Representation Theory of Finite Groups

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Representation Theory of Finite Groups . MARTIN BURROW COURANT INSTITUTE OF MATHEMATICAL SCIENCES NEW YORK UNIVERSITY NEW YORK, NEW YORK

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Preface

The old representation theory of finite groups by matrices over the complex field was largely the work of G. Frobenius, together with significant contributions by I. Schur. Many of the important results of Frobenius were again found independently by W. Burnside whose book, "Theory of Groups" (rgrr), is now a classic. Burnside's intriguing use of group characters to obtain results on abstract groups earned great attention for the theory. The next decisive influence on the development of representation theory was E. N oether's shift of emphasis to the study of the representation module ( 1929). Her point of view has produced valuable gains in an algebraic direction. A major extension of the subject in the last 25 years has been the study of modular representations initiated by R. Brauer. This theory too has provided significant applications to the theory of finite groups. For example, Thompson and Feit have recently aroused great interest by their use of the modular theory in the study of the solvability of groups of odd order. Until recently much of the material on modular representations has been accessible only through research articles and the lectures of its principal developers. There was no systematic account of the modular theory until the publication ( 1962) of Curtis and Reiner's "Representation Theory of Finite Groups and Associative Algebras." A treatment of the modular theory which is due to R. Brauer is given in the final chapter of this book. The remaining chapters contain the standard material of representation theory which is here treated consistently from v

vi

PREFACE

the point of view of the representation module. The rudiments of linear algebra and a knowledge of the elementary concepts of group theory are useful, if not entirely indispensable, prerequisites for reading this book. A graduate student so equipped who wishes to acquire some knowledge of representation theory should not find the work too difficult to master on his own. Also the book might prove useful as supplementary reading for a course in group theory or in the applications of representation theory to Physics. The author wishes to express heartfelt thanks to Miss Kate Winter in particular and to Mr. ]. Koppelman for their invaluable assistance in proofreading.

New York !965

MARTIN BURROW

Contents v

PREFACE

Chapter I.

Foundations 1.

2.

3. 4. 5.

Chapter I I.

Introduction Group Characters Representation Modules Application of Ideas and Results from Group Theory The Regular Representation Exercises

I2

13

17 23 24

Representation Theory of Rings with Identity 6. 7. 8. g. 10.

11.

12.

13.

Some Fundamental Lemmas Exercise The Principal Indecomposable Representations The Radical of a Ring Semisimple Rings The Wedderburn Structure Theorems for Semisimple Rings Intertwining Numbers Multiplicities of the Indecomposable Representation The Generalized Burnside Theorem Exercises

vii

28 34 34 37 40 42

47

53 55

s6

viii

Chapter Ill.

CONTENTS

The Representation Theory of Finite Groups 14· IS. 16. 17· 18. 19. 20. 21

2 2. 23

Chapter IV.

s8 6o 6o 64 6s 67 71 72 7S 76

77 77 83

Applications of the Theory of Characters 24 2S. 26.

27

Chapter V.

The Group Algebra The Regular Representation of a Group Semisimplicity of the Group Algebra The Center of the Group Algebra The Number of Inequivalent Irreducible Represen lations Relations on the Irreducible Characters The Module of Characters over the Integers The Kronecker Product of Two Representations Exercises Linear Characters Exercises Induced Representations and Induced Characters Exercises

Algebraic Numbers Some Results from the Theory of Characters Normal Subgroups and the Character Table A. The Existence of Normal Subgroups B. The Determmation of All Normal Subgroups Some Classical Theorems Exercises

8s 87 92 92 93 94 98

The Construction of Irreducible Representations 28. 29.

Primitive Idempotents Some examples of Group Representations 1. Cyclic Groups 2. Abelian Groups 3. The Symmetric Groups 5" Exercises

99 IOS IOS I06 106 Il9

CONTENTS Chapter VI.

ix

Modular Representations 30. 3I. 3 2. 3 3· 34· 35-

36.

3 7·

38. 39· 40. 41. 42.

General Remarks P-Regular Elements of a Finite Group Conditions for Two Representations to Have the Same Composition Factors The Brauer Characters Integral Representations Exercise Ordinary and Modular Representations of Algebras r. Arithmetic in an Algebra Exercise 2. Connection with Integral Representations P-Adic Fields r. General Definition and Properties 2. Ordinary Valuation. Metrical Properties 3. Completion of a P-Adic Field 4. p-Adic Valuation of the Rational Field 5. Extension of the P-Adic Valuation to Algebraic Number Fields Algebras over a P-Adic Field r. Notation 2. Preliminary Results A Connection between the Intertwining Numbers Modular Representations of Groups Cartan Invariants and Decomposition Numbers Character Relations Modular Orthogonality Relations

120 122 I 25 128 130 134 134 134 136 136 137 13 7 139 140 143 143 148 148 149 157 159 r6r 164 169

Appendix I. 2.

Groups Rings, Ideals, and Fields

173 q6

BIBLIOGRAPHY

I80

SuBJECT INDEX

I83

CHAPTER I

Foundations 1. Introduction Nowadays it is natural for us to think of a group abstractly as a set of elements {a, b, c, ... },which is closed under an associative multiplication and which permits a solution, for x and y, of any equations: ax = b, and ya = b. On the other hand, we regard a group, which is given in some concrete way, as a realization of an abstract group. This point of view is an inversion of the historical development of group theory which won the abstract concept from particular modes of representation. Group theory began with finite permutation groups. Any arrangement of n objects in a row is called a permutation of the objects. If we select some arrangement as standard, then any other arrangement can be regarded as achieved from it by an operation of replacements: each object in the standard being replaced by that object which takes its place in the new arrangement. Thus if I23 is standard and 312 is another arrangement, then the replacements are 1 ---+ 3, 2---+ I, and 3---+ 2. We write compactly for this operation

( 31 21 3) 2 . If the replacements of two operations are performed in succession, we get an arrangement which could be achieved directly by a third operation, called the product of the two operations. For example,

(I

2 3)(1

2 3

1 2

21 3)3 = (11 32 3)2 . 1

I. FOUNDATIONS

2

Here we have proceeded from left to right. The product of operations is associative and any set of operations which form a group is a permutation group. If we haven objects I, 2, ... , n, then there are n! arrangements and hence n! operations are possible, including the identity:

(~

2 2

They form a group Sn , the symmetric group on n symbols. Every permutation group on n symbols is a subgroup of Sn . In a remarkable application of a group theory in its infancy Galois showed that every algebraic equation possesses a certain permutation group on whose structure its properties depend. Cayley discovered the abstract group concept. A theorem of his asserts that every abstract group with a finite number of elements can be realized as a group of permutations of its elements. Thus if G ={a, b, c, ... , g, ... } is the abstract group, then the element of the permutation group P which corresponds to g is the set of replacements a ---+ ag, b ---+ bg, c ---+ cg, ... , or compactly: x running over G.

The groups G and Pare isomorphic (see Appendix). A generalization of the permutation group, and the next step historically, is the group of linear substitutions on a finite number of variables. In this case, if the variables are x1 , ••• , Xn , the group operation consists of replacing each variable X; by a linear combination of the variables; thus

The coefficients a;; are numbers, real or complex. Substitutions are multiplied by carrying them out in succession. As an example let us find the product of the substitutions x'

(1.2)

y'

= =

2x- 3y

x"

=

y"

=

and X-

y

+ 4y' 2x' + 3y'. 3x'

1. INTRODUCTION

3

Here we have written the first substitution as replacing unprimed variables by primed, and the second as replacing primed by double primed variables. Now in the second system substitute for x', y' from the first and get

+ 4(x 3y) + 3(x -

x"

=

3(2x - 3y)

y)

y"

=

2(2x -

y)

and so

x" =lOx- l3y

(t.3) y"

=

?x- 9y.

This is the product substitution. Note that the use of primes is for distinction only. For instance (1.3) means that xis to be replaced by lOx- l3y and y by ?x- 9y, irrespective of the single symbols, x" and y", which we use to denote these replacements. Going back to the general substitution {1.1} we see that it is entirely determined by the numbers a;; , that is, by the matrix

( 1.4)

The first row consists of the coefficients of x 1 , taken in succession from the first, second, ... , nth equations, and in a similar way the ith row is formed from the coefficients of X;. Note that (1.4} is the transpose of the array as it appears in (1.1). For example, in (1.2) we have the matrices

A

=

(

2 -1, I)

-3

Now

AB = (

10

-13

7)

-9,

which is the matrix of the product substitution, so that in this case the correspondence between substitutions and their matrices

I. FOUNDATIONS

4

is preserved by multiplication. It is easy to see that this is true in general. Thus, let X denote the row vector, or 1 X n matrix, (x 1 , x2 , ••• , xn) and let X' be the same with primed variables. If (1.4) is denoted by A, then (1.1) can be written X' = XA. Now let X" = X'B be another substitution with matrix B. The product is the substitution X" = XAB and its matrix is AB. If the substitutions form a group, the inverse substitution must exist and Eqs. (1.1) are solvable for the X; in terms of the This means that the determinant of the matrix (1.4) is not zero. Then A-1 exists and X= X'A- 1 • We now see that a substitution group on a given set of variables is abstractly identical to a group of nonsingular matrices. It is clear that substitutions, or matrices, admit a greater freedom of algebraic treatment than do permutations. For instance, matrices automatically generate a ring. Also, whereas permutation groups on a finite number of symbols are necessarily finite, substitutions allow us to deal with infinite groups. For example,

x; .

(~ ~).

n

= 0, ±I, ±2, ...

is an infinite discrete group. Again, the substitutions x' = x cos 6 - y sin 6

y'

=

x sin 6

+ y cos 6

+

which leave the expression x2 y 2 invariant is an infinite continuous group. This is the orthogonal group. It is the group of rotations of the Cartesian plane about its origin. Because of their algebraic flexibility it is natural to use matrices to represent abstract groups. Let us call a homomorphism of a group G into a group of n X n matrices a representation of G of degree n. This means that to each elementg of G there corresponds a matrix a(g) and if x andy are any elements of G: a(xy)

=

a(x)a(y).

1. INTRODUCTION

5

The representation is called faithful when the homomorphism a is an isomorphism. When this is the case the correspondence is one-to-one and a(g) = I, the identity matrix, if and only if g = I, the identity of the group. Frobenius proposed the question: Find all matrix representations of a finite group G. Let us make the following observations: I. There always is a representation. For as we have seen there invariably is the permutation representation

x running through G.

But, giving some fixed order x 1 , x 2 , ... , Xn to the group elements, this merely expresses the linear substitutions

where x;g =xi, and i =I, 2, ... , n. Thus to eachg corresponds a linear substitution and G is represented by a group of linear substitutions. Since x;g = X; if and only if g = I, the representation is faithful. The corresponding matrices give a faithful matrix representation of G. This is the regular representation. At the other extreme, so to speak, we have the one-representation i for which i(g) = 1, Vg E G. 2. Given any matrix representation we can find infinitely many. Thus if a is a matrix representation and Tis any fixed invertible matrix we can define T(x) = Ta(x)T-1, Vx E G. Since

T(xy) = Ta(xy)T- 1 = Ta(x)a(y)T- 1 = Ta(x)T- 1 Ta(y)T- 1 =

T(X)T(y),

is a matrix representation of G. The new representation is brought about by a mere change of variable in the corresponding substitutions. To see this let the original variables be X = (x 1 , x2 , ••. , xn) and let new variables Y = (y 1 , y 2 , ••• , Yn) be related by

T

n

X;

=

k t ;Y 1

}=1

1,

or in matrices:

X=YT.

I. FOUNDATIONS

6

Then any substitution X' = XA, with matrix A, in the old variables becomes when expressed in the new: Y'T = YTA, and hence Y' = YTAT-1, with matrix TAT- 1 • Representations related as are a and T are said to be equivalent and are regarded as essentially the same representation. All representations equivalent to a are clearly equivalent to each other and form an infinite class. Since they are all the same representation the task proposed by Frobenius can be narrowed to the survey of all inequivalent representations. We will write a,..._, T or a ,.,., T according as the two representations are or are not equivalent. 3. If a and T are two matrix representations of G of degrees s and t, respectively, and if g E G, the matrix

p.(g)

=

0 ) ( a(g) 0 T(g)

of degrees + tis the direct sum of the matrices a(g) and T(g). The 0 in the first row and that in the second represent, respectively, an s X t and at X s matrix of zeros. Now p.(x)p.(y)

=

(a(0x) 0 )(a(y) 0 ) (a(x)a(y) 0 ) T(X) 0 T(y~ 0 T(X)T(y) =

( a(xy) 0

0 )

T(xy)

=

p.(xy).

Hence p. is a representation. It is called the direct sum of a and T and we write p. = a 8J T. Thus, given two representations, possibly the same, we can find a third by adding them directly. Conversely, let p. be a representation of degree s + t and Vx E G suppose that the matrix p.(x) has the form

( A(x)

0

0 ) B(x)

where A(x) and B(x) are, respectively, s X s and t X t matrices depending on the element x. As before, the O's are s X t and

1. INTRODUCTION

t X s zero matrices. We now define u and T by: u(x) T(x) = B(x), 'Vx E G. Since !L(X.Y) = f'(X)f'(r), ( A(0xy)

0 ) B(xy)

=

(A(x)

0

7 =

A(x),

0 )(A(y) 0 ) 0 B(y)

B(x)

Therefore u(xy) = u(x)u(y) and T(xy) = T(x)T(y) so that u and T are representations. The representation IL is said to be decomposable and u and T are its components. If v "' f', v is also said to be decomposable. A representation which is not decomposable is indecomposable. The basic question can now be more sharply put: Find all inequivalent indecomposable representations of a finite group G. Are there only a finite number of different ones? If a component of a decomposable representation is itself decomposable we get a further decomposition. Continuing in this way we can decompose any given representation into a finite number of indecomposable components. Are these unique, or can a different set be obtained by carrying out the decomposition differently? In due course we shall see these natural questions answered completely by the theory. Historically, Frobenius started representation theory with the group determinant. Though it will play no role in our treatment we mention it now. If A 1 =I, A 2 , ••• ,An are the elements of a finite matrix group in any fixed order and x1 , x 2 , ••• , Xn are an equal number of indeterminate variables, we can form the group matrix

Its coefficients consists of linear combinations of the xi; for example, at the intersection of the ith row and jth column we have

I. FOUNDATIONS

8

where a;~ is the coefficient in the same position in the matrix Ak. The determinant of this matrix is the group determinant. It is a polynomial of the same degree as the representation. It can be shown that this polynomial is factorable if and only if the group is decomposable and that then its irreducible factors are the group determinants of the indecomposable components. Moreover, the same transformation T that decomposes the group will transform M into the corresponding group matrices (see the Exercises). 4. The property of being indecomposable depends on the field of the representation. For example, the matrix group

1) (0 (-1 -1 0 ' 1 is indecomposable over the rational field: there is no matrix T with rational coefficients so that

r(-1 1)r-1 0

1

=(ex0

0)f3 •

This is easily seen by equating for the trace and determinant on both sides; since transforming a matrix as on the left of the equation does not alter these quantities, we have ex f3 = -1, exf3 = 1, leading to the quadratic equation ex2 + ex + 1 = 0, ""' whose roots are cube roots of unity. On the other hand, if complex · values are permitted,

+

1)(-1 1)(w 1)1 -1 0 w 1

1

2

(w w0)

= 0

2

+

where w is the cube root of unity w = ( -1 v'3i)j2. It is interesting to note that in this example the group determinant z 2 - xy - xz - yz which is irreducible in the is x2 y 2 rational or real field but can be factored thus:

+ +

(x in the

co~plex

+ wy + w2z)(x + w2y + wz)

field.

1. INTRODUCTION

9

Again consider the group

with its coefficients in the prime field of charactenstlc 1., that 1s, m the field of two elements 0, I where addition and multiplication is performed modulo 2. If T

(I1 0)I T-

1

=

(a0 0)b '

then on equating the trace and determinant on both sides we should have a b = 0 ab = 1, yielding a = b = 1. Thus the right side is the identity matrix. This is impossible and so G is indecomposable. Moreover, unlike the previous example, G stays indecomposable even if the field be extended. Such a representation which remains indecomposable in any extension of the field is called absolutely indecomposable. If at the outset we consider the field of representation to be algebraically closed (and so incapable of further algebraic extension) or at any rate sufficiently large, then we may confine our attention to absolutely indecomposable representations. In general we will adopt this simplification. It turns out that the representation theory in fields of characteristic pis very different from the case at characteristic zero, if the order of the group is divisible by p.

+

5. There is a weaker Col)cept than decomposability. Thus let J-Ibe a matrix representation which for every x E G has the form

(1.5)

J-1-(x) =

( A(x) l(x)

0 ) B(x)

in which A(x), B(x) are, respectively, s X s and t X t matrices depending on x. l(x) is at X s matrix depending on x and 0 is the s X t zero matrix. Defining a(x) = A(x) and T(x) = B(x) it is easy to see as in Observation 3 that a and T are representations of G. The representation J-1-, and any equivalent representation, is said to be reducible. A representation that is not reducible is irreducible. The representations a and T are constituents of J-1-· If either is irreducible it is an irreducible constituent. A representation

I. FOUNDATIONS

10

which is decomposable is clearly reducible [I(x) = 0 in (1.5)]. On the other hand, the second example of observation 4 shows that a representation which is reducible need not be decomposable. If the characteristic of the field is zero, or is not a divisor of the order of the group, this cannot happen. We shall see that in these cases every reducible representation is decomposable provided the field is taken large enough, and then the constituents are components as well. If a and -r are themselves reducible, they yield constituents of smaller degree. Continuing in this way we obtain a set of irreducible representations a 1 , a 2 , ••• , ak as constituents of p.. It will be shown that these are uniquely determined up to equivalence for each representation p.. Moreover, we shall see that the set of distinct irreducible representations of a finite group in a given field is an invariant of the group. They are finite in number and any irreducible constituent of a representation must be one of them. They have been called the building blocks for the representations of the group. If the indecomposable components of a representation are also irreducible the representation is called completely reducible. The symmetric group S 3 of order 6 is generated by all products of the permutations

Example.

l 2 3) a= ( 2 I 3

and

Expressed as substitutions this gives

and X~=

X~=

x3

x2

so that the corresponding matrix representation is

v(a)

=

(~ ~' -~)• 0 0

I

v(b)

=

(~ ~ ~)· I 0 0

1. INTRODUCTION

11

Now v is reducible, for if

and

Vg

J.L(g) = Tv(g)T-1,

E

S3

,

then

1 .

J.L(a) =

Thus p"' a

(~ ~=~ ~)·

ttl -r,

a(a) = a(b) ==

0 0

I,

1

J.L(b) =

0 0

I

(a0 11-1 1). -1 0

where

-1 -r(a) = ( -1

~).

and

-r(b) =

(=~ ~)·

Now ,. is irreducible, for if there were a matrix L such that -1 ( L -1

~)L-1

=

-1 ( L -1

~)L-1

=

e ~) (:

~)

then, by equating traces and determinants, we would get x == ± 1, y == =f1, u == w or w 2 , v = w 2 or w. Here w is a cube root of unity. But on adding Eqs. (*)we have

( -2

L -2

a = w

f3

=

w2

or or

w2 w.

This is impossible since the determinant of the matrix on the left is zero and that of the matrix on the right is not. Thus there are no relations ( *) and ,. is irreducible. Cl~~rly a is irreducible and so v is completely reducible. The representation ,. lies in the rational field. We have seen that it remains irreducible even when transformed by a matrix L which is allowed to have complex coefficients. A representation

12

I. FOUNDATIONS

which, like T, remains irreducible in any extension of the field of representation is called absolutely i"educible. The remark of Observation 4 concerning absolutely indecomposable representations applies here as well. The setLn(F) of all nonsingular n X n matrices with coefficients in a field F form a multiplicative group called a linear group. We conclude this section by restating formally: (1.6) Definition. A matrix representation of degree n of a group G is a homomorphism a of G into the linear group Ln(F). This means that to each x E G there corresponds an n X n matrix a(x) with entries in F and, for all x, y E G,

a(xy)

= a(x)a(y).

The representation is faithful if a is an isomorphism.

2. Group Characters The matrices representing a group, besides the readiness with which they admit calculation, allow the introduction of numerical functions on the group. These functions, called characters, play an important part in the theory. We recall that if A = (a;1), i, j = I, 2, ... , n, is any matrix, then tr A, the trace of A, is n

tr A= ~aii. i=l

It can be seen by direct calculation that if T is any other matrix, tr AT= tr TA and then, if T is nonsingular so that T-1 exists, tr T- 1AT

= tr A.

