CHARACTER THEORY OF FINITE GROUPS Chapter 1: REPRESENTATIONS › ~falk › classes › 511 › Isaacs... · A linear character is a character such that the degree (1) = 1. Note: The

CHARACTER THEORY OF FINITE GROUPS

Chapter 1:

REPRESENTATIONS

G is a finite group andK is a field.

A K-representation of G is a homomorphism

X : G ! GL(n,K) ,

where GL(n,K) is the group of invertible n⇥ nmatrices over K.

The positive integer n is the degree of X .

A K-representation X of G of degree n is irre-ducible if there does NOT exist a nonzero propersubspace U of the n-dimensional row space overK such that UX (g) ✓ U for all g 2 G.

To say that X is irreducible thus means that itis NOT similar to a representation Y of the form

Y(g) =

U(g) 0⇤ V(g)

�

Note: Here, U and V are representations.

The group algebra K[G] is the vector space overK with basis G.

Multiplication is defined by the group producttogether with the distributive law.

If X is a K-representation of G with degree n,then X can be extended linearly to K[G].

The result is an algebra homomorphism

X : K[G] ! Mn(K) ,

where Mn(K) is the algebra of n⇥ n matrices.

Fact: Assuming that K is algebraically closed,the K-representation X of G is irreducible i↵ Xmaps K[G] onto the full algebra Mn(K).

Note: The “if” direction is obvious but “only if”requires algebraically closed.

If X is irreducible and K is algebraically closed,it follows that the only matrices that central-ize X (G) are the scalar matrices. This is calledSchur’s lemma.

EXERCISE (1.1): Let K be an arbitrary field,and let z be the sum of all elements of G inthe group algebra K[G]. Show that the one-dimensional subspace Kz of K[G] is an ideal.Also, prove that z is nilpotent i↵ the character-istic of K divides |G|.

EXERCISE (1.2): With the notation as above,show that K[G] has a proper ideal I such thatKz + I = K[G] i↵ the characteristic of K doesnot divide |G|.

From now on, we take K = C, though any alge-braically closed characteristic zero field will workas well.

Fact: C[G] is a direct sum of full matrix algebras:

C[G] =kX

i=1

Mni(C) .

Note:

|G| = dim(C[G]) =kX

i=1

(ni)2.

What is k?

Since each matrix algebra has center of dimen-sion 1, we have

k = dim(Z(C[G])) .

Also, the sums in C[G] of the elements of eachconjugacy class of G form a basis for Z(C[G]),and thus

k = number of classes of G .

Observe that the composite map

G ! C[G] ! Mni(C)

is an irreducible C-representation of G.

Here, the second map is the projectionto the i th summand.

Write Xi to denote this representation.

Fact: Every irreducible C-representation of G issimilar to one of the Xi.

Chapter 2:

CHARACTERS

Given a C-representation X of G, the associatedcharacter � is the function G ! C given by:

�(g) = trace(X (g)) .

We say X a↵ords �.

Note: Similar representations a↵ord equal char-acters.

Note: If X a↵ords � then �(1) is thedegree of X . Thus �(1) is a positive integer; itis called the degree of �.

Note:�(x�1gx) = trace(X (x�1gx))

= trace(X (x)�1X (g)X (x))

= trace(X (g))

= �(g) .

so characters are constant on conjugacy classes.

The characters of G thus lie in the C-space cf(G)of class functions of G.

Note: We have dim(cf(G)) = k, the numberof classes of G.

If ↵ and � are characters of G a↵orded by U andV respectively, then the representation X givenby

X (g) =

U(g) 00 V(g)

�

a↵ords the character � = ↵+ �.

Thus sums of characters are characters.

A character that is not a sum of two charactersis said to be irreducible.

Note: Every character is a sum of irreduciblecharacters.

This follows since character degrees are positiveintegers, and thus one cannot keep decomposingindefinitely.

