Solutions to Exercises of Representations and Characters of Groups by Gordon James and Martin Liebeck

7/28/2019 Solutions to Exercises of Representations and Characters of Groups by Gordon James and Martin Liebeck

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Solutions to Exercises of Representations and Characters of Groups by Gordon

James and Martin Liebeck

Solutions by Jay Taylor† (University of Aberdeen)Last Updated 22/07/2011

This document contains solutions to the exercises in the book “Rep-

resentations and Characters of Groups” written by Gordon James and

Martin Liebeck. Throughout we will freely use the notation used in

this book. All non explicit references will refer to results from the

book. A table of contents is given on the next page and at the

head of each chapter/section all exercises from that chapter/section

which have a complete solution are listed. There may be solutions

to other exercises in this document, (which are not contained in this

list), but such exercises are considered to have only a partial solution.

† email [email protected].

mailto:[email protected]

mailto:[email protected]



Contents

3 Group Representations 1

4 F G-modules 6

5 F G-submodules and reducibiity 10

6 Group algebras 14

7 F G-homomorphisms 18

8 Maschke’s Theorem 22

9 Schur’s Lemma 26

10 Irreducible modules and the group algebra 31

11 More on the group algebra 35

12 Conjugacy classes 38

13 Characters 41

14 Inner products of characters 45

15 The number of irreducible characters 47

16 Character tables and orthogonality relations 49

17 Normal subgroups and lifted characters 53

18 Some elementary character tables 74

19 Tensor products 76

i



Contents ii

20 Restriction to a subgroup 81

21 Induced modules and characters 88

22 Algebraic integers 95

23 Real representations 106

25 Characters of groups of order pq 116

26 Characters of some p -groups 127

27 Character table of the simple group of order 168 143



Chapter 3. Group Representations

Exercise 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Exercise 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Exercise 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Exercise 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Exercise 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 3.1. Assume ρ is a representation of G then because am = 1 we must have

In = 1ρ = amρ = (aρ)m = Am. Now if Am = In then we have (ai )ρ = Ai for all i ∈ Z.

Therefore

(ai a j )ρ = ai + j ρ = Ai + j = Ai A j = (ai ρ)(a j ρ).

Exercise 3.2. We can see that

A3 = I2 B3 =

1 0

0 e 2πi/3

3=

1 0

0 e 2πi

= I2,

C 3 =

0 1

−1 −1

3=

0 1

−1 −1

0 1

−1 −1

0 1

−1 −1

=

−1 −1

1 0

0 1

−1 −1

= I2.

Hence ρ1, ρ2 and ρ3 are representations by the above. Clearly ρ1 is not faithful as Gρ1 = I2.

However ρ2 and ρ3 are faithful. We can see in the calculation of C 3 that C 2 = I2, hence

ρ3(g ) = I2

⇒g = 1. Also B2

= I2 as e 4πi/3

= 1 and so ρ2 is faithful.

Exercise 3.3. Now we know that every element of G can be written ai b j for some 0 i

n − 1 and 0 j 1, using the relations. We certainly have that (1)n = (−1)2 = (1) and

(−1)(1)(−1) = (1), so defining the map ρ : G → GL(1, F ) by (ai b j )ρ = (−1) j we have ρ

is a representation by Example 1.4 and aρ = (1), bρ = (−1) as required.

Exercise 3.4. We are showing that the definition of equivalent representations is an equiv-

alence relation.

Reflexive clearly we have gρ = I−1n (gρ)In for all g ∈ G so ρ is equivalent to ρ.

1



Chapter 3 2

Symmetric assume ρ is equivalent to σ then there exists T such that, for all g ∈ G,

we have gσ = T −1(gρ)T . However this gives us gρ = (T −1)−1(gσ)T −1 and so σ is

equivalent to ρ.

Transitive assume ρ is equivalent to σ and σ is equivalent to τ then there exists T, S

such that, for all g ∈ G, we have gσ = T −1(gρ)T and gτ = S−1(gσ)S. Then this

gives us that, for all g ∈ G, we have gτ = S−1(T −1(gρ)T )S = (T S)−1(gρ)(T S). In

other words ρ is equivalent to τ .

Exercise 3.5. We first check that the matrices satisfy the relations of G. So,

A6 = e iπ/3 0

0 e −iπ/3

6

= e 2iπ 0

0 e −2iπ = I2

B2 =

0 1

1 0

0 1

1 0

= I2,

B−1AB =

0 1

1 0

e iπ/3 0

0 e −iπ/3

0 1

1 0

,

=

0 e −iπ/3

e iπ/3 0

0 1

1 0

,

= e −iπ/3 0

0 e iπ/3 ,

= A−1

C 2 =

1/2

√ 3/2

−√ 3/2 1/2

1/2

√ 3/2

−√ 3/2 1/2

=

−1/2

√ 3/2

−√ 3/2 −1/2

C 4 =

−1/2

√ 3/2

−√ 3/2 −1/2

−1/2

√ 3/2

−√ 3/2 −1/2

=

−1/2 −√

3/2√ 3/2 −1/2

C 6 = −1/2 −√ 3/2

√ 3/2 −1/2 −1/2√

3/2

−√ 3/2 −1/2 = 1 0

0 1D2 =

1 0

0 −1

1 0

0 −1

=

1 0

0 1

D−1CD =

1 0

0 −1

1/2

√ 3/2

−√ 3/2 1/2

1 0

0 −1

=

1/2 −√

3/2√ 3/2 1/2

= C −1

By Example 1.4 the above calculations are enough to show that ρ1 and ρ4 are representations

of D12. Let ar b s , ap b q ∈ D12 then we have, (recalling that the calculations above show that



Chapter 3 3

the matrices respect the rules of D12),

(ar

b s

ap

b q

)ρ2 = (ar

−p

b s +q

)ρ2 (ar

b s

ap

b q

)ρ3 = (ar

−p

b s +q

)ρ,= A3(r −p )(−B)s +q = (−A)r −p Bs +q ,

= A3r (−1)s A−3p Bs (−B)q = (−A)r (−1)−p A−p Bs Bq ,

= A3r (−B)s A3p (−B)q = (−A)r Bs (−A)p Bq ,

= (ar b s )ρ2(ap b q )ρ2 = (ar b s )ρ3(ap b q )ρ3.

Recall that (−1)−p = (−1)p (−1)−2p = (−1)p [(−1)2]−p = (−1)p . Therefore ρ2 and ρ3 are

representations of D12.

We first of all consider the other powers of A and C . We can see the following holdtrue

A3 =

e iπ 0

0 e −iπ

= −I2 C 3 =

−1/2

√ 3/2

−√ 3/2 −1/2

1/2

√ 3/2

−√ 3/2 1/2

= −I2,

which also gives us C 5 = C 3C 2 = −C 2. Now if ai b j ∈ ker(ρ1) then we have (ai b j )ρ1 = I2

but this gives us Ai B j = I2 ⇒ Ai = B− j . However no power of A, other than 6, is equal

to I2 or B, therefore we must have the kernel is trivial and the representation is faithful. A

similar argument shows that if ai b j ∈ ker(ρ4) then we must have C i = D− j but no power of

C , other than 6, is equal to I2 or D so the representation is faithful. Clearly ρ2 is not faithfulas a2ρ2 = A6 = I2. Finally we also have ρ3 is not faithful as a3ρ3 = (−A)3 = −A3 = I2.

We have that the representations ρ1 and ρ4 are in fact equivalent. This is a little

more enlightening if we write e iπ/3 = 12

+√ 32 i . It is then maybe easier to see that A is

the diagonalised form of C . What are the eigenvalues of C ? Well this has characteristic

polynomial λ2−λ+ 1, whose solutions are precisely λ1 = e iπ/3 = 12

+√ 32

i and λ2 = e −iπ/3 =12

−√ 32 i . What are the corresponding eigenvectors?

1/2√

3/2

−√ 3/2 1/2 v 1

v 2 = 1

2 +

√ 3

2 i v 1

v 2 ⇒ −

√ 3/2i

√ 3/2

−√ 3/2 −√ 3/2i v 1

v 2 = 0

0 ,

⇒

i −1

1 i

v 1

v 2

=

0

0

.

Now the second line is just −i times the first and so we have v 2 = i v 1. Similarly for the



Chapter 3 4

second we get v 2 = −i v 1. Then let T be the matrix

T = 1 1−i i ⇒ T −1 = 1

2i i −1

i 1 .

Using this we see that

T −1(bρ4)T =1

2i

i −1

i 1

1 0

0 −1

1 1

−i i

,

=1

2i

i 1

i −1

1 1

−i i

,

= 12i

0 2i 2i 0

,

= bρ1.

Similarly we have

T −1(aρ4)T =1

2i

i −1

i 1

1/2

√ 3/2

−√ 3/2 1/2

1 1

−i i

,

= 12i 1/2i + √ 3/2 −1/2 + √ 3/2i

1/2i − √ 3/2 1/2 +

√ 3/2i

1 1−i i ,

=1

2i

i +

√ 3 0

0 i − √ 3

,

= aρ1.

As the matrices are equivalent on the generators it is enough to see that in fact ρ4 is

equivalent to ρ1. None of the other representations are equivalent. This can be seen by

checking the characteristic polynomial of aρ2 and aρ3 and seeing that they are different,

(recall equivalent matrices have the same characteristic polynomial).

Exercise 3.6. We define the following permutation matrices in GL(3, F )

A =

0 1 0

0 0 1

1 0 0

B =

0 1 0

1 0 0

0 0 1

.



Chapter 3 5

We check that these satisfy the relations of D8, so

A3 =0 1 0

0 0 1

1 0 0

0 1 00 0 1

1 0 0

0 1 00 0 1

1 0 0

=0 0 1

1 0 0

0 1 0

0 1 00 0 1

1 0 0

=1 0 0

0 1 0

0 0 1

,

B2 =

0 1 0

1 0 0

0 0 1

0 1 0

1 0 0

0 0 1

=

1 0 0

0 1 0

0 0 1

,

B−1AB =

0 1 0

1 0 0

0 0 1

0 1 0

0 0 1

1 0 0

0 1 0

1 0 0

0 0 1

=

0 0 1

0 1 0

1 0 0

0 1 0

1 0 0

0 0 1

=

0 0 1

1 0 0

0 1 0

= A−1

By Example 1.4 this means that the map ρ : D8 → GL(3, F ) given by (ai b j )ρ = Ai B j , for

all 0 i 3 and 0 j 1 is a representation. Is it faithful? Well if ai b j ∈ ker(ρ) then we

have Ai B j = I2 ⇒ Ai = B− j but no power of A is equal to I3 or B, except for A3 and so

our representation is faithful.

Exercise 3.7. Assume ρ is a representation of degree 1 then it is a map from G to

GL(1, F ) ∼= F . Therefore G/ ker(ρ) ∼= im(ρ) ⊆ F and as F is a field we have G/ ker(ρ) is

abelian.

Exercise 3.8. If we have (gρ)(hρ) = (hρ)(gρ) then this gives us that (gh)ρ = (hg )ρ.

However this only implies gh = hg if ρ is an isomorphism. A classic counter-example

is the trivial representation ρ : G → GL(1, F ) given by gρ = I for all g ∈ G. Clearly

(gρ)(hρ) = (hρ)(gρ) for all g, h ∈ G but G isn’t necessarily abelian so we don’t necessarily

have gh = hg .



Chapter 4. F G-modules

Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 4.1. Let u 1 = v 1 + v 2 + v 3, u 2 = v 1 − v 2 and u 3 = v 1 − v 3. Then the change of

basis matrix from B 1 = v 1, v 2, v 3 to B 2 = u 1, u 2, u 3 is given as followsv 1

v 2

v 3

=

1

3

1 1 1

1 −2 1

1 1 −2

u 1

u 2

u 3

Now the matrices of S3 = 1, (12), (13), (23), (123), (132) with respect to the basis B 1are given by the standard permutation matrices and so

[1]B1=

1 0 0

0 1 0

0 0 1

[(12)]B1 =

0 1 0

1 0 0

0 0 1

[(13)]B1

=

0 0 1

0 1 0

1 0 0

,

[(23)]B1=

1 0 0

0 0 1

0 1 0

[(123)]B1

=

0 1 0

0 0 1

1 0 0

[(132)]B1

=

0 0 1

1 0 0

0 1 0

.

So to calculate the matrices in the new basis B 2 we just multiply by the change of basis

matrix on the right

[1]B2 =

1

31 1 1

1 −1 01 0 −1

1 0 0

0 1 00 0 1

1 1 1

1 −2 11 1 −2

= 1 0 0

0 1 00 0 1

,

[(12)]B2=

1

3

1 1 1

1 −1 0

1 0 −1

0 1 0

1 0 0

0 0 1

1 1 1

1 −2 1

1 1 −2

=

1 0 0

0 −1 0

0 −1 1

,

[(13)]B2=

1

3

1 1 1

1 −1 0

1 0 −1

0 0 1

0 1 0

1 0 0

1 1 1

1 −2 1

1 1 −2

=

1 0 0

0 1 −1

0 0 −1

,

6



Chapter 4 7

[(23)]B2=

1

3

1 1 1

1 −1 0

1 0 −1

1 0 0

0 0 1

0 1 0

1 1 1

1 −2 1

1 1 −2

=

1 0 0

0 0 1

0 1 0

,

[(123)]B2=

1

3

1 1 1

1 −1 0

1 0 −1

0 1 0

0 0 1

1 0 0

1 1 1

1 −2 1

1 1 −2

=

1 0 0

0 −1 1

0 −1 0

,

[(132)]B2=

1

3

1 1 1

1 −1 0

1 0 −1

0 0 1

1 0 0

0 1 0

1 1 1

1 −2 1

1 1 −2

=

1 0 0

0 0 −1

0 1 −1

.

What we notice about the matrices in the basisB 2 is that they fall into block form, namely

a 1 × 1 block and a 2 × 2 block. In Chapter 5 we will come to see that this means the

representation is reducible.

Exercise 4.2. We check the definition for V to be an F G module. The following statements

are for all v ∈ V .

(a) We clearly have v g ∈ V as ±v ∈ V because V is a vector space.

(b) Let g, h ∈ Sn then we have

v (gh) = v if gh is even,−v if gh is odd.

=

v if g is even and h is even,

−v if g is odd and h is even,

−v if g is even and h is odd,

v if g is odd and h is odd.

= v g if h is even,

−v g if h is odd.

= (v g )h.

(c) Recall 1 ∈ Sn is even so we clearly have v 1 = v .



Chapter 4 8

(d) Let λ ∈ F then

(λv )g =λv if g is even,

−λv if g is odd.= λ

v if g is even,−v if g is odd.

= λ(v g ).

(e) For all u ∈ V we have

(u + v )g =

u + v if g is even,

−u − v if g is odd.

= u if g is even,

−u if g is odd.

+ v if g is even,

−v if g is odd.

= ug + v g.

Therefore V certainly is an F G-module.

Exercise 4.3. We check the conditions of Proposition 4.6 to show that this is indeed an

RQ8-module.

(a) Clearly we have v i g ∈ V for all 1 i 4 and g ∈ G.

(b) This is clear as we’re defining the action of of Q8 on V by acting one letter at a time,

i.e. v k (ab ) = (v k a)b .

(c) We have that 1 = a4 and so we can see that the following hold

v 11 = v 1a4 = −v 2a3 = v 1a2 = −v 2a = v 1,

v 21 = v 2a4 = −v 1a3 = v 2a2 = −v 1a = v 2,

v 31 = v 3a4 = −v 4a3 = −v 3a2 = v 4a = v 3,

v 41 = v 4a4 = v 3a3 = −v 4a2 = −v 3a = v 4.

(d) This is clear by how we’re defining the module.

Therefore, by Proposition 4.6, we have that V is an RQ8 module.

Exercise 4.4. Let E ij be the standard basis for the vector space of all matrices over F .

Then for some aij ∈ F and σ ∈ Sn we have

A =

ni,j =1

aij E ij B =

ni,j =1

aij E σ(i ) j .



Chapter 4 9

Now consider the matrix P defined by P =n

k =1 E σ(k )k . We can see that this is a permu-

tation matrix, as there is only one non-zero entry in each row and column. Now, this gives

us that

P A =

nk =1

E σ(k )k

ni ,j =1

ai j E ij =

ni,j,k =1

aij E σ(k )k E ij =

ni,j,k =1

δik aij E σ(k ) j =

ni,j =1

aij E σ(i ) j = B.

Similarly if τ ∈ Sn such that B =n

i ,j =1 ai j E iτ ( j ), then we define a permutation matrix

P =n

k =1 E kτ (k ) which gives us

AP =

n

i,j =1aij E ij

n

k =1E kτ (k ) =

n

i,j,k =1aij E ij E kτ (k ) =

n

i,j,k =1δ jk aij E iτ (k ) =

n

i ,j =1ai j E iτ ( j ) = B.



Chapter 5. F G-submodules and reducibiity

Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Exercise 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Exercise 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Exercise 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercise 5.1. It is clear that the properties of Proposition 4.6 hold and so V is an F G-

module. Now V is of dimension 2 so there is only one possible non-trivial F G submoduleof dimension 1. Consider U = span(α, β) ⊆ V , when is this an F G submodule of V ?

It’s clearly a 1-dimensional subspace of V as (0, 0) ∈ U and for all k 1, k 2 ∈ F we have

k 1(α, β) + k 2(α, β) = (k 1 + k 2)(α, β) ∈ U .

What about the action of G on U ? Well we have (α, β)1 ∈ U and (α, β)a = ( β, α) ∈U if ( β, α) = k (α, β) for some k ∈ F . We can see that letting k = ±1 turns U

into an F G module and so U is either span(1, 1) or span(1, −1). Hence U is an

F G submodule of V . Now as vector spaces are classified by their dimension we have

0

, span

(1, 1)

, span

(1,

−1)

and V are all the F G submodules of V .

Exercise 5.2. Assume ρ : G → GL(n, F ) is a reducible representation, then the corre-

sponding F G-module F n given by v g = v (gρ) for all v ∈ F n and g ∈ G is reducible. In

other words there exists a proper submodule W of V . Now assume σ : G → GL(n, F ) is

equivalent to ρ, then there exists a matrix T ∈ GL(n, F ) such that gσ = T −1(gρ)T for all

g ∈ G.

We claim that W T is also a proper submodule of F n under σ, in other words (w T )g =

(w T )(gσ) ∈ W T for all w T ∈ W T and g ∈ G. This is clear to see as for g ∈ G and

w ∈ W we have

(w T )g = (w T )(gσ) = (w T )(T −1(gρ)T ) = w (gρ)T ⊆ W T

because w (gρ) ∈ W . Therefore W T is a submodule of F n under σ. Note that because T

is invertible we must have the dimension of W T is the same as the dimension of W . In

other words W = 0 or F n implies W T = 0 or F n and we’re done.

Exercise 5.3. We break the four representations down individually. In each case we’re

looking for a 1-dimensional submodule of C2, lets call it W with basis element w = (α, β).

10



Chapter 5 11

Recall that e ikπ/3 = e −ikπ/3 unless e ikπ/3 is real, in other words k = 3n for some n ∈ Z.

(a) Let H = b G the cyclic subgroup of order 2. Then C2 is also a CH -module.

Consider U to be a 1-dimensional submodule of C2 then by question 1 we haveeither (1, 1) ∈ U or (1, −1) ∈ U . However we have (1, 1) and (1, 1)a are linearly

independent, as are (1, −1) and (1, −1)a. Therefore dim U 2 and so U cannot be

a 1-dimensional submodule. Hence ρ1 is an irreducible representation.

(b) We have the actions of a and b are

w a =

α β −1 0

0 −1

=−α − β

= −w ∈ W,

w b = α β 0 11 0

= β α ∈ W ⇒ β = kβ,

for some k ∈ C. Therefore W = span(1, 1) and W = span(1, −1) are submod-

ules of C2, which means ρ2 is reducible.

(c) We can see that ρ3 is irreducible by a virtually identical argument to that of ρ1.

(d) We have that ρ4 is equivalent to ρ1, therefore ρ1 irreducible implies ρ4 irreducible.

Exercise 5.4. Let G = (123), (456), (23)(45).

(a) We can see that

a3 = (123)3 = (132)(123) = 1,

b 3 = (456)3 = (465)(456) = 1,

c 2 = (23)(45)(23)(45) = 1,

ab = (123)(456) = (456)(123) = ba,

c −1ac = (23)(45)(123)(23)(45) = (132) = a−1,

c −1bc = (23)(45)(456)(23)(45) = (465) = b −1.

In a similar fashion to the dihedral groups this gives us that any element of G can be

written as ai b j c k with 0 i , j 2 and 0 k 1. Therefore |G| 18. However

|a, b | = 9 and by Lagrange’s theorem we have 9 | |G| but a, b = G and so we

must have |G| = 18.

(b) Let ω be a primitive cube root of unity, i.e. ω3 = 1 and ω = 0, ω2 = 0. Then all the

primitive cubed roots of unity are 1, ω and ω2. Let ε = ω and η = ω2 then we have ρ

is a representation of G. As in previous cases, this is true if the matrices satisfy the



Chapter 5 12

relations of the group. So,

(aρ)3 = ω 00 ω−1

3 = ω3

00 ω−3

= I2 (bρ)3 = ω2

00 ω−2

3 = ω6

00 ω−6

= I2,

(cρ)2 =

0 1

1 0

0 1

1 0

= I2,

(cρ)−1(aρ)(cρ) =

0 1

1 0

ω 0

0 ω−1

0 1

1 0

,

=

0 ω−1

ω 0

0 1

1 0

,

=

ω−1 0

0 ω

,

= (aρ)−1,

(cρ)−1(bρ)(cρ) =

0 1

1 0

ω2 0

0 ω−2

0 1

1 0

,

=

0 ω−2

ω2 0

0 1

1 0

,

=ω−2 0

0 ω2

,

= (aρ)−1.

Hence it is a representation.

(c) We must have that ε and η are cubed roots of unity because ε3 = η3 = 1. If these

are not primitive then, for example, we have ε = 1 or ε2 = 1 which gives us a or

a2 ∈ ker(ρ). Hence ρ is not faithful. Now assume that they are primitive cubed roots

of unity as above. Then we have (aρ)(bρ) = I2

, which means ab ∈

ker(ρ). Therefore

there are no values for which ρ is a faithful representation.

(d) Consider the action of ρ on C2. We want to find a non-trivial, in other words 1-

dimensional, submodule of C2 say U . Consider the cyclic subgroup H = c G of

order 2. Now if U is a CG submodule then it’s also a CH module and by question 1



Chapter 5 13

we must have that U contains either u 1 = (1, 1) or u 2 = (1, −1).

u 1a = 1 1 ε 00 ε−1

= ε ε−1 ∈ U ⇒ ε = 1,

u 1b =

1 1 η 0

0 η−1

=

η η−1

∈ U ⇒ η = 1,

u 2a =

1 −1 ε 0

0 ε−1

=

ε −ε−1

∈ U ⇒ ε = 1,

u 2b =

1 −1

η 0

0 η−1

=

η −η−1

∈ U ⇒ η = 1.

Therefore we have ρ is irreducible as long as (ε, η) = (1, 1). If ε = η = 1 then the

subspace U = span(1, 1) is a submodule of C2.

Exercise 5.5. Let V = 0 and 0g = 0 for all g ∈ G. Then V is a CG-module which is

neither reducible nor irreducible.



Chapter 6. Group algebras

Exercise 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 6.1. Let G = D8 = a, b | a4 = b 2 = 1, b −1ab = a−1.(a) We have x = a + 2a2 and y = b + ab − a2, which gives us

xy = (a + 2a2)(b + ab − a2),

= −2 − a3 + ab + 3a2b + 2a3b,

y x = (b + ab − a2)(a + 2a2),

= ba + 2ba2 + aba + 2aba2 − a3 − 2a4,

= a3b + 2a2b + b + 2a3b − a3 − 2,

= −2 − a3 + b + 2a2b + 3a3b,

x 2 = (a + 2a2)(a + 2a2),

= a2 + 2a3 + 2a3 + 4a4,

= 4 + a2 + 4a3.

(b) Recall that G = 1, a , a2, a3,b,ab,a2b, a3b . Then if z = b + a2b we have

z a = (b + a2b )a = ba + a2ba = a3b + ab = a(a2b + b ) = az ,

z b = (b + a

2

b )b = b

2

+ a

2

b

2

= b

2

+ ba

2

b = b (b + a

2

b ) = bz ,z a2 = (z a)a = a(z a) = a2z ,

za3 = az a2 = a2z a = a3z ,

z ab = az b = abz,

z a2b = azab = a2zb = a2bz ,

z a3b = az a2b = a2z ab = a3z b = a3bz .

Now as the elements of G form a basis for the group algebra, it is clear that z r = r z

14



Chapter 6 15

for all r ∈ CG.

Exercise 6.2. Let G = C 2×

C 2 =

(x i , y j )|

0 i , j 1 and x 2 = y 2 = 1

. Consider the

action of C 2 × C 2 on the group algebra basis, then we have

(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(1, 1) = λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ),

(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(1, y ) = λ2(1, 1) + λ1(1, y ) + λ4(x, 1) + λ3(x, y ),

(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(x, 1) = λ3(1, 1) + λ4(1, y ) + λ1(x, 1) + λ2(x, y ),

(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(x, y ) = λ4(1, 1) + λ3(1, y ) + λ2(x, 1) + λ1(x, y ).

So with respect to the basis G of F G we can see that the matrices of the regular represen-

tation are going to be

[(1, 1)]G =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

[(1, y )]G =

0 1 0 0

1 0 0 0

0 0 0 1

0 0 1 0

,

[(x, 1)]G =

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

[(x, y )]G =

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

.

Exercise 6.3. In short, no. Take r = 1−a and s = 1+ a then we have r s = (1−a)(1+a) =

1 + a − a − a2 = 0. In other words, the group algebra is not in general an integral domain.

Exercise 6.4. We let G = g 1, . . . , g n be a finite group and c =n

i =1 g i .

(a) Let h ∈ G, then h = g j for some 1 j n. Recall that given any 1 i n there

exists a unique 1 k, n such that g i g j = g k and g j g i = g . Then we have

ch = n

i =1

g i g j =

n

i =1

g i g j =

n

i =1

= g k = c,

hc = g j

n

i =1

g i

=

ni =1

g j g i =

n=1

= g = c.

(b) Therefore we have

c 2 = (g 1 + · · · + g n)c = c + · · · + c n times

= |G|c.



Chapter 6 16

(c) So let’s examine what’s happening to the basis G. We have g i ϑ = g i c = c for all

1 i n. Therefore the matrix of ϑ with respect to G is

[ϑ]G =

1 . . . 1

.... . .

...

1 . . . 1

.

Exercise 6.5. Recall that as a consequence of the definition of an F G module we have

0g = (u − u )g = ug − ug = 0 for all g ∈ G. Therefore, let r =n

i =1 αi g i ∈ F G then we

have

0r = 0n

i =1

αi g i =

n

i =1

0(αi g i ) =

n

i =1

(0αi )g i =

n

i =1

0g i = 0.

Also, for any v ∈ V we have v 0 = v (1 − 1) = v 1 − v 1 = v − v = 0.

Let G = 1, g 2, . . . , g n, with n > 1, and let V = span1 ⊆ F G be a 1-dimensional

vector subspace of the group algebra, such that 1g i = −1 for all 2 i n and 1.1 = 1.

Then this is certainly a submodule of F G. However we have 1(1+g i ) = 1.1+1g i = 1−1 = 0

but neither 1 nor 1 + g i are zero.

Exercise 6.6. We have to check that W is a sub module. Let w 1 = 1 + ω2a + ωa2 and

w 2 = b + ω2ab + ωa2b We have G =

1, a , a2,b,ab,a2b

and we can see that

w 1a = a + ω2a2 + ω w 2a = ba + ω2aba + ωa2ba,

= ω(1 + ω2a + ωa2) = ω2(b + ω2ab + ωa2b ),

w 1a2 = a2 + ω2 + ωa w 2a2 = ba2 + ω2aba2 + ωa2ba2,

= ω2(1 + ω2a + ωa2) = ω(b + ω2ab + ωa2b ),

w 1b = b + ω2ab + ωa2b w 2b = b 2 = ω2ab 2 + ωa2b 2,

= 1(b + ω2ab + ωa2b ) = 1(1 + ω2a + ωa2),

w 1

ab = (ab + ω2a2b + ωb ) w 2

ab = bab + ω2abab + ωa2bab,

= ω(b + ω2ab + ωa2b ) = ω2(1 + ω2a + ωa2),

w 1a2b = (a2b + ω2b + ωab ) w 2a2b = ba2b + ω2aba2b + ωa2ba2b,

= ω2(b + ω2ab + ωa2b ) = ω(1 + ω2a + ωa2).

Hence W is a 2-dimensional submodule of CG. We check its irreducibility, to show this

we just have to show that there is no 1-dimensional submodule of W . Let 0 = U =

spanαw 1 + βw 2 be a 1-dimensional vector subspace of W . How does G act on the basis



Chapter 6 17

of U ? Well, recalling ω2 = ω,

(αw 1 + βw 2)a = αw 1a + βw 2a = αωw 1 + βω2

w 2 ∈ U ⇒ α = 0 or β = 0,(αw 1 + βw 2)b = αw 1b + βw 2b = αw 2 + βw 1 ∈ U ⇒ β = α,

for some ∈ C. However if both conditions are satisfied then α = β = 0 and U is the 0module. Hence W is an irreducible module.



Chapter 7. F G-homomorphisms

Exercise 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 7.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 7.1. For all u 1, u 2 ∈ U, λ ∈ F and g ∈ G we have

(u 1 + u 2)(ϑφ) = ((u 1 + u 2)ϑ)φ = (u 1ϑ + u 2ϑ)φ = (u 1ϑ)φ + (u 2ϑ)φ = u 1(ϑφ) + u 2(ϑφ),

(λu 1)(ϑφ) = ((λu 1)ϑ)φ = (λ(u 1ϑ))φ = λ((u 1ϑ)φ) = λ(u 1(ϑφ)),

(ug )(ϑφ) = ((uϑ)g )φ = (uϑφ)g.

because uϑ ∈ V and φ is an F G-homomorphism. Therefore ϑφ is an F G homomorphism.

Exercise 7.2. Let V = spanv 1, . . . , v 5 be the permutation module defined by v i g = v ig

for all g ∈ (12345) S5. Let σ = (12345) and define a map ϑ : V → F G byv i ϑ = σi and extend linearly, we claim this is an F G-isomorphism. It is clear to see that

(v i σ)ϑ = v i +1ϑ = σi +1 = σi σ = (v i ϑ)σ. As σ generates G we have that ϑ is an F G-

homomorphism. It’s clear that it’s then a bijection.

Exercise 7.3. We first show that V 0 is a vector subspace of V . We clearly have 0 ∈ V 0

as 0g = 0 for all g ∈ G. Now if v , v ∈ V 0 and λ1, λ2 ∈ F then for all g ∈ G we

have (λ1v 1 + λ2v 2)g = (λ1v 1)g + (λ2v 2)g = λ1(v 1g ) + λ2(v 2g ) = λ1v 1 + λ2v 2 and so

λ1v 1 + λ2v 2 ∈ V 0, hence V 0 is a vector subspace of V . It’s clearly then closed under the

action of G by definition and so is an F G-submodule.Now let v 1, v 2 ∈ V , λ ∈ F and h ∈ H then we have

(v 1 + v 2)ϑ =g ∈G

(v 1 + v 2)g =g ∈G

v 1g +g ∈G

v 2g = v 1ϑ + v 2ϑ,

(λv 1)ϑ =g ∈G

(λv 1)g =g ∈G

λ(v 1g ) = λg ∈G

v 1g = λ(v 1ϑ),

(v 1h)θ =g ∈G

v 1gh =

g ∈G

v 1g

h = (v 1ϑ)h.

18



Chapter 7 19

It’s clear that the image is contained in V 0 as summing over gh ∈ G is the same as summing

over g ∈ G. Let v 0 ∈ V 0 then we havev 0|G|

ϑ =

g ∈G

v 0g

|G| =g ∈G

v 0|G| = v 0,

which gives us the map is surjective.

Exercise 7.4. Let φ : V → W be the isomorphism between V and W . Now let w 0 ∈ W 0 then

there exists v ∈ V such that w 0 = v φ. We have, for all g ∈ G, that w 0g = w 0 ⇒ (v φ)g =

v φ ⇒ v g = v ⇒ v ∈ V 0 because φ is an isomorphism. So W 0 ⊆ V 0φ and for v 0 ∈ V 0 we

have, for all g ∈ G, that (v 0φ)g = (v 0g )φ = v 0φ. Therefore V 0φ ⊆ W 0 ⇒ W 0 = V 0φ, which

gives us the restriction map φ|V 0 : V 0 → W 0 is an isomorphism.

