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

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 “Representations 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].

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. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

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Group Representations . . . . . . . .

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1 1 1 1 2 4 5 5

Exercise 3.1. Assume ρ is a representation of G then because am = 1 we must have In = 1ρ = am ρ = (aρ)m = Am . Now if Am = In then we have (ai )ρ = Ai for all i ∈ Z. Therefore (ai aj )ρ = ai+j ρ = Ai+j = Ai Aj = (ai ρ)(aj ρ).

Exercise 3.2. We can see that # #3 " 1 0 1 0 = I2 , = A3 = I2 B3 = 0 e 2πi 0 e 2πi/3 " #3 " #" #! " # " #" # 0 1 0 1 0 1 0 1 −1 −1 0 1 C3 = = = = I2 . −1 −1 −1 −1 −1 −1 −1 −1 1 0 −1 −1 "

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 6= I2 , hence ρ3 (g) = I2 ⇒ g = 1. Also B 2 6= I2 as e 4πi/3 6= 1 and so ρ2 is faithful. Exercise 3.3. Now we know that every element of G can be written ai bj for some 0 6 i 6 n − 1 and 0 6 j 6 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 bj )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 equivalence relation. Reflexive clearly we have gρ = In−1 (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, #6 " # iπ/3 2iπ e 0 e 0 A6 = = = I2 0 e −iπ/3 0 e −2iπ " #" # 0 1 0 1 B2 = = I2 , 1 0 1 0 #" # " #" iπ/3 0 1 0 1 e 0 B −1 AB = , 1 0 0 e −iπ/3 1 0 " #" # 0 e −iπ/3 0 1 = iπ/3 , e 0 1 0 # " e −iπ/3 0 , = 0 e iπ/3 "

= A−1 " #" # " # √ √ √ 1/2 1/2 −1/2 3/2 3/2 3/2 √ √ √ C2 = = − 3/2 1/2 − 3/2 1/2 − 3/2 −1/2 " #" # " # √ √ √ −1/2 3/2 −1/2 3/2 −1/2 − 3/2 √ √ C4 = = √ − 3/2 −1/2 − 3/2 −1/2 3/2 −1/2 " # " # " # √ √ −1/2 − −1/2 1 0 3/2 3/2 √ C6 = √ = 3/2 −1/2 − 3/2 −1/2 0 1 " #" # " # 1 0 1 0 1 0 D2 = = 0 −1 0 −1 0 1 " #" # " # " # √ √ 1 0 1/2 3/2 1 0 1/2 − 3/2 √ D−1 CD = = √ = C −1 0 −1 − 3/2 1/2 0 −1 3/2 1/2 By Example 1.4 the above calculations are enough to show that ρ1 and ρ4 are representations of D12 . Let ar bs , ap bq ∈ D12 then we have, (recalling that the calculations above show that

Chapter 3

3

the matrices respect the rules of D12 ), (ar bs ap bq )ρ2 = (ar −p bs+q )ρ2

(ar bs ap bq )ρ3 = (ar −p bs+q )ρ,

= A3(r −p) (−B)s+q

= (−A)r −p B s+q ,

= A3r (−1)s A−3p B s (−B)q

= (−A)r (−1)−p A−p B s B q ,

= A3r (−B)s A3p (−B)q

= (−A)r B s (−A)p B q ,

= (ar bs )ρ2 (ap bq )ρ2

= (ar bs )ρ3 (ap bq )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 hold true " # " #" # √ √ e iπ 0 −1/2 3/2 1/2 3/2 3 3 √ √ = −I2 , A = = −I2 C = 0 e −iπ − 3/2 −1/2 − 3/2 1/2 which also gives us C 5 = C 3 C 2 = −C 2 . Now if ai bj ∈ ker(ρ1 ) then we have (ai bj )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 bj ∈ 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 faithful as a2 ρ2 = A6 = I2 . Finally we also have ρ3 is not faithful as a3 ρ3 = (−A)3 = −A3 = I2 . We have that the representations ρ1 and ρ4 are in fact equivalent. This is a little √ 1 3 iπ/3 more enlightening if we write e = 2 + 2 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 1 2 iπ/3 polynomial λ −λ+1, whose solutions are precisely λ1 = e = 2 + 23 i and λ2 = e −iπ/3 = √ 1 3 2 − 2 i . What are the corresponding eigenvectors? "

1/2 √ − 3/2

√

#" #  " √ #" # " # √ √ " # 3/2 v1 v1 − 3/2i 3/2 v1 0 1 3 √ √ = = , + i ⇒ 2 2 1/2 v2 0 v2 − 3/2 − 3/2i v2 " #" # " # i −1 v1 0 ⇒ = . 1 i v2 0

Now the second line is just −i times the first and so we have v2 = iv1 . Similarly for the

Chapter 3

4

second we get v2 = −iv1 . Then let T be the matrix " # " # 1 1 1 i −1 T = ⇒ T −1 = . 2i i 1 −i i Using this we see that T

−1

" 1 i (bρ4 )T = 2i i " 1 i = 2i i " 1 0 = 2i 2i

#" #" # −1 1 0 1 1 , 1 0 −1 −i i # #" 1 1 1 , −1 −i i # 2i , 0

= bρ1 . Similarly we have # #" #" " √ 1/2 1 1 i −1 3/2 1 √ T −1 (aρ4 )T = , 2i i 1 −i i − 3/2 1/2 " # #" √ √ 1 1 1 1/2i + 3/2 −1/2 + 3/2i √ √ = , 2i 1/2i − 3/2 1/2 + 3/2i −i i " # √ 0 1 i+ 3 √ , = 2i 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 )   0 1 0   A = 0 0 1 1 0 0

  0 1 0   B = 1 0 0 . 0 0 1

Chapter 3 We check that  0  3 A = 0 1

these satisfy the relations of D8 , so      1 0 0 1 0 0 1 0 0 0 1 0      0 1 0 0 1 0 0 1 = 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 1     0 1 0 0 1 0 1 0     2 B = 1 0 0 1 0 0 = 0 1 0 0 1 0 0 1 0 0       0 1 0 0 1 0 0 1 0 0 0 1 0       B −1 AB = 1 0 0 0 0 1 1 0 0 = 0 1 0 1 0 0 1 1 0 0 0 0 1 1 0 0 0

5

  1 0 1   0 1 = 0 0 0 0  0  0 , 1   1 0 0   0 0 = 1 0 1 0

 0 0  1 0 , 0 1

 0 1  0 0 = A−1 1 0

By Example 1.4 this means that the map ρ : D8 → GL(3, F ) given by (ai bj )ρ = Ai B j , for all 0 6 i 6 3 and 0 6 j 6 1 is a representation. Is it faithful? Well if ai bj ∈ 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 However this only implies gh = hg if ρ is an isomorphism. A classic is the trivial representation ρ : G → GL(1, F ) given by gρ = I for all (gρ)(hρ) = (hρ)(gρ) for all g, h ∈ G but G isn’t necessarily abelian so we have gh = hg.

(gh)ρ = (hg)ρ. counter-example g ∈ G. Clearly don’t necessarily

Chapter 4. Exercise Exercise Exercise Exercise

4.1 4.2 4.3 4.4

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F G-modules

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6 7 8 8

Exercise 4.1. Let u1 = v1 + v2 + v3 , u2 = v1 − v2 and u3 = v1 − v3 . Then the change of basis matrix from B1 = {v1 , v2 , v3 } to B2 = {u1 , u2 , u3 } is given as follows      v1 1 1 1 u1   1   v2  = 1 −2 1  u2  3 v3 1 1 −2 u3 Now the matrices of S3 = {1, (12), (13), (23), (123), (132)} with respect are given by the standard permutation matrices and so      1 0 0 0 1 0 0 0      [1]B1 = 0 1 0 [(12)]B1 = 1 0 0 [(13)]B1 = 0 1 0 0 1 0 0 1 1 0      0 0 1 0 1 0 0      [(23)]B1 = 0 0 1 [(123)]B1 = 0 0 1 [(132)]B1 = 1 0 1 0 0 0 1 0

to the basis B1

 1  0 , 0  0 1  0 0 . 1 0

So to calculate the matrices in the new basis B2 we just multiply by the change of basis matrix on the right

[1]B2

[(12)]B2

[(13)]B2

  1 1 1 1 1  = 1 −1 0  0 3 1 0 −1 0   0 1 1 1 1  = 1 −1 0  1 3 1 0 −1 0   1 1 1 0 1  = 1 −1 0  0 3 1 0 −1 1

    1 0 0 0 0 1 1 1     1 0 1 −2 1  =  0 1 0  , 0 1 1 1 −2 0 0 1     1 0 0 1 0 1 1 1     0 0 1 −2 1  =  0 −1 0  , 0 1 1 1 −2 0 −1 1     1 0 0 0 1 1 1 1     1 0 1 −2 1  =  0 1 −1  , 0 0 1 1 −2 0 0 −1 6

Chapter 4

[(23)]B2

[(123)]B2

[(132)]B2

7  1 1 = 1 3 1  1 1 = 1 3 1  1 1 = 1 3 1

 1 1 1  −1 0  0 0 −1 0  1 1 0  −1 0  0 0 −1 1  0 1 1  −1 0  1 0 −1 0

 0 0 1  0 1 1 1 0 1  1 0 1  0 1 1 0 0 1  0 1 1  0 0 1 1 0 1

  1 1 1   −2 1  =  0 1 −2 0   1 1 1   −2 1  =  0 1 −2 0   1 1 1   −2 1  =  0 1 −2 0

 0 0  0 1 , 1 0  0 0  −1 1  , −1 0  0 0  0 −1  . 1 −1

What we notice about the matrices in the basis B2 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   v     −v

if gh is odd. if g is even and h is even, 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, v = + −u if g is odd. −v

if g is even, 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 vi g ∈ V for all 1 6 i 6 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. vk (ab) = (vk a)b. (c) We have that 1 = a4 and so we can see that the following hold v1 1 = v1 a4 = −v2 a3 = v1 a2 = −v2 a = v1 , v2 1 = v2 a4 = −v1 a3 = v2 a2 = −v1 a = v2 , v3 1 = v3 a4 = −v4 a3 = −v3 a2 = v4 a = v3 , v4 1 = v4 a4 = v3 a3 = −v4 a2 = −v3 a = v4 . (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 Eij be the standard basis for the vector space of all matrices over F . Then for some aij ∈ F and σ ∈ Sn we have A=

n X i,j=1

aij Eij

B=

n X i,j=1

aij Eσ(i)j .

Chapter 4

9

Pn Now consider the matrix P defined by P = k=1 Eσ(k)k . We can see that this is a permutation matrix, as there is only one non-zero entry in each row and column. Now, this gives us that PA =

n X

Eσ(k)k

k=1

n X

aij Eij =

i,j=1

n X

aij Eσ(k)k Eij =

i,j,k=1

AP =

n X i,j=1

aij Eij

n X k=1

Ekτ (k) =

n X i,j,k=1

δik aij Eσ(k)j =

i,j,k=1

Similarly if τ ∈ Sn such that B 0 = P P 0 = nk=1 Ekτ (k) which gives us 0

n X

Pn

i,j=1

n X

aij Eσ(i)j = B.

i,j=1

aij Eiτ (j) , then we define a permutation matrix

aij Eij Ekτ (k) =

n X i,j,k=1

δjk aij Eiτ (k) =

n X i,j=1

aij Eiτ (j) = B 0 .

Chapter 5. Exercise Exercise Exercise Exercise Exercise

5.1 5.2 5.3 5.4 5.5

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F G-submodules and reducibiity . . . . .

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10 10 10 11 13

Exercise 5.1. It is clear that the properties of Proposition 4.6 hold and so V is an F Gmodule. Now V is of dimension 2 so there is only one possible non-trivial F G submodule of 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 k1 , k2 ∈ F we have k1 (α, β) + k2 (α, β) = (k1 + k2 )(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 corresponding 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 6= {0} or F n implies W T 6= {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 6= e −ikπ/3 unless e ikπ/3 is real, in other words k = 3n for some n ∈ Z. (a) Let H = hbi 6 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 have either (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 " h i −1 wa = α β 0 " h i 0 wb = α β 1

# h i 0 = −α −β = −w ∈ W, −1 # h i 1 = β α ∈ W ⇒ β = kβ, 0

for some k ∈ C. Therefore W = span{(1, 1)} and W = span{(1, −1)} are submodules 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 = h(123), (456), (23)(45)i. (a) We can see that a3 = (123)3 = (132)(123) = 1, b3 = (456)3 = (465)(456) = 1, c 2 = (23)(45)(23)(45) = 1, ab = (123)(456) = (456)(123) = ba, c −1 ac = (23)(45)(123)(23)(45) = (132) = a−1 , c −1 bc = (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 bj c k with 0 6 i, j 6 2 and 0 6 k 6 1. Therefore |G| 6 18. However |ha, bi| = 9 and by Lagrange’s theorem we have 9 | |G| but ha, bi = 6 G and so we must have |G| = 18. (b) Let ω be a primitive cube root of unity, i.e. ω 3 = 1 and ω 6= 0, ω 2 6= 0. Then all the primitive cubed roots of unity are 1, ω and ω 2 . Let ε = ω and η = ω 2 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, "

ω 0 (aρ)3 = 0 ω −1

#3

# " #3 " # 2 6 ω3 0 ω 0 ω 0 = = I2 (bρ)3 = = = I2 , 0 ω −3 0 ω −2 0 ω −6 # #" " 0 1 0 1 = I2 , (cρ)2 = 1 0 1 0 # #" #" " ω 0 0 1 0 1 , (cρ)−1 (aρ)(cρ) = 1 0 0 ω −1 1 0 # #" " 0 ω −1 0 1 , = 1 0 ω 0 # " ω −1 0 , = 0 ω "

= (aρ)−1 , # #" #" " 2 0 1 ω 0 0 1 , (cρ)−1 (bρ)(cρ) = 0 ω −2 1 0 1 0 # #" " 0 ω −2 0 1 , = 1 0 ω2 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 = η 3 = 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 1dimensional, submodule of C2 say U. Consider the cyclic subgroup H = hci 6 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 u1 = (1, 1) or u2 = (1, −1). " # h i ε 0 h i −1 u1 a = 1 1 = ε ε ∈ U ⇒ ε = 1, 0 ε−1 " # h i η 0 h i −1 u1 b = 1 1 = η η ∈ U ⇒ η = 1, 0 η −1 " # h i ε 0 h i −1 u2 a = 1 −1 = ε −ε ∈ U ⇒ ε = 1, 0 ε−1 " # h i η 0 h i −1 u2 b = 1 −1 = η −η ∈ U ⇒ η = 1. 0 η −1 Therefore we have ρ is irreducible as long as (ε, η) 6= (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. Exercise Exercise Exercise Exercise Exercise Exercise

6.1 6.2 6.3 6.4 6.5 6.6

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14 15 15 15 16 16

Exercise 6.1. Let G = D8 = ha, b | a4 = b2 = 1, b−1 ab = a−1 i. (a) We have x = a + 2a2 and y = b + ab − a2 , which gives us xy = (a + 2a2 )(b + ab − a2 ), = −2 − a3 + ab + 3a2 b + 2a3 b, y x = (b + ab − a2 )(a + 2a2 ), = ba + 2ba2 + aba + 2aba2 − a3 − 2a4 , = a3 b + 2a2 b + b + 2a3 b − a3 − 2, = −2 − a3 + b + 2a2 b + 3a3 b, x 2 = (a + 2a2 )(a + 2a2 ), = a2 + 2a3 + 2a3 + 4a4 , = 4 + a2 + 4a3 . (b) Recall that G = {1, a, a2 , a3 , b, ab, a2 b, a3 b}. Then if z = b + a2 b we have za = (b + a2 b)a = ba + a2 ba = a3 b + ab = a(a2 b + b) = az, zb = (b + a2 b)b = b2 + a2 b2 = b2 + ba2 b = b(b + a2 b) = bz, za2 = (za)a = a(za) = a2 z, za3 = aza2 = a2 za = a3 z, zab = azb = abz, za2 b = azab = a2 zb = a2 bz, za3 b = aza2 b = a2 zab = a3 zb = a3 bz. Now as the elements of G form a basis for the group algebra, it is clear that zr = r z 14

Chapter 6

15

for all r ∈ CG. Exercise 6.2. Let G = C2 × C2 = {(x i , y j ) | 0 6 i, j 6 1 and x 2 = y 2 = 1}. Consider the action of C2 × C2 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 representation are going to be  1 0  [(1, 1)]G =  0 0  0 0  [(x, 1)]G =  1 0

0 1 0 0 0 0 0 1

0 0 1 0

 0 0   0

1 0 0 0



 0 1  [(1, y )]G =  0 0  0 0  [(x, y )]G =  0 1

1 0 1   0 0

1 0 0 0

0 0 0 1

 0 0  , 1

0 0 1 0

0 1 0 0

 1 0  . 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. Pn Exercise 6.4. We let G = {g1 , . . . , gn } be a finite group and c = i=1 gi . (a) Let h ∈ G, then h = gj for some 1 6 j 6 n. Recall that given any 1 6 i 6 n there exists a unique 1 6 k, ` 6 n such that gi gj = gk and gj gi = g` . Then we have ch =

n X

! gi

gj =

i=1

hc = gj

n X i=1

! gi

=

n X

gi gj =

n X

i=1

i=1

n X

n X

i=1

gj gi =

= gk = c, = g` = c.

`=1

(b) Therefore we have c 2 = (g1 + · · · + gn )c = c| + ·{z · · + c} = |G|c. n times

Chapter 6

16

(c) So let’s examine what’s happening to the basis G. We have gi ϑ = gi c = c for all 1 6 i 6 n. Therefore the matrix of ϑ with respect to G is  1 ...  .. . . [ϑ]G =  . .

 1 ..  . .

1 ... 1 Exercise 6.5. Recall that as a consequence of the definition of an F G module we have P 0g = (u − u)g = ug − ug = 0 for all g ∈ G. Therefore, let r = ni=1 αi gi ∈ F G then we have ! n n n n X X X X 0r = 0 αi gi = 0(αi gi ) = (0αi )gi = 0gi = 0. i=1

i=1

i=1

i=1

Also, for any v ∈ V we have v 0 = v (1 − 1) = v 1 − v 1 = v − v = 0. Let G = {1, g2 , . . . , gn }, with n > 1, and let V = span{1} ⊆ F G be a 1-dimensional vector subspace of the group algebra, such that 1gi = −1 for all 2 6 i 6 n and 1.1 = 1. Then this is certainly a submodule of F G. However we have 1(1+gi ) = 1.1+1gi = 1−1 = 0 but neither 1 nor 1 + gi are zero. Exercise 6.6. We have to check that W is a sub module. Let w1 = 1 + ω 2 a + ωa2 and w2 = b + ω 2 ab + ωa2 b We have G = {1, a, a2 , b, ab, a2 b} and we can see that w1 a = a + ω 2 a2 + ω = ω(1 + ω 2 a + ωa2 ) w1 a2 = a2 + ω 2 + ωa = ω 2 (1 + ω 2 a + ωa2 ) w1 b = b + ω 2 ab + ωa2 b = 1(b + ω 2 ab + ωa2 b) w1 ab = (ab + ω 2 a2 b + ωb) = ω(b + ω 2 ab + ωa2 b) w1 a2 b = (a2 b + ω 2 b + ωab) = ω 2 (b + ω 2 ab + ωa2 b)

w2 a = ba + ω 2 aba + ωa2 ba, = ω 2 (b + ω 2 ab + ωa2 b), w2 a2 = ba2 + ω 2 aba2 + ωa2 ba2 , = ω(b + ω 2 ab + ωa2 b), w2 b = b2 = ω 2 ab2 + ωa2 b2 , = 1(1 + ω 2 a + ωa2 ), w2 ab = bab + ω 2 abab + ωa2 bab, = ω 2 (1 + ω 2 a + ωa2 ), w2 a2 b = ba2 b + ω 2 aba2 b + ωa2 ba2 b, = ω(1 + ω 2 a + 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} 6= U = span{αw1 + βw2 } be a 1-dimensional vector subspace of W . How does G act on the basis

Chapter 6

17

of U? Well, recalling ω 2 6= ω, (αw1 + βw2 )a = αw1 a + βw2 a = αωw1 + βω 2 w2 ∈ U ⇒ α = 0 or β = 0, (αw1 + βw2 )b = αw1 b + βw2 b = αw2 + βw1 ∈ U ⇒ β = `α, for some ` ∈ C. However if both conditions are satisfied then α = β = 0 and U is the {0} module. Hence W is an irreducible module.