3. REPRESENTATION MODULES

13

(2.1) Definition. The character of the representation a of a group G is the function x" on G to the field of representation F given by VgEG. x"(g) = tr a(g),

If p. ,.._ a, then by definition, Vg we have p.(g) some fixed matrix T. But then

x '(g) 1

=

tr p.(g)

=

tr Ta(g)T- 1

=

tr a(g)

=

Ta(g)T- 1 for

= x"(g).

This shows that equivalent representations have the same character. Now if p. is a reducible representation, for some fixed matrix T

Tp.(x)T

_1 _

-

(a(x)

](x)

0 ) T(x) '

VxEG.

Then taking the trace on both sides we get

x"(x)

=

x"(x)

+ XT(x),

VxEG,

showing that the character of a reducible [decomposable, if ](x) 0] representation is the sum of the characters of its constituents (components). Thus every character is the sum of irreducible characters, that is, characters of irreducible representations. A group can be partitioned into equivalence classes of conjugate elements, where g and h are in the same class if and only if there is an element t such that g = t- 1ht. Since

=

x"(g) = x"(t- 1ht) = tr a(t-1ht) = tr(a(t))- 1a(h)a(t)

=

tr a(h)

=

x"(h),

we see that the character is a class function. It has the same value for elements of the same class.

3. Representation Modules Representations can be given a more algebraic formulation and thus fitted into a wider context by means of the concept of a

I. FOUNDATIONS

representation module. This point of view is due to E. Noether. At the same time we will consider the more general question of the representations of a ring. First we introduce the double module. (3.1) Definition. Let F be a commutative field and Raring with an identity. A set M of elements is an F-R mot{u/f! if: \.

1. 2.

3. 4. 5. 6. 7. 8. 9.

.:.r. ,·

...

M is an additive abelian group. Vf E F, \fm EM, \fr E R, there are unique elements fm and mrEM. f(m + m') = fm + fm'. (f + f')m = fm + f'm. (ff')m = f(f'm). (m + m')r = mr + m'r. m(r + r') = mr + mr'. m(rr') = (mr)r'. 1m = m = ml, 1 and I being the identities ofF and R.

In the language of group theory M is an abelian group with two distinct sets of operators. If we disregard the operators R, then M is an F-module or a vector space over F. When R is an algebra over F, so that fr = rf E R is defined for f E F and r E R, there is the added condition:

10.

m(fr) = (fm)r.

Finally, throughout most of our discussion we will require: 11.

M has a finiteF-basis.

Since M is a finite-dimensional vector space over F there is a close connection between linear transformations in M and matrices. Let us recall:

(3.2) Definition. An F-endomorphism of an F-module M is a mapping a of M into itself which assigns to each m E M a unique element ma E M and which satisfies the rules (fm)a

(3.3)

(3.4)

(m

+ m')a

=

f(ma)

= ma

+ m'a.

3. REPRESENTATION MODULES

15

If a and -r are two endomorphisms their sum and product are defined by (3.5)

m(a

+ -r)

=

ma

+ m-r

and

j

The set of all endomorphisms of M form a ring E. The ring E of F-endomorphisms of M is actually an algebra over F for we can v define fa = af as the mapping of M into M given by m(fa) = (fm)a. Then fa E E. E is the ring of all linear transformations of the vector space Mover F. If M is of finite dimension n over F, E can be represented analytically by the ring Wln of all n X n matrices over F in the following way. Let B = {m 1 , m2 , ... , mn} be an ordered basis of Mover F. Then n

(3.6)

m;a

= ~f/;m; ,

i = I, 2, ... , n,

j~l

and the n2 elements f~; E F. Thus a is associated with the n X n matrix S = (/~;). Conversely, if Sis given, a mapping a of the basis can be defined by (3.6) and extended toM by using (3.3) and (3.4) as definitions. Thus there is a one-to-one correspondence between linear transformations a and n X n matrices S. If a +--+ S and -r +--+ T, it is easily checked that a + -r +--+ S + T · and a-r +--+ ST, so that E "" Wln . (3.7) Definition. A representation of a ring R is a ring homomorphism p of R into E, the ring of linear transformations of an F-module M. ·

This means that to each r

E

R there corresponds a unique element

p(r) E E such that

+ r')

(3.8)

p(r

(3.9)

p(rr')

=

=

p(r)

+ p(r')

p(r)p(r').

On taking a basis for M, p(r) is associated with a matrix M(r) and these relations hold for the corresponding matrices. An F-R module may be called a representation module or, representation space, of R, for we have:

16

I. FOUNDATIONS

(3.10) Lemma. To each F-R module M there is a unique representation of the ring Rand conversely.

Proof.

Let r

E

R. Define the mapping J.L(r) : M---+ M as follows: mJ.L(r)

=

VmEM.

mr,

Then

(fm)J.L(r)

=

(fm)r = f(mr) = f(mJ.L(r))

and

(m

+ m')J.L(r)

= (m

+ m')r =

mr

+ m'r =

mJ.L(r)

+ m'J.L(r).

Thus J.L(r) satisfies (3.3) and (3.4) and is therefore an F-endomorphism of Mas an F-module. Hence J.L(r) E -2. Moreover, Vm EM,

mJ.L(r

+ r')

+ r') = mr + mr' = mJ.L(r) + mJ.L(r') the final step by (3.5), m(J.L(r) + J.L(r')),

= m(r

= and

mJ.L(rr') = m(rr') = (mr)r' = (mJ.L(r))J.L(r')

= m(J.L(r)J.L(r')),

the final step by (3.5).

This shows that fL(r

+ r')

=

J.L(rr')

= J.L(r)J.L(r')

J.L(r)

+ J.L(r')

and hence that J.L is a representation of R. Conversely, if Vr E R, J.L(r) is an F-endomorphism of an F-module M and satisfies the last two equations, we can turn Minto an F-R module by defining: mr = mJ.L(r). The proof is now complete. This result enables us to turn our attention from the representation to the corresponding representation module. Since this is an abelian group with operators, certain fundamental results from group theory can be applied. This is done in the next section.

4. APPLICATION OF IDEAS AND RESULTS

17

4. Application of Ideas and Results from Group Theory Since a representation module is an abelian group with operators it is natural to investigate the meaning, for representations, of such group theoretical notions as: admissable subgroup, factor group, direct product (sum). First we have the natural:

(4.1) Definition. Two representations p. and v of a ring R are equivalent (or similar) if their corresponding representation modules M and N are operation isomorphic; that is, if there is an isomorphism T : M ---+ N such that 'Vm E M there exists a unique mTEN and (m

+ m')T

= mT + m'T

(f mr)T = f(mT)r,

'VjEF,

'Vr

E

R.

We are going to show that this definition of the equivalence of representations is in accord with the one introduced earlier. Now'VmEM

and hence 'Vr

(4.2)

E

R.

Because of (4.2) T is said to intertwine the representations p. and v and is called an intertwining mapping. Since T is one-to-one onto, T-1 exists and p.(r) = Tv(r)T- 1 • Now let B = {m1 , m2 , ... , mk} and B' = {n1 , n 2 , ... , nk} be ordered bases of M and N. Then 1r

m;p.(r) = m;r = !,JiJ(r)m; , J~l

lc

n;v(r) = n;r = !,J;;(r)n;, i~l

J't;(r)

E

F,

I. FOUNDATIONS

18

and in each case i

= I, 2, ... , k. Thus we have the correspondences

p.(r) f--+ M(r) = (f/1(r))

and

v(r) f.-+ R(r)

=

(f;;(r))

where M(r) and R(r) are each k X k matrices depending on r. If now m;T = I::~ I t; 1n1 , i = I, 2, ... , k, then Tis associated with a matrix T = (t;1). Applying T to the first equation of(*) and using (4.2): k

m;p.(r)T

=

m;Tv(r)

k

k

= ~ t;1n1v(r) = ~ ~ t;_;/j1(r)n 1 J~l

i~l !~l

k

k

k

= ~ff;(r)m 1 T = ~ ~ft(r)ti!n!. J=l l=l

J=l

Comparing the fourth term with the last term we see that the entry at the intersection of the ith row and /th column of TR(r) is the same as the corresponding entry of M(r)T. This is true Vi, I and so M(r)T

=

TR(r),

or

M(r) = TN(r)T- 1 ,

VrER,

yielding our former definition of the equivalence of matrix representations. Let us recall that S is an admissable submodule of an F-R module M if S is a subgroup of the group M and if Vf E F, Vs E S, Vr E R, fsr E S. Then S is itself an F-R module and provides a representation. Moreover, since M is a~elian, S is a normal subgroup. The factor group MfS is also anF-R module if we define the action of the operators thus: jmr = fmr, where m denotes the class of m modulo S. How are the representations provided by Sand MfS related to that provided by M? We can find a basis {m1 , m2 , ••• , ma, m~, m~, ... , m~} of M such that the first a terms are a basis of S. Such a basis is said to be adapted

4. APPLICATION OF IDEAS AND RESULTS

19

(or accommodated) to S. With this basis we find the corresponding matrix representation:

+ ··· + faama m1 + ··· + /1ama + f~1m~ + ··· + f{bmb

(4.3) mar= fa1m1

m~r = / 11

This gives the matrix representation (4.4)

p.(r)

=

0 ) ( p.1(r) ](r) P-2(r)

in which p.1(r) = (/; 1), i, j = 1, 2, ... , a, and p. 2(r) = (!; 1), i, j = 1, 2, ... , b. 11-2 is the representation corresponding to the representation module MfS. This can be seen by taking the last b equations of (4.3) modulo S. If M is the direct sum of two admissable submodules, M = S EB T, then a basis adapted to this decomposition will yield a matrix (4.4) having J(r) == 0. Here p.1 and p.2 are representations arising from the modules S and T, respectively. We can now state formally: (4.5) Definition. (a) A representation p. of a ring R and its corresponding module M are reducible if M possesses a proper ( *0, M) admissable submodule S. The representation p.1 corresponding to S is a top constituent of p.. The representation p. 2 corresponding to MfS is a bottom constituent of p.. If M has no proper admissable submodule it is irreducible and p. is an irreducible representation. (b) A representation p. of a ring R and its corresponding module M are decomposable if M is a direct sum M = S EB T, and S, T are proper F-R submodules. The modules S, T and their correspon-

20

I. FOUNDATIONS

ding representations p. 1 and p. 2 are called components of M and p., respectively. Indecomposable means not decomposable.

(c) A representation p. of a ring Rand its corresponding module M is completely reducible if M is a direct sum M = M 1 EEl · · · EEl M k and the components M; are irreducible. A group G with operators is said to satisfy the descending chain condition if every sequence

of admissable subgroups G; contains only a finite number of distinct terms. Thus, if the inclusions are proper, such a sequence must terminate. G satisfies the ascending chain condition if every sequence G 1 c G 2 c G 3 c ... of admissable subgroups G; contains only a finite number of distinct terms. Again, if the inclusions are proper, the chain must end with Gat most. Since our representation modules M, and their admissable submodules, are finite-dimensional vector spaces over F, any submodule must have a smaller dimension than that of another in which it is properly contained. Hence both chain conditions are satisfied by the modules M. It is not hard to show that the chain conditions are equivalent, respectively, to the maximum (minimum) condition: every nonempty set of admissable subgroups has a maximal (minimal) member; that is, a member which is contained in (contains) no other member of the set. Now let M be a reducible F-R module. Then there is a proper F-R submodule M" such that M~M"~O.

If M" is itself reducible there is a proper submodule M'" such that M" ~ M'" ~ 0. If MjM" is reducible it contains a submodule K: MfM" ~ K. This implies the existence of a submodule M' such that M ~ M' ~ M", and K ~ M'jM".

4. APPLICATION OF IDEAS AND RESULTS

21

Then the original series can be extended to M "J M' "J M" "J M'" "J 0.

In the same way we may insert a submodule between any two consecutive modules in this series as long as the factor module of the one by the other is reducible. Repeated insertions will finally give a senes (4.6) which is incapable of further refinement. It is called a composition series of length k. The M;/M;_1 are factors of the series. Let some other reduction of M give a second composition series: 0

(4.7)

=

M~

c

M{

c M~ c ... c

M~

=

M.

The Jordan-Holder theorem for groups with operators for which the chain conditions are valid states that k = s and that the factors M 1/Mi-l of (4.6) are operator isomorphic to the factors MjjMj_ 1 of (4.7) in some order. Since the irreducible factors M;/M;_ 1 are the representation modules for the irreducible constituents ILt , IL2 , ... , ILk of the representation IL which correspondes to the module M we have:

(4.8) Theorem. Any representation IL of a ring R has a fixed number k of iffeducible constituents ILl , IL2 , ••• , ILk . They are unique up to equivalence and order of affangement. The case is similar if M is decomposable. Then M = M 1 EEJ M 2 and repeated decomposition of decomposable summands will give a Remak decomposition (4.9) in which each M 1 is indecomposable. Let

(4.10) M

=

M{

EEl M~ EEl ... EEl M;,

M; indecomposable,

be any other decomposition. The Remak-Krull-Schmidt theorem for groups with operators for which the chain conditions are valid asserts that k = s and that the M 1 are operator isomorphic to theM; in some order.

22

I. FOUNDATIONS

Since theM; are representation modules for the indecomposable components !J-; of the representation !J-, this result gives:

(4.11) Theorem. Each representation JJ- of a ring R has a fixed number s of indecomposable components JJ-1 , J.J- 2 , ••. , JJ-s • These are unique up to equivalence and order of arrangement. It is instructive to look at a matrix representation of R which exhibits the irreducible constituents. Let B be an ordered basis of the representation module M accommodated to the series (4.6): B

= {m~ , ... , m~ 1

,

m; , ... , m~ 2

, ••• ,

m~ , ... , m~.}

where the set of the first n 1 elements form a basis of M 1 and in general the set of the first n1 n2 n; elements are a basis of M;, fori= I, 2, ... , k. We have now

+ + ··· +

* 2

mn2r=

/c

mr= nk

* *

*

*

*

In each line an asterisk indicates a linear combination over F of earlier basis elements. This g1ves the matrix representation

(4.12)

5. THE REGULAR REPRESENTATION

where p.;(r) stands for then;

X

23

n; matrix

This is the matrix for p.;(r) corresponding to the basis for M; modulo M;_ 1 , that is, for the module M;/M;_ 1 • A basis forM adapted to the decomposition (4.9) will give for p. a matrix of the form (4.12) with the asterisk replaced by zero. In this case the diagonal blocks are indecomposable. They may be reducible. Let x'' be the character (Definition 2.1) of the representation p.. From the matrix (4.12) we get

x'"(r) = x'"'(r)

(4.13)

+ ... + x'"•(r).

This gives: (4.14) Lemma. The character of a representation is the sum of the characters of its irreducible constituents. '- R 1 be an operator homomorphism of an R-module M onto R1 , then M = M 1 8j M~, where M 1 " " R1 and M 2 is the kernel of v.

6. SOME FUNDAMENTAL LEMMAS

31

Proof. Let p of Lemma 6.3 be the identity map i : R 1 and hence determine T : R 1 - M such that 'TV = i:

R1

Let R1T = M 1 and let M 2 be the kernel of v. Since Tv = i is an isomorphism, therefore T is an isomorphism and R1 ~ M 1 . If m' E M 1 n M 2 , then 3r1 E R 1 such that m' = r 1T. But 0. = m'v = r1Tv = r 1 , so that m' = 0. Thus M 1 n M 2 = 0. Again let mE M and put mv = r 1 E R 1 • Then and Now (m - r 1T)v = mv - r1Tv (m - r 1T) E M 2 • Hence

= r1

-

r1

= 0,

so

that

It will be convenient for later use to have Corollary 6.6 reformulated in matrix terms as: (6.6)

Corollary. Let p.

(x)

=

(p.C(x) (x) 1

0 )

Pl(x)

be a reducible matrix representation of the ring R. If the bottom constituent p 1(x) is a component of the regular representation p(x), then there is a matrix T

=

(~ ~)

such that

T-lp.(x)T = (P.I(x) 0 ). 0 PI(x)

32

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Proof. Let M be the representation module, with basis which affords the representa{m1 , m2 , ... , m1 1 , m1 1 +l• ... , m1 1+I}, 2 tion p.. The first / 1 elements are a basis of the representation submodule M 1 which gives the representation p.1 . Then M/M 1 affords p1{x), which by hypothesis is afforded by a direct summand R 1 of R. Thus

and Corollary 6.6 gives

M{

(6.7) For each mi ,/1

M 2 :> M 3 :> ...

contain only a finite number of distinct terms. Let

N; = {m: mE M, m8i = 0}. Then N 1 C N 2 C ... and by (a) 3s such that N, = N,+l . Thus m8s+ 1 = 0 ~ m8• = 0. Now(} induces an automorphism of MjN, = M, for if mE M: ffz(}

= 0

~m(J

=0

EN.~

(m8)8• = m8B+1 = 0

~

m8

=>

m8• = 0 =>mEN,=>

m = 0.

Moreover, (} is onto M, for (b) holding in M also holds in M and so M :> M(J :> · · · :> M(Jk = M(Jk+l for some k. Hence Vm E M, 3m1 EM such that fii{}k = m1(Jk+l. Since(} is an automorphism this implies m = m18, so that every element of M is an image of another under (} and M = M8 = M8 2 = .. · = M(J•. But then Vm, 3m 2 such that m = m28" = m28•. This implies that m = m28• + n, n EN• . Therefore

M

=

M(J•

+ N,.

Finally if x E M(J• n N. , we have x8• = 0 and x = m8•, for some m E M. Therefore

0

x8•

=

=

m82•,

and thus mEN, giving x = m8• = 0. Therefore

M

=

M(J• EEJ N,

Since M is indecomposable either N, = 0, making(} an automorphism of M onto itself, or M = N,'in which case M(J• = 0 and so (}• = 0. As the length 'of a series in M, s ~ l, and hence 81 = 0. This proves the lemma.

34

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Note. It is easy to see that the converse of Lemma 6 9 holds, viz. if every operator endomorphism of an R-module M is either an automorphism or is nilpotent, then M is indecomposable. For if M = M 1 E8 M 2 then the decomposition operator S1 is neither nilpotent nor an automorphism, since S~ = S1 and, Vx, (xS2 )S 1 = 0. This result gives a useful criterion for deciding whether or not a module is decomposable.

Example. Let R be the ring of integers modulo 8. If 8 is an endomorphism let 18 = t. Then Vs E R, sO =(I I + ... 1)8 = (IB)s = ts. Now if t $ 0 (Mod 2), 8 is an automorphism, and if t ~ 0 (Mod 2), 8 is nilpotent. Hence R is indecomposable.

+

+

EXERCISE Let A be the group algebra of the Klein 4-group over F = {0, I} (see Exercise 4, Chapter II). Show that the ideal generated by (I + a) is indecomposable but reducible.

7. The Principal Indecomposable Representations Let R be a ring. An indecomposable component of the regular representation of R is called a principal indecomposable representation. It is the representation provided by a minimal right ideal R 1 which is a direct summand: R = R 1 EB R (R is a right ideal). We shall assume that the double chain condition holds for right ideals of R: every ascending chain of right ideals R1

c R 2 c R3 c ...

and every descending chain of right ideals

R1 :::> R2 :::> Ra :::> ... contains only a finite number of distinct ideals.

7. INDECOMPOSABLE REPRESENTATIONS

35

These are equivalent, respectively, to the condition that every set of right ideals of R has a maximal and a minimal right ideal. Under these conditions R will have a Remak decomposition:

in which the R; are indecomposable right ideals. We have encountered this case before [in (4.9)] for F-R modules, where the finite F-basis condition implies the double chain condition for admissable submodules. We rephrase the result there (Theorem 4.11) as: (7.1) Theorem. A ring R, with the double chain condition on right ideals, has a finite number k of principal indecomposable representations. They are unique up to equivalence.

(7.2) Theorem. Let R;, i ideals of a ring R and let

= I, ... , k, be indecomposable right

R = R 1 8j R 2 8j · · · 8j Rk .

Then (a)

There is a unique maximal right ideal R;

C

R; , i

= I, ... , k.

(b)

R; "" R; if and only if R;/ R; ~ R;/ R; .

(c)

Every irreducible representation of R is provided by a representation module R;/R; for some i.

(d)

There are exactly as many inequivalent irreducible representations of R as there are principal indecomposable representations.

Proof.

(d) Clearly (b) and (c)

(a) Let X be one of the ideal of R in X, and T =I= X Suppose T if. S, then S C X = S + T. Consider the (Section 6). Let

lo = e'

=

s

~

(d).