We will see that a character � is uniquely a sumof irreducible characters. These are called theirreducible constituents of �.

Notation: Irr(G) is the set of irreducible charac-ters of G.

Recall: The C-representations Xi for 1 i kcome from the decomposition of the group alge-bra C[G] as a direct sum of matrix rings.

Also recall: Up to similarity, these Xi are all ofthe irreducible C-representations of G.

THEOREM: The characters a↵orded by the Xi

are distinct and linearly independent, and they

form the set Irr(G).

Proof: Since the Xi come from projections intodirect summands of C[G], we can find aj 2 C[G]such that Xi(aj) = 0 if i 6= j, but Xi(ai) is anarbitrary matrix of the appropriate size.

Let �i be the character a↵orded by Xi, and view�i as being defined on the whole of C[G]. Thenwe can choose the ai such that �i(aj) = 0, but�i(ai) = 1. It follows that the �i are distinctand linearly independent.

Now let � 2 Irr(G) be arbitrary, and let X be aC-representation a↵ording �. We argue that Xis irreducible.

Otherwise, X is similar to a representation Ysuch that

Y(g) =

U(g) 0⇤ V(g)

�

and � is a↵orded by Y.

Then � = ↵+�, where U afords ↵ and V a↵ords�. This contradicts the irreducibility of �.

We now know that each character � 2 Irr(G)is a↵orded by an irreducible representation, andhence � is a↵orded by some Xi. It follows that� = �i for some i.

What remains is to show that each character �j

is irreducible.

Otherwise, since �j is a sum of irreducible char-acters, it is a sum of characters �i with i 6= j,and this contradicts the linear independence.

COROLLARY:X

�2Irr(G)

�(1)2 = |G|.

COROLLARY: |Irr(G)| = k, the number of con-

jugacy classes of G.

COROLLARY: G is abelian i↵ every member of

Irr(G) has degree 1.

Proof: G is abelian i↵ k = |G|.

COROLLARY: The set Irr(G) is a basis for thespace cf(G) of class functions.

A linear character is a character � such that thedegree �(1) = 1.

Note: The linear characters of G are exactly thehomomorphisms from G into the group C⇥.

The principal character 1G of G is the trivialhomomorphism, with constant value 1.

Note: The set of linear characters of G forms agroup under pointwise multiplication.

The identity is 1G and ��1 = �, the complexconjugate of �.

COROLLARY: The number of linear characters

of G is |G : G0|.Proof: Since a linear character of G is a homo-morphism of G into the abelian group C⇥, thederived subgroup G0 is contained in the kernelof every such character.

The linear characters of G, therefore, are exactlythe linear characters of the abelian group G/G0,and the number of these is |G/G0|.Fact: The group of linear characters of G is iso-morphic to G/G0.

EXERCISE (2.1): Compute the degrees of theirreducible characters of S3, A4, S4 and A5.

EXERCISE (2.2) : If � is a character of G and �is a linear character, define �� to be the functiondefined by (��)(g) = �(g)�(g). Show that ��is a character, and that it is irreducible i↵ � isirreducible.

In fact, if � and are arbitrary characters ofG then the function � defined by (� )(g) =�(g) (g) is a character. We sketch a proof.

Let A be an m⇥m matrix with entries ai,j , andlet B be an n ⇥ n matrix with entries bk,l. TheKronecker product of A and B, denoted A⌦B,is an mn⇥mn matrix, defined as follows.

The rows and columns of A⌦B are indexed bythe set of ordered pairs (u, v) with 1 u mand 1 v n. The ((i, k), (j, l))-entry of A⌦Bis ai,jbk,l.

Note: To write A ⌦ B explicitly as a matrix,one must define some specific order on the indexset. The choice of order is essentially irrelevant,however.

It is easy to compute that

trace(A⌦B) = trace(A)trace(B) .

Also, if C and D are m ⇥ m and n ⇥ n matri-ces, respectively, it follows from the definition ofmatrix multiplication that

(A⌦B)(C ⌦D) = AC ⌦BD .