Chapter 7. Exercise Exercise Exercise Exercise Exercise Exercise

7.1 7.2 7.3 7.4 7.5 7.6

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18 18 18 19 19 19

Exercise 7.1. For all u1 , u2 ∈ U, λ ∈ F and g ∈ G we have (u1 + u2 )(ϑφ) = ((u1 + u2 )ϑ)φ = (u1 ϑ + u2 ϑ)φ = (u1 ϑ)φ + (u2 ϑ)φ = u1 (ϑφ) + u2 (ϑφ), (λu1 )(ϑφ) = ((λu1 )ϑ)φ = (λ(u1 ϑ))φ = λ((u1 ϑ)φ) = λ(u1 (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 = span{v1 , . . . , v5 } be the permutation module defined by vi g = vig for all g ∈ h(12345)i 6 S5 . Let σ = (12345) and define a map ϑ : V → F G by vi ϑ = σ i and extend linearly, we claim this is an F G-isomorphism. It is clear to see that (vi σ)ϑ = vi+1 ϑ = σ i+1 = σ i σ = (vi ϑ)σ. As σ generates G we have that ϑ is an F Ghomomorphism. It’s clear that it’s then a bijection. Exercise 7.3. We first show that V0 is a vector subspace of V . We clearly have 0 ∈ V0 as 0g = 0 for all g ∈ G. Now if v , v 0 ∈ V0 and λ1 , λ2 ∈ F then for all g ∈ G we have (λ1 v1 + λ2 v2 )g = (λ1 v1 )g + (λ2 v2 )g = λ1 (v1 g) + λ2 (v2 g) = λ1 v1 + λ2 v2 and so λ1 v1 + λ2 v2 ∈ V0 , hence V0 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 v1 , v2 ∈ V , λ ∈ F and h ∈ H then we have (v1 + v2 )ϑ =

X

(v1 + v2 )g =

g∈G

(λv1 )ϑ =

X

X

v1 g +

g∈G

(λv1 )g =

g∈G

X

X

v2 g = v1 ϑ + v2 ϑ,

g∈G

λ(v1 g) = λ

g∈G

X

v1 g = λ(v1 ϑ),

g∈G

! (v1 h)θ =

X g∈G

v1 gh =

X g∈G

18

v1 g

h = (v1 ϑ)h.

Chapter 7

19

It’s clear that the image is contained in V0 as summing over gh ∈ G is the same as summing over g ∈ G. Let v0 ∈ V0 then we have   X v0 g X v0 v0 ϑ= = = v0 , |G| |G| |G| g∈G g∈G which gives us the map is surjective. Exercise 7.4. Let φ : V → W be the isomorphism between V and W . Now let w0 ∈ W0 then there exists v ∈ V such that w0 = v φ. We have, for all g ∈ G, that w0 g = w0 ⇒ (v φ)g = v φ ⇒ v g = v ⇒ v ∈ V0 because φ is an isomorphism. So W0 ⊆ V0 φ and for v0 ∈ V0 we have, for all g ∈ G, that (v0 φ)g = (v0 g)φ = v0 φ. Therefore V0 φ ⊆ W0 ⇒ W0 = V0 φ, which gives us the restriction map φ|V0 : V0 → W0 is an isomorphism. Exercise 7.5. Let {v1 , v2 , v3 , v4 } be the standard basis for the permutation module V . Then we have that v1 (12) = v2

v2 (12) = v1

v3 (12) = v3

v4 (12) = v4 ,

v1 (34) = v1

v2 (34) = v2

v3 (34) = v4

v4 (34) = v3 ,

v1 (12)(34) = v2

v2 (12)(34) = v1

v3 (12)(34) = v4

v4 (12)(34) = v3 .

These calculations tell us that V0 = span{v1 + v2 , v3 + v4 }. Alternatively, by exercise 3, we P have that (F G)0 = span{ g∈G g}. By exercise 4, we have that if V and F G are isomorphic then V0 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 = hx | x 2 = 1i. (a) We check the properties of an F G-homomorphism. For all (α1+βx), (γ1+δx) ∈ F G, λ ∈ F we have ((α1 + βx) + (γ1 + 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 + (γ1 + δx)ϑ.

(λ(α1 + βx))ϑ = ((λα)1 + (λβ)x)ϑ = (λα − λβ)(1 − x)

((α1 + βx)x)ϑ = (β1 + αx)ϑ, = (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 " # 1 −1 [ϑ]G = . −1 1 What are the eigenvalues of this matrix? There the λ2 ) − 1 = λ(λ − 2) = 0, which are λ = 0 and λ = with respect to λ = 2 gives us # " " h i h i −1 i 1 −1 h = 2 v1 v2 ⇒ v1 v2 v1 v2 −1 1 −1

solutions of the equation (1 − 2. Calculating the eigenvector # h i −1 = 0 0 ⇒ v2 = −v1 . −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 " # " # 1 −1 1 1 1 T = and T −1 = 2 −1 1 1 1 we have T [ϑ]G T −1

" #" #" # 1 −1 1 1 1 −1 1 = , 2 1 1 −1 1 −1 1

Chapter 7

21 # #" " 1 1 1 2 −2 = , 2 0 0 −1 1 # " 2 0 , = 0 0 = [ϑ]B

Then T is our change of basis matrix and we have # # " " h i h i h i 1 −1 h i 1 −1 = 1 1 . = 1 −1 , 0 1 1 0 1 1 1 1

Chapter 8. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

8.1 8.2 8.3 8.4 8.5 8.6 8.7

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22 22 23 23 24 24 25

Exercise 8.1. Let B = {v1 , v2 } then we can see that the following is true " # 0 1 [x]B = . −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 # # " " h i h i h i −λ h i 0 1 1 i = 0 0 , = λi v1 v2 ⇒ v1 v2 v1 v2 −1 −1 − λi −1 −1  −λ v = v , i 1 2 ⇒ v1 + (−1 − λi )v2 = 0, ⇒ (1 + λ + λ2 )v1 = 0, which means v1 can be chosen arbitrarily in each case. Therefore we have a direct sum V = span{v1 − λ1 v2 } ⊕ span{v1 − λ2 v2 }. Exercise 8.2. Now G = {1, x, y , xy } ∼ = C2 × C2 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 + δxy for some α, β, γ, δ ∈ R. We consider the action of the group on this basis element (α1 + βx + γy + δxy )1 = α1 + βx + γy + δxy , 22

Chapter 8

23 (α1 + βx + γy + δxy )x = β1 + αx + δy + γxy , (α1 + βx + γy + δxy )y = γ1 + δx + αy + βxy , (α1 + βx + γy + δxy )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 r1 , r2 and r3 ∈ R such that  α = r1 β = r2 γ = r3 δ      β = r1 α = r2 δ = r3 γ  ⇒ r12 = r22 = r32 = 1.  γ = r1 δ = r2 α = r3 β     δ = r1 γ = r2 β = r3 α 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 = span{1 + x + y + xy } ⊕ span{1 + x − y − xy } ⊕ span{1 − x + y − xy } ⊕ span{1 − 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 = span{v1 , v2 } ⊆ 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 (αv1 + βv2 )ϑ = αv2 . Now this is a CG homomorphism because (α1 v1 + β1 v2 + α2 v1 + β2 v2 )ϑ = α1 v2 + α2 v2 = (α1 v1 + β1 v2 )ϑ + (α2 v1 + β1 )ϑ, (λ(αv1 + βv2 ))ϑ = λ(αv2 ) = λ((αv1 + βv2 )ϑ), ((αv1 + βv2 )g)ϑ = (αv1 + βv2 )ϑ = αv2 = (αv2 )g = (αv1 + βv2 )ϑg. However we can see that ker(ϑ) = im(ϑ) = span{v2 }. Exercise 8.4. Assume ρ : G → GL(2, C) is reducible. Maschke’s theorem we have, for all g ∈ G " # λg 0 gρ = , 0 µg

Then by the matrix form of

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 " # h i h i 1 0 = 1 0 , 1 0 n 1 which means U is a 1-dimensional submodule of C2 . Let’s try and find a submodule U 0 such that C2 = U ⊕ U 0 . Now any vector (v1 , v2 ) ∈ C2 will be linearly independent of (1, 0) if v2 6= 0, so consider U 0 = span{(v1 , v2 )}. This gives us " # h i 1 0 h i = v1 + nv2 v2 . v1 v2 n 1 If this is in U 0 then there exists k ∈ C with v1 = k(v1 + nv2 ) and v2 = kv2 . The second condition gives k = 1 but if this is the case then the first condition gives us nv2 = 0 ⇒ n = 0. Hence there can be no submodule U 0 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 u1 , u2 ∈ U, v ∈ V and λ1 , λ2 ∈ C [u1 , v ] =

X X X (u1 x, v x) = (v x, u1 x) = (v x, u1 x) = [v x, u1 x], x∈G

x∈G

x∈G

X ((λ1 u1 + λ2 u2 )x, v x) = λ1 [u1 , v ] + λ2 [u2 , v ], [λ1 u1 + λ2 u2 , v ] = x∈G

[u, u] =

X x∈G

(ux, ux) > 0 if u 6= 0. | {z } >0

(b) We first show a property of the inner product. For any g ∈ G we have [ug, v g] =

X (uxg, v xg) = x∈G

X

(ux 0 , v x 0 ) = [u, v ].

x 0 =xg∈G

Therefore, for every g ∈ G and v ∈ U ⊥ we have [u, v g] = [ug −1 , v gg −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 = V1 ⊕ · · · ⊕ Vk for some irreducible CG modules. Now CG is faithful as a CG module, therefore we have v x 6= v for some v ∈ Vi . Consider the following set Ki = {x ∈ G | v x = v for all v ∈ Vi }, we claim this is a normal subgroup of G. Clearly 1 ∈ K as v 1 = v for all v ∈ Vi , also for all x, y ∈ Ki we have v (xy ) = (v x)y = v y = v and so xy ∈ Ki . Finally if x ∈ Ki then v x = v ⇒ v = v x −1 and so x −1 ∈ Ki , which gives us K is a subgroup. If x ∈ Ki and g ∈ G then v (g −1 xg) = (v g −1 )xg = v (g −1 g) = v and so Ki is a normal subgroup of G. However we clearly have Ki 6= G and so, as G is simple, we have Ki = {1} which means Vi is a faithful irreducible CG submodule.

Chapter 9. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

9.1 9.2 9.3 9.4 9.5 9.6 9.7

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26 26 27 27 29 29 29

Exercise 9.1. For C2 = hx | x 2 = 1i we have the irreducible representations over C are given by ρi : C2 → GL(1, C) such that xρ1 = (1)

xρ2 = (−1).

Let ω = e 2πi/3 . For C3 = hy | y 3 = 1i, the irreducible representations over C are given by ρi : C3 → GL(1, C) such that y ρ1 = (1)

y ρ2 = (ω)

y ρ3 = (ω 2 ).

For C2 × C2 ∼ = ha, b | a2 = b2 = 1 and ab = bai we have the irreducible representations over C are given by ρi : C2 × C2 → 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 = C4 × C4 ∼ = ha4 = b4 = 1 and ab = bai. (a) Recall that every irreducible representation of C4 = hx | x 4 = 1i over C is given by ρi : C4 → GL(1, C) such that xρ1 = (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 bj and so clearly (ai bj )2 ρ = (a2 )i (b2 )j ρ = (a2 ρ)i (b2 ρ)j = (1)i (1)j = (1).

26

Chapter 9

27

(b) The elements of order 2 in G are 1, a2 , b2 and a2 b2 . If σ existed then a2 σ = b2 σ = (−1) but (a2 b2 )σ = (a2 σ)(b2 σ) = (−1)(−1) = (1). Hence σ cannot exist. Exercise 9.3. Let g1 , . . . , gr 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  i1 ξ1 . . .  .. . . i1 ir (g1 . . . gr )ρ =  . . 0

 0 ..  .

. . . ξrir

is a faithful irreducible representation of G. Yes. Let r = 2 and n1 6= n2 . Then let ξ1 and ξ2 be primitive n1 th and n2 th roots of unity respectively. Then if g1 , g2 generate G = Cn1 × Cn2 we have the map τ : G → GL(1, C) given by (g1i1 g2i2 )τ = (ξ1i1 ξ2i2 ) 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 #2 #4 " −1 0 −7 10 = I2 , = (aρ)4 = 0 −1 −5 7 # #" " −5 6 −5 6 = I2 , (bρ)2 = −4 5 −4 5 " #" #" # −5 6 −7 10 −5 6 (bρ)−1 (aρ)(bρ) = , −4 5 −5 7 −4 5 " #" # 5 −8 −5 6 = , 3 −5 −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 have " #" # " #" # a b −7 10 −7 10 a b = c d −5 7 −5 7 c d " # " # −7a − 5b 10a + 7b −7a + 10c −7b + 10d ⇒ = , −7c − 5d 10c + 7d −5a + 7c −5b + 7d

Chapter 9

28 #" # " #" −5 6 a −5 6 = −4 5 c −4 5 # " " −5a + 6c −5a − 4b 6a + 5b = ⇒ −4a + 5c −5c − 4d 6c + 5d "

a b c d

b d

#

# −5b + 6d . −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. " #4 " #2 5 −6 1 0 (aσ)4 = = = I2 , 4 −5 0 1 # #" " −5 6 −5 6 = I2 , (bρ)2 = −4 5 −4 5 # #" #" " −5 6 5 −6 −5 6 , (bρ)−1 (aρ)(bρ) = −4 5 4 −5 −4 5 " #" # −1 0 −5 6 = , 0 −1 −4 5 " # 5 −6 = , 4 −5 = (aρ)−1 . We comment on how many unique matrices there are. Now a2 σ = I2 and a3 σ = aρ. Also we can see that bρ = −aρ and a−1 ρ = aρ. Therefore we only have one condition on a matrix which commutes with gσ for every g ∈ G. In particular this is " #" # " #" # " # " # a b 5 −6 5 −6 a b 5a + 4b −6a − 5b 5a − 6c 5b − 6d = ⇒ = . c d 4 −5 4 −5 c d 5c + 4d −6c − 5d 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. P Exercise 9.5. Clearly z = g∈G g 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 = ha, b | a3 = b2 = 1 and bab = a−1 i. (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−1 b+ab = (a+a−1 )b.

(b) Consider u1 = 1 + ω 2 a + ωa2 and u2 = b + ω 2 ab + ωa2 b then we have u1 (a + a−1 ) = (a + a−1 ) + ω 2 a(a + a−1 ) + ωa2 (a + a−1 ), = (ω + ω 2 )1 + (1 + ω)a + (1 + ω 2 )a2 , = (ω + ω 2 )u1 , u2 (a + a−1 ) = b(a + a−1 ) + ω 2 ab(a + a−1 ) + ωa2 b(a + a−1 ), = (ω + ω 2 )b + (1 + ω)ab + (1 + ω 2 )a2 b, = (ω + ω 2 )u2 . Therefore, given any w ∈ W we have w (a + a−1 ) = (ω + ω 2 )w . Exercise 9.7. We consider the centres of the following groups and use Proposition 9.16. (a) Consider Cn = hx | x n = 1i. Let ω = e 2πi/n then ω is a primitive nth root of unity. We know that the map ρ : Cn → GL(1, C) given by xρ = (ω) is an irreducible representation of Cn and is clearly faithful as all powers of ω are distinct. (b) Let D8 = ha, b | a4 = b2 = 1 and bab = a3 i = {1, a, a2 , a3 , b, ab, a2 b, a3 b}. Now it’s quick to check that Z(D8 ) = {1, a2 } ∼ = C2 . Consider the representation σ : D8 → GL(2, C) given by " # " # i 0 0 1 aσ = bσ = 0 −i 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(C2 ×D8 ) ∼ = C2 ×C2 , which is not cyclic. Therefore there cannot exist a faithful irreducible C(C2 × D8 ) module by Proposition 9.16.

Chapter 9 (d) Now Z(C3 × D8 ) ∼ = C6 and so is cyclic. Let ξ = e 2πi/3 then = C3 × C2 ∼ representation π : C3 × D8 → GL(2, C) given by # " " # " 0 ξ 0 i 0 (1, b)π = (x, 1)π = (1, a)π = −1 1 0 ξ 0 −i is a faithful irreducible representation of C3 × D8 .

30 we have the

1 0

#

Chapter 10. Exercise Exercise Exercise Exercise Exercise Exercise

10.1 10.2 10.3 10.4 10.5 10.6

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Irreducible modules and the group algebra . . . . . .

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31 31 31 32 33 33

P Exercise 10.1. Let V = span{ g∈G g ∈ G} be the trivial submodule of CG. We now want to see if there exists a 1-dimensional submodule, say U, of CG such that U ∼ = V . Let U = span{u}, then for U to be the trivial submodule we need ug = u for all g ∈ G. In P P 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 = hx | x 4 = 1i. We consider i as a fourth root of unity and define the following elements of CG, v0 = 1 + x + x 2 + x 3 , v1 = 1 + ix − x 2 − ix 3 , v2 = 1 − x + x 2 − x 3 , v3 = 1 − ix − x 2 + ix 3 . Then Vi = span{vi } are irreducible CG modules because v0 x = v0 , v1 x = iv1 , v2 x = −v2 and v3 x = −iv3 . Therefore we have CG = V0 ⊕ V1 ⊕ V2 ⊕ V3 . Exercise 10.3. Consider u1 = 1 + a + a2 + a3 − b − ab − a2 b − a3 b ∈ CG. Then we have the following u1 a = a + a2 + a3 + 1 − a3 b − b − ab − a2 b = u1 , u1 b = b + ab + a2 b + a3 b − 1 − a − a2 − a3 = −u1 . Hence U1 = span{u1 } is the required 1-dimensional CG submodule. Alternatively let u2 = 1 − a + a2 − a3 + b − ab + a2 b − a3 b and u3 = 1 − a + a2 − a3 − b + ab − a2 b + a3 b then

31

Chapter 10

32

we have that u2 a = a − a2 + a3 − 1 + a3 b − a + ab − a2 b = −u2 , u2 b = b − ab + a2 b − a3 b + 1 − a + a2 − a3 = u2 , u3 a = a − a2 + a3 − 1 − a3 b + b − ab + a2 b = −u3 , u3 b = b − ab + a2 b − a3 b − 1 + a − a2 + a3 = −u3 . Therefore U2 = span{u2 } and U3 = span{u3 } are the required 1-dimensional CG modules. Exercise 10.4. Let G = D8 = ha, b | a4 = b2 = 1 and bab = a3 i then define the following elements of CG v0 = 1 + a + a2 + a3

w0 = bv0 = b + ba + ba2 + ba3 ,

v1 = 1 + ia − a2 − ia3

w1 = bv1 = b + iba − ba2 − iba3 ,

v2 = 1 − a + a2 − a3

w2 = bv2 = b − ba + ba2 − ba3 ,

v3 = 1 − ia − a2 + ia3

w3 = bv3 = b − iba − ba2 + iba3 .

Now as in question 2 we have that span{vi } are Chai modules and it’s clear to see that span{wi } are also Chai modules. Now b acts on these elements in the following way v0 b = b + ba3 + ba2 + ba = w0

w0 b = 1 + a3 + a2 + a = v0 ,

v1 b = b + iba3 − ba2 − iba = w3

w1 b = 1 + ia3 − a2 − ia = v3 ,

v2 b = b − ba3 + ba2 − ba = w2

w2 b = 1 − a3 + a2 − a = v2 ,

v3 b = b − iba3 − ba2 + iba = w1

w3 b = 1 − ia3 − a2 + ia = v1 .

Therefore this gives us that span{v0 , w0 }, span{v1 , w3 }, span{v2 , w2 } and span{v3 , w1 } are Chbi modules. Now we have that span{v0 , w0 } is reducible as U0 = span{v0 + w0 }, (the trivial module), and U1 = span{v0 − w0 } are CG submodules. Also, by question 3, we have span{v2 , w2 } is reducible as U2 = span{v2 + w2 } and U3 = span{v2 − w2 } are CG submodules. We quickly check that U4 = span{v1 , w3 } and U5 = span{v3 , w1 } are irreducible. Now a acts on a linear combination of the basis in the following way (αv1 + βw3 )a = αiv1 − βiw3

(γv3 + δw1 )a = −γi v3 + δi w1 .

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 = −`iγ, δ = `iδ. However no such k and ` exist, therefore the modules are irreducible and we have CG = U0 ⊕ U1 ⊕ U2 ⊕ U3 ⊕ U4 ⊕ U5 . | {z } | {z } 1-dim

2-dim

because v0 + w0 , v0 − w0 , v2 + w2 , v2 − w2 , v1 , v3 , w1 , w3 form a basis for CG. Note that non of the 1-dimensional modules are isomorphic but U4 ∼ = U5 by the isomorphism sending v1 7→ w1 and w3 7→ v3 . Now by Theorem 10.5 we have that every 1-dimensional irreducible representation of D8 is equivalent to one of the following ρ0 : a 7→ (1) b 7→ (1)

ρ1 : a 7→ (1) b 7→ (−1)

ρ2 : a 7→ (−1)

ρ3 : a 7→ (−1)

b 7→ (1)

b 7→ (−1)

and any 2-dimensional irreducible representation of D8 is equivalent to # # " " 0 1 i 0 . b→ 7 ρ4 : a 7→ 1 0 0 −i Exercise 10.5. Let ψ : U1 → U2 be the CG isomorphism between these modules. Let 0 6= λ ∈ C then we define a map ϑ : U1 → V such that uϑ = u + λ(uψ). We claim this is a CG homomorphism. We check the following properties for all u1 , u2 ∈ U1 , µ ∈ C and x ∈G (u1 + u2 )ϑ = (u1 + u2 ) + λ((u1 + u2 )ψ) = (u1 + λ(u1 ψ)) + (u2 + λ(u2 ψ)) = u1 ϑ + u2 ϑ, (µu1 )ϑ = µu1 + λ((µu1 )ψ) = µ(u1 + λ(u1 ψ)) = µ(u1 ϑ), (u1 x)ϑ = u1 x + λ(u1 xψ) = u1 x + (λ(u1 ψ))x = u1 ϑx. 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 U1 ∼ = im(ϑ). This argument works equally well for U2 and we’re done. Exercise 10.6. We first show that this is irreducible. Let U = span{αv1 + βv2 } ⊆ V be a

Chapter 10

34

1-dimensional submodule, then we have (αv1 + βv2 )a = iαv1 − iβv2 ∈ U ⇒ α = k1 iα, β = −k1 iβ, (αv1 + βv2 )b = αv2 − βv1 ∈ U ⇒ α = k2 α, β = −k2 β, for some k1 , k2 ∈ 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 u1 = 1 − ia − a2 + ia3 + i b − ab − ia2 b + a3 b and u2 = −i + a + ia2 − a3 + b − iab − a2 b + ia3 b. Then we have u1 a = a − ia2 − a3 + i + ia3 b − b − iab + a2 b = iu1 , u1 b = b − iab − a2 b + ia3 b + ia2 − a3 − i + a = u2 , u2 a = −ia + a2 + ia3 − 1 + a3 b − ib − ab + ia2 b = −iu2 , u2 b = −ib + ab + ia2 b − a3 b + a2 − ia3 − 1 + ia = −u1 . Therefore U is an irreducible CG submodule which is isomorphic to V .