Exercise 7.5. Let v 1, v 2, v 3, v 4 be the standard basis for the permutation module V . Then

we have that

v 1(12) = v 2 v 2(12) = v 1 v 3(12) = v 3 v 4(12) = v 4,

v 1(34) = v 1 v 2(34) = v 2 v 3(34) = v 4 v 4(34) = v 3,

v 1(12)(34) = v 2 v 2(12)(34) = v 1 v 3(12)(34) = v 4 v 4(12)(34) = v 3.

These calculations tell us that V 0 = spanv 1 + v 2, v 3 + v 4. Alternatively, by exercise 3, wehave that (F G)0 = spang ∈G g . By exercise 4, we have that if V and F G are isomorphic

then V 0 and (F G)0 are isomorphic. However V has dimension 2 and F G has dimension 1,

so they can’t be isomorphic.

Exercise 7.6. Let G = x | x 2 = 1.

(a) We check the properties of an F G-homomorphism. For all (α1+ βx ), ( γ 1+δx ) ∈ F G,

λ ∈ F we have

((α1 + βx ) + ( γ 1 + δx ))ϑ = ((α + γ )1 + ( β + δ)x )ϑ,

= (α + γ − β − δ)(1 − x ),

= (α − β)(1 − x ) + ( γ − δ)(1 − x ),

= (α1 + βx )ϑ + ( γ 1 + δx )ϑ.

(λ(α1 + βx ))ϑ = ((λα)1 + (λβ)x )ϑ ((α1 + βx )x )ϑ = ( β1 + αx )ϑ,

= (λα − λβ)(1 − x ) = ( β − α)(1 − x ),



Chapter 7 20

= λ(α − β)(1 − x ) = (α − β)(x − 1),

= λ((α1 + βx )ϑ) = (α

− β)(1

−x )x,

= (α1 + βx )ϑx.

Therefore ϑ is an F G-homomorphism.

(b) Now for any α1 + βx ∈ F G we have

(α1 + βx )ϑ2 = ((α − β)(1 − x ))ϑ,

= (α − β)((1 − x )ϑ),

= (α − β)(1 − (−1))(1 − x ),

= 2(α − β)(1 − x ),

= 2((α1 + βx )ϑ).

Therefore ϑ2 = 2ϑ.

(c) What is [ϑ] with respect to the standard basis G? This is

[ϑ]G =

1 −1

−1 1

.

What are the eigenvalues of this matrix? There the solutions of the equation (1 −λ2) − 1 = λ(λ − 2) = 0, which are λ = 0 and λ = 2. Calculating the eigenvector

with respect to λ = 2 gives us

v 1 v 2

1 −1

−1 1

= 2

v 1 v 2

⇒

v 1 v 2

−1 −1

−1 −1

=

0 0

⇒ v 2 = −v 1.

Clearly any vector is an eigenvector of λ = 0. Therefore we have that B = 1−x , 1 +

x form a basis such that the matrix [ϑ]B has the required form. Note that we can

see that by setting

T =

1 −1

1 1

and T −1 =

1

2

1 1

−1 1

we have

T [ϑ]GT −1 =1

2

−1 1

1 1

1 −1

−1 1

1 1

−1 1

,



Chapter 7 21

=1

2

2 −2

0 0

1 1

−1 1

,

=

2 0

0 0

,

= [ϑ]B

Then T is our change of basis matrix and we have

1 0

1 −1

1 1

=

1 −1

,

0 1 1 −1

1 1

=

1 1

.



Chapter 8. Maschke’s Theorem

Exercise 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Exercise 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Exercise 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Exercise 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Exercise 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 8.1. Let B = v 1, v 2 then we can see that the following is true

[x ]B =

0 1

−1 −1

.

We now try and diagonalise this matrix. What are the eigenvalues? Well these are the

solutions of the characteristic polynomial −λ(−1 − λ) + 1 = λ2 + λ + 1 = 0, which gives

us solutions λ1 = −e iπ/3 or λ2 = −e −iπ/3. Therefore the eigenvectors for λ1 and λ2 have

the form

v 1 v 2

0 1

−1 −1

= λi

v 1 v 2

⇒

v 1 v 2

−λi 1

−1 −1 − λi

=

0 0

,

⇒−λi v 1 = v 2,

v 1 + (−1 − λi )v 2 = 0,

⇒ (1 + λ + λ2)v 1 = 0,

which means v 1 can be chosen arbitrarily in each case. Therefore we have a direct sum

V = spanv 1 − λ1v 2 ⊕ spanv 1 − λ2v 2.

Exercise 8.2. Now G = 1, x , y , x y ∼= C 2 × C 2 such that x 2 = y 2 = 1 and xy = y x . We

want to find all 1-dimensional submodules of RG, say U . Now taking G as the standard

basis we must have that the basis element of a 1-dimensional submodule has the form

α1 + βx + γy + δx y for some α, β , γ, δ ∈ R. We consider the action of the group on this

basis element

(α1 + βx + γy + δx y )1 = α1 + βx + γy + δxy,

22



Chapter 8 23

(α1 + βx + γy + δx y )x = β1 + αx + δy + γxy,

(α1 + βx + γy + δx y ) y = γ 1 + δx + αy + βxy,

(α1 + βx + γy + δx y )xy = δ1 + γx + βy + αxy.

Clearly the first element will always lie in U . For the rest of the elements to lie in U we

must have there exists r 1, r 2 and r 3 ∈ R such that

α = r 1 β = r 2 γ = r 3δ

β = r 1α = r 2δ = r 3 γ

γ = r 1δ = r 2α = r 3 β

δ = r 1 γ = r 2 β = r 3α

⇒ r 21 = r 22 = r 23 = 1.

Now the only linearly independent solutions to these equations are the quadruples (1, 1, 1, 1),

(1, 1, −1, −1), (1, −1, 1, −1) and (1, −1, −1, 1). Therefore we have

RG = span1 + x + y + x y ⊕ span1 + x − y − x y ⊕ span1 − x + y − xy ⊕ span1 − x + y − xy .

Note that we could also have got this information by finding all the eigenvectors of x in the

regular representation.

Exercise 8.3. Let G be any non-trivial finite group and consider a 2-dimensional vector

subspace V = spanv 1, v 2 ⊆ CG. By defining v g = v for all v ∈ V and g ∈ G we have V

is a submodule of CG. Then define a homomorphism ϑ : V → V by (αv 1 + βv 2)ϑ = αv 2.

Now this is a CG homomorphism because

(α1v 1 + β1v 2 + α2v 1 + β2v 2)ϑ = α1v 2 + α2v 2 = (α1v 1 + β1v 2)ϑ + (α2v 1 + β1)ϑ,

(λ(αv 1 + βv 2))ϑ = λ(αv 2) = λ((αv 1 + βv 2)ϑ),

((αv 1 + βv 2)g )ϑ = (αv 1 + βv 2)ϑ = αv 2 = (αv 2)g = (αv 1 + βv 2)ϑg.

However we can see that ker(ϑ) = im(ϑ) = spanv 2.

Exercise 8.4. Assume ρ : G → GL(2,C) is reducible. Then by the matrix form of

Maschke’s theorem we have, for all g ∈ G

gρ =

λg 0

0 µg

,



Chapter 8 24

for some suitable λg , µg ∈ C determined by g . Now if this is true then (gρ)(hρ) = (hρ)(gρ)

for all g, h ∈ G because diagonal matrices commute but this implies G is abelian. Therefore

we must have ρ is irreducible.

Exercise 8.5. Consider the subspace U = span(1, 0) ⊆C2. Now clearly we have

1 0

1 0

n 1

=

1 0

,

which means U is a 1-dimensional submodule of C2. Let’s try and find a submodule U such

that C2 = U ⊕ U . Now any vector (v 1, v 2) ∈ C2 will be linearly independent of (1, 0) if

v 2

= 0, so consider U = span

(v 1, v 2)

. This gives us

v 1 v 2

1 0

n 1

=

v 1 + nv 2 v 2

.

If this is in U then there exists k ∈ C with v 1 = k (v 1 + nv 2) and v 2 = kv 2. The second

condition gives k = 1 but if this is the case then the first condition gives us nv 2 = 0 ⇒ n = 0.

Hence there can be no submodule U and so C2 is not completely reducible.

Exercise 8.6. Let ( , ) be a complex inner product on U × V .

(a) We first check that [ , ] is also a complex inner product on U

×V . By definition 14.2

this involves checking the following, for all u 1, u 2 ∈ U , v ∈ V and λ1, λ2 ∈ C

[u 1, v ] =x ∈G

(u 1x , v x ) =x ∈G

(v x , u 1x ) =x ∈G

(v x , u 1x ) = [v x , u 1x ],

[λ1u 1 + λ2u 2, v ] =x ∈G

((λ1u 1 + λ2u 2)x , v x ) = λ1[u 1, v ] + λ2[u 2, v ],

[u, u ] =x ∈G

(ux,ux ) >0

> 0 if u = 0.

(b) We first show a property of the inner product. For any g ∈ G we have

[ug,vg ] =x ∈G

(uxg,vxg ) =

x =xg ∈G(ux , v x ) = [u, v ].

Therefore, for every g ∈ G and v ∈ U ⊥ we have

[u , v g ] = [ug −1, v g g −1] = [ug −1, v ] = 0



Chapter 8 25

because v ∈ U ⊥ and ug −1 ∈ U because U is a submodule. Hence U ⊥ is indeed a

submodule of V .

(c) Clearly Maschke’s thereom holds, as given any submodule U of V we have U ⊥ is also

a submodule such that V = U ⊕ U ⊥.

Exercise 8.7. By Maschke’s theorem we have the group algebra is completely reducible

and so CG = V 1 ⊕ · · · ⊕ V k for some irreducible CG modules. Now CG is faithful as a CG

module, therefore we have v x = v for some v ∈ V i . Consider the following set

K i = x ∈ G | v x = v for all v ∈ V i ,

we claim this is a normal subgroup of G. Clearly 1 ∈ K as v 1 = v for all v ∈ V i , also forall x, y ∈ K i we have v (xy ) = (v x ) y = v y = v and so x y ∈ K i . Finally if x ∈ K i then

v x = v ⇒ v = v x −1 and so x −1 ∈ K i , which gives us K is a subgroup. If x ∈ K i and

g ∈ G then v (g −1x g ) = (v g −1)xg = v (g −1g ) = v and so K i is a normal subgroup of G.

However we clearly have K i = G and so, as G is simple, we have K i = 1 which means V i

is a faithful irreducible CG submodule.



Chapter 9. Schur’s Lemma

Exercise 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 9.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 9.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 9.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 9.1. For C 2 = x | x 2 = 1 we have the irreducible representations over C are

given by ρi : C 2 → GL(1,C) such that

x ρ1 = (1) x ρ2 = (−1).

Let ω = e 2πi/3. For C 3 = y | y 3 = 1, the irreducible representations over C are given by

ρi : C 3 → GL(1,C) such that

y ρ1 = (1) y ρ2 = (ω) y ρ3 = (ω2).

For C 2 × C 2 ∼= a, b | a2 = b 2 = 1 and ab = ba we have the irreducible representations

over C are given by ρi : C 2 × C 2 → GL(1,C) such that

xρ1 = (1) xρ2 = (1) xρ3 = (−1) xρ4 = (−1),

y ρ1 = (1) y ρ2 = (−1) y ρ3 = (1) y ρ4 = (−1).

Exercise 9.2. We have G = C 4 × C 4 ∼= a4 = b 4 = 1 and ab = ba.

(a) Recall that every irreducible representation of C 4 = x | x 4 = 1 over C is given byρi : C 4 → GL(1,C) such that x ρ1 = (1), xρ2 = (i ), xρ3 = (−1) and xρ4 = (−i ).

Therefore we have ρ : G → GL(1,C) given by aρ = (−1) and bρ = (−1) is an

irreducible representation of G over C. Now every element in G is of the form ai b j

and so clearly

(ai b j )2ρ = (a2)i (b 2) j ρ = (a2ρ)i (b 2ρ) j = (1)i (1) j = (1).

26



Chapter 9 27

(b) The elements of order 2 in G are 1, a2, b 2 and a2b 2. If σ existed then a2σ = b 2σ =

(−1) but (a2b 2)σ = (a2σ)(b 2σ) = (−1)(−1) = (1). Hence σ cannot exist.

Exercise 9.3. Let g 1, . . . , g r generate G and let ξ1, . . . , ξr be distinct primitive ni th roots

of unity. Then the following map ρ : G → GL(r,C) given by

(g i 11 . . . g i r r )ρ =

ξi 11 . . . 0

.... . .

...

0 . . . ξi r r

is a faithful irreducible representation of G.

Yes. Let r = 2 and n1 = n2. Then let ξ1 and ξ2 be primitive n1th and n2th roots of unity

respectively. Then if g 1, g 2 generate G = C n1 × C n2 we have the map τ : G → GL(1,C)given by (g i 11 g i 22 )τ = (ξi 11 ξi 22 ) is a faithful irreducible representation of degree 1 < r = 2.

Exercise 9.4. We have this is a representation if the matrices satisfy the relations of the

group. It is quick to check that

(aρ)4 =

−7 10

−5 7

4=

−1 0

0 −1

2= I2,

(bρ

)2 = −5 6

−4 5−5 6

−4 5 =I2,

(bρ)−1(aρ)(bρ) =

−5 6

−4 5

−7 10

−5 7

−5 6

−4 5

,

=

5 −8

3 −5

−5 6

−4 5

,

=

7 −10

5 −7

,

= (aρ)−1

.

Therefore ρ is indeed a representation of D8. Now consider a matrix which commutes with

gρ for all g ∈ D8, then in particular we havea b

c d

−7 10

−5 7

=

−7 10

−5 7

a b

c d

⇒−7a − 5b 10a + 7b

−7c − 5d 10c + 7d

=

−7a + 10c −7b + 10d

−5a + 7c −5b + 7d

,



Chapter 9 28

a b

c d

−5 6

−4 5

=

−5 6

−4 5

a b

c d

⇒−5a − 4b 6a + 5b

−5c − 4d 6c + 5d

=

−5a + 6c −5b + 6d

−4a + 5c −4b + 5d

.

This tells us that −5b = 10c and −4b = 6c , which means b = c = 0. Using this we have

10a = 10d ⇒ a = d and so the matrix M is just a scalar multiple of the identity. In other

words, ρ is an irreducible representation of D8.

We now do exactly the same for the representation σ of D8. We first check that this is

a representation by checking the relations.

(aσ)4 =5 −6

4 −5

4=1 0

0 1

2= I2,

(bρ)2 =

−5 6

−4 5

−5 6

−4 5

= I2,

(bρ)−1(aρ)(bρ) =

−5 6

−4 5

5 −6

4 −5

−5 6

−4 5

,

= −1 0

0−

1−5 6

−4 5 ,

=

5 −6

4 −5

,

= (aρ)−1.

We comment on how many unique matrices there are. Now a2σ = I2 and a3σ = aρ. Also

we can see that bρ = −aρ and a−1ρ = aρ. Therefore we only have one condition on a

matrix which commutes with gσ for every g ∈ G. In particular this is

a b c d

5 −64 −5

=5 −6

4 −5

a b c d

⇒ 5a + 4b −6a − 5b 5c + 4d −6c − 5d

=5a − 6c 5b − 6d

4a − 5c 4b − 5d

.

This means the only conditions on our matrix are that 4b = −6c and 10b = 6(d − a).

Therefore the following matrix commutes with all (gσ)10 −6

4 0

,



Chapter 9 29

which means σ is not an irreducible representation.

Exercise 9.5. Clearly z = g ∈Gg is in the centre of CG. By Proposition 9.14 we have

there exists λ ∈ C such that v z = λv for all v ∈ V .

Exercise 9.6. Let G = D6 = a, b | a3 = b 2 = 1 and bab = a−1.

(a) We want to show that g (a + a−1) = (a + a−1)g for g = a or b , then it follows that

a + a−1 ∈ Z (CG). We have that

a(a+a−1) = a2+1 = (a+a−1)a b (a+a−1) = ba+ba−1 = a−1b +ab = (a+a−1)b.

(b) Consider u 1 = 1 + ω

2

a + ωa

2

and u 2 = b + ω

2

ab + ωa

2

b then we have

u 1(a + a−1) = (a + a−1) + ω2a(a + a−1) + ωa2(a + a−1),

= (ω + ω2)1 + (1 + ω)a + (1 + ω2)a2,

= (ω + ω2)u 1,

u 2(a + a−1) = b (a + a−1) + ω2ab (a + a−1) + ωa2b (a + a−1),

= (ω + ω2)b + (1 + ω)ab + (1 + ω2)a2b,

= (ω + ω2)u 2.

Therefore, given any w ∈ W we have w (a + a−1) = (ω + ω2)w .

Exercise 9.7. We consider the centres of the following groups and use Proposition 9.16.

(a) Consider C n = x | x n = 1. Let ω = e 2πi/n then ω is a primitive nth root of

unity. We know that the map ρ : C n → GL(1,C) given by x ρ = (ω) is an irreducible

representation of C n and is clearly faithful as all powers of ω are distinct.

(b) Let D8 = a, b | a4 = b 2 = 1 and bab = a3 = 1, a , a2, a3,b,ab,a2b, a3b . Now it’s

quick to check that Z (D8) = 1, a2 ∼= C 2. Consider the representation σ : D8 →

GL(2,C

) given byaσ =

i 0

0 −i

bσ =

0 1

1 0

then because i is a primitive fourth root of unity we have this is a faithful irreducible

representation of D8.

(c) It’s clear to see that the Z (G×H ) = Z (G)×Z (H ) and therefore Z (C 2×D8) ∼= C 2×C 2,

which is not cyclic. Therefore there cannot exist a faithful irreducible C(C 2 × D8)

module by Proposition 9.16.



Chapter 9 30

(d) Now Z (C 3 × D8) ∼= C 3 × C 2 ∼= C 6 and so is cyclic. Let ξ = e 2πi/3 then we have the

representation π : C 3 × D8 → GL(2,C) given by

(x , 1)π =

ξ 0

0 ξ−1

(1, a)π =

i 0

0 −i

(1, b )π =

0 1

1 0

is a faithful irreducible representation of C 3 × D8.



Chapter 10. Irreducible modules and the group algebra

Exercise 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Exercise 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Exercise 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Exercise 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Exercise 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Exercise 10.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Exercise 10.1. Let V = spang ∈G g ∈ G be the trivial submodule of CG. We now wantto see if there exists a 1-dimensional submodule, say U , of CG such that U ∼= V . Let

U = spanu , then for U to be the trivial submodule we need ug = u for all g ∈ G. In

other words |G|u = u (

g ∈G g ) = (

g ∈G g )u ⇒ u ∈ V and so U = V . Therefore there is

only one unique trivial CG module.

Exercise 10.2. Let G = x | x 4 = 1. We consider i as a fourth root of unity and define

the following elements of CG,

v 0 = 1 + x + x 2

+ x 3

,v 1 = 1 + i x − x 2 − i x 3,

v 2 = 1 − x + x 2 − x 3,

v 3 = 1 − i x − x 2 + i x 3.

Then V i = spanv i are irreducible CG modules because v 0x = v 0, v 1x = i v 1, v 2x = −v 2

and v 3x = −iv 3. Therefore we have CG = V 0 ⊕ V 1 ⊕ V 2 ⊕ V 3.

Exercise 10.3. Consider u 1 = 1 + a + a2 + a3 − b − ab − a2b − a3b ∈ CG. Then we have

the following

u 1a = a + a2 + a3 + 1 − a3b − b − ab − a2b = u 1,

u 1b = b + ab + a2b + a3b − 1 − a − a2 − a3 = −u 1.

Hence U 1 = spanu 1 is the required 1-dimensional CG submodule. Alternatively let u 2 =

1 − a + a2 − a3 + b − ab + a2b − a3b and u 3 = 1 − a + a2 − a3 − b + ab − a2b + a3b then

31



Chapter 10 32

we have that

u 2a = a − a2

+ a3

− 1 + a3

b − a + ab − a2

b = −u 2,u 2b = b − ab + a2b − a3b + 1 − a + a2 − a3 = u 2,

u 3a = a − a2 + a3 − 1 − a3b + b − ab + a2b = −u 3,

u 3b = b − ab + a2b − a3b − 1 + a − a2 + a3 = −u 3.

Therefore U 2 = spanu 2 and U 3 = spanu 3 are the required 1-dimensional CG modules.

Exercise 10.4. Let G = D8 = a, b | a4 = b 2 = 1 and bab = a3 then define the following

elements of CG

v 0 = 1 + a + a2 + a3 w 0 = bv 0 = b + ba + ba2 + ba3,

v 1 = 1 + i a − a2 − ia3 w 1 = bv 1 = b + i ba − ba2 − iba3,

v 2 = 1 − a + a2 − a3 w 2 = bv 2 = b − ba + ba2 − ba3,

v 3 = 1 − i a − a2 + ia3 w 3 = bv 3 = b − i ba − ba2 + iba3.

Now as in question 2 we have that spanv i are Ca modules and it’s clear to see that

spanw i are also Ca modules. Now b acts on these elements in the following way

v 0b = b + ba3 + ba2 + ba = w 0 w 0b = 1 + a3 + a2 + a = v 0,

v 1b = b + iba3 − ba2 − iba = w 3 w 1b = 1 + i a3 − a2 − i a = v 3,

v 2b = b − ba3 + ba2 − ba = w 2 w 2b = 1 − a3 + a2 − a = v 2,

v 3b = b − iba3 − ba2 + iba = w 1 w 3b = 1 − i a3 − a2 + i a = v 1.

Therefore this gives us that spanv 0, w 0, spanv 1, w 3, spanv 2, w 2 and spanv 3, w 1 are

Cb modules.

Now we have that spanv 0, w 0 is reducible as U 0 = spanv 0 + w 0, (the trivial module),

and U 1 = spanv 0 − w 0 are CG submodules. Also, by question 3, we have spanv 2, w 2 isreducible as U 2 = spanv 2 + w 2 and U 3 = spanv 2 − w 2 are CG submodules. We quickly

check that U 4 = spanv 1, w 3 and U 5 = spanv 3, w 1 are irreducible. Now a acts on a linear

combination of the basis in the following way

(αv 1 + βw 3)a = αiv 1 − βi w 3 ( γv 3 + δw 1)a = − γi v 3 + δi w 1.

So if these are bases of 1-dimensional submodules then there exists k and such that



Chapter 10 33

α = ki α, β = −kiβ and γ = −i γ , δ = iδ. However no such k and exist, therefore the

modules are irreducible and we have

CG = U 0 ⊕ U 1 ⊕ U 2 ⊕ U 3 1-dim

⊕ U 4 ⊕ U 5 2-dim

.

because v 0 + w 0, v 0 − w 0, v 2 + w 2, v 2 − w 2, v 1, v 3, w 1, w 3 form a basis for CG. Note that

non of the 1-dimensional modules are isomorphic but U 4 ∼= U 5 by the isomorphism sending

v 1 → w 1 and w 3 → v 3.

Now by Theorem 10.5 we have that every 1-dimensional irreducible representation of

D8 is equivalent to one of the following

ρ0 : a → (1) ρ1 : a → (1) ρ2 : a → (−1) ρ3 : a → (−1)

b → (1) b → (−1) b → (1) b → (−1)

and any 2-dimensional irreducible representation of D8 is equivalent to

ρ4 : a →

i 0

0 −i

b →

0 1

1 0

.

Exercise 10.5. Let ψ : U 1

→U 2 be the CG isomorphism between these modules. Let

0 = λ ∈ C then we define a map ϑ : U 1 → V such that uϑ = u + λ(uψ). We claim this

is a CG homomorphism. We check the following properties for all u 1, u 2 ∈ U 1, µ ∈ C and

x ∈ G

(u 1 + u 2)ϑ = (u 1 + u 2) + λ((u 1 + u 2)ψ) = (u 1 + λ(u 1ψ)) + (u 2 + λ(u 2ψ)) = u 1ϑ + u 2ϑ,

(µu 1)ϑ = µu 1 + λ((µu 1)ψ) = µ(u 1 + λ(u 1ψ)) = µ(u 1ϑ),

(u 1x )ϑ = u 1x + λ(u 1xψ) = u 1x + (λ(u 1ψ))x = u 1ϑx.

What is the kernel of ϑ? Well we require that

u ∈ ker(ϑ) ⇔ uϑ = 0 ⇔ u + λuψ = 0 ⇔ u = 0

as the sum is direct. Now im(ϑ) is a submodule of V and U 1 ∼= im(ϑ). This argument

works equally well for U 2 and we’re done.

Exercise 10.6. We first show that this is irreducible. Let U = spanαv 1 + βv 2 ⊆ V be a



Chapter 10 34

1-dimensional submodule, then we have

(αv 1 + βv 2)a = i αv 1 − iβv 2 ∈ U ⇒ α = k 1iα,β = −k 1i β,(αv 1 + βv 2)b = αv 2 − βv 1 ∈ U ⇒ α = k 2α, β = −k 2 β,

for some k 1, k 2 ∈ C but these only exist if α = β = 0 and so U is the 0 module. Therefore

V is irreducible.

Now consider a 2-dimensional CG submodule, say U , with basis u 1 = 1 − i a − a2 + i a3 +

ib − ab − ia2b + a3b and u 2 = −i + a + ia2 − a3 + b − i ab − a2b + ia3b . Then we have

u 1a = a − i a2 − a3 + i + i a3b − b − iab + a2b = iu 1,

u 1b = b − i ab − a2b + ia3b + i a2 − a3 − i + a = u 2,

u 2a = −ia + a2 + i a3 − 1 + a3b − i b − ab + i a2b = −i u 2,

u 2b = −ib + ab + ia2b − a3b + a2 − ia3 − 1 + ia = −u 1.

Therefore U is an irreducible CG submodule which is isomorphic to V .



Chapter 11. More on the group algebra

Exercise 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Exercise 11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Exercise 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Exercise 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Exercise 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Exercise 11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Exercise 11.1. If G is non-abelian then not all CG modules have dimension 1, however weknow we always have the trivial module with dimension 1. Therefore we have 1+

5i =2 d 2i =

6 then we must have one module of dimension 2 and two modules of dimension 1.

Exercise 11.2. If G is of order 12 then there are four possibilities for the degrees of the

irreducible representations. Either 112, 1102, 1422 or 133.

By exercise 3.5 we have that ρ1 and ρ3 are irreducible representations of D12 with degree

2. We recall that these representations are not isomorphic because ρ1 is faithful but ρ3 is

not. Therefore we must have the degrees of the irreducible representations of D12 are 1422.

Exercise 11.3. We have that G = g 1, . . . , g n forms a basis for the group algebra CG.

We define a family of CG homomorphisms by r φi = g i r for all r ∈ CG and 1 i n. Let

φ : CG → CG be any CG-homomorphism then

1φ = λ1g 1 + · · · + λng n,

for some λi ∈ C. Therefore, for any r ∈ CG we have

r φ = (1r )φ = (1φ)r,

= (λ1g 1 + · · · + λng n)r,

= λ1g 1r + · · · + λng nr,

= r (λ1φ1 + · · · + λnφn)

and so φi span HomCG(CG,CG). We check that they’re linearly independent by evaluating

both sides of the following equation at 1

0 = 1(λ1φ1 + · · · + λnφn),

35



Chapter 11 36

= λ1g 1 + · · · + λng n.

Now the g i ’s are linearly independent and so we must have λ1 = · · · = λn = 0 and we’redone.

Exercise 11.4. Let v 1, . . . , v n be the natural basis for the permutation module V . Then

spanv 1 + · · · + v n is the unique trivial submodule module of CG. Therefore by Corollary

11.6 we have dim(HomCG(V, U )) = 1.

Exercise 11.5. We can construct a basis for HomCG(CG, U 3) in roughly the same way as

in the proof of Proposition 11.8. We have v 1, w 2 form a basis for U 3. Now define two maps

φ1 : CG

→U 3 and φ2 : CG

→U 3 by r φ1 = v 1r and r φ2 = w 2r . Then consider a CG

homomorphism φ : CG → U 3. For some λ1, λ2 ∈ C we have

1φ = λ1v 1 + λ2w 2.

Hence for any r ∈ CG we have

r φ = (1r )φ = (1φ)r = λ1v 1r + λ2w 2r = r (λ1φ1 + λ2φ2).

So φ1, φ2 span HomCG(CG, U 3) and they’re a basis because if 0 = µ1φ1 + µ2φ2 then

0 = µ1(1φ1) + µ2(1φ2) = µ1v 1 + µ2w 2,

which gives us µ1 = µ2 = 0.

We wish to now find a basis for the vector space HomCG(U 3,CG). Recall from the proof

of Proposition 11.4 that

HomCG(U 3,CG) ∼= HomCG(U 3, U 1) ⊕ HomCG(U 3, U 2)

⊕HomCG(U 3, U 3)

⊕HomCG(U 3, U 4),

∼= HomCG(U 3, U 3) ⊕ HomCG(U 3, U 4),

as vector spaces because CG = U 1 ⊕ U 2 ⊕ U 3 ⊕ U 4. Now ϑ1 : U 3 → U 3 and ϑ2 : U 3 → U 4

given by uϑ1 = u and uϑ2 = bu for all u ∈ U 3 are basis elements for HomCG(U 3, U 3) and

HomCG(U 3, U 4) respectively. Therefore they form a basis for HomCG(U 3,CG).

Exercise 11.6. This means that we can express V as a direct sum V ∼= k i =1 V d i i , where

V d i i = V i ⊕ · · · ⊕ V i for d i copies. Also we can express W as a direct sum W ∼=

k i =1 V e i i .



Chapter 11 37

Therefore we have

dim(HomCG(V, W )) = dim(HomCG(

k i =1

V d i i ,

k i =1

V e i i )),

=

k i =1

d i e i dim(HomCG(V i , V i )),

=

k i =1

d i e i .



Chapter 12. Conjugacy classes

Exercise 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Exercise 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Exercise 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Exercise 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Exercise 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Exercise 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Exercise 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Exercise 12.1. Clearly we have 1 ∈ C G(x ) as 1x = x 1. Now if g ∈ C G(x ) then gx = x g ⇒xg −1 = g −1x ⇒ g −1 ∈ C G(x ). Finally, if g, h ∈ C G(x ) then (gh)x = gx h = x (gh) which

gives us gh ∈ C G(x ) and so C G(x ) is a subgroup of G. If g ∈ Z (G) then gy = y g for all

y ∈ G, in particular for x ∈ G and so Z (G) ⊆ C G(x ).

Exercise 12.2. Now g ∈ C G(x ) ⇔ gx = x g ⇔ g (xz ) = (xz )g ⇔ g ∈ C G(xz ). Therefore

we have |g G| = |G|/|C G(x )| = |G|/|C G(xz )| = |(gz )G|.

Exercise 12.3. Let G = Sn.

(a) Well (12) is conjugate to σ ∈ Sn if and only if they have the same cycle type. Now

there aren2

2-cycles in Sn which gives us |(12)Sn | =

n2

. Now if g ∈ C G((12)) then

g −1(12)g = (12), which gives us (1g 2g ) = (12). Therefore we have g = y or (12) y ,

where y is an element which fixes 1 and 2. There are (n −2)! such elements y and so

|C G((12))| = 2 · (n − 2)! which gives us |(12)Sn | = |G|/|C G((12))| = n!/2!(n − 2)! =n2

as required.

(b) Choose a 3 element subset from 1, . . . , n, say i , j , k . Then (i jk ) and (ikj ) are

the only distinct 3-cycles from this set. In otherwords there are 2n3 3-cylces in

Sn and so |(123)Sn | = 2n3. Now choose a 4 element subset from 1, . . . , n, say

i , j , k , . Then (i j )(k), (ik )( j) and (i)( jk ) are the only distinct 2-2-cycles you

can make from this set. Hence there are 3n4

distinct 2-2-cycles in Sn, which gives

us |(12)(34)Sn | = 3n4

.

(c) We consider the sizes of the conjugacy classes in S6. We express this in table 12.1.

Exercise 12.4. Now the cycle types appearing in A6 are: (1), (2,2), (2,4), (3), (3,3) and

(5). Now (12)(34) commutes with (56), which is odd. We have (12)(3456) commutes

38



Chapter 12 39

Cycle Shape Representative |g G| Reason

(1) 1 1 –(2) (12) 15

62

(2,2) (12)(34) 45

62

42

/2

(2,2,2) (12)(34)(56) 1562

42

/3 × 2

(2,3) (12)(345) 120 262

43

(2,4) (12)(3456) 90

62

× 3!