R; . Let S =I= X be a maximal right be an arbitrary right ideal of R in X. T + S, and since S is maximal: decomposition operator o: R -++ X

+ t,

s E S,

t E T.

36

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Define i',a,T:

x- X

xi'= x,

by

xa

=

sx,

XT =

fX

Since

XO

=

x,

Vx EX: xi' = x = xo = (lx)o = (lo)x

= i'

(s

+ t)x

=

xa

+

XT.

=a+ T.

Moreover a -=I- 0, otherwise X =Xi' = XT = tX CT. Similarly -=;1= 0. Now (xr)a = s(xr) = (sx)r = xar, Vr E R. Hence, since Xa C S, a and likewise T are operator endomorphisms of X into X. By Lemma 6.9 they are nilpotent. Thus 3m, n > 1 such that am = 0, am- 1 -=;1= 0, Tn = 0, Tn- 1 -=;1= 0. But then

T

glVlng a contradiction .. Therefore T C S, so that S = unique maximal right id~al in X = R; .

R~

is a

(b) Assume R;/R; ~ Ri/ R; . The following scheme of mappings is evident:

By Corollaries 6.4 and 6.5 with p = v' i', 3T : R; ---+4- Ri. Then by Corollary 6.6, since R; is indecomposable, R; ~ Ri . On the other hand, if T is the isomorphism R; ~ Ri , then R;T = R; which leads to the isomorphism R;/R; ~ Ri/R;. (c) Let M be an irredicible R-module. Since MR -=I- 0, 3i such that MR; -=I- 0. Then 3m E M such that mR; -=I- 0. Since M is irreducible and mR; is a submodule, therefore mR; = M. Let T : R; ---+ M be defined by rT = mr, Vr E R; . Let R; be the kernel

8. THE RADICAL OF A RING

37

of .,., Then R;(R.; ,..., M. Now M is irreducible, hence R.; is maximal and so by (a) R; = R~. Thus R;/R;"' M. The proof is now complete. Example. Let R be the ring of integers modulo 8. Now R is indecomposable (Example, Section 6) and by Theorem 7.2 has a unique maximal ideal R'. Here R' = {0, 2, 4, 6}. The principal indecomposable representation maps each element a E R onto the endomorphism p(a) given by xp(a)

=

'Vx

xa,

E

R.

The irreducible representation a provided by RjR' is given by xa(a)

=

'VxE R/R'.

xa,

The study of principal indecomposable representations will be resumed in Section 12.

8. The Radical of a Ring

Let R be a ring with an identity in which the double chain condition holds for right ideals. Theorems 7.1 and 7.2(c) show that there are only a finite number of distinct irreducible representations p1 , p2 , ... , Ps of R. (8.1) Definition. The radical N of R is the set of all elements r E R for which p;(r) = 0, for every irreducible representation P; of R:

N = {r: r E R, (8.2)

Lemma.

p;(r)

=

0,

i

=

1, 2, ... , s}.

N is a two-sided ideal of R.

Proof. If n1 , n2 , n EN, r E R, and p; is any irreducible representation, then 0 = p;(n1 ± n2 ) = p;(rn) = p~(nr). Hence n1 ± n2 EN nr EN, rn EN, showing that N is a two-sided ideal of R.

38

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Let us recall that the product of two subsets S 1 and S 2 of a ring R is the set of all finite sums ~ sls2 , sl E sl , s2 E s2 . Then, for example, N 1 as a set is the set of all finite sums ~ n1 n2 · · · n 1 , n; EN. Under the addition and multiplication of the ring this set forms a two-sided ideal which we denote by N 1• (8.3) Lemma. so that N 1 = 0.

N is nilpotent, that is there is a positive integer l

Let R 1 :J R 2 :J · · · :J R 1 = 0 be a composition series for R. Now Vn E N, R;n C R;+l , since n must be represented by zero in the i_~reducible representation afforded by the module R;/R;+l . Thus

Proof

and since 1 E R,

that is, N 1 = 0. (8.4) Theorem. Let R = R 1 EB ··· EB Rk, R; indecomposable and let R; be the unique maximal [Theorem 7.2(a)] subideal of R;, then

N = R~

E8 ·· · E8 R~

where N is the radical of R.

Proof (a) Let n EN. Now R;n s; R; , since n must be represented by zero through the irreducible representation module R;/R; . Hence

Since 1 E R,

n E R~

EB ··· EB R~ ,

and hence

N s; R~

E8 · ·· E8 R~ .

(b) Since the R; are right ideals R;R; s;R;. Now if we prove that R;R; s; R;, Vi, j then p;(R;) = 0 for all irreducible repre-

8. THE RADICAL OF A RING

39

+ ··· +

sentations so that Rj C N, Vj. Hence N:) R~ R~. Together with the final inequality of part (a) this would give

N = R~

EB ··· EB R~ .

To complete the proof we must show that R 1Rj -=1= R;; for then, since R;Rj is a proper right ideal of R; and since R; is the unique maximal ideal in R 1 , we get R;Rj £ R; . Suppose R 1Rj = R; . Then 3z E R; such that zRj = R 1 (otherwise xRj C R 1 , Vx E R 1 and since the xRj are right ideals and R; is unique maximal: xRj C R; so that R;Rj £ R; -=1= R 1). But then also zR1 = R; . Hence Vx E R; , 3x' E Rj such that zx = zx' or z(x- x') = 0. Now remark that A ={a: a E R;, za = 0}, the set of right annihilators of z, is a right ideal in R; and A -=1= R; since zR; = R; -=1= 0. Therefore A C Rj , since Rj is unique maximal in R; . Then x - x' E Rj; but x' E Rj , therefore x E Rj , therefore R; £ Rj giving a contradiction. This establishes the lemma. (8.5)

Lemma.

N contains all nilpotent right and left ideals.

Proof. (a) Let D be a, right ideal of R and suppose D1 = 0. If d ¢'. N there is some, irreducible representation p; such that p;(d) -=1= 0 and this means that in the associated representation module which must occur as some R;/R; (Theorem 7.2(c)] we must have R 1d $ R; . Then R;D $ R; . Since R;D is a right ideal and R; is unique and maximal: R 1D = R;. But then R; = R 1D = R;D 1 = 0, which is impossible. Therefore D £ N. (b) Let L be a nilpotent left ideal. Remark that LR is a nilpotent right ideal so that by (a) LR £ N. Since R has an identity L s;;LR £Nand the proof is complete.

Remark.

Since N is nilpotent every right or left ideal in N is nilpotent. Hence from Lemma 8.5 the radical can be characterised as the largest nilpotent right or left ideal of a ring.

Example.

Let R =

1(a 0) I c b '

I• a, b, creal numbers 1

40

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Then A =

!(a 0)1 I o oI

and

are right ideals and

R =A EBB,

A, B indecomposable.

A is irreducible but

Cis the maximal right ideal of R in B. It is two-sided and nilpotent and is the radical. As representation modules, A gives the irreducible representation

and BjC the irreducible representationO

There are no other irreducible representations.

9. Semisimple Rings (9.1) Definition.

A ring is semi'simple

if

its radical N is zero.

From Theorem 8.4 we see that R is semisimple if and only if the indecomposable components of R are irreducible, or in other words, if R is completely reducible. This can be stated as:

(9.2) Theorem. R is semisimple sentation is completely reducible.

if and only if its

regular repre-

(9.3) Theorem. Every representation p. of a semisimple ring R is completely reducible.

9. SEMISIMPLE RINGS

Proof.

41

Let M be the representation module for JL and let

M; indecomposable. We must show that theM; are also irreducible. Now since R has an identity by assumption and M is unitary (condition 9, Definition 3.1) M;R i:- 0, Vi. Also since R is semisimple R = ~~~~ EB R;, R; irreducible right ideals. Let M; be a maximal representation submodule of M;. Then M;/M; is irreducible and by Theorem 7.2(c) there is an R; such that M;/M; ~ R 1 (R; = 0 by the semisimplicity). Consider the scheme of mappings:

where v is the natural homomorphism and i is the assumed isomorphism. Then vi maps M; onto R 1 so that by Corollary 6.6, since M; is indecomposable, M; ~ R 1 • This implies that vis an isomorphism and hence M; = 0 so that the lemma is proved. (9.4) Definition. A ring is simple other than itself and the zero ideal. (9.5) Lemma.

R

simple~

if it

has no two-sided ideals

R semisimple.

Proof. Let N be the radical of R. Since N is a two-sided ideal either N = 0 or N = R. If N = R the nilpotentcy of N gives 0 = N 1 = R 1 which is impossible, since by assumption R has an identity. Hence N = 0 and R is semisimple. Let us recall that if B is a two-sided ideal of a ring R then the elements taken modulo B form a ring, the difference ring, denoted by R -B. Its elements are the equivalence classes of R modulo B. If 1_!:...] denotes the equivalence class of r, addition and multiplication are defined:

It is easy to verify that these class operations are independent of the class representatives and makeR - B into a ring. Moreover of R onto R -B. there is a natural homomorphism v: r-->-

l!J,

42

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

(9.6) Theorem.

R- N is a semisimple ring.

Proof. The existence of identity and of a composition series for R - N follows from their existence for R. Now if p is any representation of R for which p(N) = 0 we see that a representation pof R - N is well defined by

P(L!:J)

=

p(r).

Conversely, given p, this relation defines p. (If M is the module for p: mp(r) = mr = m(r + n), Vn EN, :. MN = 0.) In this way there is a one-to-one correspondence between the irreducible representations p1 , p2 , ... , Ps of R and the corresponding irreducible representations p1 , p2 , ••• , Ps of R - N. If now L!..l E R - N and Ml!:_l) = 0, Vi, then p;(r) = 0, Vi, and r E N so that L!:J = 0. Hence if N denote the radical of R - N, N = 0, and R - N is semisimple.

10. The Wedderburn Structure Theorems for Semisimple Rings (10.1) Lemma.

A ring with identity which is the direct sum of isomorphic minimal right ideals is simple. Proof. Let R = R 1 8j · · · 8j R 1 , R; irreducible, R 1 ~ R 1 , Vi. Let B be a two-sided ideal of R. If B -=1=- 0, since 1 E R, 0 -=I= RB = (R 1 8j ··· 8j R 1)B = R 1 B + ··· + R 1B. Hence one of the summands, say R1B, -=/=-0. Since R 1 is a minimal right ideal R 1B = R 1 C B. But now R;B -=1=- 0, otherwise 0 = R 1Ba1 = R 1a 1B = R 1B = R 1 , where a; is the operator isomorphism of R; -... R 1 . Therefore R1B

= R1 s B,

R

= ~R; s

B =R,

B.

and so R is simple.

10. WEDDERBURN STRUCTURE THEOREMS

43

Now let R be semisimple. Since R is completely reducible: R = (Ru

(10.2)

EB .. · EB R1"',) EB .. · EB (R;t EB .. · EB R;n;) EB .. · EB (Rst EB .. · EB R.n.).

Here the summation is arranged so that then; minimal right ideals in the ith block are mutually isomorphic, for each i, and R;; ~ Rtk if i -=I= t. If I = 1:;~ 1 (eil + "· + e;n) then eil + ··· + e;ni is the identity for the ring A; = Ril + ··· + R;n;· Now Lemma 10.1, using the relation ( 10.2), gives: R;;

(10.3) Theorem. simple rings. (10.4)

Every semisimple ring R is the direct sum of

R = A1

EB .. · EB A •.

Moreover, the A; are two-sided ideals and A;A; The decomposition is unique.

= 0, i

-=I=

j.

Proof. Only the last two statements need proof. Since the R;; are right ideals A; = R; 1 EB ··· EB R;n; is a right ideal. Let r E R. If rR;; ¢.A; then for some t, k, t -=I= i: rR;; n R 1k -=I= 0. Let o1k be the decomposition operator mapping R--+-+ R 1k. Now the mapping T : R;;- Rtk given by XT = (rx)otk = rot~. X E R;; ' is a homomorphism -=/=-0, and since R 1k is irreducible it is an isomorphism. This is a contradiction. Therefore rR;; C A;, Vr E R. Then A; is a left ideal, and so a two-sided ideal. Finally, A;A; C A; n A; = 0. If B is any two-sided ideal, since the A; are simple: B n A; = 0 or A; . This shows that the decomposition (10.4) is actually unique. Since a simple ring is semisimple we have as a converse to Lemma I0.1 the: (10.5) Corollary. right ideals.

A simple ring is the direct sum of isomorphic

Proof. If there were more than one summand in (10.4) R would have a proper two-sided ideal. Thus there is only one summand and it is the direct sum of isomorphic minimal ideals by (10.2).

44

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

This result shows that a simple ring can have only one irreducible representation up to equivalence. A ring R is simple if and only if it isomorphic to a complete matrix algebra of degree m over a division ring. The number m and the division ring are uniquely determined by R.

(10.6) Theorem.

Proof.

(a) only if.

Since R is simple: R = R1

(10.7)

EEl ·.. EEl Rm

in which the R; are irreducible right ideals and R; ,...__ R;'-+ R; . Recall (Section 6) that

i =I= j. We consider Hom(R, R), the ring of all R-endomorphisms of R as an R-module. Thus a E Hom(R, R) implies that (r + r')a = ra + r' a; (rr')a = (ra)r'. Vj let T; be the fixed isomorphism R 1 ~ R; . For every a E Hom(R, R) define the homomorphisms a;; : R 1 --->- R 1 as follows: {10.8)

Since R 1 is irreducible all endomorphisms, if not zero, are automorphisms and have inverses and the set of all of them form a skew field. Thus to each a E Hom(R, R) there corresponds an m X m matrix (a;;) with elements a;; from a skew field. Now the a;; can be prescribed arbitrarily and the corresponding a found again, for from (10.8) we have the relation (10.9)

ka;;T; 1

=

kT;aO; J

=

T;a kO;

=

T;al =

T;a

j

which, since R 1T; = R;, defines a on R;, Vi, if we know the a;;. But a on the R; gives a on R through ( 10.7). Thus there is a unique correspondence at--+ (a;;) between elements of Hom(R, R) and all m X m matrices over a certain

10. WEDDERBURN STRUCTURE THEOREMS

45

skew field. If J-1- ~ (J-1-ii), consider the element at the intersection of the ith row and kth column of the product matrix ( a;;)(J-1-;;). It is

!- a;;J-1-jk = !- T;aO;T j J

1

'T;J-1-0kTl/

=

!- T;aO;J-1-0kTkl

j

=

T;a

;

k O;J.I-OkTk

1

=

T,aJ-1-0kT//.

i

But this is the similarly placed element of the matrix corresponding to aJ-1-. Therefore aJ-1- ~ (a;;)(J-1-;;).

Similarly a

+ J-1- ~ (a;;) + (P-;;)·

Hom (R, R)

~

9Jl,. •

9Jlm is the set of all m X m matrices over the skew field K, uniquely

determined as automorphisms of R 1 • The number m is the number of summands in (10.7) and it is unique by the KrullSchmidt theorem. Again, if a E Hom(R, R), let Ia = c E R. Then ra

=

(lr)a

Conversely, given c E R, a by

E

= (la)r =

cr.

Hom(R, R) is uniquely determined

ra

=

cr,

which gives again

Ia =c. Thus there is a one-to-one correspondence between elements a E Hom(R, R) and elements c E R. If a~ c, T ~ t we have l(a

+ T)

= la

+ IT

= c+t

and laT

= (Ja)T =

CT

=

(Jc)T

= (lT)C = tc,

Hence and

aT~

tc.

46

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Thus R is anti-isomorphic to Hom(R, R). Taking the transposes of the matrices in 9Rm we get m;,. anti-isomorphic to 9Rm. Therefore R is isomorphic tom;,.. This proves (a). (b) if. Let R be the set of all m X m matrices over a skew field K. Then R has the unit matrix. Moreover R has a finite K-basis of m2 elements so that the double chain condition holds. Denote by e;k the matrix with I E K at the intersection of the ith row and kth column and zeros elsewhere. Then Va E R there is a unique expression a= ~a;;e;;,

a;; E K.

'·' Now let B i= 0 be any two-sided ideal of R. Let a E B. If a i= 0, some ars i= 0, a,. E K. Since B is two sided it contains

Using the fact that emnekf = 0 or emf according as n i= k or n = k we have a' = bife;; . But i and j are arbitrary and h;; is an arbitrary element of K. Therefore B = R and R is simple. In this context we give the following proof of: (10.10) Theorem (Frobenius-Schur theorem). If p1 , p2 , ... , Ps are the distinct representations of a ring Rand if a 1 , a 2 , .:.,a. are arbitrary elements of R, then there exists an element a E R so that

p;(a)

=

p;(a;),

i

I, ... , s.

=

Proof. It suffices to consider R as semisimple, for if ~ E R - N were found so that ,0;(1~) = ,0;(1~), then since ,B(W) = p(b) (see Theorem 9.6) we would have the result in R. Accordingly let R = A 1 EEJ .. · EEJ A. , where the A; are mutually annihilating two-sided ideals, each the direct sum of minimal right ideals (Theorem 10.3). Then for each a;: i

a; = a 1

+ ai2 + .. · + a.i ,

Let a

= a~

+ a~ + .. · + a~

a~

E

A;, j = I, ... , s.

47

11. INTERTWINING NUMBERS

Suppose now that the minimal right ideal R; 1 C A; is the representation module for p;. Since

p;(a)

=

p;(a;)

and the proof is completa.

11. Intertwining Numbers In this section all modules considered are F-R modules with a finite F-basi~. This includes the ring R itself, as a module. Hence R is an algebra. This means that R has a basis consisting of a finite number of elements of R, and every element of R is a unique linear combination of these basis elements with coefficients from F. As a consequence the double chain condition holds for ideals in R. We assume again that R has an identity. (11.1) Definition. The set of allF-R homomorphisms of MintoN is denoted by Hom(M, N). Every a E Hom(M, N) is called an intertwining of M and N.

(11.2) Lemma.

Proof.

For a,

T

Hom(M, N) is a module with a finite F-basis. E

Hom(M, N) define a

+ T)

=

+ m')(a + T)

=

m(a

Now (m

ma

+ mT

+

T:

VmEM.

+ m')a + (m + m')T = ma + m' a + mT + m' = m(a + T) + m'(a + T), homomorphism. Clearly a + = + a. (m

T

so that a + T is a Moreover, iff E F, r E R: fmr(a

+ T)

= fmra

+ fmrT

T

+ f(mT)r f(m(a + T))r.

= f(ma)r

=

a+

T

E

Hom(M, N).

= f(ma

T

+ mT)r

48

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

Now define fa

=

uf for f E F,u E Hom(M, N) as follows: m(fa) = (fm)a

=

f(mu).

Then (m

+ m')fa = f(m + m')u = fm + fm')u = m{fa) + m'(fu) . .".

fa

E

= (fm)a

+ {fm')a

Hom(M, N).

We have now proved that Hom(M, N) is an F module. Let {m1 , .•• , m1} and {n 1 , •.. , n.} be bases of M and N, respectively. Let ex € Hom(M, N), then m;ex = L;_ 1 a;1n 1 , i = 1, ... , I, a;1 EF. Thus ex corresponds to an I X s matrix A = (a;1). If f3 E Hom(M, N) corresponds to B = (h;;) it is clear that ex = f3 if and only if A = B. Moreover every linear relation among the ex subsists among the corresponding A and conversely. rank Hom{M, N)

~Is.

(11.3) Definition. The intertwining number of M and N, denoted by i(M, N), is the rank(F-dimension) of Hom(M, N). {11.4) Lemma.

EB N 2) = Hom(M, Nt) EB Hom(M, N 2 ). EB M 2 , N) = Hom(M1 , N) EB Hom(M2 ,N).

(a)

Hom(M, N 1

(b)

Hom(M1

Proof. (a) Hom(M, N 1), Hom(M, N 2), as homomorphisms of Minto submodules of N 1 EB N 2 , can be regarded as submodules of Hom{M, N 1 EB N 2). Let S; be the decomposition operators: N ~ N;, i = 1, 2. Now for a E Hom(M, N 1 EB N 2) define a 1 E Hom(M, N 1), u2 E Hom(M, N 2) thus: VmEM,

This implies

11. INTERTWINING NUMBERS

49

Conversely, given a 1 , a 2 we get back a by using the last relation. Therefore a

= a1 + a2 and Hom(M, N 1 EEJ N 2 ) = Hom{M, N 1) + Hom(M, N 2).

If p E Hom{M, N 1) n Hom(M, N 2 ), then Vm E M. Therefore

p=O

mp

EN1 n N2

and the sum is direct.