Chapter 11. Exercise Exercise Exercise Exercise Exercise Exercise

11.1 11.2 11.3 11.4 11.5 11.6

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35 35 35 36 36 36

Exercise 11.1. If G is non-abelian then not all CG modules have dimension 1, however we P5 know we always have the trivial module with dimension 1. Therefore we have 1+ i=2 di2 = 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 , 110 2, 14 22 or 13 3. 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 14 22 . Exercise 11.3. We have that G = {g1 , . . . , gn } forms a basis for the group algebra CG. We define a family of CG homomorphisms by r φi = gi r for all r ∈ CG and 1 6 i 6 n. Let φ : CG → CG be any CG-homomorphism then 1φ = λ1 g1 + · · · + λn gn , for some λi ∈ C. Therefore, for any r ∈ CG we have r φ = (1r )φ = (1φ)r, = (λ1 g1 + · · · + λn gn )r, = λ1 g1 r + · · · + λn gn r, = r (λ1 φ1 + · · · + λn φn ) 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(λ1 φ1 + · · · + λn φn ), 35

Chapter 11

36 = λ1 g1 + · · · + λn gn .

Now the gi ’s are linearly independent and so we must have λ1 = · · · = λn = 0 and we’re done. Exercise 11.4. Let v1 , . . . , vn be the natural basis for the permutation module V . Then span{v1 + · · · + vn } 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, U3 ) in roughly the same way as in the proof of Proposition 11.8. We have v1 , w2 form a basis for U3 . Now define two maps φ1 : CG → U3 and φ2 : CG → U3 by r φ1 = v1 r and r φ2 = w2 r . Then consider a CG homomorphism φ : CG → U3 . For some λ1 , λ2 ∈ C we have 1φ = λ1 v1 + λ2 w2 . Hence for any r ∈ CG we have r φ = (1r )φ = (1φ)r = λ1 v1 r + λ2 w2 r = r (λ1 φ1 + λ2 φ2 ). So φ1 , φ2 span HomCG (CG, U3 ) and they’re a basis because if 0 = µ1 φ1 + µ2 φ2 then 0 = µ1 (1φ1 ) + µ2 (1φ2 ) = µ1 v1 + µ2 w2 , which gives us µ1 = µ2 = 0. We wish to now find a basis for the vector space HomCG (U3 , CG). Recall from the proof of Proposition 11.4 that HomCG (U3 , CG) ∼ = HomCG (U3 , U1 ) ⊕ HomCG (U3 , U2 ) ⊕ HomCG (U3 , U3 ) ⊕ HomCG (U3 , U4 ), ∼ = HomCG (U3 , U3 ) ⊕ HomCG (U3 , U4 ), as vector spaces because CG = U1 ⊕ U2 ⊕ U3 ⊕ U4 . Now ϑ1 : U3 → U3 and ϑ2 : U3 → U4 given by uϑ1 = u and uϑ2 = bu for all u ∈ U3 are basis elements for HomCG (U3 , U3 ) and HomCG (U3 , U4 ) respectively. Therefore they form a basis for HomCG (U3 , CG). Lk d Exercise 11.6. This means that we can express V as a direct sum V ∼ = i=1 Vi i , where Lk e Vi di = Vi ⊕ · · · ⊕ Vi for di copies. Also we can express W as a direct sum W ∼ = i=1 Vi i .

Chapter 11

37

Therefore we have dim(HomCG (V, W )) = dim(HomCG (

k M i=1

=

=

k X i=1 k X i=1

Vi d i ,

k M

Vi ei )),

i=1

di ei dim(HomCG (Vi , Vi )), di ei .

Chapter 12. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

12.1 12.2 12.3 12.4 12.5 12.6 12.7

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38 38 38 38 39 39 39

Exercise 12.1. Clearly we have 1 ∈ CG (x) as 1x = x1. Now if g ∈ CG (x) then gx = xg ⇒ xg −1 = g −1 x ⇒ g −1 ∈ CG (x). Finally, if g, h ∈ CG (x) then (gh)x = gxh = x(gh) which gives us gh ∈ CG (x) and so CG (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) ⊆ CG (x). Exercise 12.2. Now g ∈ CG (x) ⇔ gx = xg ⇔ g(xz) = (xz)g ⇔ g ∈ CG (xz). Therefore we have |g G | = |G|/|CG (x)| = |G|/|CG (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 are n2 2-cycles in Sn which gives us |(12)Sn | = n2 . Now if g ∈ CG ((12)) then g −1 (12)g = (12), which gives us (1g2g) = (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 |CG ((12))| = 2 · (n − 2)! which gives us |(12)Sn | = |G|/|CG ((12))| = n!/2!(n − 2)! =
n 2 as required. (b) Choose a 3 element subset from {1, . . . , n}, say {i, j, k}. Then (ijk) and (ikj) are
the only distinct 3-cycles from this set. In otherwords there are 2 n3 3-cylces in
Sn and so |(123)Sn | = 2 n3 . Now choose a 4 element subset from {1, . . . , n}, say {i , j, k, `}. Then (ij)(k`), (ik)(j`) and (i`)(jk) are the only distinct 2-2-cycles you
can make from this set. Hence there are 3 n4 distinct 2-2-cycles in Sn , which gives
us |(12)(34)Sn | = 3 n4 . (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

–
6

(2)

(12)

15

(2,2)

(12)(34)

45

(2,2,2)

(12)(34)(56)

15

(2,3)

(12)(345)

120

(2,4)

(12)(3456)

90

(3)

(123)

40

(3,3)

(123)(456)

40

(4)

(1234)

90

(5)

(12345)

144

/2
4
2 2 /3 × 2

2 62 43
6 2 × 3!
2 63
2 63
6 64
4 × 3 × 2 65

(6)

(123456)

120

5!

6 2


24


6 2

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 6= (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 = ha, b | a4 = 1, b2 = a2 , b−1 ab = a−1 i. Now straight away we have b−1 ab = 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 a3 ba = a2 b and b−1 (ab)b = a3 b. Therefore the conjugacy classes are {1}, {a2 }, {a, a3 }, {ab, a3 b} and {b, a2 b}. Therefore a basis for Z(CQ8 ) is 1, a2 , a + a3 , ab + a3 b, b + a2 b.

P Exercise 12.7. Now we have from the class formula that |G| = |Z(G)| + xi 6∈Z(G) |xiG |. Now p | |G| and p | |xiG | for each xi 6∈ Z(G) so we must have p | |Z(G)| and so Z(G) 6= {1}.

Chapter 12

40

Now assume G does not have a conjugacy class of size p. Then we have p 2 | |xiG | for each xi 6∈ 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. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10

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41 42 42 42 42 43 43 43 44 44

Exercise 13.1. Now we have G = {1, a, a2 , a3 , a4 , a5 , b, ab, a2 b, a3 b, a4 b, a5 b}. We first see that the conjugacy classes of G are {1}

{a3 }

{a, a5 }

{a2 , a4 }

{b, a2 b, a4 b}

{ab, a3 b, a5 b}

by (12.12). Now the matrices of the representations are as follows # # " # " " 2 1 0 ω 0 ω 0 a5 ρ1 a 3 ρ1 = a4 ρ1 = a 2 ρ1 = 0 1 0 ω −1 0 ω −2 " # " # " # 2 0 ω 0 ω 0 1 abρ1 = a2 bρ1 = a3 bρ1 = a4 bρ1 ω −1 0 ω −2 0 1 0 # " # " # " 2 0 ω 1 0 −1 0 a3 ρ2 = a4 ρ2 a 2 ρ2 = a5 bρ1 = 0 1 0 1 ω −2 0 " # " # " # −1 0 −1 0 1 0 a 5 ρ2 = abρ2 = a2 bρ2 = a3 bρ2 0 1 0 −1 0 −1 " # " # 1 0 −1 0 a4 bρ2 = a5 bρ2 = . 0 −1 0 −1

# ω2 0 , = 0 ω −2 " # 0 ω = , ω −1 0 # " 1 0 = , 0 1 " # −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 ω 2 + 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 C4 = hx | x 4 = 1i which are all linear. Consider i as a primitive fourth root of unity then the irreducible representations are xρj = i j−1 for 1 6 j 6 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

x2

x3

χ1 (g)

1

1

1

1

χ2 (g)

1

i

−1

−i

χ3 (g)

1

−1

1

−1

χ4 (g)

1

−i

−1

i

Table 13.2: The irreducible characters of C4 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) 6= 0 because it’s the 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 zρ ∈ Z(GL(n, C)), which means zρ = λIn for some λ ∈ C. We have z m = 1 ⇒ λm In = In ⇒ λm = 1 and so λ is an mth root of unity. Now for all g ∈ G we have (zg)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 zρ is a scalar of λn . If the representation 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 ⇔ (gρ)(hρ) = (hρ)(gρ)

for some λ ∈ C, 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) 6⇒ 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 homomorphism. 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−1 BA) = 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 −1 hg) 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. 1 Finally if x ∈ N then δ(x −1 ) = δ(x) > 0 and so x −1 ∈ N, which means N is a subgroup. Let x ∈ N and g ∈ G then δ(g −1 xg) = 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 matrices associated with some permutation σg ∈ Sn . Now we have det([gρr eg ]G ) = sgn(σg ). Consider, as in question 7, the linear character δ : g 7→ det([gρr eg ]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 on the 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 0 for some s 0 ∈ N and r − s = r + s − 4s 0 ≡ 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 7→ 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. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

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45 45 45 46 46 46 46 46

Exercise 14.1. We have that 32 (−1)2 02 32 (−1)2 hχ, χi = + + + + = 2, 24 4 3 8 4 3 · 3 (−1) · 1 0 · 0 (−1) · 3 (−1) · (−1) 32 32 hχ, ψi = + + + + = − = 0, 24 4 3 8 4 24 24 32 12 0 (−1)2 (−1)2 hψ, ψi = + + + + = 1. 24 4 3 8 4 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 " # " # " # " # −1 0 0 i −1 0 i 0 a2 ρ1 = abρ1 = a2 ρ2 = abρ2 = , 0 −1 i 0 0 −1 0 −i " # " # 1 0 −1 0 2 a ρ3 = abρ3 = , 0 1 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 6= χ2 and χ3 6= χ1 means ρ3 is not equivalent to either χ1 or χ2 . Exercise 14.3. Now we have that tr(gσ) = tr(Tg−1 (gρ)Tg ) = tr(gρ) for all g ∈ G. Therefore 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

0

χ2 (g)

2

0 −2

0

0

χ3 (g)

2

0

−2

0

2

Table 14.1: The characters of ρ1 , ρ2 and ρ3 Exercise 14.4. We consider the inner product of χ with itself, hχ, χi =

1 X 1 X 1 X |G| χ(g)χ(g) = |χ(g)| = χ(g) > = 1. |G| |G| g∈G |G| g∈G |G|

Therefore χ must be reducible as we cannot have hχ, χi = 1. Exercise 14.5. Expanding the inner product of hχreg , χi we see hχreg , χi =

1 X 1 χreg (g)χ(g) = |G|χ(1) = χ(1), |G| g∈G |G|

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)) = hπ, 1Sn i. Alternatively, a more direct approach is to use the Orbit-Stabiliser theorem to show that P g∈G | fix(g)| = |G|. Exercise 14.7. Now we know that hψ, ψi = d12 + · · · + dk2 . Now if hψ, ψi = 1 then we must have that di = 1 for just one i and so ψ is an irreducible character. If hψ, ψi = 2 then we must have two di = 1 and the rest zero, therefore ψ = χi1 + χi2 . If hψ, ψi = 3 then we must have ψ = χi1 + χi2 + χi3 . Finally if hψ, ψi = 4 then either ψ = χi1 + χi2 + χi3 + χi4 or ψ = 2χi1 . 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 6= 2φ.

Chapter 15. Exercise Exercise Exercise Exercise

15.1 15.2 15.3 15.4

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47 47 47 48

Exercise 15.1. We consider the inner product of χ with the irreducibles 1 12 (19 − 3 + 2 · (−2)) = = 2, 6 6 1 18 hχ, χ2 i = (19 + 3 + 2 · (−2)) = = 3, 6 6 42 1 = 7. hχ, χ3 i = (2 · 19 + 0 + 2 · (−2) · (−1)) = 6 6

hχ, χ1 i =

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 1 6 1 hψ1 , χ2 i = 6 2 1 hψ1 , χ3 i = = 6 3 hψ1 , χ1 i =

3 1 = 6 2 −1 · 3 1 hψ2 , χ1 i = =− 6 2 hψ2 , χ1 i =

hψ2 , χ1 i = 0

2 1 = , 6 3 2 1 hψ3 , χ1 i = = , 6 3 −2 1 hψ3 , χ1 i = =− . 6 3 hψ3 , χ1 i =

Therefore we have ψ1 =

1 1 1 χ1 + χ2 + χ 3 6 6 3

ψ2 =

1 1 χ1 − χ2 2 2

ψ3 =

1 1 1 χ1 + χ2 − χ3 . 3 3 3

Exercise 15.3. Examining the inner products we have 6 0 3 3 1+i −1 + i + + − − + = 0, 12 12 6 6 4 4 6 0 3 3 −i + 1 −i − 1 hψ, χ2 i = + − − + + = 0, 12 12 6 6 4 4 6 0 3 3 1+i 1−i hψ, χ3 i = + + − + + = 1, 12 12 6 6 4 4 hψ, χ1 i =

47

Chapter 15

48 6 0 3 3 i − 1 −1 − i + − − + + = −1, 12 12 6 6 4 4 12 0 3 3 0 0 hψ, χ5 i = + − + + + = 1, 12 12 6 6 4 4 12 0 3 3 0 0 hψ, χ6 i = + + + + + = 2. 12 12 6 6 4 4 hψ, χ4 i =

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) t xZ(G) for some x ∈ G. However consider xz ∈ xZ(G) then for g ∈ Z(G) we clearly have g(xz) = (xz)g. Now consider xz 0 ∈ xZ(G) then (xz)(xz 0 ) = (xz 0 )(xz) 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 6 therefore |Z(G)| 6 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 , 110 2, 14 22 , 13 3. Therefore the group has either 12, 6 or 4 conjugacy classes. Now the cyclic group C12 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. Exercise Exercise Exercise Exercise Exercise Exercise

16.1 16.2 16.3 16.4 16.5 16.6

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49 49 50 50 51 52

Exercise 16.1. Recall that C2 × C2 = {(1, 1), (1, y ), (x, 1), (x, y )} such that x 2 = y 2 = 1 then 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 C2 × C2 Exercise 16.2. Using the column orthogonality relations we get the character table of G to be as in table 16.2. g

g1

g2

g3

g4

g5

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 g4 is an element of order 2. This is because g4 ∈ CG (g4 ) and o(g4 ) | |CG (g4 )| ⇒ o(g4 ) = 2. The column relation on g4 says 12 + 02 + |χ3 (g4 )|2 + |χ4 (g4 )|2 = 2, which gives us |χ3 (g4 )|2 + |χ4 (g4 )|2 = 1. We know from Corollary 13.10 that χ4 (g4 ) ≡ 0 (mod 2) so χ4 (g4 ) = 0. Finally the column relations on g1 and g4 give us 1 + χ3 (g4 ) = 0 so the character table so far is as in table 16.3 g

g1

g2

g3

g4

χ1 (g)

1

1

1

1

χ2 (g)

2

α

β

0

χ3 (g)

1

x1

x2

−1

χ4 (g)

2

y1

y2

0

Table 16.3: Partial character table of G, |G| = 10 We now use the column orthogonality relations of g2 , g4 and g3 , g4 to get 1 − x1 = 0 ⇒ x1 = 1

1 − x2 = 0 ⇒ x2 = 1.

Finally we use the column orthogonality of g1 , g2 and g1 , g3 to get the values 1 + 2α + 1 + 2y1 = 0 ⇒ y1 = −1 − α = β, 1 + 2β + 1 + 2y2 = 0 ⇒ y2 = −1 − β = α. Therefore the final character table of G is as in table 16.4 g

g1

g2

g3

g4

χ1 (g)

1

1

1

1

χ2 (g)

2

α

β

0

χ3 (g)

1

1

1

−1

χ4 (g)

2

β

α

0

Table 16.4: The character table of G from exercise 16.3 Exercise 16.4. We have the character table as in the question.

Chapter 16

51

(a) From the column orthogonality of g2 we have 3 + |ζ|2 + |ζ|2 = 7 ⇒ 2|ζ|2 = 4 ⇒ |ζ|2 = 2. Now using the orthogonality of g1 and g2 we have 1 3 + 3ζ + 3ζ = 0 ⇒ ζ + ζ = −1 ⇒ 2 Re(ζ) = −1 ⇒ Re(ζ) = − . 2 Using this and combining it with the first equation we have  2 1 1 − + Im(ζ)2 = 2 ⇒ + Im(ζ)2 = 2, 2 4 ⇒ 1 + 4 Im(ζ)2 = 8, ⇒ 4 Im(ζ)2 = 7, 7 ⇒ Im(ζ)2 = , √4 7 ⇒ Im(ζ) = . 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 that χ4 (g2 ) and χ5 (g2 ) are complex, means that g2−1 6∈ g2G . Now we know χi (g2−1 ) = 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 xg2 = g2 x ⇔ g2−1 x = xg2−1 . Therefore |CG (g2−1 )| = |CG (g2 )| = 7. This leaves us with the third column of the character table, which is given in table 16.5. g

g1

g2

g2−1

χ1 (g)

1

1

1

χ2 (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

Pk Exercise 16.5. Let X = {g ∈ G | i=1 χi (g)χi (g) = |G|}. Recall from the column orthogonality relation that k X χi (z)χi (z) = |CG (z)|. i=1

Then we have x ∈X⇔

k X

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

i=1

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. P We consider the matrix C T C. Now we can see that (C T C)ij = k`=1 χ` (gi )χ` (gj ) = δij |CG (gi )| by the column orthogonality relations. Therefore we have 2

T

| det C| = (det C)(det C) = det(C C) =

k Y

|CG (gi )|.

i=1 √ 1+ 3i 2

√ 1− 3i 2 ,

Finally for G = C3 , recall that ω = and ω 2 = then we have the determinant of the character table to be 1 1 1 1 ω ω 2 = (ω 2 − ω) − (ω − ω 2 ) + (ω 2 − ω), 1 ω 2 ω = 3(ω 2 − ω), √ = −3 3i.