(3) (123) 40 263

(3,3) (123)(456) 40 263(4) (1234) 90 66

4

(5) (12345) 144 4 × 3 × 265

(6) (123456) 120 5!

Table 12.1: Conjugacy Classes of S6

with (12), which is odd. We have (123) commutes with (45), which is odd. We have

(123)(456) commutes with (14)(25)(36), which is odd. However there is no odd element

which commutes with a cycle of shape (5). Therefore (12345)A6

= (12345)S6.

Exercise 12.5. Any normal subgroup of A5 must be a union of conjugacy classes and

the order must divide the order of the group, namely 60. Now the divisors of 60 are

2,3,4,5,6,10,12,15,20 and 30. Now the conjugacy class sizes of A5 are 1,12,15 and 20.

However as we must always include the identity it’s clearly impossible to construct a normal

subgroup of order 15, 20 or 30. Therefore A5 is simple.

Exercise 12.6. We have Q8 = a, b | a4 = 1, b 2 = a2, b −1ab = a−1. Now straight away we

have b −1ab = a3 and so a, a3 are conjugate. We have a2 is the only element of order 2 so

it’s in a conjugacy class on its own. We have a3ba = a2b and b −1(ab )b = a3b . Thereforethe conjugacy classes are 1, a2, a, a3, ab,a3b and b, a2b . Therefore a basis for

Z (CQ8) is

1, a2, a + a3, ab + a3b, b + a2b.

Exercise 12.7. Now we have from the class formula that |G| = |Z (G)| +

x i ∈Z (G) |x Gi |.Now p | |G| and p | |x Gi | for each x i ∈ Z (G) so we must have p | |Z (G)| and so Z (G) = 1.



Chapter 12 40

Now assume G does not have a conjugacy class of size p . Then we have p 2 | |x Gi | for

each x i ∈ Z (G) and clearly p 2 | |G| = p n. If n 3 this gives us p 2 | |Z (G)| but |Z (G)| = p

and so we must have G has a conjugacy class of size p .



Chapter 13. Characters

Exercise 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Exercise 13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Exercise 13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Exercise 13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Exercise 13.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Exercise 13.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Exercise 13.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Exercise 13.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Exercise 13.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Exercise 13.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Exercise 13.1. Now we have G = 1, a , a2, a3, a4, a5,b,ab,a2b, a3b, a4b, a5b . We first

see that the conjugacy classes of G are

1 a3 a, a5 a2, a4 b, a2b, a4b ab,a3b, a5b

by (12.12). Now the matrices of the representations are as follows

a2ρ1 =

ω2 0

0 ω−2

a3ρ1 =

1 0

0 1

a4ρ1 =

ω 0

0 ω−1

a5ρ1 =

ω2 0

0 ω−2

,

abρ1 =

0 ω

ω−1 0

a2bρ1 =

0 ω2

ω−2 0

a3bρ1 =

0 1

1 0

a4bρ1 =

0 ω

ω−1 0

,

a5bρ1 =

0 ω2

ω−2 0

a2ρ2 =

1 0

0 1

a3ρ2 =

−1 0

0 1

a4ρ2 =

1 0

0 1

,

a5ρ2 =−1 0

0 1

abρ2 =

−1 0

0 −1

a2bρ2 =

1 0

0 −1

a3bρ2 =

−1 0

0 −1

,

a4bρ2 =

1 0

0 −1

a5bρ2 =

−1 0

0 −1

.

Recall that X 3 = 1 can be written as (X − 1)(X 2 + X + 1) = 0. So if ω is a root of

unity we have ω2 + ω = −1 ⇒ ω + ω−1 = −1. Therefore the characters associated

41



Chapter 13 42

are the ones given in table 13.1. It’s easy to see that ker(ρ1) = ker(χ1) = 1, a3 and

ker(ρ2) = ker(χ2) = 1, a2, a4.

g 1 a3 a a2 b ab

χ1(g ) 2 2 −1 −1 0 0

χ2(g ) 2 0 0 2 0 −2

Table 13.1: Two characters of D12

Exercise 13.2. Well we know that there are four irreducible characters of C 4 = x | x 4 = 1which are all linear. Consider i as a primitive fourth root of unity then the irreduciblerepresentations are xρ j = i j −1 for 1 j 4. Therefore the irreducible characters are given

in table 13.2. From the table we can see χreg(g ) = χ1(g ) + χ2(g ) + χ3(g ) + χ4(g ) is

4 = |G| if g = 1 and 0 otherwise.

g 1 x x 2 x 3

χ1(g ) 1 1 1 1

χ2(g ) 1 i −1 −i

χ3(g ) 1 −1 1 −1χ4(g ) 1 −i −1 i

Table 13.2: The irreducible characters of C 4

Exercise 13.3. Well χ(g ) = | fix(g )|, in particular we can easily see that the character

values are then given by χ((12)) = 5 and χ((16)(235)) = 2.

Exercise 13.4. Let χ : G → C be a non-zero character. If χ is a homomorphism then

χ(1) = χ(12

) = χ(1)2

⇒ χ(1)(χ(1) − 1) = 0 ⇒ χ(1) = 1, (note χ(1) = 0 because it’sthe dimension of the module), in other words χ is a linear character.

Exercise 13.5. Let ρ : G → GL(n,C) be the corresponding irreducible represenation of χ.

If z ∈ Z (G) then z ρ ∈ Z (GL(n,C)), which means z ρ = λIn for some λ ∈ C. We have

z m = 1 ⇒ λmIn = In ⇒ λm = 1 and so λ is an mth root of unity. Now for all g ∈ G we

have

(z g )ρ = (zρ)(gρ) = λIn(gρ) = λ(gρ).



Chapter 13 43

Therefore taking traces gives us χ(zg ) = λχ(g ) for all g ∈ G.

We comment that irreducibility is required to say z ρ is a scalar of λn. If the represen-

tation is reducible then you could have different scalars of different block identities.

Exercise 13.6. Now using Theorem 13.11 we have for some λ ∈ C and all h ∈ G we have

|χ(g )| = χ(1) ⇔ gρ = λIn for some λ ∈ C,

⇔ (gρ)(hρ) = (hρ)(gρ) for all h ∈ G,

⇔ (gh)ρ = (hg )ρ,

⇔ gh = hg because ρ is faithful,

⇔g

∈Z (G).

Therefore Z (G) = g ∈ G | |χ(g )| = χ(1). Note that if χ is reducible then (gρ)(hρ) =

(hρ)(gρ) ⇒ gρ = λIn.

Exercise 13.7. Let ρ : G → GL(n,C) be a representation of a finite group G.

(a) Define a map ψ : G → GL(1,C) by gψ = (det(gρ)). We claim this is a homo-

morphism. For all g, h ∈ G we have (gh)ψ = (det((gh)ρ)) = (det((gρ)(hρ))) =

(det(gρ))(det(hρ)) = (gψ)(hψ). Clearly δ is then a character as δ(g ) = tr(gψ) and

it’s linear because δ(1) = tr(1ψ) = tr(1) = 1.

(b) Recall that matrices A, B ∈ GL(n,C) we have

det(A−1BA) = det(A)−1 det(B) det(A) = det(B).

Now ker(δ) = g ∈ G | δ(g ) = δ(1) = g ∈ G | det(gρ) = 1. By the above matrix

fact we have that similar matrices are in the same left coset of G/ ker(ρ). Let g, h ∈ G

then we have the left cosets (g −1hg ) ker(δ) = h ker(δ) ⇔ (hg ) ker(δ) = (gh) ker(δ),

which means G/ ker(δ) is abelian.

(c) Let N = g ∈ G | δ(g ) > 0, we claim this is a normal subgroup of G. Now 1 ∈ N because δ(1) = 1 > 0. Let x , y ∈ N then δ(xy ) = δ(x )δ( y ) > 0 and so xy ∈ N .

Finally if x ∈ N then δ(x −1) = 1δ(x )

> 0 and so x −1 ∈ N , which means N is a subgroup.

Let x ∈ N and g ∈ G then δ(g −1xg ) = δ(g −1)δ(x )δ(g ) = δ(x ) > 0, which gives us

N is a normal subgroup of G. If g, h ∈ G then gN = hN ⇔ gh−1 ∈ N ⇔ δ(gh−1) >

0 ⇔ δ(g )δ(h)

> 0 ⇔ δ(g ), δ(h) > 0 or δ(g ), δ(h) < 0. Therefore as there exists an

element h ∈ G with δ(h) = −1 we have N is a normal subgroup of index 2 in G.



Chapter 13 44

Exercise 13.8. Let G be the standard basis of CG, with dim(CG) = |G| = n, and

ρreg : G → CG be the regular representation. We know that [gρreg]G are permutation ma-

trices associated with some permutation σg ∈ Sn. Now we have det([gρreg ]G) = sgn(σg ).Consider, as in question 7, the linear character δ : g → det([gρreg ]G). We have 2 | |G| and

so there exists an element of order 2 in G. In other words there exists an element g ∈ G

such that σg is a transposition and so δ(g ) = −1. Therefore by part c of question 7 we

have there exists a normal subgroup of index 2 in G.

Exercise 13.9. Let g ∈ G be an element of order 2. Let ρ : G → GL(n,C) be the

corresponding representation to χ. We have from position 9.11 that there exists a basis

B such that [gρ]B is a diagonal matrix, whose entries are square roots of unity. Therefore

χ(g ) = r − s for some r, s ∈ N where r is the number of +1’s and s the number of −1’s onthe diagonal of gρ. Now χ(1) = r + s and so we have r − s = r + s − 2s ≡ r + s (mod 2).

If s is even then s = 2s for some s ∈ N and r − s = r + s − 4s ≡ r + s (mod 4),

in other words χ(g ) ≡ χ(1) (mod 4). If s is not even then det([gρ]B) = (−1)s = −1.

Therefore considering the linear character δ : g → det([gρ]B) from question 7, we have

δ(g ) = −1 which gives us there exists a normal subgroup of index 2 by part c.

Exercise 13.10. Consider the regular character χreg = χ1+· · ·+χn where χi are irreducible

characters. Now χreg(1) = |G| but χreg(x ) = 0 in other words

χ1(1) + · · · + χn(1) = |G| χ1(x ) + · · · + χn(x ) = 0.

Therefore we cannot have χi (x ) = χi (1) for all i because |G| > 0.



Chapter 14. Inner products of characters

Exercise 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Exercise 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Exercise 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Exercise 14.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Exercise 14.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Exercise 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Exercise 14.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Exercise 14.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Exercise 14.1. We have that

χ, χ =32

24+

(−1)2

4+

02

3+

32

8+

(−1)2

4= 2,

χ, ψ =3 · 3

24+

(−1) · 1

4+

0 · 0

3+

(−1) · 3

8+

(−1) · (−1)

4=

32

24− 32

24= 0,

ψ, ψ =32

24+

12

4+

0

3+

(−1)2

8+

(−1)2

4= 1.

Therefore ψ is an irreducible character but χ is not.

Exercise 14.2. We calculate the characters χi corresponding to the representations ρi .

Using the following matrix calculations

a2ρ1 =

−1 0

0 −1

abρ1 =

0 i

i 0

a2ρ2 =

−1 0

0 −1

abρ2 =

i 0

0 −i

,

a2ρ3 = 1 0

0 1 abρ3 = −1 0

0−

1 ,

we give the values of the characters in table 14.1.

As χ1 = χ2 we have ρ1 and ρ2 are equivalent representations. Also χ3 = χ2 and χ3 = χ1

means ρ3 is not equivalent to either χ1 or χ2.

Exercise 14.3. Now we have that tr(gσ) = tr(T −1g (gρ)T g ) = tr(gρ) for all g ∈ G. There-

fore σ and ρ have the same character and by Theorem 14.21 this gives us σ and ρ are

equivalent representations.

45



Chapter 14 46

g 1 a a2 ab b

χ1(g ) 2 0 −2 0 0χ2(g ) 2 0 −2 0 0

χ3(g ) 2 0 2 −2 0

Table 14.1: The characters of ρ1, ρ2 and ρ3

Exercise 14.4. We consider the inner product of χ with itself,

χ, χ =1

|G

| χ(g )χ(g ) =

1

|G

| g ∈G|χ(g )| =

1

|G

| g ∈Gχ(g ) >

|G|

|G

|= 1.

Therefore χ must be reducible as we cannot have χ, χ = 1.

Exercise 14.5. Expanding the inner product of χreg, χ we see

χreg, χ =1

|G|g ∈G

χreg(g )χ(g ) =1

|G| |G|χ(1) = χ(1),

because χ(1) ∈ N \ 0.

Exercise 14.6. In Exercise 11.4 we showed that if V is the permutation module and U is

the trivial module then dim(HomCG(V, U )) = 1. Now by Theorem 14.24 we have

1 = dim(HomCG(V, U )) = π, 1Sn.

Alternatively, a more direct approach is to use the Orbit-Stabiliser theorem to show thatg ∈G | fix(g )| = |G|.

Exercise 14.7. Now we know that

ψ, ψ = d 21 + · · · + d 2k .

Now if ψ, ψ = 1 then we must have that d i = 1 for just one i and so ψ is an irreducible

character. If ψ, ψ = 2 then we must have two d i = 1 and the rest zero, therefore

ψ = χi 1 + χi 2. If ψ, ψ = 3 then we must have ψ = χi 1 + χi 2 + χi 3. Finally if ψ, ψ = 4

then either ψ = χi 1 + χi 2 + χi 3 + χi 4 or ψ = 2χi 1.

Exercise 14.8. No. For example consider the trivial character and the sign character of

Sn. Then we have χ(g ) = 1Sn(g ) + χsgn(g ) = 2 or 0. Now the trivial and sign characters

are irreducible so χ = 2φ.



Chapter 15. The number of irreducible characters

Exercise 15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Exercise 15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Exercise 15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Exercise 15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Exercise 15.1. We consider the inner product of χ with the irreducibles

χ, χ1 =

1

6 (19 − 3 + 2 · (−2)) =

12

6 = 2,

χ, χ2 =1

6(19 + 3 + 2 · (−2)) =

18

6= 3,

χ, χ3 =1

6(2 · 19 + 0 + 2 · (−2) · (−1)) =

42

6= 7.

Therefore we have χ = 2χ1 + 3χ2 + 7χ3. All the coefficients are positive integers and so

χ is a character of S3.

Exercise 15.2. We check the appropriate inner products

ψ1, χ1 =1

6ψ2, χ1 =

3

6=

1

2ψ3, χ1 =

2

6=

1

3,

ψ1, χ2 =1

6ψ2, χ1 =

−1 · 3

6= −1

2ψ3, χ1 =

2

6=

1

3,

ψ1, χ3 =2

6=

1

3ψ2, χ1 = 0 ψ3, χ1 =

−2

6= −1

3.

Therefore we have

ψ1 =1

6χ1 +

1

6χ2 +

1

3χ3 ψ2 =

1

2χ1 − 1

2χ2 ψ3 =

1

3χ1 +

1

3χ2 − 1

3χ3.

Exercise 15.3. Examining the inner products we have

ψ, χ1 =6

12+

0

12+

3

6− 3

6− 1 + i

4+

−1 + i

4= 0,

ψ, χ2 =6

12+

0

12− 3

6− 3

6+

−i + 1

4+

−i − 1

4= 0,

ψ, χ3 =6

12+

0

12+

3

6− 3

6+

1 + i

4+

1 − i

4= 1,

47



Chapter 15 48

ψ, χ4 =6

12+

0

12− 3

6− 3

6+

i − 1

4+

−1 − i

4= −1,

ψ, χ5 =12

12 +0

12 −3

6 +3

6 +0

4 +0

4 = 1,

ψ, χ6 =12

12+

0

12+

3

6+

3

6+

0

4+

0

4= 2.

Therefore ψ = χ3− χ4 + χ5 + 2χ6. So ψ is not a character of G because χ4 has a negative

coefficient.

Exercise 15.4. Let G be a group with |G| = 12.

(a) We show that the centre of a group never has index 2. If the centre does have index

2 then G = Z (G)

x Z (G) for some x

∈G. However consider xz

∈xZ (G) then

for g ∈ Z (G) we clearly have g (x z ) = (xz )g . Now consider x z ∈ x Z (G) then

(xz )(xz ) = (xz )(x z ) and so xz ∈ Z (G) means xZ (G) = Z (G) ⇒ G = Z (G).

Therefore Z (G) cannot have index 2 in the group. If G is abelian then there are

12 conjugacy classes as |G| = 12. Assume G is non-abelian then |Z (G)| | |G| but

we’ve just shown |Z (G)| = 6 therefore |Z (G)| 4. So there are at most 4 conjugacy

classes with order 1. All the other conjugacy classes have order at least 2 which leaves

us with |G| 4 · 1 + 5 · 2 = 14, which isn’t possible. Therefore G cannot have exactly

9 conjugacy classes.

(b) By Exercise 11.2 we have the dimensions of the irreducible modules of a group of order 12 are either 112, 1102, 1422, 133. Therefore the group has either 12, 6 or 4

conjugacy classes. Now the cyclic group C 12 has 12 conjugacy classes. The dihedral

group D12 has precisely 6 conjugacy classes by (12.12). Finally by Example 13.25 we

have A4 has precisely 4 conjugacy classes.



Chapter 16. Character tables and orthogonality relations

Exercise 16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Exercise 16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Exercise 16.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Exercise 16.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Exercise 16.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Exercise 16.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Exercise 16.1. Recall that C 2 × C 2 = (1, 1), (1, y ), (x, 1), (x, y ) such that x 2 = y 2 = 1then the character table is as in table 16.1.

g (1, 1) (1, y ) (x, 1) (x , y )

χ1(g ) 1 1 1 1

χ2(g ) 1 −1 1 −1

χ3(g ) 1 1 −1 −1

χ4(g ) 1

−1

−1 1

Table 16.1: The character table of C 2 × C 2

Exercise 16.2. Using the column orthogonality relations we get the character table of G

to be as in table 16.2.

g g 1 g 2 g 3 g 4 g 5

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 1 −1 −1

χ3(g ) 1 1 −1 1 −1

χ4(g ) 1 1 −1 −1 1

χ5(g ) 2 −2 0 0 0

Table 16.2: The character table of G from exercise 16.2

49



Chapter 16 50

Exercise 16.3. We first use the column othogonality relations to find χ3(1) and χ4(1).

We can see that g 4 is an element of order 2. This is because g 4 ∈ C G(g 4) and o (g 4) ||C G(g 4)| ⇒ o (g 4) = 2. The column relation on g 4 says 12 + 02 + |χ3(g 4)|2 + |χ4(g 4)|2 = 2,which gives us |χ3(g 4)|2 + |χ4(g 4)|2 = 1. We know from Corollary 13.10 that χ4(g 4) ≡ 0

(mod 2) so χ4(g 4) = 0. Finally the column relations on g 1 and g 4 give us 1 + χ3(g 4) = 0

so the character table so far is as in table 16.3

g g 1 g 2 g 3 g 4

χ1(g ) 1 1 1 1

χ2(g ) 2 α β 0

χ3(g ) 1 x 1 x 2 −1

χ4(g ) 2 y 1 y 2 0

Table 16.3: Partial character table of G, |G| = 10

We now use the column orthogonality relations of g 2, g 4 and g 3, g 4 to get

1 − x 1 = 0 ⇒ x 1 = 1 1 − x 2 = 0 ⇒ x 2 = 1.

Finally we use the column orthogonality of g 1, g 2 and g 1, g 3 to get the values

1 + 2α + 1 + 2 y 1 = 0 ⇒ y 1 = −1 − α = β,

1 + 2 β + 1 + 2 y 2 = 0 ⇒ y 2 = −1 − β = α.

Therefore the final character table of G is as in table 16.4

g g 1 g 2 g 3 g 4

χ1(g ) 1 1 1 1

χ2(g ) 2 α β 0

χ3(g ) 1 1 1 −1

χ4(g ) 2 β α 0


Exercise 16.4. We have the character table as in the question.



Chapter 16 51

(a) From the column orthogonality of g 2 we have

3 + |ζ |2

+ |ζ |2

= 7 ⇒ 2|ζ |2

= 4 ⇒ |ζ |2

= 2.

Now using the orthogonality of g 1 and g 2 we have

3 + 3ζ + 3ζ = 0 ⇒ ζ + ζ = −1 ⇒ 2Re(ζ ) = −1 ⇒ Re(ζ ) = −1

2.

Using this and combining it with the first equation we have

−

1

22

+ Im(ζ )2 = 2

⇒

1

4

+ Im(ζ )2 = 2,

⇒ 1 + 4 Im(ζ )2 = 8,

⇒ 4Im(ζ )2 = 7,

⇒ Im(ζ )2 =7

4,

⇒ Im(ζ ) =

√ 7

2.

Therefore ζ = −1+√ 7i

2.

(b) Recall that we have χ(g ) is real valued if g is conjugate to g −1

. Now the fact thatχ4(g 2) and χ5(g 2) are complex, means that g −12 ∈ g G2 . Now we know χi (g −12 ) = χi (g )

and so the column of the character table reads 1, 1, 1, ζ , ζ . Also recall that for all

x ∈ G we have xg 2 = g 2x ⇔ g −12 x = xg −12 . Therefore |C G(g −12 )| = |C G(g 2)| = 7.

This leaves us with the third column of the character table, which is given in table

16.5.

g g 1 g 2 g −12

χ1(g ) 1 1 1χ2(g ) 1 1 1

χ3(g ) 1 1 1

χ4(g ) 3 ζ ζ

χ5(g ) 3 ζ ζ

Table 16.5: Three columns of the character table of G from exercise 16.4



Chapter 16 52

Exercise 16.5. Let X = g ∈ G | k i =1 χi (g )χi (g ) = |G|. Recall from the column

orthogonality relation thatk

i =1

χi (z )χi (z ) = |C G(z )|.

Then we have

x ∈ X ⇔k

i =1

χi (z )χi (z ) = |G| ⇔ |C G(z )| = |G| ⇔ z ∈ Z (G),

in other words Z (G) = X as required.

Exercise 16.6. Now let C be the character table of G. Now we know that χ(g ) = χ(g −1)

and so C is obtained from C be rearranging the columns. If we swap two columns in a

matrix then we multiply the determinant by −1. Therefore det C = det C = ± det(C ). If

det C = det C then det C is real and if det C = − det(C ) then det C is purely imaginary.

We consider the matrix C T C . Now we can see that (C T C )ij =k

=1 χ(g i )χ(g j ) =

δij |C G(g i )| by the column orthogonality relations. Therefore we have

| det C |2 = (det C )(det C ) = det(C T C ) =

k i =1

|C G(g i )|.

Finally for G = C 3, recall that ω = 1+√ 3i 2 and ω2 = 1−√ 3i

2 , then we have the determinant

of the character table to be1 1 1

1 ω ω2

1 ω2 ω

= (ω2 − ω) − (ω − ω2) + (ω2 − ω),

= 3(ω2 − ω),

= −3√

3i .



Chapter 17. Normal subgroups and lifted characters

Exercise 17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Exercise 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Exercise 17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Exercise 17.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Exercise 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Exercise 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Exercise 17.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Exercise 17.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Exercise 17.1. Let G = Q8 = a, b | a4 = 1, b 2 = a2, b −1ab = a−1.

(a) From Exercise 12.6 we have the conjugacy classes are 1, a2, a, a3, ab,a3b and b, a2b .

(b) We can see that a−1b −1ab = a−2 = a2. Therefore we can see that G = a2 ∼= C 2.

Now, let H = a2 then we have G/H = H,aH,bH,abH and (aH )2 = (bH )2 = 1.

Therefore it’s easy to see that G/H ∼= C 2 × C 2. Using table 16.1 we can easily see

the character table of G/H is given in table 17.1.

gH H aH bH abH

χ1(g ) 1 1 1 1

χ2(g ) 1 −1 1 −1

χ3(g ) 1 1 −1 −1

χ4(g ) 1 −1 −1 1

Table 17.1: The character table of G/H

We can now use this to obtain the linear characters of Q8. We see that a2H = H

and so the linear characters are given in table 17.2.

(c) Clearly there is one more character to find and it has degree 2. Therefore using the

column orthogonality relations it’s easy to see the character table of Q8 is as in table

17.3. Comparing the two we can see that Q8 has the same character table as D8 but

Q8 ∼= D8. Therefore two groups with the same character table are not necessarily

isomorphic.

53



Chapter 17 54

g 1 a2 a ab b

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 −1 1 −1

χ3(g ) 1 1 1 −1 −1

χ4(g ) 1 1 −1 −1 1

Table 17.2: The linear characters of Q8

g 1 a2 a ab b

|C G(g )| 8 8 4 4 4

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 −1 1 −1

χ3(g ) 1 1 1 −1 −1

χ4(g ) 1 1 −1 −1 1

χ5(g ) 2 −2 0 0 0

Table 17.3: The character table of Q8



Chapter 17 55

Exercise 17.2. It’s clear that a7 = b 3 = 1 by the elements cycle shape. Now,

b −1

ab = (1b 2b 3b 4b 5b 6b 7b ) = (1357246) = (1234567)(1234567) = a2

.

Therefore the relations hold and G = a, b | a7 = b 3 = 1, b −1ab = a2.

(a) It’s clear from the relations that every element can be written in the form b i a j for

some 0 i 6 and 0 j 2. So we can see that |G| 21. However G contains

the cyclic subgroups a and b , which have orders 7 and 3 respectively. Therefore

7 | |G| and 3 | |G|, which means |G| = 21.

(b) First we note that the conjugacy classes either have size 3 or 7. Now clearly a

and a

2

are conjugate by the relation b −1

ab = a

2

. Using this relation we get theconjugacy classes a, a2, a4 and a3, a5, a6. Now using this relation again we get

b −1(ba)b = ab = ba2 and so we potentially get the conjugacy classes ba,ba2, ba4and ba3, ba5, ba6. Again using the relation we get b −1(b 2a)b = bab = b 2a2 and

so we potentially get the conjugacy classes b 2a, b 2a2, b 2a4 and b 2a3, b 2a5, b 2a6.

What about b and b 2? Recall from the relations that we have a−1b = ba−2 and

a−1b 2 = b 2a4. Then a−1ba = ba5a = ba6 and a−3ba3 = ba−6a3 = ba4. Also

a−1b 2a = b 2a5 and a−3b 2a3 = b 2a12a3 = b 2a. Therefore the conjugacy classes of G

are

1, a, a2, a4, a3, a5, a6,

b,ba,ba2, ba3, ba4, ba5, ba6, b 2, b 2a, b 2a2, b 2a3, b 2a4, b 2a5, b 2a6.

(c) What’s the derived subgroup of G? Well [a, b ] = a−1b −1ab = a and so G = a.

Therefore G/G = G , bG , b 2G ∼= C 3 because (bG )3 = G. Therefore the charac-

ter table of G/G is as in table 17.4. Note that in table 17.4 ω = e 2πi/3 = 1+√ 3i

2is a

3rd root of unity.

gG G bG b 2G

χ1(g ) 1 1 1

χ2(g ) 1 ω ω2

χ3(g ) 1 ω2 ω

Table 17.4: The character table of G/G from exercise 17.2

We can see that aG = a3G = G and so we can lift the above character table





Chapter 17 57

g 1 a a3 b b 2

|C G(g )

|21 7 7 3 3

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 1 ω ω

χ3(g ) 1 1 1 ω ω

χ4(g ) 3 α β 0 0

χ5(g ) 3 α β 0 0

Table 17.6: The partial character table of G from exercise 17.2

From the row orthogonality we get

3

21+

α

7+

β

7= 0 ⇒ 3 + 3α + 2 β = 0 ⇒ β = −1 − α.

If α = 1+√ 7i

2 then β = −1 − −1+√ 7i

2= −1−√

7i 2

= α. In other words the character

table for G is as in table 17.7.

g 1 a a3 b b 2

|C G(g )| 21 7 7 3 3

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 1 ω ω

χ3(g ) 1 1 1 ω ω

χ4(g ) 3 α α 0 0

χ5(g ) 3 α α 0 0


Exercise 17.3. Recall that |G/G | = |G|/|G| counts the number of linear characters. Now

|G | | |G| and so |G/G | = 1, 2, 3, 4, 6 or 12. From Exercise 15.4(b) we have that a group

of order 12 has either 4, 6 or 12 conjugacy classes, in other words it has either 4, 6 or 12

irreducible characters. Assume that G has 12 conjugacy classes then G is abelian and so

we must have G has 12 linear characters.

Assume G is not abelian and has 6 irreducible characters, say χ1, χ2, χ3, χ4, χ5 and χ6



Chapter 17 58

then

χ1(1)2 + χ2(1)2 + χ3(1)2 + χ4(1)2 + χ5(1)2 + χ6(1)2 = 12.

In this case we cannot have 6, 3, 2 or 1 linear characters. However we can have 4 linear

characters with χ5(1) = χ6(1) = 2.

Assume G is not abelian and has 4 irreducible characters, then we must have

χ1(1)2 + χ2(1)2 + χ3(1)2 + χ4(1)2 = 12.

In this case we cannot have 4, 3 or 2 linear characters. However we can have 3 linear

characters by letting χ4(1) = 3.

Therefore we have the following options for G. Either G is abelian and so has 12

conjugacy classes and 12 linear characters, G is not abelian and has 6 conjugacy classes

with 4 linear characters or G has 4 conjugacy classes and 3 linear characters. Note that

this shows there is no perfect group, (i.e. a group G such that G = G), of order 12. Also

clearly there is no simple group of order 12 as G is always a non-trivial normal subgroup in

this case.

Exercise 17.4. We use proposition 17.14 to construct a new character from the linear

character given. From the above Exercise we have that because G is of order 12 and G

has 6 conjugacy classes then G has four linear characters. Let χ4 = χ and χ5 = φ thenχ6 = χ5χ4. We have χ3 = χ4 because χ4 isn’t real, which means χ4 is another irreducible

character. Finally we get χ2 by using the column orthogonality relations with χ1. Hence

we have the character table is given as in table 17.8.

g 1 g 2 g 3 g 4 g 5 g 6

χ1(g ) 1 1 1 1 1 1

χ2(g ) 1 −1 −1 1 1 1

χ3(g ) 1 i −i 1 −1 −1χ4(g ) 1 −i i 1 −1 −1

χ5(g ) 2 0 0 −1 −1 2

χ6(g ) 2 0 0 −1 1 −2


An alternative approach would be to consider powers of χ with itself. By Proposition

17.14 we have χ, χ2, χ3, χφ are all irreducible characters of G. In fact χ1 = 1G, χ2 = χ2,



Chapter 17 59

χ3 = χ3, χ4 = χ, χ5 = φ, χ6 = χφ. We can now determine the order of the centralisers

and hence the conjugacy classes using the column orthogonality relations and the orbit

stabiliser theorem. We give these in table 17.9.

g 1 g 2 g 3 g 4 g 5 g 6

|C G(g )| 12 4 4 6 6 12

|g G| 1 3 3 2 2 1

Table 17.9: The conjugacy classes of G from exercise 17.4

Exercise 17.5. We can read off from the character table what the kernels are. Thereforethe normal subgroups of D8 are

ker(χ1) = D8 ker(χ2) = 1, a , a2, a3 ker(χ3) = 1, a2, b , a2b ,

ker(χ4) = 1, a2,ab,a3b ker(χ5) = 1 ker(χ2) ∩ ker(χ3) = 1, a2.

Exercise 17.6. Let G = T 4n = a, b | a2n = 1, an = b 2, b −1ab = a−1.

(a) Let ρ : T 4n → GL(2,C) be the map defined in the question. To show that this is a

representation it is enough to show that the relations hold.

(aρ)2n =

ε 0

0 ε−1

2n=

ε2n 0

0 ε−2n

= I2,

(aρ)n =

εn 0

0 ε−n

=

εn 0

0 εn

=

0 1

εn 0

0 1

εn 0

= (bρ)2,

(bρ)−1(aρ)(bρ) =

0 εn

1 0

ε 0

0 ε−1

0 1

εn 0

,

= 0 εn−1

ε 0 0 1

εn 0 ,

=

ε2n−1 0

0 ε

,

= (aρ)−1.

Therefore this is indeed a representation of T 4n.

(b) We start by considering the conjugacy classes for T 4n. We claim that, for 1 r



Chapter 17 60

n − 1, the conjugacy classes are as follows

1 |C G(1)| = |G| = 4n,an |C G(an)| = |G| = 4n,

ar , a−r |C G(a)| = |a| = 2n,

ba2 j | 0 j n − 1 |C G(b )| = |b | = 4,

ba2 j +1 | 0 j n − 1 |C G(ab )| = |b | = 4.