(b) Va E Hom(M1 EEJ M 2 , N) define a 1 , a2 : m1a 1 Vm 1 E M 1 , and M 2a 1 = 0; similarly M 1a 2 = 0, m2a 2 Vm 2 E M 2 • Then

Vm

=

=

+ m2 M EEJ M {m + m ){a + a m a + m2a 2 = {m + m2 )a. a + a =a. m1

= 0,

E

2 :

1

1 1

1

2

1

= m1a, = m2a,

2)

1

1

2

Moreover, given a 1 , a 2 we get back a by using the same relation. Since a; can be regarded as an element of Hom(M;, N), i = l, 2, this gives Hom(M1 EEJ M 2 , N)

= Hom(M1 , N) + Hom{M2 , N)

If p E Hom(M1 , N) n Hom(M2 , N), then (m1 Vm E M 1 EEl M 2 • Therefore

p=O

+ m2 )p

0,

and the sum is direct.

(11.5) Corollary.

EEl N 2) = i(M, N 1) + i(M, N 2) i(M1 EEl M 2 , N) = i(M1 , N) + i(M2 , N). i(M, N 1

(11.6) Lemma.

Hom(R,M)""' Mandhencei(R,M)

= dimM.

Proof. 'Ita E Hom{R, M) define the mapping?T: a--la= m0 EM. Now ra = (lr)a = lar = mor. Conversely, given m0 EM we have the mapping 6 : r6 = m0 r, Vr E R, so that 6 E Hom(R, M) and 6 = a. Thus 1T is a one-one mapping of Hom(R, M) onto M.

50

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

But since (a + T)7T = l(a an isomorphism and

+ T)

= Ia

Hom(R, M) ,....., M,

+

IT = (a)7T

+ (T)7T,

7T is

i(R, M) = dim M.

(I 1.7) Lemma (Shur's Lemma). Hom(M, M) is a ring, and M is irreducible, a skew field over F.

if

Proof. Lemma I 1.2 showed that Hom(M, M) is a module over F. If a, T E Hom(M, M) define aT: m(aT) = (ma)T, Vm EM. Since (m

+ m')aT

+ m')a)T

=

((m

=

(ma)T

=

+ (m'a)T and

arE Hom(M, M)

(ma

+ m'a)T

m(aT)

=

+ m'(aT).

Hom(M, M)

is a ring.

Now let M be irreducible, a E Hom(M, M). If a =I= 0, Ma =I= 0 and is a submodule of M. Thus Ma = M, i.e., a is onto. If M 0 =kernel a= {ml, ma = 0}, then M 0 is a submodule. Since M 0 =I= M, therefore M 0 = 0 and a is an automorphism. Thus a =I= 0 => a- 1 exists arid so Hom(M, M) is a skew field. Let p. be a representation of R and M its associated representation module. Let 7T E Hom(M, M). Then since

(mr)1r = (m1r)r, mp.(r)1r

=

(m1r)p.(r)

p.(r)1r

=

1rp.(r),

=

m(1rp.(r))

VrER,

and this holds V1r E Hom(M, M). For this reason we have: (11.8) Definition. Hom(M, M) is called the commuting ring of the representation p.. We denote it by C(p.) or C(M). More generally, if 7T E Hom(M, N), p.(r)1r = ?TV(r) where p., v are the representations associated with M and N, respectively. For this reason Hom(M, N) is sometimes called the intertwining module of M and N, and the corresponding representations p. and

11. INTERTWINING NUMBERS

51

v are said to be intertwined. Recall that a field F is algebraically

closed if every equation

a;EF, has all its roots in F. (11.9) Lemma. If p. is an irreducible representation of a ring R over an algebraically closed field F, then rank C(p.) = I.

Proof. Let M be the representation module for p.. Since M is irreducible C(p.) is a skew field by Lemma 11.7. Let 0 =I= -r E C(p.). Now VfEF, VmEM, VrER:

m(fl - -r)p.(r) = m(fl - -r)r = (fm -- m-r)r = fmr- m-rr

= fmr- mr-r = mr(fl - -r) = mp.(r)(JI - -r). {fl - -r)p.(r)

=

p.(r){fl - -r),

Vr E R.

{fl - -r) E C{p.). Now since F is algebraically closed 3/' E F, m' EM, m' =I= 0 such that

m'(f'I - -r) = 0, [otherwise m(If1 - -r){lf2 - -r) ···(If.- -r) =I= 0, Vm, Vf;, whereas it must =0 if the f; are roots of the polynomial p(x) for which p(-r) expresses the linear dependence which must subsist between m, m-r, m-r2 , ••• , m-r•, for some s]. Thus, since m' =I= O,f'I - -r = 0, therefore T

=f'J,

proving that rank C{p.) = l. Let R1 be an indecomposable component of the ring R, R = R 1 EEJ •· • . Let R~ be the unique maximal right ideal in R 1 [Theorem 7.2(a)]. Suppose that the representation module M of a representation p. has the composition series: M = M 0 ~ M 1 ~ •·• ~ M 1 = 0. Then the factors M;_ 1JM; are representation modules for the irreducible constituents of p.. We have:

52

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

(ll.IO) Theorem. i(R1 , M)

= qk

where q is the number of factors of M of the skew field C(R 1 /R~).

Proof.

~ R 1 /R~

and k is the rank

Consider the scheme of homomorphic mappings:

where v is the natural homomorphism of Mi-l onto M;_ 1 fM;. If T is any homomorphism of R 1 into M;_1 , then Tv = a is a homomorphism of R 1 into M;_ 1 fM;. Conversely, by Lemma 6.3 a there is aT such that TV = a. Hence the mapping r: T- TV is a mapping of Hom{R1 , Mi_1) onto Hom(R 1 , M;_ 1 fM;). Since (T T')F = (T T')v = T V + T'v = Tr TT, r is a homomorphism. Its kernel is all T which map R 1 ---+ M; that is all T E Hom(R 1 , M;). Hence

v

+

+

+

Taking ranks on both sides we have

and summing over j: I

i(R1 , M)

=

ki(R 1 , M;_ 1 fM;). i~l

Now either i(R1 , M;_ 1 fM;) = 0, or 3a E Hom{R1 , M;_ 1 jM;) such that R 1a c# 0. But then R 1a = M;_ 1 fM; since the latter is irreducible. If R~ = Kernel of a, then R 1 jR~ ~ M;_ 1 fM;, implying that R~ is maximal in R 1 which in turn implies that R~ = R~ by Theorem 7.2(a). Thus a induces an isomorphism 6: R 1 /R~ ~ M;_ 1 fM;. Conversely, given 6 we can find

53

12. INDECOMPOSABLE COMPONENTS

a : R 1 -->- M;_ 1 fM; with kernel R~, from a

natural homomorphism R 1 ---~

Hom(R1 , M;_1 /M;)

R 1 fR~.

=

a where v

v1

1

is the

Thus

~ Hom(R 1 /R~,

M;_ 1 jM1)

~ Hom(R 1 /R~, R 1 /R~).

Then i(R1 , M;_ 1 /M;) = i(R 1 /R~, R 1 /R~) = k, the rank of C(R 1 /R~). We have now proved that i(R 1 , M;_ 1 fM1 ) = 0 or k, according as R 1 fR~ is or is not isomorphic to M 1_ 1 fM1 • Applied to ( *) this result gives the theorem.

12. Multiplicities of the Indecomposable Components in the Regular Representation The results of the last section can be used to prove: (12.1) Theorem. Let R be an algebra over afield F. Suppose that

where the R;; are indecomposable right ideals and the summation (direct) is arranged so that the q; terms in the jth block are isomorphic andj = 1, 2, ... , s. Thus R;k ~ R;m if and only ifj = i. If n

=

the rank of R

d;

=

the rank of Ril

n1

=

the rank of R 11/Rj 1 , R;1 being the unique maximal right ideal in R 11

k;

=

the rank of the skew field C(Ril/R;t).

then

n

=

t (n ;~1

1fk 1)d;

.

.

Remark. Theorem 7.2(b) and (c) shows that the Ril/R; 1 , j = 1, 2, ... , s, are representation modules for all the distinct

54

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

irreducible representations. In the language of representations the n; are the degrees of the irreducible representations, the d1 the degrees of the principal indecomposable representations and n is the degree of the regular representation. Proof. Thus

By Lemma 11.6 i(R, R 11 /R;d n; = i(R, R; 1 /Rj 1) = i

rank(Ril/R; 1)

=

=

n1

•

(~ Rab, R; 1/R;1 ) a,b

where the Rab run through the terms of (12.2). Therefore n;

=

~ i(Rab, Ril/R;l)

a,b

by Corollary 11.5. But i(Rab, Ril/R; 1) Rab ~ R 11 or not. Therefore

=

k1 or 0 according as

q,

12.3)

n1

= ~ i(R;b , R11 /R;1 ) = q1k;. b-1

Now counting ranks on both sides of (12.2) we get n so that (12.4)

n =

i

=

L;_

1

q;d;,

(n1fk 1)d;,

i-l

proving the theorem.

(l2.5) Corollary. Proof. rank R 11 /R; 1 from (12.4).

(12.6) Corollary.

n ~ L;_ 1 n~/k;. ~

n

rank R 11 or n1

=

L;_

1

n~/k;,

~

d1 and the result follows

if and only if R

Proof. R semisimple ~radical N = 0 ~ R; 1 Hence d1 = n; and (12.4) gives the result.

is semisimple.

= 0 (Theorem 8.4).

(12.7) Corollary. If R is over an algebraically closed field F, then n ~ 1 n; and equality holds if and only rf R is semisimple.

L;_

13. THE GENERALIZED BURNSIDE COMPONENTS

55

Proof. This follows from Corollaries 12.5 and 12.6 and the fact that k1 = 1 by Lemma 11.9. q1

In the course of proving Theorem 12.1 it was shown that = n1fk 1 • We state this result as:

(12.8) Corollary. The number q1 of indecomposable components of the regular representation of a ring R which are isomorphic to a particular indecomposable component R 1 is given by q1 = n1fk 1 where n1 = rank R 1fR;, k 1 = rank of the skew field C(R1/Ri), and Rj is the unique maximal right ideal of R contained in R 1 .

13. The Generalized Burnside Theorem

{13.1) Theorem. Let R be an irreducible algebra of n1 X n 1 matrices over a field F. Let R have rank n. Then R has an identity and n = n~/k 1 , where k1 is the rank of C(R), the commuting algebra of R. Moreover R is isomorphic to a complete matrix algebra of degree nfn1 = n1/k1 over a skew field which itself has rank k 1 over F.

Proof. If the n1 X n1 identity matrix If/= R let R* be the algebra generated by I and R. Its elements are finite sums of finite products of elements taken from R or the set of scalar matricesfi, f E F. Clearly R* is irreducible, otherwise R C R* would be reducible. Also R* is its own faithful matrix representation. This makes R* simple (see Lemma 13.2). But now R is a two-sided ideal of R* and we have a contradiction. Hence IE R and R as its own faithful irreducible representation is simple:

and all R; are isomorphic right ideals (Corollary 10.5).

i(R, R 1)

= i (iR 1 , R 1 ) = mk 1 • J=l

56

II. REPRESENTATION THEORY OF RINGS WITH IDENTITY

and, counting ranks on both sides of(*):

so that

as required. (Also follows directly from Corollary 12.6.) Finally from the structure Theorem 10.6 and (*) above it follows that R is isomorphic to a complete matrix algebra of degree m over the skew field C(R). Since m = nfn 1 and rank C(R) = k1 this is the required result. It remains to prove:

(13.2) Lemma. If a ring R has a faithful irreducible representation p. it is simple. Proof. If n EN, the radical of R, then p.(n) = 0 and so n Hence R is semisimple and so by Theorem 10.3:

R

=

= 0.

A 1 (B .. · (B A. ,

where A; are simple two-sided ideals, each the direct sum of minimal right ideals and, moreover, A;A; = 0, i =I= j. Suppose R; C Ai is the minimal right ideal which affords the representation p.. Then R;A 1 = 0, Vj =I= i, and so p.(A;) = 0, Vj =I= i. But then by the faithfulness of p., A; = 0, Vj =I= i: R=A;

and R is simple. EXERCISES

1.

Determine the principal components of the ring R

of matrices with real coefficients and find the unique maximal right ideal contained in each component. What is the radical of R? [Cf. Example, Section 8.]

57

EXERCISES

2.

Let F be the field of real numbers and let

Verify the Burnside theorem for this case. What is the skew field ?

3. Exercise 2 with

R=

(-:

b

c

a

-d

d -c -d -c

b

a

-I)

Compare with

over the complex field. 4. Let R be as in Exercise 2 but let F be the field of complex numbers. Show that now R is not simple but that

J=

I(I

.z iz)l -IZ

z \

is an ideal. Is R semisimple ? 5.

Let R

=

\r

I

=(ab

0)

c '

a b c EFI • •

1·

Is R its own regular representation? Observing that p 1(r) = a, are two irreducible representations and that F = C(p 1 ) = C(p 2 ) use the relations (12.5) and (12.6) to prove that R is not semisimple and that there are no other irreducible representations of R. p2 (r) = c

6. If R 1 is a direct summand of the ring R and M 1 is an admissable submodule of the R-module M prove that the intertwining numbers are related as follows:

CHAPTER Ill

The Representation Theory of Finite Groups 14. The Group Algebra The concept of the representation of a group was discussed in Seotion I of Chapter I. In the present chapter the representation theory of algebras which was developed in Chapter II will be applied to the theory of the representations of finite groups. The connection between the theories is provided by the concept of the group algebra.

(14.1) Definition. The group algebra. Let G = {g1 ,g2 , ••• ,g,.} be a group of finite order n, and let F be an arbitrary field. Denote by A(G) an n-dimensional vector space over F in which the elements of a basis are labeled g 1 , g 2 , ••• , gn . Thus every element a

E

A(G) can be written uniquely: n

a=

""'fJ;g;,

f; EF.

i~l

Now define a multiplication in A( G) as follows: if a' then

=

L;~ 1 flg;,

n

aa'

=

kfdjg;;,

g;i

=

the group element g,gi .

i,j=l

It is a routine matter to check the validity of the rules:

(aa')a"

=

a"(a +a')

a(a'a");

(a+ a')a"

= a"a +a" a' 58

and

=

aa"

+

a'a";

ag1 = g 1a,

59

14. THE GROUP ALGEBRA

if g 1 is the identity of G. Thus A(G) is an algebra, with an identity, of rank n over F. It is the group algebra of G over F. The relation between representations of G and of A(G) is given by: (14.2) Lemma. Every representation a of A(G) induces a representation a of G and conversely. The correspondence a - a is unique. Moreover a is reducible (decomposable) if and only if a is reducible (decomposable).

Proof. Given a take a to be the restriction of a to the elements of G. Since a(g;g;) = a(g;)a(g;) we have the same relation for a so that is a representation of G. Conversely, if is given, define a(a) = "i:.:=d;a(g;), when a = "£:= 1 f;g; . Then if a' = 1 f;g; , we have

a

a

"£;=

a(a)a(a') =

(tf;a(g;*~>;a(g;)) z=l

= .tf;J;a(g;)a(g;)

:1=1

z,J=l

n

=

~fJ;a(g;g;)

=

a(aa').

i,j=l

+

+

a-

In the same way a(a a') = a(a) a(a'). If this a is now a uniquely. Finally restricted to G we get back a, so that observe that if Sis a submodule of an F-A(G) module M, then S is admissable under G if and only if Sis admissable under A( G). This remark proves the last statement of the lemma. As a consequence of Lemma 14.2 all the results with respect to reducibility and decomposability which were proved for the representation of an algebra carry over to the representation of a finite group. In particular we have: (14.3) Theorem. The irreducible constituents p.1 , p. 2 , ••• , P.k of a representation p. of a group G are fixe¢ in number and are unique up to order and equivalence. The same statement holds for the indecomposable components.

Proof.

The proof follows from Theorems 4.8 and 4.11.

60

Ill. REPRESENTATION THEORY OF FINITE GROUPS

15. The Regular Representation of a Group Recall that the regular representation of A( G) is the representation p afforded by A(G) as its own representation module. The regular representation p of G is the restriction of p to G. Taking g 1 = l, g 2 , ••• , gn as a basis of A(G) we get

i = l, 2, ... , n. Thus, expressed as a matrix:

where the element 8g;,g 1a at the intersection of the ith row and jth column is the Kronecker 8: a =F- b,

and

Now for the character xP (Definition 2 .l) of the regular representation of G we have, since g;g = g; if and only if g = l, ( 15.1)

xP(g) = 0,

g=F-l;

xP(l) = n.

Finally, note that the regular representation is faithful, smce

'V g; => g = I. From Theorem 7.2(c) and (d), valid for A(G), we get directly: (15.2) Theorem. (a) Every irreducible representation of a group G occurs as an irreducible constituent of the regular representation. (b) The number of irreducible representations of G is the same as the number of principal indecomposable representations.

16. Semisimplicity of the Group Algebra (16.1) Theorem. Let G be a finite group of order n. Let F be a field of characteristic p. The group algebra A(G) of G over F is semisimple if and only if p does not divide n, or p = 0.

16. SEMISIMPLICITY OF THE GROUP ALGEBRA

Proof (a) suppose

61

p = 0 or p 1 n. Let N be the radical of A( G) and n

(16.2)

r

=

ZJigi'

rEN.

;~1

If pis the regular representation of A(G) we get p(r)

"

n

;~1

;~1

= ZJip(g;) = kf;{J(g;),

p

the regular representation of G. Taking the traces of the matrices on both sides, (15.1) gives

(16.3) However Lemma 4.14 showed that

p; all irreducible constituents. But since r EN, p;(r) = 0 and XP'(r) = 0; henCe XP(r) = 0 and (16.3) becomes

( 16.4) Since p = 0 or p 1 n we have / 1 = 0. Now N is an ideal and rgj 1 c= N. Again,

so that by the same argument /; = 0, Vi. Therefore r = 0 and hence N = 0, showing that A(G) is semisimp1e. (b) pjn. Let r = L7~ 1 g;. Now Vi, rg; = L7~digi = r, since the terms of the summation are again the groups elements in some order. Similarly g;r = r. Thus Fr is a two-sided ideal. Now r2 =

i

i~l

(ig;) g; = r + r + .. · + r = nr = 0. i~1

62

Ill. REPRESENTATION THEORY OF FINITE GROUPS

Therefore Fr is a nilpotent two-sided ideal. By Lemma 8.5: N:) Fr so that N =f=. 0 and A( G) is nonsemisimple. (16.5) Corollary. A representation p. of a finite group G of order n is completely reducible over a field of characteristic p zf p = 0 or p -r n.

Proof. The proof follows from Theorem 9.3, since A(G) is semisimple. (16.6) Corollary. If G is of order n and n1 , n2 , ••. , n. are the degrees of the distinct irreducible representations p 1 , p2 , ••• , Ps of G over an algebraically closed field F of characteristic p, p = 0, or p -r n, then (16.7)

n1

= n~ + n~ + ··· + n~ .

Proof. The proof follows from Corollary 12.7, since A(G) is semisimple. Remark. Every group G has the trivial irreducible representation p1{g) =I, 'VgEG. Thus in (16.7) we always have n 1 =I. (16.8) Corollary. In the regular representation p of a finite group G of order n over an algebraically closed field F of characteristic p, p = 0, or p -r n, each irreducible representation p1 occurs as often as its degree n1 •

Proof. A( G) is semisimple by Theorem 16.1 so that the indecomposable components of A(G) are irreducible. Let R1 be the right ideal which affords the irreducible representation p;, then according to Corollary 12.8 the number of irreducible constituents of p which are equivalent to PJ is nJ

k;

=

rank of R; = degree of Pi

= rank of the skew field C(R;).

Since F is algebraically closed, k; = I by Lemma 11.9. This proves the result. An immediate consequence is:

16. SEMISIMPLICITY OF THE GROUP ALGEBRA

63

(16.9) Corollary. have

Under the conditions of Corollary 16.8 we

(16.10)

=

s

xP(r)

~ nixj(r),

'Vr

E

A( G),

i~l

where x" denotes the character of the representation a.

