relations In the present section we shall examine the relationship between the principal indecomposable kG-modules and the irreducible kG-modules. Our starting point is the material of $1.8. Write the kG-module kG as a sum of indecomposable submodules (principal indecomposables)

kG = U , @ . . . @ U,,

(1)

where the U i are enumerated so that each U iis isomorphic to exactly one of U 1, . . ., U , . Then by the Krull-Schmidt Theorem, each principal indecomposable module of kG is isomorphic to exactly one of U , , ..., U , . By Theorem 1.8 and its corollary, each U i has a unique maximal submodule Ui = U i n rad kG and each irreducible kG-module is isomorphic to exactly one of Ul/U;, . . . , U , / U i . We shall use the notation Ui

- Ccij(Uj/UJ)

( i = 1,

..., r).

j= 1

where cij denotes the multiplicity of U j / U Jas an irreducible constituent of U i . The nonnegative integers cij are called the Cartan inoariants of kG, and the r x r matrix r := [cij] is called the Cartan matrix of kG.

80

111


Note 1 It follows from Theorem 1.5 that r is equal to the number of classes of p’elements of G,and Theorem 1.9A shows that cij = i ( U j , Ui) for all i, j. The Cartan matrix r is only determined up to a permutation of its rows and columns (depending on the numbering of U , , . . ., U r ) . Now by Theorem 3.3 there exist A-free AG-modules V,, . . ., V, with the property that each irreducible KG-module is isomorphic to exactly one of . .., the KG-modules 6 @ A K (i = 1, .. ., s). Consider the kG-modules P,. Let c’, ..., (? be the (ordinary) irreducible characters of G over K such that Ci is afforded by 6 @ A K for each i. Then the restriction of to the set Goof p’elements of G is the modular character afforded by Pi, and Theorem 3.6 shows that the kG-composition factors of Piand their multiplicities are uniquely determined by c’. We write

vl,

ci

-C r

(3)

Pi

d,(U,/U;)

( i = 1, 2,

..., s)

j= 1

where d, denotes the multiplicity of the irreducible kG-module U,/U>as an irreducible constituent of Pi.The nonnegative integers d i i are called the decomposition numbers of kG, and the s x r matrix A := [d,] is the decomposition matrix. Note 2 It follows from Theorem 1.5 that s is the number of conjugate classes of G,and Theorem 1.9A shows that d, = i ( U j , q) for all i, j . The decomposition matrix A is only determined up to independent permutations of its rows and columns (depending on the ordering of the characters cl, . . ., and the modules U1,.. ., Ur). The first important relation between A and Theorem 3.7A With the notation above spose of A. In particular, r is symmetric.

r is the following.

r = AT A, where Ar

is the tran-

Proof Let fl, ..., f, be a complete set of orthogonal idempotents in kG such that U i = f , ( k G ) in (1) (see $1.6). Then by Theorem 3.4A these can be lifted to a complete set of orthogonal idempotents el, . . ., e, of AG with Pi = A for each i. Write a

(4)

ei(AG)- x u i j ?

( i = 1,

..., t)

j= 1

where a,, denotes the multiplicity of the irreducible KG-module V, 0 A K as K N e,(KG). Since KG is semisiman irreducible constituent of ei(AG) ple, each principle indecomposable KG-module is irreducible, and hence

3.7

81

CARTAN INVARIANTS, NUMBERS, A N D RELATIONS

isomorphic to one of the irreducible KG-modules Theorems 1.9A and 1.9C have alj = i( t$ 0 A K, e,(KG)) = i(e,(KG), Since %e, E

@A

58

A

K. Hence by

K) = dim,( Ye, 8 K).

5 is an A-free AG-module, this shows that aIj = rank,(Ye,) = dim, VjF1= i(Fl(kG), V j )

by Theorem 1.9C. Since zi(kG) =f,(kG) = U i , Note 2 above shows that aij = dji. If we substitute this in (4) and go over to the associated kGmodules, we finally obtain

- 2 PI- 2 1 s

S

Ui = Fi(kG)

r

dli

I= 1

dlid,( U,/U;)

j=l I=1

for i = 1, . . .,r using (3). Comparing this with (2) gives cij = & dIi d,, and so r = A' A as asserted. The inner product ( , ) G defined on the vector space Class,(G) of class functions defined from G to K was introduced in $2.4. Here we shall also be interested in the vector space Class,(G") of class functions 8: Go-+ K (where Go is the set of p'-elements of G).In particular, each modular character of G is an element of Class,(G"). In the notation introduced above there are r classes of p'elements of G, and so dim, ClaSS,(G") = r. We define the inner product ( , on Class,(G") by

where g:= (GI. Clearly ( , )c. is a symmetric, bilinear form. NOTATION Let U , , ..., U r be a full set of nonisomorphic principal indecomposable kG-modules [see (1) above]. Then the modular characters ql, . . ., qr afforded by these modules are called the principal indecomposable modular characters. Associated with each principal indecomposable modular character qi there is the irreducible modular character 4i afforded by U i / U i (i = 1, . . ., r). By Theorem 1.7A and the note in $3.6 we know that @', .. ., #f are all distinct and include every irreducible modular character of G. We shall also let c', ..., be the irreducible (ordinary) characters of G, where in the notation above ci is afforded by 0" K . Finally enumerate the conjugate classes of G, V 2 , . .., V, so that Vl, V 2 , ..., 5f?* are the classes of p'-elements. We shall put hi:= ( i = 1, 2, .. ., s), and if 8 is a class function on G (or GO),we shall write 8, to denote the value of 8 on gi. We shall also use the notation V, for the class {x- Ix E Vi}.

cs


82

Note 3 As we saw in the proof of Theorem 3.7A, each U i can be written where W, := e,(AG) is an A-free AG-module. Therefore the in the form modular character qi afforded by U i is the restriction of an ordinary characto Go. ter (afforded by

q,

w.)

Note 4 It follows from (2), (3), and (4) that , q'= x c i j @ ( i = 1 , ..., r ) j= 1

Idij,#+

[i=

on Go (i = 1, ..., s)

j= I

and X

o n G o ( i = l , ..., r )

qi= j= 1

since aij = dji. Theorem 3.7B (Orthogonality relations) With the notation above, the following hold.

(i) r is nonsingular (we shall write (ii) For all i, j = 1, . . ,, r we have

r-' = [cij]).

+pi, 4 J ) G 0 = Cij (q', rlJ)Go = cij

and (+', q j ) G o = ( q j , 4')G. = 6 , .

We have three matrices of character values on the classes in I = [t;j], the r x r matrix F := [4;], and the r x r matrix E := [qj]; and also the s x r decomposition matrix A = [d,] and the r x r Cartan matrix r = [cij].Using these we can rewrite the relations in Note 4 in matrix form: Proof

Go: the s x r matrix Z

(5)

E = TF,

Z = AF,

and

E = ATZ.

We also have the ordinary orthogonality relations 5

[:[:.=

g

Sij/hi

for i, j = 1, . . ., r

I= 1

by Theorem 2.4A. These can be expressed in, say, the form (6)

ZTZ= [g dij*/hi]= L.

3.7

CARTAN INVARIANTS, NUMBERS, A N D RELATIONS

Note that L is nonsingular and L-' = [hi S,,/g].

83

Since

FTrF= (AF)T(AF)= ZTZ = L

(7)

it follows that r is invertible, and so (i) is proved. Moreover from (7) and (5) and the symmetry of r we have the relations:

F(F-'T-'(FT)-')FT

(8 )

FL-'F'=

(9)

EL-

(10)

FL- 'ET = (FL- lFT)TT= I

~ E T= qFL- 1 F T p - T

=

=

r-'

r (the identity matrix).

These are precisely the matrix forms of the relations in (ii). The irreducible modular characters c$', . . .,6 and the principal indecomposable modular characters q l , . . ., f for G both form K-bases for Class,( G O ) . Corollary

Proof As we saw above dim, Class,(G") = r, so it is enough to show that the two sets of characters are linearly independent sets. In terms of the matrices defined in the proof of the theorem this is equivalent to the conditions that F and E are nonsingular; and the nonsingularity of F and E follows at once from (10). EXAMPLE Let F = ei(AG) be a principal indecomposable AG-module such that affords the principal indecomposable modular character qi, and let i,hi be the (ordinary) character afforded by (see Note 3). Then @' = qi on Goby the definition of modular character. We shall now show that i,hi = 0 on G\Go. Indeed t j i = tiji(' by (4) since uij = dji. Therefore by dji. On the other hand, by Note 4 above, Theorem 2.4A, ( $ i , $i)c = we have on Go that

If=

Therefore by Theorem 3.7B

Thus

and so I,$~(x)= 0 for all x in G\Go.

84

111 MODULAR REPRESENTATIONS A N D CHARACTERS EXERCISES

1. Suppose that G is a cyclic group of order n. Describe the ordinary irreducible characters, the modular irreducible characters, the principal indecomposable modular characters, the decomposition matrix, and the Cartan matrix. 2. If G is a group with a Sylow p-subgroup P of order p,' show that p' divides qi( 1) for each principal indecomposable modular character q'. [Hint: With the notation of the example above, calculate ($;, lp)p.]

3.8 Modular characters of p-solvable groups Recall that a group G is called p-soluuble for a given prime p provided it has a normal series G = Go z G1 z * - * 2 G , = 1 with each G i Q G such that each factor G i / G i + l is either a p-group or a p'-group. EXAMPLE 1 A solvable group is p-solvable (for any prime p). Conversely, since every 2-group is solvable (by a theorem of Feit and Thompson [2]), every 2-solvable group is solvable.

EXAMPLE

2 Any subgroup or factor group of a p-solvable group is also

p-solvable. Note 1 In any group G there exists a unique maximal normal psubgroup (usually denoted by O,(G)) and a unique maximal normal p'-subgroup (usually denoted by OJG)); and G is p-solvable iff for each proper normal subgroup N of G, at least one of O,(G/N) and O,,(G/N) is nontrivial. Note 2 Suppose that G is a p-solvable group with O,(G) = 1, then N I= OJG) contains C,(N). Indeed otherwise, by the definition of O,,(G), C,(N) has a p-subgroup Q # 1such that the normal subgroup C , ( N ) N / N of G / N contains the normal p-subgroup Q N / N of GIN. But €hen Q N = Q x N since Q c C,(N), and so Q = O,(QN). Thus Q is a characteristic subgroup of the normal subgroup QN and hence Q 4G. This contradicts the hypothesis that O,(G) = 1.

The object of the present section is to prove the following theorem (Theorem 3.8) about characters of p-solvable groups. The original proof was given by Swan [l] using results of Fong [l]; our proof is based on a modification due to Serre [ 13. Before stating the theorem we first shall prove a lemma which will be needed in the proof. If N is a normal subgroup of a group G, and T is a representation of N,then for each y E G we define the conjugate representation TY of N by P ( XI=) T ( y x y - l ) for all x E N (see $2.2).

3.8

MODULAR CHARACTERS OF

85

p-SOLVABLE GROUPS

Lemma 3.8 Let K* be an algebraic closure of K, let N be a normal subgroup of a group G, and let V b,e an irreducible K*N-module which affords the representation T . Assume that T is equivalent to Ty for each y E G. Put D := (det T ( x )I x E N } and define

H

:= ((y, c) E

G x Aut,,( V) I det c E D

and cT(x)c-' = T y ( x ) for all x

E

N}.

Then H is a finite subgroup of G x Aut,,( V) with the following properties. (i) There is an injective homomorphism N -+ H given by X H (x, T ( x ) ) ; we shall identify N with its image under this mapping. (ii) There is a surjective homomorphism H -+ G, given by (y, c ) ~ y , which maps N (embedded in H)onto N (embedded in G). (iii) There is a representation of H on V given by (y, C)I+C, and its restriction to N is equal to T . Proof It is readily verified that H is a subgroup of G x Aut,,( V), and that (i) follows immediately. The mapping in (ii) is clearly a homomorphism which maps N onto N ; the only nonobvious point is that it is surjective. However, for each y E G, T is equivalent to Ty, and so for some b E Aut,,( V ) , bT(x)b- = T y ( x ) for all x E N. Since K* is algebraically closed, det b = Pd for some fi E K* (and d := dim,. V), then (y, /3- ' b ) E H. This proves (ii). Moreover, the kernel of the homomorphism in (ii) is ((1, c)ldet c E D}. Clearly D is a finite group; let 1, = 1, A,, . .., 1, be the elements of D. If H , := {( 1, c) I det c = l}, then H, is a subgroup of H and ((1, c)ldet c E D } = H , u H , c2 u u H , c,, where ci is an element of Aut,.( V) such that det ci = li for i = 2, 3, ..., n. To prove the finiteness of H it suffices to prove that H , is finite. By Schur's lemma (see $1.2),

H , = ((1, c) I det c = 1 and c = y * 1 for some y E K*}

I

= ((1, c) c = y

*

1 and yd = l},

and so the order of H o divides d. Thus H , , and hence H,is finite. The assertion (iii) is obvious. Remark The proof above shows that the kernel of the homomorphism in (ii) is {( 1, c) I det c E D}. This has order dividing d I D 1, and hence dividing dINL Theorem 3.8 (Fong-Swan) Let G be a p-solvable group and $ an irreducible (Frobenius) character of G of degree d over k. Then there exists an irreducible ordinary character [ of G of degree d such that 1 = $.

86

111


Remark In general ( is not uniquely determined by I), nor does each irreducible ordinary character of G correspond in this way to an irreducible character over k. (See the exercises at the end of this section.) Recall that by our standing hypothesis (43.3). all fields involved are splitting fields for G,so I) and are absolutely irreducible. Proof We shall proceed by induction on the degree d of I).By 82.7 there is no loss in generality in supposing that K (and hence k) is a splitting field for XB“ - 1 where g := I G I, and we may also assume that G acts faithfully on a kG-module V of dimension d which affords I).Note that by the corollary of Clifford’s theorem (Theorem 2.2A), O,(G) = 1. Put N := O,,(G). Case I (d = 1) In this case G is abelian. Since O,(G) = 1, G is a p’-group, and so the result follows from Theorem 3.5. Case 2 (d > 1 and the irreducible constituents of VN are not all conjugate) In this case by Clifford’s theorem (Theorem 2.2A) there exists a subgroup S of G and an irreducible kS-module U of dimension less than d such that V 1: U c . By the induction hypothesis there exists an irreducible ordinary character x of S such that U affords 2. Then zG is an ordinary = $, and zG is irreducible because I) is (see the character of G such that note of 43.3). Case 3 (d > 1 and all irreducible constituents of V, are isomorphic) Let V, be an irreducible constituent of V, . Since N is a p’-group, it follows from Theorem 3.5 that there exists an A-free AN-module Wo such that Wo‘v V, . Put U o:= W, @ A K, and U,* := U oO KK*, where K* is an algebraic closure of K. Then U o is an irreducible KN-module (because V, is an irreducible kN-module). Since K is a splitting field for K N by the standing hypothesis of §3.3, this shows that U,* is also an irreducible K*N-module. Let T be the representation of N afforded by U,*,and x the character afforded by U,+.For each y E G, TY affords the character xy. On the other hand, Vo affords the character 2 and the kN-module Vo y affords the character Since V, y is an irreducible constituent of V,, V i 2: V, by hypothesis, so = xy. Hence x = x y by Theorem 3.5. This shows that T is equivalent to T y for all y E G (by Theorem 2.3 and its corollary). Hence with the notation of Lemma 3.8 there exists a group H in which N is embedded together with an extension of the action on U,*to H so that U,* is a K*H-module. Since H has order dividing g2d (see the remark following Lemma 3.8), the character afforded by the irreducible K*H-module U,+ is afforded by some irreducible KH-module U1, say. Now by Theorem 3.3 there exists an A-free AH-module W such that U1 z W @ A K. Since WN K ‘v ( U , ) Nit follows from Theorem 3.5 that W N Z VO.

x

2.

3.8

MODULAK CHARACTERS OF P-SOLVABLE GROUPS

87

Now define on the k-space Horn,(@,' V) an action of H by W J ( ~ - ~ ) : = [ ( w c - ' ) f ] y for each w E W , f € Hom,,(W, V) and (y, c) E H; clearlyf(Y*c) is k-linear, and it is a kN-homomorphism because for each x E N, ~

-~

~

-_

iiif(Y~ "x = [ ( w c - ' ) f ( y x y - ' ) ] y = [(wc- ' y x y - ' ) f l y ____~.

= [ ( w c - ' T Y ( x ) ) f ] y= [ ( w T ( x ) c - ' ) f ] y= ( w X ) p c )

by the definition of H (see Lemma 3.8). Moreover, it is clear that this action of H is k-linear, and so we have defined a kH-module structure on Horn,,( W,V). On the other hand we can define a kH-module structure on V using the homomorphism H -, G described in part (ii) of Lemma 3.8. We now claim that Horn,,( W,V ) 81, W 'v V as kH-modules under the mapping 8 given by ( J w)8 := W JIt is straightforward to verify that 8 is a kH-homomorphism, and it is injective because Wf= 0 for f # 0 implies w = 0 (since W is an irreducible kN-module). Finally, by Clifford's theorem (Theorem 2.2A), V, is completely reducible, and so by hypothesis it is a direct sum, say VN = V, 0 V, 0 * * * 0 V,, where each = W,.Then as kspaces Horn,,( W,V ) 'v 0=: Horn,,( W, is isomorphic to in copies of k because k is a splitting field. Hence dim, Hom,,(W, V ) = rn = dim V/dim W.Thus the k-spaces Horn,,( W,V )@ k W and V have the same dimension, and so 8 is a k-isomorphism. Since 8 is a kH-homomorphism, this shows that 8 is a kti-isomorphism as asserted. Note that dim, W > 1. Otherwise, V, is a sum of isomorphic one-dimensional kN-submodules and that implies N acts as a group of scalars on V. However N = O,,(G) 2 C , ( N ) by Note 2 above. Since G acts faithfully on V and N acts as a group of scalars on V, this would imply G E N, and so G is abelian. But d > 1 by hypothesis, so the corollary of Theorem 1.5 gives a contradiction. Hence dim, W > 1, and hence dim, Horn,,( FV, V) < dim, V = d. Now the kH-rnodule Horn,,( W,V) is irreducible since otherwise V would be reducible as a kH-module and hence as a kG-module, which is contrary to hypothesis. Since dim, Horn,,( ,'$I V) < d, the induction hypothesis shows that there exists an irreducible ordinary character i l of H such that Horn,,( W ,V) affords the k-character Tl of H. Again we have seen above that of H (afforded by W) such there exists an irreducible ordinary character i2 of H. Then c := c, 1, is an irreducible that W affords the k-character ordinary character of H such that V 'v Horn,,( W,V) 0,W affords the character cz = Finally, we shall show that the kernel of i2 contains the kernel L of the homomorphism H -.+ G defined in part (ii) of Lemma 3.8. Indeed we know from Example 2 of $3.3 that if U is any A H module, then Ker U/Ker U is always a p-group. However, by the remark following Lemma 3.8, I L I divides d I N I. Since N is a p'-subgroup of G, and d I IN I by the corollary of Theorem 2.4B. this shows that L is a p'-subgroup

v)

r2

~

el

rl r2.

111

88

MODULAR REPRESENTATIONS A N D CHARACTERS __

of H. On the other hand Ker I/ = Ker c , l2 (where V is considered as a kti-module) contains L by the definition of the action of H ; hence L C Ker l2 = Ker c. Thus we can consider ( as an irreducible ordinary character of G 2 H/L, and this completes the proof.

el

EXERCISES

1. Let G be an abelian group of order p" and i',iz. ..., ip"be the irreducible ordinary characters of G. Then show that = 5' for all i = I, 2, . . ., p". 2. Let G be the nonabelian group of order 6. Show that G has an irreducible character of degree 2 over a field of characteristic 2 and that G has no irreducible character of degree 2 over a field of characteristic 3.

ci

3.9 Notes and comments The connection between the ordinary and modular representations of a finite dimensional algebra was first recognized by Brauer (see Artin, et al. [l]) who developed the relation between the decomposition numbers and the Cartan invariants. This was later applied to the study of modular representations of a group by Brauer and Nesbitt in an important paper [l] which set the stage for the subsequent development of the theory. Our treatment of the relation between the ordinary and modular representations follows closely a paper of Dade [5]. For other references to this topic see Artin et al. [l], Curtis and Reiner [I], and Dornhoff [l]. Theorem 3.3 was first proved by Burnside when A is the ring of rational integers. The basic properties of modular characters were proved by Brauer and Nesbitt [1,2] and Brauer [ 1,8]. Note that our definition of modular characters is equivalent to the one defined by Brauer and Nesbitt. There are a number of papers on the representations of p-solvable groups. For further results on modular representations of p-solvable groups, see Berman [4], Dade [4], Fong and Gaschutz [l], Green and Hill [l], Isaacs [2], It6 [1,2], Nagao [2], Rukolaine [l], and Winter [l].

CHAl’TEli

1v

Blocks of Group Algebras

The theory of blocks of a group algebra attempts to classify the irreducible representations. This chapter presents several of the characterizations of the blocks and their properties. The main topics are blocks and their characterization in terms of ideals, irreducible modular and ordinary characters, irreducible modular and ordinary representations, and principal indecomposable modules. We then study the defect groups of a block and by making a further analysis of the Cartan matrix and the matrix of decomposition numbers we obtain bounds on the number of irreducible characters in a block of given defect. Finally, in the last section we discuss blocks of small defect.

4.1

Blocks

For the present section let k denote an arbitrary field (possibly of characteristic 0) and let G be a group. In $1.8 we considered the direct decomposition of the group algebra kG as a kG-module into principal indecomposable modules. We shall consider here the decomposition of kG as an algebra into a direct sum of indecomposable (two-sided) ideals. In many ways the situation here is simpler. We shall use it as a basic tool in the classification of kG-modules. 89

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BLOCKS OF GROUP ALGEBRAS

Definition A block of kG is a nonzero ideal B such that (a) kG = B 0 B’ for some ideal B’, and (b) B cannot be written as a direct sum of two nonzero ideals of kG. Since kG is Artinian, we can certainly write

kG = B , @

(1)

@ B,

as a direct sum of blocks. Associated with (1) we have the decomposition 1 =f,

(2)

+ ... +f,,

wheref,, . . .,f, is a complete set of orthogonal idempotents of kG with& E Bi and B, =f;:(kG) for each i (see $1.6). These idempotents are completely determined by the decomposition (l), and we call 4 the idempotent associated with the block B , . Note 1 The ideals B,, . . ., B, in (1) are the only blocks of kG, and so the decomposition (1) is unique. Indeed, suppose that E is a block of kG. Then since B and Biare ideals, BE, E B n Bifor each i. Thus from (1) we have

B = BE,

+ + BE, c ( B n Bi)0 .. @ ( B n B,) E B. * *

Hence B = (B n B , ) @ @ (B n B,), and since B cannot be written as a sum of two nonzero ideals, there existsj such that B 5 B, and B n Bi= 0 for all i # j . Since B is a block, there is an ideal B such that kG = B @ B’. Therefore Bj = BE, + B E , G ( B n E,) @ (B’ n B,) G B,, and so B, = ( B n B,)@ (B’ n B,). Since B, is not the sum of two nonzero ideals, this shows that B, = B n B, = B. Thus B,, . . ., B, are the only blocks of kG. Note 2 Let a

E

kG. Then by (2) we have

+ +

a = af1 *.. af, =f1a + ... +Ja. (3) Since B, is an ideal, afi andfia both lie in Bi and so by (1) and (3) we conclude a& =fia for each a E kG. Hence the idempotentsfi in (2) all lie in the center Z ( k G ) of kG. We shall call them the block idernpotents (or “central primitive idempotents”) of kG. Thus kG has exactly t block idempotents and these are given by (2).

Lemma 4.1 Letf,. decomposition (4)

.. .,f, be the block idempotents of kG. Then we have the Z(kG) =fi Z(kG) @

* * a

@f,Z(kG)

of the center Z ( k G ) of kG into a sum of ideals. Moreover, eachfiZ(kG) is

4.1

91

BLOCKS

indecomposable as a Z(kG)-module. If k is a splitting field for Z(kG), then for each i we have f,Z(kC)/rad[f,Z(kG)] N k. Remark Both Z(kG) and 1;. Z(kG) are commutative k-algebras (in the latter case 1;. is the unity element). Thus by Theorem 1.3A(i), their radicals consist of all their nilpotent elements. In particular, rad[& Z(kG)] = &Z(kG) n rad Z(kG) =1;. rad Z(kG). Proof Since 1;.Z(kG) z & ( k G ) = Bi,the decomposition (4) follows from (1) and (2). To show that each &Z(kG) is indecomposable as a Z(kG)module, it follows from 41.6 that it is enough to prove that the idempotentJ of Z(kG) cannot be written 1;. = e + e', where e and e' are idempotents of Z(kG) and ee' = 0. Suppose on the contrary that such a decomposition existed. Then Bi= e(kG) + e'(kG)is a sum of two ideals in kG (since e and e' lie in the center of kG). Moreover, e(kG) n e'(kG) = 0; indeed, if ea = e'a', where a, a' E kG, then ea = e(e'a) = 0. Thus Bi= e(kG) 0 e'(kG) is a sum of two nonzero ideals, contrary to the definition of a block. This shows that & Z(kG) is an indecomposable Z(kG)-module. Thus we are in the situation of Theorem 1.8 with R = Z(kG), and the & Z(kG) are the principal indecomposable R-modules. In particular, rad[&Z(kG)] is the unique maximal R-submodule of &Z(kG). Since R is commutative, this shows that& Z(kG)/rad[& Z(kG)] is a simple commutative algebra over k. If k is a splitting field for Z(kG), then the latter algebra has dimension 1 over k and so the final assertion of the lemma follows. Definition A central character of kG is a k-algebra homomorphism w : Z(kG) -+ k. [By our convention, since w is an algebra homomorphism it maps the unity of Z(kG) onto the unity of k ; hence w is surjective.]

Theorem 4.1 Let k be an arbitrary splitting field of the group algebra kG (possibly k has characteristic 0), and let fl, . . .,S, be the block idempotents of kG. Then kG has exactly t distinct central characters wl, .. . , 0,. These are characterized by the equations

1 wi(fi) = / O

if i = j otherwise.

Remark We shall call wi the central character associated with the block

J W *

Proof The composition wi of the projection Z(kG) -1;Z(kG) and the canonical mapping 1;. Z ( k G ) +fi Z(kG)/rad[A Z(kG)] is a k-algebra homomorphism. Since k is a splitting field for kG, w i :Z(kG) -,k by Lemma 4.1 and so is a central character. Since thefj are orthogonal, it is clear that ( 5 )

92

Iv


holds for wi. On the other hand, suppose that w is any central character of kG. If a E rad Z(kG), then a is nilpotent and so w(a)"' = w(am)= 0 in k for some integer m 2 1; hence w ( a ) = 0. Thus rad Z(kG) lies in the kernel of every central character. Moreover, since k is a splitting field for kG, Lemma 4.1 shows thatf;.Z(kG)= kh 0 rad[f;.Z(kG)] as k-spaces. Hence we have as k-spaces. Z(kG) = kfi 0 * . 0 kf, 0 rad Z(kG) (6) Now any central character w satisfies w(1) = 1. Since 1 =f, + +ft in Z(kG), this shows that w(&) # 0 for at least one j . Since in k we have

this shows that a(&)= 1 and w ( A ) = 0 for all i # j . Thus the two central characters w j and w agree on f l , . . . , A , and each contains rad Z(kG) in its kernel. Therefore it follows from (6) that w = oj. Hence wl, . . . , w , are the only central characters of kG. EXAMPLE 1 (Constructing central characters from irreducible characters) Under the hypotheses of Theorem 4.1 let { be an irreducible character of G over k. Let V be a kG-module which affords c and let T be the representation afforded by V . Let W l , . . ., W, be the conjugate classes of G, let cl, . . ., c, be the class sums in kG, and put hi := 1 Wi I ( i = 1, . . ., s). We know from $2.3 that cl, . . ., c, is a k-basis of Z(kG). Since ci centralizes kG, and k is a splitting field for kG, T ( c i )= yi 1 for some yi E k ( i = 1, ..., s). Taking traces we obtain h i t i = yi{(l), where ci is the value on %,. Now suppose that c(1) # 0 in k (so either k has characteristic 0, or k has characteristic p > 0, where p does not divide the degree of {). Then yi = hi ci/{( l), so we can define a k-linear mapping w : Z(kG) + k by w(ci) := hici/[(l) ( i = 1, . . ., s). Since T is a k-algebra homomorphism, it is clear that w is a central character of kG.

EXAMPLE 2 (The central characters and block idempotents when the characteristic is 0 ) Now suppose that K is a field of characteristic 0 and that K is a splitting field for KG. In this case rad K G = 0 (Theorem 1.3B). Since the nilpotent elements of Z(KG) all generate nilpotent ideals in KG, this means that 0 is the only nilpotent element in Z(KG), and so rad Z(KG) = 0. Then from Lemma 4.1, since K is a splitting field for KG,eachf;. KG 2 K , and so dim, Z ( K G ) = t (the number of block idempotents). However (see Example l), the class sums cl, . .., c, form a K-basis of Z ( K G ) , and so t = s (the number of conjugate classes of G). Furthermore KG has exactly s central characters by Theorem 4.1, and exactly s irreducible characters . . ., over K by Theorem 2.4A.

c',

cs

4.2 CLASSIFYING MODULES, CHARACTERS, AND IDEMFQTENTS

93

Using Example 1 (and the notation there), we can construct all s central characters ol, . . ., 0,. They are the K-linear mappings defined by

(I

oj(c,):=h,@p(l)

(7)

= 1,

..., s).

(Note that these are distinct because S

0=

1hlgc;*#

1=1

c hJ{C{* F

=g

I= 1

for i # j by Theorem 2.4A). It follows from Theorem 4.1 that the block idempotenth associated with lj,c,, say, with l j ,E K, w j satisfies the relations (5). Then writingfj = ,Ijl h,(i/Ci( 1) = hi,, and so by Theorem 2.4A we find we have S

S

S

Hence (in the notation of Theorem 2.4A)

W )15'I ,cI

f . = -J

9

for j

=

1, .. ., s.