Chapter 17. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8

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53 55 57 58 59 59 62 65

Exercise 17.1. Let G = Q8 = ha, b | a4 = 1, b2 = a2 , b−1 ab = a−1 i. (a) From Exercise 12.6 we have the conjugacy classes are {1}, {a2 }, {a, a3 }, {ab, a3 b} and {b, a2 b}. (b) We can see that a−1 b−1 ab = a−2 = a2 . Therefore we can see that G 0 = ha2 i ∼ = C2 . 2 2 2 Now, let H = ha i then we have G/H = {H, aH, bH, abH} and (aH) = (bH) = 1. Therefore it’s easy to see that G/H ∼ = C2 × C2 . 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

χ ˜1 (g)

1

1

1

1

χ ˜2 (g)

1

−1

1

−1

χ ˜3 (g)

1

1

−1

−1

χ ˜4 (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 a2 H = 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 ∼ 6 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

|CG (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 = b3 = 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 = ha, b | a7 = b3 = 1, b−1 ab = a2 i. (a) It’s clear from the relations that every element can be written in the form bi aj for some 0 6 i 6 6 and 0 6 j 6 2. So we can see that |G| 6 21. However G contains the cyclic subgroups hai and hbi, 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 a2 are conjugate by the relation b−1 ab = a2 . Using this relation we get the conjugacy 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 , ba4 } and {ba3 , ba5 , ba6 }. Again using the relation we get b−1 (b2 a)b = bab = b2 a2 and so we potentially get the conjugacy classes {b2 a, b2 a2 , b2 a4 } and {b2 a3 , b2 a5 , b2 a6 }. What about b and b2 ? Recall from the relations that we have a−1 b = ba−2 and a−1 b2 = b2 a4 . Then a−1 ba = ba5 a = ba6 and a−3 ba3 = ba−6 a3 = ba4 . Also a−1 b2 a = b2 a5 and a−3 b2 a3 = b2 a12 a3 = b2 a. Therefore the conjugacy classes of G are {1}, {a, a2 , a4 }, {a3 , a5 , a6 }, {b, ba, ba2 , ba3 , ba4 , ba5 , ba6 }, {b2 , b2 a, b2 a2 , b2 a3 , b2 a4 , b2 a5 , b2 a6 }. (c) What’s the derived subgroup of G? Well [a, b] = a−1 b−1 ab = a and so G 0 = hai. Therefore G/G 0 = {G 0 , bG 0 , b2 G 0 } ∼ = C3 because (bG 0 )3 = G 0 . Therefore the charac√ ter table of G/G 0 is as in table 17.4. Note that in table 17.4 ω = e 2πi/3 = 1+2 3i is a 3rd root of unity. gG 0

G0

bG 0

b2 G 0

χ ˜1 (g)

1

1

1

χ ˜2 (g)

1

ω

ω2

χ ˜3 (g)

1

ω2

ω

Table 17.4: The character table of G/G 0 from exercise 17.2 We can see that aG 0 = a3 G 0 = G 0 and so we can lift the above character table

Chapter 17

56

to the character table of G. Before this we work out the degrees of the reamining characters. We know there are 2 more irreducible characters of G because there are 5 conjugacy classes in total. Now by the column orthogonality of the identity we have 1 + 1 + 1 + x 2 + y 2 = 21 ⇒ x 2 + y 2 = 18 ⇒ x = y = 3. Therefore we have the character table so far to be as in table 17.5. g

1

a

a3

b

b2

|CG (g)|

21

7

7

3

3

χ1 (g)

1

1

1

1

1

χ2 (g)

1

1

1

ω

ω2

χ3 (g)

1

1

1

ω2

ω

χ4 (g)

3

α

β

γ

γ

χ5 (g)

3

α

β

γ

γ

Table 17.5: The partial character table of G from exercise 17.2 Remember that b−1 = b2 and that χi (b−1 ) = χi (b2 ) = χi (b) for all 1 6 i 6 5. Also, because b is not conjugate b−1 we have χi (b) is not real for some 1 6 i 6 5. Now recall Proposition 13.15 that say if χi is an irreducible character then so is χi . Also recall Corollary 15.6 which says g is conjugate to g −1 if and only if χ(g) = χ(g) for all characters χ. Now a−1 6∈ aG and so we must have χi (a) 6= χi (a), in other words we must have χ5 = χ4 . Using the column orthogonality of b and b2 , and recalling ω = ω 2 , we get 1 + ωω 2 + ω 2 ω + |γ|2 + |γ|2 = 0 ⇒ 1 + ω 2 + ω + 2|γ|2 = 0, ⇒ 2|γ|2 = 0, ⇒ γ = 0. Therefore, using this information, our character table is as in table 17.6. Using the column orthogonality relations we get ) 1 + 1 + 1 + 3α + 3α = 0 1 + 1 + 1 + αα + αα = 7 Therefore Re(α) = − 21 and

1 4

⇒

2 Re(α) = −1, |α|2 = 2.

+ Im(α)2 = 2 ⇒ Im(α)2 = 74 . Therefore α =

√ −1± 7i . 2

Chapter 17

57

g

1

a

a3

b

b2

|CG (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 α β + + = 0 ⇒ 3 + 3α + 2β = 0 ⇒ β = −1 − α. 21 7 7 √

If α = 1+2 7i then β = −1 − table for G is as in table 17.7.

√ −1+ 7i 2

=

√ −1− 7i 2

= α. In other words the character

g

1

a

a3

b

b2

|CG (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

Table 17.7: The character table of G from exercise 17.2 Exercise 17.3. Recall that |G/G 0 | = |G|/|G 0 | counts the number of linear characters. Now |G 0 | | |G| and so |G/G 0 | = 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 0 = G), of order 12. Also clearly there is no simple group of order 12 as G 0 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 then χ6 = χ5 χ4 . 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

g2

g3

g4

g5

g6

χ1 (g)

1

1

1

1

1

1

χ2 (g)

1

−1

−1

1

1

1

χ3 (g)

1

i

−i

1

−1

−1

χ4 (g)

1

−i

i

1

−1

−1

χ5 (g)

2

0

0

−1

−1

2

χ6 (g)

2

0

0

−1

1

−2

Table 17.8: The character table of G from exercise 17.4 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

g2

g3

g4

g5

g6

|CG (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. Therefore the normal subgroups of D8 are ker(χ1 ) = D8

ker(χ2 ) = {1, a, a2 , a3 }

ker(χ4 ) = {1, a2 , ab, a3 b} ker(χ5 ) = {1}

ker(χ3 ) = {1, a2 , b, a2 b}, ker(χ2 ) ∩ ker(χ3 ) = {1, a2 }.

Exercise 17.6. Let G = T4n = ha, b | a2n = 1, an = b2 , b−1 ab = a−1 i. (a) Let ρ : T4n → 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. # #2n " 2n ε 0 ε 0 = I2 , = (aρ)2n = 0 ε−2n 0 ε−1 # #" # " # " " n n 0 1 0 1 ε 0 ε 0 = n = (bρ)2 , = (aρ)n = n −n n 0 ε ε 0 ε 0 0 ε #" # " #" n 0 1 0 ε ε 0 (bρ)−1 (aρ)(bρ) = , 1 0 0 ε−1 εn 0 " #" # 0 εn−1 0 1 = , ε 0 εn 0 " # ε2n−1 0 = , 0 ε "

= (aρ)−1 . Therefore this is indeed a representation of T4n . (b) We start by considering the conjugacy classes for T4n . We claim that, for 1 6 r 6

Chapter 17

60

n − 1, the conjugacy classes are as follows {1}

|CG (1)| = |G| = 4n,

{an }

|CG (an )| = |G| = 4n,

{ar , a−r }

|CG (a)| = |hai| = 2n,

{ba2j | 0 6 j 6 n − 1}

|CG (b)| = |hbi| = 4,

{ba2j+1 | 0 6 j 6 n − 1}

|CG (ab)| = |hbi| = 4.

Note that there are n + 3 conjugacy classes. We now argue why these are indeed the conjugacy classes of T4n . However before doing this we make a passing comment on the multiplication in T4n . We have b−1 ab = a−1 ⇔ ab = ba−1 ⇔ b = a−1 ba−1 ⇔ ba = a−1 b. • {1}. This is clear. • {an }. It’s clear that hai 6 CG (an ) but because an = b2 we also have hbi 6 CG (an ). In other words |CG (an )| = |G| = 4n, which means |(an )G | = [G : CG (an )] = 1. • {ar , a−r }. We have b−1 ar b = a−r from the relations and so {ar , a−r } ⊆ (ar )G . What about the centraliser? Well hai 6 CG (ar ) and so we have 2 6 |(ar )G | = [G : CG (ar )] 6 [G : hai] = 2. • {ba2j | 0 6 j 6 n − 1}. We have a−1 (ba2j )a = (a−1 b)a2j+1 = ba2j+2 = ba2(j+1) from the relations, so certainly {ba2j | 0 6 j 6 n − 1} ⊆ bG . What about the centraliser of b? Well hbi 6 CG (b) which gives us n 6 |bG | = [G : CG (b)] 6 [G : hbi] =

4n = n. 4

• {ba2j+1 | 0 6 j 6 n − 1}. We have a−1 (ba2j+1 )a = (a−1 b)a2j+2 = ba2j+3 = ba2(j+1)+1 from the relations, so certainly {ba2j+1 | 0 6 j 6 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 6= ±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 6= ±1 then the character is as in table 17.10.

Chapter 17

61

g

1

an

ar

b

ba

|CG (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. n−1

(−2)2 X (εr + ε−r )(εr + ε−r ) 22 + + , hχ, χi = 4n 4n 2n r =1 n−1

1 1 X |εr |2 + εr ε−r + ε−r εr + |εr |2 . = + + n n r =1 2n Recall that εr = ε−r then the summand simplifies to n−1

2 X 2 + ε2r + ε−2r = + , n r =1 2n n−1

2 n − 1 X ε2r + , = + n n n r =1 =1+

1 1 − , n n

= 1. Therefore the character is irreducible. However note that this is not true if ε = ±1 Pn−1 because j=1 ε2j = n − 1 in this case. P 2j We briefly explain why n−1 j=1 = ε = −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 6= 1 ⇔ ε 6= ±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 relations we have b−1 (a−1 b)a = b−1 (ba)a = a−2 , so G 0 = ha2 i. This means G/G 0 = {G 0 , aG 0 , bG 0 , baG 0 } and so there are |G/G 0 | = 4 linear characters. If n is even we have G/G 0 ∼ = C2 × C2 because b2 = an ∈ G 0 . If this is the case then the character table for G is as in table 17.11. g

1

an

ar 16r 6n−1

b

ba

|CG (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) 06k6n−1

2

2(−1)k

εkr + ε−kr

0

0

Table 17.11: The character of T4n , when n is even If n is odd then G/G 0 ∼ = C4 because b2 G 0 = an G 0 = aG 0 and b3 G 0 = bb2 G 0 = ban G 0 = baG 0 . Therefore we lift the four linear characters of C4 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 6 k 6 n − 1 when ε is a primitive 2nth root of unity, for example ε = e 2πi/2n . Note that taking n 6 k 6 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 = U6n = {a, b | a2n = b3 = 1, a−1 ba = b−1 }. (a) Let ρ : U6n → GL(2, C) be the map defined in the question. To show this is a representation it is enough to show that the relations hold #2n " #n " # 0 ε ε2 0 ε2n 0 = = = = I2 , ε 0 0 ε2 0 ε2n " #3 " # 3 ω 0 ω 0 (bρ)3 = = = I2 , 0 ω2 0 ω6 "

(aρ)2n

Chapter 17

63

g

1

an

ar 16r 6n−1

b

ba

|CG (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) 16k6n−1

2

0

0

2(−1)

k

kr

ε +ε

−kr

Table 17.12: The character of T4n , when n is odd #" #" # −1 ω 0 0 ε 0 ε , (aρ)−1 (bρ)(aρ) = −1 0 ω2 ε 0 ε 0 " #" # 0 ε−1 ω 2 0 ε = −1 , ε ω 0 ε 0 " # ω2 0 = , 0 ω "

= (bρ)−1 . Therefore ρ certainly is a representation of U6n . (b) We start by considering the conjugacy classes of U6n . We claim that, for 0 6 j 6 n−1, the conjugacy classes are as follows {a2j }

|CG (1)| = |G| = 6n,

{a2j b, a2j b2 }

|CG (b)| = |hb, a2 i| = 3n,

{a2j+1 , a2j+1 b, a2j+1 b2 }

|CG (a)| = |hai| = 2n.

Note that there are n + n + n = 3n conjugacy classes. We now argue why these are indeed the conjugacy classes of U6n . However before doing this we make a passing comment on the multiplication in U6n . We have a−1 ba = b−1 ⇔ a−1 b−1 a = b ⇔ b−1 a = ab and a−1 ba = b−1 ⇔ ba = ab−1 . Therefore baj = aj b if j is even and aj b−1 if j is odd. • {a2j }. We have b−1 (a2j )b = a2j b−1 b = a2j from the relations. This means hbi 6 CG (a2j ) and so CG (a2j ) = G, which gives us |(a2j )G | = [G : CG (aj )] = 1.

Chapter 17

64

• {a2j b, a2j b2 }. We have a−1 (a2j b)a = a2j−1 (ab−1 ) = a2j b2 , so certainly we have {a2j b, a2j b2 } ⊆ (a2j b)G . What about the centraliser? We have hbi 6 CG (a2j b) because hbi 6 CG (aj ) and CG (b). However we also have a−2 (a2j b)a2 = a−1 (a2j b2 )a = a2j−1 (ab−2 ) = a2j b. Therefore ha2 i 6 CG (a2j b), which leaves us with 6n 2 6 |(a2j b)G | = [G : CG (a2j b)] 6 [G : ha2 , bi] = = 2. 3n • {a2j+1 , a2j+1 b, a2j+1 b2 }. We have and

b−1 a2j+1 b = a2j+1 b2

b−2 a2j+1 b2 = a2j+1 b4 = a2j+1 b

from the relations, so certainly we have {a2j+1 , a2j+1 b, a2j+1 b2 } ⊆ (a2j+1 )G . What about the centraliser? Clearly hai 6 CG (a2j+1 ), which gives us 3 6 |(a2j+1 )G | = [G : CG (a2j+1 )] 6 [G : hai] =

6n = 3. 2n

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

a2j

a2j+1

a2j b

|CG (g)|

6n

2n

3n

χ(g)

2ε2j

0

−ε2j

Table 17.13: The character of the representation from Exercise 17.7(a) These values come from the following matrix calculations "

ε2j 0 (a2j ρ) = 0 ε2j

#

" (a2j+1 ρ) =

0 ε2j+1

ε2j+1 0

#

"

ωε2j 0 (a2j bρ) = 0 ω 2 ε2j

and the fact that ω 2 + ω = −1. The inner product of the character is hχ, χi =

n−1 X 2ε2j 2ε2j j=0

6n

+

n−1 X −ε2j −ε2j j=0

3n

,

#

Chapter 17

65

=

n−1 X 2|ε2j |2 j=0

3n

+

n−1 X |ε2j |2 j=0

3n

,

2 1 ·n+ · n, 3n 3n = 1. =

We now consider the derived subgroup of G. We know from the relations that b−1 a−1 ba = b−2 = b, which gives us that G 0 = hbi. Therefore |G/G 0 | = 6n/3 = 2n and so there are 2n linear characters lifted from G 0 . We have that G/G 0 = {G 0 , aG 0 } ∼ = C2n because (aG 0 ) = a2n G 0 = G 0 . It’s easy to see that a2j G 0 = a2j G 0 , a2j+1 G 0 = a2j+1 G 0 and a2j bG 0 = a2j G 0 . Therefore the character table of G is as in table 17.14. g

a2j 06j6n−1

a2j+1 06j6n−1

a2j b 06j6n−1

|CG (g)|

6n

2n

3n

φk (g)

ε2kj

ε2k(j+1)

ε2kj

2ε2`j

0

−ε2`j

06k62n−1

χ` (g) 06`6n−1

Table 17.14: The character table of U6n 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 = V8n = ha, b | a2n = b4 = 1, ba = a−1 b−1 , b−1 a = a−1 bi, for n odd. (a) Let ρ : V8n → GL(2C) be the map defined in the question. To show this is a representation it is enough to show that the relations hold #2n " # 2n ε 0 ε 0 (aρ)2n = = = I2 , 0 −ε−1 0 (−)2n ε−2n #2 " #4 " 0 1 −1 0 = = I2 , (bρ)4 = 0 −1 −1 0 "

Chapter 17

66 #" # 0 0 1 ε (bρ)(aρ) = , −1 0 0 −ε−1 # " 0 −ε−1 , = −ε 0 # #" " 0 −1 ε−1 0 , = 0 −ε 1 0 "

= (aρ)−1 (bρ)−1 , " #" # 0 −1 ε 0 (bρ)−1 (aρ) = , 1 0 0 −ε−1 " # −1 0 ε = , ε 0 " #" # ε−1 0 0 1 = , 0 −ε −1 0 = (aρ)−1 (bρ). Therefore this is indeed a representation of V8n . (b) We start by considering the conjugacy classes for V8n . We claim, for 1 6 r 6 0 6 s 6 n − 1, the conjugacy classes are as follows {1}

|CG (1)| = |G| = 8n,

{b2 }

|CG (b2 )| = |G| = 8n,

{a2r , a−2r }

|CG (a2 )| = |ha, b2 i| = 4n,

{a2r b2 , a−2r b2 }

n−1 2

and

|CG (a2 b2 )| = |ha, b2 i| = 4n,

{a2s+1 , a−2s−1 b2 }

|CG (a)| = |ha, b2 i| = 4n,

{a2` b, a2` b3 | 0 6 ` 6 n − 1}

|CG (b)| = |hbi| = 4,

{a2`+1 b, a2`+1 b3 | 0 6 ` 6 n − 1}

|CG (ab)| = |hbi| = 4.

Note that 2 | n−1 because n is odd. We now argue that these are indeed the conjugacy classes of V8n . Before doing this we note some incredibly useful multiplication formulae for the group ba = a−1 b−1 ba2r = a−2r b

b−1 a = a−1 b b−1 a2r = a−2r b−1

ba−1 = ab−1 ba2s+1 = a−2s−1 b−1

b−1 a−1 = ab, b−1 a2s+1 = a−2s−1 b.

Chapter 17

67

• {1}. This is clear. • {b2 }. We have ab2 a−1 = aa−1 b−2 = b2 , so hai 6 CG (b2 ). Clearly we have hbi 6 CG (b2 ), which gives us CG (b2 ) = G so |(b2 )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 hai 6 CG (a2r ) but we also have b2 (a2r )b−2 = ba−2r bb−2 = a2r bb−1 = a2r so hb2 i 6 CG (a2r ). This gives us 2 6 |(a2r )G | = [G : CG (a2r )] 6 [G : ha, b2 i] =

8n = 2. 4n

• {a2r b2 , a−2r b2 }. We have b(a2r b2 )b−1 = a−2r bb = a−2r b2 , so certainly we have {a2r b2 , a−2r b2 } ⊆ (a2r b2 )G . What about the centraliser? By the previous point we have hb2 i 6 CG (a2r ) so hb2 i 6 CG (a2r b2 ). Now a(a2r b2 )a−1 = a2r +1 b2 a−1 = a2r +1 a−1 b−2 = a2r b2 . Therefore hai 6 CG (a2r b2 ) and so we have 2 6 |(a2r b2 )G | = [G : CG (a2r b2 )] 6 [G : ha, b2 i] = 2. • {a2s+1 , a−2s−1 b2 }. We have ba2s+1 b−1 = a−2s−1 b−1 b−1 = a−2s−1 b2 , so certainly {a2s+1 , a−2s−1 b2 } ⊆ (a2s+1 )G . What about the centraliser? Clearly hai 6 CG (a2s+1 ) but we also have b2 (a2s+1 )b−2 = ba−2s−1 b−3 = a2s+1 b−4 = a2s+1 so hb2 i 6 CG (a2s+1 ). This gives us 2 6 |(a2r )G | = [G : CG (a2r )] 6 [G : ha, b2 i] = 2. • {a2` b, a2` b3 | 0 6 ` 6 n − 1}. We have aba−1 = aab−1 = a2 b3 and ab3 a−1 = ab−1 a−1 = a2 b. Recall that n is odd so when we conjugate by an we get an ba−n = an an b−1 = b3 and an b3 a−n = an b−1 a−n = an an b = b. So certainly we have {a2` b, a2` b3 | 0 6 ` 6 n − 1} ⊆ bG . What about the centraliser? Clearly hbi 6 CG (b) so we have 2n 6 |bG | = [G : CG (b)] 6 [G : hbi] =

8n = 2n. 4

• {a2`+1 b, a2`+1 b3 | 0 6 ` 6 n − 1}. We have a(ab)a−1 = a2 ab−1 = a3 b3 and a(ab3 )a−1 = a2 b−1 a−1 = a3 b. Recall that n is odd so when we conjugate by an we get an (ab)a−n = an+1 ba−n = an+1 an b−1 = ab3 and an (ab3 )a−n = an+1 b−1 a−n = an+1 an b = ab. So certainly we have {a2`+1 b, a2`+1 b3 | 0 6 ` 6 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 Exercise 8.4. 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

b2

a2r

a2r b2

a2s+1

b

ab

|CG (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

Table 17.15: The character of the representation from Exercise 17.8(a) These values come from the following matrix calculations "

#

"

ε2r 0 (a2r )ρ = −2r 0 ε # " 2s+1 ε 0 (a2s+1 )ρ = 0 −ε−2s−1

−1 0 (b2 )ρ = 0 −1

# 2r −ε 0 , (a2r b2 )ρ = 0 −ε−2r # " 0 ε . (ab)ρ = −1 ε 0

#

"

Recall that for an nth root of unity εk = ε−k . Then the inner product of the character with itself is as follows n−1

n−1

2 2 X 2 (−2) (ε2r + ε−2r )(ε2r + ε−2r ) X (ε2r + ε−2r )(ε2r + ε−2r ) hχ, χi = + + + 8n 8n 4n 4n r =1 r =1

2

+

2

n−1 X (ε2s+1 − ε−2s−1 )(ε2s+1 − ε−2s )