Note that there are n + 3 conjugacy classes. We now argue why these are indeed the

conjugacy classes of T 4n. However before doing this we make a passing comment on

the multiplication in T 4n. We have b −1

ab = a−1

⇔ ab = ba−1

⇔ b = a−1

ba−1

⇔ba = a−1b .

• 1. This is clear.

• an. It’s clear that a C G(an) but because an = b 2 we also have b C G(an). In other words |C G(an)| = |G| = 4n, which means |(an)G| = [G :

C G(an)] = 1.

• ar , a−r . We have b −1ar b = a−r from the relations and so ar , a−r ⊆ (ar )G.

What about the centraliser? Well a C G(ar ) and so we have

2 |(ar )G| = [G : C G(ar )] [G : a] = 2.

• ba2 j | 0 j n − 1. We have a−1(ba2 j )a = (a−1b )a2 j +1 = ba2 j +2 = ba2( j +1)

from the relations, so certainly ba2 j | 0 j n − 1 ⊆ b G. What about the

centraliser of b ? Well b C G(b ) which gives us

n |b G| = [G : C G(b )] [G : b ] =4n

4= n.

• ba2 j +1

| 0 j n − 1. We have a−1

(ba2 j +1

)a = (a−1

b )a2 j +2

= ba2 j +3

=ba2( j +1)+1 from the relations, so certainly ba2 j +1 | 0 j n − 1 ⊆ (ba)G. So

far we’ve used 1 + 1 + 2(n − 1) + n = 3n elements and there are n elements in

this set. Therefore this accounts for all the elements and we’re done.

We claim that the degree 2 representation from part (a) is irreducible. This in

fact follows from Exercise 8.4, if ε = ±1, but we will show the inner product of the

character is equal to 1 for verification. Recall that ε2n = 1 ⇒ εn = ±1. We choose

ε = ±1 then the character is as in table 17.10.



Chapter 17 61

g 1 an ar b ba

|C G(g )

|4n 4n 2n 4 4

χ(g ) 2 −2 εr + ε−r 0 0

Table 17.10: The character of the representation from Exercise 17.6(a)

We now check the inner product of this character with itself to verify that ρ is an

irreducible representation.

χ, χ

=

22

4n

+(−2)2

4n

+

n−1

r =1(εr + ε−r )(εr + ε−r )

2n

,

=1

n+

1

n+

n−1r =1

|εr |2 + εr ε−r + ε−r εr + |εr |22n

.

Recall that εr = ε−r then the summand simplifies to

=2

n+

n−1r =1

2 + ε2r + ε−2r

2n,

= 2n

+ n − 1n

+

n−1r =1

ε2r

n,

= 1 +1

n− 1

n,

= 1.

Therefore the character is irreducible. However note that this is not true if ε = ±1

becausen−1

j =1 ε2 j = n − 1 in this case.

We briefly explain why

n−1 j =1 = ε2 j = −1. Recall ε is a solution of the polynomial

X 2n

− 1 = 0. Now we have

X 2n − 1 = 0 ⇔ (X n + 1)(X n − 1) = 0,

⇔ X n + 1 = 0 or X n − 1 = 0,

⇔ X n + 1 = 0 or (X − 1)(X n−1 + · · · + X 2 + X + 1) = 0.

Consider ε2, then this is a solution of X n − 1 = 0 because (ε2)n = ε2n = 1. If

ε2 = 1 ⇔ ε = ±1 then we have ε2 is a solution of X n−1 + · · · + X + 1 = 0. This



Chapter 17 62

gives us the required identity for the character calculation. Note, this explains why if

ε = ±1 we have the character is not irreducible

The linear characters are the lifts from the derived subgroup. Using the rela-tions we have b −1(a−1b )a = b −1(ba)a = a−2, so G = a2. This means G/G =

G, aG , bG ,baG and so there are |G/G | = 4 linear characters. If n is even we

have G/G ∼= C 2 × C 2 because b 2 = an ∈ G . If this is the case then the character

table for G is as in table 17.11.

g 1 an ar 1r n−1

b ba

|C G(g )| 4n 4n 2n 4 4

1G 1 1 1 1 1

φ1 1 1 1 −1 −1

φ2 1 1 (−1)r 1 −1

φ3 1 1 (−1)r −1 1

χk (g )0k n−1

2 2(−1)k εkr + ε−kr 0 0

Table 17.11: The character of T 4n, when n is even

If n is odd then G/G ∼= C 4 because b 2G = anG = aG and b 3G = bb 2G =banG = baG . Therefore we lift the four linear characters of C 4 to the group G and

we have the character table is as in table 17.12.

Notice that we get a character χk for each 1 k n − 1 when ε is a primitive

2nth root of unity, for example ε = e 2πi/2n. Note that taking n k 2n−1 wouldn’t

give us any more unique characters. This gives us (n − 1) + 4 = n + 3 irreducible

characters and so these are all the irreducible characters of G.

Exercise 17.7. Let G = U 6n = a, b | a2n = b 3 = 1, a−1ba = b −1.

(a) Let ρ : U 6n → GL(2,C) be the map defined in the question. To show this is arepresentation it is enough to show that the relations hold

(aρ)2n =

0 ε

ε 0

2n=

ε2 0

0 ε2

n=

ε2n 0

0 ε2n

= I2,

(bρ)3 =

ω 0

0 ω2

3=

ω3 0

0 ω6

= I2,



Chapter 17 63

g 1 an ar 1r n−1

b ba

|C G(g )| 4n 4n 2n 4 4

1G 1 1 1 1 1

φ1 1 −1 (−1)r i −i

φ2 1 1 1 −1 −1

φ3 1 −1 (−1)r −i i

χk (g )1k n−1

2 2(−1)k εkr + ε−kr 0 0

Table 17.12: The character of T 4n, when n is odd

(aρ)−1(bρ)(aρ) =

0 ε−1

ε−1 0

ω 0

0 ω2

0 ε

ε 0

,

=

0 ε−1ω2

ε−1ω 0

0 ε

ε 0

,

=

ω2 0

0 ω

,

= (bρ)−1.

Therefore ρ certainly is a representation of U 6n.

(b) We start by considering the conjugacy classes of U 6n. We claim that, for 0 j n−1,

the conjugacy classes are as follows

a2 j |C G(1)| = |G| = 6n,

a2 j b, a2 j b 2 |C G(b )| = |b, a2| = 3n,

a2 j +1, a2 j +1b, a2 j +1b 2 |C G(a)| = |a| = 2n.

Note that there are n + n + n = 3n conjugacy classes. We now argue why these are

indeed the conjugacy classes of U 6n. However before doing this we make a passing

comment on the multiplication in U 6n. We have a−1ba = b −1 ⇔ a−1b −1a = b ⇔b −1a = ab and a−1ba = b −1 ⇔ ba = ab −1. Therefore ba j = a j b if j is even and a j b −1

if j is odd.

• a2 j . We have b −1(a2 j )b = a2 j b −1b = a2 j from the relations. This means

b C G(a2 j ) and so C G(a2 j ) = G, which gives us |(a2 j )G| = [G : C G(a j )] = 1.



Chapter 17 64

• a2 j b, a2 j b 2. We have a−1(a2 j b )a = a2 j −1(ab −1) = a2 j b 2, so certainly we

have a2 j b, a2 j b 2 ⊆ (a2 j b )G. What about the centraliser? We have b C G(a2 j b ) because b C G(a j ) and C G(b ). However we also have a−2(a2 j b )a2 =a−1(a2 j b 2)a = a2 j −1(ab −2) = a2 j b . Therefore a2 C G(a2 j b ), which leaves us

with

2 |(a2 j b )G| = [G : C G(a2 j b )] [G : a2, b ] =6n

3n= 2.

• a2 j +1, a2 j +1b, a2 j +1b 2. We have

b −1a2 j +1b = a2 j +1b 2 and b −2a2 j +1b 2 = a2 j +1b 4 = a2 j +1b

from the relations, so certainly we have a2 j +1, a2 j +1b, a2 j +1b 2 ⊆ (a2 j +1)G.

What about the centraliser? Clearly a C G(a2 j +1), which gives us

3 |(a2 j +1)G| = [G : C G(a2 j +1)] [G : a] =6n

2n= 3.

We claim that the representation defined in part (a) is irreducible. This is true by

Exercise 8.4 for any 2nth root of unity ε. We will also show the inner product is equal

to 1 for verification. The character of the representation is given in table 17.13.

g a2 j a2 j +1 a2 j b

|C G(g )| 6n 2n 3n

χ(g ) 2ε2 j 0 −ε2 j


These values come from the following matrix calculations

(a2 j ρ) =

ε2 j 0

0 ε2 j

(a2 j +1ρ) =

0 ε2 j +1

ε2 j +1 0

(a2 j bρ) =

ωε2 j 0

0 ω2ε2 j

and the fact that ω2 + ω = −1. The inner product of the character is

χ, χ =

n−1 j =0

2ε2 j 2ε2 j

6n+

n−1 j =0

−ε2 j −ε2 j

3n,



Chapter 17 65

=

n−1

j =02|ε2 j |2

3n+

n−1

j =0|ε2 j |2

3n,

=2

3n· n +

1

3n· n,

= 1.

We now consider the derived subgroup of G. We know from the relations that

b −1a−1ba = b −2 = b , which gives us that G = b . Therefore |G/G | = 6n/3 =

2n and so there are 2n linear characters lifted from G. We have that G/G =

G, aG ∼= C 2n because (aG ) = a2nG = G. It’s easy to see that a2 j G = a2 j G ,

a2 j +1G = a2 j +1G and a2 j bG = a2 j G. Therefore the character table of G is as in

table 17.14.

g a2 j 0 j n−1

a2 j +10 j n−1

a2 j b 0 j n−1

|C G(g )| 6n 2n 3n

φk (g )0k 2n−1

ε2kj ε2k ( j +1) ε2kj

χ(g )0n−1

2ε2j 0 −ε2j

Table 17.14: The character table of U 6n

for ε some primitive 2nth root of unity, for example ε = e 2πi/2n. This gives us

2n + n = 3n irreducible characters, which is the number of conjugacy classes and

hence all the irreducible characters.

Exercise 17.8. Let G = V 8n = a, b | a2n = b 4 = 1, ba = a−1b −1, b −1a = a−1b , for n odd.

(a) Let ρ : V 8n → GL(2C) be the map defined in the question. To show this is a

representation it is enough to show that the relations hold

(aρ)2n =

ε 0

0 −ε−1

2n=

ε2n 0

0 (−)2nε−2n

= I2,

(bρ)4 =

0 1

−1 0

4=

−1 0

0 −1

2= I2,



Chapter 17 66

(bρ)(aρ) =

0 1

−1 0

ε 0

0 −ε−1

,

=

0 −ε−1

−ε 0

,

=

ε−1 0

0 −ε

0 −1

1 0

,

= (aρ)−1(bρ)−1,

(bρ)−1(aρ) =

0 −1

1 0

ε 0

0 −ε−1

,

=0 ε−1

ε 0

,

=

ε−1 0

0 −ε

0 1

−1 0

,

= (aρ)−1(bρ).

Therefore this is indeed a representation of V 8n.

(b) We start by considering the conjugacy classes for V 8n. We claim, for 1 r n−12 and

0 s n − 1, the conjugacy classes are as follows

1 |C G(1)| = |G| = 8n,

b 2 |C G(b 2)| = |G| = 8n,

a2r , a−2r |C G(a2)| = |a, b 2| = 4n,

a2r b 2, a−2r b 2 |C G(a2b 2)| = |a, b 2| = 4n,

a2s +1, a−2s −1b 2 |C G(a)| = |a, b 2| = 4n,

a2b, a2b 3 | 0 n − 1 |C G(b )| = |b | = 4,

a2+1b, a2+1b 3 | 0 n − 1 |C G(ab )| = |b | = 4.

Note that 2 | n−1 because n is odd. We now argue that these are indeed the conjugacy

classes of V 8n. Before doing this we note some incredibly useful multiplication formulae

for the group

ba = a−1b −1 b −1a = a−1b ba−1 = ab −1 b −1a−1 = ab,

ba2r = a−2r b b −1a2r = a−2r b −1 ba2s +1 = a−2s −1b −1 b −1a2s +1 = a−2s −1b.



Chapter 17 67

• 1. This is clear.

•

b 2

. We have ab 2a−1 = aa−1b −2 = b 2, so

a

C G(b 2). Clearly we have

b C G(b 2), which gives us C G(b 2) = G so |(b 2)G| = 1.

• a2r , a−2r . We have ba2r b −1 = a−2r bb −1 = a−2r , so certainly a2r , a−2r ⊆(a2r )G . What about the centraliser? Clearly a C G(a2r ) but we also have

b 2(a2r )b −2 = ba−2r bb −2 = a2r bb −1 = a2r so b 2 C G(a2r ). This gives us

2 |(a2r )G| = [G : C G(a2r )] [G : a, b 2] =8n

4n= 2.

• a2r b 2, a−2r b 2. We have b (a2r b 2)b −1 = a−2r bb = a−2r b 2, so certainly we have

a2r

b 2

, a−2r

b 2

⊆ (a2r

b 2

)G

. What about the centraliser? By the previous pointwe have b 2 C G(a2r ) so b 2 C G(a2r b 2). Now a(a2r b 2)a−1 = a2r +1b 2a−1 =

a2r +1a−1b −2 = a2r b 2. Therefore a C G(a2r b 2) and so we have

2 |(a2r b 2)G| = [G : C G(a2r b 2)] [G : a, b 2] = 2.

• a2s +1, a−2s −1b 2. We have ba2s +1b −1 = a−2s −1b −1b −1 = a−2s −1b 2, so cer-

tainly a2s +1, a−2s −1b 2 ⊆ (a2s +1)G. What about the centraliser? Clearly a C G(a2s +1) but we also have b 2(a2s +1)b −2 = ba−2s −1b −3 = a2s +1b −4 = a2s +1 so

b 2 C G(a2s +1). This gives us

2 |(a2r )G| = [G : C G(a2r )] [G : a, b 2] = 2.

• a2b, a2b 3 | 0 n − 1. We have aba−1 = aab −1 = a2b 3 and ab 3a−1 =

ab −1a−1 = a2b . Recall that n is odd so when we conjugate by an we get

anba−n = ananb −1 = b 3 and anb 3a−n = anb −1a−n = ananb = b . So certainly we

have a2b, a2b 3 | 0 n − 1 ⊆ b G. What about the centraliser? Clearly

b

C G(b ) so we have

2n |b G| = [G : C G(b )] [G : b ] =8n

4= 2n.

• a2+1b, a2+1b 3 | 0 n − 1. We have a(ab )a−1 = a2ab −1 = a3b 3 and

a(ab 3)a−1 = a2b −1a−1 = a3b . Recall that n is odd so when we conjugate

by an we get an(ab )a−n = an+1ba−n = an+1anb −1 = ab 3 and an(ab 3)a−n =

an+1b −1a−n = an+1anb = ab . So certainly we have a2+1b, a2+1b 3 | 0

n − 1 ⊆ (ab )G. There are 2n elements in this set and we’ve used 1 + 1 +



Chapter 17 68

(n − 1) + (n − 1) + 2n + 2n = 6n elements so far. Therefore we must have

|(ab )G| = 2n and we’re done.

We know that the representation defined in part (a) is irreducible by Exercise8.4. However we also check the inner product of the character for verification. In

table 17.15 we give the character of this representation with respect to the conjugacy

classes defined above.

g 1 b 2 a2r a2r b 2 a2s +1 b ab

|C G(g )| 8n 8n 4n 4n 4n 4 4

χ(g ) 2 −2 ε2r + ε−2r −ε2r − ε−2r ε2s +1 − ε−2s −1 0 0


These values come from the following matrix calculations

(b 2)ρ =

−1 0

0 −1

(a2r )ρ =

ε2r 0

0 ε−2r

(a2r b 2)ρ =

−ε2r 0

0 −ε−2r

,

(a2s +1)ρ = ε2s +1 0

0 −ε−2s −1 (ab )ρ =

0 ε

ε−1

0 .

Recall that for an nth root of unity εk = ε−k . Then the inner product of the

character with itself is as follows

χ, χ =22

8n+

(−2)2

8n+

n−12

r =1

(ε2r + ε−2r )(ε2r + ε−2r )4n

+

n−12

r =1

(ε2r + ε−2r )(ε2r + ε−2r )4n

+

n−1

s =0(ε2s +1 − ε−2s −1)(ε2s +1 − ε−2s )

4n,

=8

8n+ 2

n−12

r =1

2|ε2r |2 + ε2r ε2r + ε−2r ε−2r

4n

+

n−1s =0

2|ε2s +1|2 − ε2s +1ε2s +1 − ε−2s −1ε−2s −1

4n,

=2

2n+

2

2n· n − 1

2+

1

2n· n +

n−12

r =1

ε4r + ε−4r

2n−

n−1s =0

ε4s +2 + ε−4s −2

4n,



Chapter 17 69

=2 + n − 1 + n

2n+

n−12

r =1ε4r + ε−4r

2n−

n−1

s =0ε4s +2 + ε−4s −2

4n,

=2n + 1

2n+

n−12

r =1

ε4r + ε−4r

2n−

n−1s =0

ε4s +2 + ε−4s −2

4n.

We investigate these sums a little further to make them more enlightening. Now

ε−4r = εn−4r = ε2n−4r . We have 1 r n−12

⇒ 4 4r 2n − 2. From this we can

see that 2n − (2n − 2) 2n − 4r 2n − 4 ⇒ 2 2n − 4r 2n − 4. So for the first

sum we have

n−12

r =1

ε4r + ε−4r =

n−12

r =1

ε4r +

n−32

r =1

ε4r +2,

= ε4 + ε8 + · · · + ε2n−2 + ε2 + ε6 + · · · + ε2n−4,

=

n−1r =1

ε2r .

Considering the second sum we have ε−4s −2 = εn−4s −2 = ε4n−4s −2 = ε4(n−s )−2. So

if 0 s n − 1 then 0 4s 4n − 4 ⇒ 2 4s + 2 4n − 2. This gives us that

4n−

(4n−

2) 4n−

(4s + 2) 4n−

2⇒

2 4(n−

s )−

2 4n−

2. Using this in

the second sum we see

n−1s =0

ε4s +2 + ε−4s −2 =

n−1s =0

ε4s +2 + ε4(n−s )−2,

= 2

n−1s =0

ε4s +2,

= 2(ε2 + ε6 + · · · + ε2n−4 + ε2n + ε2n+4 + · · · + ε4n−2),

= 2(ε2 + ε6 +

· · ·+ ε2n−4 + 1 + ε4 +

· · ·+ ε2n−2),

= 2

n−1s =0

ε2s .

Recall that ε is an nth root of unity and so is a solution to the polynomial X n−1 =

0. We have that 1 is a solution of this polynomial so we get a factorisation

X n − 1 = (X − 1)(X n−1 + · · · + X 2 + X + 1) = 0.



Chapter 17 70

If ε is an nth root of unity then so is ε2 because (ε2)n = (εn)2 = 12 = 1. Now

ε2 = 1 ⇒ ε = ±1 but ε = −1 because n is odd therefore εn = (−1)n = −1.

Assume ε = 1 then ε2 is a solution of X n − 1 = 0 but not a solution of X − 1 = 0.Therefore ε2 is a solution of X n−1 + · · · + X + 1 = 0. In other words

n−1s =0 ε2s = 0

andn−1

r =1 ε2r = −1. Using all this information in the calculation of our inner product

we have

χ, χ =2n + 1

2n+

n−1r =1

ε2r

2n− 2

n−1s =0

ε2s

4n=

2n + 1 − 1

2n= 1.

Finally, to show this we had to make the assumption that ε = 1. Assume ε = 1

then we have the inner product to be

χ, χ =2n + 1

2n+

n−1r =1

ε2r

2n− 2

n−1s =0

ε2s

4n=

(2n + 1) + (n − 1) − n

2n=

2n

2n= 1.

Therefore χ is an irreducible character of V 8n for any nth root of unit ε.

We now consider the linear characters of G. The derived group is generated by

[a, b ] = a−1b −1ab = a−2b 2 and hence also its inverse b −2a2 = b 2a2. Now (b 2a2)2 =

b 2a2b 2a2 = a4, so a4 G . Recall n = 2v + 1 is odd so a2n = 1 ⇒ a4v +2 = 1 ⇒a4v = a−2, which means a2 a4 G . Using the fact that a−2 ∈ G we have

(b 2a2)a−2 = b 2

∈G . Therefore G =

b 2, a2

, which gives us

|G/G

|= 8n/2n = 4.

In other words G/G = G, aG , bG ,abG with (aG )2 = a2G = G and (bG )2 =

b 2G = G . Clearly G/G ∼= C 2 × C 2 so we can lift the character table of C 2 × C 2

to the whole of G. Checking the cosets of G we can see a2r G = 22r b 2G = G and

a2s +1G = aG . Therefore the character table so far is given in table 17.16. Note

that in table 17.16 ε a primitive nth root of unity and 0 k n − 1.

g 1 b 2 a2r a2r b 2 a2s +1 b ab

|C G(g )| 8n 8n 4n 4n 4n 4 4

λ1(g ) 1 1 1 1 1 1 1

λ2(g ) 1 1 1 1 1 −1 −1

λ3(g ) 1 1 1 1 −1 1 −1

λ4(g ) 1 1 1 1 −1 −1 1

χk (g ) 2 −2 ε2kr + ε−2kr −ε2kr − ε−2kr ε(2s +1)k − ε−(2s +1)k 0 0

Table 17.16: The partial character table of V 8n



Chapter 17 71

There are 2n + 3 conjugacy classes but in the above table we only have n + 4

irreducible characters. Therefore there are n − 1 irreducible characters left. We could

try and combine a linear character with the χk but this in fact will not give us anew character. So, there must be more irreducible representations. Consider the

representation σ : G → GL(2,C) given by

aσ =

ω 0

0 ω−1

bσ =

0 1

1 0

,

for some 2nth root of unity ω. We check that this indeed a representation of V 8n by

checking the relations.

(aσ)2n =

ω 0

0 ω−1

2n=

ω2n 0

0 ω−2n

= I2,

(bσ)4 =

0 1

1 0

4=

1 0

0 1

2= I2,

(bσ)(aσ) =

0 1

1 0

ω 0

0 ω−1

=

0 ω−1

ω 0

=

ω−1 0

0 ω

0 1

1 0

= (aσ)−1(bσ)−1,

(bσ)−1(aσ) = 0 1

1 0ω 0

0 ω−1 = 0 ω−1

ω 0 = ω−1 0

0 ω0 1

1 0 = (aσ)−1(bσ).

This representation is irreducible, as long as ω = ±1, by Exercise 8.4. We also

check the inner product of the character for verification. The character for σ is given

in table 17.17.

g 1 b 2 a2r a2r b 2 a2s +1 b ab

|C G(g )| 8n 8n 4n 4n 4n 4 4

φ(g ) 2 2 ω2r + ω−2r ω2r + ω−2r ω2s +1 + ω−2s −1 0 0

Table 17.17: The character of σ for V 8n

Assume ω = ±1 then we have the inner product to be

φ, φ =22

8n+

22

8n+ 2

n−12

r =1

(ω2r + ω−2r )(ω2r + ω−2r )4n



Chapter 17 72

+

n−1

s =0(ω2s +1 + ω−2s −1)(ω2s +1 + ω−2s −1)

4n,

=2

2n+

n−12

r =1

2 + ω4r + ω−4r

4n+

n−1s =0

2 + ω4s +2 + ω−4s −2

4n,

=2

2n+

2

2n· n − 1

2+

2n

4n+

n−1r =1

ω2r

2n+ 2

n−1s =0

ω2s

4n,

=2 + (n − 1) + n − 1

2n,

=2n

2n,

= 1.

Most of the arguments for the above calculation are identical to the previous inner

product calculation, however the arguments for the sums are slightly different. Recall

ω is a 2nth root of unity, hence it is a solution of the polynomial X 2n − 1 = 0. We

have a factorisation X 2n − 1 = (X n +1)(X n − 1) = 0. Now ω2 is an nth root of unity

because (ω2)n = ω2n = 1, hence it is a solution of the polynomial X n − 1 = 0. We

have a factorisation

X n

− 1 = (X − 1)(X n−1

+ · · · + X 2

+ X + 1) = 0.

If ω2 = 1 ⇔ ω = ±1 then ω2 is a solution of the polynomial X n−1 + · · · + X + 1 = 0

and this gives us the desired results as before.

Assume ω = ±1 then we have the inner product calculation gives

φ, φ =2n + 1

2n+

n−1r =1

ω2r

2n+ 2

n−1s =0

ω2s

4n=

(2n + 1) + (n − 1) + n

2n=

4n

2n= 2

and so the character is not irreducible in this case. This confirms what we’re told by

Exercise 8.4 that the representation is irreducible as long as ω = ±1.

Let η be a primitive 2nth root of unity then η2 is a primitive nth root of unity.

Now letting ω = η and ε = η2 we have the character table of V 8n is as in table 17.18.



Chapter 17 73

g 1 b 2 a2r 1r n−1

2

a2r b 21r n−1

2

a2s +10s n−1

b ab

|C G(g )| 8n 8n 4n 4n 4n 4 4

λ1(g ) 1 1 1 1 1 1 1

λ2(g ) 1 1 1 1 1 −1 −1

λ3(g ) 1 1 1 1 −1 1 −1

λ4(g ) 1 1 1 1

−1

−1 1

χk (g )0k n−1

2 −2 η4kr + η−4kr −η4kr − η−4kr η2k (2s +1) − η−2k (2s +1) 0 0

φ(g )1 n−1

2

2 2 η2r + η−2r η2r + η−2r η(2s +1) + η−(2s +1) 0 0

Table 17.18: The character table of V 8n



Chapter 18. Some elementary character tables

Exercise 18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Exercise 18.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Exercise 18.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Exercise 18.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Exercise 18.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Exercise 18.1. We need one rotation symmetry of order 4, so we use the obvious a =

(1234). We also need a reflection of order 2 such that b −1(1234)b = (1234)−1 = (1432).Letting b = (14)(23) we have b −1(1234)b = (1b 2b 3b 4b ) = (4321) = (1432). Therefore

D8∼= H = (1234), (14)(23) S4.

Explicitly we have the elements of the subgroup and the conjugacy classes are

H = 1, (13), (24), (12)(34), (13)(24), (14)(23), (1234), (1432),

1

(13)(24)

(1234), (1432)

(24), (13)

(14)(23), (12)(34)

.

Using this information and the table in Example 16.3(3), (page 161), we have the character

table of D8 is as in table 18.1.

g 1 (13)(24) (13) (12)(34) (1234)

|C G(g )| 8 8 4 4 4

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 −1 −1 1

χ3(g ) 1 1 −1 1 −1

χ4(g ) 1 1 1 −1 −1

χ5(g ) 2 −2 0 0 0

π(g ) 4 0 2 0 0

Table 18.1: The character table of D8

Clearly π is not the sum of the four linear characters as the values on (13) don’t agree.

74



Chapter 18 75

Therefore it is the sum of χ5 and two linear characters. The dimension of the trivial module

in the permutation module is always one, so we know one of the linear characters is the

trivial character. Therefore it’s clear to see that π = χ1 + χ4 + χ5.Note: the decomposition of the permutation character is dependent upon the choice

of D8 as a subgroup of S4. Choosing b to be a reflection in a diagonal of the square will

result in π = χ1 + χ3 + χ5, (see solutions - page 419).

Exercise 18.2. By section 18.3 we have the character tables for all dihedral groups. Ex-

plicitly the character table of D12 is as in table 18.2. Recall ε = e 2πi/6 is a primitive 6th

root of unity. Explicitly we have ε = 1+√ 3i

2 , ε2 = −1+√ 3i

2 and ε4 = −ε = −1−√ 3i

2 .

g 1 a3

a a2

b ab |C G(g )| 12 12 6 6 2 2

χ1(g ) 1 1 1 1 1 1

χ2(g ) 1 1 1 1 −1 −1

χ3(g ) 1 −1 −1 1 1 −1

χ4(g ) 1 −1 −1 1 −1 1

ψ1(g ) 2 −2 1 −1 0 0

ψ2(g ) 2 2−

1−

1 0 0


It’s now simple to read off from the table what the kernels of all the characters are, by

comparing the character values to that of the identity.

ker(χ1) = G ker(χ4) = a2, ab ,

ker(χ2) = a ker(ψ1) = 1,

ker(χ3) =

a2, b

ker(ψ2) =

a3

.

Using Proposition 17.5 we have that these are nearly all the normal subgroups of D12. In

fact the only subgroup that is missing is ker(χ3) ∩ ker(χ4) = a2. Therefore the seven

distinct normal subgroups of D12 are 1, a3, a2, a, a2, b , a2, ab and G.

Exercise 18.3. See Exercise 17.6





Chapter 19. Tensor products

Exercise 19.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Exercise 19.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Exercise 19.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Exercise 19.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Exercise 19.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Exercise 19.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Exercise 19.1. Examining the inner products we see

χψ,φ =1

|G|g ∈G

χψ(g )φ(g ) χψ,φ =1

|G|g ∈G

χψ(g )φ(g ),

=1

|G|g ∈G

χ(g )ψ(g )φ(g ) =1

|G|g ∈G

χ(g )ψ(g )φ(g ),

=1

|G|g ∈G

χ(g )ψ(g )φ(g ) =1

|G|g ∈G

ψ(g )χ(g )φ(g ),

= g ∈G χ(g )ψφ(g ) =

1

|G| g ∈G ψ(g )χφ(g ),

= χ,ψφ = ψ,χφ.

Therefore χψ,φ = χ,ψφ = ψ,χφ as required.

Exercise 19.2. Let χ and ψ be irreducible characters of G then, using Exercise 1, we have

χψ, 1G = χ, ψ1G = χ, ψ =

1 if χ = ψ,

0 if χ

= ψ.

Recall that if ψ is an irreducible character then so is ψ, by Proposition 13.15. Therefore

the final equality comes from Theorem 14.12.

Exercise 19.3. Let U be a CG module which affords the unfaithful character χ. As χ is

not faithful we know there exists a non-identity element g ∈ ker(χ) such that

ug = u for all u ∈ U.

76



Chapter 19 77

The element g is not conjugate to 1, therefore there exists an irreducible character ψ such

that ψ(g ) = ψ(1), (by Proposition 15.5).

Let n 0 be an integer and consider the CG module constructed as the n-fold tensorproduct of U , say W = U ⊗ · · · ⊗ U . Clearly

w g = (u 1 ⊗ · · · ⊗ u n)g = u 1g ⊗ · · · ⊗ u ng = u 1 ⊗ · · · ⊗ u n = w for all w ∈ W.

Consider any submodule W ⊆ W and let the character afforded by this submodule be φ.

Then it’s clear that φ(g ) = φ(1) because w g = w for all w ∈ W . So if ψ(g ) = ψ(1) we

must have χn, ψ = 0 because ψ cannot correspond to a submodule of W .

Exercise 19.4. We express the characters χS, χA, φS and φA in table 19.1. We also calcu-late what g 2 is for each representative of the conjugacy classes. Then we use the formulae

for χS and χA given in the summary of Chapter 19.

g 1 (123) (12)(34) (12345) (13452)

g 2 1 (132) 1 (13524) (14235)

χ(g ) 5 −1 1 0 0

χS(g ) 15 0 3 0 0

χA(g ) 10 1 −2 0 0

φ(g ) 3 0 −1 1+√ 5

21−√

52

φS(g ) 6 0 2 1 1

φA(g ) 3 0 −1 1+√ 5

21−√

52

Table 19.1: Decomposing squares of characters of A5

It is worthwhile commenting that (12345) is conjugate to (13524) by (2354) ∈ A5

and (13452) is conjugate to (14235) by (3425)∈A5

. Therefore the squared five cycles

lie in the opposite conjugacy class to where they started. We explain in more depth the

calculations in the above table. For a start we have1 +

√ 5

2

2

=1 + 2

√ 5 + 5

4=

3 +√

5

2

1 − √

5

2

2

=1 − 2

√ 5 + 5

4=

3 − √ 5

2

Using this information we can then show the explicit calculations of φS and φA on the five



Chapter 19 78

cycles.