Proof. Apply Lemma 4.14 and the fact that equivalent representations have the same character. Remark. Let F be a subfield of a field E. Then E is called an extension field of F. Let M be an E-R module for the ring R. The example on p. 64 shows that M regarded as an F-R module may be irreducible while being reducible as an E-R module. In such a case we say that the representation splits on extension of the field. ( 16.11) Definition. A representation p. of a ring R in a field F is absolutely irreducible if it remains irreducible (does not split) in any extension E of thegroundfieldF. Now let G be a finite group of order nand letF be an algebraically closed field of characteristic p, p = 0, or p -r n. IfF C Ewe know from the theory of fields that the field E can be extended to an algebraically closed field E*, and all three fields necessarily have the same characteristic p. Thus F C E C E* and any representation a in F can be considered as a representation in E*. Moreover, because of the condition on p, A( G) is semisimple over each field. By Corollary 16.8 the irreducible representation p; of A( G) in F occurs n1 times (as often as its degree) in the regular representation p of A(G) in F. Now if Pi is reducible in E* andY; is an irreducible constituent of p;, then Yi occurs at least n; times. But this is more than its degree (since degree y; .g

+ p.g2) =

+ /#2 , /; EF,

[>.(fl - /2) - P./t]et

1

prescribed/,/~

. Thus mR

~

M and M has no proper submodule. (2) M is reducible as an E-R module.

Proof. Let m = e1 + (I - i)e 2 • It is easily verified that mg = im. Hence Em is a proper R-submodule of M and M is reducible.

17. The Center of the Group Algebra ( I7 .I) Definition.

The set of all elements z of the group algebra A(G) which commute with each element of A(G) is called the Center of the Group Algebra. If Z denote the center of A(G) we have

Z

= {z: xz = zx,

Since z, z' E Z, jEF => zz', z subalgebra of A(G).

'Vx E A( G)}.

+ z',

fz E Z, therefore Z is a

18. INEQUIVALENT IRREDUCIBLE REPRESENTATIONS

65

Now let zEZ: n

(17.2) z = '"iJ;g;'

f;EF,

g;EG,

G: 1

=

n.

i=1

'rig E G we have "

n

i=1

j=1

gzg-1 = z = "'£Ji.gi = ~/;gg;g-1,

and equating coefficients we see that

fi =f;

if and only if

Thus all group elements which are conjugate have the same coefficient in (17.2). We may write (17.3) where Ci denotes the sum of all the elements in a certain class of conjugate group elements. Conversely, any element z given by (17.3) belongs to Z since gCig-1 = Ci,

the left-hand side merely rearranging the terms m the sum. This gives: (17.4) Lemma. Z is a subalgebra of ranks where sis the number of classes of conjugate elements of G. It has as a basis C 1 , C 2 , ..• , C•; each Ci is the sum of the elements in a class of conjugate elements.

18. The Number of Inequivalent Irreducible Representations (18.1) Theorem. The number of inequivalent irreducible representations of the group G in an algebraically closed field F of characteristic p, p = 0 or p f n = G: 1, is equal to the number s of classes of conjugate elements in G.

66

Ill. REPRESENTATION THEORY OF FINITE GROUPS

Proof. Let p1 , p2 , •.. , Pk be the k distinct irreducible representation of A( G). Let z E Z. Since xz = zx: Pi(x)pi(z)

=

v X E A(G).

Pi(z)p;(x),

Thus p;(z) E C(p;), the commuting algebra of Pi (Definition 11.8). Since Pi is irreducible and F algebraically closed, C{p;) is a skew field of rank I over F (Lemma I 1.9). Thus V Pi, i = I, 2, ... , k, (18.2)

j;EF.

p;(z) =/;I,

Then v z E z we have a mapping r: z - (/1 ' /2' ... , fk). Since p;(fz+f'z') = fp;(z)+J'p;(z') =//;I +J'/;1 = (//;+J'/:)1, r is a homomorphism of z as a module into vk' a k-dimensional vector space over F. If z--->- (0, 0, ... , 0), then (18.2) gives p;(z) = 0, Vi so that zEN, the radical of A( G). Since A( G) is semisimple, N = 0 and hence z = 0. Thus r is an isomorphism. Finally, let (/1 , /2 , ••• ,A) be an arbitrary vector of Vk. If I is the identity of A(G), then z; =/;I E Z and p;(z;) =/;I. However, by the Frobenius-Shur theorem (10.10) 3z such that

p;(z) = p;(z;) =/;I,

Vi,

and hence

r is an isomorphism of Z onto Vk, and so dim Vk = rank Z s, the number of classes of conjugate elements of G (Lemma

Thus

=

17.4). Therefore k=s and the theorem is proved. (18.3) Corollary. The irreducible representations of an abelian group are all of degree one. Moreover, their number equals the order of the group.

Proof.

By (16.7), n

= n~ +

··· + n~.

19. RELATIONS ON IRREDUCIBLE CHARACTERS

67

Since the group is abelian, each element is its only conjugate and s = n. Therefore n; = 1.

19. Relations on the Irreducible Characters Notation. x; is the character of the irreducible representation p; of degree n;, i = I, 2, ... , s:

cl' c2' ... ,c. denote the classes of conjugate elements of G

cl is the class consisting of only the identity

I

E

G.

Ci is the sum of the elements of the class C; .

x:

xi(g,), g, being a fixed representative of the class c, (xi is constant on C1).

=

h; is the number of elements in the class C; .

Since Ci

E

Z, the center of A( G), we have as in (18.2):

JfEF,

1 is the basis of C(p 1) over F.

Thinking in terms of the matrix representations and taking traces on both sides, (19.1) gives (19.2) n1

=

the degree of Pt , a positive integer. But

xt(Ci) = xt (~g) = h;xl' geCi

so that (19.3) Now since the C' are a basis of Z (Lemma (19.4)

CiCi

=

if~'Ck, k~l

1~.4)

68

Ill. REPRESENTATION THEORY OF FINITE GROUPS

Actually f ~; = !fc1 • I, I the identity ofF and z~i integer. This is because

a nonnegative

=

and when the group elements which are the terms of the final sum are collected to form the classes, only integer coefficients can occur. Applyingp 1 to (19.4) we get

..

Pt(CiCJ) = Pt(Ci) · Pt(C1) = ""fJ~ip1 (Ck) k-1

and using ( 19 .I)

// ·j{ · I =

(±t:J~) I

=

k-1

so that (19.5)

t/tt =

(± L!J~)

I

k-1

t z:r~. k-1

Now substitute from (19.3) to get (19.6)

(h;h}nDx~x;

R

=

~ l~i(hkfnt)Xt . k-1

If we multiply (19.6) by n~ and sum over t results

k

=

I, 2, ... , s there

(19.7)

:E:_

However, we saw (16.10) that x1(gk) = xP(gk), xP being the 1 n1 character of the regular representation p. But then by (I5.I) xP(gk) = 0 unless gk = g1 = I E Gin which case xP(g1) = n, the order of G. Thus, since h1 = I we get from (19.7) (19.8)

19. RELATIONS ON IRREDUCIBLE CHARACTERS

69

It remains to compute l~;. This is the number of times the identity of G occurs in the product CiCi. Since the inverses of the elements of a class form a class, called the inverse class,

z:j

= o,

if C; and C; are not inverse classes

= h; = h;,

if C; and C; are inverse classes.

Introduce the symbol oij-1 = I, 0 according as C; and C; are, or are not, inverse classes. Then we can write for (19.8)

t xlx~

(19.9)

=

(njh;)o;;-1.

1~1

This is the first orthogonal relation on the irreducible characters of a group. Consider the s X s matrix !f = (x)) formed by the values of the s characters xi on the s classes C;:

(19.10)

Let

!f'

be the transpose matrix of

P

=

!f'!f

=

!f.

Using (19.9) we calculate

(P;;);

p;;, the element at the intersection of the ith row andjth column, will be (nj h;)o;;-• . Now put Pab = (h 0 jn)oab-• and form the matrix P = (fiab)· Now

t

j~1

P;;P;k =

t

(njh;)ou-•(hjjn)o;k-• = o;k

j~1

because all summands are zero unless C; is the inverse class of both C; and Ck; that is, unless i = k, in which case there is only the term 1. Thus PP = I so that P = p-1. Then P- 1 = !f-1!f'-1

70

Ill. REPRESENTATION THEORY OF FINITE GROUPS

if/ P- 11/1 = I. Equating the elements at the intersection of the ith row and jth column on both sides we have

or

or {1/n)

:± x~hal>ab-1X~ =

(1/n)

±.hax~x~- 1 =

Si;.

a=l

a,b~i

If we use ia , the fixed representative of the class Ca we can write (19.1 \)

(I /n)

:±

ha/(ia)x1(g;;1) = Sii.

a= I

Since there are hn elements in the class Ca and since same on each, (19.11) can be written (19.12)

(1/n) ~ xi(g)xi(g-1)

=

l

is the

l>u.

yEG

This is the second orthogonal relation on the irreducible characters. By means if (19.12) we prove: (19.13) Lemma. Two representations a and T of a group G are equivalent if and only if the corresponding characters are equal.

Proof. If a and T are equivalent we have already seen (Section 2) that xo = x'. Conversely, assume that these characters are equal. If each irreducible character p; is contained in a with multiplicity a; and in T with multiplicity b1 , then R

XT

= ~b;X1 • i=1

But then (1/n)

~xa{g)x~·(g-1) = gEG

t

(1/n)

i=1

~a;xi{g)x"(g-1) gEG

= ±a1S1k = ak= (1/n) j-1

~xT{g)xk(g-1)=bA-. geG

20. MODULE OF CHARACTERS OVER INTEGERS

71

Thus a and T contain the same irreducible representations with the same multiplicity and so, because of their complete reducibility, they are equivalent.

20. The Module of Characters over the Integers If M and N are two F-R modules we can form their direct sum M C:B N. This is the set S of pairs (m, n), mE M, n EN, made into an F-R module by the rules (m, n)

+ (m', n')

=

(m

+ m', n + n')

f(m, n)

=

(fm,jn),

fEF,

(m, n)r

=

(mr, nr),

rER.

S contains theF-R submodules M' = {(m, 0)}, N' = {(0, n)} and it is clear that M ~ M', N ~ N'. A basis for Sis: (m 1 , 0), (m2 0), ... , (ma , 0), (0, n1), ••• , (0, nb) where the first a elements are a basis of M' and the last b a basis of N'. As a representation module S provides a representation a and if IL and v are the representations afforded by M and N we write a = IL C:B v. If M(r) and N(r) are the matrices for IL(r) and v(r),

M(r) a(r) ___. ( 0

0 ) N(r)

and so

This shows that any linear combination of characters with nonnegative integer coefficients is the character of some representation. Now let '}( = p::;_ 1 a1xi : a1 arbitrary rational integers}. The xi, j = I, ... , s are the irreducible characters of the group G. (20.1) Definition. ~(is the module of generalized characters over the integers with a basis of irreducible characters. Since every character is the sum of irreducible characters (4.14) we see that '}( contains all characters. Let us define on ~ a symmetric bilinear form (x, ,P).

72

Ill. REPRESENTATION THEORY OF FINITE GROUPS

(20.3)

(x, ,P)

(1/n) ~ x(g),P(g- 1),

=

n the order of G.

gEG

Because of (19.12) s

s

s

i,i=l

i=l

(x, ,P) = (l/tz) ~ ~ a;h1xi(g)x1(g- 1) = ~ a;h/>;; = ~ a1b1 • i,i=l geG

Note that (xi, x;) = l>;; so that the irreducible characters are an orthonormal basis of '2£. Since (x, x) = 1 t1;, (x, x) = I if and only if 3i such that a~ = I and a; = 0 for j -=1= i. But then either X = Xi or X = -xi. This gives the following criterion:

1::_

x E '2I

is an irreducible character if and only if (x, x) x(l)

>

=

I and

o.

Hence, given a number of characters, we can form arbitrary integral combinations and check for irreducible characters.

Example.

The symmetric group S 3 is generated by a, b with the defining relations a 2 = b3 = (ab) 2 = 1. We can see (final example of Section I) that a character x is x(I) = 3, x(a) = x(ab) = x(ba) = I, x(h) = x(h 2) = 0. Now x 1 being the onecharacter, x1(g) = 1 Vg E G. We can form ,p = mx + nx' and impose the conditions (,P, ,P) = 1, ,P(l) > 0 to find that 1/J = X - x1 is an irreducible character of degree 2.

Remark. Note that if x"" is an irreducible character and ,P = :Ea. aa.x"" is any character, then (,P, x"") = tla. is the multiplicity with which xa. occurs in ,P. Moreover two irreducible characters x"" and xP are equal if and only if (x"', xP) = 1. In the next section a multiplication of characters is defined which makes the module '2I into a ring of characters ooer the integers. 21. The Kronecker Product of Two Representations Let p. and v be two representations of a ring R afforded by the F-R modules M and N, respectively. We are going to construct an

21. THE KRONECKER PRODUCT

73

F-R module M

X N which will be the representation module for a representation denoted by p. X v and called the Kronecker product of p. by v. Let m1 , ... , m. and n1 , ... , n 1 , be a basis of M and N, respectively. Denote by M X N an arbitrary F-module with a basis of st elements labeled (m;, n;), i = 1, 2, ... , s, j = 1, 2, ... , t. Suppose that for r E R, m;r = L.~_ 1 ffm 1 and n;r = L.~_ 1 j;knk. Then M X N can be made into anF-R module by defining the effect of ron each basis element (m;, n;) as follows:

The effect of r is then extended linearly to a general element u EM X N. The relations

(1)

(fu)r

=

f(ur),

jEF,

(2)

u(rr')

=

(ur)r',

r, r' ER,

are readily checked. The relation (2) means, for instance,

[p.

X

v](rr ')= [p.

X

v](r) · [p.

X

v](r')

so that p. X v is a representation of R. Let us arrange the basis elements {(m;, n;)} in t blocks of s elements each:

Using this ordered "basis we wish to find the matrix for [p. Now

(m~, n;)[p.

s

X

v](r) = (m., n;)r = (m.r, n;r) = (I,f~m13 , /3-1

X f

v](r).

I, g~nk) k-1

74

Ill. REPRESENTATION THEORY OF FINITE GROUPS

Viewing the st X st matrix [I£ X v](r) as a t X t matrix of blocks of s X s matrices, the relation above shows that we have the element J'/,_g~ E Fat the intersection of the cxth row and Pth column of the block occupying the jth row and the kth column of blocks. Keepingj and k fixed we have the matrix (f~)gJ as the j, kth block matrix. Since 1-'-(r) = (/~) and v(r) = (g~), the rule for writing down the matrix [I£ X v](r) can be stated: replace each entry g~ in v(r) by the matrix IL(r) with each of its elements multiplied by For the character x"x" of 1£ X v we get

g:.

t

t

.•

x x. (r) = ~ (~/:) .(; i~l

trace /L(r)

=

cx~l

~g~ = trace 1-'-(r) trace v(r) i~l

= x"(r)x"(r). This shows that the product of two characters is again a character. Since every character is the sum of irreducible characters,

(21.1) where e:,_•;? 0 are rational integers. It is clear that M X N and N X M are operator isomorphic under the correspondence (m;, n;) ~ (n;, m;). Similarly (MxN) X K"' M x (NxK), under ((m;, n;), k 1 )~ (m;, (n;, kz)). Hence p. X v ,...,_, v X 1£ and (I£ X v) X K'""p. X (v X K). The same results follow, by Lemma 19.13, from the equality of the characters:

Finally it is easy to verify the following rules: 1£ X (v 1

8j v2)

(p. 1 8j 1-'-2)

X

,....,

1£ X v1

v ,....., 1£1

X

8j 1£

X v2 ,

v 8j 1£2 X v,

where 8j denotes the direct sum as usual.

75

21. THE KRONECKER PRODUCT

EXERCISES I. If xis a character show that X• the conjugate complex of x. is also a character. 2. Let xa. and xP be irreducible characters of a group. Show that x"xfl does not contain the one-character, unless xa. = x{J in which case it contains it exactly once. 3. Can a nonabelian group have all its irreducible characters of degree one?

If xa.. xfl, xr are irreducible characters of a group show that xa.:XfJ contains xr with the same multiplicity with which xP occurs

4.

inxa.xr;

Remark. Let a be a representation of a group over the complex field and let x be its character. We will frequently need the result (21.2) Since x{g) is the sum of the characteristic roots of the matrix a(g) the result follows directly from matrix theory. On the other hand, it can be deduced from our results as follows: if a(g) denotes the matrix obtained by taking the conjugate complex of the entries of the matrix a(g), it is clear that a(g) = a(g) provides a representation with character X· Moreover :X is irreducible if and only if xis irreducible. Again a(g) = ((a(g))- 1)' {that is, taking the inverse transpose of the matrix a(g)) provides a representation with character such that x(g) = x(g- 1 ), and is irreducible if and only if x is irreducible. Now let x be irreducible and consider {:X, x). which is zero unless =X· But

x

x

x

(:X. x) = (1/n) k:X(g)x(g- 1) = (1/n) :_tx{g)x(g) yeG

x

>0

geG

so that =X· Hence x(g) = x(g) = x{g- 1) and our result is proved for irreducible characters. Since every character is the sum of irreducible characters the general case follows easily.

76

n.

Ill. REPRESENTATION THEORY OF FINITE GROUPS

Linear Characters

The degree of a character is defined as the degree of the corresponding representation. Characters of degree one are called linear. Since a representation of degree one is a number it coincides with its character. Thus a linear character is a representation whose values, as field elements, commute: x(ab) = x(a)x(b) = x(b)x(a). How many linear characters does a finite group G have? If G is abelian all its representations are of degree one and so all its characters are linear. Moreover, by Corollary 18.3 their number equals the order of the group. In the general case the answer depends on the commutator subgroup of G. (22.1) Definition. An element a- 1b- 1ab, where a and b are any elements of G is called a commutator. The subgroup G' generated by all the commutators of G is called the commutator subgroup. Since (a-1b- 1ab)- 1 = b- 1a- 1ba, the inverse of a commutator is a commutator. Also if t E G,

so that the transform of a commutator is a commutator. This shows that G' is a normal subgroup of G. Finally GfG' is abelian, since a- 1b- 1ab E G' means ab = ba(a- 1b- 1ab) == ba mod G'. Note that if G is abelian G' = I. We can now state: (22.2) Lemma. If G is a group and G' its commutator subgroup then the linear characters of G and GfG' are in one-one correspondence, and the corresponding characters have the same set of values. The number of linear characters of GfG', then also of G, equals G: G', the index of G' in G.

Proof.

Let

X be

x(g) Since

a linear character of GfG'. Define

= xL~J>

x on

L!J the class of g modulo G'.

G as

23. INDUCED REPRESENTATIONS AND CHARACTERS

77

therefore x is a linear character on G. Conversely if x is a linear character on G, define X on G/G' by (*)· To see that X is well defined we need only check that if g E G', then x(g) = 1, and for this it suffices to have g a commutator, g = a- 1b- 1ab. But x(a-lb- 1ab) = (x(a))- 1(x(b))- 1x(a)x(b) = 1. Thus X• X are mutually defined by ( *) and we have that the linear characters of G and GjG' are in one-one correspondence and that corresponding characters have the same set of values. However, GjG' is abelian so that (Corollary 18.3) all its characters are linear and their number equals the order of GfG' = G: G'. EXERCISES

1. Show that the symmetric group Sn , consisting of all permutations on n symbols, has exactly two linear characters. 2. Let F be a finite field of characteristic p having q = pn elements. The set of all nonsingular 2 X 2 matrices over F form a group denoted by GL(2, q). Show that this group has q - 1 linear characters over the complex field. [Note: the order of GL(2, q) is q(q - l)(q2 - 1). See Dickson [11], p. 77.] 23. Induced Representations and Induced Characters

Let G be a group and H a subgroup of G. If a is a representation of G then a restricted to H clearly gives a representation of H. We consider now the opposite question: given a representation a of H can a representation a* of G be constructed? For convenience we assume that a is irreducible. It will be seen that the final formula is independent of this assumption. We proceed as follows: Let A(H) be the group algebra of H over the field of representation F. Then A(H) is a subalgebra of A( G) the group algebra of G. Let the minimal right ideal ~ of A(H) be the representation module for a. Suppose that a 1 , a 2 , ••• , ar is a basis of~ over F. Form the rs elements (23.1)

i = 1, ... , r; j = 1, ... , s

78

Ill. REPRESENTATION THEORY OF FINITE GROUPS

where t 1 = 1, t 2 decomposition (23.2)

, ... ,

G

=

t 8 are right coset representatives m the

Ht 1

+ Ht 2 + ··· + Ht • .