4n

s=0

,

n−1 2 X 8 2|ε2r |2 + ε2r ε2r + ε−2r ε−2r = +2 8n 4n r =1

+

n−1 X 2|ε2s+1 |2 − ε2s+1 ε2s+1 − ε−2s−1 ε−2s−1 s=0

4n n−1

,

n−1

2 X 2 2 n−1 1 ε4r + ε−4r X ε4s+2 + ε−4s−2 = + · + ·n+ − , 2n 2n 2 2n 2n 4n r =1 s=0

Chapter 17

69 n−1

n−1 2 2+n−1+n X ε4r + ε−4r X ε4s+2 + ε−4s−2 = + − , 2n 2n 4n r =1 s=0 n−1

n−1

2 ε4r + ε−4r X ε4s+2 + ε−4s−2 2n + 1 X + − . = 2n 2n 4n r =1 s=0

We investigate these sums a little further to make them more enlightening. Now ε−4r = εn−4r = ε2n−4r . We have 1 6 r 6 n−1 2 ⇒ 4 6 4r 6 2n − 2. From this we can see that 2n − (2n − 2) 6 2n − 4r 6 2n − 4 ⇒ 2 6 2n − 4r 6 2n − 4. So for the first sum we have n−1

n−1

n−3

2 X

2 X

2 X

ε4r + ε−4r =

r =1

ε4r +

r =1 4

ε4r +2 ,

r =1

= ε + ε + · · · + ε2n−2 + ε2 + ε6 + · · · + ε2n−4 , =

8

n−1 X

ε2r .

r =1

Considering the second sum we have ε−4s−2 = εn−4s−2 = ε4n−4s−2 = ε4(n−s)−2 . So if 0 6 s 6 n − 1 then 0 6 4s 6 4n − 4 ⇒ 2 6 4s + 2 6 4n − 2. This gives us that 4n − (4n − 2) 6 4n − (4s + 2) 6 4n − 2 ⇒ 2 6 4(n − s) − 2 6 4n − 2. Using this in the second sum we see n−1 X s=0

ε4s+2 + ε−4s−2 =

n−1 X

ε4s+2 + ε4(n−s)−2 ,

s=0 n−1 X

=2

ε4s+2 ,

s=0

= 2(ε2 + ε6 + · · · + ε2n−4 + ε2n + ε2n+4 + · · · + ε4n−2 ), = 2(ε2 + ε6 + · · · + ε2n−4 + 1 + ε4 + · · · + ε2n−2 ), =2

n−1 X

ε2s .

s=0

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 6= −1 because n is odd therefore εn = (−1)n = −1. Assume ε 6= 1 then ε2 is a solution of X n − 1 = 0 but not a solution of X − 1 = 0. P 2s Therefore ε2 is a solution of X n−1 + · · · + X + 1 = 0. In other words n−1 =0 s=0 ε Pn−1 2r and r =1 ε = −1. Using all this information in the calculation of our inner product we have n−1 n−1 2s X 2n + 1 X ε2r ε 2n + 1 − 1 hχ, χi = + −2 = = 1. 2n 2n 4n 2n r =1 s=0 Finally, to show this we had to make the assumption that ε 6= 1. Assume ε = 1 then we have the inner product to be n−1 n−1 2s X 2n + 1 X ε2r ε (2n + 1) + (n − 1) − n 2n hχ, χi = + −2 = = = 1. 2n 2n 4n 2n 2n r =1 s=0

Therefore χ is an irreducible character of V8n for any nth root of unit ε. We now consider the linear characters of G. The derived group is generated by [a, b] = a−1 b−1 ab = a−2 b2 and hence also its inverse b−2 a2 = b2 a2 . Now (b2 a2 )2 = b2 a2 b2 a2 = a4 , so ha4 i 6 G 0 . Recall n = 2v + 1 is odd so a2n = 1 ⇒ a4v +2 = 1 ⇒ a4v = a−2 , which means ha2 i 6 ha4 i 6 G 0 . Using the fact that a−2 ∈ G 0 we have (b2 a2 )a−2 = b2 ∈ G 0 . Therefore G 0 = hb2 , a2 i, which gives us |G/G 0 | = 8n/2n = 4. In other words G/G 0 = {G 0 , aG 0 , bG 0 , abG 0 } with (aG 0 )2 = a2 G 0 = G 0 and (bG 0 )2 = b2 G 0 = G 0 . Clearly G/G 0 ∼ = C2 × C2 so we can lift the character table of C2 × C2 to the whole of G. Checking the cosets of G 0 we can see a2r G 0 = 22r b2 G 0 = G 0 and a2s+1 G 0 = aG 0 . 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 6 k 6 n − 1. g

1

b2

a2r

a2r b2

a2s+1

b

ab

|CG (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 V8n

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 a new character. So, there must be more irreducible representations. Consider the representation σ : G → GL(2, C) given by " # " # ω 0 0 1 aσ = bσ = , 0 ω −1 1 0 for some 2nth root of unity ω. We check that this indeed a representation of V8n by checking the relations. #2n " # 2n ω 0 ω 0 (aσ)2n = = = I2 , 0 ω −1 0 ω −2n " #4 " #2 0 1 1 0 (bσ)4 = = = I2 , 1 0 0 1 # " # " #" # " #" 0 ω −1 ω −1 0 0 1 0 1 ω 0 = = = (aσ)−1 (bσ)−1 , (bσ)(aσ) = −1 ω 0 0 ω 1 0 1 0 0 ω # " # " #" # " #" −1 −1 0 ω ω 0 0 1 0 1 ω 0 = = = (aσ)−1 (bσ). (bσ)−1 (aσ) = ω 0 0 ω 1 0 1 0 0 ω −1 "

This representation is irreducible, as long as ω 6= ±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

b2

a2r

a2r b2

a2s+1

b

ab

|CG (g)|

8n

8n

4n

4n

4n

4

4

φ(g)

2

2

ω 2r + ω −2r

ω 2r + ω −2r

ω 2s+1 + ω −2s−1

0

0

Table 17.17: The character of σ for V8n Assume ω 6= ±1 then we have the inner product to be n−1 2 X 2 2 (ω 2r + ω −2r )(ω 2r + ω −2r ) hφ, φi = + +2 8n 8n 4n r =1

2

2

Chapter 17

72

+

n−1 X (ω 2s+1 + ω −2s−1 )(ω 2s+1 + ω −2s−1 ) s=0 n−1

4n

,

n−1

2 X 2 2 + ω 4r + ω −4r X 2 + ω 4s+2 + ω −4s−2 = + + , 2n r =1 4n 4n s=0

n−1 n−1 X 2 2 n − 1 2n X ω 2r ω 2s = + · + + +2 , 2n 2n 2 4n r =1 2n 4n s=0

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 ω 2 is an nth root of unity because (ω 2 )n = ω 2n = 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 ω 2 6= 1 ⇔ ω 6= ±1 then ω 2 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 hφ, φi =

n−1 n−1 X 2n + 1 X ω 2r (2n + 1) + (n − 1) + n 4n ω 2s + +2 = = =2 2n 2n 4n 2n 2n r =1 s=0

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 6= ±1. Let η be a primitive 2nth root of unity then η 2 is a primitive nth root of unity. Now letting ω = η and ε = η 2 we have the character table of V8n is as in table 17.18.

Chapter 17

73

g

1

b2

16r 6 n−1 2

|CG (g)|

8n

λ1 (g)

a2r

16r 6 n−1 2

a2r b2

a2s+1 06s6n−1

b

ab

8n

4n

4n

4n

4

4

1

1

1

1

1

1

1

λ2 (g)

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) 06k6n−1

2

−η 4kr − η −4kr

η 2k(2s+1) − η −2k(2s+1)

0

0

φ` (g)

2

η 2`r + η −2`r

η `(2s+1) + η −`(2s+1)

0

0

−2 η 4kr + η −4kr 2

η 2`r + η −2`r

16`6 n−1 2

Table 17.18: The character table of V8n

−1 −1

Chapter 18. Exercise Exercise Exercise Exercise Exercise

18.1 18.2 18.3 18.4 18.5

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74 75 75 75 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 = h(1234), (14)(23)i 6 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)

|CG (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. Explicitly 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+2 3i , ε2 = −1+2 3i and ε4 = −ε = −1−2 3i . g

1

a3

a

a2

b

ab

|CG (g)|

12

12

6

6

2

2

χ1 (g)

1

1

1

1

1

1

χ2 (g)

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

−1 −1

Table 18.2: The character table of D12 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(χ4 ) = ha2 , abi,

ker(χ1 ) = G ker(χ2 ) = hai

ker(ψ1 ) = {1},

2

ker(ψ2 ) = ha3 i.

ker(χ3 ) = ha , bi

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 ) = ha2 i. Therefore the seven distinct normal subgroups of D12 are {1}, ha3 i, ha2 i, hai, ha2 , bi, ha2 , abi and G. Exercise 18.3. See Exercise 17.6 Exercise 18.4. See Exercise 17.7 Exercise 18.5. See Exercise 17.8

Chapter 19. Exercise Exercise Exercise Exercise Exercise Exercise

19.1 19.2 19.3 19.4 19.5 19.6

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Tensor products . . . . . .

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76 76 76 77 78 79

Exercise 19.1. Examining the inner products we see 1 X χψ(g)φ(g) |G| g∈G 1 X = χ(g)ψ(g)φ(g) |G| g∈G 1 X χ(g)ψ(g)φ(g) = |G| g∈G X χ(g)ψφ(g) =

1 X χψ(g)φ(g), |G| g∈G 1 X χ(g)ψ(g)φ(g), = |G| g∈G 1 X ψ(g)χ(g)φ(g), = |G| g∈G 1 X = ψ(g)χφ(g), |G| g∈G

hχψ, φi =

hχψ, φi =

g∈G

= hχ, ψφi

= hψ, χφi.

Therefore hχψ, φi = hχ, ψφi = hψ, χφi as required. Exercise 19.2. Let χ and ψ be irreducible characters of G then, using Exercise 1, we have  1 if χ = ψ, hχψ, 1G i = hχ, ψ1G i = hχ, ψi = 0 if χ 6= 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) 6= ψ(1), (by Proposition 15.5). Let n > 0 be an integer and consider the CG module constructed as the n-fold tensor product of U, say W = U ⊗ · · · ⊗ U. Clearly w g = (u1 ⊗ · · · ⊗ un )g = u1 g ⊗ · · · ⊗ un g = u1 ⊗ · · · ⊗ un = w

for all w ∈ W.

Consider any submodule W 0 ⊆ W and let the character afforded by this submodule be φ. Then it’s clear that φ(g) = φ(1) because w 0 g = w 0 for all w 0 ∈ W 0 . So if ψ(g) 6= ψ(1) we must have hχn , ψi = 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 calculate 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)

g2

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 2

√ 1− 5 2

φS (g)

6

0

2

1

1

φA (g)

3

0

−1

√ 1+ 5 2

√ 1− 5 2

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 have √ 2 √ √ √ 2 √ √   1+ 5 1+2 5+5 3+ 5 1− 5 1−2 5+5 3− 5 = = = = 2 4 2 2 4 2 Using this information we can then show the explicit calculations of φS and φA on the five

Chapter 19

78

cycles. φS ((12345)) = φA ((12345)) = φS ((13452)) = φA ((13452)) =

√  1 3+ 5 2 2 √  1 3+ 5 2 2 √  1 3− 5 2 2 √  1 3− 5 2 2

+ − + −

√  1− 5 = 2 √  1− 5 = 2 √  1+ 5 = 2 √  1+ 5 = 2

1 4 · = 1, 2 2 √  √  1 2+2 5 1+ 5 = , 2 2 2 1 4 · = 1, 2 2 √  √  1 2−2 5 1− 5 = . 2 2 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 152 32 225 + 135 360 hχS , χS i = + = = = 6, 60 4 60 60 102 12 (−2)2 100 + 20 + 60 180 hχA , χA i = + + = = = 3. 60 3 4 60 60 Therefore we have d12 + d22 + d32 + d42 + d52 = 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. gi

g1

g2

g3

g4

g5

g6

g7

gi2

g1

g1

g2

g5

g4

g4

g5

|CG (gi )|

24

24

4

6

6

6

6

χ(g)

2

−2

0

−ω 2

−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 32 32 (−1)2 9+9+6 24 + + = = = 1, 24 24 4 24 24 12 12 12 ωω ω 2 ω 2 (−1)2 ωω hχA , χA i = + + + + + + , 24 24 4 6 6 6 6 1+1+6+4+4+4+4 = , 24 24 = , 24 = 1. hχS , χS i =

Recalling that ωω = ω 2 ω 2 = |ω|2 = 1. Therefore the characters are irreducible. However it is not clear that χ itself is an irreducible character. So we check the inner product hχ, χi =

22 (−2)2 ω 2 ω 2 ωω ωω ω 2 ω 2 4+4+4+4+4+4 24 + + + + + = = = 1. 24 24 6 6 6 6 24 24

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 χχ2A is another irreducible character. Therefore the character table for G is as in table 19.3. gi

g1

g2

g3

g4

g5

g6

g7

|CG (gi )|

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

−ω 2

−ω

ω

ω2

χχA (g)

2

−2

0

−1

−1

−ω

1

χχ2A (g)

2

−2

0

−ω

−ω 2

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. gi

(10 , 1)

(10 , a)

(10 , b)

(a0 , 1)

(a0 , a)

(a0 , b)

(b0 , 1)

(b0 , a)

(b0 , 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

0

χ3 × 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. Exercise Exercise Exercise Exercise Exercise

20.1 20.2 20.3 20.4 20.5

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Restriction to a subgroup . . . . .

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81 82 84 85 86

Exercise 20.1. We have G = S4 and H = h(1234), (13)i 6 G. (a) Let ϕ : H → D8 = ha, b | a4 = b2 = 1, bab = a−1 i 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)

|CG (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] = 24 8 = 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

|CG (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

Table 20.2: The character table of D8 h

1

(13)(24)

(1234)

(13)

(14)(23)

|CH (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

χ7 ↓ A5 = χ8 ↓ A5

χ9 ↓ A5 = χ10 ↓ A5 .

χ5 ↓ A5 = χ6 ↓ A5 ,

Chapter 20

83

Class

(1)

(2)

(3)

(2, 2)

(4)

(3, 2)

(5)

(2, 2, 2)

(3, 3)

(4, 2)

(6)

|CG (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

Table 20.4: The character table of S6 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)

|CH (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

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 + 8β1 = 0     1 + 5 − 20 + 9 + 5 + 8α2 + 8β2 = 0     1 + 0 + 0 − 9 + 0 + 8α3 + 8β3 = 0 ⇒ 1 + 0 + 0 − 9 + 0 + 8α4 + 8β4 = 0      1 − 5 + 10 + 0 + 10 + 8α5 + 8β5 = 0     1 − 5 + 0 + 9 − 5 + 8α6 + 8β6 = 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 + β12 = 9 α21 + β12 = 2,     1 + 1 + 4 + 1 + 1 + α22 + β22 = 8 α22 + β22 = 0,     1 + 0 + 0 + 1 + 0 + α23 + β32 = 5 α23 + β32 = 3, ⇒ 2 1 + 0 + 0 + 1 + 0 + α24 + β42 = 5 α4 + β42 = 3,      1 + 1 + 1 + 0 + 4 + α25 + β52 = 9 α25 + β52 = 2,     1 + 1 + 0 + 1 + 1 + α26 + β62 = 4 α26 + β62 = 0. Recall that the inverse of any element in A6 is of the same cycle shape as the original element. Therefore every element in A6 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 + β32 = 3 ⇒ α23 + (1 − α3 )2 = 3, ⇒ 2α23 − 2α3 + 1 = 3, ⇒ α23 − α3 − 1 = 0, √ 1± 5 ⇒ α3 = . 2 √

√

Now choose α3 = 1+2 5 then β3 = 1−2 5 . 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)

|CH (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

2

−1

ψ6 (h)

8

−1

0

−1

0

ψ7 (h)

8

−1

0

−1

0

0

√ 1+ 5 2 √ 1− 5 2

√ 1− 5 2 √ 1+ 5 2

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|. Pm From Proposition 20.5 we have that i=1 e 2 = e 2 m 6 [G : H] = n. However, as e and m are positive integers we clearly have em 6 e 2 m 6 [G : H] = n ⇒ χ(1) 6 n. Exercise 20.4. Let χ be an irreducible character of G. Let ψ1 , . . . , ψm be the irreducible characters of H. Then χ ↓ H = e1 ψ1 + · · · + em ψm . If H is a subgroup of index 3 then by Propotisition 20.5 we have 2 e12 + · · · + em 6 3. Therefore the only possibilities are that one of the ei is 1, two of the ei are 1 or three of the ei are 1. In other words 2 hχ ↓ H, χ ↓ HiH = e12 + · · · + em = 1, 2 or 3.

We want to find examples of each of these cases occurring. Consider Exercise 1 of this chapter. We have D8 ∼ = h(1234), (13)i 6 G is a subgroup of index 3 because [G : |G| 24 h(1234), (13)i] = |D8 | = 8 = 3. Keeping the notation of that exercise we have χ1 ↓ H = ψ1 and χ3 ↓ H = ψ1 + ψ4 which shows the possibilities of hχ ↓ H, χ ↓ Hi = 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)

|CG (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 /V4 where V4 = {1, (12)(34), (13)(24), (14)(23)} is the derived subgroup of A4 . Note that |A4 /V4 | = 3, so V4 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 V4 ∼ = C2 ×C2 . Therefore we know the character table of V4 , which we give in table 20.8 along with the restricted character χ4 ↓ V4 . h

(1)

(123)

(132)

|CV4 (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 ↓ V4

3

−1

−1

−1

Table 20.8: The character table of V4 By inspection we can see that χ4 ↓ V4 = 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 χ0 of S7 such that χ ↓ A7 = χ0 ↓ 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. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

21.1 21.2 21.3 21.4 21.5 21.6 21.7

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88 89 90 91 91 92 92

Exercise 21.1. We have G = D8 = ha, b | a4 = b2 = 1, b−1 ab = a−1 i and H = ha2 , bi 6 G. Note that H ∼ = C4 , which means |H| = 4. (a) Let u = 1 − a2 + b − a2 b then ua2 = (1 − a2 + b − a2 b)a2 = a2 − 1 + ba2 − a2 ba2 = −1 + a2 − b + a2 b = −u ∈ U, ub = (1 − a2 + b − a2 b)b = b − a2 b + b2 − a2 b2 = 1 − a2 + b − a2 b = u ∈ U. Therefore U is a submodule of CH. (b) We have the induced module to be U ↑ G = span{1 − a2 + b − a2 b, a − a3 + ba − a2 ba, b − a2 b + b2 − a2 b2 }, = span{1 − a2 + b − a2 b, a − a3 − ab + a3 b}. 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

|CH (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, a2 b} and {ab, a3 b}. Only {1}, {a2 } and {b, a2 b} lie in H but {b, a2 b} splits in two. Therefore the values 88

Chapter 21

89

of the induced character are |G| χ(1) = 2 |H| |CG (a2 )| χ(a2 ) = −2 (χ ↑ G)(a2 ) = |CH (a2 )|   ψ(b) ψ(a2 b) 1−1 (χ ↑ G)(b) = |CG (b)| + = 4 · =0 |CH (b)| |CH (a2 b)| 4 (χ ↑ G)(1) =

In particular the induced character is as in table 21.2. We know from the character 2 2 table of D8 that χ ↑ G is irreducible but we also have hχ ↑ G, χ ↑ Gi = 28 + (−2) = 1. 8 g

1

a2

a

b

|CG (h)|

4

4

4

4

(χ ↑ G)(g)

2

−2

0

0

ab

0

Table 21.2: The induced character of the CH module U Exercise 21.2. Let G = S4 and H = h(123)i ∼ = C3 . (a) We have the character tables of S4 and C3 = ha | a3 = 1i, (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)

|CG (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 21.3: The character table of S4 The restricted character table of S4 to H ∼ = C3 is then given in table 21.5. Using the character table of C3 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

|CG (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 C3 g

1

(123)

(132)

|CG (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 6 G and let U be the corresponding CH module that affords ψ as a character. Now U ∼ = U1 ⊕ · · · ⊕ Um for some irreducible CH-submodules Ui of CH. Let ψi be the irreducible character of U afforded by the submodule Ui , then ψ = ψ1 + · · · + ψm . In definition 21.9 we define U ↑ G = (U1 ↑ G) ⊕ · · · ⊕ (Um ↑ 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 Uk . Then we have that Uk has a basis {v1 , . . . , v` }, for some v1 , . . . , v` ∈ CH. We define the induced module to be Uk (CG) = span{xg | x ∈ Uk , 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 {vj s | 1 6 j 6 `, s ∈ S} is a basis for the

Chapter 21

91

induced module Uk (CG). In other words (ψk ↑ G)(1) = dim(Uk ↑ G) = dim(Uk (CG)) = |S| dim(Uk ) = |S|ψk (1) =

|G| ψk (1). |H|

Now extending this to the whole of ψ we get (ψ ↑ G)(1) = (ψ1 ↑ G)(1) + · · · + (ψm ↑ G)(1), |G| |G| ψ1 (1) + · · · + ψm (1), |H| |H| |G| (ψ(1) + · · · + ψ(m)), = |H| |G| = ψ(1), |H| =

as required. Exercise 21.4. Let φ be any irreducible character of G then taking the inner product we have h(ψ(χ ↓ H)) ↑ G, φiG = hψ(χ ↓ H), φ ↓ HiH , = hψ, χ ↓ Hφ ↓ HiH , = hψ, χφ ↓ HiH , = hψ ↑ G, χφiG , = h(ψ ↑ G)χ, φiG . 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)

ab2 = (157)(364),

a2 = (1234567)

a2 b = (156)(273)

a2 b2 = (126)(475),

a3 = (1357246)

a3 b = (175)(346)

a3 b2 = (165)(237),

a4 = (1473625)

a4 b = (124)(365)

a4 b2 = (134)(276),

a5 = (1526374)

a5 b = (143)(267)

a5 b2 = (173)(245),

a6 = (1765432)

a6 b = (162)(457)

a6 b2 = (142)(356),

b = (235)(476)

b2 = (253)(467).

a7 = 1

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 |G| ψ(1), |H|   ψ(a) ψ(a3 ) (ψ ↑ S7 )((1234567)) = |CG (a)| + , |CH (a)| |CH (a3 )|   ψ(b2 ) ψ(b) + (ψ ↑ S7 )((123)(456)) = |CG (b)| |CH (b)| |CH (b2 )| (ψ ↑ S7 )(1) =

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) = d1 χ1 (1) + · · · + dk χk (1). Using Frobenius Reciprocity we have di = hψ ↑ G, χi iG = hψ, χi ↓ HiH . Now χi ↓ H = di 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) = di ψ(1) + φ(1) > di ψ(1). Using this we get d12 ψ(1) + · · · + dk2 ψ(1) 6 d1 χ1 (1) + · · · + dk χk (1) ⇒ (d12 + · · · + dk2 )ψ(1) 6 [G : H]ψ(1), ⇒

k X

di2 6 [G : H]

i=1

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 d1 , . . . , dk such Pk that ψ ↑ G = d1 χ1 + · · · + dk χk . By the above theorem we have i=1 di2 6 [G : H] = 2. Now the di are positive integers and so we have only two options. Either ψ ↑ G = χi or ψ ↑ G = χi + χj for some i, j with i 6= 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 = hψ ↑ G, χi iG = hψ, χi ↓ HiH , 1 = hψ ↑ G, χj iG = hψ, χj ↓ HiH . Therefore ψ is a constituent of χi ↓ H and χj ↓ H. This tells us that χi (1) = (χi ↓ H)(1) > ψ(1) and χj (1) = (χj ↓ H)(1) > ψ(1). However 2ψ(1) = χi (1) + χj (1), so we must have χi (1) = χj (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) 6= 0. Therefore we have φ is an irreducible character of H. Using the Frobenius Reciprocity theorem we have 1 = hφ, χ ↓ HiH = hφ ↑ G, χiG . 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 = hψ ↑ G, χ1 iG = hψ, χ1 ↓ HiH

1 = hφ ↑ G, χ1 iG = hφ, χ1 ↓ HiH .