I=1

Remark The concept of block can be extended as follows. Suppose we use the notation of 93.3 and consider the group algebra AG. First note that Z(AG) = Z(kG) since in both cases the class sums form a basis. The block idempotentsf,, . . .,f; of kG form a complete set of orthogonal idempotents of Z ( k G ) and so may be lifted to a complete set of orthogonal idempotents el, ..., e, ofZ(AG) (Theorem 3.4A). It can now be proved that the eiZ(AG) ( i = 1, . . . , r ) are uniquely determined by the fi and are indecomposable Z(AG)-modules. (See the exercise below). The ideals ei(AG) (i = 1, . .., t) cannot be written as sums of two nonzero ideals of AG [since the same is true of ei(AG) =fi(kG)] and are called the blocks of AG. However in general ei(KG)is not a block of K G , but splits further into a sum of nonzero ideals in KG. (This is the point of much of the next section.) EXERCISE

With the notation of 43.3, let f be a block idempotent of kG and let e be an idempotent of AG such that 2 = J Prove that eZ(AG) is indecomposable.

4.2 Classifying modules, characters, and idempotents into blocks We now return to the notation of 93.3 for the rest of this chapter. In the present section we shall describe how to classify the modules and characters of G with respect to the blocks of kG.

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94


(a) Indecomposable kG-modules lying in blocks

,’

Let V be an indecomposable kG-module, and suppose that f,, . , .,f, are the block idempotents of kG (see $4.1). Sincef,, . . . is a complete set of orthogonal idempot ents V = Vf, @ ... 0 V ’

as k-spaces

(see §1.6), and since thef, are all central, this is a direct sum of kG-modules. Because V is indecomposable, this means that for somej we have (1)

V’=

V

and

Vf,= O

for all i#j.

In this case we associate V with the block B, :=f,(kG)of kG, and say V lies in the block B,. Clearly all kG-modules isomorphic to V lie in the same block. This gives a classification of all indecomposable (and in particular, all irreducible) kG-modules into blocks. Note I It follows at once from (1) that if an indecomposable kG-module V lies in a block B,, then so does each of its irreducible constituents. Moreover, considering any block Bi = f , ( k G ) of kG as a kG-module, we see that the indecomposable components and the irreducible constituents of Bi all lie in the block Bi(we have Bif, = Bibecause the block idempotentf, is central). Note 2 Since 1 = f l + +A, it follows from (1) that if V lies in the block B,, then u = v l = vfi for all u E V.

The block of kG in which the trivial kG-module lies is called the principal block. This block will play an especially important role; partly because it is the block which is most easily analyzed. The relation (1) classifies indecomposable kG-modules into blocks using the block indempotents. For irreducible kG-modules it is often easier to use the following criterion, which uses the associated central character. Theorem 4.2A Let B be a block of kG and let w be the associated central character of kG. Then an irreducible kG-module V lies in the block B iff

(2)

uz = o ( z ) u

for all u E V , z E Z(kG).

In particular, the central character associated with the principal block of kG is given by C a x x H ax for all a x x E Z(kG). xcG

xeG

1

xcG

Proof Since k is a splitting field for kG (see $3.3), End,,( V ) = k. However each element z E Z ( k G ) acts as an element of End,,( V) on V, so there is a function $: Z ( k G ) + k such that uz = $ ( z ) u for all LI E V, z E Z ( k G ) .Clearly, $

4.2

CLASSIFYINGMODULES, CHARACTERS, AND IDEMPOTENTS

95

is a central character of k G ; we must prove that it is the central character associated with B. Letfbe the block idempotent of kG such that B = f ( k G ) . Since I/ lies in B, uf = u for all u E V (Note 2 above). On the other hand f~ Z(kG), and so uf= IC/(f)u for all u E V . Hence we conclude $(f)= 1. It now follows from Theorem 4.1 that w ( f ) = 1 and that w = i(/ as asserted. Finally, in the case that B is the principal block, we can take I/ as the trivial kG-module. Then

for all u E I/ and

a,x E kG. Thus for the principal block

by (2). Suppose that the irreducible kG-module V lies in the block B and affords the character 4 over k. Then with the notation of Example 1 of 94.1 we have from (2) that Corollary

for i = 1, . .., r, hiq5i = w(ci)4(1) where w is the central character of kG associated with B. In particular, if B is the principal block, then

hiq5i = h i 4 ( l )

for i = 1,

.. ., r .

(b) Characters lying in blocks Each irreducible character x of G over k is afforded by an irreducible kG-module, and this kG-module is uniquely determined up to isomorphism (Corollary 1 of Theorem 2.3). Since isomorphic irreducible kG-modules lie in the same block [see (a)], this means that x determines a unique block of kG. We shall say that x lies in a block B of kG if the irreducible kG-modules that affords x lie in B. Similarly, if 4 is the modular character afforded by an irreducible or principal indecomposable kG-module, then 4 determines the kG-module up to isomorphism (Theorem 3.7B). Again we shall say that 4 lies in the block B of kG if the kG-modules that afford 4 lie in this block. To extend this classification of characters into blocks to the ordinary irreducible characters of G over K we need the following lemma. Lemma 42 Let V be an irreducible KG-module, and choose an A-free AG-module W such that V 2: W 0 A K (see Theorem 3.3). Let W denote the

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96


reduction of W modulo R. Then the irreducible constituents of the kGmodule Wall lie in the same block of kG, and this block depends only on I/ and not on the choice of W .

Proof Let f,,. . ., S, be the block idempotents of kG. Since this is a complete set of orthogonal idempotents of Z(kG), ’Theorem 3.4A shows that there exists a complete set of orthogonal idempotents e , , . . ., e, of Z ( A G ) with Pi =f;. ( i = 1, . . ., t ) . [Note that Z ( A G ) = Z(kG) because in each case the class sums of G form a basis.] Thus we can write W = W e , @ . . * @ We,

as a direct sum of AG-modules. Since V is irreducible, W is indecomposable (Theorem 3.4B), and so for some j we have W = W e j and We, = 0 for all i # j. Thus W = W P j = WJ and 0 = W P , = W’ for all i # j. Hence from (I), all irreducible constituents of W lie in the block Bj :=&(kc). By Theorem 3.6 the irreducible constituents of W are uniquely determined up to isomorphism by V. Therefore the block B j depends only on V . It is now clear how to classify the irreducible KG-modules and the ordinary irreducible characters of G into blocks of kG. If c is an irreducible character of G over K, and c is afforded by an (irreducible) KG-module V , then choose an A-free AG-module W such that V c W 0 A K.Then c and V will be said to lie in the block B containing the irreducible constituents of W. The block B is well defined and uniquely determined by c (or V ) by Lemma 4.2. Remark This classification of characters into blocks does not really depend on the particular splitting fields K and k that we have chosen, but only on the characteristic p of k. Sometimes the terminology “p-blocks of characters is used to describe this classification without specific reference to the fields involved. Once again it is often easier to describe the classification of the irreducible characters of G over K into blocks in terms of the associated central characters. This is given in the following theorem. ”

Theorem 4 3 B Let W,, . . . , V, be the conjugate classes of G with class sums . . ., c, and put hi := I ( i = 1, . . ., s).

cl,

I%,

(i) To each irreducible KG-module V there is associated a central character w of K G such that (3)

uz = w(z)u

for all u E V , z E Z ( K G ) .

If V affords the character [, then w is the K-linear mapping determined by ~ ( q=)h i c i / c ( l ) ( i = 1, . . ., s), where is the value of on the class S i . (ii) In (i) the values w(ci) of the class sums all lie in A. The k-linear

ci

c

4.2

CLASSIFYING MODULES, CHARACTERS, AND IDEMPOTENTS

97

~

mapping Q: Z ( k G )-+ k defined by Q(ci) := w(ci)( i = 1, . . ., s) is a central character of kG. The KG-module V lies in the block of kG associated with Q. (iii) If [ and x are two irreducible characters of G over K , then c and x lie in the same block of kG iff

(4)

hici/c(l)= hixi/x(l)

(mod

IT)

( i = 1, 2,

.. ., s).

Remark See the corollary of Theorem 4.2D. Proof (i) This follows from Example 1 of $4.1. (ii) It follows from Theorem 2.4B that each hici/c(l)is an algebraic integer in K. and so lies in A (Theorem 3.2). Thus o(q)E A for each i, and 61 is properly defined. Since the restriction of o to Z ( A G ) E Z ( K G ) is an A-algebra homomorphism, it is clear that Q is a k-algebra homomorphism, and hence a central character of kG. Choose Was an A-free AG-module such that V 2: W 0 A K. Then from the action of G on V we have wci = o ( c i ) w for all w E W and each i. Thus wci = Q(c,)it, for all W E W and each i. Hence by Theorem 4.2A the irreducible constituents of W all lie in the block of kG associated with the central character a.By definition V lies in the same block. (iii) Let V and U be two irreducible KG-modules that afford the characters 4 and x, respectively. Let w and $ be the central characters of K G associated with V and U as in (i). Then by (ii) we know that V and U lie in the same block of kG iff 0 = $, and the latter is equivalent to (4). EXAMPLE Let B, be the principal block of kG. Then the ordinary character 1, lies in Bo and so it follows from (4) that an ordinary irreducible character c of G lies in Bo iff hiei/[(1) = hi (mod n) for i = 1, . . ., s. Similarly, it follows from Theorem 4.2A that an irreducible kG-module V lies in Bo iff

(5)

uci=hiu

forall U E V and i = l , ..., s;

that is, iff in the representation afforded by V , each class sum ci is represented by the scalar hil. This latter condition gives the following interesting result. Suppose that V affords the character x over k. Let 0 be a field automorphism of k, and let U be a kG-module affording the irreducible character x" (see the example of 92.7).Since the values hi( i = 1, . . ., s) lie in the prime subfield of k, they are fixed under 0.Therefore it follows from ( 5 ) that ifx lies in the principal block B , , then x" also lies in B, for each automorphism d of k. Warning The corresponding result is not true for other blocks of kG. As an example of the usefulness of classifying characters into blocks we have the following orthogonality relations due to Osima.

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Theorem 43C (Block orthogonality relations) Let c', c2, . . ., ["be the set of all ordinary irreducible characters of a group G that lie in a given block B. Suppose that x is a p'element of G and y is not a p'-element of G. Then we have

c

Ci(x)Ci(y) = 0.

i= 1

Proof Let c', C2, . . ., Cs denote the set of all ordinary irreducible characters of G and let 4', @, . . ., @ denote the set of all irreducible modular characters of G enumerated so that @, 42,. . ., #' are those lying in the block B. Then from the way in which the block of characters c' is defined, d, 4 j on Go, where d, = 0 if either i I m and j > n or we have Ci = i > m and j I n. Thus the decomposition matrix A = [dij] has the form

z=

n

r - n

(see $3.7). Let Wl, . . ., W, denote the classes of p'-elements of G and let 2 := [Cf] and F := [4f] denote, respectively, the s x r and r x r matrices of the values of the characters on these classes. Since y is not a p'-element, the orthogonality relations for ordinary characters give "'(y), CZ(y),

f

a

*

1

C"y)lz.

= 0.

But 2 = AF and F is invertible (see the corollary to Theorem 3.7B). Therefore we conclude [('(y), ..., Cm(y)]A l = 0. Multiplying this equation on the right by the column [4'(x), c$'(x), .. ., d"(x)], we obtain m

n

m

as required. (c) Block idempotents of K G lying in blocks of kG

In Example 2 of $4.1 we constructed the block idempotents of K G and showed how each block idempotent of K G is associated with one of the irreducible characters of G over K, and can in fact be expressed in terms of that character [Eq. (8) of $4.11. This gives a classification of the block idempotents of K G into the blocks of kG.The relation is described further in the following theorem.

4.2

CLASSIFYING MODULES, CHARACTERS, AND IDEMPOTENTS

99

Theorem 4.20 Let c', ..., ern be the irreducible ordinary characters of G over K lying in a block B, and let

be the associated block idempotents of K G [see (8) of 44.11. Then e : = e l + + em is the unique central idempotent of AG such that B = Z(kG). Moreover we can write e = p i c i , where

I;= s

m

..., s)

(7)

pi := 5 Cr'(l)c:*

and

pi = 0 whenever g i is not a class of p'elements.

9

(i = 1,

1=I

Remark An interesting consequence of (7) is that the right-hand side lies in A because e E AG. Proof Since Z ( A G ) = Z(kG), it follows from Theorem 3.4A that there is a central idempotentf of AG such thatfis the block idempotent associated with B. Let e l , . . ., e, be the block idempotents of K G (with the first m lying in B), and wl, . . ., o,the associated central characters of KG. Since el, . .., e, are the primitive idempotents of Z ( K G ) and f~ Z ( A G ) c Z ( K G ) , we can write f = e j , + ... + e j , , say, as a sum of certain e j . Then m JfJ = 1 if J E {jl,. . ., j.} and is 0 otherwise. On the other hand Q j ( J )= wj(f) = 1 iff the irreducible character c j associated with wi lies in B = f ( k G ) (Theorem 4.2B). It follows from our choice of notation that the latter occurs only for j = 1, .. ., m ;therefore we conclude n = m a n d f = e l + ... + e m .This shows that f = e, and hence e is the unique central idempotent of AG such that B = a( kG). Since cl, . .., c, is an A-basis of Z ( A G ) , e can be written in the form e= p i ci with pi E A and the values (7) of the coefficients come directly from (6). The final assertion in the lemma follows directly from Theorem 4.2C.

z:=

Corollary In order to show that two ordinary irreducible characters of G lie in the same block of kG it is enough to verify (4) in Theorem 4.2B only in the cases where V iis a class of p'-elements.

Proof With the notation of (i) ofTheorem 4.2B we know that c lies in the block B = u(kG) iff iii(2) = 1 (Theorem 4.1); that is, iff o ( e ) = 1 (mod n). The latter condition can be written in the form 1 = z p i w ( c i )= c p i h i c i / c ( l ) i= 1

i= 1

(mod n).

100

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Similarly S

1

CI P i h i X i l X ( 1 )

(mod

i=

iff x lies in B. Since pi = 0 whenever g i is not a class of p’-elements, the corollary follows.

(d) Blocks and principal indecomposable kG-modules Each block B of kG is a kG-module and a direct summand of kG. Thus it is possible to write B as a direct sum of principal indecomposable kG-modules; and it is evident that these principal indecomposables and all their irreducible constituents “lie i n ” B in the sense of (a) above. It is possible to describe this relation in a different way which is sometimes illuminating. Define the relation on the set ofall principal indecomposable kG-modules by writing U V iff there exist principal indecomposable kG-modules U = U , , U,, ..., U r n = V (for some rn2 1) such that for each i = 1, ..., m - 1 the modules U iand U i + ,have at least one irreducible constituent in common. Then clearly is an equivalence relation and the definitions in (a) show that U V implies that U and V lie in the same block. On the other hand, let U be a principal indecomposable kG-module and let I be the sum of all principal indecomposable kG-modules V such that V U . Since V .u xV as kG-modules for all x E G, therefore V U implies x V U ; hence X Ic I for all x E G. This shows that I is a (two-sided) ideal of kG. Since all V U lie in the same block, sayf(kG); it follows from (1) that I E I f ( k G ) c f ( k G ) . Now let I,, . . ., I,, be the ideals of kG that correspond in this way to the different (-)-classes of principal indecomposable kG-modules; then kG = I, + ... + I, . On the other hand, it follows from the definition of that principal indecomposable modules in different classes have no irreducible constituents in common; therefore Ii and I j have no common irreducible constituent when i # j. But then the Jordan-Holder theorem shows that (I, + ... + I j - ,) n I j = 0 for j = 2, ..., n, and so kG = I , 0 ... 0 I,,. However, as we saw above, each I j lies in a block of kG. Thus by the definition of block we conclude that each I j is a block, and V, V, iff V, and V, lie in the same block. Hence is equivalent to the relation “lies in the same block of kG.”

--

- -

--

-

-

-

-

4.3

-

Defect groups

In this section we shall show how to associate with each conjugate class of

G and each kG-block a class of conjugate p-subgroups of G called defect groups.

4.3

101

DEFECT GROUPS

First let % be a conjugate class of G.A p-subgroup D of G is a defect group of V if for some x E W , D is a Sylow p-subgroup of C,(x). Note I If D is a Sylow p-subgroup of C,(x), then y-' Dy is a Sylow p-group of y - ' C , ( x ) y = C,(y- ' x y ) . Since the Sylow p-subgroups of C,(x) are all conjugate, it follows that the set of defect groups of 5' is a single conjugate class of p-subgroups in G. In particular, the order pd := ID I is uniquely determined by V. We call d the defect of V. If Q is a p-subgroup in the center Z ( G ) of G,then Q is contained in each defect group of every class of G.

In working with defect groups we shall frequently use the following combinatorial lemma. Lemma 4 3 A Let V,, . . ., V, be the conjugate classes of G with class sums clr . . , , c, in kG. Let Q be a p-subgroup of G and suppose that the class g i has a defect group Di which is contained in Q.

(i) If z E G and C,(z) has no Sylow p-subgroup conjugate to a subgroup of Q, then for each j the set S, := {(x, y) E W i x W j 1 xy = z} has cardinality IS,/ = 0 (mod p). yI cI in Z(kG),where y1 = 0 if the class (ii) For each j we have ci c j = W l has no defect group contained in Q. Proof (i) Let P be a Sylow p-subgroup of C,(z). Since P centralizes z, it acts on S, by conjugation (x, y)" := ( u - l x u , u - ' y u ) for all u E P. Moreover, for each x E V i , P $2 C,(x) since the defect groups of g i are conjugate to a subgroup of Q. Hence each (x, y) E S, lies in a nontrivial orbit under P . Since P is a p-group, all nontrivial orbits have orders that are multiples ofp, and so IS, I = 0 (mod p) as asserted. (ii) Since c i c j E Z ( k G ) and c l r . . . , c, is a k-basis of Z(kG), we can write ci cj = y, c l . Since

we see that yI is the value of 1 S, I in k, where z because k has characteristic p.

E

V 1 Thus . (ii) follows from (i)

In order to define what we mean by the defect group of a block we shall need the result to be proved in the next lemma. However we first introduce a little notation. Let Q be any p-subgroup of G,and let W1, ..., W, be those conjugate classes of G that have subgroups of Q as defect groups. We denote by I, the k-subspace of Z ( k G ) spanned by the corresponding class sums C], ..., c,. Note 2 It follows at once from Lemma 4.3A(ii) that I, is always an ideal in the ring Z ( k G ) . If Q is a Sylow p-subgroup of G, then I , = Z(kG).

Iv

102


Lemma 43B Let f be a block idempotent of kG. Then there exists a p subgroup D of G with the following properties. (j)

f

ID*

(ii) If Q is a p-subgroup of G thenfc I, iff D is conjugate to a subgroup of Q. (iii) Let f = y,c,, where cl, c2, ..., c, are the class sums of V,, Vg2 , .. ., Vs,and let 9 ' denote the set of defect groups of all classes V, for which y, # 0. Then each subgroup in 9' is conjugate to a subgroup of D and the maximal elements of 9' are precisely the subgroups of G that are conjugate to D.

&,

Proof Let Vl, V z , .. ., V, be the conjugate classes of G and c,, c2, . . ., c, the corresponding class sums in kG. Since f~ Z(kG) we can write f= y, cI for some y, E k. Let o be the central character of kG associated with the blockf(kG). By Theorem 4.1 we have w(f)= 1, and so for some j , o(c,) # 0 and y, # 0. Let D be a defect group of the class V, . We claim that D has the required properties.

c;=

(i) Since o(fc,) = o(f)w(c,) # 0, fc, is not nilpotent, and so fc, $ rad[jZ(kG)]. By Lemma 4.1, j Z ( k G ) is a local ring, and therefore f ~ j Z ( k G=JZ(kG). ) But C, E I D , s o f ~ f ~ j Z ( kGG I)D . (ii) If D is conjugate to a subgroup of Q, then I D c I, by definition, and s o f e I,. Conversely, suppose thatfe I, for some p-subgroup Q of G.Since cl, ..., c, are linearly independent, this implies that c, E I, whenever y, # 0; in particular, c, E I,. Hence by the definition of I,, Q contains a defect group of V, and hence a conjugate of D. (iii) This is a consequence of (i) and (ii).

It is now clear how we may define the defect groups of a block B of kG.Let fbe the block idempotent associated with B. Then define the defect groups of B to be the p-subgroups D of G that satisfy the conditions of Lemma 4.3B. It is clear from part (ii) that the defect groups of B form a single conjugacy class of p-subgroups of G.A slightly simpler description of the defect groups is given by the following theorem. Theorem 4.3 Let B = f ( k G ) be a block of kG and let o be the associated central character of kG. Let W,, .. ., V, be the classes of G with class sums c,, .. ., cs, respectively, and let 9 denote the set of all defect groups of all %?, for which ~ ( c , # ) 0. Then each subgroup in 9 contains a defect group of B and the minimal elements of 9 are precisely the defect groups of B. Moreover, there exists a class V, of p'-elements of G such that the defect are the same as those of B. groups , of %?,

Proof

Let Q be a defect group of V, and suppose that o(c,) # 0. Then the

4.3 DEFECT GROUPS

103

proof of part (i) of Lemma 4.3B shows that fci is not nilpotent, and so f ~ f Z ( k G=fciZ(kG) ) C 1,. Hence by Lemma 4.3B, Q contains a defect group of B = f ( k G ) . This shows that each subgroup Q in 9 contains a defect group of B. On the other hand, the definition of D in the proof of Lemma 4.3B shows that D and its conjugates lie in 9, and so the defect groups of B all lie in D and are therefore the minimal elements of 9. To prove the final statement of the theorem we use Theorem 4.2D. From that theorem we know that the block idempotent f for B has the form f= pici, where jiE A and fii = 0 whenever Wi is not a class of p’-elements of G. This shows that the class W j chosen in the proof of Lemma 4.3B must be a class ofp’elements (since yj # 0), and the defect group D of B obtained there is the defect group of the class W j . EXAMPLE (Characters whose degrees are not divisible by p ) Suppose [ is an irreducible ordinary character of G such that p X c(l), and let B be the block ofkG in which c lies. Let w be the central character of K G corresponding to c.Then it follows from Theorem 4.2B (with the notation there) that w(ci)= h i c i / c (1) for i = 1, ..., s. Let P be a Sylow p-subgroup of G. If P is not a defect group of the class g i , then the Sylow p-subgroups of C,(X) (x E W i ) have orders less than I P I . Thus in this case p I hi; and so Q(ci)= 0 because p Xc( 1).Thus a(ci) = 0 except for the classes W i that have P a s a defect group (or equivalently, for which W i n C,(P) # 0).By Theorem 4.3 this shows that B has P as a defect group. Now suppose that c’ is a second irreducible ordinary character with p Xc’( 1).Let c’ lie in the block B’ of kG and let w‘ be the central character of KG corresponding to c‘. It then follows from Theorem 4.2B, the corollary of Theorem 4.2D, and what we have just shown that

(1)

B = B’

iff hici/c(l)= hici/c‘(l) (mod n)

for all i such that Wi is a class of p‘elements and W i n C,(P) # (zr. However, when gi n C,(P) # 0, then p X h i , and so we can cancel the hi in (1). Finally, since the characters are class functions, this yields the following very simple condition [valid when pXc(1) and p$c’(l)]: B = B’

iff c(x)/c(l) = c’(x)/c’(l) (mod n)

for all p’elements x E C,(P). In particular (taking c‘ = I,), c lies in the principal block of kG iff c(x) = c(l) (mod n) for all p’-elements x E C,(P). An analogous argument (appealing to the corollary of Theorem 4.2A in place of Theorem 4.2B) shows that if 4 and @ are two irreducible characters

Iv

104

BLOCKS OF G R O U P ALGEBRAS

of G over k, and their degrees are not divisible by p, then 4 and @ lie in the same block of kG iff for all p'elements x E CG(P).

4(x)/4(1) = @(x)/#(1)

In particular, 4 lies in the principal block of kG iff for all p'elements x E C,(P).

$(x) = 4(1)

4.4 Further analysis of the Cartan matrix and decomposition matrix Using the concepts of block and defect group we can now analyse more carefully the decomposition matrix A and the Cartan matrix r = AT A (see 83.7). We shall need the following construction of generalized characters. (i) Let V be a conjugate class of p'elements of G. Then there exists A: G -+ K which is an A-linear combination of irreducible ordinary characters of G such that A takes integer values, A(y) f 0 (mod p) for y E %', and A(y) = 0 for each p'element y $ W. (ii) Suppose that the Sylow p-subgroups of G have order p". If is any modular character of G then the function 8: G -,K defined by Lemma 4.4

+

is an element of Char(G).

Proof (i) Take x E W and choose Q as a Sylow p-subgroup of C,(x). Since x is a p'element, the subgroup H generated by Q and x has the form H = Q x (x). Let x', xz, ..., xn be those irreducible characters of H over K that have Q in their kernels. Since H/Q is abelian it follows from the corollary of Theorem 1.5 that each xi has degree 1 and n = 1 H : Q 1. We define p as the A-linear combination of characters of H given by A

(see Theorem 2.4B); and then i=1

is an A-linear combination of irreducible ordinary characters of G. Because xi for each i, the ordinary character relations (Theorem 2.4A) show that

Q G Ker

i

P(Y) =

n

Iiz1

xi(x- ')xi(x) = n

10

if y E QX

otherwise.

4.4

FURTHER ANALYSIS OF THE CARTAN DECOMPOSITION MATRIX

105

(x', ..., x" correspond to a complete set of irreducible characters of HIQ). This shows that 1= pG takes integer values. On the other hand, since x is the only p'-element in Qx we have z - 'xz E Qx only if z E C,(x). Therefore from the definition of induced class function ($2.5) and if y is not conjugate to an element of Qx. I(y)=0 Since 1 is a class function and x is the only p'-element in Qx, this proves (i). (ii) By Brauer's theorem on the characterization of characters (Theorem 2.6B) it is enough to show that OE E Char(E) for each elementary subgroup of G. In particular, it is enough to do this for each subgroup E of the form Q x S, where Q is a p-subgroup and S is a p'-subgroup since each elementary q-subgroup (for any prime q) can be written in this form. However, since S is a p'-subgroup the restriction l(ls of the modular character l(l to S is an (ordinary) character (see $3.5). Therefore, by the definition of 8 8, = p"-l(l(ls)E

where p'

I I

:= Q Ip".

EXAMPLE

E

Char(E),

This proves (ii).

Let p" be the order of the Sylow p-subgroups of G, and put

I I . Then p" divides the degree of each principal indecomposable modular character q. Indeed, let (,',C2, . . ., (,' and +', c $ ~.., ., @ be irreducible

g := G

z=

ordinary characters and irreducible modular characters, respectively. We can write (,' = d, @ on the set Goof p'-elements ( i = 1,2, . .., s), where the integers d, are the decomposition numbers. Now define 1as in Lemma 4.4(i) with V = (1). Since 1 is an A-linear combination of (,',C2, . . ., t;", we can write A = & ui@' on Gowith the uj E A. Then by Theorem 3.7B, since 1is 0 on all p'elements except 1, 1

S

Since p is not a unit in A, this shows that p" I q( l)1( 1). But p X A( 1) by the construction of 1,so p" I q(1). (See also Exercise 2 of 53.7.) NOTATION Let (,',(,', . . ., t;" be the irreducible (ordinary) characters of the group G over K, +2, . .., @ the irreducible modular characters of G, and A = [dij] the s x r decomposition matrix (so the d , are nonnegative integers and (,' = d, @ on the set Goof p'-elements of G). Let r = AT A be the Cartan matrix (ATdenotes the transpose of A). Moreover, suppose B is a block of kG and that the characters have been enumerated so that (,',

+',

z=

Iv

106

c2, . . ., r1and 4', 4'. ..., c#fl


are those characters that lie in B. This means

that A can be written

[;\01

3

where Al is an s1 x rl submatrix. We shall use A and A1 to denote the matrices obtained from A and Al by reducing the ehtries modulo IL. Note that since r = AT A, r is an r x r matrix of the form

where rl is an rl x r1 submatrix equal to AT A l . Note Let +'and @ be irreducible modular characters of G,and let qi and

# be the corresponding principal indecomposable modular characters. Suppose that # and @ lie in different blocks. Then from the above decomposi-

tion of r we see that c,, = c;, = 0 [where r = [ci,] and r-' = [c;,]]. Therefore by Theorem 3.7B, (#i, = ( q l , qJ)Go= 0. Similarly, if c' and 'C are irreducible ordinary characters of G lying in different blocks, then (c*, c')GO = 0. (Express ci and c' in terms of the 4' and the decomposition numbers.) Theorem 4R With the notation above,

(i) The s x r matrix Z has rank r (the largest possible), and A1 has rank r l ; (ii) det r I is a power of p ; (iii) There exist rational integers b,, such that forj = 1,2, ..., rl we have @=E L bjiCion Go.

I

Proof (i) Let Vl, V,,

..., Vr be the conjugate classes of p'-elements of

rl,ez,

. . ., to Go span the full r-dimensional space of k-valued class functions on Go.Indeed, for each i = 1, 2, ..., r we can construct by Lemma 4.4(i), Ai E Char(G) such that 4 ( Y ) # 0 for Y E g, A,(y) = 0 for all y E Go\%,. G. We first show that the restrictions of

Clearly the restrictions of I,, I,, . . ., Ir form a basis of the space of k-valued class functions on Go. Since each A* is an A-linear combination of cl, ..., c, this means that ..., p" also span the space of k-valued class functions on Go. Now let and [4;] denote respectively the s x r and r x r matrices of character values on Vl, Wz, ..., V, . By definition we have = A[+:];

cz,

r', e2,

[[a

[cf]

4.5

107

THE CHARACTERS IN A BLOCK OF GIVEN DEFECT

r;]

span hence [Ifi] = A[$;]. But from what we have just shown, the rows of an r-dimensional space. Hence 1511, and consequently A as well, has rank r. Since A has rank r, its columns are linearly independent, hence so are the columns of A,; so A, has rank r , . (ii) For each modular character 4i, we define 8' as in Lemma 4.4(ii) taking II/ = 4i.Since 8' e Char(G), (Oi, 0') = p2"(4', @ ) G o is an integer. By Theorem 3.7B we know that r-' = [c\J, where (#, 4')Go= cij. Hence if L denotes the r x r matrix [(@, then LT = p 2 " . 1, and so det L det r = p2". Since det L is an integer, 1 det r I is a power of p. (iii) By (i), A, has an r , x r , submatrix A. with det A. # 0; we claim that det A. = f 1. Indeed otherwise there exists a prime q # p such that q 1 det do.But then the rank of A(mod q ) is less than or equal to r - 1, and so the rank of the r x r matrix r = AT A (mod q ) is less than or equal to r - 1. But by (ii) det r f 0 (mod q), so we have a contradiction. Hence det A. = 1 and A; = [b,,] has integer entries (by Cramer's rule). Without loss in generality we may suppose that A. consists of the first rl rows of Al. Then we have

on Go for i = 1, 2,

..., r,, which yields i= 1

on Go for j = 1, 2, .. ., r , as required.