Now let X and Y be representations of G, af-fording characters � and , respectively. DefineZ by Z(g) = X (g)⌦ Y(g).

It follows from the above remarks that Z is arepresentation, and that the character a↵ordedby Z is the product � . This shows that prod-ucts of characters are characters.

Note: If either � or is reducible, it is easyto see that � is reducible. Even if � and areirreducible, however, their product is usually notirreducible.

EXERCISE (2.3): Suppose � is the unique mem-ber of Irr(G) having degree d for some integer d.Show that �(x) = 0 for all elements x 2 G�G0.

EXERCISE (2.4): Let � be a character of G,and let X a↵ord �. Define the function � on Gby �(g) = det(X (g)). Show that � is a linearcharacter of G and that it does not depend onthe choice of the representation X a↵ording �.

Notation: � = det(�). Also, the determinantalorder o(�) is the order of det(�) in the group oflinear characters of G.

Chapter 3:

CHARACTER VALUES

Let X be a representation of G with degree dand let g 2 G have order n. Then X (g)n = I,the d⇥ d identity matrix.

It follows by linear algebra that the matrix X (g)is similar to a diagonal matrix whose diagonalentries are n th roots of unity.

If X a↵ords the character �, then

�(g) = trace(X (g)) = ✏1 + · · ·+ ✏d ,

where the ✏i for 1 i d are n th roots of unity.

Also

�(g�1) = trace(X (g)�1) = ✏1 + · · ·+ ✏d = �(g) ,

where the overbar is complex conjugation.

THEOREM: Let � be a character of a group G,

where � is a↵orded by a representation X , and

let g 2 G. Then:

(a) |�(g)| �(1).

(b) |�(g)| = �(1) i↵ X (g) is a scalar matrix.

(c) �(g) = �(1) i↵ X (g) is the identity matrix.

COROLLARY: Let � be a character of G. Then

{g 2 G | �(g) = �(1)} is the kernel of every

representation a↵ording �. In particular, this

subset is a normal subgroup of G.

Notation: We write

ker(�) = {g 2 G | �(g) = �(1)} .

This normal subgroup is called the kernel of �,and � is faithful if ker(�) = 1.

EXERCISE (3.1): Let N / G and 2 Irr(G/N).Define � : G ! C by setting �(g) = (Ng).Show that � 2 Irr(G) and that N ✓ ker(�).Also, show that the map 7! � from Irr(G/N)into Irr(G) defines a bijection from Irr(G/N) tothe set {� 2 Irr(G) | N ✓ ker(�)}.

Note: It is customary to identify with �, sowe usually pretend that

Irr(G/N) = {� 2 Irr(G) | N ✓ ker(�)} .

EXERCISE (3.2): Show that\

{ker(�) | � 2 Irr(G)} = 1 ,

the trivial subgroup of G.

EXERCISE (3.3): Let N / G. Show that

N =\

{ker(�) | N ✓ ker(�)} .

Thus, knowing the irreducible characters of Gdetermines the set of all normal subgroups of G.

COROLLARY: Let � be a character of G af-

forded by X . Then {g 2 G | |�(g)| = �(1)}is the preimage in G of the normal subgroup of

X (G) consisting of scalar matrices. In particu-

lar, this set is a normal subgroup of G.

Notation: We write

Z(�) = {g 2 G | |�(g)| = �(1)} .

This normal subgroup is the center of �.

EXERCISE (3.4): Let � be a character of G.Prove that Z(�)/ker(�) is a cyclic subgroup ofZ(G/ker(�)).

EXERCISE (3.5): If � 2 Irr(G), show that

Z(�)/ker(�) = Z(G/ker(�)) .

EXERCISE (3.6): Let P be a p-group for someprime p. Show that P has a faithful irreduciblecharacter if and only if Z(P ) is cyclic.