φS((12345)) = 123 +

√ 52 + 1 −

√ 52 = 12 · 42 = 1,

φA((12345)) =1

2

3 +

√ 5

2− 1 − √

5

2

=

1

2

2 + 2

√ 5

2

=

1 +√

5

2,

φS((13452)) =1

2

3 − √

5

2+

1 +√

5

2

=

1

2· 4

2= 1,

φA((13452)) =1

2

3 − √

5

2− 1 +

√ 5

2

=

1

2

2 − 2

√ 5

2

=

1 − √ 5

2.

Using the character table for A5 from Example 20.14, page 221, it’s easy to spot that

φA = ψ4 and φS = ψ1 + ψ3. Now χS and χA are a little more difficult to spot. We first

consider the inner products with themselves, this gives us

χS, χS =152

60+

32

4=

225 + 135

60=

360

60= 6,

χA, χA =102

60+

12

3+

(−2)2

4=

100 + 20 + 60

60=

180

60= 3.

Therefore we have d 21 + d 22 + d 23 + d 24 + d 25 = 6 or 3. In the first case our only option is 2, 1, 1

and in the second case our only option is 1, 1, 1. Using this information and working on the

values of χS(1) and χA(1) it’s easy to see that χA = ψ2+ ψ4+ ψ5 and χS = ψ1+ ψ2+ 2ψ3.

Exercise 19.5. We express all the information in the question in table 19.2 and use the

formulas for χS and χA to complete the table.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

g 2i g 1 g 1 g 2 g 5 g 4 g 4 g 5

|C G(g i )| 24 24 4 6 6 6 6

χ(g ) 2 −2 0 −ω2 −ω ω ω2

χS(g ) 3 3 −1 0 0 0 0

χA(g ) 1 1 1 ω ω2 −1 ω

Table 19.2: Symmetric and alternating components of χ2 in Exercise 19.5

To show that χS and χA are irreducible characters we consider the inner products with



Chapter 19 79

themselves. This gives us

χS, χS =32

24 +32

24 +(−

1)2

4 =9 + 9 + 6

24 =24

24 = 1,

χA, χA =12

24+

12

24+

12

4+

ωω

6+

ω2ω2

6+

(−1)2

6+

ωω

6,

=1 + 1 + 6 + 4 + 4 + 4 + 4

24,

=24

24,

= 1.

Recalling that ωω = ω

2

ω2

= |ω|2

= 1. Therefore the characters are irreducible. Howeverit is not clear that χ itself is an irreducible character. So we check the inner product

χ, χ =22

24+

(−2)2

24+

ω2ω2

6+

ωω

6+

ωω

6+

ω2ω2

6=

4 + 4 + 4 + 4 + 4 + 4

24=

24

24= 1.

Therefore we have three irreducible characters of G and the trivial character gives us a

fourth. Now χA is a linear character, so we know straight away that χχA and χ2A will be

irreducible characters and as we can see below in the table, they are distinct from the other

characters. Finally χ2A is another irreducible linear character so we have χχ2

A is another

irreducible character. Therefore the character table for G is as in table 19.3.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

|C G(g i )| 24 24 4 6 6 6 6

1G 1 1 1 1 1 1 1

χA(g ) 1 1 1 ω ω2 −1 ω

χ2A(g ) 1 1 1 ω2 ω 1 ω2

χ(g ) 2

−2 0

−ω2

−ω ω ω2

χχA(g ) 2 −2 0 −1 −1 −ω 1

χχ2A(g ) 2 −2 0 −ω −ω2 ω ω

χS(g ) 3 3 −1 0 0 0 0

Table 19.3: The character table of G from Exercise 19.5

Exercise 19.6. Recall the character table of D6 from Example 16.3(1) on page 160. There

are 9 conjugacy classes in D6 × D6, represented by pairs of conjugacy classes from each



Chapter 19 80

copy of D6. Therefore the character table is as in table 19.4.

g i (1, 1) (1, a) (1, b ) (a, 1) (a, a) (a, b ) (b , 1) (b , a) (b , b )

χ1 × χ1(g ) 1 1 1 1 1 1 1 1 1

χ1 × χ2(g ) 1 1 −1 1 1 −1 1 1 −1

χ1 × χ3(g ) 2 −1 0 2 −1 0 2 −1 0

χ2 × χ1(g ) 1 1 1 1 1 1 −1 −1 −1

χ2 × χ2(g ) 1 1 −1 1 1 −1 −1 −1 1

χ2 × χ3(g ) 2 −1 0 2 −1 0 −2 1 0

χ3 × χ1(g ) 2 2 2 −1 −1 −1 0 0 0χ3 × χ2(g ) 2 2 −2 −1 −1 1 0 0 0

χ3 × χ3(g ) 4 −2 0 −2 1 0 0 0 0

Table 19.4: The character table of D6 × D6



Chapter 20. Restriction to a subgroup

Exercise 20.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Exercise 20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Exercise 20.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Exercise 20.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Exercise 20.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Exercise 20.1. We have G = S4 and H = (1234), (13) G.

(a) Let ϕ : H → D8 = a, b | a4 = b 2 = 1,bab = a−1 be the map defined by (1234)ϕ =a and (13)ϕ = b . Clearly (1234)4 = 1 and (13)2 = 1. Also (13)(1234)(13) =

(3214) = (1432) = (1234)−1. Therefore ϕ is certainly a homomorphism. It’s clear

that ker(ϕ) = 1 and Im(ϕ) = D8 therefore ϕ is an isomorphism.

(b) Recall the conjugacy classes of H are given by 1, (13)(24), (1234), (1432),

(13), (24), (14)(23), (12)(34). The character tables of S4 and D8, (taken from

Section 18.1 on page 180 and Example 16.3(3) on page 162 respectively), are given

in tables 20.1 and 20.2.

g 1 (12) (123) (12)(34) (1234)

|C G(g )| 24 4 3 8 4

χ1(g ) 1 1 1 1 1

χ2(g ) 1 −1 1 1 −1

χ3(g ) 2 0 −1 2 0

χ4(g ) 3 1 0 −1 −1

χ5(g ) 3 −1 0 −1 1

Table 20.1: The character table of S4

Note that [G : H ] = 248

= 3 and H is not normal as it contains only 2 of the 6 four

cycles in S4, so it is not a union of conjugacy classes. Now the restricted character

table of S4 to H is as in table 20.3.

Just by inspection it’s easy to see that

χ1 ↓ H = ψ1 χ4 ↓ H = ψ3 + ψ5,

81



Chapter 20 82

g 1 a2 a b ab

|C G(g )

|8 8 4 4 4

ψ1(g ) 1 1 1 1 1

ψ2(g ) 1 1 1 −1 −1

ψ3(g ) 1 1 −1 1 −1

ψ4(g ) 1 1 −1 −1 1

ψ5(g ) 2 −2 0 0 0


h 1 (13)(24) (1234) (13) (14)(23)

|C H(h)| 8 8 4 4 4

(χ1 ↓ H )(h) 1 1 1 1 1

(χ2 ↓ H )(h) 1 1 −1 −1 1

(χ3 ↓ H )(h) 2 2 0 0 2

(χ4 ↓ H )(h) 3 −1 −1 1 −1

(χ5 ↓ H )(h) 3 −1 1 −1 −1

Table 20.3: The characters of S4 restricted to H

χ2 ↓ H = ψ4 χ5 ↓ H = ψ2 + ψ5,

χ3 ↓ H = ψ1 + ψ4.

Therefore ψ1, ψ2, ψ3, ψ4 and ψ5 all occur as components of the restricted characters.

Exercise 20.2. From Exercise 12.4 we know that the cycle types appearing in A6 are (1),

(2, 2), (4, 2), (3), (3, 3) and (5). We also know that the only class that splits are the (5)

cycles. From Exercise 12.3 we know that these conjugacy classes have respective orders 1,

45, 90, 40, 40, 72 and 72. The character table of S6, (taken from Example 19.17 on page

205), is given in table 20.4. In the table we have shaded the columns which correspond to

cycle types in A6.

It’s easy to see from the shaded columns of the character table that

χ1 ↓ A5 = χ2 ↓ A5 χ3 ↓ A5 = χ4 ↓ A5 χ5 ↓ A5 = χ6 ↓ A5,

χ7 ↓ A5 = χ8 ↓ A5 χ9 ↓ A5 = χ10 ↓ A5.



Chapter 20 83

Class (1) (2) (3) (2, 2) (4) (3, 2) (5) (2, 2, 2) (3, 3) (4, 2) (6)

|C G(g )

|720 48 18 16 8 6 5 48 18 8 6

χ1(g ) 1 1 1 1 1 1 1 1 1 1 1

χ2(g ) 1 −1 1 1 −1 −1 1 −1 1 1 −1

χ3(g ) 5 3 2 1 1 0 0 −1 −1 −1 −1

χ4(g ) 5 −3 2 1 −1 0 0 1 −1 −1 1

χ5(g ) 10 2 1 −2 0 −1 0 −2 1 0 1

χ6(g ) 10 −2 1 −2 0 1 0 2 1 0 −1

χ7(g ) 9 3 0 1 −1 0 −1 −3 0 1 0

χ8(g ) 9

−3 0 1 1 0

−1

−3 0 1 0

χ9(g ) 5 1 −1 1 −1 1 0 −3 2 −1 0

χ10(g ) 5 −1 −1 1 1 −1 0 3 2 −1 0

χ11(g ) 16 0 −2 0 0 0 1 0 −2 0 0


Now A6 is a normal subgroup of index 2 in S6. All the pairs of characters listed above are

non-zero outside A6 so they all restrict to an irreducible character of A6 by 20.13(1). Now

χ11 is zero everywhere outside of A6 so χ11

↓A6 = ψ6 + ψ7, where ψ6(1) = ψ7(1) = 8 by

20.13(2). Therefore the character table of A6 is, so far, as in table 20.5.

Class (1) (3) (2, 2) (5) (5) (3, 3) (4, 2)

|C H(h)| 360 9 8 5 5 9 4

ψ1(h) 1 1 1 1 1 1 1

ψ2(h) 5 2 1 0 0 −1 −1

ψ3(h) 10 1 −2 0 0 1 0

ψ4(h) 9 0 1 −1 −1 0 1ψ5(h) 5 −1 1 0 0 2 −1

ψ6(h) 8 α1 α2 α3 α4 α5 α6

ψ7(h) 8 β1 β2 β3 β4 β5 β6

Table 20.5: The partial character table of A6

Using the column orthogonality relations with the first column and each of the remaining



Chapter 20 84

columns we get a system of equations

1 + 10 + 10 + 0 − 5 + 8α1 + 8 β1 = 01 + 5 − 20 + 9 + 5 + 8α2 + 8 β2 = 0

1 + 0 + 0 − 9 + 0 + 8α3 + 8 β3 = 0

1 + 0 + 0 − 9 + 0 + 8α4 + 8 β4 = 0

1 − 5 + 10 + 0 + 10 + 8α5 + 8 β5 = 0

1 − 5 + 0 + 9 − 5 + 8α6 + 8 β6 = 0

⇒

α1 + β1 = −2,α2 + β2 = 0,

α3 + β3 = 1,

α4 + β4 = 1,

α5 + β5 = −2,

α6 + β6 = 0.

Using the column orthogonality relations of the columns with themselves we get a system

of equations

1 + 4 + 1 + 0 + 1 + α21 + β21 = 9

1 + 1 + 4 + 1 + 1 + α22 + β2

2 = 8

1 + 0 + 0 + 1 + 0 + α23 + β2

3 = 5

1 + 0 + 0 + 1 + 0 + α24 + β2

4 = 5

1 + 1 + 1 + 0 + 4 + α25 + β2

5 = 9

1 + 1 + 0 + 1 + 1 + α26 + β2

6 = 4

⇒

α21 + β21 = 2,

α22 + β2

2 = 0,

α23 + β2

3 = 3,

α24 + β2

4 = 3,

α25 + β2

5 = 2,

α26 + β2

6 = 0.

Recall that the inverse of any element in A6 is of the same cycle shape as the original

element. Therefore every element inA6 is conjugate to its inverse. This tells us that all

the entries in the character table of A6 are real. Therefore we can straight away say that

α2 = β2 = α6 = β6 = 0 and α1 = β1 = α5 = β5 = −1. Now β3 = 1 − α3 and

α23 + β2

3 = 3 ⇒ α23 + (1 − α3)2 = 3,

⇒ 2α23 − 2α3 + 1 = 3,

⇒ α23 − α3 − 1 = 0,

⇒ α3 =1 ± √

5

2.

Now choose α3 = 1+√ 5

2then β3 = 1−√

52

. It’s clear that α4 and β4 are solutions of the

same equations as α3 and β3. We also know that ψ6 and ψ7 are different characters so

must have α4 = β3 and β4 = α3. This gives us the character table of A6 is as in table 20.6.

Exercise 20.3. If H is an abelian subgroup of G then every irreducible character, say ψi ,

of H is linear. Certainly, as H is abelian, we have H is a normal subgroup of G. Therefore

by Clifford’s Thereom we have χ ↓ H = e (ψ1 + · · · + ψm) for some positive integer e . Now



Chapter 20 85

Class (1) (3) (2, 2) (5) (5) (3, 3) (4, 2)

|C H(h)

|360 9 8 5 5 9 4

ψ1(h) 1 1 1 1 1 1 1

ψ2(h) 5 2 1 0 0 −1 −1

ψ3(h) 10 1 −2 0 0 1 0

ψ4(h) 9 0 1 −1 −1 0 1

ψ5(h) 5 −1 1 0 0 2 −1

ψ6(h) 8 −1 0 1+√ 5

21−√

52

−1 0

ψ7(h) 8

−1 0 1−√

52

1+√ 5

2

−1 0

Table 20.6: The character table of A6

it’s clear that

χ(1) = (χ ↓ H )(1) = e (ψ1(1) + · · · + ψm(1)) = em,

where m = |H |.From Proposition 20.5 we have that

mi =1 e 2 = e 2m [G : H ] = n. However, as e and

m are positive integers we clearly have

em e 2m [G : H ] = n ⇒ χ(1) n.

Exercise 20.4. Let χ be an irreducible character of G. Let ψ1, . . . , ψm be the irreducible

characters of H . Then χ ↓ H = e 1ψ1 + · · · + e mψm. If H is a subgroup of index 3 then by

Propotisition 20.5 we have

e 21 + · · · + e 2m 3.

Therefore the only possibilities are that one of the e i

is 1, two of the e i

are 1 or three of

the e i are 1. In other words

χ ↓ H, χ ↓ H H = e 21 + · · · + e 2m = 1, 2 or 3.

We want to find examples of each of these cases occurring. Consider Exercise 1 of

this chapter. We have D8∼= (1234), (13) G is a subgroup of index 3 because [G :

(1234), (13)] = |G||D8| = 24

8= 3. Keeping the notation of that exercise we have χ1 ↓ H =

ψ1 and χ3 ↓ H = ψ1 + ψ4 which shows the possibilities of χ ↓ H, χ ↓ H = 1 or 2.



Chapter 20 86

For the final example consider the group A4. The character table of A4, (taken from

18.2 on page 181), is given in table 20.7.

g (1) (12)(34) (123) (132)

|C G(g )| 12 4 3 3

χ1(g ) 1 1 1 1

χ2(g ) 1 1 ω ω2

χ3(g ) 1 1 ω2 ω

χ4(g ) 3 −1 0 0

Table 20.7: The character table of A4

Now the linear characters are lifts from A4/V 4 where V 4 = 1, (12)(34), (13)(24), (14)(23)is the derived subgroup of A4. Note that |A4/V 4| = 3, so V 4 is a subgroup of index 3 in A4.

Also we have (12)(34)2 = (13)(24)2 = 1 and (12)(34)(13)(24) = (14)(23) so V 4 ∼= C 2×C 2.

Therefore we know the character table of V 4, which we give in table 20.8 along with the

restricted character χ4 ↓ V 4.

h (1) (123) (132)

|C V 4(h)| 12 4 3 3

ψ1(h) 1 1 1 1

ψ2(h) 1 1 −1 −1

ψ3(h) 1 −1 1 −1

ψ4(h) 1 −1 −1 1

χ4 ↓ V 4 3 −1 −1 −1

Table 20.8: The character table of V 4

By inspection we can see that χ4 ↓ V 4 = ψ2 + ψ3 + ψ4.

Exercise 20.5. If an irreducible representation, say χ, of S7 restricts to an irreducible

representation of A7 then there is another irreducible representation, say χ of S7 such

that χ ↓ A7 = χ ↓ A7. Also if χ restricts to a sum of two irreducible representations of A7

then they have the same degree and so we must have 2 | χ(1).

Using this information we can immediately see that 1, 15, 21 and 35 are four of the

degrees of the irreducible characters of A7. As 20 occurs on its own we must have that



Chapter 20 87

this character splits and so two of the degrees of the irreducible representations are 10 and

10. There are only three degrees left so this must mean that the remaining characters are

in pairs and we have 6, 14 and 14 are the remaing degrees.Therefore the degrees of the irreducible characters of A7 are: 1, 6, 14, 14, 15, 10, 10,

21 and 35.



Chapter 21. Induced modules and characters

Exercise 21.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Exercise 21.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Exercise 21.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

Exercise 21.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Exercise 21.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Exercise 21.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Exercise 21.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Exercise 21.1. We have G = D8 = a, b | a4 = b 2 = 1, b −1ab = a−1 and H = a2, b G.

Note that H ∼= C 4, which means |H | = 4.

(a) Let u = 1 − a2 + b − a2b then

ua2 = (1 − a2 + b − a2b )a2 = a2 − 1 + ba2 − a2ba2 = −1 + a2 − b + a2b = −u ∈ U,

ub = (1 − a2 + b − a2b )b = b − a2b + b 2 − a2b 2 = 1 − a2 + b − a2b = u ∈ U.

Therefore U is a submodule of CH .

(b) We have the induced module to be

U ↑ G = span1 − a2 + b − a2b, a − a3 + ba − a2ba,b − a2b + b 2 − a2b 2,

= span1 − a2 + b − a2b, a − a3 − ab + a3b .

Therefore u and ua are a basis for U ↑ G.

(c) Now the character of the CH module U is given in table 21.1.

h 1 a2

b a2

b |C H(h)| 4 4 4 4

χ(h) 1 −1 1 −1

Table 21.1: The character of the CH module U

We have the conjugacy classes of D8 to be 1, a2, a, a3, b, a2b and ab,a3b .

Only 1, a2 and b, a2b lie in H but b, a2b splits in two. Therefore the values

88



Chapter 21 89

of the induced character are

(χ ↑ G)(1) = |G

||H |χ(1) = 2

(χ ↑ G)(a2) =|C G(a2)||C H(a2)|χ(a2) = −2

(χ ↑ G)(b ) = |C G(b )|

ψ(b )

|C H(b )| +ψ(a2b )

|C H(a2b )|

= 4 · 1 − 1

4= 0

In particular the induced character is as in table 21.2. We know from the character

table of D8 that χ ↑ G is irreducible but we also have χ ↑ G, χ ↑ G = 22

8+ (−2)2

8= 1.

g 1 a2 a b ab |C G(h)| 4 4 4 4

(χ ↑ G)(g ) 2 −2 0 0 0

Table 21.2: The induced character of the CH module U

Exercise 21.2. Let G = S4 and H = (123) ∼= C 3.

(a) We have the character tables of S4 and C 3 = a | a3 = 1, (taken from 18.1 on

page 180 and Example 16.3(2) on page 160 respectively), are given in tables 21.3

and 21.4. In the tables ω = e 2πi/3 is a primitive 3rd root of unity.

g 1 (12) (123) (12)(34) (1234)

|C G(g )| 24 4 3 8 4

χ1(g ) 1 1 1 1 1

χ2(g ) 1 −1 1 1 −1

χ3(g ) 2 0 −1 2 0

χ4(g ) 3 1 0 −1 −1χ5(g ) 3 −1 0 −1 1


The restricted character table of S4 to H ∼= C 3 is then given in table 21.5. Using the

character table of C 3 it is easy to see that

χ1 ↓ H = χ2 ↓ H = ψ1 χ3 ↓ H = ψ2 + ψ3 χ4 ↓ H = χ5 ↓ H = ψ1 + ψ2 + ψ3.



Chapter 21 90

g 1 a a2

|C G(g )

|3 3 3

ψ1(g ) 1 1 1

ψ2(g ) 1 ω ω2

ψ3(g ) 1 ω2 ω

Table 21.4: The character table of C 3

g 1 (123) (132)

|C G(g )

|24 4 3

(χ1 ↓ H )(g ) 1 1 1

(χ2 ↓ H )(g ) 1 1 1

(χ3 ↓ H )(g ) 2 −1 −1

(χ4 ↓ H )(g ) 3 0 0

(χ5 ↓ H )(g ) 3 0 0

Table 21.5: The characters of S4 restricted to H

(b) Using Frobenius reciprocity it is easy to see that

ψ1 ↑ G = χ1 + χ2 + χ4 + χ5 ψ2 ↑ G = ψ3 ↑ G = χ3 + χ4 + χ5.

Exercise 21.3. Let ψ be a character of H G and let U be the corresponding CH module

that affords ψ as a character. Now U ∼= U 1 ⊕· · ·⊕ U m for some irreducible CH -submodules

U i of CH . Let ψi be the irreducible character of U afforded by the submodule U i , then

ψ = ψ1

+· · ·

+ ψm

. In definition 21.9 we define

U ↑ G = (U 1 ↑ G) ⊕ · · · ⊕ (U m ↑ G) ⇒ ψ ↑ G = ψ1 ↑ G + · · · + ψm ↑ G.

Therefore we have (ψ ↑ G)(1) = (ψ1 ↑ G)(1) + · · · + (ψm ↑ G)(1).

Consider one of the irreducible CH submodules, say U k . Then we have that U k has

a basis v 1, . . . , v , for some v 1, . . . , v ∈ CH . We define the induced module to be

U k (CG) = spanxg | x ∈ U k , g ∈ G. Let v ∈ U then we have v g ∈ U if and only if g ∈ H .

So, let S be a right transversal for H in G. Then v j s | 1 j , s ∈ S is a basis for the



Chapter 21 91

induced module U k (CG). In other words

(ψk ↑ G)(1) = dim(U k ↑ G) = dim(U k (CG)) = |S| dim(U k ) = |S|ψk (1) = |G

||H |ψk (1).

Now extending this to the whole of ψ we get

(ψ ↑ G)(1) = (ψ1 ↑ G)(1) + · · · + (ψm ↑ G)(1),

=|G||H |ψ1(1) + · · · +

|G||H |ψm(1),

=|G||H |(ψ(1) + · · · + ψ(m)),

= |G||H |ψ(1),

as required.

Exercise 21.4. Let φ be any irreducible character of G then taking the inner product we

have

(ψ(χ ↓ H )) ↑ G, φG = ψ(χ ↓ H ), φ ↓ H H,

=

ψ, χ

↓Hφ

↓H

H,

= ψ,χφ ↓ H H,

= ψ ↑ G,χφG,

= (ψ ↑ G)χ, φG.

Therefore as φ was an arbitrary irreducible character of G we must have that ψ(χ ↓ H ) ↑G = (ψ ↑ G)χ. Note the results on inner products used above come from Exercise 19.1.

Exercise 21.5. For a start we have the elements of H have the form

a = (1234567) ab = (137)(254) ab 2 = (157)(364),

a2 = (1234567) a2b = (156)(273) a2b 2 = (126)(475),

a3 = (1357246) a3b = (175)(346) a3b 2 = (165)(237),

a4 = (1473625) a4b = (124)(365) a4b 2 = (134)(276),

a5 = (1526374) a5b = (143)(267) a5b 2 = (173)(245),

a6 = (1765432) a6b = (162)(457) a6b 2 = (142)(356),

a7 = 1 b = (235)(476) b 2 = (253)(467).



Chapter 21 92

Therefore every element is either the identity a 7 cycle or a 3-3 cycle. Recall the conjugacy

classes of S7 are determined by their cycle type. Then using the formula in Proposition

21.23 we have

(ψ ↑ S7)(1) =|G||H |ψ(1),

(ψ ↑ S7)((1234567)) = |C G(a)|

ψ(a)

|C H(a)| +ψ(a3)

|C H(a3)|

,

(ψ ↑ S7)((123)(456)) = |C G(b )|

ψ(b )

|C H(b )| +ψ(b 2)

|C H(b 2)|

If σ ∈ S7 is not the identity, a 7 cycle or a 3-3 cycle then we will have (ψ ↑ S7)(σ) = 0.

Therefore this gives us that the induced character is as in table 21.6.

Class (1) (3, 3) (7)

(φ ↑ S7)(g ) 240 12 2

(ψ ↑ S7)(g ) 720 0 −1

Table 21.6: The characters φ ↑ S7 and ψ ↑ S7 of S7

Exercise 21.6. By Corollary 21.20 or Exercise 21.3, we have (ψ ↑ G)(1) = [G : H ]ψ(1)or, in other words, [G : H ]ψ(1) = d 1χ1(1)+ · · · + d k χk (1). Using Frobenius Reciprocity we

have

d i = ψ ↑ G, χi G = ψ, χi ↓ H H.

Now χi ↓ H = d i ψ +φ where φ is a character of H or 0. Therefore, as a restricted character

as the same degree as the original character, χi (1) = (χi ↓ H )(1) = d i ψ(1) + φ(1)

d i ψ(1). Using this we get

d 21ψ(1) + · · · + d 2k ψ(1) d 1χ1(1) + · · · + d k χk (1) ⇒ (d 21 + · · · + d 2k )ψ(1) [G : H ]ψ(1),

⇒k

i =1

d 2i [G : H ]

because ψ(1) > 0.

Exercise 21.7. Supose that H is a normal subgroup of index 2 in G and let ψ be an

irreducible character of H .



Chapter 21 93

Proposition 21.1 (20.9). We have that either

(a) ψ ↑ G is an irreducible character with degree 2ψ(1), or

(b) ψ ↑ G is the sum of two distinct irreducible characters of G, which both have degree

ψ(1).

Proof. Let χ1, . . . , χk be the irreducible characters of G then there exists d 1, . . . , d k such

that ψ ↑ G = d 1χ1 + · · · + d k χk . By the above theorem we havek

i =1 d 2i [G : H ] = 2.

Now the d i are positive integers and so we have only two options. Either ψ ↑ G = χi or

ψ ↑ G = χi + χ j for some i , j with i = j . In the first case you know, by Proposition 21.20

or Exercise 3, that (ψ ↑ G)(1) = [G : H ]ψ(1) = 2ψ(1). Therefore χi (1) = 2ψ(1).

For the second case we have by Frobenius Reciprocity that

1 = ψ ↑ G, χi G = ψ, χi ↓ H H,

1 = ψ ↑ G, χ j G = ψ, χ j ↓ H H.

Therefore ψ is a constituent of χi ↓ H and χ j ↓ H . This tells us that χi (1) = (χi ↓ H )(1)

ψ(1) and χ j (1) = (χ j ↓ H )(1) ψ(1). However 2ψ(1) = χi (1) + χ j (1), so we must have

χi (1) = χ j (1) = ψ(1).

Proposition 21.2 (20.11). Assume ψ ↑ G is irreducible. Aside from ψ there is only one

other irreducible character of H , say φ, such that φ↑

G = ψ↑

G.

Proof. If ψ ↑ G is irreducible then ψ ↑ G = χ for some irreducible character χ of G. By

Frobenius Reciprocity we know that χ ↓ H = ψ + φ where φ is an irreducible character of

H or φ = 0. Assume φ = 0. We know that a restriction of an irreducible character has the

same degree as the original character. Therefore,

2ψ(1) = (ψ ↑ G)(1) = ((ψ ↑ G) ↓ H )(1) = (χ ↓ H )(1) = ψ(1).

However this is a contradiction as ψ(1) = 0.

Therefore we have φ is an irreducible character of H . Using the Frobenius Reciprocitytheorem we have

1 = φ, χ ↓ H H = φ ↑ G, χG.

In other words φ ↑ G = χ = ψ ↑ G and we’re done.

Proposition 21.3 (20.12). Assume ψ ↑ G = χ1+χ2 for some irreducible characters χ1, χ2

of G. If φ is an irreducible character of H such that φ ↑ G has χ1 or χ2 as a constituent

then φ = ψ.



Chapter 21 94

Proof. If χ1 is a constituent of φ ↑ G then χ1 has coefficient 1 by the above Proposition

20.9 for induction. Therefore, using Frobenius Reciprocity, we have

1 = ψ ↑ G, χ1G = ψ, χ1 ↓ H H 1 = φ ↑ G, χ1G = φ, χ1 ↓ H H.

Also ψ(1) = χ1(1) and φ(1) = χ1(1), so ψ = χ1 = φ and we’re done.



Chapter 22. Algebraic integers

Exercise 22.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Exercise 22.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Exercise 22.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Exercise 22.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Exercise 22.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Exercise 22.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Exercise 22.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Exercise 22.1. Let χi be the irreducible characters of G. Then we know that

|G| = |G/G | +

χ(1)>1

χ(1)2

where the sum is over irreducible characters of G. We note that |G| | |G|, so our only

possibilities are that |G/G | = 1, 3, 5 or 15. Now 52 = 25 so no character can have

χ(1) = 5. So,

•

|G/G

| = 1 because 1 + 32 = 10 <

|G

|and 1 + 32 + 32 = 19 >

|G

|.

• |G/G | = 3 because 3 + 32 = 12 < |G| and 3 + 32 + 32 = 21 > |G|.• |G/G | = 5 because 5 + 32 = 14 < |G| and 5 + 32 + 32 = 23 > |G|.

Therefore |G/G | = 15, which means there are 15 linear characters and so G is abelian.

Exercise 22.2. Every irreducible character of G has degree which is a divisor of 16, i.e.

1, 2, 4, 8 or 16. However 82 = 64 > |G| and 162 = 256 > |G|, also 42 = 16 = |G| but we

always have at least one linear character, so our only possibilities are 1 or 2.

Recall that the number of linear characters is |G/G | and the only possibilities for this

are 1, 2, 4, 8 or 16. Now as χ(1)2

= |G| we have the only possibilities for the degrees of the irreducible characters are

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

1, 1, 1, 1, 1, 1, 1, 1, 2, 2

1, 1, 1, 1, 2, 2, 2.

In other words the number of conjugacy classes in G is either 7, 10 or 16.

95



Chapter 22 96

Exercise 22.3. Note that G is non-abelian so |G/G | = pq . Now the only possibilities for

|G/G | are 1, p or q .

(a) We have χ(1) | |G| for all irreducible characters χ of G, so either χ is linear orχ(1) = p or q .

(b) Recall that q < p so pq < p 2, which means no character can have degree p . Therefore

we have

|G| = |G/G | +

χ(1)>1

χ(1) ⇒ |G| = |G/G | + q 2n.

Now q | |G| ⇒ q | |G/G | but the only possibilities for |G/G | are 1, p or q so we

must have |G/G | = q . In other words pq = |G| = q |G| ⇒ |G| = p .

(c) We now know that pq = q + q 2n ⇒ p = 1 + qn ⇒ n = p −1q . Now we know n must bean intger, so q | p −1 as required. Finally we can see the group has q linear characters

and p −1q

irreducible characters of order q . In other words G has q + p −1q

conjugacy

classes.

Exercise 22.4. Let φ be a character of a group G such that φ(g ) = φ(h) for all non-identity

elements g, h ∈ G.

(a) We check the multiplicity of the trivial module in φ. Now taking inner products we

see

φ, 1G =1

|G|g ∈G

φ(g )1G(g ) =1

|G|g ∈G

φ(g ) =1

|G|(φ(1) + (|G| − 1)φ(h)),

= φ(h) +φ(1) − φ(h)

|G| ,

for some non-identity element h ∈ G. Now assume φ = a1G + bχreg and recall that

χreg, 1G = 1. Therefore the above inner product calculation has taken into account

the multiplicity of 1G inside χreg. So

φ = φ(h) + φ(1) − φ(h)|G| 1G + b (χreg − 1G).

Now evaluating at the identity we get

φ(1) =

φ(h) +

φ(1) − φ(h)

|G|

+ b (|G| − 1) ⇒ (φ(1) − φ(h))|G| − 1

|G| = b (|G| − 1),

⇒ b =φ(1) − φ(h)

|G| .



Chapter 22 97

Therefore φ = a1G + bχreg with a = φ(h) and b = φ(1)−φ(h)|G| for some non-identity

element h ∈ G.

(b) We know that φ is a character and hence the multiplicity of any irreducible character

in φ is an integer. We notice from above that a + b is the multiplicity of the trivial

character and hence is an integer. Also φ(1) is an integer, which means a1g (1) +

bχreg(1) = a + b |G| is an integer.