Also ψ(1) = χ1 (1) and φ(1) = χ1 (1), so ψ = χ1 = φ and we’re done.



Chapter 22. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

22.1 22.2 22.3 22.4 22.5 22.6 22.7

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Algebraic integers . . . . . . .

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. 95 . 95 . 96 . 96 . 97 . 97 . 104

Exercise 22.1. Let χi be the irreducible characters of G. Then we know that |G| = |G/G 0 | +

X

χ(1)2

χ(1)>1

where the sum is over irreducible characters of G. We note that |G 0 | | |G|, so our only possibilities are that |G/G 0 | = 1, 3, 5 or 15. Now 52 = 25 so no character can have χ(1) = 5. So, • |G/G 0 | = 6 1 because 1 + 32 = 10 < |G| and 1 + 32 + 32 = 19 > |G|. • |G/G 0 | = 6 3 because 3 + 32 = 12 < |G| and 3 + 32 + 32 = 21 > |G|. • |G/G 0 | = 6 5 because 5 + 32 = 14 < |G| and 5 + 32 + 32 = 23 > |G|. Therefore |G/G 0 | = 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 0 | and the only possibilities for this P 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 0 | 6= pq. Now the only possibilities for |G/G 0 | are 1, p or q. (a) We have χ(1) | |G| for all irreducible characters χ of G, so either χ is linear or χ(1) = p or q. (b) Recall that q < p so pq < p 2 , which means no character can have degree p. Therefore we have X |G| = |G/G 0 | + χ(1) ⇒ |G| = |G/G 0 | + q 2 n. χ(1)>1

Now q | |G| ⇒ q | |G/G 0 | but the only possibilities for |G/G 0 | are 1, p or q so we must have |G/G 0 | = q. In other words pq = |G| = q|G 0 | ⇒ |G 0 | = p. (c) We now know that pq = q + q 2 n ⇒ p = 1 + qn ⇒ n = p−1 q . Now we know n must be an intger, so q | p −1 as required. Finally we can see the group has q linear characters p−1 and p−1 q irreducible characters of order q. In other words G has q + q 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 hφ, 1G i =

1 X 1 X 1 φ(g)1G (g) = φ(g) = (φ(1) + (|G| − 1)φ(h)), |G| g∈G |G| g∈G |G| = φ(h) +

φ(1) − φ(h) , |G|

for some non-identity element h ∈ G. Now assume φ = a1G + bχreg and recall that hχreg , 1G i = 1. Therefore the above inner product calculation has taken into account the multiplicity of 1G inside χreg . So  φ=

φ(1) − φ(h) φ(h) + |G|

 1G + b(χreg − 1G ).

Now evaluating at the identity we get   φ(1) − φ(h) |G| − 1 φ(1) = φ(h) + + b(|G| − 1) ⇒ (φ(1) − φ(h)) = b(|G| − 1), |G| |G| φ(1) − φ(h) ⇒b= . |G|

Chapter 22

97

Therefore φ = a1G + bχreg with a = φ(h) and b = element h ∈ G.

φ(1)−φ(h) |G|

for some non-identity

(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 6= 1G be an irreducible character of G then hφ, χi = ha1G + bχreg , χi = ah1G , χi + bhχreg , χi = bχ(1) because the multiplicity of any irreducible character in χreg is the degree of the character. 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 the identity. (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|−1 2 and define a set Ω = {g1 , . . . , gk } such that 1 6∈ Ω and if gi ∈ Ω then gi−1 6∈ Ω. Then the set Ω−1 = {g1−1 , . . . , gk−1 } is such that Ω ∩ Ω−1 = ∅. Then X 1 X 1 X 1 χ(g)1G (g) = χ(g) = (χ(1) + 2 χ(gi )) |G| g∈G |G| g∈G |G| gi ∈Ω P Now χ(gi ) is an algebraic integer for each gi ∈ Ω and so α = gi ∈Ω χ(gi ) is an algebraic integer. hχ, 1g i =

(c) Assume hχ, 1g i = 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 hχ, 1G i = 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 |CG (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 the column 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 6 χi (g) 6 2. Assume χi (g) = ±2 for some i and χj (g) = 0 for all j 6= 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 any irreducible 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.

gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

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

γ1

γ2

γ3

γ4

γ5

−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

|CG (gi )|

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 column of g4 . We have χ(g4 ) ≡ χ(1) (mod 3) for each irreducible character and using the column orthogonality relations we have α23 + β32 + γ32 + δ32 + 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 , γ3 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 = γ3 = 1 and δ3 = ε3 = −1. So this gives us that the character table is as in table 22.2. Consider the column of g5 . We have the order of g5 is 22 and so χ(g5 ) ≡ χ(1) (mod 2) for all irreducible characters χ. Also using the column orthogonality on g5 we have α24 + β42 + γ42 + δ42 + ε24 + κ24 = 3. It’s clear that all these entries must be 6 1. Therefore α4 , δ4 and ε4 are equal to ±1 and β4 = γ4 = κ4 = 0. Using the column orthogonality relation with g7 we get

Chapter 22

100

gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

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

γ1

γ2

1

γ4

γ5

−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

|CG (gi )|

Table 22.2: The partial character table of G from Exercise 22.6 1 + α4 = 0 ⇒ α4 = −1. Using the column orthogonality relation with g4 we get −δ4 − ε4 = 0 ⇒ ε4 = −δ4 . Therefore we have the character table so far to be as in table 22.3. gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

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

γ1

γ2

1

0

γ5

−1

χ5 (g)

5

δ1

δ2

−1

1

δ5

0

χ6 (g)

5

ε1

ε2

−1

−1

ε5

0

χ7 (g)

6

κ1

κ2

0

0

κ5

1

|CG (gi )|

Table 22.3: The partial character table of G from Exercise 22.6 Consider the column of g3 . We have χ(g3 ) ≡ χ(1) (mod 2) for every irreducible

Chapter 22

101

character χ. Now using the column orthogonality relation on g3 we get α22 + β22 + γ22 + δ22 + ε22 + κ22 = 7. Now every entry in the column must be 6 2. Now three of the entries are ±1, which gives us β22 + γ22 + κ22 = 4. Therefore two of these entries are zero and one is ±2. Using the column orthogonality relations between g3 and g7 we have 1 + α2 − β2 − γ2 + 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 g3 and g5 we have 1 − 1 + δ2 + −ε2 = 0 ⇒ δ2 = ε2 . Using the column orthogonality of g3 with g4 we get 2 + β2 + γ2 − 2δ2 = 0. Using this equation together with the relation from g3 and g7 we can see that 4 − 2δ2 + κ2 = 0 ⇒ κ2 = 2δ2 − 4. Now δ2 = ±1, which gives us κ2 = −2 or −8. Clearly κ2 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 g2 . We have g2 is an element of order 2 and so χ(g2 ) ≡ χ(1) (mod 2) for all irreducible characters χ. Using the column orthogonality relation on g2 we see α21 + β12 + γ12 + δ12 + ε21 + κ21 = 11. Therefore it’s clear that every entry in the column is 6 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 = δ12 = ε21 = 1 means that β12 + γ12 + κ21 = 8. Therefore only one entry is 0 and the other two are ±2.

Chapter 22

102

gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

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

γ1

0

1

0

γ5

−1

χ5 (g)

5

δ1

1

−1

1

δ5

0

χ6 (g)

5

ε1

1

−1 −1

ε5

0

χ7 (g)

6

κ1

−2

κ5

1

|CG (gi )|

0

0

Table 22.4: The partial character table of G from Exercise 22.6 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 + γ1 ) + 5(δ1 + ε1 ) + 6κ1 = 0, 1 + α1 + δ1 + ε1 − 2κ1 = 0, 1 + α1 + β1 + γ1 − δ1 − ε1 = 0, 1 − α1 + δ1 − ε1 = 0, 1 + α1 − β1 − γ1 + 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 + γ1 ) + 5(δ1 + ε1 ) + 6κ1 = 0, 1 + α1 + β1 + γ1 − δ1 − ε1 = 0, 1 − α1 + δ1 − ε1 = 0, 2 + α1 − β1 − γ1 + 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 6= 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 + γ1 ) + 5δ1 + 6κ1 = 0, 1 − β1 − γ1 + δ1 = 0, 1 + δ1 = 0. Clearly δ1 = −1 ⇒ κ1 = 0. Then the remaining relations tell us γ1 = −β1 . Now the characters χ3 and χ4 are currently identical so there is no harm in choosing β1 = 2 and γ1 = −2. This gives us the character table to be as in table 22.5. gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

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

γ5

−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

|CG (gi )|

Table 22.5: The partial character table of G from Exercise 22.6 To get the final column of g6 it is easiest to use the row orthogonality relations with the trivial representation. This gives us  1 − 10 + 15 + 20 − 30 + 20α5 + 24 = 0 20 + 20α5 = 0,     4 + 20 + 20 + 20β5 − 24 = 0 20 + 20β5 = 0,     4 − 20 + 20 + 20γ5 − 24 = 0 −20 + 20γ5 = 0, ⇒ 5 − 10 + 15 − 20 + 30 + 20δ5 = 0 20 + 20δ5 = 0,      5 + 10 + 15 − 20 − 30 + 20ε5 = 0 −20 + 20ε5 = 0,     6 − 30 + 20κ5 + 24 = 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. gi

g1

g2

g3

g4

g5

g6

g7

Order

1

2

2

3

4

6

5

120

12

8

6

4

6

5

χ1 (g)

1

1

1

1

1

1

1

χ2 (g)

1

−1

1

1

χ3 (g)

4

2

0

1

0

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

1

|CG (gi )|

0

1

−1 −1

0

−1 −1

Table 22.6: The partial character table of G from Exercise 22.6 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 + cn−1 X n−1 + · · · + c1 X + c0 , with c0 = (−1)n det(A) and cn−1 = − tr(A). All ci ∈ Z and so λ is a solution of such a polynomial. (⇐) Let λ be the solution of the polynomial X n + an−1 X n−1 + · · · + a1 X + 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−1 X n−1 + · · · + a1 X + 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−1 X n−1 + · · · + a1 X + a0 ). We check that

Chapter 22

105

this is the case for A a 2 × 2 matrix # " −X 1 = −X(−a1 − X) + a0 = X 2 + a1 X + a0 . det −a0 −a1 − X Therefore this is certainly true for n = 2. Now assume that this is true for n = k then the k + 1th case is   −X 1 ... 0  ..  .. ..  .  . 1 .   det  .  . . . .  .  1 −a0 −a1 . . . −ak − X     −X 1 ... 0 0 1 ... 0   ..   .. .. .. ..  .   .  . 1 . −X 1 .  − det  , = (−X) det   ..   ..  .. ... .   .   . 1 1 −a1 −a2 . . . −ak − X −a0 −a2 . . . −ak − X = (−X)(−1)k (X k + ak X k−1 + · · · + a2 X + a1 ) − (−1)k a0 , = (−1)k+1 (X k+1 + ak X k + · · · + a2 X 2 + a1 X + 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. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

23.1 23.2 23.3 23.4 23.5 23.6 23.7 23.8 23.9 23.10

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Real representations . . . . . . . . . .

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106 106 106 107 108 110 111 112 113 114

Exercise 23.1. Let G be a group of odd order. If x ∈ G is real then x = x −1 ⇒ x 2 = 1, so if x 6= 1 this means G contains an element of order 2. However this would imply 2 | |G| but this is not possible as G has odd order, so x = 1. Exercise 23.2. Let G ∼ = Cn1 × · · · × Cnr be a finite abelian group. Let g1 , . . . , gr be generators for the respectively cyclic components of G and λ1 , . . . , λn be respectively n1 , . . . , nr primitive roots of unity. Then every irreducible character of G is linear and equivalent to one of χ(g1i1 , . . . , grir ) = (λi11 . . . λirr ). Now assume χ is a real irreducible character of G. Recall that any linear character of G is also a homomorphism of G. Therefore we have χ(g) = χ(g) ⇔ χ(g) = χ(g −1 ) ⇔ χ(g 2 ) = χ(1) ⇔ χ(g)2 = 1 ⇔ χ(g) = ±1. i

In other words we have λi11 . . . λirr = ±1 but this happens if and only if λjj = ±1, which n /2 happens if and only if 2 | nj . If 2 | nj then the only values which take ±1 are λ0j and λj j . If ` is the number of nj which are even then there are 2` such irreducible characters. Conversely, how many involutions are there? If g 2 = 1 then either we have ij = 0 or 2 | nj and we have ij = nj /2. Therefore there are 2` such elements and we’re done. Exercise 23.3. We have the character tables of D2n , for n odd and even, to be as in tables 23.1 and 23.2. 106

Chapter 23

107 ar

b

2n

n

2

χ1 (g)

1

1

1

χ2 (g)

1

1

−1

ψj (g)

2

εjr + ε−jr

0

gi

1

|CG (gi )|

16r 6 n−1 2

16j6 n−1 2

Table 23.1: The character table of D2n , for n odd gi

1

am

ar 16r 6m−1

b

ab

|CG (gi )|

2n

2n

n

4

4

χ1 (g)

1

1

1

1

1

χ2 (g)

1

1

χ3 (g)

1

χ4 (g) ψj (g)

1

(−1)

m

1

(−1)

m

2

2(−1)j

−1

−1

(−1)

r

1

−1

(−1)

r

−1

1

0

0

εjr + ε−jr

16j6m−1

Table 23.2: The character table of D2n , for n even Clearly the χi are real characters in both tables but what about the ψj ? 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 ψj 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 representation ρ. Let B = {v1 , v2 } be the basis of V then we have B 0 = {v1 ⊗ v2 − v2 ⊗ v1 } is a basis of A(V ⊗ V ). Let the matrix of g with respect to B be " # a b gρB = . c d Therefore we have v1 g = av1 + bv2 and v2 g = cv1 + dv2 . We now consider the action of g on the antisymmetric tensor product module A(V ⊗V ).

Chapter 23

108

We have (v1 ⊗ v2 − v2 ⊗ v1 )g = v1 g ⊗ v2 g − v2 g ⊗ v1 g, = (av1 + bv2 ) ⊗ (cv1 + dv2 ) − (cv1 + dv2 ) ⊗ (av1 + bv2 ), = ad(v1 ⊗ v2 ) + bc(v2 ⊗ v1 ) − cb(v1 ⊗ v2 ) − ad(v2 ⊗ v1 ), = (ad − bc)(v1 ⊗ v2 − v2 ⊗ v1 ). 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 = T4n = ha, b | a2n = 1, an = b2 , b−1 ab = a−1 i and ε 6= ±1 a 2nth 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 # " # " 0 1 ε 0 aρ = bρ = n ε 0 0 ε−1 Now clearly det(aρ) = εε−1 = 1 and det(bρ) = −εn . Recall the determinant is multiplicative and every element in T4n can be written in the form ai bj with 1 6 i 6 2n and 0 6 j 6 1. Now ιχ = −1 if and only if det(xρ) = 1 for all x ∈ T4n . 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(v1 a, v1 a) = β(εv1 , εv1 ) = ε2 β(v1 , v1 ) = 0 = β(v1 , v1 ), β(v2 a, v2 a) = β(ε−1 v2 , ε−1 v2 ) = ε−2 β(v2 , v2 ) = 0 = β(v2 , v2 ), β(v1 a, v2 a) = β(εv1 , ε−1 v2 ) = εε−1 β(v1 , v2 ) = β(v1 , v2 ), β(v2 a, v1 a) = β(ε−1 v2 , εv1 ) = ε−1 εβ(v2 , v1 ) = β(v2 , v1 ). Therefore β is invariant under the action of a. Now considering the action of b on the bilinear form β(v1 b, v1 b) = β(v2 , v2 ) = 0 = β(v1 , v1 ), β(v2 b, v2 b) = β(εn v1 , εn v1 ) = ε2n β(v1 , v1 ) = 0 = β(v2 , v2 ), β(v1 b, v2 b) = β(v2 , εn v1 ) = εn β(v2 , v1 ) = ε2n = 1 = β(v1 , v2 ),

Chapter 23

109 β(v2 b, v1 b) = β(εn v1 , v2 ) = εn β(v1 , v2 ) = εn = β(v2 , v1 ).

Therefore β is invariant under the action of b, hence invariant under the action of all elements x ∈ T4n . Using Theorem 23.16 we have ιχ = 1 if there exists a symmetric bilinear form on V , which happens if β(v2 , v1 ) = β(v1 , v2 ) = 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(v2 , v1 ) = −β(v1 , v2 ), in other words εn = −1. (c) Every element in T4n has the form ai bj for some 1 6 i 6 2n and 0 6 j 6 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 b2 = an 6= 1. Therefore the only element of order 2 in T4n is an . (d) Now the character table for T4n with n even is as in table 23.3. g

1

an

b

ba

1

ar 16r 6n−1 2r

g2

1

a

an

an

|CG (g)|

4n

4n

2n

4

4

1G

1

1

1

1

1

φ1

1

1

1

−1

−1

φ2

1

1

(−1)

1

−1

φ3

1

1

(−1)r

−1

1

χk (g)

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

r

06k6n−1

Table 23.3: The character table of T4n , with n even It’s obvious that ι1G = ιφ1 = ιφ2 = ιφ3 = 1 because 12G = φ21 = φ22 = φ23 = 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 = 2 Re(εkr ). Now in the table above we have considered χ2k and decomposed it into its symmetric 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 X

(ιχ)χ(1) = 1 + 1 + 1 + 1 +

χ

n−1 X

2(−1)k = 4 − 2 = 2,

k=1

because n is even so the sum equals −2. If n is odd then we have the character table of T4n to be as in table 23.4. g

1

an

b

ba

1

ar 16r 6n−1 2r

g2

1

a

an

an

|CG (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) 16k6n−1

2

2(−1)

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

k

kr

ε +ε

−kr

Table 23.4: The character table of T4n , 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 X χ

(ιχ)χ(1) = 1 + 0 + 1 + 0 +

n−1 X

2(−1)k = 2 + 0 = 2,

k=1

because n is odd so the sum equals 0. This confirms that there is only one involution in T4n . Exercise 23.6. Let χ be an irreducible character of G such that ιχ = −1. Let V be the irreducible CG 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 v1 , . . . , vm be a basis for the CG module V . We consider the vector subspace U = {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, v g) = 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(vi , vj ) = xij for all 1 6 i 6 j 6 m then we can construct a matrix X with 

0

x1,2 .. .

