MATH 101B: ALGEBRA II PART D: REPRESENTATIONS OF FINITE...

MATH 101B: ALGEBRA IIPART D: REPRESENTATIONS OF FINITE GROUPS

For the rest of the semester we will discuss representations of groups, mostly finitegroups. An elementary treatment would start with characters and their computation.But the Wedderburn structure theorem will allow us to start at a higher level whichwill give more meaning to the character tables which we will be constructing. This isfrom Lang, XVIII, §1-7 with additional material from Serre’s “Linear Representations ofFinite Groups” (Springer Graduate Texts in Math 42) and Alperin and Bell’s “Groupsand Representations” (Springer GTM 162).

Contents

1. The group ring KG 21.1. Representations of groups 21.2. Modules over KG 31.3. Semisimplicity of KG 51.4. Idempotents 81.5. Center of CG 92. Characters 102.1. Basic properties 102.2. Irreducible characters 132.3. formula for idempotents 172.4. character tables 182.5. orthogonality relations 223. Induction 303.1. induced characters 303.2. Induced representations 343.3. Artin’s theorem 39

1

2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS

1. The group ring KG

The main idea is that representations of a group G over a field K are “the same” asmodules over the group ring KG = K[G]. Also, left module are “the same” as rightmodules: KG-mod ⇠= mod-KG. First I defined both terms.

1.1. Representations of groups.

Definition 1.1. A representation of a group G over a field K is defined to be a grouphomomorphism

⇢ : G ! AutK(V )

where V is a vector space over K.

Here AutK(V ) is the group of K-linear automorphisms of V . This is also written asGLK(V ). This is the group of units of the ring EndK(V ) = HomK(V, V ) which, as Iexplained before, is a ring with addition defined pointwise and multiplication given bycomposition. If dimK(V ) = d then AutK(V ) ⇠= AutK(Kd) = GLd(K) which can also bedescribed as the group of units of the ring Matd(K) or as:

GLd(K) = {A 2 Matd(K) | det(A) 6= 0}

d = dimK(V ) is also called the degree of the representation ⇢.

1.1.1. examples.

Example 1.2. The trivial representation. This is defined to be the one dimensionalrepresentation V = K with trivial action of the group G (which can be arbitrary).Trivial action means that ⇢(g) = 1 = idV for all � 2 G.

In the next example, I pointed out that the group G needs to be written multiplica-tively no matter what.

Example 1.3. Let G = Z/3. Written multiplicatively, the elements are 1, �, �2. LetK = R and let V = R2 with ⇢(�) defined to be rotation by 120� = 2⇡/3 and ⇢(�2) isrotation by 240� = 4⇡/3. I.e.,

⇢(�) =

✓�1/2 �

p3/2p

3/2 �1/2

◆, ⇢(�2) =

✓�1/2

p3/2

�p3/2 �1/2

◆

Example 1.4. The regular representation is defined to be V = KG with G acting byleft multiplication. The degree of this representation is equal to n = |G|, the order of G.

Example 1.5. Suppose that E is a normal (Galois) field extension of K and G =Gal(E/K). Then G acts on E by K-linear transformations. This gives a representation:

⇢ : G ,! AutK(E)

Note that this map is an inclusion by definition of “Galois group.” This representationalso has degree n = |G|.

Is this isomorphic to the regular representation?

MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 3

1.1.2. axioms. The axioms for a group representation just say what it means for

⇢ : G ! AutK(V )

to be a group homomorphism:

(1) ⇢(1) = 1 = idV 1v = v 8v 2 V

(2) ⇢(�⌧) = ⇢(�)⇢(h) 8�, ⌧ 2 G (�⌧)v = �(⌧v) 8v 2 V

(3) ⇢(�) is K-linear 8� 2 G �(av + bw) = a�v + b�w 8v, w 2 V, a, b 2 k

The first two conditions say that ⇢ is an action of G on V . Actions are usually writtenby juxtaposition:

�v := ⇢(�)(v)

Condition (3) says that the action is K-linear. So, together, the axioms say that arepresentation of G is a K-linear action of G on a vector space V .