(c) Let χ = 1G be an irreducible character of G then

φ, χ = a1G + bχreg, χ = a1G, χ + b χreg, χ = bχ(1)

because the multiplicity of any irreducible character in χreg is the degree of the char-acter. Now φ is a character and hence the multiplicity of any irreducible character in

φ is an intger, which means bχ(1) is an integer.

(d) We have that bχ(1) is an integer but so is χ(1) so we must have b is an integer. We

have a + b is an integer but we’ve just determined b is an integer, so we must have

a is an integer as well. In otherwords the character φ is integer valued.

Exercise 22.5. Let G be a group of odd order.

(a) Assume g ∈ G such that g = g −1 then g 2 = 1. If g is not the identity then g has

order 2 and 2 | |G| but this cannot happen as G has odd order. Therefore g is theidentity.

(b) If χ = χ then we have χ(g ) = χ(g −1) for all g ∈ G. Recall |G| is odd, then 2 | |G|−1

and from part (a) we know there is no non-identity element of order 2. Let k = |G|−12

and define a set Ω = g 1, . . . , g k such that 1 ∈ Ω and if g i ∈ Ω then g −1i ∈ Ω. Then

the set Ω−1 = g −11 , . . . , g −1k is such that Ω ∩ Ω−1 = ∅. Then

χ, 1g =1

|G| g ∈Gχ(g )1G(g ) =

1

|G| g ∈Gχ(g ) =

1

|G|(χ(1) + 2

g i ∈Ωχ(g i ))

Now χ(g i ) is an algebraic integer for each g i ∈ Ω and so α =

g i ∈Ω χ(g i ) is an

algebraic integer.

(c) Assume χ, 1g = 0, then χ(1) + 2α = 0 ⇒ χ(1) = −2α. However χ(1) | |G| and

2 | χ(1) ⇒ 2 | |G| but G has odd order so this cannot happen. Therefore χ, 1G = 1

and so χ = 1G.

Exercise 22.6. Let G be a group such that |G| = 120 and G has exactly seven conjugacy

classes. Assume further that g contains an element of order 5 such that |C G(g )| = 5 and



Chapter 22 98

g, g 2, g 3 and g 4 are all conjugate in G. Also, let χ1, . . . , χ7 be the irreducible characters of

G and, without loss of generality, assume χ1 = 1G.

(a) For a start we know that, by Theorem 22.16, that χ(g ) is an integer. So, using thecolumn orthogonality relations we have

χ2(g )2 + χ3(g )2 + χ4(g )2 + χ5(g )2 + χ6(g )2 + χ7(g )2 = 4.

By Corollary 22.27 we have χ(g ) ≡ χ(1) (mod 5) for any irreducible character χ, so

−2 χi (g ) 2. Assume χi (g ) = ±2 for some i and χ j (g ) = 0 for all j = i . Then

we have

|G

| 12 + 22 + 52

×5 = 130,

which is a contradiction. Therefore our only possibility is that for every irreducible

character χ either χ(g ) = 0, 1 or −1.

(b) Using Corollary 22.27 we know that χ(g ) ≡ χ(1) (mod 5) and χ(g ) = 0 for exactly

two characters. We know that χ(1) | |G| = 120 and χ(1)2 < |G| ⇒ χ(1) < 11,

which means our only options for χ(1) in this case is 5 or 10. However we can’t have

χ(1) = 10 because then we’d have |G| 102 + 52 = 125, which is a contradiction.

Therefore we have exactly two characters of degree 5.

(c) All the divisors of 120 which are less than 10 are 1, 2, 3, 4, 5, 6 and 8. Now for anyirreducible character χ, we have from part (a) that χ(1) ≡ χ(g ) ≡ 0, 1, 4 (mod 5)

and so we can’t have χ(1) = 2, 3 or 8. So our only options are 1, 4, 5 or 6. We know

there are exactly two characters of degree 5 by part (b) and so

χ4(1)2 + χ5(1)2 + χ6(1)2 + χ7(1)2 = 69.

Now 43 = 64 < 69, so we must have at least one character of degree 6. This

leaves us with χ5(1)2 + χ6(1)2 + χ7(1)2 = 33. Finally the only solution to this is

χ5(1) = χ6(1) = 4 and χ7(1) = 1. Reordering the characters according the degree

we have

χ1(1) = 1 χ1(g ) = 1,

χ2(1) = 1 χ2(g ) = 1,

χ3(1) = 4 χ3(g ) = −1,

χ4(1) = 4 χ4(g ) = −1,



Chapter 22 99

χ5(1) = 5 χ5(g ) = 0,

χ6(1) = 5 χ6(g ) = 0,

χ7(1) = 6 χ7(g ) = 1.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2

(g ) 1 α1

α2

α3

α4

α5

1

χ3(g ) 4 β1 β2 β3 β4 β5 −1

χ4(g ) 4 γ 1 γ 2 γ 3 γ 4 γ 5 −1

χ5(g ) 5 δ1 δ2 δ3 δ4 δ5 0

χ6(g ) 5 ε1 ε2 ε3 ε4 ε5 0

χ7(g ) 6 κ1 κ2 κ3 κ4 κ5 1

Table 22.1: The partial character table of G from Exercise 22.6

So far we have the character table for G is as in table 22.1. Consider the columnof g 4. We have χ(g 4) ≡ χ(1) (mod 3) for each irreducible character and using the

column orthogonality relations we have

α23 + β2

3 + γ 23 + δ23 + ε23 + κ23 = 5.

Therefore each entry in this column is either 0, ±1 or ±2. This immediately tells

us that κ3 = 0. Now α3, β3, γ 3 are either 1 or −2 and δ3, ε3 are either −1 or 2.

Therefore no entry in the column is ±2 and we must have α3 = β3 = γ 3 = 1 and

δ3 = ε3 = −1. So this gives us that the character table is as in table 22.2.Consider the column of g 5. We have the order of g 5 is 22 and so χ(g 5) ≡ χ(1)

(mod 2) for all irreducible characters χ. Also using the column orthogonality on g 5

we have

α24 + β2

4 + γ 24 + δ24 + ε24 + κ24 = 3.

It’s clear that all these entries must be 1. Therefore α4, δ4 and ε4 are equal to

±1 and β4 = γ 4 = κ4 = 0. Using the column orthogonality relation with g 7 we get



Chapter 22 100

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 α1 α2 1 α4 α5 1

χ3(g ) 4 β1 β2 1 β4 β5 −1

χ4(g ) 4 γ 1 γ 2 1 γ 4 γ 5 −1

χ5(g ) 5 δ1 δ2 −1 δ4 δ5 0

χ6(g ) 5 ε1 ε2

−1 ε4 ε5 0

χ7(g ) 6 κ1 κ2 0 κ4 κ5 1


1 + α4 = 0 ⇒ α4 = −1. Using the column orthogonality relation with g 4 we get

−δ4 − ε4 = 0 ⇒ ε4 = −δ4. Therefore we have the character table so far to be as in

table 22.3.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 α1 α2 1 −1 α5 1

χ3(g ) 4 β1 β2 1 0 β5 −1

χ4(g ) 4 γ 1 γ 2 1 0 γ 5 −1

χ5(g ) 5 δ1 δ2 −1 1 δ5 0

χ6(g ) 5 ε1 ε2 −1 −1 ε5 0χ7(g ) 6 κ1 κ2 0 0 κ5 1


Consider the column of g 3. We have χ(g 3) ≡ χ(1) (mod 2) for every irreducible



Chapter 22 101

character χ. Now using the column orthogonality relation on g 3 we get

α2

2 + β2

2 + γ 2

2 + δ2

2 + ε2

2 + κ2

2 = 7.

Now every entry in the column must be 2. Now three of the entries are ±1, which

gives us β22 + γ 22 + κ2

2 = 4. Therefore two of these entries are zero and one is ±2.

Using the column orthogonality relations between g 3 and g 7 we have

1 + α2 − β2 − γ 2 + κ2 = 0.

We can’t have α2 = −1 as only one of the remaining values is non-zero, therefore

α2 = 1. Using the column orthogonality relations between g 3 and g 5 we have

1 − 1 + δ2 + −ε2 = 0 ⇒ δ2 = ε2.

Using the column orthogonality of g 3 with g 4 we get

2 + β2 + γ 2 − 2δ2 = 0.

Using this equation together with the relation from g 3 and g 7 we can see that 4 −2δ2 + κ2 = 0 ⇒ κ2 = 2δ2 − 4. Now δ2 = ±1, which gives us κ2 = −2 or −8. Clearlyκ2 can’t be −8 so this gives us δ2 = ε2 = 1 and κ2 = −2. Therefore the remaining

entries must be zero and we have the character table so far to be as in table 22.4.

Consider the column of g 2. We have g 2 is an element of order 2 and so χ(g 2) ≡χ(1) (mod 2) for all irreducible characters χ. Using the column orthogonality relation

on g 2 we see

α21 + β2

1 + γ 21 + δ21 + ε21 + κ21 = 11.

Therefore it’s clear that every entry in the column is 3. Say one entry is

±3

then that means there are three of the remaining entries equal to ±1 and the rest 0.

However this would imply 4 entries are congruent to 1 (mod 2) but there are clearly

only 3. Therefore every entry in the column is either 0, ±1 or ±2. In fact knowing

that α21 = δ21 = ε21 = 1 means that

β21 + γ 21 + κ2

1 = 8.

Therefore only one entry is 0 and the other two are ±2.



Chapter 22 102

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 α1 1 1 −1 α5 1

χ3(g ) 4 β1 0 1 0 β5 −1

χ4(g ) 4 γ 1 0 1 0 γ 5 −1

χ5(g ) 5 δ1 1 −1 1 δ5 0

χ6(g ) 5 ε1 1

−1

−1 ε5 0

χ7(g ) 6 κ1 −2 0 0 κ5 1


Now we check the column orthogonality relations to determine the values explicitly.

Taking all the column relations we can we get

1 + α1 + 4( β1 + γ 1) + 5(δ1 + ε1) + 6κ1 = 0,

1 + α1 + δ1 + ε1

−2κ1 = 0,

1 + α1 + β1 + γ 1 − δ1 − ε1 = 0,

1 − α1 + δ1 − ε1 = 0,

1 + α1 − β1 − γ 1 + κ1 = 0.

Adding the second and third relations gives us 2 + 2δ1 − 2κ1 = 0 ⇒ κ1 = 1 + δ1.

Therefore we can’t have κ1 = −2 as δ1 = ±1. Using this in the relations we reduce

our system of equations to

1 + α1 + 4( β1 + γ 1) + 5(δ1 + ε1) + 6κ1 = 0,1 + α1 + β1 + γ 1 − δ1 − ε1 = 0,

1 − α1 + δ1 − ε1 = 0,

2 + α1 − β1 − γ 1 + δ1 = 0.

Adding the second and fourth relations again we get 3 + 2α1 − ε1 = 0 ⇒ ε1 =

2α1 + 3 but this gives us ε1 = 5 or 1. However ε1 = 5 by the congruence relations

and so we have ε = 1 and α = −1. Putting this into our relations, (and removing



Chapter 22 103

any redundant equations), we get

5 + 4( β1 + γ 1) + 5δ1 + 6κ1 = 0,1 − β1 − γ 1 + δ1 = 0,

1 + δ1 = 0.

Clearly δ1 = −1 ⇒ κ1 = 0. Then the remaining relations tell us γ 1 = − β1. Now the

characters χ3 and χ4 are currently identical so there is no harm in choosing β1 = 2

and γ 1 = −2. This gives us the character table to be as in table 22.5.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7

Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 −1 1 1 −1 α5 1

χ3(g ) 4 2 0 1 0 β5 −1

χ4(g ) 4 −2 0 1 0 γ 5 −1

χ5(g ) 5 −1 1 −1 1 δ5 0

χ6(g ) 5 1 1 −1 −1 ε5 0

χ7(g ) 6 0 −2 0 0 κ5 1


To get the final column of g 6 it is easiest to use the row orthogonality relations

with the trivial representation. This gives us

1 − 10 + 15 + 20 − 30 + 20α5 + 24 = 0

4 + 20 + 20 + 20 β5 − 24 = 04 − 20 + 20 + 20 γ 5 − 24 = 0

5 − 10 + 15 − 20 + 30 + 20δ5 = 0

5 + 10 + 15 − 20 − 30 + 20ε5 = 0

6 − 30 + 20κ5 + 24 = 0

⇒

20 + 20α5 = 0,

20 + 20 β5 = 0,−20 + 20 γ 5 = 0,

20 + 20δ5 = 0,

−20 + 20ε5 = 0,

20κ5 = 0.

From this it is now easy to see that the complete character table for G is as follows in

table 22.6. Comparing this to the character table of S5 on page 201 we can indeed



Chapter 22 104

see that these are the same.

g i g 1 g 2 g 3 g 4 g 5 g 6 g 7Order 1 2 2 3 4 6 5

|C G(g i )| 120 12 8 6 4 6 5

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 −1 1 1 −1 −1 1

χ3(g ) 4 2 0 1 0 −1 −1

χ4(g ) 4 −2 0 1 0 1 −1

χ5(g ) 5−

1 1−

1 1−

1 0

χ6(g ) 5 1 1 −1 −1 1 0

χ7(g ) 6 0 −2 0 0 0 1


Exercise 22.7. (⇒) Assume λ is an algebraic integer, then it is the solution of det(A−XIn)

for some integer valued matrix A. Recall that

det(A − XIn) = X n

+ c n−1X n

−1

+ · · · + c 1X + c 0,

with c 0 = (−1)n det(A) and c n−1 = − tr(A). All c i ∈ Z and so λ is a solution of such a

polynomial.

(⇐) Let λ be the solution of the polynomial

X n + an−1X n−1 + · · · + a1X + a0 ∈ Z[X ].

We now wish to construct an integer valued matrix A such that λ is a solution of det(A −XIn) = X n + an−1X n−1 + · · · + a1X + a0. Let A be the matrix

0 1 . . . 0...

. . . 1...

.... . .

...

−a0 −a1 . . . −an−1

.

We now want to discover what the determinant of A − XIn is. We proceed by induction.

We claim that det(A − XIn) = (−1)n(X n + an−1X n−1 + · · · + a1X + a0). We check that



Chapter 22 105

this is the case for A a 2 × 2 matrix

det−X 1−a0 −a1 − X = −X (−a1 − X ) + a0 = X 2 + a1X + a0.

Therefore this is certainly true for n = 2. Now assume that this is true for n = k then the

k + 1th case is

det

−X 1 . . . 0...

. . . 1...

.... . . 1

−a0

−a1 . . .

−ak

−X

= (−X )det

−X 1 . . . 0...

. . . 1...

.... . . 1

−a1 −a2 . . . −ak − X

− det

0 1 . . . 0... −X 1

......

. . . 1

−a0 −a2 . . . −ak − X

,

= (−X )(−1)k (X k + ak X k −1 + · · · + a2X + a1) − (−1)k a0,

= (−1)k +1(X k +1 + ak X k + · · · + a2X 2 + a1X + a0).

Therefore the result holds by induction. Now det(A−

λIn

) = 0, so λ is an eigenvalue of A

and hence an algebraic integer.





Chapter 23 107

g i 1 ar 1r n−1

2

b

|C G(g i )| 2n n 2

χ1(g ) 1 1 1

χ2(g ) 1 1 −1

ψ j (g )1 j n−1

2

2 ε j r + ε− j r 0

Table 23.1: The character table of D2n, for n odd

g i 1 am ar

1r m

−1b ab

|C G(g i )| 2n 2n n 4 4

χ1(g ) 1 1 1 1 1

χ2(g ) 1 1 1 −1 −1

χ3(g ) 1 (−1)m (−1)r 1 −1

χ4(g ) 1 (−1)m (−1)r −1 1

ψ j (g )1 j m−1

2 2(−1) j ε j r + ε− j r 0 0

Table 23.2: The character table of D2n, for n even

Clearly the χi are real characters in both tables but what about the ψ j ? Well for any

complex number z we have z + z = 2 Re(z ). Also it’s clear that ε = e −2πi/n = ε−1 and

so ψ j is also a real character of D2n. Therefore all the conjugacy classes of D2n are real

conjugacy classes. This means that every element of D2n is real and so there are 2n real

elements. Clearly every irreducible character is real, which gives us ιχ = 1 for all irreducible

characters χ.

Exercise 23.4. Let V be the 2-dimensional irreducible CG module associated to the repre-sentation ρ. Let B = v 1, v 2 be the basis of V then we have B = v 1 ⊗ v 2 − v 2 ⊗ v 1 is a

basis of A(V ⊗ V ). Let the matrix of g with respect to B be

gρB =

a b

c d

.

Therefore we have v 1g = av 1 + bv 2 and v 2g = cv 1 + dv 2.

We now consider the action of g on the antisymmetric tensor product module A(V ⊗V ).



Chapter 23 108

We have

(v 1 ⊗ v 2 − v 2 ⊗ v 1)g = v 1g ⊗ v 2g − v 2g ⊗ v 1g,= (av 1 + bv 2) ⊗ (cv 1 + dv 2) − (cv 1 + dv 2) ⊗ (av 1 + bv 2),

= ad (v 1 ⊗ v 2) + bc (v 2 ⊗ v 1) − cb (v 1 ⊗ v 2) − ad (v 2 ⊗ v 1),

= (ad − bc )(v 1 ⊗ v 2 − v 2 ⊗ v 1).

Therefore χA(g ) = ad − bc = det(gρ). Now ιχ = −1 if and only if 1G is a constituent

of χA but this happens if and only if χA = 1G, (as χA is a linear character and hence

irreducible). In other words det(g ) = 1 for all g ∈ G.

Exercise 23.5. We let G = T 4n = a, b | a2n = 1, an = b 2, b −1ab = a−1 and ε = ±1 a2nth root of unity.

(a) Let ρ : G → GL(2,C) be the irreducible matrix represenation associated to V . It’s

clear that for a and b we have

aρ =

ε 0

0 ε−1

bρ =

0 1

εn 0

Now clearly det(aρ) = εε−1 = 1 and det(bρ) = −εn. Recall the determinant is

multiplicative and every element in T 4n can be written in the form ai b j with 1 i 2n

and 0 j 1. Now ιχ = −1 if and only if det(x ρ) = 1 for all x ∈ T 4n. Therefore

this only happens if −εn = 1 ⇒ εn = −1. So we have ιχ = 1 if εn = 1 and ιχ = −1

if εn = −1.

(b) We check the action of a on the bilinear form

β(v 1a, v 1a) = β(εv 1, εv 1) = ε2 β(v 1, v 1) = 0 = β(v 1, v 1),

β(v 2a, v 2a) = β(ε−1v 2, ε−1v 2) = ε−2 β(v 2, v 2) = 0 = β(v 2, v 2),

β(v 1a, v 2a) = β(εv 1, ε−1v 2) = εε−1 β(v 1, v 2) = β(v 1, v 2),

β(v 2a, v 1a) = β(ε−1v 2, εv 1) = ε−1εβ(v 2, v 1) = β(v 2, v 1).

Therefore β is invariant under the action of a. Now considering the action of b on

the bilinear form

β(v 1b, v 1b ) = β(v 2, v 2) = 0 = β(v 1, v 1),

β(v 2b, v 2b ) = β(εnv 1, εnv 1) = ε2n β(v 1, v 1) = 0 = β(v 2, v 2),

β(v 1b, v 2b ) = β(v 2, εnv 1) = εn β(v 2, v 1) = ε2n = 1 = β(v 1, v 2),



Chapter 23 109

β(v 2b, v 1b ) = β(εnv 1, v 2) = εn β(v 1, v 2) = εn = β(v 2, v 1).

Therefore β is invariant under the action of b , hence invariant under the action of allelements x ∈ T 4n.

Using Theorem 23.16 we have ιχ = 1 if there exists a symmetric bilinear form on

V , which happens if β(v 2, v 1) = β(v 1, v 2) = 1, in other words εn = 1. Alternatively

the theorem tells us that ιχ = −1 if there exists a skew-symmetric bilinear form on

V , which happens if β(v 2, v 1) = − β(v 1, v 2), in other words εn = −1.

(c) Every element in T 4n has the form ai b j for some 1 i 2n and 0 j 1. Now

assume j = 0 then (ai )2 = a2i = 1 ⇔ i = n or 2n, however if i = 2n then ai = 1, which

is not an element of order 2. Assume j = 1 then (ai b )2 = ai bai b = ai a−i b 2 = an

= 1.

Therefore the only element of order 2 in T 4n is an.

(d) Now the character table for T 4n with n even is as in table 23.3.

g 1 an ar 1r n−1

b ba

g 2 1 1 a2r an an

|C G(g )| 4n 4n 2n 4 4

1G 1 1 1 1 1

φ1 1 1 1 −1 −1

φ2 1 1 (−1)r 1 −1

φ3 1 1 (−1)r −1 1

χk (g )0k n−1

2 2(−1)k εkr + ε−kr 0 0

χ2k (g ) 4 4 2 + ε2kr + ε−2kr 0 0

(χk )S(g ) 3 3 1 + ε2kr + ε−2kr (−1)k (−1)k

(χk )A(g ) 1 1 1 (−1)k +1 (−1)k +1

Table 23.3: The character table of T 4n, with n even

It’s obvious that ι1G = ιφ1 = ιφ2 = ιφ3 = 1 because 12G = φ2

1 = φ22 = φ2

3 = 1G and

(1G)S = 1G, (1G)A = 0. Now all that’s left to consider is the indicator of the χk . It’s

clear that χk is real because εkr + ε−kr = εkr + εkr = 2Re(εkr ).

Now in the table above we have considered χ2k and decomposed it into its sym-

metric and alternating parts. If k is odd then k + 1 is even and (χk )A = 1G, which

gives us ιχk = −1. If k is even then k + 1 is odd and (χk )A = φ1. Therefore 1G must



Chapter 23 110

be a constituent of (χk )S and so (ιχk ) = 1. So we have the Frobenius-Schur Count

of Involutions is

χ

(ιχ)χ(1) = 1 + 1 + 1 + 1 +

n−1k =1

2(−1)k = 4 − 2 = 2,

because n is even so the sum equals −2.

If n is odd then we have the character table of T 4n to be as in table 23.4.

g 1 an ar 1r n−1

b ba

g 2 1 1 a2r an an

|C G(g )| 4n 4n 2n 4 4

1G 1 1 1 1 1

φ1 1 −1 (−1)r i −i

φ2 1 1 1 −1 −1

φ3 1 −1 (−1)r −i i

χk (g )1k n−1

2 2(−1)k εkr + ε−kr 0 0

χ2k (g ) 4 4 2 + ε2kr + ε−2kr 0 0

(χk )S(g ) 3 3 1 + ε2kr + ε−2kr (−1)k (−1)k

(χk )A(g ) 1 1 1 (−1)k +1 (−1)k +1

Table 23.4: The character table of T 4n, with n odd

It’s clear that 12G = φ22 = 1G so ι1G = ιφ2 = 1 and φ1, φ3 aren’t real so ιφ1 = ιφ3 = 0.

Now all that’s left to consider is the indicator of the χk . However the situation here

is clearly identical to the situation for when n is even. Therefore the Frobenius-Schur

Count of Involutions is

χ

(ιχ)χ(1) = 1 + 0 + 1 + 0 +

n−1k =1

2(−1)k = 2 + 0 = 2,

because n is odd so the sum equals 0. This confirms that there is only one involution

in T 4n.

Exercise 23.6. Let χ be an irreducible character of G such that ιχ = −1. Let V be the

irreducibleCG module which affords χ as a character. By Theorem 23.16 we have ιχ = −1



Chapter 23 111

if and only if there exists a non-zero G-invariant skew-symmetric bilinear form on V . Let’s

call this bilinear form β. Now let v 1, . . . , v m be a basis for the CG module V . We consider

the vector subspaceU = u ∈ V | β(u, v ) = 0 for all v ∈ V .

It’s clear that this is a vector subspace but it is also a CG module. Recall that V G = V

and so given u ∈ U, v ∈ V and g ∈ G we have β(ug,vg ) = β(u, v ) = 0 and so UG = U .

Now V is an irreducible CG submodule and β is non-zero so we must have U = 0. Now

let β(v i , v j ) = x ij for all 1 i j m then we can construct a matrix X with

X =

0 x 1,2 . . . x 1,m−1 x 1,m

−x 1,2 . . . x 2,m...

. . ....

−x 1,m−1. . . x m−1,m

−x 1,m −x 2,m . . . −x m−1,m 0

.

We can see that the matrix X is skew-symmetric, i.e. X T = −X , because it’s coming

from a skew-symmetric bilinear form. By the fact that U = 0 we know that non of the

off diagonal entries in X are 0, hence the determinant of X is non-zero. We consider the

determinant of X T in two ways. We can see that we can get from X T to X by applying

m row and column swaps and multiplying by −1. Every time we make a row or column

swap we multiply the determinant by −1 and so det(X T ) = (−1)m+1 det(X ). Alterantively

det(X T ) = det(−X ) = − det(X ), so this gives us det(X ) = (−1)m det(X ). Therefore m is

even and we’re done.

Exercise 23.7. Let V be a vector space over R and let v 1, . . . , v n be any basis of V .

Recall that given a vector subspace U ⊆ V we define U ⊥ with respect to β1 by

U ⊥ = v ∈ V | β1(u, v ) = 0 for all u ∈ U .

Also, as vector spaces, we have V = U ⊕U ⊥. We now wish to construct a basis e 1, . . . , e nof V which is orthonormal with respect to β1.

We start by letting f 1 = v 1. Consider the vector subspace U 1 = spanv 1 and the

standard projection map π1 : V → U 1. Define f 2 = v 2 − π1(v 2) ∈ U ⊥1 . Note that f 2 = 0

because if f 2 = 0 then v 2 = π1(v 2) ⇒ v 2 ∈ spanv 1. However this cannot happen because

v 1, . . . , v n is a basis for V .

In a similar fashion we consider the vector subspace U 2 = spanv 1, v 2 and the standard



Chapter 23 112

projection map π2 : V → U 2. Define f 3 = v 3 − π2(v 3) ∈ U ⊥2 . Again we have f 3 = 0 because

if f 3 = 0 then v 3 = π2(v 3) ⇒ v 3 ∈ spanv 1, v 2. However this cannot happen because

v 1, . . . , v n is a basis for V .Carrying on with this process gives us a set f 1, . . . , f n such that β(f i , f j ) = 0 for all

i = j . Now we define f i = f i /

β1(f i , f i ), then e 1, . . . , e n is an orthonormal basis of V

with respect to β1. Note that

β(f i , f i ) ∈ R because β(w , w ) > 0 for all non-zero w ∈ V .

Note that in general this process is known as the Gram Schmidt algorithm.

We now want to show that f 1, . . . , f n can be transformed into a basis which is also

orthogonal on β. Let X be an n × n matrix such that X i j = β(f i , f j ) then X is a real

symmetric matrix. By general matrix theory there exists a real orthogonal matrix Q, (i.e.

QT = Q−1), such that QXQ−1 = D where D is a diagonal matrix. Note that this a special

case of the usual diagonalisation process for complex valued matrices.

The above process corresponds to a change of basis for V . Let Q = (q i j ) then we define

a new basis

e i =

n j =1

q ij f j .

Now because QXQ−1 is diagonal we have β(e i , e j ) = 0 for any i = j . Also the ma-

trix representing β1(f i , f j ) is the identity so conjugating by Q does nothing. This means

β1(e i , e j ) = δi j as required.

Exercise 23.8. Let V and W be RG-modules.

(a) Let ϑ : V → W be an RG-homomorphism. Now im(ϑ) is an RG submodule of W

and ker(ϑ) is an RG submodule of V . We have V and W are irreducible RG modules

so either Im(ϑ) = W and ker(ϑ) = 0 or Im(ϑ) = 0 and ker(ϑ) = V . This gives

you either ϑ is an RG isomorphism or ϑ is the zero map.

(b) Now if ϑ : V → V is an RG isomorphism then ϑ is also a CG isomorphism. If V

is an irreducible CG module then by Schur’s Lemma there exists λ ∈ C such that

ϑ = λ1V . However ϑ is also an RG homomorphism so we must have λ

∈R otherwise

im(ϑ) = V is not an RG-module.

(c) Consider G = C 3 = x | x 3 = 1. We have a 2-dimensional irreducible RG submodule

V = span1 − a, 1 − a2. Let v 1 = 1 − a and v 2 = 1 − a2 then we have the action of

a on these basis elements to be

v 1a = (1 − a)a = a − a2 = v 2 − v 1 v 2a = (1 − a2)a = a − 1 = −v 1.

We can define a map ϑ : V → V by v ϑ = av . Note that multiplication on the left



Chapter 23 113

is well defined as V is a submodule of the group algebra. Also, as G is commutative,

we will have the action of the group on the left to be the same as the action of the

group on the right.So, with respect to the basis B = v 1, v 2 we have

[a]B =

−1 −1

1 0

.

Now considering the eigenvalues of this matrix we have the characteristic polynomial

to be

det([a]B − x I2) = −1

−x

−1

1 −x = −x (−1 − x ) + 1 = x

2

+ x + 1.

This polynomial has eigenvalues −1±3i 2

∈ R. Therefore the map ϑ cannot be a real

scalar multiple of the identity map.

Exercise 23.9. Let Ω = Hx 1, . . . , H x n. The map ρg : Ω → Ω is a permutation of Ω if it

is a bijection. It’s clearly 1-1 because

(Hx i )ρg = (Hx j )ρg ⇒ Hx i g = Hx j g ⇒ Hx i = Hx j ⇒ x i = x j

because the right cosets of H partition the group G. It is also surjective as given Hx i ∈ Ω

we have Hx i g −1 ∈ Ω and (Hx i g −1)ρg = Hx i g −1g = Hx i . Therefore ρg is a permutation of

Ω.

We need to show that for all g, h ∈ G we have (gh)ρ = (gρ)(hρ) or in other words

ρgh = ρg ρh. For any Hx i ∈ Ω we have

(Hx i )ρgh = Hx i gh = (Hx i g )ρh = (Hx i )ρg ρh.

Now our choice of Hx i

was arbitrary and so ρgh

= ρg

ρh

, which gives us ρ is a homomorphism

from G to Sym(Ω).

Now we have ker(ρ) = g ∈ G | ρg = idΩ. If ρg is such a map then for all Hx i ∈ Ω we

have

(Hx i )ρg = (Hx i ) idΩ ⇔ Hx i g = Hx i ,

⇔ H = Hx i gx −1i ,

⇔ x i gx −1i ∈ H,



Chapter 23 114

⇔ g ∈ x −1i Hx i .

Therefore ker(ρ) = ∩x ∈Gx −1

Hx as required.We have a bijection ϕ : Ω → 1, . . . , n given by (Hx i )ϕ = i . We have a map from

ψ : Sym(Ω) → Sn given by σ → ϕ−1σϕ. Now this is in fact an isomorphism, we can see

it’s a homomorphism as for any σ, τ ∈ Sym(Ω) we have

(στ )ψ = ϕ−1(στ )ϕ = (ϕ−1σϕ)(ϕ−1τϕ) = (σψ)(τψ).

Also we can define an inverse map ψ−1 : Sn → Sym(Ω) by πψ−1 = ϕπϕ−1. Note that the

composition of homomorphisms is again a homomorphism. Therefore if H is a subgroup

of index n in G then we have a homomorphism ρ : G → Sym(Ω), hence a homomorphismρψ : G → Sn. It’s clear that ker(ρψ) = ker(ρ) = ∩x ∈Gx −1Hx ⊆ H .

Exercise 23.10. Let G be a finite group with an involution t ∈ G such that C G(t ) ∼= C 2.

Now let χ1, . . . , χk be the irreducible characters of G and ψ1, ψ2 be the irreducible characters

of C 2. Recall that all the irreducible characters of C 2 are linear.

As t 2 = 1 we have t −1 ∈ t G, which means χi (t ) is an integer for each 1 i k .

Furthermore we know that χi (t ) ≡ χi (1) (mod 2) for each 1 i k . From the column

orthogonality relations we have

k i =1

χi (1)2 = |G|k

i =1

χi (t )2 = 2.

As the χi (t ) are integers we must have χi (t ) = ±1 for two i , j ∈ 1, . . . , k with i = j

and χi (t ) = 0 for all other irreducible characters. As χi (1) ≡ χi (t ) (mod 2) for every

irreducible character this means that only two of the character degrees are odd and the rest

are even.

We now consider the restriction of the irreducible characters to the centraliser C G(t ).

We know that χi ↓

C G

(t ) = d 1

ψ1

+ d 2

ψ2

for each 1 i k and that d 21

+ d 22 [G :

C G(t )] = 2. Therefore the only possibilities are that d 1, d 2 are 0 or 1. We have that

χi (1) = (χi ↓ C G(t ))(1) = d 1ψ1(1) + d 2ψ2(1) = d 1 + d 2.