4.5

The characters in a block of given defect

Recall that a block of kG has defect d if pd is the order of its defect groups. In the present section we shall show how to characterize the defect d in terms of the degrees of the irreducible characters lying in the block, and how to find an upper bound for the number of ordinary (and modular) irreducible characters in a block of given defect. Theorem 4 5 A Suppose that the Sylow p-groups of G have order pm, and let E be a block of defect d of kG. Then

(i) for at (ii) for at

pm-d 1 c( 1) for every irreducible ordinary character 5 lying in E, and least one of those characters p m - d + l Xc(1); pm-d 1 &( 1) for every irreducible modular character 4 lying in E, and least one of those characters pm-d+ X 4( 1).

Iv

108


[',

be the irreducible ordinary [', . . ., yS1 and 4', +2, . .., Proof Let characters and the irreducible modular characters, respectively, lying in B. Let qi be the principal indecomposable modular character associated with c # ~ ~ ( i = 1, 2, ..., rl). Then from 53.7B and Theorem 4.4 ri

(1)

ci= 11 d i j @

on Go (i = 1, 2,

..., sl)

j=

(3) where the dij and bij are integers. It follows from (1) and (3) that the largest power of p dividing all ci( 1) (i = 1,2, . . ., sl) is equal to the largest power of p dividing all 4i(l)(i = 1, 2, ..., rl). Hence (i) implies (ii) (and conversely). Thus it will be sufficient to prove (i). Using the notation of Theorem 4.2D, let e be the central idempotent of AG such that P is the block idempotent associated with B. Then Theorem 4.2D Pici, where shows that e =

Cf=

1

SI

(4) and Pi = 0 whenever W i is not a class of p'elements. Suppose that Vi is a class of p'-elements. Then so is Wi+ , and hence by (4), (l), and (2) we obtain

Since pmI q j ( 1) by the example of $4.4, this shows that q'( 1)/g E A for each j, and so Pi = C;L uj &* for suitable uj E A. In particular, since Z # 0, there exists a class Wi of p'elements with PI # 0, and some j such that +{* # n A ; it then follows from (3) that there exists some 1 such that cf+ I I A(1 I1 Isl). Choose this class W i so that the defect of W i is as large as possible. Then Lemma 4.3B(iii) shows that W i has the same defect a! as B has, and so the largest power of p dividing hi:= IW i 1 = IWi* I is pm-d from the definition of defect group of W,. Since hi &/['( 1) E A (Theorem 2.4B) and # pA, we conclude that $ [ I ( 1). Finally we show that pm-d I [(l) for each irreducible ordinary character lying in B. By Theorem 4.3 we may choose j so that o ( c j ) # 0 and the defect groups of the class W, are the defect groups of B ; this implies pm-dlhJ but pm- *+ X hi. However, since [ lies in B, Theorem 4.2B shows that w(cj)= hjcj/[(l), and so hjt,/c(l) gl nA. Since cj E A by Theorem 2.4B, and f$

[:*

4.5


109

p"-d I hi, we have p m - d I c(1). This completes the proof of (i) and so the theorem is proved. EXAMPLE Suppose that 4' is an irreducible ordinary character of G such that p '+c(,1). Then it follows from Theorem 4.5A that 5 lies in a block B of defect rn where p" is the order of the Sylow p-subgroup of G. The defect groups of B are the Sylow p-subgroups of G. Such a block is sometimes said to havefull defect. In particular, taking c = lGwe see that the principal block of kG has full defect.

In order to estimate the number of irreducible characters in a block of given defect we shall use the following result. (i) Let q l , q2, ..., qr be the principal indecomposable modular characters of G. Suppose 8 E Char(G) has the property that 8(x) = 0 for all x E G\G". Then there exist integers a,, a 2 , .. ., a, such that 8 = ai qi on Go. (ii) Suppose that the Sylow p-subgroups of G have order p" and let B be a block of defect d of kG. Let [ be an irreducible ordinary character lying in B. Define the class function 8: G + K by

Lemma 4.5

rl

Then 8 E Char(G), and if pm-d+ X[( l), then (l/p)B rj! Char(G). Moreover, 8 is a Z-linear combination of irreducible characters lying in B. Proof (i) By the corollary of Theorem 3.7B, q', q2, .. ., qr is a K-basis for the vector space Class,(G"). Thus there exist aie K such that 8 = C;= Q~qi. It remains to show that the ai are integers. Let e l , c2, . . ., p and &', &', . . ., $f be the irreducible ordinary characters and the irreducible modular characters, respectively, of G. Then by Theorem 3.7B and Theorem 4.4(iii) there exist integers bjl such that for each j aj =

r

S

i=1

1=1

C ai(qi,4 j ) G 0= (8, 4i)G0= C bjl(8, cl)Go .

cf)co

However (8, = (8, which is an integer since 8 E Char(G). Therefore aj is an integer as asserted. (ii) Let W,, V,, ..., Wr be the classes of p'elements of G and put hi := I%', I and g := I G I. Then for each irreducible ordinary character c' of G we have

where o is the central character of K G associated with [ (see Example 2 of

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110

BLOCKS OF G R O U P ALGEBRAS

54.1). Now p"-dO E Char(G) by Lemma 4.4 because the restriction of [ to Go is a modular character; therefore n,:=(p"-dO, C') is an integer. Also pdC(l)/g E A because pm-dI c(l),and o ( c ) - h'C,/C(l) and CIS lie in A; so by (l), n,/p"-d E A. This shows that ,"-'I and so 0 = f i Z l (0, C1)C1 = (n,/p"-')C' E Char(G). On the other hand, if pmWd+' X [(l) then (l/p)0 4 Char(G). Indeed, othera, qi on Gofor some integers a,. But wise it follows from (i) that (l/p)0 = by the example of 54.4, pmlqi(l)for each i, and so p"lO(l)/p. Since 0(l) = pdC( l), this implies pm-d+ 1 c( 1) contrary to the choice of C. Finally, suppose that the irreducible character c' does not lie in the block B. Then by the note preceding Theorem 4.4 we know that (C, = 0. Hence (0, C') = p"-*(C, C1)Go = 0, and so C1 is not one of the constituents of 0. This completes the proof of (ii).

i,

'

Theorem 45B Suppose that the Sylow p-subgroups of G have order p". Let B be a block of kG of defect d. Then

(i) at most t p Z d+ 1 irreducible ordinary characters lie in B; (ii) if C is an irreducible ordinary character lying in B, then for d I2 we have p"-d l ( ( 1 ) but p m - d + l s < ( l ) ,and for d > 2 we have p"-' $(,(l); (iii) an irreducible ordinary character C of G lies in a block of defect 0 if p" 1 C( 1) and in a block of defect 1 if pm- I C( 1) and p" X C( 1).

'

Proof (i) Let C1, C2, . . ., yS1 be the irreducible ordinary characters that lie in B; and for i = 1, 2, . . ., sl, define 0' E Char(G) as in Lemma 4.5, taking C = C'. Then for i, j = 1, 2, . .., s1 we have pd(ei, r j ) G = (ei,

ej),

=P

2d

(C i, cj )GO

and using Theorem 2.4B we see

Since Ci and 'C lie in the same block

h&/ci( 1) = h,C//C'( 1) (mod K )

for all

t

by Theorem 4.2B. Therefore

for all i, j, 1 = 1, 2, . . ., sl. [Note that both sides of (2) lie in A by (l).] Using Theorem 4.5A we may assume that the Ci have been enumerated so that pn-d+l XC'(1). We claim that (el, C') .$ K A . Indeed otherwise, p"-d(O', t;')/['(l) E K A because p m - d + l $C'(l). Then by (2) we have

4.5

111


p"-d(B1, cj)/cj(l) E n A ; and so p divides the integer ( e l , cj) for . . ., s1 because p"-d I cj( 1) by Theorem 4.5A. But by Lemma 4.5 we have

j = 1, 2,

j =1

and so we conclude ( l/p)O1 E Char(G). This contradicts Lemma 4.5, and we have (el, e l ) q! nA. Since (el, [ l ) $ n A , (2) shows that for each j = 1, 2, ..., slf pm-d(81,[j)/cj( 1) $ nA; in particular, aj := (el, cj) # 0. Then each of the lying in B is a constituent of 8'. Hence we conclude that

cJ

pdal = pd(O1, cl) = (el,

Sl

el)

=

C (el, ~

1af. 81

j

=)

i= 1

~

i= 1

+

+

Since the right-hand side is at least (sl - 1) a:, therefore 0 2 (s, - 1) + (a, - +pd)'. Hence s1 - 1 I $pZd and this proves (i). (ii) Now we turn to (ii). If s1 = 1 this follows at once from Theorem 4.5A; so suppose that s1 > 1. Suppose p"-" I [j(l), where n < d. Then since pm-"(O1, cJ)/[j(l) E A, we conclude as above that (el, cj)c = 0 (mod p). Since (el, (j)G = (el, cl)c by definition, (2) now yields (with 1 = j , i = 1)

a 2, - pdal = (sl - 1) - $pZd

0 = ~ " - ~ ( 8 'cj)/cj(l) ,

(mod n)

and so p d - " + l I (0' cj). As we saw in the proof (i), (@, c') = (el, cj) # 0, and so 8' = ZibiCi, say, where b, 2 1. Since b: = (OJ, ej), = p Z d ( c j , cj)GO IpZd(cj,[j) = p2*, this shows that bj < pd. But bj = (Oj, cj), and so pd-"+l < p d ; hence n 2 2 if s1 > 1. Thus n < d implies n 2 2, and hence d 2 3. Since m - n < m - d, this proves (ii). (iii) This follows immediately from (ii).

ELl

Corollary 1 With the notation of the theorem, at most 52d + 1 irreducible modular characters lie in B.

Proof

This is a consequence of Theorems 4.4(i) and 4.5B(i).

Corollary 2 If is an irreducible ordinary character of G such that [(x) = 0 for each p-element x of G with x # 1, then c lies in a block of defect 0. Proof

Let P be a Sylow p-subgroup of G with I P I = p"'. Then

is an integer and the conclusion follows from the theorem.

112

Iv


EXERCISES

1. Let [ be an irreducible ordinary character of the group G. Show that [ lies in the principal block of kG iff ([, lG)co # 0. 2. In contrast to Theorem 4.5B(ii), if d > 2 then a block of defect d may have an ordinary irreducible character lying in it such that pm-*+' I [(l); give an example for each d 2 3. [Hint:If G is a nonabelian p-group of order pd and d 2 3, then G has an ordinary irreducible character of degree p.]

4.6

Blocks of small defect

The situation where a block has defect 0 is especially simple to describe. Theorem 4 b A Let B be a block of defect 0 of kG. Then there is exactly one irreducible ordinary character [, one irreducible modular character 4, and one principal indecomposable modular character q of G lying in B. Moreover, 4 = q and

for X E Go otherwise. Proof Theorem 4.5B(i) shows that B contains a single irreducible ordinary character [ and Corollary 1 of Theorem 4.5B shows that B contains a single irreducible modular character 4, and so [ = 64 on Gofor some integer 6 > 0. The decomposition matrix and the Cartan matrix of B are the one-byone matrices A 1 = [6]and rl = [b2], respectively. By Theorem 4,4(ii), I rl I is a power of p , and so 6 is a power of p. But Theorem 4.5A(ii) then shows that 6 = 1 and hence [ = 4 on Go, = [13, and so 4 = q. Since [ is irreducible, we have using Theorem 3.7B that ([, = (4, q)Go = 1. Hence

and so [(x) = 0 for all x E G\G". This proves the theorem. The situation where a block has defect 1 is considerably more complicated. However this case has been analyzed (unlike the cases of defects greater than 1)by Brauer. In order to state the result of Brauer on blocks of defect 1, we define the concept of p-conjugate characters. Let g I = I G I = p"g', where (p, 9') = 1, and let E and F be the extensions of the rational field generated by a primitive gth root of unity and a primitive g'th root of unity, respectively. Then K 2 E 2 F and by 62.7, E is a splitting field for G. The Galois group Gal(E/F) of E over F has order p"(p - l), and it can be considered as a group of automorphisms of the F-algebra E. If [ is a character of G over E, then for any o in Gal(E/F), (? is also a

4.6 BLOCKS OF SMALL DEFECT

113

character of G over E (see $2.7). Two characters c and x are called pconjugate if ( = x" for some in Gal(E/F). Clearly ru(x) = ((x) for all p'elements x of G and hence two p-conjugate irreducible ordinary characters lie in the same block of G and they have the same irreducible modular characters as constituents. Let B be a p-block of G and let A, be the decomposition matrix corresponding to B. Let c', c2, ..., be a full set of representatives of the p conjugate classes of characters lying in B. The u-rowed submatrix AT of A, corresponding to the characters is called the reduced decomposition matrix of B. Brauer [2] has proved the following result on blocks of defect 1.

ci

Theorem 4.6B Let B be a block of defect 1, c', (', ..., the irreducible ordinary characters, and A, the decomposition matrix with A; being the reduced decomposition matrix. Suppose ti is the number of p-conjugates of and di the degree of c'. If p" is the order of the Sylow p-subgroups of G, then

ci

(i) the irreducible ordinary characters lying in B can be partitioned into two disjoint classes S and T such that tidi E t , d , (mod p") for all (' in S and l i d i = - t , d,, (mod p") for all in T ; (ii) if and [ J are not p-conjugate and belong to the same partition class in (i), then they have no irreducible modular character in common; (iii) each coefficient of A, is 1 or 0; (iv) each column of AT contains exactly two nonzero coefficients, one in a row corresponding to a member of S, and the other corresponding to a member of T.

ci

ci

The proof (which we shall not give) may be found in Brauer [2]. However see Theorem 8.5 of Chapter 8 for a special case. We conclude this section by giving examples of how the theorems of this chapter can be applied to deduce the existence of irreducible characters of certain degrees, and an elementary way in which they help classification of simple groups. EXAMPLE 1 Suppose G is a group of order g = p"g0, where rn = 1 or 2 and p $ g o . Suppose further that g o has exactly two different prime divisors 41 and 4 2 . Then G has nontrivial irreducible ordinary characters (in the principal block of k G ) whose degrees divide go and are powers of 4, and qZ, respectively. For, suppose that c' = l,, (', . . ., yS1 are the irreducible ordinary characters of G lying in the principal block Bo of kG. By the example of $4.5 the block B, has defect m. Since m I2, Theorem 4.5B shows that pXc'(1) for i = 1, 2, ..., s,. However, by Theorem 4.2C we have

114

Iv


for each y E: G\Go (since c'(1) = ( ' ( y ) = 1); in particular,s, > 1. Ifq, and q 2 are the prime factors ofg,, then it follows from (1) that for somej (2 5 j 5 s,) we have q2 SCj(1). However ci( 1) 1 g by Theorem 2.4B; so we conclude from the hypothesis on g that ( j ( 1) is a power of q , dividing go. Similarly there is a character 5' lying in Bo whose degree is a power of q 2 . EXAMPLE 2 It follows from a classical theorem of Burnside that a simple group whose order is divisible by at most two primes is cyclic of prime order (see Curtis and Reiner [ 11). On,the other hand, there are eight known simple groups whose orders are divisible by exactly three distinct primes. Under further restriction it is possible to make a partial classification of these latter groups. Let G be a simple group whose order g is divisible by exactly three primes and which has one of the forms 3p"qn, 4p"qn, 5p"qn, or 7p"q" (p, q primes). Suppose further that m = 1 or 2. In these cases it follows from Example 1 that G has a nontrivial irreducible ordinary character whose degree divides 3, 4, 5, or 7, respectively. Since G is simple, the character is faithful. It is not difficult to show that no simple group has an irreducible ordinary character of degree 2, and so in the respective cases G has a character of degree equal to 3, 4, 5, or 7; equivalently, G is isomorphic to a subgroup of G a d , C),where d = 3, 4, 5, or 7. The (finite)subgroups of these general linear groups have all been classified and consequently have lead to a classification of the simple groups of these types.

EXERCISES

1. Suppose that the group G has a faithful irreducible ordinary character of degree 2. Prove that either G # G (the commutator subgroup) or Z ( G ) # 1; in particular, G is not simple. [Hint:2 1 I G I by Theorem 2.4B. Show that if z is an element of order 2 in G, then either z q! G' or z E Z(G).] 2. Let G be a group of order pg, ,where (p, go) = 1, and let ko be the prime subfield of the field k. Show that each irreducible (Fr0benius)character of G over k lying in the principal block of kG is afforded by some k,G-module. 3. Show that the number of irreducible ordinary characters lying in a block B is equal to the number of irreducible modular characters lying in B iff B is a block of defect 0.

4.7 Notes and comments

There are purely ring theoretic proofs available for some of the results of this chapter. For example see Michler [1,2,3]. The central characters of kG are the irreducible representations of Z(kG) and easily generalize for any finite dimensional algebra over a field. In this connection see Michler [l].

4.7

NOTES AND COMMENTS

115

Theorem 4.2B was proved by Brauer and Nesbitt [l] and the Theorems 4.2C and 4.2D are essentially due to Osima [l]. The bound ipPzd+ 1 for the number of irreducible ordinary characters in a p-block of defect d was obtained by Brauer and Feit [l]. It is conjectured that there are at most pd irreducible ordinary characters in a p-block of defect, and this has been shown to be true in case the defect groups are cyclic (Dade [2]). The blocks of defect 1 were analyzed by Brauer [2] and the information was applied to determine the simple groups of orders 4p"q' ( a I2) and 3paqb (a I 2). The blocks of defect 1 bear a close resemblance to the blocks with cyclic defect groups (which will be studied in Chapter 8). For further results in this direction, see Brauer [5,8], Brauer and Fowler [l], and Carleson [l].

CHAPTER

v

The Theory of Indecomposable Modules

Further applications of modular representation theory depend on the work of Green [ 1,2], which makes an incisive analysis of the indecomposable modules of a group algebra. The purpose of this chapter is to develop this theory of the structure of indecomposable modules. The topics covered include the following: relatively projective modules, vertices and sources, absolutely indecomposable induced modules, the degrees of indecomposable modules, the relation between vertices and defect groups, and the restriction of indecomposable modules to subgroups. Finally, we use some of these results to prove a theorem of Brauer and Feit on the existence of normal abelian subgroups in linear groups.

5.1 Relatively projective modules The Frobenius reciprocity theorem (Theorem 2.5A) is a basic tool in the theory of ordinary characters. Unfortunately, in the modular case the full reciprocity relation no longer holds. However as Green first showed in [ 1,2], something can be salvaged, and this is the theory of vertices and sources which we shall look at in the next few sections. Let G be a group and A a Noetherian integral domain. The cases in which we shall be interested later will be where A is a p-adic algebra of the type we considered in $3.3 or a field. Consider a short exact sequence 1

8

o+u+v+w+o I16

5.1

117

RELATIVELY PROJECTIVE MODULES

of AG-modules (so K e r f = 0, I m f = Ker g, and Im g = W). Recall that (1) splits if there exists an AG-homomorphism h: W + V such that hg is the identity 1, on W. An equivalent condition is that (1) splits if V = U , 0 W,, where U , := Im f (which is isomorphic to U ) and W, is a complementary AG-submodule (which is isomorphic to W). NOTATION If U and V are AG-modules and I/ can be written as a direct sum V = V, 0 V' of AG-submodules with U z V,, then we write U I V (as A G-modules).

Definition Let W be an AG-module and H a subgroup of G. Then W is called H-projective if, whenever there is an exact sequence of the form (1) with W as the right-hand AG-module such that under restriction to H the exact sequence f

e

0 + UH + VH --+ WH + 0 (2) splits, then (1) also splits. In other words, the existence of an AHhomomorphism h: W + V with hg = 1, (the identity on W) implies that there is an AG-homomorphism with the same property. The following characterization of H-projectivity is due to Higman [ 1,2]. Theorem 5.1 Let H be a subgroup of the group G and let x 1 = 1, x 2 , .. ., x , , be a right transversal of H in G. Then each of the following properties of an AG-module W implies the others: (a) W is H-projective; (b) YI (wH)"; (c) W I U" for some AH-module U ; and (d) there exists f~ End,,( W) such that C;=

, ,

, x; ' f x i = 1,

.

Remark In (d), x; ' f x i should be interpreted as the endomorphism of W given by w H C;= wx; f x i . Proof Assume that (a) holds; we shall prove (b). By definition (W,)" = WH0 AH AG and each element of ( W,)" can be written uniquely in the form w i0 xi for suitable wi E W. Define g: ( WH)' + W by

Clearly g is a surjective AG-homomorphism. Moreover the exact sequence 0 + Ker g + (WH)' + W + O (3) splits as a sequence of AH-modules ; the required AH-homomorphism h: W + ( W,)" is given by wh := w 0 1. Since W is H-projective by (a), we

v

118

THE THEORY OF INDECOMPOSABLE MODULES

conclude that (3) splits as a sequence of AG-modules. Thus (W,,)G = Ker g @ W,, where Wl z W. Thus W I (W,,)'and (b) is proved. Trivially (b) implies (c). Assume (c) holds; we shall prove (d). Without loss in generality we may assume that UG = W @ V for some AG-modules U and V. Let h denote the projection of UG onto W with kernel V, and define g: UG+ UG by I

"

\

( ~ l w i @ x i ) ~ ' = w l @ x l= w l @ l

(with each wi E W). It is readily verified that g is an AH-homomorphism. Moreover for each u E UG we can write u = &l u, @ x, for some u, E W and then n

n

n

n

n

c;=

since x,x; # H if i # j. Thus x; lgx, acts as the identity on UG. Finally since h is an AG-homomorphism, for all w E W we have n

I

n

\

Thus if we take f as the restriction of g h to W, then f E End,,( W) and satisfies (d). To complete the proof, assume that (d) holds; we shall prove that (a) holds. Let PI

O+U-,V+W-,O be an exact sequence of AG-modules which splits as a sequence of AHmodules; and let h: W + V be an AH-homomorphism for which hg, = 1,. We claim that h,: W + V defined by hl I= x; 'fhx,, where x; !x 'i = lw, is an AG-homomorphism such f E End,,( W )satisfies this will prove (a). Indeed, since g, is an that hlgl = 1,; AG-homomorphism,

c;=

c;= \

n

n

since Im f r Wand hg, = lw . On the other hand, for each y

E

G and each xi

5.1

119

RELATIVELY PROJECTIVE MODULES

there exists zi E H and xi, such that xi y = zi x i , . Then for all w

c wy c

w(h, y) = w

x;

i= 1

=

c

E

W

n

n

'fhZiXi,

=w

x;

1zi

fhx,

i= 1

x; 'fhx,, = (wy)h,,

i= 1

since iH i' is a permutation of { 1,2, . . .,n}. Thus h, is an AG-homomorphism and (a) is proved. Corollary

Let H and L be subgroups of the group G.

(i) If x- 'Hx E L for some x E G, then each H-projective AG-module is Lprojective. (ii) IFH G L, I/ is an H-projective AL-module, and W is an AC-module such that W I p,then W is an H-projective AG-module.

Proof (i) Let W be an H-projective AG-module. By part (c) of the theorem there exists an AH-module U such that W I UG.Then U €3 x is an A(x- 'Hx)-submodule of U" and it is easily seen that ( U €3 x)" = UG as AG-modules. Let V be the induced ALmodule (U@I x)". Since inducing is a transitive operation, p = (U €3 x)" = UGand so W I p.Thus the theorem shows that W is Lprojective. (ii) Since I/ is an H-projective ALmodule, part (c) of the theorem shows that V 1 U" for some AH-module U . Then VG 1 ( U")" = U", so W I UGand hence W is H-projective. EXAMPLE 1 Let A = k be a field. Then every exact sequence (1) of kGmodules splits as a sequence of k-modules (that is, vector spaces). Thus over a field 1-projectivity of W simply means that each exact sequence (1) of AG-modules with W as the right-hand term splits; in other words, W is a projective kG-module in the usual sense of projectivity. EXAMPLE 2 Suppose A has the property that each integer not divisible by a prime p has an inverse in A. (This is the case if A is a p-adic algebra described in $3.3, or A is a field of characteristic p.) Suppose H contains a Sylow p-group of G. Then pX 1 C : H 1 so there exists a E A such that a 1 G : H I = 1. Takingfas the scalar a * 1 in part (d) of the theorem we see that CXEx- yx = a 1 G : H I = 1, where T is a right transversal of H in G, and so in this case every AG-module W is H-projective. EXAMPLE 3 Suppose that A is either a field or the ring of p-adic integers described in $3.3. Then an indecomposable A-free AG-module W is 1projective iff W is isomorphic to a principal indecomposable AG-module. Indeed, first suppose that W is an indecomposable 1-projectiveAG-module,

120

v


then by Theorem 5.1, W 1 where W, is the restriction of W to 1. Since the bull-Schmidt theorem holds because of our hypothesis on A (see Theorem 1.6A and Theorem 3.4B) we can choose an indecomposable component U of W, such that W I U G .Because W, is A-free, U is therefore the unique A-free indecomposable (trivial) A{1}-module. Thus U c = U 0 , AG 2 AG as AGmodules. Hence W 1 AG, and so is isomorphic to a principal indecomposable AG-module. Conversely, if W I AG, then W 1 V G ,where U is the trivial A-free A{ 1)-module. Therefore Theorem 5.1 shows that W is 1-projective. EXAMPLE 4 Suppose U and V are A-free AG-modules. If V is H-projective for some subgroup H of G, then V I W Gfor the A-free AH-module W := . ), Then by Theorem 2.1B, (U,, 0 W)" z U 0 p.Hence U Q V 1 ( V , 0 WV% which shows that U 0 V is also H-projective.

EXERCISE

Let k be a field and G a group. If the characteristic of k is either 0 or a prime not dividing I G I, show that for each subgroup H of G, each kG-module is H-projective.

5.2 Vertices and sources We shall now restrict ourselves to the case where A is a Noetherian integral domain such that each A-submodule of an A-free A-module is A-free, and such that the conclusion of the Krull-Schmidt theorem holds for A-free AH-modules for all subgroups H of G. The cases in which we are really interested are (a) A is a field, and (b) A is a the p-adic algebra described in $3.3. It follows from Theorem 1.6 and 3.4B and Example 4 of $1.1 that the hypotheses hold in these two cases. Note 1 Under these additional hypotheses the AH-module U in Theorem 5.l(c) may be taken as an indecomposable A-free AH-module whenever W is A-free (for example, take U as a suitable indecomposable summand of W,,).

Let V be an indecomposable AG-module. Then a subgroup Q of G is a vertex of V if V is Q-projective but V is not H-projective for any proper subgroup I€ ofQ. It follows from Theorem 5.l(c)that corresponding to any vertex Q of V there is at least one indecomposable AQ-module U such that VI U G ; such an AQ-module is called a source of I/: Note 2 Part (i) of the corollary to Theorem 5.1 shows that if Q is a vertex of V , then so is any conjugate x-'Qx (x E G). These definitions of vertex and source and their basic properties (which

5.2

121

VERTICES AND SOURCES

we shall give in this and the following sections) were given in a paper of Green [l].In order to prove the latter properties we shall need a simple lemma.

Lemma 5.2 Suppose that A satisfies the hypotheses stated above. Let H and L be subgroups of the group G and let V be an A-free Lprojective AG-module; so V I U G for some A-free ALmodule U. Write V, = Wl 0 W, 0 ... 0 W, as a direct sum of indecomposable AH-submodules. Then I ( U 0xi)", there exist elements xl, x,, . . ., x, of G such that for each i, where U 0xi is considered as an A(x; 'Hx,)-module. In particular, is (x; ',!,xi n H)-projective.

w

w.

Proof By Mackey's theorem (Theorem 2.1A)

8 yl), 0 ( U 8 y,)" 0 '.. 0 ( U 6ym),, where y, = 1, y,, ..., y, is a set of representatives for the (L, H)-double cosets and U 8 y j is considered as an A(y,: Lyj n H)-module. Since V I UG, it follows that each H( I ( U G ) ,. Because of the Krull-Schmidt theorem and the indecomposability of H(,we get H( I ( U 0 yj)" for some j. The lemma (UG)H'Z ( U

'

now follows. Our first main result is to prove that both vertices and sources are uniquely determined up to conjugacy.

Theorem 5.2A Suppose that A satisfies the hypotheses stated above. Let V be an indecomposable A-free AG-module, and let Q be a vertex of V. (i) If V is H-projective for some subgroup H of G, then H 2 x- 'Qx for some x E G. In particular, every vertex of V is conjugate to Q in G. (ii) Let U 1 and U , be A-free @-modules that are both sources of V. Then U , 'Z U , 0z for some z E N,(Q). Remark By Note 2 above, every conjugate of a vertex of V is also a vertex for V. Thus the vertices for V form a single class of conjugate subgroups of G.

Proof (i) By Lemma 5.2, V, = W, 0 W, 0 ... 0 W,, where each is an indecomposable AH-module which is (x; 'Qxi n H)-projective for some xi E G. Since V is H-projective we also have V I (V,)" = 0 @0 ... 0 Wf ; and hence by the Krull-Schmidt theorem, V1 Wc for some i because V is indecomposable. Hence by part (ii) of the corollary of Theorem 5.1 we conclude that V is (x; 'Qxi n H)-projective, and so x; 'Qxi n H = x; ' Q x i because x; 'Qxi is a vertex of V. Thus H 2 x; 'Qxi as asserted. (ii) Write V, = W, 0 W, 0 * . . 0 W,, where the are indecomposable AQ-modules. Since V is Q-projective, V I ( VQ)" = 0 0 ... 0 and so VI for some i because V is indecomposable. Since VI U y ,

w

c,

122

v


w.