1.2. Modules over KG. The group ring KG is defined to be the set of all formal finiteK linear combinations of elements of G:

Pagg where ag 2 k for all g 2 G and ag = 0

for almost all g. This is a vector space of K with basis G.For example, R[Z/3] is the set of all linear combinations

x+ y� + z�2

where x, y, z 2 R. I.e., R[Z/3] ⇠= R3.Multiplication in KG is given by

⇣Xa��

⌘⇣Xb⌧⌧

⌘=⇣X

c��

⌘

where c� 2 G can be given in three di↵erent ways:

c� =X

�⌧=�

a�b⌧ =X

�2G

a�b��1� =X

⌧2G

a�⌧�1b⌧

Proposition 1.6. KG is a K-algebra.

This is straightforward and tedious. So, I didn’t prove it. But I did explain what itmeans and why it is important.

Recall that an algebra over K is a ring which contains K in its center. The centerZ(R) of a (noncommutative) ring R is defined to be the set of elements of R whichcommute with all the other elements:

Z(R) := {x 2 R | xy = yx 8y 2 R}Z(R) is a subring of R.

The center is important for the following reason. Suppose thatM is a (left) R-module.Then each element r 2 R acts on M by left multiplication �r

�r : M ! M, �r(x) = rx

This is a homomorphism of Z(R)-modules since:

�r(ax) = rax = arx = a�r(x) 8a 2 Z(R)

Thus the action of R on M gives a ring homomorphism:

⇢ : R ! EndZ(R)(M)


Getting back to KG, suppose that M is a KG-module. Then the action of KG onM is K-linear since K is in the center of KG. So, we get a ring homomorphism

⇢ : KG ! EndK(M)

This restricts to a group homomorphism

⇢|G : G ! AutK(M)

In general, any ring homomorphism � : R ! S will induce a group homomorphismU(R) ! U(S) where U(R) is the group of units of R. By definition, AutK(M) is thegroup of units of EndK(M). G is contained in the group of units of KG. (An interestingunanswered question is: Which finite groups occur as groups of units of rings?)

This discussion shows that a KG-module M gives, by restriction, a representation ofthe group G on the K-vector space M . Conversely, suppose that

⇢ : G ! AutK(V )

is a group representation. Then we can extend ⇢ to a ring homomorphism

⇢ : KG ! EndK(V )

by the simple formula

⇢

⇣Xagg

⌘=X

ag⇢(g)

When we say that a representation of a group G is “the same” as a KG-module weare talking about this correspondence. The vector space V is also called a G-module.So, it would be more accurate to say that a G-module is the same as a KG-module.

Corollary 1.7. (1) Any group representation ⇢ : G ! AutK(V ) extends uniquely toa ring homomorphism ⇢ : KG ! EndK(V ) making V into a KG-module.

(2) For any KG-module M , the action of KG on M restricts to give a group repre-sentation G ! AutK(M).

(3) These two operations are inverse to each other in the sense that ⇢ is the restrictionof ⇢ and an action of the ring KG is the unique extension of its restriction to G.

There are some conceptual di↵erences between the group representation and the cor-responding KG-module. For example, the module might not be faithful even if thegroup representation is:

Definition 1.8. A group representation ⇢ : G ! AutK(V ) is called faithful if only thetrivial element of G acts as the identity on V . I.e., if the kernel of ⇢ is trivial. AnR-module M is called faithful if the annihilator of M is zero where

ann(M) := {r 2 R | rx = 0 8x 2 M}.These two notions of “faithful” do not agree. For example, take the representation

⇢ : Z/3 ,! GL2(R)which we discussed earlier. This is faithful. But the extension to a ring homomorphism

⇢ : R[Z/3] ! Mat2(R)is not a monomorphism since 1 + � + �

2 is in its kernel.Another thing I did in class:


Proposition 1.9. The categories of left and right KG-modules are isomorphic.

Proof. Given a left KG-module M , the corresponding right KG-module M 0 is the samevector space as M but with right action of G given by

xg := g�1x

This makes M into a right KG module since

(xg)h = (g�1x)h = h

�1g�1x = (gh)�1

x = x(gh)

It is straightforward to show that this gives an isomorphism of categories:

KG-mod ⇠= mod-KG

as claimed. ⇤1.3. Semisimplicity of KG. The main theorem about KG is the following.