Now the elements (23.1) are linearly independent over F, for if A;; E F,

then each summand (~:::~ 1 A;;a;)t; = 0, since the terms in the different summands belong to different cosets and so cannot cancel. But then

Therefore A;; = 0, since the a; are linearly independent. We assert that the set (23.1) is the basis of a right ideal ~* in A( G). It suffices to show that for any g E G, a;t;g is a linear combination of the elements of the set. Now from (23.2), t;g = htk for some unique h E Hand some unique tk so that r

(23.3)

a;t;g = a;htk =

(~t:;a;) tk i~l

r

=

kf~a;lg, ;~1

Thus ~* is a right ideal of A(G). It is then a representation module for a representation a* of G. If we write the set (23.1) ins blocks: a 1 t 1 , ... , art 1 ; a 1 t 2 , ... , art2 ; ... ; a 1 t 8 , ... , arts we see from (23.3), with i = 1, ... , r, that the rs X rs matrix a(g) can be viewed as an s X s matrix with entries which are r X r matrices and that at the intersection of the jth row and kth column of a*(g) there stands the r X r matrix a(h) = {f~i). Other entries along the jth row of a*(g) are then just the r X r zero matrix. Recalling the connection between g, h, j, k, viz., t; gt; 1 = h, we see that the matrix at the intersection of the jth row and lth column of a*(g) is the r X r zero

23. INDUCED REPRESENTATIONS AND CHARACTERS

79

matrix, if t;g ¢ Ht 1 but is a(t;gt-1 ) if t;gt! 1 E H. In summary we have

kth column (23.4) where

a'(t;gt//)

if

a(t;gfi/),

=

= 0, the r

X

r zero matrix, otherwise.

If we adopt (23.4) as the definition of a*(g) irrespective of whether a is irreducible or not, it is a simple matter of matrix multiplication to show that a*(g~2 ) = a*(g 1 )a*(g2 ). Thus a*(g) is a representation of G if a is any representation of H. Let x and x* be the characters of a and a*, respectively. By (23.4) 8

x*(g) = ~trace a'(t;gtj1 ) =

!- x'(t;gtj •

1)

i=1

;~1

where x'(x) = x(x) or 0 according as x ht1gtj1h- 1 E H ~ tgtj 1 E Hand since

E

H or x ¢H. Since

we have s

x*(g) = (I/H: I) ~ ~ x'(ht1gt; 1h- 1 ), 1~1

Because every element x

E

= order of H.

G is realized uniquely as an ht; , we get

(23.5) x*(g) = (I/H: I)~ x'(xgx- 1 ), xeG

H :I

hEH

and

x'(x) = x(x),

=0,

X E H, X¢ H.

This formula is independent of the choice of ~oset representatives t1 • Since a representation is determined by its character we see that a* as defined in (23.4) does not depend (up to equivalence) on the choice of the coset representatives of H in G. The result

80

Ill. REPRESENTATION THEORY OF FINITE GROUPS

(23.5) can be put in a different form. As x runs through the elements of G there will be repetitions: in fact, for each particular x 0 , x 0gx0 1 will occur 9l{g) : I times, that is, as often as the order of the normalizer 9l(g) ofg [if bE 9l{g): (xoh)g(xoh)- 1 = x 0 (bgb- 1 )x0 = x~x0 1, and conversely, if the first and last expressions are equal b permutes with g and thus bE 9l(g)]. We now have x*(g)

= (~(g) : I /H: I)~ x(g')

where the sum is over elements g' similar tog but in H. If hg denote the number of elements in the conjugate class of g then h9 = G: l/9l(g): I, so that if m = G: I/H: 1 is the index of H in G, we can write (23.6)

x*{g) = (mfho) ~ x{g'),

C0

= class of elements conjugate tog

Y'EHf\C 9

(23.7) Theorem (Frobenius reciprocity theorem). Let p and a be absolutely irredudble representations of a group G and a subgroup H, respectively. Then a*, the representation induced in G by a, contains p with the same multiplicity with which a occurs in p* (i.e., p restricted to H).

Proof. Let us recall (Remark, Section 20) that if xis any character and xa. an irreducible character of a group G, then a

= {1/G: I)~ x(g)xa.{g- 1 ) gEG

gives the multiplicity with which xa. occurs in X· Denote by x*, x the characters of a* and a, respectively, and let .p be the character of p. Let a be the multiplicity If pin a* and let b be that of a in p* . Then a= (1/G: I) ~x*(g).p{g- 1 ), 'i'l1oV

and using (23.5) a= (IJG:

I)~ ((1/H: I) ~x'(xgx- 1 ))1/J(g- 1 ) YEG

XEG

23. INDUCED REPRESENTATIONS AND CHARACTERS

Since ifl(xg- 1x- 1) =

81

.p(g-1),

a= (I/H: l)(l/G: I)~

k x'(xgx-1)1/J(xgi-1).

xEG YcG

For each fixed x the inner sum runs through all elements of G and since x'(y) = 0 if y ¢= H we have a= (1/G: I)

k ((l!H: I) k x(h)!/J(h-

xEG

1 )).

hEH

Here the inner sum gives the multiplicity b of the irreducible character x in .p restricted to H. Thus a= (1/G: I)

k b = (1/G: l)b(G: I) =

b

XEG

and the theorem is proved•

Remarks.

There are a great number of interesting results on induced characters in the literature. (See, for example Curtis and Reiner [10], Chapter VI and references there.) We shall have a few more results later in this book but reasons of space prevent any extended treatment. Since it is a simple matter to find the characters of a cyclic group it would be desirable if, for example, every character of a group G could be obtained as a linear combination with integral coefficients of characters induced by cyclic subgroups. This is not possible but the strongest result in this direction is the following theorem of Brauer.

Brauer's Theorem on Induced Characters. Every character of G is a rational integral combination of characters induced from linear characters of elementary subgroups of G. A linear character is the character of a representation of degree one, and an elementary subgroup of G is one which is a direct product of a cyclic group and a p-group for some prime p. If a group G has a pair of subgroups R and C with linear characters 8, , respectively, such that the induced characters 8*, * of G have exactly one irreducible character in common,

82

Ill. REPRESENTATION THEORY OF FINITE GROUPS

each with multiplicity one, then this irreducible character can be calculated using the formula

x(g)

=

(nf[i(Rn C: 1)]) ~ fJ(r)c{J(c)

where i is the number of elements in Cu , the conjugate class of g, n is the degree of the irreducible representation, and the summation is over all r E R, c E C for which rc E Cy [8]. For the symmetric group Sn every irreducible character is accessable in this way from a pair of subgroups [see Section 28]. It is this result which underlies the method of the Young tableau: the permuted symbols 1, 2, ... , n when written in rows of nonincreasing length form a shape, or tableau. If we now take for R the subgroup of Sn permuting only the symbols within the rows and for C the subgroup permuting only those within the columns then this pair, with (J = 1 on R and cP = ± 1 on the even or odd permutations of C, yield an irreducible character of S'n . Each shape yields such a pair and there are as many shapes as irreducible characters. For example, for S 3 there are the tableaux

I~ ~I with the pairs R = S 3 , C = I; R = {1, (12)}, C = {I, (13)}, and R = I, C = S 3 , respectively. A tableau is called a standard tableau if the numbers in each row and each column increase in magnitude. It can be shown that the degree of the irreducible character associated with a given shape is equal to the number of standard tableau of that shape (Rutherford [20], p. 68). In the example given above all three shapes are standard tableaux but the second shape has also the standard tableau

fl3l ~-· Hence it is associated with an irreducible character of degree 2.

EXERCISES

83

These results make routine the construction of the character table of a symmetric group.

EXERCISES I.

Construct the character table of the symmetric group S 4

•

2. Let G be a group and H a subgroup. Let x be any character of G and ,P any character of H. Prove that (see Definition 20.2)

where XH = x I H and ,PG is the character of G induced by ,P. [When 1/J, x are irreducible this is Frobenius theorem.] 3. Let G-:JK-:JH. Prove that (1/JK)G=,PG, if ,Pis a character of H. [Hint: show that both sides contain an irreducible character x equally often. Use Exercise 2.] 4. Let N be a normal subgroup of G and let ,P be an irreducible character of N. Let T = {g E G : ,p(g-1ng) = 1/J(n), V n EN}. (We say that T fixes ,P or that ,P is invariant under T.) Show that T is a group. Prove that ,pG I N contains ,P with multiplicity T : N. Show that if T = N, then ,PG is irreducible. 5.

Can the group ring of a nontrivial group be simple?

6. Let E be a finite extension field of degree d over F. Show that E is isomorphic to a matrix algebra A over F and that every automorphism of E which leaves F fixed corresponds to a transformation of A by a matrix T. [Hint: A is simple and hence has only one representation up to equivalence.] 7. Let G be a group and let M and N beF-A(G) modules which afford representations p. and v, respectively. Let be any linear mapping (F-mapping) of M into N. Then p.(g)1TV(g- 1) is a linear mapping of M into N. Prove that £X = :EaeG p.(g)1TV(g- 1) is an operator homomorphism (F-A(G) mapping) of MintoN. 8.

Deduce from Exercise 7 that if p.(x) = (p.;;(x)) and

84

Ill. REPRESENTATION THEORY OF FINITE GROUPS

v(x) = (v;k(x)) are nonequivalent irreducible matrix representations then I-1-'-ii(x)vrs(x-1)

=

0,

XEG

I. /k;;(x)lkrs(x- 1) =

nS;_/);r/d,

XEG

where n denotes the order of the group G, dis the degree of the representation ~-'-• and Sab are Kronecker deltas. [This result is also known as Schur's lemma. Hint: take 1T = 1T;; where for each basis element mk EM, mk1T;; = m;Ski and use (11.7) and

(ll.9).] 9. Deduce the character relation (19.12) from the results in Exercise 8. 10. Let E = 1/G : l I:.,.G xx(x- 1 ), where xis a character of the finite group G. Let A(G) denote the group algebra of G over the complex field F. Show that E is in the center of A(G). If xis irreducible prove that E 2 = E and that E is the identity of the minimal two-sided ideal containing the minimal right ideal of A(G), which provides the representation with character X·

CHAPTER

IV

Applications of the of Characters

Theory

24. Algebraic Numbers

Any root x = a of an equation, (24.1)

xn + a^-1

+ - + ««-i* + an = 0,

in which the a{ are rational numbers is called an algebraic number. If the coefficients at are rational integers the algebraic number is called an algebraic integer. We shall need a few facts concerning algebraic numbers and algebraic integers. These will be proved in the lemmas of this section. (24.2) Lemma. An algebraic integer which is a rational number is a rational integer. Proof. Let a = r/sy r, s rational integers with no common divisor. Suppose that a satisfies (24.1) in which a t are rational integers. Then r*

= — (a^-h + ··· + ansn).

Since s is a factor on the right side of this equation any prime p dividing s divides rn and hence r. But r and s are relatively prime and hence s = 1, and a is a rational integer as required. 85

86

IV. APPLICATIONS OF THE THEORY OF CHARACTERS

(24.3) Lemma. Let α, β, ...,λ be algebraic numbers (integers). A polynomial f(a, ßy ...,λ) with coefficients which are rational numbers {integers) is itself an algebraic number (integer). Proof. Let a be a root of A(x) = 0 of degree a, β a root of B(x) = 0 of degree by ..., and λ a root of L(x) = 0 of degree / where the coefficients of all equations are rational numbers (integers) and the leading coefficient is unity. If N = ab - · · / form the N numbers

0 < a! < a -

1,

0 < V < b-

1,..., 0 < / ' < / - 1,

which, arranged in some fixed order, we denote by ωχ, ω2 , ..., ωΝ. Using A(OL) = 0, αα and higher powers of a can be expressed as polynomials in a of degree "'(1),

cfo"'(y)

=

x¢

U H;,

1 =/= y E U Hi , y'EH, y' conjugate toy.

t/P(y'),

We will now show that cfo"' is an irreducible character of G. Already c/>"'(1) > 0, and (cfo"',cfo"')

= 1/G: 1 ~'/J"'(g)cfo"'(g) geG

= 1/G: 1 [Cc/>"'(1)) 2

+ ~ c/>"'(x)cP"'(x) + ~ x~UH,

cP"'(Y)cP"'(y)]

l,PyeUH;

= 1/G: 1 [(c/>"'(1)) 2 + (cP"'(1)) 2 (n-l)

+n ~

.fo"'(y').P"'(y')].

l,PY'EH

Now because .P"' is an irreducible character on H we have (1/H: 1) ~u·eH .fo"'(y').fo"'(y') = 1 so that the last summation in the bracket yields H: 1 - .fo"'(l).fo"'(I) = H: I - (cP"'(1)) 2• (cP"'•cP"')

+ (cP"'(l)) 2(n- I) + n(H: 1 - (cP"'(I))2)]

=

(1/G: l)[(cP"'(1)) 2

=

(1/G: 1)n(H: 1)

=

Thus cP"' is an irreducible character of G.

1.

98

IV. APPLICATIONS OF THE THEORY OF CHARACTERS

Since ~a.(x) = ~a.(l), x E U H; we have by Lemma 25.8 that G - U H; belongs to the kernel of Ta. , the representation whose character is cfoa.. Now this is true for each Ta. , corresponding to each irreducible character .pa. on H. On the other hand, if 1 -=1= y E U H; we cannot have cfoa.(y) = ~a.(l) for all Ta., for then by (27.5) 3y' E H, y' -=1= I, such that .pa.(y') = .pa.(l), for every ex. But then we should have the element I - y' of the group algebra A(H) of H mapped onto zero by every irreducible representation ex of H. Thus I - y' -=1= 0 belongs to the radical of A(H). As A(H) is semisimple this is impossible. Hence (G H;) I is equal to the intersection of the kernels of theTa. and as such is a normal subgroup of G. This proves the theorem.

U

U

EXERCISES I.

Let x be a character of the group G such that x(g) = 0, x is reducible. If G is a p-group and x is an irredcible character of G show that the set of all elements x for which x(x) = 0 is a normal subgroup of G.

Vg -=1= I. Prove that

2. Show that a p-group has a linear character whose kernel is a subgroup N of index p. Prove that xI N (x restricted to N) is reducible. 3. Use the relation (16.7) and Theorem 25.3 to show that a group of order p 2 is abelian; also that a group of order p" has a normal subgroup of order ~ pn-1. 4. If xis an irreducible character of a group G, not of degree one, show that x(g) = 0 for some element g E G. (Burnside [7], p. 319, Ex. 5.)

CHAPTER V

The Construction of Irreducible Representations 28. Primitive ldempotents Let A be the group algebra of a finite group Gover the complex field F. We have seen from the structure theorems (Section 10) that there.~is, up to arrangement of summands, exactly one decomposition

in which the simple rings All are mutually annihilating simple two-sided ideals of A. Moreover, each Aa., called a block, is the direct sum of nil isomorphic right ideals of A and is besides isomorphic to a complete matrix algebra of degree na. over F. Thus there are n! elements ejk E A,. , j, k = I, ... , na. , satisfying

(28.1)

o;k

the Kronecker delta,

and such that any a E Aa. can be written uniquely as n,

(28.2)

a= ~ /;;eii,

/;;EF.

i,j=l

Furthermore, since the Aa. are mutually annihilating

(28.3)

ex -=/:=

99

{3,

100

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

it is easy to see from the relations (28.1) and (28.3) that 1 the identity of A,

and that the inner sum Erx is the unique identity element of A" . Now Rrx; = efjA, j = 1, 2, ... , na., are right ideals of A and we have

By the uniqueness of the Remak decomposition the Rrx; , j = 1, ... , na., must be isomorphic minimal right ideals. Any one of them, say Rrx 1 , is a representation module for an irreducible representation a" . It can be seen from the relations (28.1) and (28.3) that ef1 , ... , e1~ is an F-basis of Rrx 1 . An element e of an algebra is an idempotent if e2 = e. It is a primitive idempotent if it cannot be written as e = e1 + e2 where ei = e1 , e~ = e2 , and e1e2 = e2e1 = 0. If e is not primitive then the right ideal eA = e1 A ffi e2 A, so that eA is not a minimal right ideal. Conversely, suppose that eA is not minimal. Then the representation module eA is reducible and, since A is semisimple, it is decomposable (Theorem 9 .3) so that eA = eA 1 ffi eA 2 • Hence: e

=

ea

+ ea',

ea

eA 1

E

,

ea'

E

eA 2

•

Now e = e2 = eae

+ ea' e =

ea

+ ea'

which implies that eae = ea,

ea'e

=

ea'.

Also ea = ea2

+ ea'a,

so that

ea'a = 0.

28. PRIMITIVE IDEMPOTENTS

101

Thus ea = eaa = (eae)a = (ea)2

and ea'a

=

(ea'e)a

=

(ea')(ea) = 0. ·

Similarly (ea') 2 = ea'

and

(ea)(ea') = 0.

This shows that e is not primitive. We state this result as: (28.4) Lemma. primitive

An idempotent e of a semisimple algebra A ts is minimal.

if and only if the right ideal eA

This result shows that the elements el; , i = I, 2, ... , n" , are primitive idempotents of the ath block and any one of them, say ef1 , generates a minimal right ideal 1 A = Rrx 1 which is a representation module for the irreducible representation a" . The problem of constructing irreducible representations can thus be reduced to that of finding a primitive idempotent for each block. It has been pointed out (Remarks, Section 23) that for the symmetric group Sn each Young tableau is associated with an irreducible representation. Actually each shape gives a primitive idempotent

er

(28.5)

e = (nafn!)

k (c)rc

where n" is the degree of the corresponding irreducible representation, the summation is over all elements r E R, c E C, and (c) = ±I according to whether c is an even or odd permutation. This follows from a combinatorial lemma of von Neumann [van der Waerden [22], Vol. 2, p. 191], the precise generalization of which can be stated for an arbitrary group as: (28.6) Lemma. Let two subgroups R and C of a group G have representations (} and , respectively, both of the first degree. Suppose that the following condition holds \fs E G: s ¢'. CR tj and only tj

102

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

3r E R, c E C, such that srs- 1 = c and 8(r) -=/:= (c). Then e = PN is a multiple of a primitive idempotent, where

Proof.

P = ~ r8(r),

N = ~c(c).

reR

ceC

First note that

Pr'B(r')

=

~ r8(r)r'8(r')

=

reR

~ rr'B(rr')

=

P

reR

where r' is any element of R. Similarly Nc'(c') = N. Consider PNsPN. If s E CR, s = cr say, then

PNsPN

=

PNcrPN

=

8- 1(r)- 1(c)P(Nc¢(c))(r8(r)P)N

=

B- 1{r)- 1{c)(PN) 2 •

On the other hand, if s ¢= CR, then by the condition of the lemma there is an r, c such that srs- 1 = c and B(r) -=/:= (c). Then

PNsPN

=

8(r)PNsrs- 1sPN

=

B(r)PNcsPN = (r)- 1(c)PNsPN.

Hence

PNsPN(I - 8(r)- 1c) Since B{r) (28.7)

-=/:=

(c) we have PNsPN eAe

=

=

=

0.

0. Writing e = PN we get

Fe 2

where A is the group algebra of G over F the complex field. Note that e -=/:= 0, for the coefficient of identity I in PN is f = ~ B(r)(c) where the summation is over all r, c for which rc = I, that is, over all r E R 11 C. But Iri- 1 = c- 1 and since IE CR the condition of the lemma gives B(r) = e2 = 0, a contradiction. Thus e is a primitive idempotent. This completes the proof. Now let G be a group satisfying the conditions' of Lemma 28.6. Then e = j'PN = f' l:reR.cec fJ(r)c{J(c)rc is a primitive idempotent where f' is some element of the field F. Let us find an expression for f'. We know that the minimal right ideal eA must he isomorphic to some right ideal 1 A. This implies that there is a unit u E A such that u- 1eu = e11 [see Appendix]. Then if xP denotes the character of the regular representation, xP(e) ='~ xP(u- 1eu) = xP(erl)· Now as a basis for the regular representation of A we can take the elements{ef;: i,j = I, ... , na.; ex= 1, ... , s}. Since ef/~ 1 = Sa.13S;kei1 we see that xP, the trace of the regular representation, is given by xP(e~z) = na. , 0 according as k = I or not. Thus xP(e) = xP(er1 ) = na.. But xP(e) = xP{f'PN) = f'xP(PN) = f'f(G: 1) where f is the coefficient of the identity of the group in PN. In the proof of Lemma 28.6 this was found to be f = R n C: I. Hence f' = na./[(R n C : l)(G : 1)], so that

er

(28.10)

e = (na/[(Rn C: l)(G: l)J)

k

fJ(r)c{J(c)rc.

reR,ceC

Note that since

e~ =

e, this expression allows us to calculate

na, , if all other terms are known. Let us return to the Young tableau [for its definition see Remarks, Section 23]. We will show that the conditions of Lemma 28.6 will be fulfilled when 8 = I on Rand c{J(c) = ±I according as c E Cis an even or odd permutation. It will then follow immediately that the element e of (28.5) is a primitive idempotent.