Lemma 5.2 shows that there exists x E G such that I ( U , (9 x ) ~and , that is (x-'Qx n Q)-projective. But part (ii) of the corollary to Theorem 5.1 now implies that V is (x- 'Qx n Q)-projective, so x- 'Qx n Q = Q because Q is a vertex of K This shows that x E NG(Q), and that K(l(Ul @ x). Since U , is an indecomposable AQ-module, the same is true of U , @ x; hence 1: U , @ x. A similar argument now shows that K ( 1: U , @ y for some y E NG(Q)- Thus U, % U , (9 z, where z = x y - E N,(Q), and the theorem is proved.

'

EXAMPLE 1 Let A = k be a field of characteristic p and let P be a Sylow p-subgroup of the group G.Then by Example 2 of $5.1 every kG-module is P-projective. Thus in this case the vertex of each indecomposable kGmodule V is a p-group, and the set of vertices of V is a class of conjugate p-subgroups of G.

Theorem 5.2B Suppose that A satisfies the hypotheses above. Let H be a subgroup of the group G and let W be an indecomposable A-free AHmodule with a vertex Q and a corresponding A-free source U. Then there exists an indecomposable A-free AG-module V 1 Wc such that V also has vertex Q and source U .

Proof Since W(9 1 z W, we have W I (w")"(as AH-modules). Therefore, since W is indecomposable, there exists an indecomposable AG-module V such that W I V, and V I W; in particular, V is H-projective. By hypothesis W 1 U H , and so V I ( UH)G= UG. Since U is an A-free AQ-module, Theorem 5.2A now shows that Q contains a vertex Qo of V ; let U o be a source for V corresponding to Q,. Since W is an indecomposable direct summand of V, and V I U g , Lemma 5.2 shows that W I ( U , (9 x)" for some x E G, where U , @ x is considered as an A(x- 'Qox n H)-module. Hence W is x- 'Q, x n H-projective. Since Q is a vertex of W, 1 Qo I 2 I x- 'Qo x n HI 2 IQI. But Qo c Q, so Q = Qo and Q is a vertex of V On the other hand, U is an indecomposable AQ-module since it is a source for Wand we saw above that V I UG.Thus U is a source for V corresponding to Q,and the theorem is proved. EXAMPLE 2 Let H be a subgroup of the group G. Let V be a principal indecomposable kG-module, where k is a field of characteristic p. Then V 1 U g , where U o is the trivial module for the group 1. (See Example 3 of $5.1.) Hence by Lemma 5.2, VH = Wl @ W, @ W,,where the are indecomposable kti-modules such that 4 I ( U , (9 xJH for some xi E G and U , (9 x, 1: U o is considered as a k - I-module. Since U t = U o BkkHz kH, each 4 is isomorphic to a principal indecomposable kH-module. In particular, in the case where H = P is a Sylow p-group there is only one principal indecomposable kP-module, namely the regular module kP (see the $ * a *

w.

5.3

123

GREEN’S THEOREM

Example of 91.9). This implies that 1 PI = dim, kP divides dim, V for each principal indecomposable kG-module. (See also the example of 94.4). EXAMPLE 3 The result noted in Example 2 leads to an interesting observation about the submatrix rl of the Cartan matrix r for kG corresponding to a block B1of kG. (See the notation of 54.4.) Suppose that Bl has defect d. denote, respectively, the principal Let q l , q2, ..., qrl and #J1, #J2, ..., indecomposable modular characters and the irreducible modular characters lying in B1.Then rl is the rl x rl matrix [cij] with integer entries where ri

qi = xcijc#+

( i = 1, 2,

..., rl).

j= 1

In particular, ri

q i ( l ) = xci,c#+(l)

( i = 1, 2,

..., rl),

j= 1

and we can solve these equations using Cramer’s rule and obtain

@( l)(det r,)=

ri

cijqi(1) i= 1

for suitable integers cij. By Example 2 above, the right-hand side of this equation is divisible by p”, the order of a Sylow p-subgroup of G. By Theorem 4.5A there exists some j such that p m - d + l Xc#+(l). Hence pd I det rl. EXERCISES

[For these exercises suppose that the notation of 93.3 holds.] 1. Let W be an indecomposable kG-module. Suppose that for some subgroup H of G, W, is isomorphic to a sum of principal indecomposable kH-modules. Show that W is isomorphic to a principal indecomposable kG-module. Hence there is an AG-module Wl such that Wl z W. 2. Show that every p-subgroup of a group G is the vertex of some indecomposable kG-module.

5.3

Green’s theorem

In general, inducing from an indecomposable module does not yield another indecomposable module. However, in an especially important case, absolute indecomposability is preserved under induction. This is the content of Green’s theorem.

v

124


In the present section k will denote an arbitrary field of characteristic p . Recall that if V is a kG-module, and E!=End,,(V), then V is absolutely indecomposable iff E/rad E = k (Theorem 1.7). We begin with a lemma which serves as the crucial step in the proof of Green's theorem. Lemma 5.3 Let H be a normal subgroup of index p in the group G, and suppose that k is a perfect field of characteristic p. Let W be an absolutely indecomposable kH-module, and suppose that the indecomposable components of ( W c ) Hare all isomorphic to W Then W c is an absolutely indecomposable kG-module.

Remark Recall that a field k of characteristic p is perfect iff each element in k is a pth power of an element in k. In particular, it is well known that k is perfect when it is finite or algebraically closed.

Proof Let y E G\H.Then 1, y, and

. .., y p -

is a right transversal of H in G

WG = ( W @ l ) @ * - @ ( W @ y y " - ' ) .

(1)

By hypothesis, W @ y z W as kH-modules, and so there exists an invertible 4 E End,( W )such that w @ y~ w 4 is a kH-homomorphism from W @ y to W . By definition, (w @ y)x = wyxy- @ y for all x E H,so the condition that the mapping from W @ y to W commutes with the action of H is given by yxy- '4 = 4 x for all x E H. More generally, since H 4 G, induction on i shows that

'

(2)

yixy-

i+i

- 4iX

forall x ~ H , i = 0 ,1,2,....

In particular, we have (3) YP4= 4YP since y p E H. Let R be the k-algebra End, W. Then we can construct a k-linear mapping $ from End,( w") into the matrix ring Mat(p, R) by defining $(a) [aij] ( j , j = 0, 1, . . ., p - l), where the ail E End,( W) are uniquely determined by the conditions I=

(4)

(w 8 $)a =

P- 1

C ( w 4 i a i j 4 - j @ yj)

(i = 0, 1, 2,

. .., p - 1 ; w E W ) .

j=o

It is readily verified that $ : End,( w") z Mat(p, R) is a k-algebra isomorphism. Now y p E H, and for all w E W we have (w 8 y i ) y = / W @ yi+ wyP@ 1

if if

i = O , 1, . . . , p - 2 i=p-1.

5.3

125

GREEN'S THEOREM

On the other hand, for x

E

H, (2) shows that for all w

E

W we have

( w @ y ' ) x = w y ' x y - ' @ y ' = w4'x(b-'@y'.

Therefore we conclude from (4) that

for all x E H. (Here the matrix entries shall be interpreted as elements of End,( W ) = R.) Let E := End,,( W) C R. Then the image of End,,( w")under $ consists of all elements of Mat(p, R) that commute with $ ( x ) ( x E H); thus $(End,,( w"))= Mat(p, E). Put S := $(End,,( w")).Then s EndkG(w") as k-algebras because $ is a k-algebra isomorphism. On the other hand, S consists of all matrices in Mat(p, E) that commute with $(y), so we must consider this latter condition. First note that we can write $ ( y ) = ab, where

with :=4 - p y p E E by (2). Note that (3) also shows that a and b commute. Since W is absolutely indecomposable, E/rad E = k, and so there is a canonical homomorphism Mat(p, E) -,Mat(p, k) (taking the entries modulo rad E); we shall denote the latter homomorphism by C H F . Note that the kernel of this homomorphism is Mat(p, rad E), which is a nilpotent ideal of Mat(p, E) (in fact, the radical). We now claim that for each c E Mat(p, E), a-'ca = C. This is equivalent to showing that for each y E E, 4-'y+ - Y E rad E. However, since Elrad E = k, for each y E E there exists 1E k such that y - 1. 1 E rad E. Then 4-'y4 - y E 4- '(1 1)4 - 1 . 1 + rad E = rad E because 4 commutes with 1.This proves our claim. Now consider S and S : = { F I c E S}. To show that WG is absolutely indecomposable we must show that S/rad S = k; and since S n Mat@, c rad S, this is equivalent to showing that Slrad S = k. Since a-lca = F for all c E Mat(p, E), and S is the centralizer of $ ( y ) = ab, it follows that S is the centralizer in Mat(p, k) of 6 (recall that b E Mat(p, E)

126

v

because E E). Let F = [yl,] valent to the conditions

E

Yij=

Yp-l.,SO

for i, J' = 0, 1, form i?=

[


Mat(p, k). Then the condition bc = cb is equi-

Yi+l.j+1,

Yi,p-I

= YO.j+l,

= Yi+l,oBo,

Yp-l.p-1SO

= YOOSO

..., p - 2, where Po i=B + rad E. Hence t? E S iff it has the

Yo

Yl

'"1

...

----.__________-______________!!I! Yo "' =yo * l+y16+...+yp-,6p-1,

BoYp-1

.................................... BOY1 *.* BoYp-1 Yo

where y o , yl, . .., yp-l E k. Since k is perfect there exists I E k such that Po = 2'. Then by the binomial theorem we have (6- I * l)P = I P . 1 = Po 1 - Po * 1 = 0; and so 6 - I - 1 is nilpotent. Since S is commutative, and 6 - I * 1 E S, this shows that f(6- I * 1) C rad S. But S(6 - I * 1) is the kernel of the algebra homomorphism S --* k defined by yo

1+y, ~6+...+yp~,6P-~Hyo+y11+.~.+yp-11P-1.

Therefore S/rad S = S/f(6- I

*

1) = k, as required. This proves the lemma.

Theorem 53 Let H be a normal subgroup of index p in a group G. Let k be a field of characteristic p, and let W be an absolutely indecomposable kHmodule. Then w" is an absolutely indecomposable kG-module.

Proof First note that it is enough to prove the theorem for the case where k is algebraically closed (and then we can drop the word "absolutely" by Theorem 1.7A). Indeed, let k* be an algebraic closure of k. Then W k* = (W k*)', and W (respectively, w") is absolutely indecomposable if and only if W k* (respectively, w" Bkk * ) is indecomposable. Consider the set I ( W ) = {x E G I W @ x 2 W as kH-modules}. Clearly I(W ) is a subgroup of G and H E I(W). [I(W) is called the " inertia group " of W.] Since H has index p in G, we have two cases.

( I ( W ) = H ) Let y € G \ H . Then ( w " ) H = W @ l @ W @ Y @ I ) as a sum of kH-modules. Using the Krull-Schmidt theorem we know that w" has an indecomposable component V such that W @ 1 I V'. But then W @y' I VHyi, and the latter equals V, since V is a kG-module. Since W is indecomposable, the kH-modules W @ yi (i = 1, 2, ..., p - 1) are indecomposable, and they are mutually nonisomorphic because I(W) = H.Therefore the Krull-Schmidt theorem shows that the sum of these modules divides V'; that is, ( w")' 1 V, . But this implies w" = V, so w" is indecomposable. Case 1

*

- @ (W @ yp-

5.4

THE DEGREES OF INDECOMPOSABLE MODULES

127

Case 2 ( I ( W) = G) In this case we are in the situation of Lemma 5.3 and the result was proved there. This proves the theorem.

Remark In Case 1 the proof does not depend on absolute indecomposability, but in Case 2 this is important. The normality of H is also important. In Green [1,2] there are examples that show that Theorem 5.3 fails to remain true if H is not assumed to be normal in G or if the words “absolutely indecomposable” are replaced by “ indecomposable.” Corollary Let G be a p-group and H any subgroup. Let k be a field of characteristic p and let W be an absolutely indecomposable kH-module. Then W G is an absolutely indecomposable kG-module.

Proof Since G is a p-group, there is a normal series of subgroups of the form G = Go 3 G I 3 ... =I G, = H with IGI/Gi+l I = p for each i.Then the corollary follows from the theorem using induction on r. EXAMPLE Let G be a p-group and k a field of characteristic p. Let H be a subgroup of G and Wo the trivial kH-module. Then @ is an absolutely indecomposable kG-module. Note that this module affords the transitive permutation representation of G with H as a stabilizer.

EXERCISES

1. Let H be a normal subgroup of a group G such that G/H is a cyclic group whose order m is prime to p . Let k be an algebraically closed field of characteristic p, and let W be an indecomposable kH-module such that the are all isomorphic to W. Show that indecomposable components of ( w)H WG is a direct sum of m nonisomorphic indecomposable kG-modules, say W,, . . ., W,, such that ( WJHz W for each i. 2. Let G be a p-group of linear transformations on a vector space V over a field k of characteristic p. If there exists a basis for V on which G acts as a transitive permutation group, show that V is an indecomposable kG-module. 5.4 The degrees of indecomposable modules

Green’s theorem (Theorem 5.3) leads at once to useful facts about the degrees of indecomposable modules and the characters that they afford. Theorem 5RA Let k be a field of characteristic p and let P be a Sylow p-subgroup of the group G. Let V be an indecomposable kG-module with vertex Q c P . Then I P : Q I divides dim, V .

128

v


Remark We know from Example 1 of $5.2 that V has at least one vertex contained in P. As a special case, it follows that if p Xdim, V, then P itself is a vertex of V. Proof First consider the case where k is algebraically closed (and so V is absolutely indecomposable). Let U be a source of V corresponding to the vertex Q. By Lemma 5.2 we have V, = W,@ W,0 ... 0 W,, where the W, are indecomposable kP-modules and for some xl, x, , . . ., x, in G we have yI(U @I xi)', where U @I x i is viewed as a k(x; 'Qxi n P)-module. By the corollary of Theorem 5.3, the indecomposable components of ( U @ xi), are modules induced from the indecomposable components of U @ xi restricted to x; 'Qxi n P ; in particular, = U: for some k(x; 'Qxi n P)-module U i . Thus dim, = I P : x; 'Qxi n P 1 dim, U i and so is divisible by 1 P : Q I . Since dim, V = dim, W, dim, W,, we conclude that I P : Q I divides dim, V as required. This proves the theorem in the case k is algebraically closed. In general, let k* denote the algebraic closure of k. Then V @, k* = V, 0 V, @ . . . @ V,, where the V;. are indecomposable k*G-modules. Since for all i we have V;. I V @ ,k* and V @ I ,k* 1 UG@,k* = ( U B kk*)G, therefore V;. I ( U @, k*)G for the k*Q-module U @I k*. This shows that each 4 is Q-projective and so has a vertex Qi E Q. By the algebraically closed field case considered above, I P : Qi I divides dim, V;. for each i. Since I P : Q I divides each I P : Qi 1 and dim, V = dim, V, - * + dim, V,, we conclude that I P : Q I divides dim, V. This proves the theorem in general.

+ +

,

+

Let k be a field of characteristic p and let W be an absolutely indecomposable kG-module. If x is an element of E = EndkG(W )of finite order relatively prime to p, then x = 1 1 for some 1E k.

Lemma 5.4

Proof Since W is absolutely indecomposable, E/rad E N k (Theorem 1.7A), and so x may be written in the form x = 1 1 r for some r in rad E and 1 in k. Now rad E is nilpotent and there is an integer n such that r" = 0. Choose an integer m so that p" 2 n. Then xPm= (1* 1 r)p" = AP" * 1, since k has characteristic p. As x has order prime to p , it is a power of xp", and the result follows.

+ +

Theorem 5.4B Let G be a group, x a p'element of G, and P a Sylow p-subgroup of C,(X). Let k be a field of characteristic p and let V be an indecomposable kG-module affording the character 4 over k. Suppose P properly contains a vertex Q of V. Then 4(x) = 0.

Proof First consider the case where k is algebraically closed. Let H := (P, x) and let V, = W,@ W,@ @ W,, where each is an indecomposable kH-module. Lemma 5.2 shows that there is an element xi of * . a

5.5

VERTICES AND DEFECT GROUPS

129

G such that

is x; ' Q x i n H = x; ' Q x i n P-projective. As Q # P, it follows that p divides 1 P: x; 'Qxi n P 1, and we deduce from Theorem 5.4Athat p divides dim, . Now since x is central in H, we may consider x as an element of End,,( K).Thus we see from Lemma 5.4 that x acts as a scalar multiple of the identity on In particular, 4 ( x ) = 0, since the multiplicity of each eigenvalue is divisible by p. In general, let k* be the algebraic closure of k. Then V* = V k* is a k*G-module. Let V* = Vl @ V' @ * @ V, be a direct sum decomposition of V* into indecomposable modules We know from the proof of Theorem 5.4A that each y has a vertex contained in Q . Thus our previous reasoning applied to each gives the required conclusion in the general case.

w

w.

v.

EXERCISE

Let k be a field of characteristic p and let U be a principal indecomposable kG-module. Suppose '1 is the modular character of G associated with U.If x is a p'-element of G with (C,(x)l = p"h, where (p, h) = 1, then show that q ( x ) = p"I(x), where I is a character of (x). 5.5 Vertices and defect groups

Suppose k is a field of characteristic p satisfying the hypothesis of 53.3. If V is an indecomposable kG-module, then the vertices of V and the defect groups of the block in which V lies both form conjugate classes of subgroups of G.There is an interesting relation between these two classes. Theorem 5.5 Let V be an indecomposable kG-module lying in a block B of kG. If D is a defect group of B, then V is D-projective, and hence D contains a vertex of V.

Proof Let w be the central character associated with B. Then by Theorem 4.3 there exists a class V of G with class sum c in kG such that D is a defect group of V and a ( c ) # 0. Choose x E V so that D is a Sylow psubgroup of C,(x), and let xl, x z , . . ., xn be a right transversal of D in G. Then n

C x; lxxi = mc i= 1

where rn := I C,(x) : D 1 f 0 (mod p).

Now mc acts on V as the nonzero scalar m ( c ) * 1 (see Theorem 4.2B). DefinefE End,( V) by uf:= ux{rna(c)}-' .Then the condition (d) of Theorem 5.1 is satisfied for V and so V is D-projective.

130

v


EXAMPLE If B is a block of defect 0 (so D = l), then each indecomposable kG-module V lying in B has vertex 1 and so is isomorphic to a principal indecomposable kG-module (Example 3 of $5.1). This gives further information about blocks of defect 0 (see also $4.6).

5.6 Restriction of indecomposable modules In the next section we shall be proving a theorem of Brauer and Feit on the existence of normal abelian subgroups. This will require a description of the way in which certain indecomposable kG-modules decompose under restriction to subgroups of G. The example offi4.3 shows how in the case that a character has degree not divisible by p , a knowledge of a rather few values of the character is sufficient to determine the block in which it lies. This motivates our first result. Theorem 5.6A Suppose that the hypotheses of $3.3 hold. Let P be a Sylow p-subgroup of the group G, and put N := N G ( P ) .Let V be an irreducible kG-module, and suppose that pXdim, V. Then

(i) V, = W, @ W, 0 * . @ W, is a direct sum of indecomposable k N modules such that V I w(i; and P is a vertex for W,. (ii) (w(i;)N has exactly one indecomposable component with P as a vertex. Moreover, for the components of V, we have pXdim W, but p l d i m Wfor i = 2 , 3 , ..., m. (iii) Let 9 and 8 be the characters over k afforded by V and W,, respectively; then + ( x ) = 8 ( x ) for all elements x E C,(P)". (iv) If W, lies in the principal block of kN, then V lies in the principal block of kG; moreover, W, is trivial only if I/ is trivial. Proof (i) Since N contains a Sylow p-subgroup, it contains a vertex of V ; hence V is N-projective. Therefore by Theorem 5.1, V 1 (V,)". Writing V, = W, @ W, @ ...@ W, as a sum of indecomposable kN-modules, it follows from the Krull-Schmidt theorem that V 1 (PI$)" for somej, and without loss in generality we can takej = 1. Let Q be a vertex of W, with Q E P, and let U be a source for W, at Q.Then W, I UN,and so V I ( UN)"= UG.This implies that Q contains a vertex of V . However, since pXdim, V, P is a vertex for V by Theorem 5.4A. Thus Q = P, and P is also a vertex for W,. (ii) Since P is the only Sylow p-subgroup of N, x - ' P x $! N for each x E G\N; thus N n x- ' N x does not contain a Sylow p-subgroup of G if x N. By the Mackey theorem (Theorem 2.1A) f$

n

(w(i;)N

=

@ (W1 B XIIN* i= 1

where W, @ x is to be viewed as a k ( x - ' N x n N)-module and x 1 = 1, x , , ..., x , is a set of representatives for the (N, N) double cosets of G. In

5.6

RESTRICTION OF INDECOMPOSABLE MODULES

131

particular, (W, @ x,), = (W, 1), z W, and has P as a vertex by (i). On the other hand, for i 2 2 we have xi # N, and the indecomposable components of (W, 0 xi), are all (x; 'Nxi n N)-projective. Since x i r$ N , therefore x; "xi n N does not contain a Sylow p-subgroup of G.This means that, except for the single component W, 8 1, the indecomposable components of ( all have vertices that are proper subgroups of the Sylow p-subgroups of G. Since V I ( Wi)" by (i), and V, = W, 0 W, O O W,, we can also conclude that each for i 2 2 has as its vertices proper subgroups of the Sylow p-subgroups of G.By Theorem 5.4A this implies that p 1 dim W i for i = 2, 3, . . ., m, and p Xdim W, because p Xdim, V. ( i = 1, 2, ..., m). (iii) Let Oi be the character of N over k afforded by Our hypothesis on x shows that we can apply Theorem 5.4B to W; with N in place of G, and it follows from (ii) that Oi = 0 on the set C,(P)" of p'-elements of C,(P) for i = 2, ..., m. Since 8 = 8,, this shows that 4 = 8, O2 ... + 8, = 8 on C,(P)" as required. (iv) Since the degree of 4 is prime to p , it lies in the principal block iff 4(x) = 4( 1) for all x E C,(P)' (by the example of $4.3).On the other hand, if 8 lies in the principal block, then so do all of its constituents. Let $ be one of the irreducible constituents. For x E C,(P)' let %be the conjugate class of G containing x, c the corresponding class sum, and h := I% I. Since P E C,(x), and P is a Sylow p-subgroup, p Xh. Let Y be a kN-module which affords $. Then by Theorem 4.2A we have uc = hu for all u E Y, and so taking traces we obtain h+(x) = h$( 1). Since p X h, this shows that +(x) = $( 1) for each x E C,(P)'. Adding these relations for the various irreducible constituents of 8 we obtain 8(x) = 8( 1) for all x E C,(P)" whenever 8 lies in the principal block. Hence it follows from (iii) that lies in the principal block of kG whenever 8 lies in the principal block of k N . Finally, suppose that W, is trivial. Then W, = (U,), when U , is the trivial kG-module, and so U o I by Theorem 5.1. If V 34 U , , then V O U , I by Krull-Schmidt theorem, . This implies that ( has at least two componand so V, 0 W, I ( ents (isomorphic to W,) with vertex P, and that is contrary to (ii). Hence V z U , , and the proof of (iv) is complete.

fl),

+ +

fl),

fl

fl

c),

Under the hypothesis of Theorem 5.6A we shall call the uniquely determined indecomposable kN-module W, the deriuatiue of the irreducible kG-module V .

Definition

Theorem 5.6B Let N be a group containing a normal Sylow p-subgroup P and a normal p'-group H. Let k be a splitting field of characteristic p for N and its subgroups, and let U be an indecomposable kN-module. Then we have a decomposition of the form

(1)'

up,

where the Xi Q

(XIO Y1)O ( X z O Yz)O . * * O ( X m O Ym), are indecomposable k(PH)-modules and

132

v


(i) H acts trivially on each X i , so ( X i ) , , is an indecomposable kP-module ; (ii) P acts trivially on each y i , so ( y ) , is an indecomposable kH-module (and therefore irreducible by Maschke's theorem); (iii) the modules Xi 0 yi (i = 1, 2, . .. , m)are conjugate under the action of N (see $2.2). Proof Since P H contains a Sylow p-subgroup of N, PH contains a vertex of U , and so U is PH-projective. Therefore by Lemma 5.2 there exists an indecomposable k(PH)-module W such that U I WN and W, = W, @ 1 W 6x i for some W, @ @ W,, where the K are indecomposable and xi E N ( i = 1,2,. . ., m).(Note that in this case; x; ' P H x i n PH = PH for all x i .) This implies that each z W @ xi, and so the are conjugate under N. It remains to prove that each has the form Xi @ yi described in (i) and (ii). Dropping indices, let W be any indecomposable k(PH)-module. Since p X I H 1, W, is completely reducible and hence a direct sum of irreducible KH-modules. Write W, = V, @ V, @ ... @ V, as a direct sum of its homogeneous components (see $2.2). Since P centralizes H, v x = for all i and all x E P. Thus the 5 are k(PH)-components of W, and so n = 1 by the indecomposability of W. This shows that all irreducible constituents of W, are isomorphic to Y, say. We extend Y to a k(PH)-module by defining the action of P on Y as the trivial action. Now put X Horn,,( Y, W,). Since each.irreducible constituent of W, is isomorphic to Y , Note 2 of 1.9 shows that dim, X = dim, W/dim, Y because k is a splitting field for k H . Hence dim,(X 8 Y ) = dim, W. Define an action of P H on X pointwise; namely, wfxY I= (wf )x for all f E X, x E P, y E H,and w E X. This defines X as a k(PH)-module on which H acts trivially. Now the mapping f 6 WH wfof X @ Y into W is clearly a k(PH)-homomorphism; and since W, is a direct sum of copies of Y, the mapping is surjective. But we showed that dim, W = dim,(X 0 Y), and so this surjective k-linear mapping must in fact be bijective. Hence X 6 Y N W a s a k(PH)-module. Clearly X is indecomposable as a kP-module (for otherwise W will not be indecomposable). This completes the proof.

w.

w.

I=

Corollary If, under the hypothesis of the theorem, U p has at least one component that is the trivial kP-module, then P acts trivially on U . In this case U is irreducible and pXdim, U .

Proof @ n,

Let n, !=dim, y i . Then from (1) we have U p 1: n, X ,@ n , X , @ X,. By hypothesis, one of the indecomposable X i is the trivial

kP-module. Hence by (iii) of the theorem, each X i is the trivial kP-module, so U p , 5 Y, @ Y2@ @ Y , and P acts trivially on U . Thus U can be

5.7

133

JORDAN’S THEOREM IN CHARACTERISTICp

I,

considered as a k(N/P)-module and since p J\ I NIP the indecomposable module U is in fact irreducible (see Theorem 1.3B). Moreover, dim, U divides 1 N/P I by the corollary of Theorem 3.5, and so p Xdim, U .

5.7 Jordan’s theorem in characteristic p A classical theorem due to C. Jordan (and proved in full generality by I. Schur) states the following: There exists a constant v, (depending only on n ) such that if G is a group with a faithful ordinary representation of degree n, then G possesses a normal abelian subgroup H such that I G : H I Iv,.

For a proof of this result see Curtis and Reiner [l], $36, or Dixon [I], §5.7. Note 1 It follows from $3.5 that the same conclusion holds if G is a p’-group and G has a faithful representation of degree n over a field k of characteristic p. The result, however, is no longer true if we drop the hypothesis that G is a p’-group (see the exercise at the end of this section). The object of the present section is to prove an analog of Jordan’s theorem for fields of characteristic p when G is not necessarily a p’-group. This result (Theorem 5.7) is due to Brauer and Feit [ 2 ] . Before we state the main theorem of this section we prove three lemmas that are needed in the proof. We shall write Triv(H) to denote the trivial irreducible kH-module. Suppose that the hypotheses of $3.3 hold, and let U1and U, be irreducible kG-modules that lie in the same block E of kG. Lemma 5.7A

(i) If pJ\dim, U i ( i = 1, 2), then some irreducible constituent of U : 0 U , lies in the principal block of kG (see $1.4 for the definition of U:). (ii) In the case where U , = U , , the trivial kG-module appears as an irreducible constituent of U : 0 U , ; and if p I dim, U , , then it has multiplicity at least 2. Proof (i) Let 4i and qi ( i = 1, 2, . . ., r) be the irreducible modular characters and the principal indecomposable modular characters of G, respectively, where they are enumerated in such a way that 4i and qi ( i = 1, 2, . . ., r,) are the characters in the block B. Let TI = [cij] be the rl x r l Cartan submatrix corresponding to the block B (see 44.4), and put r; = [cij].Then by Theorem 3.7B (using the notation above) we have r

(1)

&j

= g(di3 4’)GC’

=

1 h, 4;4;.

I=1

for i, j = 1, 2, . . ., rl, where g := 1 G 1. In particular, since the left-hand side of (1) is rational and the right-hand side of (1) is an algebraic integer, gcjj is an

v

134


integer for all i, j. Let w be the central character of kG associated with B, and let w, denote the value of w on the Ith class sum. Then by the definition of the modular characters, reduction modulo II of any of the irreducible modular characters 4' gives a character which is equal (on Go)to an irreducible character $' over k in the same block. Since $i (i = 1, 2, . . .,rl) lie in B, the corollary of Theorem 4.2A therefore shows that $$f= $Ji(l)a, for 1 = 1,2, ...,r and all 4' lying in B. Hence from (1) we have (2)

QC;=4'(1)c~,& =V(1)pJ,& I= 1

I= 1

for i, j = 1, 2, . . ., r l . Without loss of generality, suppose that 4l is the irreducible modular character afforded by U 1 . Then p $ #'( 1)by hypothesis, w, $;. Now (2) shows that and so there exists a E k such that a4'( 1) = a4'(1) = C;=lo,$t for i = 1, 2, . .., r l . Hence it follows from (2) that gc; = rji(l)@(l)a for all i, j = 1, 2, ..., r l . We now show that a # 0. Indeed, from the definition of rl we have

xri

for i = 1 , 2 , ..., rl

q ' ( ~ ) =f c l j @ ( l ) j= 1

and hence

4i(1)=

fcij$(l)

for i = 1,2, ..., rl.