Theorem 1.10 (Maschke). If G is a finite group of order |G| = n and K is a field withchar k - n (or char k = 0) then KG is semisimple.

Instead of saying char k is either 0 or a prime not dividing n, I will say that 1/n 2 k.By the Wedderburn structure theorem we get the following.

Corollary 1.11. If 1/|G| 2 K then

KG ⇠= Matd1(Di)⇥ · · ·⇥Matdb(Db)

where Di are finite dimensional division algebras over K.

Example 1.12.R[Z/3] ⇠= R⇥ C

In general, if G is abelian, then the numbers di must all be 1 and Di must be finite fieldextensions of K.

1.3.1. homomorphisms. In order to prove Maschke’s theorem, we need to talk abouthomomorphisms of G-modules. We can define these to be the same as homomorphismsof KG-modules. Then the following is a proposition. (Or, we can take the followingas the definition of a G-module homomorphism, in which case the proposition is thatG-module homomorphisms are the same as homomorphisms of KG-modules.)

Proposition 1.13. Suppose that V,W are KG-modules. Then a K-linear mapping� : V ! W is a homomorphism of KG-modules if and only if it commutes with theaction of G. I.e., if

�(�(v)) = �(�v)

for all � 2 G.

Proof. Any homomorphism of KG-modules will commute with the action of KG andtherefore with the action of G ⇢ KG. Conversely, if � : V ! W commutes with theaction of G then, for any

Pag� 2 KG, we have

�

X

�2G

ag�v

!=X

�2G

ag�(�v) =X

�2G

ag��(v) =

X

�2G

ag�

!�(v)

So, � is a homomorphism of KG-modules. ⇤


We also have the following Proposition/Definition of a G-submodule.

Proposition 1.14. A subset W of a G-module V over K is a KG-submodule (and wecall it a G-submodule) if and only if

(1) W is a vector subspace of V and(2) W is invariant under the action of G. I.e., �W ✓ W for all � 2 G.

I pointed out that “invariant” is defined as gW ✓ W for all g 2 G. This implies thatgW = W since g

�1W ✓ W implies W ✓ gW .

We also had a discussion about what happens when char K = p and p divides n = |G|.The simplest example is K = Fp and G = Z/p. Then

Fp[Z/p] ⇠= Fp[x]/(xp)

To see this, take x = 1�� where � is the generator of Z/p. Then xp = 1p��p = 1�1 = 0.

So, Fp is a local ring with maximal ideal m = (x).In general, the way to handle the “modular” case when p = charK divides |G| is

to find a DVR R so that R/m = K (with characteristic p) but QR has characteristiczero. For example, R = Z(p) which has factor ring R/m = Z/p and QR = Q. Since anyring homomorphism R ! S induces a functor S-mod ! R-mod, the homomorphismsR ⇣ K and R ,! QR induce functors the other way giving the following diagram:

KG-mod

((

QR[G]-mod

uuRG-mod

This gives a lot of information about the “bad” case of KG-modules in terms of thewell-behaved QR[G]-modules. Perhaps we can discuss this more after we finish with theeasy case when K = C.


Proof of Maschke’s Theorem. Suppose that V is a finitely generated G-module and W

is any G-submodule of V . Then we want to show that W is a direct summand of V .This is one of the characterizations of semisimple modules. This will prove that all f.g.KG-modules are semisimple and therefore KG is a semisimple ring.

Since W is a submodule of V , it is in particular a vector subspace of V . So, there isa linear projection map � : V ! W so that �|W = idW . If � is a homomorphism ofG-modules, then V = W � ker� and W would split from V . So, we would be done. If �is not a G-homomorphism, we can make it into a G-homomorphism by “averaging overthe group,” i.e., by replacing it with

=1

n

X

g2G

�g�1 � � � �g

First, I claim that |W = idW . To see this take any w 2 W . Since W is G-invariant,gw 2 W for any g 2 G. So, �(gw) = gw and

(w) =1

n

X

g2G

g�1�(gw) =

1

n

X

g2G

g�1(gw) = w

Next I claim that is a homomorphism of G-modules. To show this take any h 2 G

and v 2 V . Then

(hv) =1

n

X

g2G

g�1�(ghv) = h

1

n

X

g2G

h�1g�1�(ghv) = h

1

n

X

k2G

k�1�(kv) = h (v)

since summing over all g 2 G is the same as summing over all k = gh 2 G. So, givesa splitting of V as required. ⇤1.3.2. R[Z/3]. The isomorphism R[Z/3] ⇠= R⇥ C is given by the mapping:

� : Z/3 ! R⇥ Cwhich sends the generator � to (1,!) where ! is a primitive third root of unity. Since(1, 0), (1,!), (1,!) are linearly independent over R, the linear extension � of � is anisomorphism of R-algebras.