...

x1,m−1

x1,m



   −x1,2 x2,m     .. ..  . . X= . . . .    ..   . xm−1,m  −x1,m−1 −x1,m −x2,m . . . −xm−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 {v1 , . . . , vn } 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 {e1 , . . . , en } of V which is orthonormal with respect to β1 . We start by letting f10 = v1 . Consider the vector subspace U1 = span{v1 } and the standard projection map π1 : V → U1 . Define f20 = v2 − π1 (v2 ) ∈ U1⊥ . Note that f20 6= 0 because if f20 = 0 then v2 = π1 (v2 ) ⇒ v2 ∈ span{v1 }. However this cannot happen because {v1 , . . . , vn } is a basis for V . In a similar fashion we consider the vector subspace U2 = span{v1 , v2 } and the standard

Chapter 23

112

projection map π2 : V → U2 . Define f30 = v3 − π2 (v3 ) ∈ U2⊥ . Again we have f30 6= 0 because if f30 = 0 then v3 = π2 (v3 ) ⇒ v3 ∈ span{v1 , v2 }. However this cannot happen because {v1 , . . . , vn } is a basis for V . Carrying on with this process gives us a set {f10 , . . . , fn0 } such that β(fi 0 , fj 0 ) = 0 for all p i 6= j. Now we define fi = fi 0 / β1 (fi 0 , fi 0 ), then {e1 , . . . , en } is an orthonormal basis of V p with respect to β1 . Note that β(fi 0 , fi 0 ) ∈ 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 {f1 , . . . , fn } can be transformed into a basis which is also orthogonal on β. Let X be an n × n matrix such that Xij = β(fi , fj ) 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 = (qij ) then we define a new basis n X qij fj . ei = j=1

Now because QXQ−1 is diagonal we have β(ei , ej ) = 0 for any i 6= j. Also the matrix representing β1 (fi , fj ) is the identity so conjugating by Q does nothing. This means β1 (ei , ej ) = δij 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 = C3 = hx | x 3 = 1i. We have a 2-dimensional irreducible RG submodule V = span{1 − a, 1 − a2 }. Let v1 = 1 − a and v2 = 1 − a2 then we have the action of a on these basis elements to be v1 a = (1 − a)a = a − a2 = v2 − v1

v2 a = (1 − a2 )a = a − 1 = −v1 .

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 = {v1 , v2 } we have # " −1 −1 . [a]B = 1 0 Now considering the eigenvalues of this matrix we have the characteristic polynomial to be −1 − x −1 det([a]B − xI2 ) = = −x(−1 − x) + 1 = x 2 + x + 1. 1 −x This polynomial has eigenvalues −1±3i 6∈ R. Therefore the map ϑ cannot be a real 2 scalar multiple of the identity map. Exercise 23.9. Let Ω = {Hx1 , . . . , Hxn }. The map ρg : Ω → Ω is a permutation of Ω if it is a bijection. It’s clearly 1-1 because (Hxi )ρg = (Hxj )ρg ⇒ Hxi g = Hxj g ⇒ Hxi = Hxj ⇒ xi = xj because the right cosets of H partition the group G. It is also surjective as given Hxi ∈ Ω we have Hxi g −1 ∈ Ω and (Hxi g −1 )ρg = Hxi g −1 g = Hxi . 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 Hxi ∈ Ω we have (Hxi )ρgh = Hxi gh = (Hxi g)ρh = (Hxi )ρg ρh . Now our choice of Hxi 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 Hxi ∈ Ω we have (Hxi )ρg = (Hxi ) idΩ ⇔ Hxi g = Hxi , ⇔ H = Hxi gxi−1 , ⇔ xi gxi−1 ∈ H,

Chapter 23

114 ⇔ g ∈ xi−1 Hxi .

Therefore ker(ρ) = ∩x∈G x −1 Hx as required. We have a bijection ϕ : Ω → {1, . . . , n} given by (Hxi )ϕ = i . We have a map from ψ : Sym(Ω) → Sn given by σ 7→ ϕ−1 σϕ. 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−1 σϕ)(ϕ−1 τ ϕ) = (σψ)(τ ψ). 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 homomorphism ρψ : G → Sn . It’s clear that ker(ρψ) = ker(ρ) = ∩x∈G x −1 Hx ⊆ H. Exercise 23.10. Let G be a finite group with an involution t ∈ G such that CG (t) ∼ = C2 . Now let χ1 , . . . , χk be the irreducible characters of G and ψ1 , ψ2 be the irreducible characters of C2 . Recall that all the irreducible characters of C2 are linear. As t 2 = 1 we have t −1 ∈ t G , which means χi (t) is an integer for each 1 6 i 6 k. Furthermore we know that χi (t) ≡ χi (1) (mod 2) for each 1 6 i 6 k. From the column orthogonality relations we have k X i=1

2

χi (1) = |G|

k X

χi (t)2 = 2.

i=1

As the χi (t) are integers we must have χi (t) = ±1 for two i, j ∈ {1, . . . , k} with i 6= 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 CG (t). We know that χi ↓ CG (t) = d1 ψ1 + d2 ψ2 for each 1 6 i 6 k and that d12 + d22 6 [G : CG (t)] = 2. Therefore the only possibilities are that d1 , d2 are 0 or 1. We have that χi (1) = (χi ↓ CG (t))(1) = d1 ψ1 (1) + d2 ψ2 (1) = d1 + d2 . 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 0 ] = 2.

Chapter 23

115

Assume G is a simple group. We have G 0 C G is a normal subgroup of G and clearly G 0 6= G so G 0 = {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 CG (t) ∼ = C2 . = C2 which means G ∼

Chapter 25. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

25.1 25.2 25.3 25.4 25.5 25.6 25.7 25.8

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116 116 117 118 120 122 122 123

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 y0 + yx 1 y 1 y0 . = 0 xx 0 0 x 0 x0 Now y 0 + y x ∈ Zp because x ∈ Z∗p ⇒ y x ∈ Zp and Zp is a group under addition. Clearly x, x 0 ∈ Z∗p ⇒ xx 0 ∈ 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 that # " # # " " # " #" 1 0 1 −y x −1 + y x −1 1 −y x 1 y 1 −y x −1 . = = ∈ G and 0 1 0 x 0 x −1 0 xx −1 0 x −1 Exercise 25.2. We have F11,5 = ha, b | a11 = b5 = 1 and b−1 ab = a3 i, 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 0 ∼ = C5 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 C5 and |Z∗p /S| = 2 irreducible 116

Chapter 25

117

characters of degree 5. The character table of F11,5 is therefore as in table 25.1, where α and β are as follows. g

1

a

a2

b

b2

b3

b4

|CG (g)|

55

11

11

5

5

5

5

χ0

1

1

1

1

1

1

1

χ1

1

1

1

ε

ε2

ε3

ε4

χ2

1

1

1

ε2

ε4

ε

ε3

χ3

1

1

1

ε3

ε

ε4

ε2

χ4

1

1

1

ε4

ε3

ε2

ε

φ1

5

α

β

0

0

0

0

φ2

5

β

α

0

0

0

0

Table 25.1: The character table of F11,5 Let η = e 2πi/11 , a primitive eleventh root of unity. We obtained the values of φ1 and φ2 using the information in Theorem 25.10, so in detail we have φ1 (ax ) =

X

x

X

η sx = η x + η 3x + η 4x + η 5x + η 9x ,

s∈S

φ2 (a ) =

η 2sx = η 2x + η 6x + η 8x + η 10x + η 7x .

s∈S

Therefore we define α := η + η 3 + η 4 + η 5 + η 9 and β := η 2 + η 6 + η 8 + η 10 + η 7 , which gives us the correct character table. Exercise 25.3. What we want to show is that we can transform G1 into G2 . Now u ∈ Z∗p and by Theorem 25.1 we know that Z∗p is cyclic. Now any non-identity element of a cyclic group is a generator of the group. Clearly u is non-identity as u r 6≡ 1 (mod p) for all 0 < r < q, so u is a generator for Z∗p . So, as v ∈ Z∗p , this means there is an integer k such that u k ≡ v (mod p). Now i k b−i abi = au as in the proof of Proposition 25.9. Therefore b−k abk = au = av , so we want to replace b by bk . Clearly bk has order q as (bk )q = (bq )k = 1k = 1 but what about the distinctness of the powers of bk ? The values k and q are coprime because both u and v

Chapter 25

118

are such that u q ≡ v q ≡ 1 (mod p). Therefore setting b0 = bk we get G1 = ha, b0 | ap = bq = 1 and b0−1 ab0 = av i = G2 . Hence G1 ∼ = G2 . Exercise 25.4. Let p 6= 2, q = (p − 1)/2 and G = Fp,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 ⇔ 2 | q, p−1 ⇔2| , 2 ⇔p−1≡0 ⇔p≡1

(mod p),

(mod 4),

(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 6 i 6 2 we have 0 = hφi , 1G i =

X φi (gj )1G (gj ) j

|CG (gj )|

=

1 + φi (a) + φi (av ) ⇒ φi (a) + φi (av ) = −1. p

Using the inner product of φi with itself we have 1 = hφi , φi i =

X φi (gj )φi (gj ) j

|CG (gj )|

⇒ p − q = φi (a)φi (a) + φi (av )φi (av ), p−1 ⇒p− = φi (a)φi (a) + φi (av )φi (av ), 2 p+1 ⇒ = φi (a)φi (a) + φi (av )φi (av ). 2

=

q + φi (a)φi (a) + φi (av )φi (av ) , p

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 p+1 ⇒ φi (a)2 + (φi (a) + 1)2 = , 2 2 p+1 ⇒ 2φi (a)2 + 2φi (a) + 1 = , 2 ⇒ 4φi (a)2 + 4φi (a) + 1 − p = 0, p −4 ± 42 − 4 · 4 · (1 − p) ⇒ φi (a) = , 2 · 4 p −4 ± 42 p ⇒ φi (a) = , 2 ·√4 −1 ± p ⇒ φi (a) = . 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, ⇒ 2 Re(φi (a)) = −1, 1 ⇒ Re(φi (a)) = − . 2 Using the second equation we have that φi (a)φi (a) + φi (av )φi (av ) =

p+1 p+1 ⇒ 2φi (a)φi (a) = , 2 2 ⇒ Re(φi (a))2 + Im(φi (a))2 =

p+1 , 4

p+1 1 − , 4 4 p 2 ⇒ Im(φi (a)) = , 4√ p ⇒ Im(φi (a)) = ± , √4 −1 ± pi ⇒ φi (a) = . 2 ⇒ Im(φi (a))2 =

Comparing with the question this settles the two possible cases. (d) Let ε = e 2πi/p be a primitive pth root of unity. Consider, as in the question, δ = ±1

Chapter 25

120

dependent upon p then we know that p−1 √ 2 X −1 ± δp k εu . = φ1 (a) = 2 k=1

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 6≡ 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 √ X −1 ± δp s ε = , 2 s∈Q 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 ha, bi 6 CE (ai bj ), for any 0 6 i, j 6 2 but not i = j = 0. This means that |CE (ai bj )| > 9, so either the conjugacy class has size 1 or 2. However we have c −1 (ai bj )c = (c −1 ac)i (c −1 bc)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, b2 }

{ab, a2 b2 }

{ab2 , a2 b}.

This only deals with half the elements of E. We have hci 6 CE (c), which means

Chapter 25

121

|c E | 6 9. However, keeping notation as before, we have a−i b−j cbj aj = a−i b−2j cai = a−i b−2j a−i c = a−2i b−2j c. Therefore it’s clear that the final conjugacy class has order 9 and explicitly is {c, ac, bc, a2 c, b2 c, abc, ab2 c, a2 bc, a2 b2 c}. Now ha, bi ∼ = C3 ×C3 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+i2 3 then every irreducible character of C3 is given by χk (ai ) = η ki , (with 0 6 k 6 2), and every irreducible character of C3 × C3 is given by the products of these characters. Note that there are 6 conjugacy classes of E and hence 6 irreducible characters. A generic irreducible character of ha, bi is given by χk χ` (ai bj ) = η ki η `j = η ki+`j with 0 6 k, ` 6 2. Notice that the conjugacy classes of ai bj split in to two conjugacy classes in the subgroup ha, bi. Therefore we have the values of the induced characters to be (χk χ` ↑ E)(c) = 0, χk χ` (a−i b−j ) χk χ` (ai bj ) + (χk χ` ↑ E)(a b ) = |CE (a b )| |Cha,bi (ai bj )| |Cha,bi (a−i b−j )| η ki+`j + η −ki−`j =9· , 9 = η ki+`j + η ki+`j ,  2 if ki + `j ≡ 0 (mod 3), = −1 otherwise. i j

i j



 ,

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 (0, 2) ⇒ 2j ≡ 0 (1, 0) ⇒ i ≡ 0

(mod 3) ⇒ (χ0 χ1 ↑ E)(a) = 2, (mod 3) ⇒ (χ0 χ1 ↑ E)(a) = 2, (mod 3) ⇒ (χ0 χ1 ↑ E)(b) = 2,

(1, 1) ⇒ i + j ≡ 0

(mod 3) ⇒ (χ0 χ1 ↑ E)(ab2 ) = 2,

(1, 2) ⇒ i + 2j ≡ 0

(mod 3) ⇒ (χ1 χ2 ↑ E)(ab) = 2

(2, 0) ⇒ 2i ≡ 0

(mod 3) ⇒ (χ0 χ1 ↑ E)(b) = 2,

Chapter 25

122 (2, 1) ⇒ 2i + j ≡ 0

(mod 3) ⇒ (χ0 χ1 ↑ E)(ab) = 2, (mod 3) ⇒ (χ0 χ1 ↑ E)(ab2 ) = 2.

(2, 2) ⇒ 2i + 2j ≡ 0

Therefore the following characters are equal χ0 χ1 ↑ E = χ0 χ2 ↑ E

χ1 χ0 ↑ E = χ2 χ0 ↑ E

χ1 χ1 ↑ E = χ2 χ2 ↑ E,

χ1 χ2 ↑ E = χ2 χ1 ↑ 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

ab2

c

|CE (g)|

18

9

9

9

9

2

χ0 χ0 ↑ E

2

2

2

2

2

0

χ0 χ1 ↑ E

2

2

−1

−1

−1

0

χ1 χ0 ↑ E

2

−1

2

−1

−1

0

χ1 χ2 ↑ E

2

−1

−1

2

−1

0

χ2 χ1 ↑ E

2

−1

−1

−1

2

0

Table 25.2: Some characters induced from ha, bi to E It’s easy to see that E 0 = h[a, c], [b, c]i = ha−2 , b−2 i = ha, bi, hence there are only |G/G 0 | = 18/9 = 2 linear characters of E. These both occur has constituents of χ0 χ0 ↑ E, hence the remaining four characters induced from E 0 must be irreducible. Therefore the character table of E is as in table 25.3. Exercise 25.6. Clearly we have Z(E) = CE (a) ∩ CE (b) ∩ CE (c) = ha, bi ∩ hci = {1}, which is 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) 6= 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 ∼ = F13,3 . To verify we clearly

Chapter 25

123

g

1

a

b

ab

ab2

c

|CE (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

0

ψ3

2

−1

−1

−1

0

ψ4

2

−1 −1

2

−1

0

ψ5

2

−1 −1 −1

2

0

−1 −1 2

Table 25.3: The character table of E have 3 | 13 − 1 = 12, now F13,3 has q = 3 linear characters and |Z∗p /S| = 12/3 = 4 characters of degree q = 3. Therefore F13,3 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 F13,3 × C2 has the character degrees that we’re looking for. Recall that the irreducible characters of F13,3 ×C2 are the products of the irreducible characters of F13,3 together with the irreducible characters of C2 . Every irreducible character of C2 is linear which means the degree of the product is the same as the degree of the irreducible character of F13,3 . Therefore there are 14 irreducible characters of F13,3 × C2 , 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 × F13,3 , (or alternatively D6 × F13,3 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 hai 6 CG (a), j which means |CG (a)| > 9 ⇒ |aG | 6 6. However for any 1 6 j 6 5 we have b−j abj = a2 , 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−2 a3 b2 = (a3 )4 = a12 = a3 , so ha, b2 i 6 CG (a3 ), which means |CG (a3 )| > 27 ⇒ |(a3 )G | 6 2. However b−1 a3 b = a6 , so we must have (a3 )G = {a3 , a6 }.

Chapter 25

124

We now comment on multiplication in the group. We know that ab = ba2 but what about ba? Well, ab = ba2 ⇒ b = aba7 ⇒ ba = aba8 ⇒ ba = ab(a2 )4 ⇒ ba = a5 b = a−4 b. Now that we have this information, what about the conjugacy class of b? Well hbi 6 CG (b) ⇒ |bG | 6 9. In fact, from the above, we know a−1 ba = a4 b and indeed for any 1 6 i 6 8 we have a−i bai = a4i b. Now as 9 isn’t even we get that bG = {ai b | 0 6 i 6 8}. We want to consider the conjugacy classes of the remaining powers of b. Remember that for any 1 6 j 6 5 we have hbi 6 CG (bj ) ⇒ |(bj )G | 6 9. For ease we write some simple multiplication formulae down for the group b2 a = ba5 b = a5×5 b2 = a7 b2

b3 a = ba7 b2 = a5×7 b3 = a8 b3 ,

b4 a = ba8 b = a5×8 b4 = a4 b4

b5 a = ba4 b4 = a5×4 b5 = a2 b5 .

Now in general this will give us that for any 1 6 i 6 8 that a−i b2 ai = a6i b2

a−i b3 ai = a7i b3

a−i b4 ai = a3i b4

a−i b5 ai = a2i b5 .

We have that 2 and 7 are coprime to 9, which means that (b3 )G = {ai b3 | 0 6 i 6 8} and (b5 )G = {ai b5 | 0 6 i 6 8}. From the above it’s clear to see that ha3 i 6 CG (bj ) for j = 2, 4 and so |(bj )G | = 3. Therefore (b2 )G = {b2 , a3 b2 , a2 b2 } and (b4 )G = {b4 , a3 b4 , a6 b4 }. Finally for any 1 6 j 6 5 we have j

b−j (ab2 )bj = a2 b2

j

b−j (ab4 )bj = a2 b4 .

Recall from the calculation of the conjugacy class of aG that this will give us six distinct powers of a. Therefore all the elements of G have been accounted for. So we have the conjugacy classes of G are {1}

{b4 , a3 b4 , a6 b4 },

{a, a2 , a4 , a5 , a7 , a8 }

{ab2 , a2 b2 , a4 b2 , a5 b2 , a7 b2 , a8 b2 }

{ai b3 | 0 6 i 6 8},

{a3 , a6 }

{ab4 , a2 b4 , a4 b4 , a5 b4 , a7 b4 , a8 b4 }

{ai b5 | 0 6 i 6 8},

{b2 , a3 b2 , a6 b2 }

{ai b | 0 6 i 6 8}.

Therefore there are 10 conjugacy classes of G, which means there are 10 irreducible characters. It’s clear to see that [a, b] = a−1 b−1 ab = a and so G 0 = hai. Therefore there

Chapter 25

125

are |G/G 0 | = 54/9 = 6 linear characters, which are the lifts from G/G 0 ∼ = C6 . Let = hbi ∼ √ 1+i 3 2πi/6 ε=e = 2 then the linear characters of G are as in table 25.4. g

1

a

a3

ab2

ab4

b

b2

b3

b4

b5

|CG (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

2

ε

−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 so we 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 = (b2 )3 = 1 and b−2 a3 b2 = (b−2 ab2 )3 = (a4 )3 = a3 . Therefore N = ha3 , b2 i ∼ = C3 × C3 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 = Nb2 = Nb4 = Na3 , Nb = Nb3 = Nb5 , Na = Nab2 = Nab4 . Therefore we have an irreducible character of degree 2 whose values are in table 25.5.

Chapter 25

126

g

1

a

a3

ab2

ab4

b

b2

b3

b4

b5

|CG (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 = ψ6 ψ1 and ψ8 = ψ6 ψ2 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 character table for G is as in table 25.6. g

1

a

a3

ab2

ab4

b

b2

b3

b4

b5

|CG (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

ε

2

−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

0

0

−2ε

ψ8 (g)

2

−1

2

−1

−1

0

−2ε

0

2ε

ψ9 (g)

6

0

−3

0

0

0

0

0

0

−1

2ε

2

2

Table 25.6: The character table of G from Exercise 25.8

0 0 0

Chapter 26. Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise

26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8

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Characters of some p-groups . . . . . . . .

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127 127 129 131 135 140 141 142

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 hχ ↓ H, ψiH 6= 0 ⇒ hχ, ψ ↑ HiG 6= 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 6 χ(1) 6 (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 6 CG (z i ) for 0 6 i 6 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−1 ab = az ⇒ b = a−1 baz ⇒ a−1 ba = bz 2 . Let 0 6 i , j, k 6 2 but such that we don’t have i = j = 0. Then in general we have that b−k (ai bj )bk = (b−k abk )i bj = (az k )i bj = ai bj z ik , a−k (ai bj )ak = ai (a−k bak )j = ai (bz k )j = ai bj z jk . Therefore we have the conjugacy classes of G to be {1}

{a, az, az 2 }

{b2 , b2 z, b2 z 2 }

{ab2 , ab2 z, ab2 z 2 },

{z}

{a2 , a2 z, a2 z 2 }

{ab, abz, abz 2 }

{a2 b2 , a2 b2 z, a2 b2 z 2 },

{z 2 }

{b, bz, bz 2 }

{a2 b, a2 bz, a2 bz 2 }.