I =1

However, if p" is the order of the Sylow p-subgroups of G,then p" 1 q j ( 1) for all j (see Example 2 of §5.2), and so #(l)/g E A. But then we have 1 4 (1) =

-c fV;,(vj(1)/9)~ r'

j= 1

where the left-hand side is nonzero because p Xrj'(1). Hence c#~'(l)q5,(1)a= # 0 for some j, and so we conclude a # 0. In particular, if 4i is the character afforded by U , , then pX$~'(l) by hypothesis and so = rjl(l)t#~'(l)a# 0; this shows that c;i # 0. Now consider the module U : @I U,. The modular character afforded by U: is (&)* where (4')*(x) I = +I(x- l ) for all x E Go.Therefore the modular character afforded by UT@I U , is ($l)*4', which we can write (t$l)*qY = xixlmI4' for some integers m,.Suppose that $' = 1, is the trivial modular character. Then by Theorem 3.7B r

cii = (4i,

41)c. = 1) lie in the principal block of kN. Moreover, p Xdim W, for each i, and since dim, U = n > 1, we have s 2 2. Step 2 There exist kN-modules W' and W" which are sums of indecomposables with vertices properly contained in P such that (W* @ W )@ W 2 (Of= m, 4)CD W".

,

5.7

137

JORDAN'S THEOREM IN CHARACTERISTIC p

By Theorem 5.6A, U N = W @ V, where V is a sum of indecomposable modules that are Q-projective for certain proper subgroups Q of P. Then (V* @ V ) , = (W* @ W ) @ W', where W := W* @ V @ W @ V* @ V @ V*. It follows from Example 4 of $5.1 that each indecomposable component of W' is Q-projective for some proper subgroup Q of P ; so these components have vertices that are properly contained in P. On the other hand, Theorem 5.6A shows that ( U J N.= 0 V;. , where the indecomposable components of V;. all have vertices properly contained in P. Thus Step 2 follows from Step 1 with W":= e iy . Step 3 W,, = X @ Y, where X is a k(PH)-module on which H acts trivially and Y is an irreducible k(PH)-module on which P acts trivially. Moreouer, Triv(PH) I (W* @ W)pHand dim Y > 1. It follows from Theorem 5.6B (using the notation there) that W,, has the form W,, = 8 ?= X i @I yi for some m 2 1. Collecting like terms we may suppose that the yi are mutually nonisomorphic (but the X i may be decomposable). Clearly the X i @ yi are still conjugate under N. We now have

and Y: @ I;. has an irreducible constituent Triv(PH) iff i = j. This shows that there is a k(PH)-module W"'such that rn

(W* @

W)pH =

@ (X: i= 1

Xi)@

v,

where Triv(PH) is not an indecomposable component of W'". Now Triv(N) = Wl has P as a vertex, so Step 2 shows that Triv(N) I W* @ W. Hence Triv(PH) I ( W * @ W),,, which shows that for some i, Triv(PH) I Xr @ X i . Since the X i@ yi are conjugate under N,it follows that Triv(PH) I Xi* @I X i for each i, and so m Triv(PH) I (W* @ W),, . Now suppose m > 1. If V is an indecomposable kN-module such that Triv(PH) 1 V p H ,then it follows from the corollary of Theorem 5.6B that p Xdim V. In particular, P is a vertex of V by Theorem 5.4A. Thus if m > 1, then by Step 2, Triv(PH) 1 (W,)pHfor some j 2 2 (recall that mI = 1). Then the corollary of Theorem 5.6B shows again that 4.is irreducible. Since ( W$)H has at least one constituent that is trivial, it follows from the corollary of Clifford's theorem (Theorem 2.2A) that H is in the kernel of W j . Since p Xdim 4 by the definition of the derivative, the example of $4.3 shows that W, lies in the principal block of kN.But then U , lies in Bo by Theorem 5.6A, which is impossible by Step 1 for j 2 2. This shows that rn = 1. We showed above that Triv(PH) 1 (W* @ W ) p H . Finally, if dim Yl = 1, then Y: @ Y, z Triv(H), and so H is contained in

v

138


the kernel of W* @ W. But by Step 2 each 4 I W* @ W, and so H lies in the kernel of each As we saw above this is impossible for i 2 2 because U i does not lie in Bo . Thus we conclude dim Y, 2 2. By the Schur-Zassenhaus theorem the normal Sylow p-subgroup P has a complement Lo in N. Let K O be the kernel of the action of Loon W. Then L , / K o is a p'-group with a faithful representation of degree less than or equal to n over k, and so by the classical theorem of Jordan stated in the beginning of this section, & / K O has a normal abelian subgroup L1/ K O of index v,. Note that L, P 4N. Put L:=L1 P n H. Then IH : L J I v, and L 4N. Moreover if K is the kernel of the action of N on W, then L/L n K is an abelian $-group.

w..

Step 4 For some j 2 2, L is contained in the kernel of

4..

Let n , = dim Y (where we write Y for Y l ) ;then n , 2 2 by Step 3. Since L/L n K is an abelian p'-group, YL is a direct sum of n , components each of dimension 1, and so Yf @ YL has the constituent Triv(L) with multiplicity greater than or equal to n , > 1. Since Y, is irreducible, Triv(H) has multiplicity 1 in ( Y * @ Y ) , ; therefore there is an irreducible constituent V of Y* @ Y such that Triv(H) is not a constituent of V,, but Triv(L) is a constituent of V,. Since P lies in the kernel of Y, P also lies in the kernel of V , and by Clifford's theorem (Theorem 2.2A), L is contained in the kernel of V. Note that V I Y* @ Y because Y* @ Y is completely reducible (P acts trivially on Y ) . Now Triv(P) I (W* 8 W), = (dim Y ) * ( X *0 X ) , so Triv(P) I ( X * 0 X ) by Step 3, and V N Triv(P) @ V 1 ( X * @ X ) @ ( Y * @ Y ) N (W* @ W ) p H . Since P lies in the kernel of V, V has P as a vertex, and so by Step 2, V I (I+$ for ,, somej; by hypothesis, j 2 2 because V # Triv(PH). However Since L 4 N, Clifford's Triv(L) I V', and so Triv(L) is a constituent of ( theorem shows that L is in the kernel of W j . The proof of the lemma is now complete if we take U j for U , and Wj for woe Lemma 5.7C Suppose the hypotheses of Lemma 5.7B hold, and that the Sylow p-subgroup P of G has order pd. Put A,, ,,d := I GL(pZdv,n4, p) 1 . Then there exists a nontrivial irreducible constituent V of U* @ U @ U* @ U such that 1 G : Ker V I I A, p d .

Proof Suppose 'that V is any irreducible kG-module with m :=dim, V. Let r j be the character afforded by V over k. Suppose that 4 has 1 algebraic conjugate characters (see the example of82.7).Then the values of the character r j must all lie in a field k , which has degree 1 over the prime subfield of k ; in particular, I k l I = p'. Hence by Theorem 2.7B, 4 is afforded by a k , G-module of dimension m, and so I G/Ker V I I 1 GL(m, p') 1 I 1 GL(ml, p) 1.

5.7

JORDAN’S THEOREM I N CHARACTERSTIC p

139

Then to prove the lemma, it is sufficient to show that there exists a nontrivial irreducible constituent V of U* Q U Q U * Q U which affords a character 4 with 1 algebraically conjugate characters, where I Ip 2 d ~(since , dim, V I (dim, U)“ = .)‘n Since the hypotheses of Lemma 5.7B hold, at least one of the conclusions (i) and (ii) is true; we consider these two possibilities. Suppose conclusion (i) of Lemma 5.7B holds. Then choose V as a nontrivial irreducible constituent of U* Q U Q U* Q U from the principal block Bo . By the example of $4.2 all algebraic conjugate characters of 4 also lie in B o . By Corollary 1 of Theorem 4.5B this means that 4 has at most ap2d+ 1 Ip 2 d algebraic conjugates, and so the lemma is proved in this case. Next suppose that conclusion (ii) of Lemma 5.7B holds, and choose I/ = U o . Since Triv(G) and U o are both constituents of U* Q U,this shows that V is a constituent of U * Q U Q U* Q U . Let 4 be the character afforded by V and 8 be the character of N , ( P ) afforded by the derivative W, of V. Then 4 = % on C,(P)O by Theorem 5.6A. In the notation of Lemma 5.7B, C,(P) = H x Z(P), and so C,(P)’ = H.Put t := 1 H : H n Ker WoI. Then the values of 8 (and hence of 4) on H will be sums of tth roots of unity in k and hence all lie in a subfield k2 of k of degree less than or equal to I over the prime subfield; in particular, 1 k, I Ip‘. Now let 4 = 4’, . . ., 4‘ be the algebraic conjugate characters of 4. If &, = &“, then 4i and 4’ lie in the same block (see the example of 94.3), and so not more than p2d of the t$i can have the same restriction to H by the corollary of Theorem 4.5B. On the other hand, since the values of 4H all lie in k,, $H has at most t algebraic conjugates; hence the number ofdifferentrestrictions 4k (i = 1,2, . . ., I) is at most t . This shows that I Ipldr. Since t Iv, by (ii) of Lemma 5.7B, the lemma is proved in this case as well. Theorem5.7 There exists a constant v,,. pd (depending only on nand p d ) such that if G is a group with a Sylow p-subgroup of order pd and G has a faithful representation of degree d In over a field k of characteristic p , then G possesses a normal abelian subgroup of index I vnep d .

Proof Without loss in generality we can suppose that k satisfies the hypotheses of $3.3. Let U be a faithful kG-module of dimension n. For each subgroup H of G we define the “type” t ( H ) to be the pair (d’, m‘),where pd’is the order of the Sylow p-subgroup of H and m‘ is the multiplicity ofTriv(H) in U g 0 U HQ U g Q U H .We order the set of types by writing (d’, m’) < (d”,m”)to mean either d’ < d” or d’ = d” and m’> m”.Since m‘ In”, there are at most dn“ types. We shall first prove that if H is a subgroup of G and t ( H ) = (d’, m’)with d’ > 0, then there is a subgroup L of H such that t(L)< t ( H ) and I H : L I I LnVpd(where is defined in Lemma 5.7C). If pI I H : H ’ l , then H has a normal subgroup L of index p, and Lclearly satisfies the requirements. Thus

v

140


suppose p $ 1 H : H' 1. If the irreducible constituents of U , all have dimension 1, then we have a kH-composition series UH =

WO 2

w1

2

' * *

2

w,, =0

t

with dim( W;- /W) = 1 for i = 1, 2, . . ., n. Then x- ly- ' x y acts trivially on each W;.- /W;. for all x, y E H,and so the same is true for each z E H'. Now the example of $1.5 shows that H'/Ker U,,, is a p-group. Since H acts faithfully on U , therefore Ker U H r= 1, and so H' is a p-group. Thus we conclude that the Sylow p-subgroup of H is normal in H. Hence by the Schur-Zassenhaus theorem, H has a p-complement L. Because L is a p'-group, @.) c t ( H ) , and because L is a p-complement, I H : LI = pd' c An,pd. This settles the case where all constituents of U H are onedimensional. If U , has an irreducible constituent of dimension greater than 1, we can apply Lemma 5.7C to this constituent to obtain a nontrivial irreducible kH-constituent I/ of U g 8 U H @ Uj$@ U H with I H : Ker I/ 1 I A,,,pc I A,,, p d . Taking L = Ker r! it follows that t ( L ) < t ( H ) , and the assertion is proved in this case as well. Using what we have just proved, we know that there exists a chain of subgroups G = H o 3 H I 3 ... 3 H , such that t ( H i - ') > t ( H i ) and [ H i - : H, 1 I A,,,pd for each i. Since there are at most dn4 types of subgroups, there exists 1 Idn4 such that r(H,) = (0,m) for some m I n4. Then H , is a p'-group, and so by the classical theorem of Jordan (see the beginning of this section), H I has an abelian subgroup So of index Iv,. Then 1 G : So I I ( A ,p d ) 4 ~ n= p,,, p d , say. Finally, the subgroup So has at most p,,, pd conjugates in G and S := X E x- 'Sox has index less than or equal to (p,,,p d ) ! in G. Putting v,,, p d ' = (p,,,pd)!, the proof is completed. EXERCISE

Let k be an algebraically closed field of characteristic p > 0 and set G, I = SL(2, p"), the special linear group of degree 2 over the finite field of p" elements. Let GY2, k) be the general linear group of degree 2 over k. Then G, E GL(2, k). Show that a normal abelian subgroup of G, has order at most 2 while the order 1 G, I of G, is arbitrarily large. 5.8 Notes and comments

The theory of indecomposable kG-modules as presented here is essentially due to Green [ 1,2]. A general setting for a part of the representation theory of finite groups, namely the part that includes the theory of vertices of modular representations and defect groups of blocks, is given in Green [3]. This theory has proved to be very useful in the theory of classification of

5.8

NOTES AND COMMENTS

141

indecomposable modules as well as in the structure theory of groups; for example, see Dade [2] and Thompson [l]. One of its uses will become apparent in Chapter 8, where the structure of a block with a cyclic defect group is studied in detail. A detailed treatment of the theory of indecomposable modules is given in Feit [2]. The material of the Section 5.7 is due to Brauer and Feit [2].

CHAPTER

VI

The Main Theorems of Brauer

Throughout this chapter we shall keep the notation introduced in 53.3. In particular, we have an integral domain A with a unique maximal ideal nA, the field K of quotients of A of characteristic 0, and the residue class field k := A/nA of characteristic p. Moreover, A is a principal ideal domain and both K and k are splitting fields for the group G and all of its subgroups. The theorems of this chapter will investigate various connections between the theory of blocks of kG and the theory of blocks of kH for certain subgroups and quotient groups H of G.

6.1

The Brauer homomorphism

The Brauer homomorphism which we describe now is an important tool in our investigation. Recall that for any subgroup Q of G, CG(Q)4 N,(Q). This observation will be used repeatedly in the following sections. Lemma 6.2A Let Q be a p-subgroup of the group G,and put N Then the mapping a defined by (1)

a

1 a,x

:=

C

:= N G ( Q ) .

a,x

XECC(Q)

(ZGG

is a k-algebra homomorphism from Z(kG) into Z(kN). Remark We shall call a the Brauer homomorphism with respect to Q.It is clear from the proof below that the analogous function from Z(AG) into 142

6.1

143

THE BRAUER HOMOMORPHISM

Z ( A N ) can also be defined and is an A-linear mapping; but the latter is not usually an algebra homomorphism. Note that when we interpret the elements Ex. c t , . ~ of Z ( k G ) as class functions x H a , from G into k, then o is simply restriction to C,(Q).

,

Proof We first show that o maps Z ( k G ) into Z ( k N ) . Let g i be any conjugate class of G with class sum c i . Then %': :=g in CG(Q)is a (possibly empty) normal subset of N since C,(Q) N. Thus %?: is a union of conjugate classes of N, and so o(ci)= x E Z ( k N ) . Since clr c 2 , ..., c,is a k-basis of Z(kG), this shows that o is a function of Z ( k G ) into Z ( k N ) . Clearly, rr is a k-linear mapping and o(1) = 1, so it remains to show that for any class sums c i , cj we have o ( c i c j )= a(ci)o(cj).However, o(cic j ) = C, v, z and o(ci)o(cj) = v:z, where v, is the number of elements of the set S, := {(x, y) E W i x Wj x y = z } and :v is the number of elements of the set S: := {(x, y) E %:' x %'f 1 xy = z } . Since S: E S , and Q

,

S,\S,* = {(x, 1') E

1,, I

(%i\w:) x g j 1 xy = z } u {(x, y ) E g:

I

x (Vj\VT) xy = z }

as a disjoint union, it follows from Lemma 4.3A that v, = 1 S , I = I S: I = vr (mod p) for all z E G. Since k has characteristic p, this shows that o(cicj)= o(ci)o(cj) as required. Lemma 6.IB Let Q be a p-subgroup of the group G and V a conjugate class in G. Put V* :=%' n CG(Q), N := N,(Q) and let CT denote the Brauer homomorphism with respect to Q. Then

(i) %?* # 4 iff Q is contained in some defect group of %?. (ii) W* is a single conjugate class in N if Q is a defect group of %'. (iii) Suppose that 59' is a conjugate class in N with %" E %'. If Q is a defect group of W, then Q is a defect group of %' and 59 = %' n CG(Q). (iv) Letfbe a block idempotent of kG with a defect group containing Q, then o(f) # 0 and is a central idempotent of k N . I f f ' is a second block idempotent of kG with a defect group containing Q, andf#f', then o(f) and a(f') are orthogonal idempotents in Z ( k N ) .

Proof (i) %?*# 4 iff x E C,(Q) for some x E V. The latter condition is equivalent to Q E C,(X) for some x E %',and this holds iff Q is contained in some defect group of %'. (ii) If Q is a defect group for V, then Q is a Sylow p-subgroup of C,(x) for some x E %'; in particular, x E C,(Q) n %? = g*.Suppose that y also lies in %'*. Then y = z - 'xz for some z E G, and Q c C,(y), so z - ' Q z c C , ( Z ~ Z - '=) C,(x). Since both Q and z - ' Q z are Sylow p-subgroups of C,(X), there exists w E C,(X) such that wzQz- ' w - = Q . Then wz E N = NG(Q), and ( w z ) - ' ~ ( w z=) y, so y is conjugate to x in N. Thus % * is a single conjugate class of N.

'

VI

144

T H E MAIN THEOREMS OF B R A U E R

(iii) Suppose that Q is not a defect group of W . Then for some x E V‘ we have Q as a Sylow p-subgroup of C,(x), but C,(X) has a Sylow p-subgroup D 3 Q. Now in any p-group each proper subgroup is properly contained in its normalizer; therefore there exists y E D\Q with y E N. Then (Q, y) G C,(x), and so Q is not a Sylow p-subgroup of C,(x), contrary to our hypothesis. This shows that Q must be a defect group of W . Since W* 2 W, it now follows from (ii) that W = W * . (iv) Let Wl, W 2 , ..., W s be the conjugate classes of G and cI, c2, .. ., c, the corresponding class sums. SincefE Z(kG) we can writef= yici for some yi E k, and by Lemma 4.3B we know that for some], y j # 0 and the class V j has a defect group containing Q.Then a(cj) # 0 by (i) and it follows (since the cj are linearly independent) that a(f) # 0. Because a is an algebra homormorphism, this shows that a ( f ) is an idempotent in Z(kN). Similarly, a(f’) is an idempotent in Z ( k N ) . Sincefandf’ are distinct block idempotents, thenf’ = 0 and therefore a(f)a(f’)= 0 so o(f) and a(f’)are orthogonal.

ci=

6.2

Blocks with normal p-subgroups

In the present section we shall examine the situation where the group G contains a normal p-subgroup Q and relate the set of blocks of kG with the set of blocks of kc“ where := G/Q. In the following section we shall show how to reduce the general case to this situation.

c

Remark The natural mapping G -+ G/Q = G induces a ring homomorphism 8: kG -, kc. We shall use this frequently and it is important to note that Ker /IG rad(kG) and so is a nilpotent ideal. Indeed, by the definition of rad(kG) it is sufficient to show that V Ker /?= 0 for each irreducible kGmodule V (see $1.3).But when V is an irreducible kG-module, Q acts trivially on the kQ-module VQ, since Q is a normal p-subgroup of G and k has characteristic p (see the corollary of Theorem 1.5 and the corollary of Theorem 2.2A). Thus V ( x - 1) = 0 for x E Q. Since Ker is generated as an ideal by the set {(x - 1) I x E Q}, V Ker 8 = 0 as required. Note In the case that Q is a normal p-subgroup of G,C,(Q) -a G, and so for each conjugate class W of G,W n CG(Q) = W or @. In this situation the results of Lemma 6.1B take an especially simple form. Lemma 6.2A Let Q be a normal p-subgroup of the group G and let I, be the ideal in kG generated by all class sums c of classes with defect groups conjugate to subgroups of Q (see $4.3).Let J, be the k-subspace of I , spanned by those c corresponding to classes that have Q (and its conjugates) as defect groups. Then

6.2

BLOCKS WITH NORMAL

SUBGROUPS

145

(i) J Q J Q G J Q ; (ii) if V is any class in G which has a defect group D not containing Q, the corresponding class sum c lies in rad Z(kG); (iii) I , E J, + rad Z(kG); (iv) Q is contained in each defect group of each block of k G ; (v) iff is a block idempotent of kG, and the block f ( k G ) has Q as its (unique) defect group, thenfE J, . Proof Let 0 denote the Brauer homomorphism with respect to Q.Then d is an endomorphism of the algebra Z(kG). By Lemma 6.1B [parts (i) and (ii)], for any class sum c E I, ,~ ( c=) c if the corresponding class V has Q as a defect group and a(c) = 0 otherwise (see the note above). (i) JQ = a(fQ)and SO JQJ? = ~ ( 1 ;G) o(IQ) = J,. (ii) By hypothesis there exists x E W such that Q is not contained in C,(x). (Actually, if this happens for one x E V it happens for all x E V because Q is normal.) Since Q 4 G, CG(X)Q is a subgroup of G and I CG(X)Q : C,(x) I = I Q : C,(X) n Q I = q, say, is a nontrivial power of p. Choose a right transversal of C,(X) in G of the form y i z , ( i = 1, 2, ..., q; j = 1, 2, ..., n), where yl, y,, . . ., y, is a right transversal for C,(X) in C,(x)Q, and zl, z2 , . . ., zn is a right transversal of C,(x)Q in G. Then

c n

c=

a

~z;ly;'xy,z,.

j=l i=l

If we apply the homomorphism fl described in the remark above we obtain B(c) =

i

j= 1

qB(zj- ' ) B ( X ) B ( Z , ) = 0,

since p I q. Thus c E Ker B c rad kG. Since c E Z(kG), this shows that c E rad Z(kG). (iii) This follows immediately from (ii) and the definition of I, and J , . (iv) Let o be the central character kG associated with a block B of kG. By Theorem 4.3 there exists a class V in G with class sum c such that o(c) # 0 and the defect groups of V are the defect groups of B. Since o ( c )is a nonzero element of k, this implies that c is not nilpotent and so c # rad Z(kG). Hence by (ii) the defect groups of W (and hence of B) all contain Q. (v) Let V1, g2,. . ., Ws be the conjugate classes of G and cl, c , , .. ., c, their class sums. By Lemma 4.3A we can writef= yi ci , where yi = 0, unless some defect group of Vi contains the defect group Q of the block f ( k G ) . Hence by definition of I,, J E I,. By (iii) we can write f = a + b, where a E J,, and b E rad Z(kG). Since f is an idempotent and Z(kG) is commutative, the binomial theorem shows thatf=fP' = up' + bP' for each

xr=

146

VI

THE MAIN THEOREMS OF BRAUER

integer r 2 0. Since b E rad Z(kG),it is nilpotent, and so for sufficiently large t, f = a"' E J , by (i). Lemma 6.2B Let Q be a normal p-subgroup of a group G and let G':= G/Q. For each conjugate class W in G we define 3 := {Qx I x E W} in G. Put H := QCG(Q) and cl' := H / Q . Then

(i) Let W be a conjugate class of p'-elements in G and suppose a defect group D of W contains Q. Then D/Q is a defect group of 0. (ii) The mapping %'-@ gives a bijection from the set of all classes of p'-elements in G with Q as defect group onto the set of all classes of p'-elements in G that lie in H and have 1 as defect group. Proof (i) Choose x E % such that D is a Sylow p-subgroup in C,(X). Then D/Q is a Sylow p-subgroup in CG(X)/Qand clearly CG(X)/QE C,(Qx). We claim that C,(x)/Q contains all p-elements of C,(Qx); it will then follow that D/Q is a Sylow p-subgroup of C,(Qx) and hence a defect group of 9. Let Qy be any p-element in C,(Qx). Then Qy commutes with Qx, and so x - ' y - ' x y E Q. This means y - ' x y = xz for some z E Q, and z x = x z since Q G D G C,(x). But x and y - ' x y are p'-elements, and z is a p-element, so we conclude that z = 1. Hence y E C,(X). This shows that Qy E CG(X)/Qas claimed, and (i) is proved. (ii) Part (i) shows that the given mapping maps a class % of p'-elements of G with a defect group Q onto a class 9 of p'-elements of G with defect group 1. Moreover, since Q G C,(X) for all x E W, therefore W E CG(Q)and 5 E I?. It remains to prove that the mapping gives a bijection between the two sets. Let % % be two classes of p'-elements of G with Q as defect group. Since Q -=I G, Q is the unique defect group, and so for each x E W,, Q E C,(x); hence x is the unique p'-element in the coset Qx. Thus if # W 2 , then 5, # 4,; hence the map is injective. On the other hand, let 5%' be a class of p'-elements in G with 1 as a defect group with v' G I?. Each element of %" can be written in the form Qx, where x is a p'-element of G and x E H because %' E I?. If p' := I Q 1, then xp' E CG(Q), since CG(Q) is a normal subgroup of index dividing p' in H.Again x is a $-element and x E ( x p ' ) , so x E CG(Q). Thus the class % of G containing x has a defect group D 2 Q because Q E C,(X).Now clearly @ = %', and by (i) this means $9' has D/Q as a defect group. By hypothesis, 1 is a defect group of v', so we conclude D = Q. This shows that the mapping is surjective, and (ii) is proved.

',

Let Q be a normal p-subgroup of G and suppose that G' :=G/Q. Let j?: kG -+ k c be the ring homomorphism induced by the canonical mapping G + G/Q = G' (see the remark above). Then j? maps the set f l , f 2 , . ..,f , of block idempotents of kG bijectively onto Theorem 6.2A

G

= QCG(Q). Put

6.2

BLOCKS WITH NORMAL

147

p-SUBGROUPS

the set of block idempotents of kG. Moreover, if the blockJ(kG) has D as a defect group, then the block p(.fi)(kc') has D/Q as a defect group ( i = 1, 2, ..., 1). Proof Since B is an algebra homomorphism of kG into kG, we have the decomposition 1 = p(f1) + p(f2) + . * .+ p(&) into central idempotents of kG. [By the remark above, Ker p is nilpotent, so each p ( J ) # 0.1 Therefore, to prove the first assertion of the theorem, it remains to show that each /I(&) is a block idempotent. Suppose on the contrary that p(fl), say, is not primitive. Then p(fl) = el e2 * . . en (n 2 2) as a sum of block idempotents. Let w 1 and w 2 be the central characters of kG associated with e l and e2, respectively. Then for i = 1, 2 we have wi(ei) = 1 and wi(ej)= wi(p(fi))= 0 for j # i and l 2 2. Thus

+ + +

This shows that w1 c p = w 2 c p is the central character of kG associated with the block fl(kG). Since p is surjective, this shows that w1 = w2 and so e l = e2 contrary to our assumption. Thus p ( f I ) (and similarly, p(f,)) is a block idempotent; the first part of the theorem is then proved. Finally suppose thatfis a block idempotent of kG. Then by Lemma 4.3B, we can write

f=

s1

~ Y i c+ i I = ]

J=SI

+1

Y7cj

9

where cl, c 2 , . . . , cSIare the class sums of classes % % 2 , . . ., %',,that have D as a defect group and c , ~ + ~..., , c, are the class sums of the classes gs1 + . .., V , that have a proper subgroup of D as a defect group. Then by Lemma 6.2B(i), has D/Q as a defect group and 3,has a proper subgroup of D/Qas a defect group. Hence D/Q is a defect group of B ( f ) k c as asserted. These results permit us to give a description of the characters that lie in a block whose defect group is contained in the center of G.

Theorem 6.2B (Reynolds) Let Q be a subgroup of order pd lying in the center of G, and let B be a block of kG with Q as a defect group. Then (i) there is only one irreducible modular character of G lying in B (we shall denote it by 4); (ii) there are exactly p d irreducible ordinary characters (', t2,. . ., (9 of G lying in B. Moreover, if A', . . ., APd are the irreducible ordinary characters of Q (all of degree 1 since Q is abelian), then

('(XI

=

lo\Ai(z)4(y)

when x = z y ( z E Q,y for x $ QG".

E

G")

VI

148


Proof We have G = C,(Q). Let f be the block idempotent for B. Then putting i: := G/Q,Theorem 6.2A shows that B := B ( f ) ( k C ) is a block of k c with 1as defect group (so 8 is a block ofdefect 0).Therefore by Theorem 4.6A there is only one irreducible modular character 6 and only one irreducible ordinary character say, lying in B, and 4 = on 6".From these we can define an irreducible modular character 4 and irreducible ordinary character ( of G by

z,

t(x) :=g(Qx) for all x E G, and 4(x):=$(Qx)

z

for all x

E

Go.

Note that since = 0 on G\6",5 = 0 on G\QGo. We first prove that if JI is an irreducible modular character of G lying in the block B, then II/ = 4; this will prove (i). Let V be an irreducible kGmodule which affords the irreducible character JI. Since Ker B annihilates V by the remark above, we can define an action of kG on V by @(a) := ua for all a E kG. Under this action V is an irreducible kC-module. Since V lies in [(kc), V f e O and so V B ( f ) # 0. Thus as a k6-module, V lies in B = /3(f)(kG).Since d, is the only irreducible modular character lying in B, V as a kc-module affords 4, and so V as a kG-module affords the modular character 4. Hence JI = 4 as required. Now the irreducible ordinary characters, say e l , t2,..., (I, lying in B are characters of the form 5' = di 4 on Gofor suitable integers di 2 1 (decomposition numbers), and conversely, each irreducible ordinary character of G of this form lies in B (see 54.4). In particular, t lies in B since 5 = 4 on G o ;put t1= t. Since K is a splitting field and Q C _ Z(G), the representation affording the irreducible character ti will represent each z E Q as a scalar, say Ai(z) . 1, where Ai(z) is a root of unity. Note that 1' is a character of Q of degree 1. Then since ti = d i 4 on Go we have t i ( z y )= diAi(z)4(y)

In particular, I ['(x) we have

1'

I

= d? t(x)

for all z E Q and y

E

Go.

l2 for all x E QG". Therefore (with g := 1 G 1 )

= d?g(t, t ) c = gdi'

because t = 0 on G\QGo. This shows that di = 1 for all i and (since the first inequality must be an equality) ti(x) = 0 for all x q! QG". Thus for all i we have if x = zy with z E Q, y E Go if x $ QG",

6.3

THE BRAUER CORRESPONDENCE: THE FIRST MAIN THEOREM

where A', A 2 , . . ., A' are (distinct) characters of degree 1 of Theorem 4.2C, for each z E Q, z # 1,

Q. Finally by

r

1

I:=I ' ( z )= 0

149

Q\{l}, and so the inner product ( A ' , 1 Q I. Since Q only has pd irreducible ordinary characters, t = pd and A', A', ..., A P d is the complete set of irreducible characters of Q as asserted.