1.3.3. group rings over C. We will specialize to the case K = C. In that case, there areno finite dimensional division algebras over C (Part C, Theorem 3.12). So, we get onlymatrix algebras:

Corollary 1.15. If G is any finite group then

CG ⇠= Matd1(C)⇥ · · ·⇥Matdb(C)In particular, n = |G| =

Pd2i.

Example 1.16. If G is a finite abelian group of order n then CG ⇠= Cn.

Example 1.17. Take G = S3, the symmetric group on 3 letters. Since this group isnonabelian, the numbers di cannot all be equal to 1. But the only way that 6 can bewritten as a sum of squares, not all 1, is 6 = 1 + 1 + 4. Therefore,

C[S3] ⇠= C⇥ C⇥Mat2(C)


This can be viewed as a subalgebra of Mat4(C) given by0

BB@

⇤ 0 0 00 ⇤ 0 00 0 ⇤ ⇤0 0 ⇤ ⇤

1

CCA

Each star (⇤) represents an independent complex variable. In this description, it is easyto visualize what are the simple factors Matdi(C) given by the Wedderburn structuretheorem. But what are the corresponding factors of the group ring CG?

1.4. Idempotents. Suppose that R = R1 ⇥ R2 ⇥ R3 is a product of three rings. Thenthe unity of R decomposes as 1 = (1, 1, 1). This can be written as a sum of unit vectors:

1 = (1, 0, 0) + (0, 1, 0) + (0, 0, 1) = e1 + e2 + e3

This is a decomposition of unity (1) as a sum of central, orthogonal idempotents ei.Recall that idempotent means that e2

i= ei for all i. Also, 0 is not considered to be an

idempotent. Orthogonal means that eiej = 0 if i 6= j. Central means that ei 2 Z(R).We had a discussion in class about what are these things Ri.

Lemma 1.18. Suppose that I1, I2 are two sided ideals in any ring R so that

(1) I1 \ I2 = 0(2) R = I1 + I2

Then we have an isomorphism of rings: R ⇠= R/I1 ⇥R/I2.

Proof. The isomorphism � : R ! R/I1 ⇥ R/I2 is given by �(r) = (r + I1, r + I2). Thekernel of � is the set of all r so that r+I1 is zero in R/I1 which means r 2 I1 and r 2 I2.So, ker� = I1\I2 = 0. Also � is onto: Take any element x = (r+I1, s+I2) 2 R/I1⇥R/I2.Since R = I1 � I2, r = r1 + r2 and s = s1 + s2 where r1, s1 2 I1 and r2, s2 2 I2. Thenr + I1 = r2 + I1 and s + I2 = s1 + I2. Then �(s1 + r2) = (s1 + r2 + I1, s1 + r2 + I2) =(r2 + I1, s1 + I2) = x. So, � is onto. ⇤

R/I1, R/I2 are factor rings of R. But they are isomorphic, as rings without unit tothe ideal I2, I1 respectively and we have the internal direct sum:

R = I1 � I2

of R as direct sum of two ideals.

Theorem 1.19. A ring R can be written as a product of b rings R1, R2, · · · , Rb i↵ 1 2 R

can be written as a sum of b central, orthogonal idempotents and, in that case, the idealRi = eiR is a ring with unity ei.

Proof. Suppose R ⇠= R1 ⇥ · · · ⇥ Rb. Let � be the isomorphism. Then 1 = e1 + · · · + eb

there ei = ��1 of the unity in Ri and these are central orthogonal idempotents of R since

they come from central orthogonal idempotents of R1 ⇥ · · · ⇥ Rb. Conversely, supposethe 1 = e1 + · · · eb. Then R is a direct sum of the two sided ideals eiR and the ideal eiRis isomorphic to the factor ring R/

Pj 6=i

ejR (the isomorphism is the restriction of theprojection map R ! R/

Pj 6=i

ejR to eiR. ⇤


A central idempotent e is called primitive if it cannot be written as a sum of twocentral orthogonal idempotents.