104

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

First note that if a permutation is written concisely as t =

p on the symbols I, 2, ... ,

n

I, 2, ... , n,

where ip is the symbol replacing i in the permutation p, then

sps-1

=

) ( is-1 ips-1 .

This means that if p is written as the product of disjoint cycles: p = ···(ii2 ··· ik) ···, then sps-1 = ··· (i1s-1i2s- 1 ··· iks- 1) ···. Since R permutes only symbols in the rows, the symbols in any cycle of an element r E R are from a single row of the tableau. Now if sRs-1 11 C -=1=- I, then s ¢'. CR: thus if, say, s = c1r1 and c1r1"1 1c;_- 1 = c -=I= 1 then r1rr1 1 = c;_- 1cc1; but this is impossible since any cycle on the left side involves symbols from a single row whereas any on the right involve symbols from a single column. Similarly it can be shown that the converse holds. Suppose now that sRs-1 11 C -=I= I: let srs-1 = c -=I= I. Let (i1 · · · ik) be one of the disjoint cycles of c. Then there is a disjoint cycle (A ···jk) of r such that As-\ ... ,jks-1 = i 1 , ••. , ik. Hence for the transposition r' = (jd2 ) we have s(j1j 2)s-1 = (ii2 ) = c'. Since c' is an odd permutation c{J(c') = -1; but O(r') = I and so O(r') -=1=- c{J(c'). We have thus established that the conditions of Lemma 28.6 are satisfied when Rand Care the subgroups of any tableau with 8 = 1 on R and cP = -1 on the odd permutations of C. Thus each tableau yields a primitive idempotent e as given by (28.5). Then the minimal right ideal eA is a representation module for an irreducible representation. It will be shown in the next paragraph that the set of different shapes provide a complete system of idempotents from which all the irreducible representations can be constructed. Let 1:1 , 1:2 denote two tableaux of different shapes and let e1 = / 1P 1N 1 , e2 = / 2P 2N 2 be their associated idempotents. Consider P 1 N 1 sP2N 2 , s E Sn . If Vr2 = (ij) E R 2 there is no c1 E C1 such that sr2s- 1 = c1 , then no pair of symbols i,j from any

29. EXAMPLES OF GROUP REPRESENTATIONS

105

row of 1:2 can be taken by s-1 into symbols within a column of 1:1 (otherwise for the transposition (ij) E R 2 we would have s(ij)s-1 = (is-1js-1 ) E C1 ). Now this implies that the rows of 1:1 have the same length as those of 1:2 so that both have the same shape, contrary to assumption. This proves that 3r2 such that sr2s-1 = (is-1js-1 ) = c1 E c1 . Now

= (ij)

P 1N 1sP2N 2 = P 1N 1sr 2s-1sP2N2 = P 1N 1c 1sP2N 2 =

cb- 1(c1)P1N 1sP2N 2 •

q,-1{c1 ))P1N 1sP2N 2 = 0 and since c 1 is a transposition I so that P 1N 1sP2N 2 = 0. This holds Vs E Sn and hence e1Ae2 = 0. Since e1 and e2 are primitive idempotents this is possible only if they belong to different blocks. This shows that the tableaux of different shapes yield primitive idempotents from distinct blocks and are thus associated with distinct irreducible representations. Now there is a one-to-one correspondence between shapes and classes of elements of Sn , which associates with each shape the class of permutations whose disjoint cycles have the lengths of its rows. For example the shape Hence (I -

q,-1(c1 )

-=I=

.' corresponds to all those permutations which, when decomposed into disjoint cycles, consist of one 3-cycle and two 2-cycles. Recalling that two permutations are conjugate if and only if their cycle structure is the same, we see that there are many shapes as classes of conjugate elements. Since there are as many irreducible representations as conjugate classes (Theorem 18.1) the tableaux provide us with a complete system of primitive idempotents from which to construct all irreducible representations. 29. Some Examples of Group Representations

1. Cyclic Groups Let G = (g : gn = 1). If£ is a primitive nth root of unity then a1{g) = £i, j = 1, ... , n gives n distinct representations of G of

106

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

degree one. Since G is abelian of order n this is the complete est of irreducible representations. 2. Abelian Groups

Let G be an abelian group of order n. By the fundamental theorem of abelian groups G is the direct product of cyclic X cl ' and for i = I' ... , l - I' groups: G = cl X c2 X n; I ni+l ' where n; is the order of C; Of course n = n~-1 n; For i = 1, ... , /let c; generate C; and let €; be a primitive n;th root of unity. Then a 11 ••. 11 , jk = I, 2, ... , nk , defined on the generators C; by 0

0

0

0

0

gives all n irreducible representations of G, and all are of degree one. 3. The Symmetric GroupS,.

We have seen in Section 28 that the Young tableaux give a routine method for the construction of all irreducible representations. In practice the work is tedious and for cases where n is small various ad hoc procedures can usually be employed. [See Example l, p. 108.] However, an irreducible representation of degree n - 1 is readily obtainable for we have: (29.1) Lemma. The natural representation of a doubly transitive permutation group is the sum of the }-representation and an irreducible representation.

A permutation group G is doubly transitive if any ordered pair of distinct symbols is taken into an arbitrary ordered pair of distinct symbols by some element of G. A representation p. of G is a natural representation if p.(g) permutes a basis of the corresponding representation module in the same way that g permutes the symbols (not elements) of G.

29. EXAMPLES OF GROUP REPRESENTATIONS

107

Proof. Let M with basis {m1 , ... , mk} be a representation module for the natural representation. Then if i=I, ... ,k,

Let where a;M; stands for the direct sum of a; isomorphic irreducible sub modules of M, and M; ';#_ M; if i -=1= j. Hence by ( 11.5) and (11.9) since Hom(M; , M;) = 0 if i -=1= j:

(29.2)

i(M, M)

=

a~

+ .. · + a~ .

We want to calculate i(M, M). Let definition,

T

E

Hom(M, M); then by VgEG.

Now if m;T

=

k

.

.

"'i:.i~d;m;, f~

~f'm 4-/ i jg

=

EF, then we get

~ " m ~lag j

'

~f'gm ~ ig jg'

=

j

j

the last step since jg runs through the same symbols as j.

Vi,j,g

:J: =!1;.

But by the double transitivity if i

-=1=

j, 3g such that ig

jg = 2.

:. !1 =J:'

i -=1= j,

and !~ =

/!! = !{ '

Vi.

=

I,

108

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

If we now define

T1

and

T2

as the operator homomorphisms

then

Thus Hom(M, M) is of rank 2, or i(M, M) = 2. Equation (29.2) implies that there are only two a; =1=- 0 and that they are both one. Suppose a 1 = a 2 = l. Then M = M 1 8j M 2 , M 1 , M 2 irreducible. Now if m = I:~_ 1 m;, then mg = m, Vg E G, so that Fm = M 1 , say, is an invariant submodule of dimension 1. This proves the lemma.

Example 1. The symmetric group S 4 • The number of conjugate classes is 5 and so there are 5 irreducible representations. The commutator subgroup is A 4 , the alternating group consisting of all even permutations (Section 22, Exercise 1). Thus the number of linear characters is S 4 : A 4 = 2. They are, of course, the !-representation and the representation which is -1 on odd permutations. By Lemma 29.1 there is a representation of degree 3. We proceed to find it. Let {m1 , ••• , m4 } be an ordered basis of the module M which gives the natural representation. Take a new basis: m~ = m1 + m2 + m3 + m4 , m; = m; - m1 , j = 2, 3, 4. Then

m~(l2) = m~, m~(12)

= (m 2

-

m1 )(12) = m1

-

m2 =

m~(12)

=

-

m1 )(12)

-

m2

(m 3

=

m3

-m~

,

=

(m 3

=

m~- m~,

=

m~-

-

m1 )

m;.

-

(m 2

-

m1 )

29. EXAMPLES OF GROUP REPRESENTATIONS

Thus

109

0 00 0) 0

-1 -1 -1 Similarly

I 0 . 0 I

00 0I 0) 0 0 0

1

°

1 0 0 Now 1-1.,....., 1 8j -r by Lemma 29.1 so that we have the irreducible representation •(12)

=

(=~ ~ ~). -1

0

-r(234) =

(00 0I 0)1

0

I 0 0

I

Since (234) and (12) generate s4 (Coxeter [9], p. 7) T can be found on the remaining elements. We have now found 3 representations of degrees n1 = n2 = 1, n3 = 3. Since the order of the group is 24, 24

= 12 + 12 + 32 + n! + n~ .

Thus n4 = 2, n5 = 3. This could also have been seen from the tableaux: the pairs of representations of equal degree correspond to the tableaux obtained by interchanging rows and columns. The unpaired representation of degree 2 arises from the symmetric tableau

1-::1 Clearly if cf> is the representation which is -1 on odd permutations and a is any irreducible representation then r = cf>-r is another irreducible representation. Accordingly

T(l2)

~ (l -~ ~). 0

-1

7(234) = -r(234) is the other irreducible representation of degree 3.

110

V. IRREDUCIBLE REPRESENTATION CONSTRUCTION

Finally we remark that S 4 has the Klein 4-group K 4 as a normal subgroup: K 4 = {I, (12)(34), (13)(24), (14)(23)}. Since S 4 /K4 ~ S 3 , the irreducible representations of S 3 provide irreducible representations of 8 4 [obvious: see also p. 93, part (2) of Lemma 26.1]. Now the irreducible representation of degree 2 derived from the natural representation of S3 gives the representation of degree 2 of 8 4 •

Example 2. The dihedral group Dm . Dm has 2m elements and is the group of rotations of a regular polygon of m sides. It is generated by a, b with the defining relations: a 2 = bm = I, a-1ba = b-1 • The element b corresponds to a rotation of the polygon through an angle of 2TT/m about a perpendicular axis through its center; a corresponds to a 180° rotation about an axis of symmetry in its plane. The commutator group n;,. = {1, b, ... , bm} 1if m is odd and is {1, b2 , ... , bmf2 } if m is even. Thus Dm: n;,. = 2, or 4 and so there are 2 or 4 linear characters according as m is odd or even. In either case the cyclic subgroup N generated by b is of index 2 in Dm and so it is a normal subgroup. Then Dm/ N is cyclic, of order 2, and its two representations of degree one are representations of Dm . If m is odd these are the only irreducible representations of degree one. Case 1 : m = 21 + 1. There are I + 2 conjugate classes: {1}, {abfl: f3 = 1, ... , m}, {bY, b-Y}, y = 1, ... , I. Thus there are I+ 2 irreducible representations of which two are of degree one, and I of degrees n;;:;:, 2, i = 1, ... , I. Hence: 2(2/ + 1) = 2 + 1 n~ ;:;:: 2 + 22 /, showing that all n; = 2. We have therefore for this case two irreducible representation of degree 1 and I of degree 2. Since Dm/ N ~ {I, a} the representations of degree one are obviously: a 1(a) = a 1 (b) = 1; a 2(a) = -1, a 2(b) = 1. The remaining a;(x) are 2 X 2 matrices which satisfy

I:!_

(a;(b))m = (a;(a)) 2 Remarking that

=

/2

and

a;(a)- 1a;(b)a;(a) = a;(b)- 1 •

29. EXAMPLES OF GROUP REPRESENTATIONS

111

is of order 2 and transforms a diagonal matrix by interchanging its elements, we can write down the a;(x) by inspection. Thus: (29.3)

a;(h) =

(

E'

0)

O ci ,

a;(a)

=

(~ ~),

i =I, 2, ... ,I,

where e is a primitive mth root of unity, satisfy the preceding relations and are surely the irreducible representations of degree 2. This can easily be checked by showing that (xa', xa·) = I. Case 2: m = 2/. We have remarked that there are 4 representations of the first degree. The conjugate classes are: {I}, {ab, ab 3 , .•• }, {ab~, ab4 , ..• ,a}, {b"', b-"'}, - 1(c)PN1g- 1 E PN1A.

Thus PN1 , PN2 , denoted by e1 , e2 belong to the same minimal right ideal of dimension 3 over F and, being independent, are elements of a basis. For convenience we shall take the remaining element of the basis to be ea = PN1w(I + c + c2 + c3 + c4 ), w = (14)(25) = a2c4 [Table 1). It is easily checked that thee; are linearly independent: any relation / 1e1 + / 2 e2 + faea = 0, multiplied in succession by N 1 , N 2 yields/1 = 0,/2 = 0, and so also fa= 0. Now

q,-; 1(c)PN1

e1c = PN1c =

e2c

= PN2c = ¢>;1(c)PN 2

so that -l

a 3 (c)

=

0 0

(wo

w

o).

0

0

I

c

=

(14325).

In general if x is any element of A 5 , e;x is a linear combination of e1 , e2 , ea: e;x = /; 1e1 + /; 2e2 + /; 3 e3 , where the/;; are complex numbers. For each i the fil , /; 2 , /;a may be found by solving the three equations obtained by . equating the coefficients of I, c, c2 on both sides of the relation. Then aa(x) = (/;;),a matrix of the third degree. Finally, to obtain the remaining representation a 2 it is only necessary to replace w by w 2· in the matrices of a 3 •

EXERCISES

119

EXERCISES I.

Prove that all irreducible characters of the symmetric group lie in the rational field, and that the representations may be obtained in this field. [Look at the primitive idempotents.)

sn

2. Show .that a doubly transitive permutation group on n symbols cllnnot have fewer than n2 - 2n 2 elements.

+

3. Find all the irreducible representations of the quaternion group G =(a, b: a4 = I, a 2 = b2 , a- 1ba = b-1 ). 4. Show that the subgroups R = A 4 , C = {(12345)k : k = I, 2, ... , 5} of A 5 with their linear characters satisfy the conditions of Lemma 28.6. Classify all the resulting primitive idempotents and give bases for the corresponding irreducible representations. 5. Use the expression (28.5) for the idempotent arising from the tableau

123 ... n- I

I,

1

~

or its transpose, to show that of degree n- I.

sn has an irreducible representation

6. Find ann X n matrix #In which permutes with all elements of a doubly transitive group of permutation matrices. [Cf. T 2 E Hom(M, M) in the proof of Lemma 29.1, for example.)

CHAPTER VI

Modular Representations 30. General Remarks In Chapter III we dealt with certain complex-valued functions on a group, namely, the ordinary characters. Recall that if a is a matrix representation of a group G which assigns to each g E G a matrix a(g) with coefficients in the field of complex numbers, then the ordinary character x" on G is defined by x"(g) = trace a(g). If a is irreducible x" is called an irreducible character. We saw that the irreducible characters on a group satisfy certain character relations and moreover that any character can be expressed uniquely in terms of them. Sections 26 and ·27 showed the importance of the ordinary characters in answering questions about the group. For further insight into the structure of the characters certain other complex-valued functions on a group, called Brauer characters, must be considered. These can be defined by means of the representation of the finite group in a modular field, that is, a field of finite characteristic p. The connection with the ordinary characters is established through the integral representations. For example, suppose that a is a representation of Gin the field P of rational numbers. It will presently be shown in a more general context that a is equivalent to another representation T with the property that Vg E G the matrix T(g) has coefficients in J, the ring of rational integers. Now T is an example of an integral representation. If A is the group algebra of G over P and if .0 is the set of all linear combinations: c1g 1 + c2g 2 + ··· + c..gn of group elements g 1 over the rational integers C; then it is easy to

120

30. GENERAL REMARKS

121

see that D is a subring of A. D is an example of an order in A. Moreover Vx ED, T(x) is a matrix with integer coefficients. If now each integer in the matrix T(x) is replaced by its residue class modulo a fixed prime p, we obtain a matrix 7 (x) with coefficients in a finite field of characteristic p which may be taken as II = {I, 2, ... , p - I, 0}. Thus f(x) = 7 (x) is a representation in II and f is called the modular representation of D, and so of G, corresponding to the integral representation T. Every integral representation gives rise to a modular representation in this way, but it should be noted that not every representation at characteristic p can be so obtained as a modular representation. It should also be understood that even if Tis irreducible, nonetheless f is in general reducible in II or some extension field of II. As a final note of caution we remark that trace f(g), which is an element in II, though a character in our former sense, is not what we will define as a Brauer character. The Brauer, or modular character if/ will be a complex-valued function derived from f in a special way. The example of an integral representation and its corresponding modular representation g1ven in the preceding paragraph is really too narrow to fit the general case. In the first place it is not possible to have all the ordinary representations of a group G lying in the rational field P. The theory of Chapter III was developed for the case where the representing field F is of characteristic p = 0 or p 1 G : I and F is algebraically closed; specifically, F is the field of all algebraic numbers: numbers which are solutions of equations of the form

(30.1) where the c; are rational numbers; moreover, every such equation has its solutions in F. For an integral representation we are again dealing with the case p = 0, but in general a(g) lies in F- P. However, since the group G is of finite order n the set of matrices {a(g)}, for all g E G and all irreducible representations a, is finite. If we now adjoin to P the finite set of coefficients of all these matrices we get a field B, in which all the representations lie, and PCB C F. Moreover,

122

VI. MODULAR REPRESENTATIONS

we know from the theory of field adjunction that B can be obtained by the adjunction of a single element 8 E F: B = P(B), and that the degree of B over Pis finite: [B : P] = m. Also 8 satisfies an irreducible equation of degree m having the form of (30.1). In the general case now described, an integral representation T has its matrices T(g) with coefficients in the integers of B. These are the numbers of B which satisfy an equation, (30 I), in which the coefficients C; are rational integers. It can be shown that these integers form a ring 1 8 and that B is the quotient field of 1a. Thus every x E B can be expressed as x = ajb, a, bE 1a . So far the relations between 1a and B parallel those between 1 and P. Now we come to a second distinction between the present case and our former simple example: whereas 1 is a principal ideal ring, in general 1a is not. This means that each ideal L C J has an element a such that L = a]. On the other hand, an ideal La C 1a may require two generators a, bE 1a so that La = a1a b1a. Now in our formulation it will be essential to work with principal ideal rings. To restore the situation we will embed Bin a p-adic field E whose integers o include 1a . In this way the local situation in 1a will be the same. However, u is a principal ideal ring having E as its quotient field. The next sections are devoted to a detailed exposition of this outline. The development of the theory of modular representations is mainly the work of R. Brauer. We follow here a treatment due to him. However, no account of the Block theory is given. For this we refer the reader to Brauer [5], Osima [17] as well as the text by Curtis and Reiner [10].

+

31. p-Regular Elements of a Finite Group Before the Brauer characters can be defined it is necessary to introduce some new terms and to prove a number of propositions. This will be done here and in the next section. (31.1) Definition. Let G be a group of finite order n = pah where p is a prime integer and (p, h) = I. An element g E G is said to

31. p-REGULAR ELEMENTS OF A FINITE GROUP

123

be p-regular if its order m is relatively prime top; that is, m is the smallest positive integer for which gm = I and (m, p) = l. Note that if an element g is p-regular then all elements which are conjugate tog, and thus have the same order, are alsop-regular. Hence we can consider p-regular classes of conjugate elements.

(31.2) Lemma. Let g be an element of order pbk where k is relatively prime top. Then there are two unique elements x andy in G such that x is p-regular, the order of y is a power of p, and g = xy = yx. x is called the p-regular factor of g.

Proof. Since (pb, k) = l there are integers ex and f1 so that cxpb + {Jk = I. Then g = g"Pb+flk = ga.v'gf3k = gflkg"Pb. Writing x = g"P' and y = gflk we have g = xy =yx. Moreover, since xk = g"~>bk = (gPbk)a. = l and yi'b = (gih)fl = 1, the order of x divides k and so is relatively prime to p while the order of y divides pb and so is a power of p. Thus x and y have the required properties. Now suppose that g = x1y 1 = y 1x1 , and that x1 is p-regular and the order of y 1 is a power of p. Let m be the order of x and m1 that of x1 . Then (m,p) = l, (m1 ,p) = l and we have (/, p) = l where I is the least common multiple of m and m1 . Then noting that y 1 permutes with g and hence with gflk = y and similarly that x1 permutes with x:

for some suitable power p•. Since (xx;- 1) 1 = l, if d is the order of xx] 1 we have d II and dIp·'. As I and p• are relatively prime this is impossible unless d = I. Hence xx) 1 = I = y- 1y 1 =:> x = x1 andy = y 1 , and the uniqueness is established.

(31.3) Lemma. If g = ~:Y. where xis the p-regular factor of g, and if a is any matrix representation of Gin a field of characteristic p, then the matrices a(g) and a(x) have the same characteristic roots.