Thus

1' + I'

+ ... +

for all z

I 1

E

= r/ Q and t 2 pd =

EXERCISES

1. Under the hypothesis of the Theorem 6.2A, prove that there is a bijection from the set ofall irreducible modular characters lying in / ? ( f ) ( k G )onto the set of all irreducible modular characters lying inf(kG) given by $I+ 4, where +(x) := $(Qx) for all p'-elements x E G. 2. Let Z(kG) be the center of kG and let a,x be an idempotent in Z(kG). Show that ax = 0 whenever x E G\G".

cxsc

6.3 The Brauer correspondence: The First Main Theorem Let H be a subgroup of G, b be a block of kH and let D be a subgroup some defect group of b. Let w be the central character of kH associated with the block b (see $4.1).We recall by Theorem 4.3 that if %' is a conjugate class of H with class sum c', then w(c') # 0 implies that D is contained in a defect group of W'; in other words, w vanishes on a class sum c' of k H except when c' corresponds to a class %' with %" n C,,(D) # 0.In this situation we can always define a k-linear mapping w c : Z(kG) -,k by (1)

wc(ci):=w(c:)

for i = 1, 2,

.. ., s,

where for each conjugate class Wi of G with class sum ci we put W: := W i n H and c r := CxErlt x. We have wG(ci)= 0 whenever S in C J D ) = (21. Note I Clearly W* is a union of conjugate classes in H, so c: E Z(kH). On the other hand, the class sums cl, c ' , . . ., c, form a k-basis of Z(kG), so wG is well defined by (1). Although, in general, wG is not a central character of kG, the following lemma gives a condition on H under which wc is a central character. Lemma 6 3 Let H be a subgroup of G and let b be a block of kH. Let w be the central character of kH associated with b and define wGas in (1).If D is a p-subgroup of G such that DC,(D) E H c N G ( D ) ,then the following hold:

VI

150

THE MAIN THEOREMS OF B R A U E R

(i) For each class W of H either %' G C G ( D )or %' n C G ( D )= fa. Consequently, wG = w o cr, where cr is the Brauer homomorphism with respect to

D.

(ii) wG is a central character of kG,and the block B of kG associated with wG has a defect group which contains D. Remark It follows from Lemma 6.2A that b has a defect group containing D because D 4H. Proof

(i) Since CG(D) 4 H, and %" is a conjugate class of H , either

%' S CG(D) or W' n CG(D) = 0. It now follows at once from (1) and the Brauer homomorphism (see 6.1) that wG = o cr as asserted. (ii) Since w and cr are both k-algebra homomorphisms (Lemma 6.1A), it 0

follows from (i) that wG is also a k-algebra homomorphism (from Z ( k G ) into k), and is therefore a central character of kG. Let B be the block of kG associated with wG.Theorem 4.3 shows that there exists a class % of p'-elements of G with class sum c such that the defect groups of 9? are the defect groups of B and wG(c)# 0. Then w(cr(c))# 0. This means that some class %' of p'elements of H with class sum c' has the property %' C _ % n H, and o ( c ' ) # 0. By Theorem 4.3 again, each defect group of b is contained in some defect group of %'. Hence by the remark above, D is contained in some defect group of W. Hence for some x E W E % we have D 5 C,(x) G CG(x),and so D is contained in some defect group of %, and hence of B. Definition Under the hypotheses of Lemma 6.3, there exists a function from the set of blocks b of kH with a fixed p-group D as a defect group into the set of blocks B of kG with a defect group containing D.We call this function the Brauer correspondence and denote the image of b under the Brauer correspondence by bG. (See the remark at the end of this section.)

Note 2 Assume that the hypotheses of Lemma 6.3 hold, and let B be a block of kG which has a defect group containing D. Suppose that f is the block idempotent of B and r~ is the Brauer homomorphism with respect to D. Then o(f) is an idempotent in Z ( k H ) by Lemma 6.1B [since Z ( k H ) 2 Z(kN,(D)) n kH], so we can write ~ ( f as ) a sum of block idempotents of k H . Thus we can enumerate the block idempotents el, e 2 , . . ., e, of kH so that

~ ( f =) e l

+ e2 + + en .**

and

+ + e, .

1 -~ ( f= ) en+,

Let wi be the central character of kH associated with the block bi := e i ( k H ) for i = 1, 2, . . ., t. Now by Lemma 6.3, w(;.,w:, . . ., wp are central characters for kG.For i = 1,2, .. ., n we have w f ( f )= (ai cr)(f) = w i ( e , )+ w i ( e 2 )+ . * .+ wi(e,) = 1, and so 07 is associated with B; hence by = b; = = b: = B. On the other hand, for i = n + 1, ..., t we have w Q ( f )= 0

- a .

6.3

THE BRAUER CORRESPONDENCE: THE FIRST MAIN THEOREM

151

C

(wi a ) ( f ) = q(1) - ... - m,(e,)= 0, and so t ~ , , +b:+2, ~ , ..., bp are all different from B (see $4.1). Theorem 6.3 (The First Main Theorem) Let Q be a p-subgroup of a group G and put N := N,(Q). Then the Brauer correspondence b H b G gives a bijection from the set of.all blocks of kN with Q as defect group onto the set of all blocks of kG with Q as defect group.

Proof Let X, (respectively,X,) denote the set ofcentral characters associated with blocks of kN (respectively, kG) that have Q as a defect group. In view of the definition of the Brauer correspondence and the correspondence between blocks and their associated central characters, it is enough to show that the mapping Y : o 4 mG [defined in (l)] is a bijection from X N onto X G. We shall do this in a series of steps. Let o be the Brauer homomorphism with respect to Q. Step 1 Let W be a class in N with Q as defect group and c’ as its class sum. Then there exists a class V in G with Q as defect group and class sum c such that a(c) = c’.

Using Lemma 6.lB(iii), there is a class V of G with Q as a defect group such that W’ = Gx n C,(Q). Then a(c) = c’. w

Step 2 For each central character o associated with a block of kN, E X N iff W‘ E X , . In particular, Y maps X N into X G .

Let b and bG be the blocks associated with o and oG,respectively. Suppose o E X , . Then by Theorem 4.3 there is a class %?‘ of N with Q as defect group such that w(c’) # 0 for the class sum c’. By Step 1 there is a class W of G with Q as defect group, and o G ( c )= o(o(c)) = o(c’) # 0 for the class sum c. Thus Theorem 4.3 again shows that Q is also contained in a defect group of bG;hence Q itself is a defect group of bc. Conversely, using Lemma 6.1B(i) a similar argument shows that if Q is a defect group of bG,then b has a defect group contained in Q by Theorem 4.3. Since Q 4 N, Lemma 6.2A now shows that Q is a defect group of b . Step 3 Y is injective.

Otherwise there exist ol, w 2 E X, with distinct associated block idempotents e l and e2 of kN such that wy = my. By Lemma 6.2A we know that e l E J, , where J, is the k-space spanned by all class sums c’ corresponding to classes W of N that have Q as a defect group. By Step 1, J , c Im a, so e l = .(a) for some a E Z ( k G ) . But then for i = 1, 2, w i ( e l )= w , ( o ( a ) )= o y ( a ) ,

so o l ( e l )= o z ( e l ) . This is impossible since o l ( e l )= 1 and w2(el)= 0. Hence Y is injective.

152

VI


Step 4 Y is surjective. Let rl/ E X G and B be the associated block of kG. Since B has Q as a defect group, it follows from Note 2 above that there is at least one central character w of k N such that wG = $I. By Step 2, w E X,. This proves Y is surjective; and the theorem is proved. Corollary Under the hypothesis of the theorem suppose that H is a subgroup of G with H z NG(Q).Then there is a bijection between the set of blocks of kH that have Q as a defect group, and the blocks of kG that have Q as a defect group.

Proof Both sets correspond bijectively with the set of blocks of k N with Q as a defect group. Remark The Brauer correspondence has been defined for blocks of kH that have a defect group D satisfying DC,(D) c H c N,(D). Theorem 6.3 enables us to extend this definition to the case where simply DC,(D) E H. Indeed, in the latter case, if B is a block of kH with D as defect group, then by Theorem 6.3 there is a unique block b of kN,(D) with D as defect group such that bH = B. Since DC,(D) E N,,(D) G N,(D), bG is already defined as a block of kG with a defect group containing D. We define BG := bG. It is important to note that the relation (1) is still valid for the central characters w and wc associated with B and BG, respectively. EXERCISES

1. Let N be a normal subgroup of G and let e(kG) andf(kN) be blocks of kG and k N , respectively. Show that e(kG) = (f(kN))' if and only if ef= e. 2. Let P be a Sylow p-subgroup of G.Prove that the number of blocks of kG with P as a defect group is equal to the number of conjugate classes of p'elements with P as a defect group. [Hint: Let H I= N G ( P )and let B be the natural algebra homomorphism from kH to k ( H / P ) . First show that the number of blocks of kH is equal to dim,(Im B) and then show the class sums of the classes of H of p'-elements form a basis of Im B.]

6.4 Extension of the First Main Theorem Let Q be a p-subgroup of the group G and let N I= N,(Q). Then the First Main Theorem (Theorem 6.3) gives a correspondence between the blocks of kG with Q as a defect group and the blocks of kN with Q as a defect group. It is sometimes useful to pursue this reduction one further step, namely down to the blocks of kH where H I = QC,(Q). Since H E N, there is no loss in generality in supposing that we have already carried out the first reduction and so take G = N.The reduction to the set of blocks of kH with Q as defect group is no longer one-to-one, but corresponds to a natural equivalence relation on this set.

6.4

EXTENSION OF THE FIRST MAIN THEOREM

153

Definition Let H be a normal subgroup of the group G. Let e and e' be block idempotents of k H . We say that the block idempotents e and e' [and the corresponding blocks e(kH)and e'(kH)]are G-conjugate ife' = y - ley for some y E G.

Note 1 If { e l , e , , . .., e,} is the set of all block idempotents of k H , then G acts on this set by conjugation (because H -=I G).The orbits under this action are the sets of G-conjugate block idempotents of k H . If D is a defect group for e(kH), then y - ' D y is a defect group for ( y - ' e y ) ( k H ) . 1 Suppose that with the notation above x is an irreducible ordinary character of H lying in a block e(kH). Then for any y E G we can define the character x Y by xY(x):= x(y- ' x y ) (see $2.2).If x is afforded by the kH-module V, then x Y is afforded by the kH-module P,which has the same underlying k-space and where the action of H on P is defined in terms of its action on V by u* := u(yxy- '). If V is a kH-module which affords x, then P is a kH-module which affords xy. Since V lies in e(kH), Ve # 0, and hence P ( y - l e y ) # 0. This shows that P lies in ( y - 'ey)(kH).Thus we have shown that G-conjugate characters x and xy lie in G-conjugate blocks e(kH) and ( y - 'ey)(kH),respectively. Clearly the analogous result holds for an irreducible modular character of H lying in e(kH). EXAMPLE

EXAMPLE 2 With the notation above, if e ( k H ) is the principal block of kH, then the only G-conjugate of e(kH) is e(kH) itself. Indeed, by definition 1, lies in e ( k H ) and by Example 1, 1& lies in y - 'ey(kH) for each y E G.Since I& = I,, this shows that y - ' e y ( k H ) = e(kH) for all y E G.

Let Q be a normal p-subgroup of G and put H I= QCG(Q).Let e and e' be block idempotents of kH, b : = e ( k H ) and b' :=e'(kH) the corresponding blocks, and o and w' the associated central characters of k H . Then the following are equivalent. Lemma 6.4

(i) b and b' are G-conjugate. (ii) w = w' on Z(kG) n Z ( k H ) . (iii) b" = (b')G. Proof First assume that (ii) holds; we shall prove (i). Indeed, suppose that e = e l , e , , ..., en are block idempotents of kH so that e,(kH) ( i = 1, 2, ..., n) are the G-conjugates of b. Then { e l , e2 , ..., en} is an orbit under conjugation by G, so a : = e , + e , + ... + enE Z ( k G ) n Z ( k H ) . Hence w'(a) = o ( a ) = o ( e , ) + .-.+ w(e,) = 1. This shows that w'(ej) # 0 for some j (1 I j 4 n), and so o'(e,) = 1 [since o'(e,') = w'(ej)' = w'(ej)].Hence o'is associated with the block e,(kH). Thus b' = ej(kH) and hence it is Gconjugate to b. Next assume that (iii) holds [so wG = (w')"];we shall prove that (ii) holds. First suppose that W is a conjugate class of H whose class sum c does not lie

VI

154


in kCc(Q). Since Q a G , this means that 45 n Cc(Q) = 0, and so Q is contained in no defect group of %. However by Lemma 6.2(iv)Q is contained in all of the defect groups of b and b’, and so w ( c ) = w’(c) = 0 by Theorem 4.3. Since the class sums of H form a basis for Z(kH),and Cc(Q) a H,this shows that w = w’ = 0 on Z(kH)\kC,(Q). Thus to prove (ii) it remains to show that w = w’ on Z ( k G ) n Z(kH) n kC,(Q). However (iii) and Lemma 6.3 together imply that w u = wc = (a’)‘= w’ o 0, where (T is the Brauer homomorphism with respect to Q.The definition of 0 shows that Im (T = Z(kG) n kCc(Q) because Q 4 G. Therefore w = w‘ on Z ( k G ) n Z(kH) n kCc(Q) as required, and we have shown that (iii) implies (ii). Finally suppose that (i) holds; we shall prove (iii). Let f be the block idempotent of bc, and let CJ be the Brauer homomorphism with respect to Q. Since Q -=I G, the defect groups of bc contain Q (Lemma 6.2A), and therefore u(f) is a central idempotent of k H (Lemma 6.1B) since Im (T E Z(kH). Write o(f) = el + e , + ... + en as a sum of distinct block idempotents of k H . Using Lemma 6.3 the central character of bC is wG = w u, and so 1 = w “ ( f )= w ( e l )+ w(e,) + + w(e,). This shows that w is the central character associated with one of the blocks e i ( k H )( i = 1,2, . . ., n) [see g4.11, and so b = e j ( k H ) , say. On the other hand, since b’ is G-conjugate to b, b = e , ( k H ) for some I, 1 5 1 In, and so w’(e,)= 1. Then (w’)‘(f) = w ’ ( u ( f )= ) w’(e,) + ..* + w’(e,) = 1. Hence the central character (of)‘ is associated with the block f ( k G ) = bC ;this proves that (b)‘ = bC. 0

1’

EXAMPLE 3 With the hypothesis and notation of Lemma 6.4, suppose that b, is the principal block of k H . If b is a different block of kH, then bG # bg. Indeed bG = bg implies b and b, are G-conjugate by Lemma 6.4, and so b = b, by Example 2. Now suppose Q is a normal p-subgroup of G and put H := QC,(Q); since Q 4 G, H is also normal in G. Let u be the Brauer homomorphism with respect to Q,and letfbe a block idempotent of kG. Since Q 4 G, the defect groups off(kG) contains Q by Lemma 6.2A. Since Im 0 C Z(kH), Lemma 6.1B then shows that ~ ( f )is a central idempotent of kH and so can be written ( ~ ( f = ) e l + e2 + + en as a sum of block idempotents of k H . We define

C f : = { e l ( k H ) ,..., e,(kH)}. (Note that C, # 0.) Note 2 If b is a block of k H , then b E Xf iff bc = f ( k G ) . Indeed, if w is the central character of kH associated with b, then b E Zf iff w ( e j )= 1 for somej (1 < j 5 n). Since w G ( f )= w 0 ( ~ ( f=) w(el) + ... + @(en),the latter condition is equivalent to w “ ( f ) = 1, and that holds iff the block bG associated with wG equalsf(kG). In particular, this shows that there is always at least

6.4

EXTENSIONOF THE FIRST MAIN THEOREM

155

one block of kH such that bG = f ( k G ) .By Lemma 6.4, Z, is a complete set of G-conjugate blocks of k H . Theorem 6.4 Let Q be a normal p-subgroup of the group G and put I I G : H 1. Then (with

H := QC,(Q). Let 1 be the largest integer such that p' X, defined by (1) above)

(i) Iff ( k G ) is a block of kG with Q as (unique) defect group, then ZJis a full set of G-conjugate blocks of kH each with Q as defect group, and /I;/1 = O (mod p'); (ii) Conversely, if R is a full set of G-conjugate blocks of kH each with Q as defect group, and 101 = 0 (mod p'), then R = C, for some block f ( k G ) which has Q as defect group. Proof

The proof will be given in several steps.

Step 1 If % is a class in G with Q as defect group, then the class sum c satisfies a(c) = c. Indeed, the hypothesis implies Q E CG(x) for some x E %'. Since Q is normal, this is then true for all x E CG, so V E C,(Q). Hence ~ ( c=) c by the definition of IS. Step 2 Iff is a block idempotent of kG and f ( k G ) has Q as defect group, then a(f )= f and Z is a full set of G-conjugate blocks.

+

It follows from Lemma 6.2A and Step 1 that a(f ) =ft and so f = el e2 ... e n .Since.fE Z ( k G ) , x - ' f x =f for all x E G, and so G acts on the set {eI,e,, ..., en}by conjugation. Suppose {el, e 2 , . .., em},say, is an orbit under this action. I f m c n, then f = (el + e2 ... em)+ (em+,+ ... + en) would give a decomposition off into a sum of two nonzero central orthogonal idempotents in k G . Since f is primitive, this is impossible. Hence G is transitive on {el, e,, . . ., en};that is, Z, is a single G-conjugate set of blocks of k H .

+ +

+ +

Step 3 Suppose %? is a class in G contained in H and that % = 59, u LA%, , where the CG, are conjugate classes of H . Then W has Q as defect group iff% (and hence each 59#)has Q as defect group and p' is the exact power of p in m, where m := I G : NG(%I ) 1. * *

Let x E Y; We first show that NG(%I ) = HC,(x). Indeed, if y E NG(%'), then yxy- E % so y x y - ' = z x z - I for some z E H. This shows that y E zC,(x) -c HC,(x). Hence N & , ) 5 H C , ( x ) and the reverse inequality is trivial. Now $$ has Q as defect group iff Q is a Sylow p-subgroup of C,(x). This is so iff Q is a Sylow p-subgroup of C,(x) and p % I C,(x) : C,(X) 1. This will be the case iff Q is a Sylow p-subgroup of C,,(x) and p X I H C , ( x ) : H 1.

156

VI


So V has Q as a defect group iff '% has Q as a defect group and pl is the exact I (by what we showed above). power of p in m = 1 G : N& S t e p 4 I f f ( k G ) is a block with Q as defect group, then Zfis a set of blocks ei(kH) with Q as defect group.

Let w i be the central character associated with ei(kH).Then w i ( e j )= 6,,, and so from o(f) = e l e2 + en we have w i ( o ( f )= ) 1. On the other hand, i f f ' is any other block idempotent of kG, then sf' = 0 and so 0 = o i ( o ( f f ' ) )= w i ( o ( f )* )q ( o (f ' ) ) , which shows that wi(o(f')) = 0. Thus wi o is the central character of kG associated withf: Sincef(kG) has Q as defect group, Theorem 4.3 shows that there is a class W of G with Q as defect group and class sum c such that (wi o)(c) # 0. With the notation of Step 3, let cj be the class sum in kH of g j . Then by Step 1 we have wi(c)# 0 so wi(cJ)# 0 for some j. Since %? has Q as a defect group, Q is also a defect group of gj,and so Theorem 4.3 shows that ei(kH)is a block with a defect group contained in Q. Hence by Lemma 6.2A, Q is the unique defect group of ei(kH).

+

+

0

0

Step 5

I f j ( k G ) is a block with Q as defect group, then 1 ZfI

= 0 (mod p i ) .

+

+ With the notation in the proof of Step 4, 0 # w , ( c ) = wl(c;) Let g := IG I and consider the g values of wI(x-Ic; x) for x E G. Since w 1 lies in a set of n G-conjugate central characters (see Note 2 above), we get each of the values wi(c;) ( i = 1,2, . . ., n) g/n times. On the other hand, since c; has m conjugates under G, we get each of the values o l ( c ; ) ( j = 1, 2, .. , , n) g / m times. Now since V has Q as defect group, Step 3 shows that p' is the exact power of p in rn and so p'/m is a unit in k. Thus from above

w,(cl,).

0 # ( P ' / m ) { w l ( c ; ) + ol(c;)

+

= (p'/n){w,(c;) w 2 ( c ; )

+ *.. + w l ( C 3 )

+ . * . + w,,(c\)}.

Since the values of oi(cj) are algebraic integers, this shows that p' I n, as required. Step 6 If R is a set of G-conjugate blocks of kH with Q as defect group, and = 0 (mod p'), then R = Zf,where f(kG) is a block with Q as defect group.

IR I

Suppose that R = { e , ( k H ) , , . ., e,(kH)}. Since R is a full set of G-conjugate blocks,fo :=el + e2 + + en is a central idempotent of kG. Let el = Zyic: with yi E k, where the ci are the class sums in k H . Let xl, x 2 , .. ., xh be a transversal of H in G;then each of e l , e , , . . ., en appears h/n times among the h conjugates x i 'el xi. Since p' is the exact power of p in h and p' I n, nlh E k and so

6.5

GENERALIZED DECOMPOSITION NUMBERS

157

Since el(kH) has Q as defect group, Lemma 6.2A shows that yi # 0 implies that the class K i of H corresponding to c: has Q as defect group. Now suppose that f , is written as a k-linear combination of class sums in kG. Suppose that c is a class sum (corresponding to the class W in G) with nonzero coefficient. This means that for some i, yi # O and xJ: ' c : x j # 0, where the class rr. W). Assume that W = WoQ3 Wl as a sum of A(zy)-submodules. Then the character x afforded by W, satisfies ~ ( z y=) 0. Remark Since z and y have relatively prime orders, ( z y ) = (2) x ( y ) . Proof It is clearly enough to prove the result in the case that W, is indecomposable. Let T be the representation afforded by W,, and let E ~ E ~ . ,. ., E, be the distinct eigenvalues of T ( y ) . By the hypothesis (see §3.3), each ei is an algebraic integer in K and so lies in A. If y has order n, then the minimal polynomial for T ( y ) divides X" - 1 E K[X]and so has n o multiple roots. Therefore the minimal polynomial must be (X - E , ) ( X - E ~ ... ) (X - E,) E A [ X ] and so we can write W, = Ker(y - E~

1) 0 ... CD Ker(y

- E, . 1).

Since zy = yz, the modules on the right hand side are all A(zy)-modules. But W, is assumed to be indecomposable, and so W, = Ker(y - E * 1) for some E E A ; that is, T ( y )= E 1. Now write U = U , @ U , @ ... @ U , as a sum of indecomposable A(zP)-submodules. Then by Lemma 6.5A(ii), U ( * )= U I Q 1 ). By Lemma 6.6 this shows that bN = (B*)N is the principal block of kN. Finally, since Q 4 H n N, Theorem 6.4 shows that there exists a block b, of kC such that by n N = B*; then b i = ( B * y is the principal block. Let bo be the principal block of kC. If Do is a defect group of 6 0 , then Q G Do by Lemma 6.3 because Q 4 C. Therefore Do CG(Do)c Do Cc(Q)C C, and so by Lemma 6.6, b: is the principal block of kN.This shows that b$ = b: and so b, = b, by Example 3 of $6.4. Thus B = (B*)" = b: = b t is the principal block ofkH by Lemma 6.6. This contradicts the choice of counterexample, so the theorem is proved.

,;

6.7 The characters in the principal block For a group G we defined O,(G) and O,,(G) as the maximal normal p-subgroup and the maximal normal p'-subgroup of G, respectively. Since the product of two normal p-subgroups (respectively, two normal p'-subgroups) is again a normal p-subgroup (p'-subgroup), O,(G) and O,.(G) are ,(G) as the normal subgroup given by uniquely determined. We define O,,, op~,p(G)/op= ~ (O,(G/O,.(G)). G) Note G has a normal pcomplement iff G/O,,,(G)is a p-group and hence iff Opt,,(G) = G . These subgroups can be characterized in the following way. Theorem 6.7A Let Bo be the principal block of kG.Then

6.7

165

THE CHARACTERS IN THE PRINCIPAL BLOCK

x E O,,(G) iff x E Ker [ for each irreducible ordinary character (i) lying in Bo . (ii) x E O,,,,(G) iff x E Ker 4 for each irreducible modular character 4 lying in Bo . c2, ..., c' be the irreducible ordinary characters lying Proof (i) Let in B,, and put H := Ker ci. Then y E H implies ['(y) = ['(l) for each i; and so Ct,, ['(y)Ci( 1) = C;= 1)2 > 0. Hence Theorem 4.2C shows that each y E H is a p'element; therefore the normal subgroup H of G is contained in O,,(G). It remains to show that O,,(G) H. Put n I= 1 O,,(G) I. Since p Y. n, n is a unit in A, and so

n:=

ci(

=

f :=-

C

XEAG.

x E Op,(G)

+ +

Clearly f~ Z ( A G ) [because OJG) -GI and f 2 =f.Write f = el e, + en as a sum of primitive idempotents in Z(AG). By Theorem 3.4A, el, e2, . . ., zn are block idempotents of kG. If V, is a trivial A-free AG-module, then uf= u for all u E V,, and so To,; # 0 for some j ; hence the principal block Bo (in which V, lies) is equal to the block Z,(kG). Now let V, ( i = 1, 2, ..., t ) be an A-free AG-module which affords the character c' (Theorem 3.3). Since c' lies in B, = Z,(kG), we have uf= 8Zj = 8 for - all-u E V,. On the other hand, for each x E O,,(G), xf=J and so 8 = uf= ( u x ) f = iix; so X E Ker for each i. Finally, we know that Ker TJKer V, is always a p-group by Example 2 of $3.3, and so Opt(G)E Ker implies that Op.(G)E Ker V, as required. (ii) Let 4 be an irreducible modular character lying in B,. Then there exists an indecomposable A-free AG-module V lying in Bo such that 4 is afforded by some irreducible kG-constituent of P. Let [ be the character afforded by V. Since the irreducible constituents of V @ " K all lie in B , , (i) shows that O,,(G) E Ker c so V may be considered as an A[G/O,,(G)]-module. Since k has characteristic p, any irreducible representation of a group G over k contains in its kernel the-normal p-subgroup O,(G) (see the corollary of Theorem 2.2A). In particular, Ker 2 O,,, ,(G). Thus if L := Ker 4 as 4 ranges over the irreducible modular character lying in B , , then L 2 Op,,,(G). It remains to show that we have equality. If we did not have equality, then the definition of O,(G) shows that there ,(G). Then $(x) = 4(1) for all x, so we exists a p'-element x E L, x # Ope, conclude that ((x) = ((1) for each irreducible ordinary character lying in Bo [see Theorem 4.4(iii)]. But this implies that x E O J G ) E O,,.,(G) by (i), contrary to the choice of x. Thus L = O,,, ,(G) and (ii) is proved. Corollury (i) The trivial character is the only irreducible modular character in the principal block of kG iff G has a normal p-complement.

166

VI


(ii) The trivial character is the only irreducible ordinary character in the principal block of kG iff G is a p'-group. EXAMPLE Let z be a pelement in the group G such that H I = C,(Z) has a normal p-complement. If [ is an irreducible ordinary character lying in the principal block of kG, then [ ( z y ) = [ ( z ) for all p'elements y E H.Indeed, by the corollary of Theorem 6.7A, the trivial character t,b, say, is the only irreducible modular character in the principal block B, of kH.If B is any block of kH with defect group D,say, then D 2 ( z ) by Lemma 6.2A, so DC,(D) E DC,(z) = H.Therefore by Theorem 6.6, Bo is the only block B of kH such that is the principal block of kG. We can apply Theorem 6.5 to conclude that there exists an algebraic integer d such that [ ( z y ) = dt,b(y) for all p'-elements y E H.But rc/ is the trivial character, so [ ( z y ) = dt,b(l) = [ ( z ) for all p'elements y E H.