EXERCISE (3.7): Let � 2 Irr(G), where G is asimple group of even order. Show that �(1) 6= 2.

HINT: Let g 2 G have order 2. Consider thevalues of �(g) and det(�)(g).

EXERCISE (3.8): Let G be a simple group andsuppose � 2 Irr(G) has degree 3. Compute �(g)for an element g 2 G of order 2.

EXERCISE (3.9): Repeat the above if � has de-gree 4.

EXERCISE (3.10): If H ✓ G and � is a charac-ter of G, it is easy to see that the restriction of� to H, denoted �H , is a character of H. Showthat H ✓ Z(�) i↵ �H has the form e�, where eis a positive integer and � is a linear characterof H.

EXERCISE (3.11): Let H ✓ G and let � be acharacter of G such that the restriction �H isirreducible. Show that CG(H) ✓ Z(�).

Chapter 4:

ORTHOGONALITY

G acts by right multiplication on C[G], so eachelement g 2 G determines a linear transforma-tion on C[G].

We get a C-representation of G by choosing abasis for C[G].

The character a↵orded by this representation isthe regular character of G, denoted ⇢ or ⇢G.

To compute ⇢, we can use the basis G for C[G].

If R is the corresponding representation, R(g)is an |G| ⇥ |G| matrix with rows and columnsindexed by the elements of G.

If x, y 2 G, then the entry of R(g) at position(x, y) is 1 if xg = y and 0 otherwise.

If g 6= 1, therefore, all diagonal entries of R(g)are 0.

We thus have

⇢(g) =

⇢0 if g 6= 1|G| if g = 1 .

Next, we wish to express the character ⇢ in termsof Irr(G).

Fact:⇢ =

X

�2Irr(G)

�(1)� .

This is proved using a di↵erent basis for C[G],one compatible with the decomposition as a di-rect sum of matrix rings. (We omit the proof.)

As a check (but not a proof) observe that

|G| = ⇢(1) =X

�2Irr(G)

�(1)2 .

Recall: C[G] is a direct sum of matrix rings. Thesummands correspond to members of Irr(G).

For � 2 Irr(G), let e� be the unit element of thedirect summand of C[G] corresponding to �.

For �, 2 Irr(G), we have

e�e =

⇢e� if � = 0 if � 6= .

The e� are the central idempotents of C[G].

Note:P

�2Irr(G)

e� = 1.

Let X� be the representation obtained by theprojection of C[G] onto the direct summand cor-responding to �, and note that X� a↵ords �.

Then X�(e�) is the �(1) ⇥ �(1) identity matrixand X�(e ) = 0 if 6= �.

For arbitrary u 2 C[G], we have

X�(ue ) = X�(u)X�(e ) =⇢X�(u) if � = 0 if � 6= .

Let X� be the representation obtained by theprojection of C[G] onto the direct summand cor-responding to �, and note that X� a↵ords �.

Then X�(e�) is the �(1) ⇥ �(1) identity matrixand X�(e ) = 0 if 6= �.

For arbitrary u 2 C[G], we have

X�(ue ) = X�(u)X�(e ) =⇢X�(u) if � = 0 if � 6= .

Thus

�(ue ) =

⇢�(u) if � = 0 if � 6= .

Write e� =Pg2G

agg with complex coe�cients ag.

Then ag is the coe�cient of 1 in g�1e�.

We have

|G|ag = ⇢(g�1e�) =X

2Irr(G)

(1) (g�1e�)

= �(1)�(g�1) .

Thus

ag =�(1)�(g)

|G| .

Now compare the coe�cients of 1 in e� = (e�)2.

�(1)2

|G| =X

g2G

(ag�1)(ag)

=X

g2G

✓�(1)�(g)

|G|◆

�(1)�(g)

|G|

!

=�(1)2

|G|2X

g2G

�(g)�(g) .

Thus 1 =1

|G|X

g2G

�(g)�(g) =1

|G|X

g2G

|�(g)|2.