So the degree of every irreducible character of G is either 1 or 2. However we know there

are two characters of odd degree and all the other characters have even degree. This means

G has two linear characters and all other irreducible characters are of degree 2. In other

words [G : G ] = 2.



Chapter 23 115

Assume G is a simple group. We have G G is a normal subgroup of G and clearly

G = G so G = 1. This means G is abelian, which means there are only two characters

and they’re both linear. So G has the character table of C G(t ) ∼= C 2 which means G ∼= C 2.



Chapter 25. Characters of groups of order pq

Exercise 25.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Exercise 25.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Exercise 25.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Exercise 25.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

Exercise 25.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Exercise 25.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

Exercise 25.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

Exercise 25.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Exercise 25.1. We only have to check that the four group axioms hold. Let G be the

group specified in the question. Clearly I2 ∈ G as 0 ∈ Zp and 1 ∈ Z∗p . Associativity is given

to us by the fact that matrix multiplication is associative. We now check closure,1 y

0 x

1 y

0 x

=

1 y + y x

0 xx

.

Now y + y x ∈Zp because x

∈Z∗p ⇒

y x ∈Zp and Zp is a group under addition. Clearly

x, x ∈ Z∗p ⇒ x x ∈ Z∗p and so the product is in G.

Finally we show there exists an inverse element. Recall that Z∗p and Zp are groups under

multiplication and addition respectively. Therefore if x ∈ Z∗p then x −1 ∈ Z∗p and y ∈ Zp

then − y ∈ Zp . This means that1 − y x

0 x −1

∈ G and

1 y

0 x

1 − y x −1

0 x −1

=

1 − y x −1 + y x −1

0 xx −1

=

1 0

0 1

.

Exercise 25.2. We have F 11,5 = a, b | a11

= b 5

= 1 and b −1

ab = a3

, note that 35

=243 = 22 × 11 + 1 ≡ 1 (mod 11). Also the remaining powers of 3 are

32 = 9 ≡ 9 (mod 11) 33 = 27 ≡ 5 (mod 11) 34 = 81 ≡ 4 (mod 11).

Therefore 3 is a primitive root modulo 11. We have S = 1, 3, 4, 5, 9 and so |Z∗p /S| = 2,

so a set of left coset representatives for S in Z∗p is given by S, 2S.

Now G/G ∼= C 5 so let ε = e 2πi/5 be a primitive fifth root of unity. By Theorem 25.10

there are q = 5 linear characters corresponding to the lifts of C 5 and |Z∗p /S| = 2 irreducible

116





Chapter 25 118

are such that u q ≡ v q ≡ 1 (mod p ). Therefore setting b = b k we get

G1 = a, b | ap

= b q

= 1 and b −1

ab = av

= G2.

Hence G1∼= G2.

Exercise 25.4. Let p = 2, q = (p − 1)/2 and G = F p,q .

(a) Now u is a generator for Z∗p as Z∗p is cyclic and u is a non-identity element. Note that

−1 is the only element of order 2 in Z∗p therefore we have there exists an integer m

such that

u m

≡ −1 (mod p )

⇔u 2m

≡u q

≡1 (mod p ),

⇔ 2 | q,

⇔ 2 | p − 1

2,

⇔ p − 1 ≡ 0 (mod 4),

⇔ p ≡ 1 (mod 4).

(b) Now a−1 ∈ aG ⇔ a−1 = as ⇔ s ≡ −1 (mod p ) for some s ∈ S, (the set of all powers

of u ). Therefore by the above result we have −1 ∈ S if and only if p ≡ 1 (mod 4).

(c) Note that q = (p − 1)/2 ⇒ 2 = (p − 1)/q , which means |Z∗p /S| = 2. Hence there

are two irreducible characters of G of degree q which we label φ1 and φ2. Consider

1, v to be a set of left coset representatives for S in Z∗p then we have two conjugacy

classes aG and (av )G. We now use the orthogonality relations to investigate the values

of φ1(a) and φ2(a). For 1 i 2 we have

0 = φi , 1G = j

φi (g j )1G(g j )

|C G(g j )| =1 + φi (a) + φi (av )

p ⇒ φi (a) + φi (av ) = −1.

Using the inner product of φi with itself we have

1 = φi , φi = j

φi (g j )φi (g j )

|C G(g j )| =q + φi (a)φi (a) + φi (av )φi (av )

p ,

⇒ p − q = φi (a)φi (a) + φi (av )φi (av ),

⇒ p − p − 1

2= φi (a)φi (a) + φi (av )φi (av ),

⇒ p + 1

2= φi (a)φi (a) + φi (av )φi (av ).



Chapter 25 119

We now proceed to break this down in to cases. Assume p ≡ 1 (mod 4) then a is

conjugate to a−1, which means φi = φi . Recall that from above we have φi (av ) =

−1 − φi (a) so we have

φi (a)2 + φi (av )2 =p + 1

2⇒ φi (a)2 + (φi (a) + 1)2 =

p + 1

2,

⇒ 2φi (a)2 + 2φi (a) + 1 =p + 1

2,

⇒ 4φi (a)2 + 4φi (a) + 1 − p = 0,

⇒ φi (a) =−4 ±

42 − 4 · 4 · (1 − p )

2 · 4,

⇒ φi (a) = −4

± 42p

2 · 4 ,

⇒ φi (a) =−1 ± √

p

2.

Now assume p ≡ −1 (mod 4) then a is not conjugate to a−1, which means

a−1 ∈ (av )G. This means that φi (av ) = φi (a−1) = φi (a). Therefore we have

φi (a) + φi (av ) = −1 ⇒ φi (a) + φi (a) = −1,

⇒ 2Re(φi (a)) = −1,

⇒ Re(φi (a)) = −12

.

Using the second equation we have that

φi (a)φi (a) + φi (av )φi (av ) =p + 1

2⇒ 2φi (a)φi (a) =

p + 1

2,

⇒ Re(φi (a))2 + Im(φi (a))2 =p + 1

4,

⇒ Im(φi (a))2 =p + 1

4− 1

4,

⇒ Im(φi (a))2 =p

4 ,

⇒ Im(φi (a)) = ±√

p

4,

⇒ φi (a) =−1 ± √

pi

2.

Comparing with the question this settles the two possible cases.

(d) Let ε = e 2πi/p be a primitive p th root of unity. Consider, as in the question, δ = ±1



Chapter 25 120

dependent upon p then we know that

−1 ± √ δp 2

= φ1(a) =

p −12

k =1

εu k

.

Now Z∗p is a cyclic group of order p − 1 and u is a generator for Z∗

p , therefore every

element in Z∗p can be expressed as a power of u .

We can choose u in the following way. Let n be the integer described in Theorem

25.1. Now p − 1 = 2q therefore we have

np −1 ≡ n2q ≡ (n2)q ≡ 1 (mod p ).

Also (n2)r ≡ n2r ≡ 1 (mod p ) for all 0 < r < q because if there was such an r then

we would have 0 < 2r < 2q = p − 1, which would invalidate the property of n. Hence

n2 is has order q modulo p . Therefore we can set u = n2, which gives us that u is a

quadratic residue modulo p . Indeed this means all powers of u are quadratic residues

modulo p .

In other words we have the desired result that

s ∈Q εs =−1 ± √

δp

2,

where Q is the set of all quadratic residues modulo p .

Exercise 25.5. First let us consider the conjugacy classes of E . Now it’s clear to see that

a, b C E (ai b j ), for any 0 i , j 2 but not i = j = 0. This means that |C E (ai b j )| 9,

so either the conjugacy class has size 1 or 2. However we have

c −1(ai b j )c = (c −1ac )i (c −1bc ) j = a−i b − j ,

so the conjugacy classes have order 2. Note that as a and b have odd order, no powers of a and b are involutions. This gives us the conjugacy classes

1 a, a2 b, b 2 ab,a2b 2 ab 2, a2b .

This only deals with half the elements of E . We have c C E (c ), which means



Chapter 25 121

|c E | 9. However, keeping notation as before, we have

a−i

b − j

cb j

a j

= a−i

b −2 j

cai

= a−i

b −2 j

a−i

c = a−2i

b −2 j

c.

Therefore it’s clear that the final conjugacy class has order 9 and explicitly is

c , a c , bc , a2c, b 2c,abc,ab 2c, a2bc,a2b 2c .

Now a, b ∼= C 3×C 3 is a normal index 2 subgroup in E , hence we can use Clifford Theory

to give us more information about the induced characters. Recall that if η = e 2πi/3 = −1+i √ 3

2

then every irreducible character of C 3 is given by χk (ai ) = ηki , (with 0 k 2), and every

irreducible character of C 3 × C 3 is given by the products of these characters. Note thatthere are 6 conjugacy classes of E and hence 6 irreducible characters.

A generic irreducible character of a, b is given by χk χ(ai b j ) = ηki ηj = ηki +j with

0 k, 2. Notice that the conjugacy classes of ai b j split in to two conjugacy classes in

the subgroup a, b . Therefore we have the values of the induced characters to be

(χk χ ↑ E )(c ) = 0,

(χk χ ↑ E )(ai b j ) = |C E (ai b j )|

χk χ(ai b j )

|C a,b (ai b j )

|

+χk χ(a−i b − j )

|C a,b (a−i b − j )

|,

= 9 · ηki +j + η−ki −j

9,

= ηki +j + ηki +j ,

=

2 if ki + j ≡ 0 (mod 3),

−1 otherwise.

Let (k, ) denote the pair of indices coming from the character χk χ. Using the above

information we can see the following values of the induced characters

(0, 1) ⇒ j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(a) = 2,

(0, 2) ⇒ 2 j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(a) = 2,

(1, 0) ⇒ i ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(b ) = 2,

(1, 1) ⇒ i + j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(ab 2) = 2,

(1, 2) ⇒ i + 2 j ≡ 0 (mod 3) ⇒ (χ1χ2 ↑ E )(ab ) = 2

(2, 0) ⇒ 2i ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(b ) = 2,



Chapter 25 122

(2, 1) ⇒ 2i + j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(ab ) = 2,

(2, 2)

⇒2i + 2 j

≡0 (mod 3)

⇒(χ0χ1

↑E )(ab 2) = 2.

Therefore the following characters are equal

χ0χ1 ↑ E = χ0χ2 ↑ E χ1χ0 ↑ E = χ2χ0 ↑ E χ1χ1 ↑ E = χ2χ2 ↑ E,

χ1χ2 ↑ E = χ2χ1 ↑ E.

Recall that by Clifford theory we know that each induced character is either irreducible

or a sum of two irreducible characters each with multiplicity 1. The induced trivial character

always contains a copy of the trivial character, hence it is a sum of two linear characters.

The non-isomorphic induced characters are then as in table 25.2.

g 1 a b ab ab 2 c

|C E (g )| 18 9 9 9 9 2

χ0χ0 ↑ E 2 2 2 2 2 0

χ0χ1 ↑ E 2 2 −1 −1 −1 0

χ1χ0 ↑ E 2 −1 2 −1 −1 0

χ1χ2

↑E 2

−1

−1 2

−1 0

χ2χ1 ↑ E 2 −1 −1 −1 2 0

Table 25.2: Some characters induced from a, b to E

It’s easy to see that E = [a, c ], [b, c ] = a−2, b −2 = a, b , hence there are only

|G/G | = 18/9 = 2 linear characters of E . These both occur has constituents of χ0χ0 ↑ E ,

hence the remaining four characters induced from E must be irreducible. Therefore the

character table of E is as in table 25.3.

Exercise 25.6. Clearly we have Z (E ) = C E (a)∩C E (b )∩C E (c ) = a, b ∩c = 1, whichis obviously cyclic. However from the character table above we can see there is no faithful

irreducible character of E because there is no character such that ψi (x ) = ψi (1) for all

x ∈ E .

Exercise 25.7. We aim the find groups with the respective character degrees.

(a) Clearly if G exists then we have |G| = 3 ·12 + 4 ·32 = 3 + 36 = 39 = 3×13. Clearly G

cannot be abelian and so by Proposition 25.7 we have G ∼= F 13,3. To verify we clearly



Chapter 25 123

g 1 a b ab ab 2 c

|C E (g )

|18 9 9 9 9 2

1E 1 1 1 1 1 1

ψ1 1 1 1 1 1 −1

ψ2 2 2 −1 −1 −1 0

ψ3 2 −1 2 −1 −1 0

ψ4 2 −1 −1 2 −1 0

ψ5 2 −1 −1 −1 2 0

Table 25.3: The character table of E

have 3 | 13 − 1 = 12, now F 13,3 has q = 3 linear characters and |Z∗p /S| = 12/3 = 4

characters of degree q = 3. Therefore F 13,3 is indeed the group we’re looking for.

(b) Clearly if G exists then we have |G| = 6 · 12 + 8 · 32 = 78 = 2 × 3 × 13. The group

F 13,3×C 2 has the character degrees that we’re looking for. Recall that the irreducible

characters of F 13,3×C 2 are the products of the irreducible characters of F 13,3 together

with the irreducible characters of C 2. Every irreducible character of C 2 is linear which

means the degree of the product is the same as the degree of the irreducible character

of F 13,3. Therefore there are 14 irreducible characters of F 13,3 × C 2, six of which are

linear and the remaining 8 have degree 3.

(c) Clearly if G exists then we have |G| = 6·12+3·22+8·32+4·62 = 234 = 2×32×13 =

6 × 39. The group S3 × F 13,3, (or alternatively D6 × F 13,3 as D6∼= S3), has the

character degrees that we’re looking for. Recall that S3 has two linear characters

and one irreducible character of degree 2. Therefore multiplying these degrees with

the character degrees from part (a) we get the desired result.

Exercise 25.8. First we aim to determine the conjugacy classes of G. Now

a

C G(a),

which means |C G(a)| 9 ⇒ |aG| 6. However for any 1 j 5 we have b − j ab j = a2 j ,

which gives us aG = a, a2, a4, a8, a16, a32 = a, a2, a4, a5, a7, a8.

Now consider the remaining powers of a. We have

b −2a3b 2 = (a3)4 = a12 = a3,

so a, b 2 C G(a3), which means |C G(a3)| 27 ⇒ |(a3)G| 2. However b −1a3b = a6, so

we must have (a3)G = a3, a6.





Chapter 25 125

are |G/G | = 54/9 = 6 linear characters, which are the lifts from G/G ∼= b ∼= C 6. Let

ε = e 2πi/6 = 1+i √ 3

2then the linear characters of G are as in table 25.4.

g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5

|C G(g )| 54 9 27 9 9 6 18 6 18 6

ψ0(g ) 1 1 1 1 1 1 1 1 1 1

ψ1(g ) 1 1 1 1 1 ε ε2 −1 −ε −ε2

ψ2(g ) 1 1 1 1 1 ε2 −ε 1 ε2 −ε

ψ3(g ) 1 1 1 1 1 −1 1 −1 1 −1

ψ4

(g ) 1 1 1 1 1−

ε ε2 1−

ε ε2

ψ5(g ) 1 1 1 1 1 −ε2 −ε −1 ε2 ε

Table 25.4: The linear characters of G from Exercise 25.8

From this we can work out the remaining character degrees of G. We know that

ψi (1) | |G| so our only options are 2, 3, 6 or 9. Therefore we have

ψ6(1)2 + ψ7(1)2 + ψ8(1)2 + ψ9(1)2 = 48.

We can see that no character has degree 9 as 92 = 81 > 48. Also 4 · 32 = 36 < 48 sowe must have a character of degree 6. Now 48 − 62 = 12 = 3 · 22, which means our only

option is to have one character of degree 6 and three characters of degree 2.

We now want to find a suitable subgroup to try and obtain the remaining irreducible

characters from. We can see from the group relations that (a3)3 = (b 2)3 = 1 and b −2a3b 2 =

(b −2ab 2)3 = (a4)3 = a3. Therefore N = a3, b 2 ∼= C 3 × C 3 is a subgroup of G. We

can see that N is a union of conjugacy classes and is therefore a normal subgroup of

G. Now |G/N | = 54/9 = 6 and clearly G/N is not abelian therefore the factor group

G/N =

Na,Nb

∼= D6

∼= S3.

As this subgroup is normal we can lift the characters of N to G. We lift the irreducible

character of degree 2 from the character table of D6 on page 125. First we note that

N = Nb 2 = Nb 4 = Na3,

Nb = Nb 3 = Nb 5,

Na = Nab 2 = Nab 4.

Therefore we have an irreducible character of degree 2 whose values are in table 25.5.



Chapter 25 126

g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5

|C G(g )

|54 9 27 9 9 6 18 6 18 6

ψ6(g ) 2 −1 2 −1 −1 0 2 0 2 0

Table 25.5: A degree 2 character of G from Exercise 25.8

We can combine this irreducible character with the linear characters we already to have

to obtain new irreducible characters. It’s clear to see using the character table above that

ψ7 = ψ6ψ1 and ψ8 = ψ6ψ2 are two distinct irreducible characters of degree 2. There is

only one irreducible character of degree 6 left to find. However a simple argument using

the column orthogonality relations determines the final character. Therefore the charactertable for G is as in table 25.6.

g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5

|C G(g )| 54 9 27 9 9 6 18 6 18 6

ψ0(g ) 1 1 1 1 1 1 1 1 1 1

ψ1(g ) 1 1 1 1 1 ε ε2 −1 −ε −ε2

ψ2(g ) 1 1 1 1 1 ε2

−ε 1 ε2

−ε

ψ3(g ) 1 1 1 1 1 −1 1 −1 1 −1

ψ4(g ) 1 1 1 1 1 −ε ε2 1 −ε ε2

ψ5(g ) 1 1 1 1 1 −ε2 −ε −1 ε2 ε

ψ6(g ) 2 −1 2 −1 −1 0 2 0 2 0

ψ7(g ) 2 −1 2 −1 −1 0 2ε2 0 −2ε 0

ψ8(g ) 2 −1 2 −1 −1 0 −2ε 0 2ε2 0

ψ9(g ) 6 0 −3 0 0 0 0 0 0 0




Chapter 26. Characters of some p -groups

Exercise 26.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Exercise 26.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Exercise 26.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

Exercise 26.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

Exercise 26.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Exercise 26.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Exercise 26.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Exercise 26.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

Exercise 26.1. Firstly, assume G is abelian then G has p n linear characters and p n−2 −p n−2 = 0 irreducible characters of degree p . Therefore this is certainly true for such an

abelian group.

Now assume G is non-abelian. Let χ be an irreducible character of G then there exists

an irreducible character ψ of H such that χ ↓ H, ψH = 0 ⇒ χ, ψ ↑ H G = 0. This just

uses the fact that every irreducible character of H is a constituent of the restriction of some

irreducible character of G and Frobenius reciprocity. Now we know that (ψ

↑G)(1) = [G :

H ]ψ(1) = [G : H ] = p because H is abelian hence every character is linear. Therefore as χ

is a constituent of (ψ ↑ G) we have

1 χ(1) (ψ ↑ G)(1) = p ⇒ χ(1) = 1 or p,

because χ(1) | |G|.So every irreducible character of G is either linear or of degree p . Now suppose there

are x linear characters and y characters of degree p then

x + p 2 y = p n ⇒ x = p n − p 2 y ⇒ x = p 2(p n−2 − y ) p 2

because x is non-negative and non-zero because we always have the trivial character as

a linear character. Therefore we have x = p m for some m 2, which leaves us with

y = p n−2 − p m−2.

Exercise 26.2. It’s clear to see from the relations that G C G(z i ) for 0 i 2, hence

we have three conjugacy classes of size 1. Now from the last defining relation of the group

127



Chapter 26 128

we get

b −1ab = az ⇒ b = a−1baz ⇒ a−1ba = bz 2.

Let 0 i , j , k 2 but such that we don’t have i = j = 0. Then in general we have that

b −k (ai b j )b k = (b −k ab k )i b j = (az k )i b j = ai b j z ik ,

a−k (ai b j )ak = ai (a−k bak ) j = ai (bz k ) j = ai b j z jk .

Therefore we have the conjugacy classes of G to be

1 a,az,az 2 b 2, b 2z , b 2z 2 ab 2, ab 2z,ab 2z 2,

z a2, a2z , a2z 2 ab,abz,abz 2 a2b 2, a2b 2z , a2b 2z 2,

z 2 b,bz,bz 2 a2b, a2bz , a2bz 2.

So there are 11 conjugacy classes of G, which means there are 11 irreducible characters.

Note that a, z G is an abelian subgroup of order 32 = 9 and so [G : a, z ] = 3. Let

K = G = z in Theorem 26.6. Now |G/G | = 9 and the relation b −1ab = az tells us

(b −1ab )G = az G ⇒ (b −1ab )G = aG ⇒ (ab )G = (ba)G .

So G/G is abelian and contains two elements of order 3, so G/G ∼= C 3 × C 3.

Let η = e 2πi/3 = −1+√ 3i

2 then we have the character table for G is as in table 26.1.

Now the final two characters are induced linear characters of a, z ∼= C 3×C 3. We note that

the conjugacy classes 1, z and z 2 do not split upon restriction to a, z . However,

for 1 j 2, we have a j , a j z , a j z 2 splits into three conjugacy classes. Therefore if ψ is

an irreducible character of a, z then

(ψ ↑ G)(z j ) = [G : H ]ψ(z j ) = 3ψ(z j ),

(ψ ↑ G)(a j ) = |C G(a j )| ψ(a j )

|C a,z (a j )| +ψ(a j z )

|C a,z (a j z )| +ψ(a j z 2)

|C a,z (a j z 2)| ,

= ψ(a j ) + ψ(a j z ) + ψ(a j z 2),

= ψ(a j )(1 + ψ(z ) + ψ(z 2)).



Chapter 26 129

g 1 z z 2 a a2 b b 2 ab a2b ab 2 a2b 2

|C G(g )

|27 27 27 9 9 9 9 9 9 9 9

χ1(g ) 1 1 1 1 1 1 1 1 1 1 1

χ2(g ) 1 1 1 1 1 η η2 η η η2 η2

χ3(g ) 1 1 1 1 1 η2 η η2 η2 η η

χ4(g ) 1 1 1 η η2 1 1 η η2 η η2

χ5(g ) 1 1 1 η η2 η η2 η2 1 1 η

χ6(g ) 1 1 1 η η2 η2 η 1 η η2 1

χ7(g ) 1 1 1 η2 η 1 1 η2 η η2 η

χ8(g ) 1 1 1 η2 η η η2 1 η2 η 1

χ9(g ) 1 1 1 η2 η η2 η η 1 1 η2

χ10(g ) 3 3η 3η2 0 0 0 0 0 0 0 0

χ11(g ) 3 3η2 3η 0 0 0 0 0 0 0 0


Now we can see that

ψ ↑ G, ψ ↑ G 32ψ(1)ψ(1)27

+ 32ψ(z )ψ(z )27

+ 32ψ(z 2)ψ(z 2)27

= 3 · 32

27= 1.

Hence we only get irreducible characters when ψ(z ) = 1 which confirms what we have in

the character table for G.

Exercise 26.3. We first need to determine the conjugacy classes of G. Now it’s clear that

for any 1 j 15 we have a C G(a j ), which means |(a j )G| 2. From the relations it’s

easy to see that b −1a j b = (b −1ab ) j = a− j . Therefore for 1 k 7 we have the following

conjugacy classes;

1

,

a8

,

ak , a−k

.

Now b C G(b ) which means |b G| 8. Now using the relations of the group we have

b −1ab = a−1 ⇒ ab = ba−1 ⇒ b = a−1ba−1 ⇒ a−1ba = a−2b.

Therefore a− j ba j = a−2 j b , which means b G = a2k b | 0 k 7. Likewise we get

a− j (ab )a j = a1−2 j b , which gives us (ab )G = a2k +1b | 0 k 7. In summary, with



Chapter 26 130

1 7, we have the conjugacy classes of G to be

1 a2k

b | 0 k 7,a8 a2k +1b | 0 k 7,

a, a−.

Now a is an abelian subgroup of order 16 in G and hence of index 2. So take H = aand K = Z (G) = a8 in Theorem 26.6. Also note that a−1b −1ab = a−2, which means

G = a2. Now G/K ∼= a, b | a8 = b 2 = 1, b −1ab = a−1 = D16. Now we know the

character table of D16 from page 305, therefore we have the lifts from G/K to be as in

table 26.2.

g 1 a8 a a2 a3 a4 a5 a6 a7 b ab

|C G(g )| 32 32 16 16 16 16 16 16 16 4 4

χ1(g ) 1 1 1 1 1 1 1 1 1 1 1

χ2(g ) 1 1 1 1 1 1 1 1 1 −1 −1

χ3(g ) 1 1 −1 1 −1 1 −1 1 −1 1 −1

χ4(g ) 1 1 −1 1 −1 1 −1 1 −1 −1 1

χ5(g ) 2 2 0 −2 0 2 0 −2 0 0 0χ6(g ) 2 2

√ 2 0 −√

2 −2 −√ 2 0

√ 2 0 0

χ7(g ) 2 2 −√ 2 0

√ 2 −2

√ 2 0 −√

2 0 0

Table 26.2: The lifted characters from G/K to G of Exercise 26.3

The remaining four characters are induced from the abelian subgroup H = a. Let

ε = e 2πi/16 be a primitive 16th root of unity. Then every irreducible representation of H

takes the form ψi (a j ) = εij for 0 i , j 15. Now, for 1 k 7, we calculate the induced

character to be

(ψi ↑ G)(a8) = [G : H ]ψi (a8) = 2ε8i = 2(−1)i

(ψi ↑ G)(a j ) = |C G(a j )|

ψi (a j )

|C H(a j )| +ψi (a− j )

|C H(a− j )|

,

= ψi (a j ) + ψi (a− j ),

= εij + ε−ij ,

= εij + εi j ,



Chapter 26 131

= 2 Re(εi j ).

Now it’s clear that Re(εk

) = Re(ε16−k

) = Re(εk

) for 1 k 7, which means ψk ↑ G =ψ16−k ↑ G for 1 k 7. So it’s sufficient to focus on the first eight characters to find

the remaining four irreducible characters of G. Recall that ε2 = e 2πi/8 =√ 2+i

√ 2

2. Straight

away we can see that ψ0 ↑ G = 2χ1 and in fact ψ8 ↑ G = χ3 + χ4 because ε8 = −1, in

other words they are reducible. Indeed, noticing that 2Re(ε2) =√

2, 2Re(ε4) = 0 and

2Re(ε6) = −√ 2, we can see that ψ2 ↑ G = χ6, ψ4 ↑ G = χ5 and ψ6 ↑ G = χ7.

Therefore we only get new distinct irreducible characters of G for i = 1, 3, 5 or 7.

Letting α = 2 Re(ε) = 2 cos(π/8) and β = 2 Re(ε3) = 2cos(3π/8), we get the character

table of G is as in table 26.3.

g 1 a8 a a2 a3 a4 a5 a6 a7 b ab

|C G(g )| 32 32 16 16 16 16 16 16 16 4 4

χ1(g ) 1 1 1 1 1 1 1 1 1 1 1

χ2(g ) 1 1 1 1 1 1 1 1 1 −1 −1

χ3(g ) 1 1 −1 1 −1 1 −1 1 −1 1 −1

χ4(g ) 1 1 −1 1 −1 1 −1 1 −1 −1 1

χ5

(g ) 2 2 0−

2 0 2 0−

2 0 0 0

χ6(g ) 2 2√

2 0 −√ 2 −2 −√

2 0√

2 0 0

χ7(g ) 2 2 −√ 2 0

√ 2 −2

√ 2 0 −√

2 0 0

χ8(g ) 2 −2 α√

2 β 0 − β −√ 2 −α 0 0

χ9(g ) 2 −2 β −√ 2 −α 0 α

√ 2 − β 0 0

χ10(g ) 2 −2 − β −√ 2 α 0 −α

√ 2 β 0 0

χ11(g ) 2 −2 −α√

2 − β 0 β −√ 2 α 0 0


Exercise 26.4. Let G = A,B,C,D where A,B,C,D are the 4×4 complex valued matrices

defined in the question.



Chapter 26 132

(a) We now check that all pairs of generators commute modulo Z .

AB =

−1 0 0 00 1 0 0

0 0 1 0

0 0 0 −1

0 0 i 00 0 0 −i

i 0 0 0

0 −i 0 0

=

0 0 −i 00 0 0 −i

i 0 0 0

0 i 0 0

= −BA,

AC =

−1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 −1

0 i 0 0

i 0 0 0

0 0 0 −i

0 0 −i 0

=

0 −i 0 0

i 0 0 0

0 0 0 −i

0 0 i 0

= −CA,

AD =

−

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 −1

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

=

0 0 0

−1

0 0 1 0

0 1 0 0

−1 0 0 0

= −DA,

BC =

0 0 i 0

0 0 0 −i

i 0 0 0

0 −i 0 0

0 i 0 0

i 0 0 0

0 0 0 −i

0 0 −i 0

=

0 0 0 1

0 0 −1 0

0 −1 0 0

1 0 0 0

= −CB,

BD =

0 0 i 0

0 0 0 −i

i 0 0 0

0 −i 0 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

=

0 i 0 0

−i 0 0 0

0 0 0 i

0 0 −i 0

= −DB,

CD =

0 i 0 0

i 0 0 0

0 0 0 −i

0 0 −i 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

=

0 0 i 0

0 0 0 i

−i 0 0 0

0 −i 0 0

= −DC.

Therefore every pair of generators commutes modulo Z . Clearly G is not abelianbut G/Z is abelian. We recall that if a quotient G/N is abelian then we must have

G N . In other words I = G Z ⇒ G = Z .

(b) We check the squares of the generating elements. It’s obvious that A2 = D2 = 1, so





Chapter 26 134

Now checking the commutation with D gives us

x 11 0 0 x 140 x 22 x 23 0

0 x 32 x 33 0

x 41 0 0 x 44

0 0 0 10 0 1 0

0 1 0 0

1 0 0 0

=

0 0 0 10 0 1 0

0 1 0 0

1 0 0 0

x 11 0 0 x 140 x 22 x 23 0

0 x 32 x 33 0

x 41 0 0 x 44

,

⇒

x 14 0 0 x 11

0 x 23 x 22 0

0 x 33 x 32 0

x 44 0 0 x 41

=

x 41 0 0 x 44

0 x 32 x 33 0

0 x 22 x 23 0

x 11 0 0 x 14

,

⇒ X =

x 11 0 0 x 14

0 x 22 x 23 0

0 x 23 x 22 0

x 14 0 0 x 11

.

Now checking the commutation with C gives us

x 11 0 0 x 14

0 x 22 x 23 0

0 x 23 x 22 0

x 14 0 0 x 11

0 i 0 0

i 0 0 0

0 0 0

−i

0 0 −i 0

=

0 i 0 0

i 0 0 0

0 0 0

−i

0 0 −i 0

x 11 0 0 x 14

0 x 22 x 23 0

0 x 23 x 22 0

x 14 0 0 x 11

,

0 i x 11 −i x 14 0

ix 22 0 0 −i x 23

ix 23 0 0 −i x 22

0 i x 14 −i x 11 0

=

0 ix 22 ix 23 0

ix 11 0 0 i x 14

−i x 14 0 0 −ix 11

0 −ix 23 −ix 22 0

,

⇒ X =

x 11 0 0 0

0 x 11 0 0

0 0 x 11 0

0 0 0 x 11

.

Therefore X is a scalar multiple of the identity and we’re done.

(d) Now we have an irreducible representation of degree 4 which means that 12 + 42 =

17 |G| 32 but G is a 2 group, which means |G| = 2x but 24 = 16 therefore we

must have |G| = 25 = 32.

Recall from part (a) that G = Z , which means |G/G | = |G|/|G| = 32/2 = 16.

So the remaining representations of G are all 1-dimensional. By part (a) we know that



Chapter 26 135

all generators commute in G/G , which means G/G ∼= C 2 × C 2 × C 2 × C 2. Now for

example an irreducible representation of C 2 is given by Aρ = −1 or Aρ = 1 = (−1)0.

So for 0 x , y , z , w 1 we get a linear representation of G given by

Ai B j C k DZ m → (−1)xr + y j +zk +w.

This gives us all the irreducible representations of G.

Exercise 26.5. Let G1, . . . , G9 be the non-abelian groups of order 16 with presentations as

given.

(a) We start by looking for a faithful irreducible representation of degree 2 for G1 = D16.

Examining the character table for G1 we can see that such an irreducible representationwould need to have trace 0 on b and trace ±√

2 on a. Recall ω = e 2πi/8 =√ 2+i

√ 2

2.