124

Proof.

VI. MODULAR REPRESENTATIONS

We know from linear algebra that there is a matrix P such

that

p-•o(g)P ~

( ,\1

"*'

'd)

n

and \ , ... , ,\k are the characteristic roots of a(g). Now from Lemma 31.1 x and y are powers of g and the order of y is a power of p. Let x = g', y = g 1, yP' = 1. Then

and we see that (.\!)P' so that

=

1. But the field is of characteristic p

Thus

p-•o(y)P

~

p-••(J:')P

~ (~ *

so that

P-1u(g)P

= P-1 u(xy)P = P- 1u(x)P · P-1u(y)P =

P- 1u(g•)P · P-1u(y)P

32. COMPOSITION FACTORS

125

whence A~ = A; . Since the A~ are characteristic roots of a(gr) = a(x) and the A; are characteristic roots of a(g) we have the result.

32. Conditions for Two Representations to Have the Same Composition Factors If two representations have the same composition factors it is clear that their characters coincide. If the field of the representations is of characteristic p and p = 0 or p 1' G : l we have seen (Lemma 19.13) that the converse is true. Moreover, because of the complete reducibility (Corollary 16.5) which holds in this case, if the characters coincide, the representations are equivalent. In the case p I G : I this is no l M. Then V(o:- am) = limn.,ro V(an - am) :'( "• m > M. Since " > 0 is arbitrary V(o: - am)---+ 0 and o: = lim am, by the definition of limit. Finally it must be shown that E is complete. Let [o:n), o:n E E, be a Cauchy sequence. Let o:, '"" [a7]. Then lim;__,ro a~ = o:n , so that i, can be chosen to ensure that

The sequence a 1,.1 , a 272

, ... ,

:'( ljm

a," , ... is a Cauchy sequence for 1l

+ l/n + "• as

m, n---+ oo.

36. p-ADIC FIELDS

Hence [a~J ,....., d E E and so limn_,oo a~n

V(ex - exn) '-'( V(ex -a~,)

143

= ex. But

+ V(a~n -

exn)


0.

36. p-ADIC FIELDS

147

Putting x = 0 we get N(y) = ±a:r, , so that l(y) = (sfn)l(am)· Therefore l(y) ;;, 0 if and only if l(am) ;;, 0, if and only if all l(a;) ;;, 0 by Lemma 36.6, if and only if y is a p-adic algebraic integer. We return now to the proof of Theorem 36.4.

Proof. We must show that l has the three properties in (ii) of Section 36.1. Now for ex, f3 E E (I) l(cxf3)

=

(1/n)l(N(cxf3))

=

(ljn)l(N(cx))

=

(1/n)l(N(cx)N(f3))

+ (l/n)l(N(f3))

=[(ex)+ [((3).

Suppose that l(cx) ~ l(f3). Then 0 ~ l(f3) -l(cx) = l(f3/cx), by (1). But then f3/cx is a p-adic algebraic integer by Lemma 36.8, and hence 1 + f3/cx is also a p-adic algebraic integer. Then, again by Lemma 36.8, l(1 + {3/cx) ;;, 0, so that (2)

l(cx

+ {3)

;;, Z(cx)

=

minimum{Z(cx), Z(f3)}.

Finally (3)

l(O)

=

(1/n)l(N(O))

=

(1/n)/(0)

=

oo.

Therefore l has the desired properties (1) to (3). Moreover, if aEQp,e;p(a) = ae;,i = 1,2, ... ,n.

p(a)

l(a)

~ =

("

· · .

J

and

N(a)

oo

a".

(lfn)l(N(a)) = (lfn)l(an) = l(a),

showing that lis an extension of l. To conclude the proof we must show that E is complete with respect to l, or equivalently, with respect to the metric V = ci, 0 < c < 1. Now Eisa vector space over QP with the metric Vand we have the following result [2, p. 54]: Any two metrics on a finite-dimensional vector space are equivalent, i.e., a sequence which

VI. MODULAR REPRESENTATIONS

148

is a Cauchy sequence with respect to either metric is also a Cauchy sequence with respect to the other. Now, resuming the proof of Theorem 36.4, let

be an arbitrary element of E. Observe that V*(cx) = max;{V(a;)} has the properties (1)-(3) of Section 36.2 and is therefore also a metric on E. Let (i be a Cauchy sequence of E with respect to sufficiently large

P'({J; - {11)

< " => V*({J; -

{11)

= 1, 2, ... )

P. Then, for i, j

< " => V(b;k - b1k) < "• k

=

1, 2, ... , n.

Thus, for all k, the sequences b1 k, b2 k, .•• are Cauchy sequences and lim;_, 00 b;k = bk E QP .If we now put f1 = b1e1 +b 2e2 + ... +bnen, then f1 E E and lim;_,oo V*(f1- {1;) = 0, so that lim;_, 00 P'({J- {1;) = 0 and this means that lim;_,co {1; = {1. Thus E is complete.

37. Algebras over a p-Adic Field 1. Notation E

A complete p-adic field; a finite extension of QP which in turn is the p-adic completion of the rational field P. o The principal ideal ring of p-adic integers of E. p The maximal ideal in o. ii ojp, the residue class ring; a field of characteristic p. p The unique rational prime in p. A An algebra, with identity, over E. D An order in A (with respect too). By Theorem 34.6 any representation of A is similar to an

37. ALGEBRAS OVER A p-ADIC FIELD

149

integral representation and we shall generally assume that any given representation p. is integral. A(x), B(x)

Matrices depending on xED, with coefficients in o.

A(x), B(x)

Corresponding matrices in ii, obtained by replacing each coefficient in A(x), B(x) by its residue class modulo p.

p(x)

The (integral) regular representation of D.

p.(x)

An arbitrary (integral) representation of D.

a(x), ji.(x)

Matrix representations of D in ii.

2. Preliminary Results These concern the implication, for an integral representation, of reducibility or decomposability in the corresponding modular representation under certain conditions. (37.1) Lemma. Let p.(x)

=

(A(x) C(x)

B(x)) D(x) '

XED,

be an integral representation of the order D in A, and let A(x)

= P,1 (.x),

D(x)

=

B(x)

= 0

a modular representation a component of the modular regular representation

P+(x),

(in other words, the modular representation p,(x) = p.(x) has a component of the modular regular representation as a bottom constituent). Then there exists a matrix H in o such that -HA(x) Proof.

+ C(x) + D(x)H =

0

(mod p).

The modular representation ii given by p,(x)

=

p.(x)

= (iil(x)

0 )

C(x)

P+(x),

150

VI. MODULAR REPRESENTATIONS

is reducible. By Corollary 6.6' there is a matrix

r

(~ ~)

=

such that

Replacing each coefficient in the matrix H by a representative of its class in o we get an integral matrix H and an integral matrix T

=

(~ ~)·

T-1

=

(-~ ~)

Moreover,

is also an integral matrix. Now T- 111-(x)T

=

( -HA(x)

A(x) + B(x)H + C(x) + HB(x)H + D(x)H

B(x) } -HB(x) D(x)

+

so that going over to the modular statement and comparing with (*)we get -HA(x)

+ C(x) + D(x)H == 0

(mod p).

(37.2) Lemma. Let 11-(x) be an integral matrix representation of the order D in A, such that the corresponding modular representation p.(x) = 11-(x) is the direct sum of a modular representation ;L 1(x) and a component P+(x) of the modular regular representation. Thus 11-(x) is of the form 11-(x)

=

( A(x) pC(x)

pB(x)} D(x) '

37. ALGEBRAS OVER A p-ADIC FIELD

151

Then there exists an integral matrix

such that

T-1 (x)T = (P-1(x) IL 0

pB(x))

P+(x)

in whzch p.1(x), P+(x) are integral representations of D and

Proof.

The required matrix will be constructed as the limit T of a sequence {T;} of matrices. Let T0 = I, the identity matrix, and assume that we already have a sequence of integral matrices:

(a:) such that

(I)

T; =

(~. ~).

L; an integral matrix,

1

j

(mod p;),

(2)

It follows from (2) that T; so

= T;_

(mod p),

(3)

1

= 1, 2, ... , k- I.

(mod p), 1

.p, = .p,

(41.3')

Dcf>

=x.

(4I.l ')

D'x =

D' =transpose of D,

Let us introduce the m X m matrix T = (t;;) where t;; = 1 or 0 according as/; and /"j 1 are or are not p-regular conjugate elements. Clearly O;; = t;AtM , since each summand is zero unless g;,....., g-; 1 and g;.,....., /"j 1, that is, unless g;,....., g; and so i = j, in which case a single summand =I. Hence T2 =I, the identity matrix, and T is nonsingular. Now the orthogonality relation (19.9) for the xi,

±, /(g;)/(fj

1)

= O;;n/h; ,

A~l

can be written rrt

,,

k k x"(g;)l(gK)tK;

=

o1;n/h;

K~l A~l

or, in matrix form: x'xT = (o;;n/h;). Since T 2 =I this is

(41.4)

x'x =

(t;;n/h;) =

T.

Taking the transpose in (41.3') and multiplying on the right by we get, after using (41.4),

x

(41.5)

Now the m X m matrix on the right is nonsingular, since its square is the diagonal matrix (n 2 /h~ , ... , n2fh!,). Therefore m = rankcf>'.P :'( rankcf>' :'( m,

and thus m = rank cf>' = rank cf>.

Since the matrix cf> has l rows we have l ;;?: m.

41. CHARACTER RELATIONS

167

Now I is the number of irreducible modular representations and m is the number of p-regular classes in G. We prove: (41.6) Lemma. The number of absolutely irreducible representations of a finite group Gin a modular field of characteristic pis equal to the number of p-regular classes of conjugate elements of G.*

Proof. We have seen that I~ m. Assume that I> m. Then there is a nontrivial relation among the rows of cfo: I

~Y;cP;(/;) = 0,

j=l,2, ... ,m,

i=l

and not all y; = 0. Recall that all elements here lie in a field B whicl) is a sufficiently large algebraic extension of the rational field P, and which is embedded in a complete p-adic field E. Moreover, the c/>;(/;), as sums of roots of unity, belong to o, the ring of integers of E. Now E is the quotient field of o so that we can ensure that they; in ( *) are integers by multiplying through by a suitable integer. Finally, we may assume that not ally; are divisible by p, since any common factor p may be canceled out. Taking (*) modulo p we write L!=l Y;cP;(g;) = 0, in the field o. But c/>;(/;) = trace 7-;{/;), since in the first place the Brauer character c/>; was obtained from the trace of 7-; by replacing roots of unity in o by the corresponding roots of unity in o. Thus I

~y;tracef-;(g;) =0,

.i =

I, 2, ... , m.

i=l

Since the matrix representation of an element g; and of its p-regular factor f; have the same characteristic roots (Lemma 31.3), trace 7-;{g;) = tracef-;{f;) and the equation above can be extended so that I

~Yi trace 7-;{g)

=

0,

VgEG.

i=l

*This theorem can be proved more directly, without the present apparatus of integral representations [4, 24, 18].

VI. MODULAR REPRESENTATIONS

168

Then by the linearity off; and of the .trace function the relation holds for any element a = ~;= 1 01.;g; of the group algebra A(G, o): I

k j; trace f;{a)

=

Va E A(G, o).

0,

i=l

Now for every i, the theorem of Frobenius and Schur shows that there is an element a such that

f;{a)

=

1 0 (~ 0

") , 0

f;(a)

=

0,

giving, by the last equation, j; = 0. This is a contradiction so that l = m and the theorem is proved. Since we had m = rank c{> = rank ,p, we have: (41.7) Corollary.

c{> and ,Pare nonsingular l

X

l matrices.

Remarks. The s X l matrix x has rank l as can be seen from (41.4), remembering that m = l. Taking the coefficients of x modulo p we get an s X l matrix X in o. An important result [4] states that has rank /. By (41.3") x = De{> and hence = D{>. Then l = rank X = rank D{> ;:;;; rank {> ;:;;; l, so that l = rank{> and determinant {> -=1= 0. Thus det c{> =I= 0 (mod p). Equating the i, jth entry on both sides of (41.5) we get

x

x

I

(41.7)

kcf>A(g;)I/JA(g;) A=l

=

t;;n/h; •

Now for fixedj and i = 1, 2, ... , l this is a system of I equations for 1/11(/;), •.. , 1/ll(g;). Writing cp;.; = cp;.(g;) and cpu for the cofactor of cp;.;, Cramer's rule gives I

det c{> ,PA(i;)

=

ki=l t;,;(n/h;)c{>M.

42. MODULAR ORTHOGONALITY RELATIONS

169

Since t; 1 = 0 unlessi;,....., il 1 in which event t;; = 1 and h; = h1 , this yields b some integer in o.

Because of ( *) the highest power of p which divides n/h1 also divides •h(ii)· In particular, if the group order n = pac, (c, p) = I, then pais a divisor of .fo.t(l) = the degree of .P.t . We have denoted the exponential valuation onE by l(x). Then if l(x) is normalized so that l(p) = 1 we have

l(dg.fo.t)

> l(n)

= a.

Finally, we mention the following:

Theorem. (Brauer [5, 6]). The elementary divisors of the Cartan matrix Care the numbers pt!nfh;), i = 1, 2, ... , l.

41. Modular Orthogonality Relations By manipulating the relations (41.2') and (41.5) certain orthogonal relations between the modular characters can be stated explicitly. Let us recall that 1' = (t; 1n/h;) where t;; equals 1 or 0 according as the p-regular elements i; and il1 belong to the same or different conjugate classes, n is the order of the group, and h; the number of elements in the conjugate class C; . Then

1'-1 = (t;;h;/n), as is easily checked. Now from (41.5) by taking inverses and rearranging

I

(42.1)

=

.p1'-1'.

Taking the transpose in (41.2') and multiplying on the left by

f-1

'

•

T-l,P' = T-l'C' = (.p)-lC',

VI. MODULAR REPRESENTATIONS

170

by (42.1). Thus

!fo1'-1ifo' = C' = C.

(42.2)

Again, taking inverses in (41.2') and (41.5) we get

which together yield

c-1 =

(42.3)

~1 A~1

!

= (1/n)

k ifJ;(gA)hAcp;(ff";1),

since tA•

=

0 if,\ -=1=- v

A=1

= (1/n)

k ifo;(g)cp;(g-1), geG'

since each character is constant on all h;. elements of the p-regular class containing g;. . We rewrite this as (42.1 ') Similarly, from (42.2) and (42.3) we obtain (42.2') (42.3')

"t ifo;(g)ifo (g-1) ( 1/n) "t cp;(g)cp1(g-1) = ( 1/n)

1

= C;; ,

i;; •

The prime indicates that the summation is only taken over the p-regular elements of G, O;; is the Kronecker o, and C;;, C;; are the elements at the intersection of the ith row and jth column of C and c-I, respectively. C is the Cartan matrix.

42. MODULAR ORTHOGONALITY RELATIONS

171

Example. Let G be the symmetric group S 8 ={I, (I2), (I3), (23), (I23), (132)}. Sa can be generated by the elements a = (I2) and b = (132). This was our first example at the end of Section I.

The equivalence classes are C 1 ={I}, C2 = {(12), (13), (23)}, and Ca = {(I23), (I32)}. Thus s = 3 and there are three irreducible representations in a field of characteristic zero. We found them to be and

aa(a)

0) (-I -I I ,

=

aa(b)

I) (-I -I O .

=

Let ll = {Q, !• ~} be the integers modulo 3. ll is a field of characteristic p = 3. Since Sa has 2 p-regular classes, there are two irreducible modular representations. Clearly,

(I)

l

a 1 (a)

=

=

l.

a 1 (b)

= 1\(b) = l

a 2(a)

= .,-2(a) =

~.

a 2(b)

=

:t(a)

f

2

(b) =

and

l

are irreducible modular r-epresentations. Then aa() must be reducible. Now and if we find

~ T-'o,(a)'f ~ (!

(2)

l

T-'o,(b)'f

~ (!

From (I) and (2) it is seen that the decomposition matrix

172

VI. MODULAR REPRESENTATIONS

and the Cartan matrix

Note that determinant C = 3. The accompanying tabulation gives the irreducible Brauer characters on the p-regular elements: a -1

Thus

cp=G _!) so that

=Ccp =(~

_!)

x~IJ+~G

-D·

ifJ and

Appendix 1. Groups {l.l) Definition.

A group is a set G together with an operation defined between pairs of elements a, b E G (the operation is denoted by ab and is called multiplication) which satisfies the following axioms: 1. For all a, b E G, abE G

(closure).

2. (ab)c = a(bc) (associativity). 3. For each a, b E G there are elements x, y E G such that ax = b and ya = b (equations are solvable in G). Consequences: I.

2.

Every group has a unique element e with the property ae = ea = a, for every a E G (e is called the identity and is denoted by I). To every element a E G there is a unique inverse, written a-I, with the property: a- 1a = aa- 1 = e.

If G is a group and if for all a, bEG, ab = ba, then G is commutative or abelian. In this case the operation is usually that is, a b instead of ab. The identity is then denoted by written as 0 and the inverse of a as -a. A module is an abelian group.

+.

+

{1.2) Definition. A subset Sofa group G is called a subgroup if its elements form a group; we must have then: a, bE S implies abE S and ax = b, ya = b must be solvable inS for x andy. This is equivalent to the conditions: I a- 1 E S.

I73

E

S, and a E S implies

APPENDIX

174

If Sis a subgroup of a group G a left (right) congruence can defined between the elements of G. Thus a, bEG are left congruent modulo S, written a ~ b (modS) if there is an element s E S such that a = sb. Similarly a ' b (modS) if s' E S and a = bs'. It is easy to show that left (right) congruence is an equivalence relation, that is: (I) a = a, (2) a = b implies b = a, and (3) a b, b = c implies a c. Then G can be separated into disjoint equivalence classes of left (right) congruent elements. However, we do not in general have that a band c - d implies ac = bd. For this property to hold S must be a normal subgroup

=

=

=

(I.3) Definition. N is a normal subgroup of G, written N N, if g- 1Ng = N for all g E G. That is, if n EN then g- 1ng = n' E N. If N a== (mod Ill)::::> Now define a(~!)= a(a). Since a -bE Ill ::::> a(a - b) = 0 ::::> a(a) = a(b), the definition is independent of the representative and is thus unambiguous. Moreover,

W ll.J

a(l~l

+ 11._1) =

a(Ia +

= a(a)

b I) = a( a + b)

+ a(b) =

a(l~l)

+ a(ll._l)

APPENDIX

178

and a(f~ll!_l) =

a([__E!_I)

=

a(ab)

=

a(a)a(b)

= a(f~l)a(l!_l)

so that a is a ring homomorphism of Rj'll onto R'. Finally a(~) = 0 = a(a) => a E 'li => ~~ = J_Q_I, the zero element of Rj'll showing that a is an isomorphism. (2.6) Theorem II. If 'll, !8 are two-sided ideals of R then so are ('ll, !8} and 'li fl !8. Moreover, ('ll, !B)/'ll !B/'ll r. !B. F'o,J

Proof.

The first statement is clear. For the second, define = Jl_liJI where the single and double "buckets" refer to classes modulo 'li and 'li fl !8, respectively. Since J a + b J =

a(J a + b I)

W L!J

~ + ~~ = LQJ + = this is the same as a(L!j) = lll__IJ and is well defined since = ~~ => b - b' E 'li => b - b' E 'li fl !8 => /WI = Jl_~l/. Now, as before, we can show that a is a ring homomorphi~~-of ('l!, !8} onto !8/'l! (J !8 and then Theorem I of 2.5 gives the result. For references, see: Artin [2]; van der Waerden [22].

3.

l!J

A Formula for the Character

The character of the representation corresponding to the idempotent derived from PN (Lemma 28.6) can be calculated by the formula

(3.1)

x(g) = (n/(Ci(g) : I )(R n C : I))

k

8(r)rf>(c)

rce\£(g)

where x is the character of the irreducible representation corresponding to the primitive idempotent formed from R and C, n is the degree of the irreducible representation, Ci(g) is the class of elements conjugate tog, and the other symbols are as in Lemma 28.6.

Proof. It is clear that ~seG s(PN)s- 1 is an element of the center of the subalgebra to which PN belongs; moreover the expression

A FORMULA FOR THE CHARACTER

179

~t~G tx(t) is the central idempotent of this subalgebra up to a multiple. Thus, since PN remains a multiple of a primitive idempotent even after extension of the ground field to an algebraically closed field so that the center of the Wedderburn component is of dimension I,

J"',£ tx(t) = k s(PN)rl tEG

sEG

Recalling that PN = ~ rcO(r)