Now suppose �, 2 Irr(G) are di↵erent, andcompare the coe�cients of 1 in 0 = e�e .

Writing e� =P

agg and e =P

bgg, we get

0 =X

g2G

(ag�1)(bg)

=X

g2G

✓�(1)�(g)

|G|◆

(1) (g)

|G|

!.

ThusX

g2G

�(g) (g) = 0.

We now have the first orthogonality relation:

1

|G|X

g2G

�(g) (g) =

⇢1 if � = 0 if � 6= ,

for �, 2 Irr(G).

Notation: For class functions ↵ and � on G, wewrite

[↵,�] =1

|G|X

g2G

↵(x)�(x) .

This is an inner product on the space cf(G) ofclass functions, and Irr(G) is an orthonormal ba-sis for this space.

Note: The form [., .] is linear in the first variableand conjugate-linear in the second.

Also [�,↵] = [↵,�] and [↵�, �] = [↵,��].

If ↵ is an arbitrary class function on G, we knowthat ↵ is a linear combination of Irr(G).

The coe�cient of � in this linear combination is[↵,�], so we can write

↵ =X

�2Irr(G)

[↵,�]� .

Observe that [⇢,�] = �(1) since ⇢ vanishes onnonidentity group elements and ⇢(1) = |G|.Thus ⇢ =

P�(1)�, which is a fact we stated

previously, but did not prove.

Of course, this is not a proof.

COROLLARY: Let ↵ be a nonzero class func-

tion on G. Then ↵ is a character i↵ [↵,�] is a

nonnegative integer for all � 2 Irr(G).

COROLLARY: Let � be a character of G. Then

� is irreducible i↵ [�,�] = 1.

EXERCISE (4.1): Let H ✓ G and � 2 Irr(G),and recall that �H is the character ofH obtainedby restricting � to H. Show that [�H ,�H ] |G : H| with equality i↵ � has the value 0 on allelements of G�H.

EXERCISE (4.2): If G is abelian, show that[�,�] � �(1) for all (not necessarily irreducible)characters � of G.

EXERCISE (4.3): Let H ✓ G, with H abelian,and let � 2 Irr(G). Show that �(1) |G : H|.

EXERCISE (4.4): Let � 2 Irr(G), where G is anontrivial group. Show that there is a noniden-tity element g 2 G such that �(g) 6= 0.

EXERCISE (4.5): Let H < G with H is abelian.If � 2 Irr(G) with |G : H| = �(1), show that Hcontains a nonidentity normal subgroup of G.

EXERCISE (4.6): Let G act on a set ⌦. Let⇡ be the corresponding permutation character,so ⇡(g) = |{t 2 ⌦ | t.g = t}|. Show that ⇡ is acharacter of G and [⇡, 1G] is the number of orbitsof G on ⌦.

EXERCISE (4.7): Let G = A5, the alternatinggroup, and let ⇡ be the permutation characterof the natural action of G on five points. Showthat � = ⇡� 1G is an irreducible character of Ghaving degree 4.

EXERCISE (4.8): Again let G = A5 and observethat G acts by conjugation on the set of six Sy-low 5-subgroups of G. Let � be the permutationcharacter of this action. Show that = � � 1Gis an irreducible character of G having degree 5.

It is customary to display the irreducible char-acters of a group G in a character table.

This is a square array of complex numbers withrows indexed by Irr(G) and columns indexed bythe set of classes of G. The entry in a given rowand column is the common value of the givencharacter on the elements of the given class.

Usually, the first column of the character tablecorresponds to the class of the identity in G, andthus we see the irreducible character degrees ofG by reading down this column.

It is also customary for the first row of the char-acter table to correspond to the principal char-acter 1G, and thus the first row consists of 1s.

EXERCISE (4.9): Show that knowing the char-acter table of G determines whether or not G issolvable or G is nilpotent.

We introduce some temporary notation now.