We shall need the following classical theorem of Burnside in the next chapter. The usual proof is based on ideas from transfer theory, but we take this opportunity to use some of the results we have just developed to give an alternative proof. Lemma 6.7 Let P be a Sylow p-subgroup of G and let S , and S2 be normal subsets of P . [In particular, we may have Si = {zi}, i = 1, 2, where zi lies in the center Z ( P ) of P.] If S, is conjugate to S2 in G,then S1 is conjugate to S , in N , ( P ) . Proof Suppose S , = y - ' S , y for some y E G.Then both P and y - ' P y are contained in NG(S1).Hence by the Sylow theorems, x - ' P x = y- ' P y for some x E NG(S1). Then x y - E N , ( P ) and ( x y - ')- 'Sl(xy- ') = S 2 . Remark Let P be a Sylow p-subgroup of G and let H be a normal subgroup of G. Put G I = G / H . If C , ( P ) = N,(P), then C e ( H P / H ) = N c ( H P / H ) . Indeed, we have C c ( H P / H )E N c ( H P / H ) = N G ( P ) H / H = C , ( P ) H / H E C c ( H P / H ) ,from which the assertion follows. Theorem 6.7B Let G be a group with a Sylow p-subgroup P such that C , ( P ) = N , ( P ) (so in particular, P is abelian). Then G has a normal p-complement . Proof We proceed by induction on I G 1. Case 2 The center Z ( G ) contains a p-subgroup Q # 1. By the remark above, G/Q satisfies the hypothesis of the theorem. By induction G/Q has a normal p-complement Lo/Q. Since Lo/Q is a p'-group, the SchurZassenhaus theorem shows that Lo has a p-complement L ; and L Lo because Q E Z ( G ) . Then L = Op,(&,) is a characteristic subgroup of the normal subgroup L o , and so L Q G. Since L is a p'-group and G / L is a p-group, L is a normal p-complement of G. Q

6.7

THE CHARACTERS IN THE PRINCIPAL BLOCK

167

Case 2 The center Z ( G ) is a p'-group. In this case, for each p-element z # 1 of G, C,(z) # G and C,(Z) contains a Sylow p-subgroup of G,SO it satisfies the hypothesis of the theorem. By induction C,(Z) has a normal p-complement. Since the theorem is trivial when G is a p'-group, by Corollary (ii) of Theorem 6.7A we may assume that the principal block Bo has more than one irreducible ordinary character. Let c # 1, be any irreducible ordinary character of degree d lying in B, . By the example above, this shows that [ ( z y ) = ( ( z )for each p-element z # 1 in G and each y E C,(z)".Note that C&)" is the normal p'-complement of C,(Z), and since P E C,(Z), I C,(Z)" 1 = I C,(Z) : P 1. Now each element of G can be written uniquely as a product zy of commuting elements (z a p-element and y a p'-element) (see 41.5). Moreover each p-element z # 1 is conjugate in G to one (and, by Lemma 6.7, only one) element of P. Then c - d . 1, is a class function which vanishes on 1 and so

The left-hand side equals

+

and the right-hand side is at least 1 G 1 (d2 1). Hence we must have equality throughout. In particular, this shows that J i ( y )- d l = 0 for all y E cG(1)" = Go,so [(y) = d for all y in Go.Since this is true for each ordinary irreducible character lying in B,,,it follows that the trivial character is the only irreducible modular character in E , . Hence by Corollary (i) of Theorem 6.7A, G has a normal p-complement. Corollary If G is a group with a cyclic Sylow 2-subgroup P, then G has a normal 2-complement.

Proof We shall prove that N , ( P ) = C,(P). Put P = (x). If x has order 2", then the a.utomorphism group Aut P has order 2"-'. Indeed for each integer m, 1 Im I 2", relatively prime to 2" there is an automorphism defined by x i w x i m( i = 0, 1, . . ., 2" - 1). On the other hand, N , ( P ) acts on P

VI

168


by conjugation, and the kernel of this action is C,(P). This gives a group homomorphism N J P ) + Aut P with kernel C,(P). Since P is abelian, P E C,(P) and so N,(P)/C,(P) has odd order. Since Aut P is even, this shows that N,(P) = C,(P) as required. EXERCISES

1. Let G be a group of order prim, (m,p) = 1, and let N be a normal subgroup of order m. Put G':= GIN. If P is any p-subgroup of G, prove that Cc(NP/rV)= C,(P)N/N. 2. Let G be a group and let p be the smallest prime factor of I G I. If the Sylow p-subgroups of G are cyclic, prove that G has a normal p-complement. 3. Let G be a group of order p"m, (m,p) = 1, and let N be a normal subgroup of order m. Let f be a block idempotent of kN and let S :={x E G I x-'fx =f}.If x1 = 1, x 2 , . . ., x, is a right transversal of S in G, show that x; 'fxi is a block idempotent of kG.

c:=

6.8

Notes and comments

Several of the results of this chapter can be proved by ring theoretic arguments (without any recourse to character theory). For an expository work in this line, see Michler [ 11. The First Main Theorem was proved in Brauer [8] and a somewhat different (although not quite complete) proof was given by Osima [ 11. Both proofs use the theory of ordinary and modular characters. The proof given here is due to Rosenberg [ 13. For other proofs of the First Main Theorem see Brauer [lo] and Scott [l]. For the relation between the blocks of normal subgroups and the blocks of groups see Fong [l], Passman [2], and Reynolds [l]; in particular, Theorem 6.2B is due to Reynolds [ 13. There are several proofs of the Second Main Theorem. The original proof of this theorem uses a combination of arithmetical and norm-theoretic arguments. Iizuka [l] simplified some of Brauer's lemmas. The proof of Nagao [ 11 is based on results of Green [ 421 concerning P-projective modules. An arithmetical proof based on the choice of a suitable order in the group ring was given by Dade [ 11. The present proof (based on Nagao [ 13) was given by Dade [5]. Also see Scott [l]. The Third Main Theorem was proved by Brauer [9] and the present proof closely resembles Brauer's proof. For a group theoretic proof of the result that the modular kernel of a block of a finite group G is p-nilpotent, see Michler [2]. Theorem 6.7B is a well-known result of Burnside, but our proof is believed to be new. For a proof by transfer theory see Hall [l] and for a character theoretic proof see Feit [l].

CHAPTER

VII

Fusion of ZGroups

An important problem in the theory of finite groups is the classification of simple groups. Feit and Thompson [2] have proved that the orders of nonabelian simple groups are even, and so the simplicity of a nonabelian group depends on its Sylow 2-subgroups. Two conjugate classes of a subgroup H of a group G arefused in G if they are contained in a single conjugate class of G. In the present chapter we shall prove three main results (Theorem 7.3B, Theorems 7.4A and 7.4B, and Theorem 7.9, each of which can be interpreted as saying that certain kinds of fusion can never occur when H is a Sylow 2-subgroup with specified structure and G is a simple group. Alternatively, they can be read as giving sufficient conditions for the group G to be nonsimple. 7.1 Further results on generalized decomposition numbers Let z # 1 be a p-element of G and put H := C,(z). Let c$', c$2,. . .,c$' be the irreducible modular characters of H (so r is the number of classes of p'-elements of H). Then c$', c$2, . . ., c$' are linearly independent elements of Class,(H") (from Theorem 3.7B) and so form a Z-basis of the free Z-module that they generate and which we shall denote by Char(H"). Since rank(Char(H")) = r, any Z-basis of Char(H"), say I)', I)2,. . ., yY, has r 169

170

VII

FUSION OF 2-GROUPS

elements and for some invertible r x r matrix [aij] with integer entries we have r

4, = C aij$j on H"

( i = 1, 2,

..., r).

j= 1

Now relative to this new Z-basis of Char@") we can define generalizations of the concept of "decomposition numbers," " Cartan matrix," and "generalized decomposition numbers." We proceed to do this. (a) Let x', x', . . ., xs be the irreducible ordinary characters of H . Then the decomposition numbers (with respect to $', $', . . ., @) are the (uniquely determined) integers d,, such that r

(2)

xi =

C

dij$j on H"

( i = 1, 2,

..., s).

j= 1

These decomposition numbers can be calculated from the ordinary decomposition numbers using (1). (b) The Cartan matrix (relative to $', $', . .., @) is defined to be the r x r matrix rH = A; A H , where A H := [d,,] is the matrix of decomposition numbers given in (a) (compare with Theorem 3.7A). (c) Let e l , . . ., be the irreducible ordinary characters of G. Then the generalized decomposition numbers at z (relative to $', $', . . ., are the (uniquely determined) algebraic integers dij satisfying

cz, e"

v)

r

(3)

for all y E H" ( i = 1, 2,

d;,$j(y)

C'(zy) =

. .., n).

j= 1

These generalized decomposition numbers can be calculated from the ordinary generalized decomposition numbers (see $6.5) using (1). In particular, suppose that B is a block of kG and the characters Ci are ordered so that c2, , .., C' are the irreducible ordinary characters lying in B. Suppose further that the irreducible modular characters of H are ordered so that +', 4', . . ., 4" are those characters that lie in blocks b with the property bG = B ; and that the Jli are chosen so that $', $*, . . .,$" form a 2-basis of the submodule . .., 4". Then it follows from Theorem 6.5 of Char(H") generated by &, (The Second Main Theorem) that d;, = 0 for all i = t + 1, . . .,n and j = 1 , 2 , ..., m.

c',

+',

Note I Suppose that z # 1 is a pelement in G and w:=xzx-l is a conjugate of z. Then CG(w)= xCG(z)x-'. Since c' is a class function, it follows from (3) that r

l'(wy) = C'(zx-'yx)

=

C j= 1

d;j$j(x-lyx)

7.1

FURTHER RESULTS ON GENERALIZED DECOMPOSITION NUMBERS

171

for all y E C,(w)" and i = 1, 2, . . ., n. Hence relative to the conjugate Z-basis (I)')~ ( j = 1, 2, . .., r ) of Char(C,(w)") we have the same generalized decomposition numbers at w as we do at z. Note 2 Using (1) and the definition of rH,it follows by a direct calculation from Theorem 3.7B that (t,bi, $ j ) H o = c;, where ril= [cij]. (Ofcourse this is not the same Cartan matrix as the one referred to in Theorem 3.7A.) Theorem 7.2 With the notation above, if r

(4)

C I=1

dfi

6= cji

rH = [cij],then

for i = 1, 2, . .., rn and all j

(for this section, the bar denotes complex conjugate). Moreover, if w is a nontrivial pelement of G not conjugate to z, and d c are the generalized decomposition numbers at w with respect to some Z-basis of Char(C,(w)"), then 1

(5)

C ei

for all i = 1, 2, ..., rn and all j .

=0

I= 1

Remark The case w is conjugate but not equal to z is covered in the Note 1 above. Proof Let Wl, g 2 ,. . .,W, be the classes of p'-elements in H,and for i = 1, 2, . . ., n, define (j := ( ' ( z y ) when y E Wj (the value of cj is independent of the choice of y because z E Z ( H ) ) .Similarly write = $'(y) when y E W j . Consider the n x r matrix X := [cj], the r x r matrix U :=[t,bj] and the n x r matrix A:=[d;,]; H r =lcjl] is an r x r matrix. By definition we have X = AU. Now ( f ( j = nj dij, where n j = I C,(ZY)( for y E W j . Note that C,(ZY) = C&) n C,(Z) = C H ( y )because z and y are, respectively, the p-part and the p'-part of zy; therefore nj = I C H ( y I) for y E W j . This shows that the complex conjugate transpose X* of X satisfies X*X = [ni Sij]. On the other hand, with g := I G I and hi :=g/ni it follows from Note 2 that

Ch,$fs

= g(t,bi,

$j)

= cijg

where

r,

= [cij],

I= 1

Thus U[n; Sij]U* = r, is invertible, and so

H r

'. Since

$I,

$', . . ., i,V are linearly independent, U

= (U*)-'[ni 6ijIU-I = ( U * ) - ' X * X U - ' = A* A.

Hence Z, d;i d;j = c i j for all i and j. However if i I rn, then the Second Main Theorem shows that dfi = 0 for 1 = t 1, . . ., n [see (c) above]. Hence (4) is proved. Consider (5). Since z and w are not conjugate, the two elements zy and wx

+

VII FUSION OF 2-GROUPS

172

are not conjugate for any y E C,(Z)" and x E C,(W)". Therefore by Theorem 2.2A we have

(',rz,

for all y E C,(z)", x E C,(W)". Let .. .,c'be any basis of Char(CC(w)"). Then substituting in (6) the values of ['(zy) and ~ ' ( W X )in terms of the generalized decomposition numbers we obtain

for all y E C,(Z)" and all x E C,(W)". The linear independence of $ I , ( I , t2, .. ., 5" now show that

fl and

dfi

=0

$2,

. . .,

for all i and j.

I= 1

Finally using the fact that dii = 0 for i = 1 , 2 , . . ., m and I = t (c) above], we have ( 5 ) .

+ 1, ..., n [see

Corollary Let z be a nontrivial p-element in G and put H I = C,(Z). Let Bo be the principal block of kG and b, be the principal block of kH.Let c', .. ., and xl, x2, . . .,x" be the irreducible ordinary characters lying in Bo and b o , respectively. Then for all y E H",

rz,

r'

I

I'

I= 1

,

I= 1

I= 1

whenever w is a p-element not conjugate to z in G. Proof (i) If b is any block of kH with defect group Q,say, then (z) G Q by Lemma 6.2A because (z) Q H. Hence H 2 QC,(Q), and so by the Third Main Theorem (Theorem 6.6), bC = Bo iff b = b, . Choose $', 1(12, , .., $" as those characters lying in the block b , . Then it follows from (3) and the theorem that for all y E H"

m

=

m

1c i=l

Cji$'(y)W

j=1

since dii = 0 for 1 = 1, 2,

..., t when i > m. On the other hand, we can write

7.1

FURTHER RESULTS ON GENERALIZED DECOMPOSITION NUMBERS

173

the x1 in terms of the $1 using the decomposition numbers dij [Eq. (2)]. Since dli = 0 for 1 = t' + 1, . . ., n and i = 1, 2, . . ., m by ordering of the $ j , the relation [cij]= rH = [dijIT[dij]shows that cji = Cr= d,, d l i . Hence for all y E H",we have

m

This proves (i). (ii) Let tl, t2,..., Then as above

m

r' be the irreducible modular characters of C,(W).

by ( 5 ) whenever w is a p-element not conjugate to z . EXAMPLE Suppose that the hypotheses of the corollary hold and suppose that H = C,(Z)has a normal p-complement. In this case, for each y E H", we have

where P is a Sylow p-subgroup of H. Indeed, since H has a normal pcomplement, H/O,,(H) N P. Let x be an irreducible ordinary character of H. If x lies in the principal block b o , then Ker x 2 O,,(H) by Theorem 6.7A. Conversely, since O,,(H) = H" in this case, Theorem 4.2B shows that if Ker x 2 O,,(H), then x lies in b o . Thus the characters xl,x2, . . ., xt'lying in b, are precisely the irreducible characters of H whose kernels contain O,.(H). These characters correspond in a natural way with the set of all irreducible ordinary characters of P N H/O,,(H). Hence by Theorem 2.2A we have for each y E H" 1'

c

I=1

1'

Ix1(y)l2 =

c Ix1(1)I2

I= 1

=

PI.

Thus (7) now follows from the corollary. EXERCISES

cz,

1. Let B be a block of kG and c', ..., c' be the irreducible ordinary characters lying in B. If z # 1 is a p-element of G, let dfj be the ordinary generalized decomposition numbers at z. For each p-element w of G, show

VII

174

FUSION OF 2-GROUPS

zi=

that dfi is a rational integer divisible by p. [Hint: For any y E C,(z) and x E C,(W), show that ('(zyM'(wx) belongs to nA.1 2. Let B be a block of defect 0 and let (' be the (unique) irreducible ordinary character lying in B. Show that the generalized decomposition number di, = 0 for a l l j and all pelements z # 1 in G. 7.2 Some technical lemmas

In the remaining theorems in this chapter we shall be interested in the Sylow 2-subgroups of the group and consequently be applying our results on modular representations in the case where the characteristic of k is p = 2. The prime 2 plays a special role in many group theoretic investigations, partly because the existence of involutions (elements of order 2) in a group enables one to analyze the structure of the group more easily. The lemmas of the present section deal with results of this type; they will find application in the succeeding sections. Lemma 7.2A (p = 2) Let z be a 2-element in the group G, let Vl, Vz be two classes of involutions in G such that no element in VIV, has z as its 2-part, and let (?, . . ., c' be the irreducible ordinary characters lying in a block B of kG.Let 1(/l, +2, . .,,fl be a Z-basis of Char(C,(z)") such that 1(/l, 1/2,. . ., 1(/" is a Z-basis of the submodule generated by the modular characters of C,(Z) that lie in blocks b with the property bG = B. Let df, be the corresponding generalized decomposition numbers at z. Then for each x E Vl, y E V, we have

(',

(1)

ci;,[i(x)ci(y)/ci(l)

=o

for j = 1,2,

..., rn

i= 1

z I

(2)

('(z)(i(x)~i(Y)/~i(l) = 0.

i= 1

Proof Let V,, V,, ..., V, be the classes of G and cl, c,, ..., c, the corresponding class sums in K G ;we have x E Vl, y E V, . Then we can write

c v,ci S

(3)

clcz =

(with vi E Z).

i= 1

If T is a representation of G over K which affords the irreducible character (, then T(c,)T(c,)= v, T(c,)by (3). However T(c,) is a scalar because K is a splitting field, and so T(c,)= 1,* 1 gives A,((l) = hi(, where h, := I%,[ and c, is the value of ( on W,, Thus we have

E=

(4)

,=

1

7.2

175

SOME TECHNICAL LEMMAS

If w is a 2'-element in C,(z), then zw # V1%,' by hypothesis, and so (zw)- # %?;'%; = V,%, (since %?,, g2 consist of involutions). Suppose that n g1g2 = @, (3) shows that Then since zw E W l and put

'.

c', c2,

cs

v l z = 0. On the other hand if .. ., are all the irreducible ordinary characters of G, then (4) together with the orthogonality relations (Theorem

2.2A) shows that S

cf

hl h, . I Cic;/ci(1) = vI*I G I = 0.

z.= i= 1

Writing

c; = ~ ' ( z w )=

&j$ j ( w ) ,

this gives

for each Z-element w in C,(z). The linear independence of $', shows that

$2,

. ..,fl now

for each j. From the ordering of the $ j it follows from the Second Main Theorem that dfj = 0 for i = t + 1, ..., s a n d j = 1, 2, ..., m (see (c) of57.1). Therefore for j = 1, 2, . . ., m we have

c I

d;jci,c$/ci(l) = 0.

i= 1

zm=l

This proves (1). Equation (2) now follows because ('(z) = cj=l dfj$j(l) = dfj$j(l) for i = 1, 2, . . ., t. Remark If z is a 2-element of the group G and g1and g2are classes of involutions in G such that for each x E W1, x- 'zx # z- I , then no element in Vl V , has z as its 2-part; so in this case (1) and (2) hold. Indeed, suppose on the contrary that x E W,, y E W, and xy = zw for some 2'-element w E C,(z). Then x- lzwx = y x = y- l x - = (xy)- = (zw)-' = w - 'z- '. But equating the 2-parts then gives x- lzx = z- l , contrary to the hypothesis. Let el, ..., be the irreducible ordinary characters of G and suppose that e l , c2, . . ., c' are the characters lying in the principal block Bo . Let Q be a p-subgroup of G. Then for each a E Char(Q), the induced generalized character a' E Char(G); suppose a' = ai (where each a, E Z). We define i(a):= [a,, a,, . . .,a,], the t-tuple of coefficients corresponding to the characters e l , c2, . . .,C' lying in Bo .If /.I E Char(Q) and i(p) = [b,, b, , . . .,b,], then we define the dot product as usual by i(a)* i(/.I) = a , bl ... arbl.

c2,

cs

xr=

ci

+ +

Note By the Frobenius reciprocity (Theorem 2.5A) we have a, = (a',

Ci)'

= (a,

c&.

VII

176

Hence if x', x2, i = 1, 2, ..., t

. . ., 2''

FUSION OF

2-GROUPS

are the irreducible ordinary characters of Q, then for

S'

Cb = ]=1ai,x',

where h(xJ)= [ a l j , a 2 j ,.. ., atj].

1

Lemma 7.2B (p = 2) Let Q be a 2-subgroup of the group G and let S be a subset of Q. Let H be a subgroup of G such that Q E H c N,(Q). Let a, /3 E Char(Q) such that a(.) = 0 for all x in Q\S and let C2, . . ., (' be the irreducible ordinary characters of G lying in the principal block of kG. Suppose that

(',

(i) if x E S and y E Q are conjugate in G, then they are conjugate in H ; and ICi(x)l' for all (ii) there is a constant y such that y l C H ( x ) l = x

E

s.

Then

Proof By the note above

Now r Z C i ( x ) C i ( y - l )= i= 1

1'

C I ( ' ( X I I'

if y is conjugate to x

i=l

0

otherwise

by the corollary ofTheorem 7.1. Therefore taking into account (i) and (ii) we conclude that

y Conjugate to x

I

a(x)

=y Q xcs

C B(z- 'x-

'z)

Z E H

Theorem 7.2 (p = 2) Let Q be a 2-subgroup of the group G, u an involution in G,and S a subset of Q such that for each x E G conjugate to an

7.2 SOME TECHNICALLEMMAS

177

element in S, u- ' x u # x- Let a E Char(Q) with a(.) = 0 for all x E Q\S. Finally, let c2, . .., be the irreducible ordinary characters in the principal block of kG with c1 = lG. Then

c',

c'

(7)

where h(a) :=[a,(a), a2(a), . . ., a,(a)]. Proof By the remark following Lemma 7.2A we see that the hypotheses of Lemma 7.2A are satisfied in our case with u in place of x and y, and any element of S - in place of z. Thus by (2)

ci)G

for all x E S. Since a,(.) = (aG, = (a, cb), = I Q lxEQ a(x)c'(x- ' ) and a(.) = 0 on Q\S, ( 5 ) follows at once from (8). On the other hand since u - l u u = u - ' , u is not conjugate in G to any x E S by the hypothesis on S. Therefore by the corollary of Theorem 7.1, Ci(l)ci(x)= 0 and ci(u)ci(x)= 0 for all x E S, since 1, x, and u are mutually nonconjugate elements of G. Again using the expression for ai(a)given above, (6) and (7) follow at once.

C:=

Corollary If t > 1 and (a, 1Q)Q # 0, then either there exist at least two q ( a ) > 0 and at least two a,(a) < 0, or else G has a proper normal subgroup

containing O,,(G).

Proof Since (a, 1,) # 0, not all a&) = 0. Suppose that the characters are ordered so that a,(.) # 0 for i I rn, but ai(a)= 0 for i = rn + 1, . .., t. By (6) not all ai(a)have the same sign. Suppose that the first alternative does not hold. Taking - a in place of a if necessary and reordering the 4' we may assume that ai(a) > 0 for i = 1, 2, . .., rn - 1 and am(a) 0. Then

-=

vI1

178

FUSION OF 2-CROUPS

which equals 0 using (5), (6), and (7). This implies that ['(u)[J( 1) = ['( l ) [ J ( u ) for all i, j = 1, 2, , . ., m - 1 and hence [using (6) and (7)J for i = m as well. In particular, taking [ j as the principal character l G ,we get [ ' ( u ) = ['( 1) for all i. Since Ker 0 2 , ( G )for all i (by Theorem 6.7A), the corollary follows.

['

7.3 Groups with Sylow 2-subgroups of type (2", 2") In the present section we investigate the structure of a group G which has a Sylow 2-subgroup P of the type (2", 2"), (that is, P is isomorphic to a direct product of two cyclic groups each of order 2"). In particular, when m > 1, Theorem 7.3 shows that a group with such a Sylow 2-subgroup cannot be simple. Remark In dealing with involutions the following fact is often useful. If u is an involution in the group G,and [ is an irreducible ordinary character of G, then [ ( u ) is an integer and ( ( u ) = [(l) (mod 2). [Proof If T is a representation which affords [, then the eigenvalues of T(u)are 1and - 1, and so [ ( u ) is an integer. The second assertion follows from the observation (( r and then a, = - 1. On the other hand, r = 4(22m- 1) 2 5 since m > 1. Therefore it follows from (6) that a , = 0 and (reordering if necessary) a,, = -gl, a,, = - c 2 , and a,+ = -c3 for suitable ci = 1 or - 1, whilst a, = 0 for all j > r + 3. Substituting these values and calculating we obtain - $ o ) = a($, - $J i($, -

+

7.3

GROUPS WITH SYLOW 2-SUBGROUPS OF TYPE

Finally, if $i is any character of P ( i = 0, 1, some x E H and some j 5 r. Then ($i

- $0

9

(2", 2")

183

. . ., 22" - l), then $f = $ j

Cb> = ($t - $0 (CLP) = = ( $ j

for

9

- $0

9

9

Cb),

for I = 1, 2, .. ., t (since cp is constant on classes of G). Thus by the definition of h we have proved that for i = 1, 2, and 3,

&+'>-

1, ( 8 ) implies that z , + ~= -ci (mod 16) for i cular, either z , + ~ = 1 or z , + ~2 15. From (9) we obtain

= 32,.

E~(~(U)*

=

1 , 2 , 3 ; in parti-

(mod 16)

and so (1 1) shows that z j = k 3 (mod 16). We saw above (in the calculation of h(t,hr - $o)) that one of the characters ( i = 1, 2, 3 ) is the principal character of G ; say = l G ,and then el = -zr+' = -1. On the other hand, since ICi(u)l = 3 by ( l l ) , these conditions on z j , z ~ + z~, +, ~ ,and z , + ~ together with (9) easily show that at least one of z , + ~ and z r + 3 is also 1. Suppose z , + ~= 1; then E , + ~ = - z , + ~ = - 1.

c+i

VII

184

FUSION OF

2-CROUPS

Finally, from (7) we conclude r + ' ( z )= - E , + , = 1 for all z E P\{l}; so P is contained in the kernel L of Since O,,(G) = 1, and O,,(L) is a characteristic subgroup of L, O,,(L) = 1. Since P is a Sylow 2-subgroup of L and L # G, the induction hypothesis shows that P 4 L;hence P 4 G and so C,(P) 4 G. Moreover, C,(P) has a normal 2-complement M by Theorem 6.7B. Since C,(P) 4 G , M is normal in G; since O,,(G) = 1, this means M = 1 and P = C,(P). By Lemma 7.3A, I G : C,(P) I = 1 or 3 and so the theorem is proved.

r+'.

Remark The above theorem shows that when rn > 1 there are no simple groups which have a Sylow 2-subgroup of type (2", 2"). In the case rn = 1 this is no longer true. Gorenstein and Walter [l] have shown that the simple groups with Sylow 2-subgroups of type (2, 2) are precisely the projective special linear groups PSL(2, q), where q is a prime such that q = 3 or 5 (mod 8). EXERCISES

1. Let G be a group with a Sylow 2-subgroup Q of type (2, 2) and let u be an involution of G.If G has no 2-complement, prove that

(d.2 + ~ J ( d + 3 ~ 3 ) ( d+ 4 ~ 4 ) IgCdQ)1' = 8dzd3d4 I C d U )

1'9

where 1, d, , d3 , d4 are the degrees of the irreducible ordinary characters in the principal 2-block and E~ = k 1. 2. Let G be a group with Sylow 2-subgroup Q of type (2, 2) and let Q = ( u , 0). i f G has a normal 2-complement M, then prove Wielandt's fixed point formula

[ M I ICdQ)12= ICM(U)II C d u ) I ICduu)I. 7.4 Groups with quaternion Sylow 2-subgroups

A generalized quaternion group of order 2"' (n 2 2) is a group P which can be generated by two elements z and w such that z has order 2", w2 = z2"-' and w - l z w = 2 - l . The following properties of generalized quaternion groups are known (see Hall [13). (a) Any generalized quaternion group has a normal cyclic subgroup of index 2 (generated by z), and posesses only one involution (namely z2n-1

- WZ).

(b) Any two generalized quaternion groups of the same order are isomorphic and any homomorphic image of a quaternion group is either of order less than or equal to 4 or is also generalized quaternion.

7.4

GROUPS WITH QUATERNION SYLOW

185

2-SUBGROUPS

(c) A 2-group possessing only one involution is either cyclic or a generalized quaternion group; consequently any subgroup of a quaternion group is either cyclic or generalized quaternion. The generalized quaternion group of order 8 is known as the quaternion group. It has the property (which we shall need) of being a nonabelian group in which each subgroup is normal. The generalized quaternion groups of order greater than 8 do not have this property. The theorems of this section investigate the structure of groups that have generalized quaternion Sylow 2-subgroups; in particular such groups are never simple. The case of the quaternion group of order 8 is treated first, and then used to give an inductive proof of the general result. Theorem 7.4A ( p = 2) Let G be a group with a Sylow 2-subgroup P which is a quaternion group of order 8. If 0 2 , ( G )= 1, then Z ( G )I = 2.

I

Proof By hypothesis, P is generated by two elements z and w of order 4 with z 2 = w2 and z - l w z = w - ' . Then u = z 2 is the unique involution in P. We shall suppose that the theorem is false and take Gas a counterexample of minimal order, and obtain a contradiction in a series of steps. Let c', 12, . . . , (' be the irreducible ordinary characters of G lying in the principal 2-block. All elements of order 4 in P are conjugate in NG(P). The elements of order 4 in P are z, z - w, w - l , zw, and (zw)- (the two other elements are 1 and u). Since z - l w z = w - l , W - I Z W = z - ' , and z - ' ( Z W ) ~= ( zw) - there are at most 3 classes of elements of order 4 in N,(P). If the elements of order 4 are not all conjugate in N,(P), then one of the sets ( z , z - I}, {w, w - ' ) , and {zw, (zw)-'} is not conjugate to either of the other. Then symmetry of P allows us to suppose that { z , z - ' } , say, is not conjugate to either {w, w - ' ] or {zw, ( z w ) - ' } in N,(P). By Lemma 6.7, {z , z - '} is not conjugate to either {w, w - '} or {zw, ( zw) - '} in G . Thus we can partition P into two disjoint sets P , := { 1, u, z, z - '1 and P - := { w, w - l , zw, (z w) - '} such that no element of P , is conjugate in G to an element of P This permits us to define a class function 8: G -,I< by 8(x) := 1 if the 2-part of x is conjugate to an element in P , and 6(x) := - 1 if the 2-part of x is conjugate to an element in P - We shall apply Theorem 2.6B to prove that 8 is an irreducible character of G. Let E be a subgroup of G of the form E = Q x S, where Q is 2-group and S is a 2'-subgroup. It is readily verified that OE E Char(E). Since every elementary subgroup of G is nilpotent and hence conjugate to a subgroup of this form, it follows that OE E CharfE) for every elementary subgroup of G. Moreover, (6, 6 ) G = 1 because I qX)l2 = 1 for all x E G, and e( 1) = 1. Therefore by Theorem 2.6B, 8 is an irreducible character of G, and clearly I G/Ker 8 1 = 2. Since the subgroups of index 2 in Step 1

',

',

,.

'.