Now ω−1 = e 2πi/8 =√ 2−i √

22 , therefore we get such an irreducible representation

ρ : G1 → GL(2,C) by defining

aρ =

ω 0

0 ω−1

bρ =

0 1

1 0

.

It’s clear that (aρ)8 = (bρ)2 = I2 and so all that’s left to check is the remaining

relation.

(bρ)−1(aρ)(bρ) =

0 1

1 0

ω 0

0 ω−1

0 1

1 0

,

=

0 ω−1

ω 0

0 1

1 0

,

=

ω−1 0

0 ω

,

= (aρ)−1.

Now consider such an irreducible representation for G2. Recall that ω4 = −1 and

so if we define aρ as for G1 we have a4ρ = −I2. Therefore we must define bρ such

that (bρ)2 = (aρ)4 = −I. We define our representation ρ : G2 → GL(2,C) by

aρ =

ω 0

0 ω−1

bρ =

0 −1

1 0

.

It’s clear to see that (aρ)8 = I2 and (bρ)2 = (aρ)4 = −I2, so we only need to check



Chapter 26 136

the remaining defining relation holds.

(bρ)−1(aρ)(bρ) = 0 1−1 0ω 0

0 ω−10 −1

1 0 ,

=

0 ω−1

−ω 0

0 −1

1 0

,

=

ω−1 0

0 ω

,

= (aρ)−1.

Now consider such an irreducible representation for G3. We notice that for G3 thetrace on a is not ±√

2 but ±i √

2. Now we can see that ω − ω−1 = i √

2 and also that

−ω−1 = ω3. So, we define our representation ρ : G3 → GL(2,C) by

aρ =

ω 0

0 ω3

bρ =

0 1

1 0

It’s clear to see that (aρ)8 = (bρ)2 = I2, so we only need to check the remaining

defining relation holds.

(bρ)−1(aρ)(bρ) =0 1

1 0

ω 00 ω3

0 11 0

,

=

0 ω3

ω 0

0 1

1 0

,

=

ω3 0

0 ω

,

= (aρ)−1.

Note that (ω3)3 = ω9 = ω.

Now consider such an irreducible representation for G4. Note that the presentation

of G4 written in the book is redundant so we rewrite this as G4 = a, b | a8 = b 2 =

1, b −1ab = a5. Note from the character table of G4 that we require the trace on a



Chapter 26 137

to be 0 and that −ω = ω5. So, we define our representation ρ : G4 → GL(2,C) by

aρ = ω 00 ω5

bρ = 0 11 0

.

It’s clear to see that (aρ)8 = (bρ)2 = I2, so we only need to check the remaining

defining relation holds.

(bρ)−1(aρ)(bρ) =

0 1

1 0

ω 0

0 ω5

0 1

1 0

,

= 0 ω5

ω 0 0 1

1 0 ,

=

ω5 0

0 ω

,

= (aρ)−1.

Note that (ω5)5 = ω25 = ω.

Finally consider such an irreducible representation for G9. Note that there is no

redundancy in the presentation of G9. Now z is an element of order 4 and recall that

i is a fourth root of unity. From the character table we know that the trace on a

should be 2i and 0 on a and b . So, we define our representation ρ : G9 → GL(2,C)

by

aρ =

1 0

0 −1

bρ =

0 1

1 0

i 0

0 i

Now it’s clear that (aρ)2 = (bρ)2 = (z ρ)4 = I2 so we only need to check that the

remaining relations hold.

(aρ)(z ρ) = 1 0

0 −1i 0

0 i = i 0

0 −i = i 0

0 i 1 0

0 −1 = (z ρ)(aρ),

(bρ)(zρ) =

0 1

1 0

i 0

0 i

=

0 i

i 0

=

i 0

0 i

0 1

1 0

= (zρ)(bρ),

(bρ)−1(aρ)(bρ) =

0 1

1 0

1 0

0 −1

0 1

1 0

=

0 −1

1 0

0 1

1 0

=

−1 0

0 1

= (aρ)(z ρ)2.

(b) Recall from Proposition 9.16 that if there exists a faithful irreducible CG module

then Z (G), (the centre of G), is cyclic. Now the centre of G is the union of all



Chapter 26 138

conjugacy classes of order 1 in the group. Reading off from the character tables we

get Z (G5) = Z (G6) = Z (G7) = Z (G8) = C 1 ∪ C 2 ∪ C 3 ∪ C 4 = 1, z , a2, a2z .

Now z = b 2 in G5 so Z (G5) = 1, b 2, a2, a2b 2 ∼= C 2 × C 2 which is not cyclic.Note that it is easier to see this is C 2 × C 2 once you know ba = ab −1 and ab = b −1a.

In G6, G7 and G8 we have az = z a so Z (G6) ∼= Z (G7) ∼= Z (G8) ∼= C 2 × C 2, which is

not cyclic.

(c) Let the morphism in the question be denoted by ρ : G5 → GL(2,C), then this gives

a representation of G5 if it satisfies the defining relations of G5. We note first that

clearly z 2 = 1. Now

(bρ)2 = i 2 0 0

0 (−i )2 0

0 0 12

= −1 0 0

0 −1 0

0 0 1

= (z ρ),

(aρ)4 =

0 1 0

1 0 0

0 0 i

0 1 0

1 0 0

0 0 i

2

=

1 0 0

0 1 0

0 0 −1

2

=

12 0 0

0 12 0

0 0 (−1)2

= I3.

Now we check the final relation

(bρ)−1(aρ)(bρ) = −i 0 0

0 i 0

0 0 1

0 1 0

1 0 0

0 0 i

i 0 0

0 −i 0

0 0 1

,

=

0 −i 0

i 0 0

0 0 i

i 0 0

0 −i 0

0 0 1

,

=

0 −1 0

−1 0 0

0 0 i

,

=

0 1 0

1 0 0

0 0 i

−1 0 0

0 −1 0

0 0 1

,

= (aρ)(z ρ).

Therefore this is indeed a representation of G5. Is it faithful? Well from the above it’s

easy to see that 1ρ,aρ,a2ρ, a3ρ,bρ,zρ,azρ,b −1aρ,abρ are all distinct elements of the

group generated by the matrices. Therefore there are more than 8 distinct elements



Chapter 26 139

in this group which means it must have order 16, hence ρ is a faithful representation.

Now let φ : G6 → GL(2,C) be the morphism of G6 defined in the question. It’s

clear that (aφ)4 = (bφ)2 = (z φ)2 = 1. We first of all check the commuting relations

(aρ)(z ρ) =

i 0 0

0 −i 0

0 0 i

−1 0 0

0 −1 0

0 0 1

=

−i 0 0

0 i 0

0 0 i

=−

1 0 0

0 −1 0

0 0 1

i 0 0

0 −i 0

0 0 i

= (z ρ)(aρ)

(bρ)(z ρ) =

0 1 0

1 0 0

0 0 1

−1 0 0

0 −1 0

0 0 1

,

=

0 −1 0

−1 0 0

0 0 1

,

=−

1 0 0

0 −1 0

0 0 1

0 1 0

1 0 0

0 0 1

,

= (z ρ)(bρ)

Now we check that the final relation holds

(bρ)−1(aρ)(bρ) =

0 1 0

1 0 0

0 0 1

i 0 0

0 −i 0

0 0 i

0 1 0

1 0 0

0 0 1

,

=

0 −i 0

i 0 0

0 0 i

0 1 0

1 0 0

0 0 1

,

=

−i 0 0

0 i 0

0 0 i

,

= i 0 0

0

−i 0

0 0 i

−1 0 0

0

−1 0

0 0 1 ,

= (aρ)(z ρ).

Therefore this indeed a representation of G6 but is it faithful? Well again we see that

we have generated more than 8 distinct elements, hence the group generated by the

matrices must have order 16 so it is a faithful representation.



Chapter 26 140

(d) Note that we get an irreducible representation of D8 by the morphism

a → i 00 −i

b → 0 11 0

.

Therefore our plan is to extend this by defining a morphism ρ : G7 → GL(3,C) by

aρ =

i 0 0

0 −i 0

0 0 1

bρ =

0 1 0

1 0 0

0 0 1

z ρ =

1 0 0

0 1 0

0 0 −1

.

It’s clear that the relations of G7 hold for these matrices. Notice that the the subgroup

generated by aρ and bρ gives us 8 distinct elements and multiplying anyone of theseby z ρ will give us another distinct element, hence the representation of faithful.

Likewise for G8 we define a morphism φ : G8 → GL(3,C) by

aρ =

i 0 0

0 −i 0

0 0 1

bρ =

0 1 0

−1 0 0

0 0 1

z ρ =

1 0 0

0 1 0

0 0 −1

.

Notice again that we have extended this idea from a 2-dimensional irreducible repre-

sentation of Q8. So it’s clear to see that this will indeed be a 3-dimensional faithful

irreducible representation of G8.

Exercise 26.6. Recall that if two groups have distinct character tables then they cannot

possibly be isomorphic. So the only possible isomorphisms are G1∼= G2, G5

∼= G6 and

G7∼= G8. Notice that we have G7

∼= D8 × C 2 and G8∼= Q8 × C 2 therefore we cannot have

G7∼= G8 otherwise this would give us D8

∼= Q8.

Now for G5 and G6 we list, in table 26.4, the representatives of the conjugacy classes

and the orders of those elements. Note that in G5 we have ab = b −aa and ba = ab −1 and

in G6 we have ab = baz and ba = abz .

g 1 z a2 a2z a a3 b a2b ab a3b

Order in G5 1 2 2 2 4 4 4 4 4 4

Order in G6 1 2 2 2 4 4 2 2 4 4

Table 26.4: The orders of elements of G5 and G6 from Exercise 26.6

Therefore it’s clear that the number of elements of order 2 and order 4 are different in G5

and G6, which means G5 ∼= G6.



Chapter 26 141

In table 26.5 we now do the same thing for G1 and G2. Note that in both G1 and G2

we have ab = ba−1 and ba = a−1b .

g 1 a4 a2 a a3 b ab

Order in G1 1 2 4 8 8 2 2

Order in G2 1 2 4 8 8 4 4

Table 26.5: The orders of elements of G1 and G2 from Exercise 26.6

Therefore the number of elements of order 2 and order 4 in G1 and G2 are different so we

cannot have G1

∼= G2.

Exercise 26.7. Let G be a non-abelian group of order p 4.

(a) Now G is a p -group so this means Z (G) = 1 so |Z (G)| = 1 and G is not abelian

so |Z (G)| = p 4. Recall from Lemma 26.1 that if K Z (G) and G/K is cyclic,

then G is abelian. Well if |Z (G)| = p 3 then |G/Z (G)| = p 4/p 3 = p , which means

G/Z (G) ∼= C p . However this would imply G abelian but this is not the case, so

|Z (G)| = p or p 2.

If |Z (G)| = p 2 then we clearly have p 2 conjugacy classes of order 1. We want

to then show that this leaves us with p 3

−p conjugacy classes. Let x 1, . . . , x k be

representatives of the conjugacy classes of G then by the class equation we have

p 4 = p 2 +

x i ∈Z (G)|x Gi | ⇒

x i ∈Z (G)

|x Gi | = p (p 3 − p ) = p 2(p 2 − 1).

Therefore there are either p 3 − p conjugacy classes of order p or p 2 − 1 conjugacy

classes of order p 2. Assume x ∈ G but x ∈ Z (G), then Z (G) C G(x ) but Z (G) =C G(x ) because x ∈ Z (G). Therefore |C G(x )| = p 3 ⇒ |x G| = p , which gives us the

desired result.

(b) Recall that we have the number of linear characters of G is [G : G] = |G/G | and wehave

|G| = [G : G ] +

χ(1)>1

χ(1)2 ⇒ p 4 = [G : G] +

χ(1)>1

χ(1)2.

Now if χ(1) > 1 then χ(1) = p because χ(1) | |G| and if χ(1) p 2 otherwise the

above equation is not satisfied as [G : G ] 1. Therefore p 2 | p 4 and p 2 divides the

sum so we must have p 2 | [G : G]. Note that |G | = 1 because G is not abelian, so

we have [G : G ] = p 2 or p 3. In other words |G| = p or p 2.



Chapter 26 142

Assume that |G | = p 2 then we have [G : G ] = p 2. Recall that any non-linear

character of G has order p so using the above equation we have

p 4 = p 2 +

χ(1)>1

p 2 ⇒

χ(1)>1

p 2 = p 2(p 2 − 1).

Therefore this tells us there are p 2 − 1 irreducible characters of order p and p 2 linear

characters. Or put another way we have p 2 + (p 2 − 1) = 2p 2 − 1 conjugacy classes

of G.

(c) Recall that G is a normal subgroup of G and that if G is a p -group then any normal

subgroup cannot intersect the centre trivially. This means p |G ∩ Z (G)| p 2

because G

∩Z (G) is a subgroup of G and Z (G). If

|G

∩Z (G)

|= p 2 then

|G

|=

|Z (G)| = p 2 but this means that the number of conjugacy classes is p 3 + p 2 − p and

2p 2 − 1. So this means p 3 + p 2 − p = 2p 2 − 1 ⇒ p 3 − p 2 − p + 1 = 0 but the only

solutions to the polynomial X 3 − X 2 − X + 1 = 0 are ±1. Therefore we must have

|G ∩ Z (G)| = p .

Exercise 26.8.

(a) Let G be any group and let Z = Z (G). Assume G/Z (G) ∼= Q8, then we can assume

there exists the following presentation for G/Z (G)

G/Z (G) = aZ,bZ | a4Z = Z, a2Z = b 2Z, (b −1ab )Z = a−1Z .

Now we have from the defining relations of Q8 that abZ = ba−1Z and baZ = a−1bZ .

Therefore

(a2b )Z = (ba−2)Z = ba2Z.

Note that the centre of G/Z must be trivial but we clearly have a2Z in the centre

of G/Z so we must have a2Z = Z , or in other words a2 ∈ Z . So we cannot have

G/Z ∼= Q8.(b) Certainly if G is an abelian group of order 16 then G = 1, so G/G ∩Z (G) ∼= Q8. Now

suppose G is non-abelian then G = 1, so by Exercise 7 we have |G ∩ Z (G)| = 2,

so either G ∩ Z (G) = G or G ∩ Z (G) = Z (G). We know that G/G is abelian so

G/G ∼= Q8 and G/Z (G) ∼= Q8 by part (a). Therefore G/(G ∩ Z (G)) ∼= Q8.



Chapter 27. Character table of the simple group of order

168

Exercise 27.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Exercise 27.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Exercise 27.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Exercise 27.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

Exercise 27.1. Let us fix a matrix

X =

x y

z w

∈ SL(2, p )

and assume that X ∈ Z (SL(2, p )). Thenx y

z w

1 1

0 1

=

1 1

0 1

x y

z w

⇒

x x + y

z z + w

=

x + z y + w

z w

,

⇒ z = 0 and x = w .

Choosing another appropriate matrix we find

x y

0 x

1 0

1 1

=

1 0

1 1

x y

0 x

⇒

x + y y

x x

=

x y

x x + y

,

⇒ y = 0.

Therefore X = xI2 but X ∈ SL(2, p ) means det(X ) = x 2 = 1 ⇒ x = ±1 and we’re done.

Exercise 27.2. We start by noting that|

SL(2, 3)|

= 3·

(32

−1) = 3

·8 = 24. We first

start by calculating the conjugacy classes of SL(2, 3). Much as for the conjugacy classes

of PSL(2, 7) we claim that the information in table 27.1 covers all the conjugacy classes of

SL(2, 3).

We now aim to confirm that the information in this table is correct. Clearly the information

about g 1 is correct. Now g 2 = 2I2 = −I2 ∈ Z (SL(2, 3)), therefore the information for g 2 is

clear.

143



Chapter 27 144

Order |C G(g i )| |g Gi |

g 1 = 1 0

0 1 1 24 1

g 2 =

2 0

0 2

2 24 1

g 3 =

0 1

2 0

4 4 6

g 4 =

1 1

0 1

3 6 4

g 5 = 1 2

0 1 3 6 4

g 6 =

2 1

0 2

6 6 4

g 7 =

2 2

0 2

6 6 4

Table 27.1: The conjugacy classes of SL(2, 3)

Start by considering X ∈ C G(g 3) then we havex y

z w

0 1

2 0

=

0 1

2 0

x y

z w

⇒

2 y x

2w z

=

z w

2x 2 y

,

⇒ X =

x y

2 y x

.

Now det(X ) = x 2 − 2 y 2 = x 2 + y 2 = 1. Our only choices are (x , y ) = (0, 1), (0, 2), (1, 0)

or (2, 0) therefore |C G(g 3)| = 4, which means |g G3 | = 24/3 = 6. Now as for the order of g 3

we have

g 43 =

0 1

2 0

0 1

2 0

2

=

2 0

0 2

2=

4 0

0 4

= I2.

Consider an element X ∈ C G(g 4) then we havex y

z w

1 1

0 1

=

1 1

0 1

x y

z w

⇒

x x + y

z z + w

=

x + z y + w

z w

,



Chapter 27 145

⇒ X =

x y

0 x

.

Now det(X ) = x 2 = 1 then our options are y = 0, 1, 2 and x = 1 or 2, hence |C G(g 4)| = 6,

which means |g G4 | = 24/6 = 4. Now as for the order of g 4 we have

g 34 =

1 1

0 1

1 1

0 1

1 1

0 1

=

1 2

0 1

1 1

0 1

=

1 0

0 1

= I2.

Consider an element X ∈ C G(g 5) then we have

x y

z w 1 2

0 1 = 1 2

0 1x y

z w ⇒ x 2x + y

z 2z + w = x + 2z y + 2w

z w ,

⇒ X =

x y

0 x

.

It’s clear from the previous calculation that |C G(g 5)| = 6, which means |g G5 | = 4. Now as

for the order of g 5 we have

g 35 =

1 2

0 1

1 2

0 1

1 2

0 1

=

1 1

0 1

1 2

0 1

=

1 0

0 1

= I2.

Consider an element X ∈ C G(g 6) then we havex y

z w

2 1

0 2

=

2 1

0 2

x y

z w

⇒

2x x + 2 y

2z z + 2w

=

2x + z 2 y + w

2z 2w

,

⇒ X =

x y

0 x

.

Hence again it’s clear from the previous calculation that we have |C G(g 6)| = 6, which means

|g G6 | = 4. Now as for the order of g 6 we have

g 66 =

2 1

0 2

2 1

0 2

2 1

0 2

2

=

1 1

0 1

2 1

0 2

2

=

2 0

0 2

2= I2.

Finally consider an element X ∈ C G(g 7) then we havex y

z w

2 2

0 2

=

2 2

0 2

x y

z w

⇒

2x 2x + 2 y

2z 2z + 2w

=

2x + 2z 2 y + 2w

2z 2w

,



Chapter 27 146

⇒ X =

x y

0 x

.

Therefore again we confirm that |C G(g 7)| = 6, which means |g G7 | = 4. Now as for the order

of g 7 we have

g 67 =

2 2

0 2

2 2

0 2

2 2

0 2

2

=

1 2

0 1

2 2

0 2

2

=

2 0

0 2

2= I2.

So the information in the table is correct. However we need to make sure that none of

the elements are conjugate. By the orders of the elements we only have to worry about

whether g 4 is conjugate to g 5 and g 6 is conjugate to g 7. Wellx y

z w

1 1

0 1

=

1 2

0 1

x y

z w

⇒

x x + y

z z + w

=

x + 2z y + 2w

z w

,

⇒ X =

x y

0 2x

.

However det(X ) = 2x 2 = 1 but x 2 = 1 for all x ∈ Z3 so X cannot be in SL(2, 3). Similarly

for g 6 and g 7 we find

x y

z w

2 1

0 2

=

2 2

0 2

x y

z w

⇒

2x x + 2 y

2z z + 2w

=

2x + 2z 2 y + 2w

2z 2w

,

⇒ X =

x y

0 2x

and such a matrix cannot lie in SL(2, 3). In other words the conjugacy classes in the table

are indeed correct.

Recall that Z = Z (SL(2, 3)) = I2, 2I2 is a non-trivial normal subgroup of SL(2, 3) andSL(2, 3)/Z = PSL(2, 3), which we know is isomorphic to A4 from the text. The elements

of PSL(2, 3) split up by conjugacy classes are0 1

2 0

Z,

2 1

1 1

Z,

2 2

2 1

Z

1 1

0 1

Z,

1 0

2 1

Z,

0 1

2 2

Z,

2 1

2 0

Z

1 2

0 1

Z,

1 0

1 1

Z,

0 1

2 1

Z,

1 1

2 0

Z

1 0

0 1

Z

.



Chapter 27 147

Now we can see that the conjugacy classes of order 4 here correspond to the 3-cycles in A4

and the conjugacy class of order 3 correspond to the 2-2-cycles in A4. It’s not important

to know exactly how the conjugacy classes match up to the conjugacy classes of 3-cycles inA4. This is because interchanging the conjugacy classes in the character table just reorders

the irreducible characters. Therefore we can lift the character table of A4, (on page 181),

to SL(2, 3).

Let ω = e 2πi/3 = 1+√ 3i

2then in table 27.2 we write down the lifted characters from

PSL(2, 3) to obtain four irreducible characters of SL(2, 3).

g g 1 g 2 g 3 g 4 g 5 g 6 g 7

|C G(g )

|24 24 4 6 6 6 6

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 1 1 ω ω2 ω ω2

χ3(g ) 1 1 1 ω2 ω ω2 ω

χ4(g ) 3 3 −1 0 0 0 0

Table 27.2: Four irreducible characters of SL(2, 3)

There are 7 conjugacy classes, which means there are 7 irreducible characters in total. We

start by working out the degrees of the remaining characters. Let d 5, d 6 and d 7 be the

remaining degrees then we know that d i | |G| = 24 and d 2i 24 ⇒ d i 4. Therefore the

degrees are either 1, 2, 3 or 4. We also know that

24 = 12 + 12 + 12 + 32 + d 25 + d 26 + d 27 ⇒ 12 = d 25 + d 26 + d 27 .

The only integer solution to this is d 5 = d 6 = d 7 = 2. Therefore the remaining three

characters are all of degree 2. Before carrying on we note that g −12 ∈ g G2 , g −13 ∈ g G3 ,

g 5 = g −14 and g 7 = g −16 . Therefore the columns of g 2 and g 3 are real and χ(g 5) = χ(g 4),

χ(g 7) = χ(g 6) for all irreducible characters χ of G.

Indeed because of the orders of the elements we will have that the columns of g 2 and g 3

will be integer valued. Also, we know that |C G(g 3)| = 4 which means the remaining entries

in the column of g 3 are zeros. Also because g 2 is an involution we know χ(g 2) is an integer

for all irreducible characters of G and χ(g 2) ≡ χ(1) (mod 2). Using this information and

the column orthogonality relations between g 1 and g 2 we can determine the column of g 2.

Reordering the characters slightly we have the character table so far to be as in table 27.3,

where η, ε are complex numbers and a, b are real numbers.



Chapter 27 148

g g 1 g 2 g 3 g 4 g 5 g 6 g 7

|C G(g )

|24 24 4 6 6 6 6

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 1 1 ω ω2 ω ω2

χ3(g ) 1 1 1 ω2 ω ω2 ω

χ4(g ) 2 −2 0 a a b b

χ5(g ) 2 −2 0 η η ε ε

χ6(g ) 2 −2 0 η η ε ε

χ7(g ) 3 3

−1 0 0 0 0

Table 27.3: The partial character table of SL(2, 3)

We explain why we have the entries for χ4, χ5 and χ6. Recall that if χ is an irreducible

character then so is χ. If all three characters were real then they would all be the same

and this cannot happen. Therefore at least one must be complex and hence comes with its

complex conjugate, we assume this to be χ5 and χ6 without loss of generality. Therefore

the final character must be real and recalling that g 5 = g −14 and g 7 = g −16 this fills in the

remaining details.

Taking the inner product of χ1 with χ4 we find

χ1, χ4 =2

24− 2

24+ 0 +

a + a + b + b

6⇒ 0 = 2a + 2b ⇒ b = −a.

Also considering the inner product of χ4 with itself gives us

χ4, χ4 =22

24+

22

24+ 0 +

4a2

6⇒ 24 = 8 + 16a2 ⇒ a2 = 1 ⇒ a = ±1.

Now g 4 is an element of order 3 and a is an integer so we must have χ(g 4)≡

χ(1) (mod 3)

for all irreducible characters, which give us a = −1. Recall that χ4χ2 and χ4χ3 must also

be irreducible characters of degree 2 therefore these must be χ4 and χ5. Hence we have

the character table of SL(2, 3) to be as in table 27.4.

Exercise 27.3. We can see from the character table that there is no non-identity element

g ∈ PSL(2, 7) such that χ(g ) = χ(1) for some irreducible character χ of G. Hence the

kernel of every character is trivial, which means there are no non-trivial normal subgroups.

Hence the group is simple.



Chapter 27 149

g g 1 g 2 g 3 g 4 g 5 g 6 g 7

|C G(g )

|24 24 4 6 6 6 6

χ1(g ) 1 1 1 1 1 1 1

χ2(g ) 1 1 1 ω ω2 ω ω2

χ3(g ) 1 1 1 ω2 ω ω2 ω

χ4(g ) 2 −2 0 −1 −1 1 1

χ5(g ) 2 −2 0 −ω −ω2 ω ω2

χ6(g ) 2 −2 0 −ω2 −ω ω2 ω

χ7(g ) 3 3

−1 0 0 0 0

Table 27.4: The character table of SL(2, 3)

Exercise 27.4.

(a) We start by working out the conjugacy classes of the subgroup. We claim that they

conjugacy classes are as in table 27.5.

Order |C G(g i )| |g Gi |

t 1

= 1 0

0 1Z 1 21 1

t 2 =

2 0

0 4

Z 3 3 7

t 3 =

3 0

0 5

Z 3 3 7

t 4 =

1 1

0 1

Z 7 7 3

t 5 = 1 6

0 1Z 7 7 3

Table 27.5: The conjugacy classes of the subgroup T PSL(2, 7)

Clearly the information about t 1 is correct. Let X ∈ C T (t 2) be a generic element thenx y

0 x −1

2 0

0 4

Z =

2 0

0 4

x y

0 x −1

Z ⇒

2x 4 y

0 4x −1

Z =

2x 2 y

0 4x −1

Z,

⇒ y = 0.



Chapter 27 150

Therefore |C T (t 2)| = 3 ⇒ |t T 2 | = 7. As for the order of t 2 we have

t 32 = 2 00 4

2 00 4

2 00 4

Z = 4 00 2

2 00 4

Z = 1 00 1

Z = Z.

Let X ∈ C T (t 3) then we havex y

0 x −1

3 0

0 5

Z =

3 0

0 5

x y

0 x −1

Z ⇒

3x 5 y

0 5x −1

Z =

3x 3 y

0 5x −1

Z,

⇒ y = 0.

Therefore|C T (t 3)

|= 3

⇒ |t T 3

|= 7. As for the order of t 3 we have

t 33 =

3 0

0 5

3 0

0 5

3 0

0 5

Z =

2 0

0 4

3 0

0 5

Z =

6 0

0 6

Z = Z.

Let X ∈ C T (t 4) then we havex y

0 x −1

1 1

0 1

Z =

1 1

0 1

x y

0 x −1

Z ⇒

x x + y

0 x −1

Z =

x x −1 + y

0 x −1

Z,

⇒ x = x −1.

So x = 1, (or 6 but this case is equivalent modulo the centre), which means |C T (t 4)| =

7 and |t T 4 | = 3. As for the order of t 4 we have

t 74 =

1 1

0 1

7Z =

1 7 × 1

0 1

Z =

1 0

0 1

Z = Z.

Let X ∈ C T (t 5) then we have

x y

0 x −11 6

0 1Z = 1 6

0 1x y

0 x −1Z ⇒ x 6x + y

0 x −1Z = x 6x −1 + y

0 x −1Z,

⇒ x = x −1.

Again this gives us |C T (t 5)| = 7 and |t T 5 | = 7. As for the order of t 5 we have

t 75 =

1 6

0 1

7Z =

1 7 × 6

0 1

Z =

1 0

0 1

Z = Z.



Chapter 27 151

Therefore the information in the table is correct. All that is left is to make sure that

t 3 ∈ t T 2 and t 5 ∈ t T 4 . Let X ∈ T thenx y

0 x −1

2 0

0 4

Z =

3 0

0 5

x y

0 x −1

Z ⇒

2x 4 y

0 4x −1

Z =

3x 3 y

0 5x −1

Z.

No such element exists in T therefore these elements are not conjugate. Similarly we

havex y

0 x −1

1 1

0 1

Z =

1 6

0 1

x y

0 x −1

Z ⇒

x x + y

0 x −1

Z =

x 6x −1 + y

0 x −1

Z,

⇒x = 6x −1.

However no such element exists in T therefore these elements are not conjugate so

the table of conjugacy classes is correct.

Note that |T | = 21 = 3 × 7 and T is certainly not abelian. Then by Proposition

25.7 we have T ∼= F 7,3. The character table for this group was calculated on page 240.

It’s clear that it’s not necessary to know exactly how these groups are isomorphic to

use the character table as swapping the conjugacy classes just permutes the irreducible

characters.

We can see that the only conjugacy class of PSL(2, 7) that splits upon restrictionto T is g G4 . It’s clear that [G : T ] = 8 so we have 1T ↑ G is as in table 27.6.

g g 1 g 2 g 3 g 4 g 5 g 6

|C G(g )| 168 8 4 3 7 7

(1T ↑ G)(g ) 8 0 0 2 1 1

χ 7 −1 −1 1 0 0

Table 27.6: The values of 1T

↑G and χ = 1T

↑G

−1G

Recall that we always have a copy of the trivial character in the induced trivial char-

acter by Frobenius Reciprocity. Let χ = 1T ↑ G − 1G then taking the inner product

we see

χ, χ =72

168+

1

8+

1

4+

1

3=

49 + 21 + 42 + 56

168=

168

168= 1.

Note we could also have seen this directly from the character table of PSL(2, 7) on





Chapter 27 153

so g 25 ∈ g G5 . Similarly we have

a b c d

1 60 1

Z = 1 50 1

a b c d

⇒ a 6a + b c 6c + d

= a + 5c b + 5d c d

,

⇒ c = 0 and d = 4a.

So again this element exists in G so g 26 ∈ g G6 . Using this information and the formula

on page 198, we see that χS and χA are as in table 27.8.

g g 1 g 2 g 3 g 4 g 5 g 6

g 2 g 1 g 1 g 2 g 4 g 5 g 6

|C G(g )| 168 8 4 3 7 7

χ 7 −1 −1 1 0 0

χS 28 4 0 1 0 0

χA 21 −3 0 0 0 0

Table 27.8: The decomposition of χ2

We inspect which of the characters we already know appear in χS. Now taking

inner products we see

χS, χ =7 × 28

168− 4

8+

1

3=

196 − 84 + 56

168=

168

168= 1,

χS, λ ↑ G =8 × 28

168− 1

3=

224 − 56

168=

168

168= 1,

χS, 1G =28

168+

4

8+

1

3=

28 + 84 + 56

168=

168

168= 1.

Therefore χ = ψ + χ + λ ↑ G + 1G for some character ψ of G. This character ψ has

values as in table 27.9.

g g 1 g 2 g 3 g 4 g 5 g 6

|C G(g )| 168 8 4 3 7 7

ψ 12 4 0 0 −2 −2

Table 27.9: The character ψ





Chapter 27 155

χ

χ(g 1)χ(g 3) = 1 − 7 ± 3 ± 3 = 0.

Recall from the calculation in the chapter that χ(g 5) is complex for some irre-

ducible character χ. We assume this is χ5 and as χ5 is non-real we have χ6 = χ5 is

another irreducible character. Also g 6 = g −15 ⇒ χi (g 6) = χi (g 5) for each i = 5, 6.

This gives us the current character table of G. We now just have determine the value

of η.

Considering the column orthogonality relation of g 5 with g 3 we obtain

0 = 1 + η + η ⇒ 2Re(η) = −1 ⇒ Re(η) = −1

2.

Using the column orthogonality of g 5 with itself we get

7 = 1 + 1 + 1 + 2|η|2 ⇒ 2|η|2 = 4 ⇒ |η|2 = 2.

In other words this tells us that

Re(η)2 + Im(η)2 = 2 ⇒ 1

4+ Im(η)2 = 2 ⇒ Im(η)2 =

7

4⇒ Im(η) = ±

√ 7

2.

Documents

Solutions to Exercises of Representations and Characters of Groups by Gordon James and Martin Liebeck