Corollary 1.20. The number of factors Ri = eiR is maximal i↵ each ei is primitive.

So, the problem is to write unity 1 2 CG as a sum of primitive, central () orthogonal)idempotents. We will derive a formula for this decomposition using characters.

1.5. Center of CG. Before I move on to characters, I want to prove one last thingabout the group ring CG.

Theorem 1.21. The number of factors b in the decomposition

CG ⇠=bY

i=1

Matdi(C)

is equal to the number of conjugacy classes of elements of G.

For, example, the group S3 has three conjugacy classes: the identity {1}, the trans-positions {(12), (23), (13)} and the 3-cycles {(123), (132)}.

In order to prove this we note that b is the dimension of the center of the right handside. Any central element of Matdi(C) is a scalar multiple of the unit matrix which weare calling ei (the ith primitive central idempotent). Therefore:

Lemma 1.22. The center ofQ

b

i=1 Matdi(C) is the vector subspace spanned by the prim-itive central idempotents e1, · · · , eb. In particular it is b-dimensional.

So, it su�ces to show that the dimension of the center of CG is equal to the numberof conjugacy classes of elements of G. (If G is abelian, this is clearly true.)

Definition 1.23. A class function on G is a function f : G ! X so that f takes thesame value on conjugate elements. I.e.,

f(hgh�1) = f(g)

for all g, h 2 G. Usually, X = C.For example, any function on an abelian group is a class function.

Lemma 1.24. For any field K, the center of KG is the set of allP

g2G agg so that agis a class function on G. So, Z(KG) ⇠= K

c where c is the number of conjugacy classesof elements of G.

Proof. IfP

g2G agg is central thenX

g2G

agg = h

X

g2G

aggh�1 =

X

g2G

aghgh�1

The coe�cient of hgh�1 on both sides must agree. So

ahgh�1 = ag

I.e., ag is a class function. The converse is also clear. ⇤These two lemmas clearly imply Theorem 1.21 (which can now be stated as: b = c if

K = C).


2. Characters

If ⇢ : G ! GLd(C) is a representation of G over C then the character of ⇢ is thefunction

�⇢ : G ! Cgiven by �⇢(g) = Tr(⇢(g)).

Example 2.1. For ⇢ the degree 2 representation of Z/3 we have

�(1) = 2, �(�) = �1

2� 1

2= �1 = �(�2)

The main property of characters is that they determine the representation uniquelyup to isomorphism. So, once we find all the characters (by constructing the charactertable) we will in some sense know all the representations. We will assume that all groupsG are finite and all representations are finite dimensional over C.2.1. Basic properties. The basic property of trace is that it is invariant under conju-gation:

Tr(ABA�1) = Tr(B)

Letting A = ⇢(g), B = ⇢(h) we get

�⇢(ghg�1) = Tr(⇢(ghg�1)) = Tr(⇢(g)⇢(h)⇢(g)�1) = Tr(⇢(h)) = �⇢(h)

for any representation ⇢. So:

Theorem 2.2. Characters are class functions. (They have the same value on conjugateelements.)

If ⇢ : G ! AutC(V ) is a representation of G over C, then the character of ⇢, alsocalled the character of V , is defined to be the function

�⇢ = �V : G ! Cgiven by

�V (g) = Tr(⇢(g)) = Tr(� � ⇢(g) � ��1)

for any linear isomorphism � : V⇡�! Cd.


There are three basic formulas that I want to explain. In order of di�culty they are:

(1) The character of a direct sum is the sum of the characters:

�V�W = �V + �W

(2) The character of a tensor product is the product of the characters:

�V⌦W = �V �W

(3) The character of the dual representation is the complex conjugate of the originalcharacter:

�V ⇤ = �V

2.1.1. direct sum. The trace of a direct sum of matrices is the sum of traces:

Tr(A� B) = Tr

✓A 00 B

◆= Tr(A) + Tr(B)

Theorem 2.3. If V,W are two G-modules then

�V�W = �V + �W

Proof. If ⇢V , ⇢W , ⇢V�W are the corresponding representations then

⇢V�W (g) = ⇢V (g)� ⇢W (g)

The theorem follows. ⇤2.1.2. character formula using dual basis. Instead of using traces of matrices, I preferthe following equivalent formula for characters using bases and dual bases.