VII

186

FUSION OF

2-GROUPS

P are cyclic, Ker 0 has a cyclic Sylow 2-subgroup and therefore it has a normal 2-complement L by the corollary of Theorem 6.7B. Since L is characteristic in Ker 8, L G.But O,,(G) = 1, so L = 1. Hence G = P contrary to the hypothesis that G is a counterexample. Q

Step 2 For each i, ('(z) is an integer congruent to ( ' ( u ) and ('(1)

(mod 2).

Let T be a representation of G which affords c'. Since z has order 4, the and eigenvalues of T ( z )are the fourth roots of unity, namely 1, - 1, with multi licities m,, m,, m,, and m4, say. Then ('(z) = m , m, (m3- m4&, ('(u) = m, m2 - m3 - m4, and ('(1) = m, m2 m, m4. Since z is conjugate z-l (even in P), ('(z) = v(z-l), and so rn, - m4 = 0. The assertion now follows immediately.

-0, + +

0

+

+ +

Step 3 Analysis at z.

Since z # Z(P), (z) is the Sylow 2-subgroup of C,(Z). Therefore by the corollary of Theorem 6.7B, C,(Z) has a normal 2-complement. The example of $7.1 now shows that zIEll('(z)1' = ((z) I = 4. Since c'(z) = lG(z)= 1, it follows (using Step 2) that c'(z) = f 1 for four of the characters c' [and for these Ci( 1) is odd] whilst ('(z) = 0 for the remaining characters (and for these c(l)is even). Step 4 Analysis at u.

Put H := C,(u) and fi I = H / ( u ) . Then there is a bijection from the set of 2'elements of H onto the set of 2'-elements of I? given by XHX(U); and P I= P / ( u ) is a Sylow 2-subgroup of H and is of type (2, 2). Since N,(P) E C,(U), Step 1 implies that all involutions in P are conjugate in I?. Thus we can apply Lemma 7.3B(iii) to I? and conclude that the submodule of Char@") corresponding to the principal block of k f i has a 2-basis consisting of three generalized modular characters $' = lR, $', and $,. The associated Cartan matrix has the form

[;i i].

Now since ( u ) is a normal 2-subgroup of H,every irreducible representation of H over k has ( u ) in its kernel (corollary to.Theorem 2.2A). Thus there is a one-to-one correspondence between the irreducible modular characters of H and those of fi; namely, to each irreducible modular character 0 of H there is the irreducible modular character of I? given by ~ ( x ( u ) )= 0(x) for all x E H.In particular, the submodule of Char@)' corresponding to the principal block on kH has a Z-basis JI' = ,l JI', and $3 of generalized modular characters ofH related to $l, $', and $3 in the same way. Since IH 1 = 2 I 1,

7.4

187

GROUPS WITH QUATERNION SYLOW 2-SUBGROUPS

we have for i, j = 1, 2, 3. Thus, by Note 2 of $7.1, the Cartan matrix associated with $I, twice the Cartan matrix above, namely 1. For any integer m we have u-"xum = xhn. Since h is odd, we know from elementary number theory that hZd-*= 1 (mod 2d) (since 2'- is the value of the Euler phi-function at 2d);thus u2'- ' E C,(x). This shows that the 2'-part of u commutes with x. Hence without loss in generality we may assume that u is a 2-element. But then (u, x) is a 2-group (with (x) as a normal subgroup) and so contained in a Sylow 2-subgroup P , , say. But in P , the element x of order greater than or equal to 4 has I P , : C,,(x)I = 2 because P , is a generalized quaternion group. Hence x can only be conjugate to x or x - l in P , ; thus u - ' x u = x - l as asserted. This shows that hypothesis (ii) of Lemma 7.2B holds. Secondly, for all x E S, Q is a cyclic Sylow 2-subgroup of C,(x). Hence C,(x) has a normal 2-complement by the corollary of Theorem 6.7B, and so by the example of $7.1 we have I

Since IC,(x)( = ( Q1 = 2"-', hypothesis (iii) of Lemma 7.2B holds with y = 1. Step 2 Application of Lemma 7.28 and Theorem 7.2. Let

zo = 1, and let x

be the irreducible ordinary character of Q = (2)

7.5

GLAUBERMAN’S

Z*-THEOREM

191

0.

defined by ~ ( z=) Then a = x - xo is identically 0 on Q\S, and so by Lemma 7.2B and Step 1

=

( x - xo x - xo + xw - X O ) 9

=3

since

Since the entries of ii(a) are nonzero integers, this shows that it has exactly three nonzero entries, each f 1. Next, the hypothesis of Theorem 7.2 is satisfied. Indeed if a 2-element x has order greater than or equal to 4 in G then u-’xu E (x) implies that ( u , x) is a 2-group; and then u- ‘xu = x since the involution u is in the center of each Sylow 2-subgroup in which it lies. Hence the corollary of Theorem 7.2 shows that G has a proper normal subgroup N, say, containing u. Since O,,(G) = 1 we have O,,(N) = 1. Step 3 Conclusion.

Since the subgroups of a generalized quaternion group are either cyclic or generalized quaternion (see the beginning of this section), the Sylow 2subgroups of N are either cyclic, quaternion (of order S), or generalized quaternion (of order greater than or equal to 16). In the second and third cases (using Theorem 7.4A and the minimality of G,respectively) we see that IZ(N)) = 2. But Z(N) 4 G, and since it has order 2, this means Z(P) = ( u ) = Z(N) E Z(G); thus Z(G) = ( u ) , contrary to the assumption on G.O n the other hand, in the first case N has a normal 2-complement by the corollary of Theorem 6.7B; and since O,,(N) = 1 by Step 2, this means N is a 2-group. As a subgroup of a generalized quaternion group, N has a unique involution, namely, u. Then ( u ) E Z(G), and since ( u ) = Z(P) this means Z(G) = ( u ) again. Thus in all cases we obtain a contradiction to the hypothesis that G is a counterexample. This proves the theorem. 7 5 Glauberman’s Z*-theorem

The theorems of the previous section give sufficient criteria for a group to have a nontrivial center. These have been generalized by Glauberman in what is referred to as the “Z*-theorem,” which we shall consider now. Definition Let G be a group. Then Z * ( G ) is defined as the subgroup of G such that Z*(G)/O,,(G) = Z(G/O,,(G)). In particular, if O,,(G) = 1, then Z*(G) = Z(G).

192

VII FUSION OF 2-GROUPS

EXAMPLE 1 If the Sylow 2-subgroups of G are generalized quaternion (of order greater than or equal to 8), then each involution of G lies in Z*(G). This follows at once from the theorems of 57.4 applied to G/O,,(G). EXAMPLE 2 If the Sylow 2-subgroups of G are cyclic, then G = Z*(G). Indeed in this case,if P is a Sylow 2-subgroup of G,then G = PO,,(G) by the corollary of Theorem 6.7B;so Z(G/O,,(G))= G/O,,(G). We shall use the following elementary group theoretic lemma in the proof of Glauberman's theorem (Theorem 7.5).

Lemma 7 5 Let P be a Sylow 2-subgroup of the group G,and let u and u be involutions in G with u E P.

(i) If uu has order n, then ( u , u ) is a subgroup of order 2n with u-'(uu)u = u-'(uu)u = (uD)-' ("dihedral group of order 2n **).Moreover, if n is odd, then u is conjugate to D in (u, 0). (ii) x- ' u - ' x u is a 2'-element for each x E G iff u is not conjugate in G to

any other element of P. Proof (i) Since u and u have order 2, we have u-'(uu)u = u - ' u - ' = (uu)-' and similarly u-'(uu)u = (IN)-'. Since ( u , u ) = ( u , uu), this proves the first assertion. The second assertion follows at once since, if n is odd, ( u ) and ( u ) are Sylow 2-subgroups of (u, u ) . (ii) First suppose that x- ' u - ' x u is a ?-element for all x E G.If for some x E G we have x - ' u x E P, then x - ' u - ' x u E P is a Z-element and so x- lu- ' x u = 1. This shows that the only conjugate x- ' u x of u lying in P is u itself. Conversely, suppose that the only conjugate of u in G lying in P is u itself. Consider u I = x- lux for x E G.Then for any integer m 2 1, it follows U (uu)-rn(uu)-rnuu= ( U U ) - ~ ~ + ' Now . supfrom (i) that y I= ( U U ) - ~ U ( L W ) ~ = pose that for some x E G,x- ' u - ' x u = LW is not a 2'element. Then we can choose the integer m so that y is a nontrivial 2-element (lying in (u, u ) ) . Then (u, y) is a 2-subgroup and so contained in a conjugate z - ' P z of P . But then zuz-' and z(yu)z-' = z(uu)-mx-lux(vu)rnz-l both lie in P and are conjugate to u in G. By hypothesis this means that they are both equal to u ; hence u = yu and so y = 1. This is contrary to the choice of m. Hence we have shown that x- ' u - ' x u is always a 2'-element in this case, and

the lemma is proved. Theorem 7 5 (Z*-theorem) Let u be an involution in the group G.Then E Z*(G) iff x- ' u - ' x u is a 2'element for each x E G.

u

Proof If u E Z*(G), then for each x E G we have x - ' u - ' x u E O,,(G) because Z*(G)/O,.(G)is the center of G/O,.(G); thus x- ' u - ' x u has odd order. This proves the theorem in one direction. We now turn to the proof in the other direction. Suppose that the assertion is false and take G as a minimal

7.5

GLAUBERMAN'S

Z*-THEOREM

193

counterexample. Then for all x E G, x - ' u - ' x u is a 2'-element, but u $ Z*(G). We shall proceed by a series of steps to obtain a contradiction. Step 1

O , , ( N ) = 1 for each normal subgroup N of G .

Otherwise, we shall have O,,(G) # 1 since O,.(N) is normal in G. But then G/O,,(G) would be counterexample of smaller order. Step 2 Z ( G ) contains no involutions.

Suppose v is an involution in Z(G); clearly u # u. Put G":= G / ( v ) and define L a G by L/( v ) = O,.(c).Then (0) is a Sylow 2-subgroup of L and so by the corollary of Theorem 6.7B, L has a normal 2-complement. But O,,(L) = 1 by Step 1, so we have L = (0). Hence Z * ( c ) = Z(G).Now since G is a minimal counterexample to the theorem, it follows easily that u ( v ) E Z*(G")= Z(G"),and so ( u , u ) a G. But then for each x E G, x - ' u - ' x u has odd order and lies in the 2-subgroup ( u , v ) ; hence x - ' u - ' x u = 1. This implies that u E Z(G) E Z*(G), contrary to our choice of G. Step 3 Let P be a Sylow 2-subgroup containing u. Then u E Z(P) and P contains an involution v diferentfrom u. Moreover, u is not conjugate to u in G .

For each x E P, X - ' U - ' x u E P and x - ' u - ' x u is a 2'-element by hypothesis; hence x - ' u - ' x u = 1. This proves that u E Z ( P ) . If u were the only involution in P, then P would be cyclic or the generalized quaternion group. Since O,.(G) = 1, the corollary ofTheorem 6.7B and Theorems 7.4A and 7.4B then show that u E Z(G) = Z*(C),contrary to hypothesis. Hence P contains at least one other involution v, say. Finally, any conjugate x - ' u x of u in G lying in P has the property that U - ' X - lux E P but u- ' x - ' u x is a Z-element by hypothesis. This implies that u- ' x - ' u x = 1;and so each conjugate x - b x of u lying in P equals u. Since u E P and u # u, we conclude that u is not conjugate in G to u. Step 4

Characters in the principal 2-block of G .

Since G is not a 2'-group we can choose an irreducible ordinary character

C in the principal 2-block of G with ( # 1, (by the corollary of Theorem 6.7A). Suppose u is an involution in the Sylow 2-subgroup P with u # u (see Step 3). Then for each z in G conjugate to u we shall show that (1)

C(uv) = C(uz).

Consider y := uz. If y were a 2'-element, then the involutions u and z are conjugate in ( u , z) by Lemma 7.5. This implies that u and u are conjugate, which is contrary to Step 3. Thus y has an even order 2n, say. Put w := y". Then w is an involution. Put H := C,(w). Since u- ' y u = z- ' y z = y - and w = y" = w - ', we have u, z E C,(W).

FUSION OF 2-GROUPS

VII

194

We shall now apply the Second and Third Main Theorems to H. First of all, if B is any 2-block of H with Q, say, as a defect group, then ( w ) E Q by Lemma 6.2A because ( w ) H. Thus QCG(Q) E QCG(w) = H, and so Theorem 6.6 shows that BG is the principal 2-block of G exactly when B is the principal 2-block of H. Let +', @, ..., +m be the irreducible modular characters of H lying in the principal 2-block of H. Then Theorem 6.5 shows that Q

m

(2)

d,"+j(x)

((wx) =

for all x

E

H",

I=1

where d," (j= 1, 2, ..., m) are the generalized decomposition numbers at w with respect to (. Since G is a minimal counterexample to the theorem, and H # G by Step 2, u E Z*(H). In particular, y2 = (uz)' = u - l z - ' u z E O,,(H), and so the order n of y z is odd. Therefore y"+' E ( y z ) C O,.(H) and so y"+' is a 2'-element of H. Applying (2) with x = yn+' = wy, we get m

( ( y ) = r ( w y n + l )=

1 d,"@(y"+')

j= 1

by Theorem 6.7A(ii), since yn+'

E

m

=

1 d," dj+j(l)

j= 1

O,,(H). With (2) this shows that

(3) C ( U 4 = C(w) by the definition of y. Next we note that since u E C,(W), ( u w ) = ~ 1 . On the other hand, uw = uy y"-' E z ( y 2 ) c_ zO,,(H). Therefore uw is an involution and ( u w ) ~ is a 2'element. By Lemma 7.5 this shows that uw isconjugate to z and hence conjugate to u in G. Choose x E G such that x - ux = uw. Finally we have x - l u x E x-'C,(o)x = CG(uw)by Step 3. Sincex-'u-'xu has odd order by hypothesis on G, we conclude from Lemma 7.5 that there exists s E ( x - l u x , u ) E C G ( u w ) such that S - ~ X - ~ U X S= u. Then ( x s ) - l u ( x s ) = s-l(uw)s = uw = (xs)-lu(xs)w, and so uu = ( x s ) w ( x s ) - l . Since ( is a class function, this latter result together with (3) proves ( 1 ) . Step 5 If u is an involution in P different from u, and ( is an irreducible ordinary character from the principal 2-block of G such that ( # lG and C(u) # 0, then ( ( u ) = -((I).

Let V l and V2 be the conjugate classes of G containing u and u, respectively, and let c 1 and cz be the corresponding class sums. Let T be a representation of G affording (. Since K is a splitting field, T ( c l ) and T(c2)are scalars; taking traces we obtain T ( c , )= (hl((u)/((l)) * 1 and T ( c , ) = (h2((u)/[(l))- 1 where hi I = Now each product x y ( x E V,, y E V2) is

7.5

GLAUBERMAN'S Z*-THEOREM

195

conjugate to an element of the form uz(z E W'), and so by Step 4 we have [ ( x y ) = [ ( u u ) . Thus taking traces in the equation T(Cl)T(C,)=

c TkY)

(summed over x E Wl, y E W z )

we obtain h , h, ( ( u ) [ ( u ) / [ ( 1) = hl h, [(uu). Therefore

l ( U ) l ( 4 = 5(1)((uo).

(4)

Similarly, replacing u by the involution uu (see Step 3) we obtain

l(uK(u4 = r(l)r(u).

(5)

By hypothesis [ ( u ) # 0, so (4) and ( 5 ) imply that [(u)' = [(I)'; that is C(u)= *C(l). It remains to show that [ ( u ) # ((1). Since ( # l,, N := Ker [ is a proper normal subgroup of G, and [ ( u ) = [(l) implies u E N. By Step 1, O,,(G) = 1 and so Z*(N) = Z ( N ) and Z(N) is a 2-group. Since G is a minimal counterexample, and N # G, we conclude that u E Z(N). But Z(N) is characteristic in N and hence normal in G. Therefore for each x E G, x - l u - ' x u E Z(N), and x - l u - l x u is a 2'element by hypothesis on G. Since Z(N) is a 2-group, this shows that x - ' u - ' x u = 1 for all x E G, and so u E Z ( G ) c Z*(G), contrary to hypothesis. This shows that ( ( u ) # [(l), and so ( ( u ) = -[(l). Step 6

The orthogonality relations.

Let [' = l,, [', . . ., [' be the irreducible ordinary characters of G that belong to the principal 2-block of G. With the notation in Step 5 we have ci(u)Ci(u) = -Ci(1)li(u) for all i 2 2. However since u is not conjugate to u (Step 3), the corollary of Theorem 7.1 together with Theorem 4.2C gives a contradiction:

o=

iC(

i= 1

I

u)[i(u) = -

X(i(l)[i(u) i= 1

+ 2[1(1)[1(u)

= 2['(1)[1(u) = 2.

Thus we have arrived at a contradiction to the assumption that G was a counterexample to the theorem. This completes the proof of the theorem.

EXERCISE

Let G be a nonabelian simple group, P a Sylow 2-subgroup of G, and u an involution in G. Then prove that there exists x E G such that x - h x E P but x - lux # u.

196

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FUSION OF

2-GROUPS

7.6 Notes and comments The general problem of fusions of 2-groups is far from complete solution. Usually there are several possibilities for the fusion of a 2-group (and in general a p-group) and this makes the problem more difficult and complicated. In addition to the groups of type (2”, 2”) and quaternion groups considered here, the problems of fusions of a dihedral group, quasidihedral group, and wreathed Sylow 2-subgroups have been completely solved. The proofs of Theorems 7.4A and 7.4B were outlined by Brauer and Suzuki [l], and detailed proofs appeared in Suzuki [ 13 and in Brauer [9, 111. We have closely followed Brauer in the proofs of Theorems 7.3, 7.4A, and 7.4B. Theorems 7.4A and 7.4B can also be proved by using ordinary characters; see Feit [l] and Glauberman [2]. The problem of fusions of a dihedral group was solved by Gorenstein and Walter [2] in a series of three papers, and the cases of quasi-dihedral group and wreathed 2-group were solved in two long papers by Alperin et al. [l, 21. The necessary character theory was developed by Brauer [9, 121. For further extensions of these results see Harada [ 13 and Gorenstein and Harada [ 1, 21. Theorem 7.5 and a generalization (proved purely by group theoretical argument) appeared in Glauberman [ 13. For another kind of generalization see Goldschmidt [ 13. For some general problems of fusion of 2-groups, see Alperin [l] and Goldschmidt [2].

CHAPTER

VIII

Blocks with Cyclic Defect Groups

The earliest attempt to determine and describe the irreducible ordinary and modular characters lying in a block was made by Brauer [l], who successfully analyzed blocks of defect 0 and 1. Although Brauer’s methods did not extend to other cases, Thompson [ 11 later gave a new proof of these results which did generalize in certain special cases. By exploiting Thompson’s technique, Dade [2] extended Brauer’s results to all blocks with cyclic defect groups. We shall not prove here the full theorem of Dade, but using his methods we shall give the analysis in an especially important case, namely, for blocks of defect 1 for a group whose order is divisible by the first power ofponly (Theorem 8.5). A number of theorems dealing more generally with blocks with cyclic defect groups lead up to this result. In the final two sections, we apply these results to prove theorems of Brauer, Feit, and Thompson on the existence of normal Sylow p-subgroups in linear groups. 8.1 Extending characters from normal subgroups

In the following sections we shall need on several occasions a method of extending characters from a normal subgroup H of a group G to functions that are characters on G. There are several results of this kind known but here we only deal with the simplest ones. Recall that if 8 is an ordinary (or a modular) character of a normal subgroup H of G, then for each y E G we define B Y as an ordinary (respectively, 197

VIII

198

BLOCKS WITH CYCLIC DEFECT GROUPS

modular) character of H by Oy(x)~=O(y-lxy)for all x in H (respectively, H"). Our first result deals with extension of a character 0 from a normal subgroup H of G up to a character of G. Clearly a necessary condition for this to be possible is that OY = 0 for all y E G. In some cases this is also sufficient. Theorem 8.1 Let H be a normal subgroup of the group G such that G / H is a cyclic group of order h.

(i) Let x be an irreducible ordinary character of H such that x Y = x for all y E G. Then there exists an irreducible ordinary character of G such that CH =

x*

(ii) Let i+h be an irreducible modular character of H such that i+hY = i+h for all y E G, and let p be an ordinary character of degree 1 of G with H = Ker p. If p Xh, then there exists an irreducible modular character 4 of G such that = i+h and i+hG = 4 p$ + . . . ph- '4, where each pi# is an irreducible modular character of G (pi denotes the ith power of p).

+

+

Proof First let T be any irreducible matrix representation of H over an algebraically closed field K O .For each y E G define the representation Ty of H by T ( x ) := T(y- 'xy) and suppose that for ally E G,T yis equivalent to T. Then there exists an invertible matrix ay such that F(x)= a; 'T(x)a,for all x E H.In particular, if we choose y so that Hy generates G / H , and write a for ayr then we get u - ~ T ( x )=u ~ Tyk(x)= T ( ~ - ~ x = y ~T)( Y - ~ ) T ( x ) T (since ~~) yh E H. Thus T ( Y ~ ) ucommutes -~ with T ( x ) for all x E H.Since T is abso~ for some scalar a. E K O .Since K Ois algelutely irreducible, T ( Y ~ ) u=-aol braically closed we can find a E K O such that ah = a. and then T(yh)= (aa)h. Put a. = aa, and define Toon G by To(xyi):= T(x)ab for all x E H, i = 0,1, . . ., h - 1. We claim that Tois a representation of G over K O . Indeed, if x,x' E H and 0 Ii, j < h, then T,(xy')T,(x'Y') = T(x)abT(x')a,'ab+j =

T(x)a'T(x')a-'ay

= T(x) T(y'x'y- ')a;' = T(xy'x'y-')p.

Since ub+j= T(yh)ab+j-hif h Ii + j c 2h, we see that To(xy')To(x'9) = T,( xy'x' 9) as required. Note that the restriction of the representation To of G to the subgroup H equals the original representation T of H; in particular, this shows that To is also irreducible. We now consider the situations described in (i) and (ii).

8.2 BLOCKS WITH NORMAL CYCLIC DEFECT GROUPS

199

(i) In this case take K O as a field of characteristic 0 and let T be a representation affording x. Since x Y = x for all y E G by hypothesis, and T y affords x y , we have Tv equivalent to T for all y E G (Corollary 1 of Theorem 2.3). Thus we obtain an irreducible representation To of G extending T and the character afforded by To satisfies cH = x. (ii) In this case take K O as a field of characteristic p, and let T be a representation of H over K Owhich affords the irreducible modular character $. Since = $ for all y E G by hypothesis, the (Frobenius) characters of Ty and T are equal, and so TY is equivalent to T. Hence we again obtain an irreducible representation To of G over K O extending T. The irreducible modular character 4 afforded by To has the property 4" = $. Define I); on G by 6 = $ on H and = 0 on G\H. Then

4

c

h- 1 $C(z):=

for all z E G,

$(y-'zy')

i=O

where y is chosen such that H y generates G / H . Hence if z i s H otherwise. On the other hand, for i = 0, 1, . . ., h - 1, pi4 is the irreducible modular character of G afforded by the representation T defined by T(z) := T0(z)p(z)'. Since Ker p = H, p(z) is a nontrivial hth root of unity for all z $I H. Therefore

This shows that

Hence $' = CFZ; p i 4 as asserted.

8.2 Blocks with normal cyclic defect groups Throughout the remainder of this chapter we shall use the notation first introduced in $3.3. Thus G will be a group of order g, A will be a p-adic algebra which is an integral domain of characteristic 0 containing a primitive 8th root of unity, and K will be the field of quotients of A. Also K A is the unique maximal ideal of A and k = A/nA is the residue field of characteristic p. Both K and k are splitting fields for G and all its quotient groups as well as subgroups.

VIII

200


Let Q be a cyclic normal p-subgroup of G. As we know from Lemma 6.2A,

Q is contained in each defect group of a block B of kG. In the present section we shall consider the case where Q is the (unique) defect group of B. The theorem below describes the set of irreducible ordinary and modular characters that lie in such a block. The proof requires the following lemma. Lemma 8.2 Let Q be a normal p-subgroup of the group G and put := QCG(Q).Let b be a block of k H . If x is an irreducible ordinary character of H lying in b, then each irreducible constituent of xG lies in bG.

H

Proof Let Wl, V 2 , . ..,Vsbe the conjugate classes of G with class sums cl, c 2 , ..., c, in kG and put h i = l V i J , V F = V i n H and c : ~ = ~ ~ ~ ~ , . x (i = 1, 2, . .., s). If o is the central character of kH associated with the

block b, then (see $6.3) the central character of kG associated with the block bG is mG given by mG(ci) :=~ ( c f ) (i = 1, 2, . .., s). (1) If %" is a class of H with class sum c' and h' := IV' 1, then by Theorem 4.2B we have

o(c') = h'x(y)/x(l) =

c x(y)/x(l)

(where the bar denotes reduction modulo n) with y E W'.

X€W'

Since H

4

G, we have for each i either Wf = Wi or VF = 0.Therefore

for i = 1, 2, ..., s. Now suppose that [ is an irreducible constituent of xG. Then (see Theorem 4.2A) the central character 1(1 associated with the block of kG in which [ lies is given by I&,.) = h i l i / [ ( l )(i = 1, 2, ..., s), where fli is the value of [ on W i . We have to prove that II/ = oG.First of all, by the Frobenius reciprocity theorem (Theorem 2.5A) x is an irreducible constituent of Moreover by Clifford's theorem (Theorem 2.2A) all irreducible constituents of l Hare of the form xy for some y E G (since H 4 G). Since (xy, [ H ) H = ( x , [k-l)H = ( x , C H ) H , it follows that each of the different conjugates of x occurs with the same multiplicity in I H , say m. Let x' = x, x2, . . ., x" be the distinct conju+ x"} and so for all x E H we have = m{X' + gates of x. Then

cH.

rH

8.2

Thus if Vi E H and z E Wi, then Ci = ( ( z ) and

(3)

20 1

BLOCKS WITH NORMAL CYCLIC DEFECT GROUPS

hi5i/5(1) = hi

C X(Y-

5( 1) = mnx( I), so

'ZY)/gX( 1)

YeG

=

1 x(x)/x(1).

XEYi

Together with (2) this shows that (4)

u G ( c i )= $(ci)

whenever W i E H.

Now suppose W i$ H. Let P = O,(H); then P -a G because it is characteristic in H. Moreover Q G P because Q -a H. Let z E W i . Since W i $ H = QCG(Q),therefore z $ CG(Q)and so Q (and therefore P ) is not contained in C,(z). This shows that the defect groups of%, do not contain P. Therefore by Lemma 6.2A, ciE rad Z ( k G ) and so ci lies in the kernel of every central character of kG.In particular, (5)

u G ( c i )= 0 = $(ci)

whenever W i$ H.

The equations (4) and ( 5 ) prove the lemma. NOTATION During the remainder of this chapter we shall be examining the following situation. The group G will contain a cyclic p-subgroup Q and we put C := CG(Q)2 Q and N := NG(Q)2 C . Let B be a block of kG with Q as a defect group. Then by the First Main Theorem (Theorem 6.3) there is a unique block B , of kN with Q as a defect group such that B = BY. By Theorem 6.4 the blocks of kC that correspond to the block B, of k N are all N-conjugate; fix b as one of these blocks of kC. Then Q is the (unique) defect group of b, bG = B, and the N-conjugates of b are the only blocks of kC that correspond to B under the Brauer correspondence. Write b = e(kC),where e is a central primitive idempotent of kC, and define F := {x E N I x-lex = e} as the stabilizer of the block b under N. We shall put 4 I F : C 1. Let ( QI = pd. Then the automorphism group Aut Q has order p d - ' ( p - 1). Since N / C is isomorphic to a subgroup of Aut Q, this shows that IN : Cl divides p"-'(p - 1); and since 41 IN:C ( we conclude that 41$-l(p - 1). On the other hand, b has IN : F I N-conjugates. Therefore if p' is the highest power of p dividing IN : C I, then Theorem 6.4 shows that p' 1 IN : F I. In particular, this shows that 4 = IN : CI/JN: F ( is relatively prime to p, and so we must have q ( p - 1. Note that F/C is a cyclic group because the subgroup of order p - 1 in Aut Q is cyclic (see Hall [l], $6.2). We next note that ify E F\C, then C,(y) = 1. Indeed, choose z as a generator of the cyclic group Q. Then y - ' z y = z'" for some integer m with 1 Im < pd. Since I F/C I divides p - 1, yP-' E C. Therefore z = y -w l )Z Y P - l = F - ' and ; so m p - l = 1 (mod pd). On the other hand, if ziE C,(y), then zi = y- ' z i y = z M i and ; so mi = i (mod pd). If zi # 1, then we I=

202

VIII


would have m = 1 (mod p). Since y # C, m # 1 and so we would have m = 1 + $1 for some integers] and 1 with 1 I j < d and p X 1. But then the binomial theorem shows that m = m m P - ' = ( l + d l ) P = 1 +$+' I + . . . =

1

(modpJ"), which contradicts the condition p XI. Hence we have proved C,(y) = 1 for all y E F\C. This implies that Q = {x- ' y - ' x y I x E Q)for each y E F\C since the elements x - ' y - ' x y (x E Q) must all be different. The cyclic group Q has pd - 1 irreducible ordinary characters (all of degree 1) different from 1,. The group F acts by conjugation on this set of characters, and no character x # 1, is fixed by any y E F\C; indeed xy = x implies 1 = xy(x)x(x)- = x(y- ' x y x - '), so from what we have just proved, Ker x = Q. Thus each orbit of F on the set of nontrivial irreducible characters of Q has length 1 F : C 1 = q; and there are (pd- l)/q orbits. We shall choose a set A of representative characters, one from each orbit. Then IA I = (pd - l)/q and each nontrivial irreducible character of Q is conjugate under F to exactly one character from A.

'

With this notation we can now state the first theorem on blocks with cyclic defect groups. Theorem 8.2 With the notation above suppose that Q is a normal psubgroup of G (so the block B has Q as its unique defect group). Let b be a block of kC such that bG = B, and using the notation of Theorem 6.2B let 4 and tl, t2,...,

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