Write Irr(G) = {�1,�2, . . . ,�k} and let gj be arepresentative of the j th conjugacy class of G.

The character table of G is thus the matrix Xwith (i, j)-entry equal to �i(gj).

Write dt to denote the size of the t th conjugacyclass of G. Then by the first orthogonality rela-tion, we have:

�i,j = [�i,�j ] =1

|G|kX

t=1

�i(gt)dt�j(gt) .

We can rewrite this in matrix notation:

I =1

|G|XDXt ,

where D = diag(d1, d2, . . . , dk).

Then

I =1

|G|XtXD .

We have|G|I = XtXD

so

|G|�i,j =kX

t=1

�t(gi)�t(gj)dj .

Now |G|/dj = |CG(gj)|, and thus for x, y 2 G,we have

X

�2Irr(G)

�(x)�(y) =

⇢ |CG(x)| if x ⇠ y0 otherwise,

where x ⇠ y means that x and y are conjugate.

This is the second orthogonality relation, alsocalled “column orthogonality”. Since it is a con-sequence of the first orthogonality relation, or“row orthogonality”, it gives no additional in-formation about the characters of G. It is, how-ever, useful, especially when constructing char-acter tables.

A consequence is that no two columns of thecharacter table can be identical.

If x, y 2 G are not conjugate, therefore, thereexists � 2 Irr(G) such that �(x) 6= �(y).

Chapter 5:

INTEGRALITY

An algebraic integer is a complex number thatis a root of a monic polynomial in Z[x].

Examples: Elements of Z, roots of unity andthings like n

pm where m,n 2 Z.

Fact: The algebraic integers are a subring of C.

COROLLARY: Character values are algebraic

integers.

Fact: Rational algebraic integers are in Z.

Fact: Suppose ui 2 C for 1 i n, and let Rbe the set of Z-linear combinations if the ui. IfR is closed under multiplication, then all ui arealgebraic integers.

Now let � 2 Irr(G) and let X a↵ord �.

Let z 2 Z(C[G]) so X (z) commutes with all ma-trices in X (C[G]). Since this is a full matrix ring,we see that X (z) is a scalar matrix.

Thus X (z) = !I for some complex number !,and we have

�(z) = trace(X (z)) = trace(!I) = !�(1) .

It follows that ! depends on � but not on X ,and we write !� for !.

We have !�(z) = �(z)/�(1).

Also X (z) = !�(z)I, and it follows that !� is analgebra homomorphism Z(C[G]) ! C.

The function !� is sometimes referred to as acentral character.

Recall that the conjugacy class sums in C[G]form a basis for Z(C[G]) so the central charactersare determined by their values on class sums.

Let K be a class of G and write bK to denote thesum of the elements of K, so bK 2 Z(C[G]).

We have �( bK) = |K|�(g), where g 2 K.

Then

!�( bK) =�( bK)

�(1)=

|K|�(g)�(1)

.

Now suppose K and L are classes of G. ThenbKbL is a linear combination of class sums, so wecan write

bKbL =X

M

aK,L,McM ,

where M runs over the classes of G.

The coe�cient of each group element on the leftof the above equation is a nonnegative integer, sothe coe�cients aK,L,M must also be nonnegativeintegers.

We have

!�( bK)!�(bL) =X

aK,L,M!�(cM) .

As K runs over the classes of G, we see thatthe set of Z-linear combinations of the complexnumbers !�( bK) is closed under multiplication.

These numbers, therefore, are algebraic integers.

THEOREM: Let g 2 G and � 2 Irr(G), and let

K be the class of g inG. Then !�( bK) =|K|�(g)�(1)

is an algebraic integer.

THEOREM: �(1) divides |G| for all � 2 Irr(G).

Proof: For each class K of G, let gK be an ele-ment of K. By the first orthogonality relation

|G| =X

g2G

�(g)�(g) =X

K

|K|�(gK)�(gK) ,

where the second sum runs over the classes of G.