If V is a G-module, we choose a basis {v1, · · · , vd} for V as a vector space overC. Then recall that the dual basis for V

⇤ = HomC(V,C) consists of the dual vectorsv⇤1, · · · , v⇤d : V ! C given by

v⇤j

dX

i=1

aivi

!= aj

I.e., v⇤jpicks out the coe�cient of vj.

Proposition 2.4.

�V (g) =dX

i=1

v⇤i(�vi)

Proof. The matrix of the linear transformation ⇢(g) has (i, j) entry v⇤i(�vj) Therefore,

its trace isP

v⇤i(�vi). ⇤

For example, the trace of the identity map is

Tr(idV ) =dX

i=1

v⇤i(vi) = d

Theorem 2.5. The value of the character at 1 is the degree of the representation:

�V (1) = d = dimC(V )


2.1.3. tensor product. If V,W are two G-modules then the tensor product V ⌦ W isdefined to be the tensor product over C with the following action of G:

�(v ⌦ w) = �v ⌦ �w

Theorem 2.6. The character of V ⌦W is the product of the characters of V and W .I.e.,

�V⌦W (g) = �V (g)�W (g)

for all � 2 G.

Proof. Choose bases {vi}, {wj} for V,W with dual bases {v⇤i}, {w⇤

j}. Then the tensor

product V ⌦ W has basis elements vi ⌦ wj with dual basis elements v⇤i⌦ w

⇤j. So, the

character is:

�V⌦W (g) =X

i,j

(v⇤i⌦ w

⇤j)�(vi ⌦ wj) =

X

i,j

(v⇤i⌦ w

⇤j)(�vi ⌦ �wj)

=X

i,j

v⇤i(�vi)w

⇤j(�wj) =

X

i

v⇤i(�vi)

X

j

w⇤j(�wj) = �V (g)�W (g)

⇤

2.1.4. dual reprentation. The dual space V⇤ is a right G-module. In order to make it a

left G-module we have to invert the elements of the group. I.e., for all f 2 V⇤ we define

(gf)(v) := f(g�1v)

Lemma 2.7.

�V ⇤(g) = �V (g�1)

Lemma 2.8.

�V (g�1) = �V (g)

Proof. The trace of a matrix A is equal to the sum of its eigenvalues �i. If A has finiteorder: A

m = Id then its eigenvalues are roots of unity. Therefore, their inverses areequal to their complex conjugates. So,

Tr(A�1) =X

��1i

=X

�i = Tr(A)

Since G is finite, the lemma follows. ⇤

Theorem 2.9. The character of the dual representation V⇤ is the complex conjugate of

the character of V :

�V ⇤(g) = �V (g)


2.2. Irreducible characters.

Definition 2.10. A representation ⇢ : G ! AutC(V ) is called irreducible if V is a simpleG-module. The character

�⇢ = �V : G ! Cof an irreducible representation is called an irreducible character.

Theorem 2.11. Every character is a nonnegative integer linear combination of irre-ducible characters.

Proof. Since CG is semisimple, any G-module V is a direct sum of simple modulesV ⇠=

LSi. So, the character of V is a sum of the corresponding irreducible characters:

�V =P�Si . ⇤

If we collect together multiple copies of the same simple module we get V =L

niSi

and

�V =rX

i=1

ni�i

where �i is the character of Si. This makes sense only if we know that there are onlyfinitely many nonisomorphic simple modules Si and that the corresponding characters�i are distinct functions G ! C. In fact we will show the following.

Theorem 2.12. (1) There are exactly b (the number of blocks) irreducible represen-tations Si up to isomorphism.

(2) The corresponding characters �i are linearly independent.

This will immediately imply the following.

Corollary 2.13. The irreducible characters �1, · · · ,�b form a basis for the b-dimensionalvector space of all class functions G ! C.

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