Since |K|�(gK) = !�( bK)�(1), we have

|G| = �(1)X

K

!�( bK)�(xK) .

so the rational number |G|/�(1) is an algebraicinteger, and hence lies in Z.

Note: A related argument can be used to provethe following stronger result.

Fact: Let � 2 Irr(G). Then |G : Z(G)| is divisi-ble by �(1).

EXERCISE (5.1): Let G be a p-group, whereG0 = Z(G) has order p. Prove the following.

(a) Each noncentral class of G has size p.

(b) G has exactly p � 1 nonlinear irreduciblecharacters.

(c) The average of �(1)2 for nonlinear� 2 Irr(G) is |G|/p.

(d) �(1)2 |G|/p for all � 2 Irr(G), and thus�(1)2 = |G|/p for nonlinear � 2 Irr(G).

Note: Groups of this type are extraspecial.

We begin work now toward a proof of Burnside’sfamous paqb-theorem.

LEMMA: Let � be a character of G, where |G| =n, and let g 2 G. Suppose |�(g)| < �(1). Let

� = Gal(Qn/Q), where Qn is the n th cyclotomic

field. Then |�(g)�| < �(1) for all � 2 �.

Proof: We know that �(g) is a sum of �(1) rootsof unity in Qn. These are not all equal since|�(g)| < �(1). It follows that �(g)� is also asum of �(1) roots of unity that are not all equal.The result follows.

THEOREM (Burnside): Let � 2 Irr(G) and g 2G, and assume that �(1) is relatively prime to

the size of the class K containing g. Then either

�(g) = 0 or g 2 Z(�).

Proof: Write a�(1) + b|K| = 1 for integers a, b.

Now ! = �(g)|K|/�(1) is an algebraic integer.

Then�(g)

�(1)=

�(g)

�(1)

�a�(1) + b|K|� = a�(g) + b!

is an algebraic integer.

Let � = Gal(Qn/Q), where n = |G|, and let

z =Y

�2�

✓�(g)

�(1)

◆�

,

so z is an algebraic integer in Q. Then z 2 Z.Assume now that g 62 Z(�). Then |�(g)| < �(1),so by the lemma, |�(g)�| < �(1) for all � 2 �.

Each factor of z has absolute value less than 1,so |z| < 1. Since z 2 Z, we have z = 0. Then�(g)� = 0 for some �, so �(g) = 0.

THEOREM (Burnside): Let G be nonabelian

and simple. Then no nontrivial conjugacy class

of G has prime power size.

Proof: Suppose |K| is a power of a prime p,where K is a nontrivial class, and let g 2 K,so g 6= 1.

If � 2 Irr(G), then Z(�) / G. If Z(�) = G, then� is linear. But G0 = G, so � is principal.

Thus if � 2 Irr(G) is nonprincipal, Z(�) = 1.

By the previous theorem, �(g) = 0 for all non-principal � where p does not divide �(1).

We have

0 = ⇢(g) =X

�2Irr(G)

�(1)�(g)

Separate � = 1G and sum only over � for whichp divides �(1).

This yields 0 = 1 + p↵, where ↵ is an algebraicinteger. Since ↵ = �1/p 2 Q, this is a contra-diction.

THEOREM (Burnside): If |G| has at most two

prime divisors, then G is solvable.

Proof: Let G be a minimal counterexample, soG is nonabelian. If N / G with 1 < N < G, thenN and G/N are solvable, so G is solvable, whichis a contradiction. Thus G is simple.

Let P > 1 be a Sylow subgroup of G. Choosez 2 Z(P ) with z 6= 1. Let K be the class ofz in G, so |K| = |G : CG(z)|, and this divides|G : P |, which is a prime power, by hypothesis.This contradicts the previous theorem.

EXERCISE (5.2): Suppose that no prime otherthan p or q divides the degree of any memberof Irr(G). Show that G cannot be a nonabeliansimple group.