So there are 11 conjugacy classes of G, which means there are 11 irreducible characters. Note that ha, zi 6 G is an abelian subgroup of order 32 = 9 and so [G : ha, zi] = 3. Let K = G 0 = hzi in Theorem 26.6. Now |G/G 0 | = 9 and the relation b−1 ab = az tells us (b−1 ab)G 0 = azG 0 ⇒ (b−1 ab)G 0 = aG 0 ⇒ (ab)G 0 = (ba)G 0 . So G/G 0 is abelian and contains two elements of order 3, so G/G 0 ∼ = C3 × C3 . √ −1+ 3i 2πi/3 Let η = e = then we have the character table for G is as in table 26.1. 2 Now the final two characters are induced linear characters of ha, zi ∼ = C3 ×C3 . We note that 2 the conjugacy classes {1}, {z} and {z } do not split upon restriction to ha, zi. However, for 1 6 j 6 2, we have {aj , aj z, aj z 2 } splits into three conjugacy classes. Therefore if ψ is an irreducible character of ha, zi then (ψ ↑ G)(z j ) = [G : H]ψ(z j ) = 3ψ(z j ),   ψ(aj ) ψ(aj z) ψ(aj z 2 ) j j (ψ ↑ G)(a ) = |CG (a )| + + , |Cha,zi (aj )| |Cha,zi (aj z)| |Cha,zi (aj z 2 )| = ψ(aj ) + ψ(aj z) + ψ(aj z 2 ), = ψ(aj )(1 + ψ(z) + ψ(z 2 )).

Chapter 26

129

g

1

z

z2

a

a2

b

b2

ab

a2 b

ab2

a2 b 2

|CG (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

η

2

2

η

1

η

η

2

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

3η 2

0

0

0

0

0

0

0

0

χ11 (g)

3

3η 2

3η

0

0

0

0

0

0

0

0

η

Table 26.1: The character table of G from Exercise 26.2 Now we can see that hψ ↑ G, ψ ↑ Gi >

32 ψ(1)ψ(1) 32 ψ(z)ψ(z) 32 ψ(z 2 )ψ(z 2 ) 3 · 32 + + = = 1. 27 27 27 27

Hence we only get irreducible characters when ψ(z) 6= 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 6 j 6 15 we have hai 6 CG (aj ), which means |(aj )G | 6 2. From the relations it’s easy to see that b−1 aj b = (b−1 ab)j = a−j . Therefore for 1 6 k 6 7 we have the following conjugacy classes; {1}, {a8 }, {ak , a−k }. Now hbi 6 CG (b) which means |bG | 6 8. Now using the relations of the group we have b−1 ab = a−1 ⇒ ab = ba−1 ⇒ b = a−1 ba−1 ⇒ a−1 ba = a−2 b. Therefore a−j baj = a−2j b, which means bG = {a2k b | 0 6 k 6 7}. Likewise we get a−j (ab)aj = a1−2j b, which gives us (ab)G = {a2k+1 b | 0 6 k 6 7}. In summary, with

Chapter 26

130

1 6 ` 6 7, we have the conjugacy classes of G to be {1}

{a2k b | 0 6 k 6 7},

{a8 }

{a2k+1 b | 0 6 k 6 7},

{a` , a−` }. Now hai is an abelian subgroup of order 16 in G and hence of index 2. So take H = hai and K = Z(G) = ha8 i in Theorem 26.6. Also note that a−1 b−1 ab = a−2 , which means G 0 = ha2 i. Now G/K ∼ = ha, b | a8 = b2 = 1, b−1 ab = a−1 i = 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

|CG (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

2

−2

0

2

2

0

0

χ7 (g)

2

2

0 √ 2 √ − 2

0

χ6 (g)

0 √ − 2 √ 2

0

0

0 −2 0 √ √ 2 0 − 2 √ √ − 2 0 2

−2 −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 = hai. Let ε = e 2πi/16 be a primitive 16th root of unity. Then every irreducible representation of H takes the form ψi (aj ) = εij for 0 6 i, j 6 15. Now, for 1 6 k 6 7, we calculate the induced character to be (ψi ↑ G)(a8 ) = [G : H]ψi (a8 ) = 2ε8i = 2(−1)i   ψi (aj ) ψi (a−j ) j j (ψi ↑ G)(a ) = |CG (a )| + , |CH (aj )| |CH (a−j )| = ψi (aj ) + ψi (a−j ), = εij + ε−ij , = εij + εij ,

Chapter 26

131 = 2 Re(εij ).

Now it’s clear that Re(εk ) = Re(ε16−k ) = Re(εk ) for 1 6 k 6 7, which means ψk ↑ G = ψ16−k ↑ G for 1 6 k 6 7. So it’s sufficient to focus on the first eight characters to find √ √ 2+i 2 2 2πi/8 the remaining four irreducible characters of G. Recall that ε = e = . Straight 2 8 away we can see that ψ0 ↑ G = 2χ1 and in fact ψ8 ↑ G = χ3 + χ4 because ε = −1, in √ other words they are reducible. Indeed, noticing that 2 Re(ε2 ) = 2, 2 Re(ε4 ) = 0 and √ 2 Re(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 ) = 2 cos(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

|CG (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

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

−2

−2

0

2

0

0

χ7 (g)

2

2

0 √ 2 √ − 2

0

2

0 √ − 2 √ 2

2

χ6 (g)

0 √ 2 √ − 2

0

0

χ8 (g)

2

−2

α

−α

0

0

χ9 (g)

2

−2

β

−β

0

0

χ10 (g)

2

−2

−β

β

0

0

χ11 (g)

2

−2

−α

α

0

0

0 0 √ 2 √ − 2 √ − 2 √ 2

0 √ −2 − 2 √ −2 2

β

0

−β

−α

0

α

α

0

−α

−β

0

β

0 0 √ − 2 √ 2 √ 2 √ − 2

−1 −1

Table 26.3: The character table of G from Exercise 26.3 Exercise 26.4. Let G = hA, B, C, Di 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

AC

AD

BC

BD

CD

 −1 0 0 0 1 0  = 0 0 1 0 0 0  −1 0 0 0 1 0  = 0 0 1 0 0 0  −1 0 0 0 1 0  = 0 0 1 0 0 0  0 0 i 0 0 0  = i 0 0 0 −i 0  0 0 i 0 0 0  = i 0 0 0 −i 0  0 i 0 i 0 0  = 0 0 0 0 0 −i

 0 0   0  0  0  i −1 0  0 0   0  i  0  0 −1 0  0 0   0  0  0  0 −1 1  0 0   −i   i  0  0 0 0  0 0   −i  0  0  0 1 0  0 0   0  0  −i  0 0 1

   0 i 0 0 0 −i 0   0 0 −i   0 0 0 −i  =  = −BA, 0 0 0  i 0 0 0  −i 0 0 0 i 0 0    i 0 0 0 −i 0 0   0 0 0  i 0 0 0  =  = −CA, 0 0 −i  0 0 0 −i  0 −i 0 0 0 i 0    0 0 1 0 0 0 −1   0 1 0  0 0 1 0 =  = −DA, 1 0 0  0 1 0 0  0 0 0 −1 0 0 0    0 0 0 1 i 0 0   0 0 0  0 0 −1 0 =  = −CB,   0 0 −i  0 −1 0 0 1 0 0 0 0 −i 0    0 i 0 0 0 0 1   0 1 0  −i 0 0 0  = −DB, = 1 0 0  0 0 0 i  0

0 0 0 0 0 1 0

0 1 0 0





0 1   0  0 = 0 −i 0

0

0 −i 0  0 i 0 0 0 i   = −DC. 0 0 0 −i 0 0

Therefore every pair of generators commutes modulo hZi. Clearly G is not abelian but G/hZi is abelian. We recall that if a quotient G/N is abelian then we must have G 0 6 N. In other words {I} = 6 G 0 6 hZi ⇒ G 0 = hZi. (b) We check the squares of the generating elements. It’s obvious that A2 = D2 = 1, so

Chapter 26

133

the only things to really check are B 2 and C 2 . Now  0 0  B2 =  i 0  0 i  C2 =  0 0

 0 i 0 0   0 0 −i  0  0 0 0  i −i 0 0 0  i 0 0 0   0 0 0  i  0 0 −i  0 0 −i 0 0

   0 i 0 −1 0 0 0  0 0 −i  0   0 −1 0  =  = −I, 0 0 0  0 0 −1 0  −i 0 0 0 0 0 −1    i 0 0 −1 0 0 0  0 0 0 0   0 −1 0  =  = −I. 0 0 −i   0 0 −1 0  0 −i 0 0 0 0 −1

It’s clear that every element g ∈ G can be written in the form Ai B j C k D` Z m for 0 6 i , j, k, `, m 6 1 and by the above relations we have g 2 = ±I. Therefore g 2 ∈ hZi for every g ∈ G. Every element has order 2 means the group is a 2 group. Clearly from the presentation we have given in this part of the question it’s clear that the order of the group is less than or equal to 25 = 32. (c) By Corollary 9.3 we have that the representation is irreducible if and only if every matrix X = (xij ) which satisfies (gρ)X = X(gρ) for all g ∈ G has the form X = λI for λ ∈ C. Now if X commutes with every element g ∈ G then it commutes with each of A, B, C and D. Checking the commutation with A gives us  x11 x  21  x31 x41

x12 x22 x32 x42

x13 x23 x33 x43

 −1 x14   x24   0  x34   0 x44 0

 −x11 −x  21 ⇒ −x31 −x41

x12 x22 x32 x42

   x11 x12 x13 −1 0 0 0 0 0     0 0   0 1 0 0  x21 x22 x23  = 1 0   0 0 1 0  x31 x32 x33 0 −1 0 0 0 −1 x41 x42 x43    x13 −x14 −x11 −x12 −x13 −x14  x22 x23 x24  x23 −x24     x21 , = x32 x33 x34  x33 −x34   x31 x43 −x44 −x41 −x42 −x43 −x44   x11 0 0 x14 0 x 0   22 x23 ⇒X= .  0 x32 x33 0  x41 0 0 x44 0 1 0 0

 x14 x24   , x34  x44

Chapter 26

134

Now checking the commutation with D gives us   x11 0 0 x14 0 0 x   0  0  22 x23    0 x32 x33 0  0 x41 0 0 x44 1  x14 0 0 x  23 ⇒  0 x33 x44 0

0 0 1 0

0 1 0 0

    1 0 0 0 1 x11 0 0 x14    0  0 0 1 0  0 x22 x23 0   , = 0 0 1 0 0  0 x32 x33 0  0 1 0 0 0 x41 0 0 x44    x11 x41 0 0 x44   0   0 x32 x33 0   = , 0   0 x22 x23 0 

0 x22 x32 0 x41

x11 0 0 x14   x11 0 0 x14 0 x 0   22 x23 ⇒X= .  0 x23 x22 0  x14 0 0 x11

Now checking the commutation with C gives us  x11 0 0 0 x  22 x23   0 x23 x22 x14 0 0  0 ix  22  ix23 0

   x11 0 0 0 i 0 0 i 0 0     0 0 0   i 0 0 0   0 x22 x23  = 0 0 −i  0 0 0 −i   0 x23 x22 x14 0 0 0 0 −i 0 0 −i 0    0 ix22 ix23 0 −ix14 0  0 0 ix14  0 −ix23     ix11 = ,     −ix14 0 0 −ix11  0 −ix22 0 −ix23 −ix22 0 −ix11 0   x11 0 0 0 0 x 0 0   11 ⇒X= . 0 0 x11 0  0 0 0 x11

 0 x14   0  i  0  0 0 x11 ix11 0 0 ix14

 x14 0  , 0 x11

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 6 |G| 6 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 0 = hZi, which means |G/G 0 | = |G|/|G 0 | = 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 0 , which means G/G 0 ∼ = C2 × C2 × C2 × C2 . Now for example an irreducible representation of C2 is given by Aρ = −1 or Aρ = 1 = (−1)0 . So for 0 6 x, y , z, w 6 1 we get a linear representation of G given by Ai B j C k D` Z m 7→ (−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 representation √ √ √ 2+i 2 2πi/8 would need to have trace 0 on b and trace ± 2 on a. Recall ω = e = . 2 √ √ 2−i 2 −1 2πi/8 Now ω = e = , therefore we get such an irreducible representation 2 ρ : G1 → GL(2, C) by defining # " # " 0 1 ω 0 bρ = . aρ = 1 0 0 ω −1 It’s clear that (aρ)8 = (bρ)2 = I2 and so all that’s left to check is the remaining relation. #" # " #" 0 1 0 1 ω 0 −1 , (bρ) (aρ)(bρ) = 1 0 0 ω −1 1 0 " #" # 0 ω −1 0 1 = , ω 0 1 0 " # −1 ω 0 = , 0 ω = (aρ)−1 . Now consider such an irreducible representation for G2 . Recall that ω 4 = −1 and so if we define aρ as for G1 we have a4 ρ = −I2 . Therefore we must define bρ such that (bρ)2 = (aρ)4 = −I. We define our representation ρ : G2 → GL(2, C) by " # " # ω 0 0 −1 aρ = bρ = . 0 ω −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. "

#" #" # 0 1 ω 0 0 −1 (bρ)−1 (aρ)(bρ) = , −1 −1 0 0 ω 1 0 " #" # −1 0 ω 0 −1 = , −ω 0 1 0 " # ω −1 0 = , 0 ω = (aρ)−1 .

Now consider such an irreducible representation for G3 . We notice that for G3 the √ √ √ trace on a is not ± 2 but ±i 2. Now we can see that ω − ω −1 = i 2 and also that −ω −1 = ω 3 . So, we define our representation ρ : G3 → GL(2, C) by # " # " 0 1 ω 0 bρ = aρ = 1 0 0 ω3 It’s clear to see that (aρ)8 = (bρ)2 = I2 , so we only need to check the remaining defining relation holds. # #" #" " 0 1 ω 0 0 1 , (bρ)−1 (aρ)(bρ) = 1 0 0 ω3 1 0 # #" " 0 ω3 0 1 , = 1 0 ω 0 " # ω3 0 , = 0 ω = (aρ)−1 . Note that (ω 3 )3 = ω 9 = 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 = ha, b | a8 = b2 = 1, b−1 ab = a5 i. Note from the character table of G4 that we require the trace on a

Chapter 26

137

to be 0 and that −ω = ω 5 . So, we define our representation ρ : G4 → GL(2, C) by " # " # ω 0 0 1 aρ = bρ = . 0 ω5 1 0 It’s clear to see that (aρ)8 = (bρ)2 = I2 , so we only need to check the remaining defining relation holds. # #" #" " ω 0 0 1 0 1 , (bρ)−1 (aρ)(bρ) = 1 0 0 ω5 1 0 # #" " 0 ω5 0 1 , = 1 0 ω 0 " # ω5 0 = , 0 ω = (aρ)−1 . Note that (ω 5 )5 = ω 25 = 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 # # " # " " i 0 0 1 1 0 bρ = aρ = 0 i 1 0 0 −1 Now it’s clear that (aρ)2 = (bρ)2 = (zρ)4 = I2 so we only need to check that the remaining relations hold. " #" # " # " #" # 1 0 i 0 i 0 i 0 1 0 (aρ)(zρ) = = = = (zρ)(aρ), 0 −1 0 i 0 −i 0 i 0 −1 " #" # " # " #" # 0 1 i 0 0 i i 0 0 1 (bρ)(zρ) = = = = (zρ)(bρ), 1 0 0 i i 0 0 i 1 0 " #" #" # " #" # " # 0 1 1 0 0 1 0 −1 0 1 −1 0 (bρ)−1 (aρ)(bρ) = = = = (aρ)(zρ)2 . 1 0 0 −1 1 0 1 0 1 0 0 1 (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 ) = C1 ∪ C2 ∪ C3 ∪ C4 = {1, z, a2 , a2 z}. Now z = b2 in G5 so Z(G5 ) = {1, b2 , a2 , a2 b2 } ∼ = C2 × C2 which is not cyclic. Note that it is easier to see this is C2 × C2 once you know ba = ab−1 and ab = b−1 a. In G6 , G7 and G8 we have az = za so Z(G6 ) ∼ = C2 × C2 , which is = Z(G8 ) ∼ = Z(G7 ) ∼ 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     −1 0 0 i2 0 0     (bρ)2 =  0 (−i )2 0  =  0 −1 0 = (zρ), 0 0 1 0 0 12   2  2   0 1 0 0 1 0 1 0 0 12 0 0        (aρ)4 = 1 0 0 1 0 0 = 0 1 0  =  0 12 0  = I3 . 0 0 i 0 0 i 0 0 −1 0 0 (−1)2 Now we check the final relation   −i 0 0 0   −1 (bρ) (aρ)(bρ) =  0 i 0 1 0 0 1 0   0 −i 0 i   =  i 0 0 0 0 0 i 0   0 −1 0   = −1 0 0 , 0 0 i   0 1 0 −1   = 1 0 0  0 0 0 0 i

  1 0 i 0 0   0 0 0 −i 0 , 0 i 0 0 1  0 0  −i 0 , 0 1

 0 0  −1 0 , 0 1

= (aρ)(zρ). Therefore this is indeed a representation of G5 . Is it faithful? Well from the above it’s easy to see that 1ρ, aρ, a2 ρ, a3 ρ, bρ, zρ, azρ, b−1 aρ, 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 = (zφ)2 = 1. We first of all check the commuting relations 

  i 0 0 −1 0 0    (aρ)(zρ) = 0 −i 0  0 −1 0 0 0 i 0 0 1   −i 0 0   =  0 i 0 0 0 i    −1 0 0 i 0 0    =  0 −1 0 0 −i 0 0 0 1 0 0 i = (zρ)(aρ)



 0 1 0 −1 0   (bρ)(zρ) = 1 0 0  0 −1 0 0 1 0 0   0 −1 0   = −1 0 0 , 0 0 1   −1 0 0 0 1   =  0 −1 0 1 0 0 0 1 0 0

 0  0 , 1

 0  0 , 1

= (zρ)(bρ)

Now we check that the final relation holds     0 1 0 i 0 0 0 1 0     (bρ)−1 (aρ)(bρ) = 1 0 0 0 −i 0 1 0 0 , 0 0 1 0 0 i 0 0 1    0 −i 0 0 1 0    =  i 0 0 1 0 0 , 0 0 i 0 0 1   −i 0 0   =  0 i 0 , 0 0 i    i 0 0 −1 0 0    = 0 −i 0  0 −1 0 , 0 0 i 0 0 1 = (aρ)(zρ). 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 " # " # i 0 0 1 a 7→ b 7→ . 0 −i 1 0 Therefore our plan is to extend this by defining a morphism ρ : G7 → GL(3, C) by 

 i 0 0   aρ = 0 −i 0 0 0 1

  0 1 0   bρ = 1 0 0 0 0 1

  1 0 0   zρ = 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 these by 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 

 i 0 0   aρ = 0 −i 0 0 0 1



 0 1 0   bρ = −1 0 0 0 0 1

  1 0 0   zρ = 0 1 0  . 0 0 −1

Notice again that we have extended this idea from a 2-dimensional irreducible representation 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 ∼ = G6 and = G2 , G5 ∼ G7 ∼ = G8 . Notice that we have G7 ∼ = D8 × C2 and G8 ∼ = Q8 × C2 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−a a and ba = ab−1 and in G6 we have ab = baz and ba = abz. g

1

z

a2

a2 z

a

a3

b

a2 b

ab

a3 b

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 ∼ 6 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−1 b. 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 6∼ = 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) 6= {1} so |Z(G)| = 6 1 and G is not abelian 4 so |Z(G)| = 6 p . Recall from Lemma 26.1 that if K 6 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) ∼ = Cp . 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 x1 , . . . , xk be representatives of the conjugacy classes of G then by the class equation we have p4 = p2 +

X

|xiG | ⇒

xi 6∈Z(G)

X

|xiG | = p(p 3 − p) = p 2 (p 2 − 1).

xi 6∈Z(G)

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 6∈ Z(G), then Z(G) 6 CG (x) but Z(G) 6= CG (x) because x 6∈ Z(G). Therefore |CG (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 0 ] = |G/G 0 | and we have X X |G| = [G : G 0 ] + χ(1)2 ⇒ p 4 = [G : G 0 ] + χ(1)2 . χ(1)>1

χ(1)>1

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 0 ] > 1. Therefore p 2 | p 4 and p 2 divides the sum so we must have p 2 | [G : G 0 ]. Note that |G 0 | = 6 1 because G is not abelian, so 0 2 3 0 we have [G : G ] = p or p . In other words |G | = p or p 2 .

Chapter 26

142

Assume that |G 0 | = p 2 then we have [G : G 0 ] = p 2 . Recall that any non-linear character of G has order p so using the above equation we have p4 = p2 +

X χ(1)>1

p2 ⇒

X

p 2 = p 2 (p 2 − 1).

χ(1)>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 0 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 6 |G 0 ∩ Z(G)| 6 p 2 because G 0 ∩ Z(G) is a subgroup of G 0 and Z(G). If |G 0 ∩ Z(G)| = p 2 then |G 0 | = |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 0 ∩ 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) = haZ, bZ | a4 Z = Z, a2 Z = b2 Z, (b−1 ab)Z = a−1 Zi. Now we have from the defining relations of Q8 that abZ = ba−1 Z and baZ = a−1 bZ. Therefore (a2 b)Z = (ba−2 )Z = ba2 Z. Note that the centre of G/Z must be trivial but we clearly have a2 Z in the centre of G/Z so we must have a2 Z = 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 0 = 1, so G/G 0 ∩Z(G) ∼ 6 Q8 . Now = 0 0 suppose G is non-abelian then G 6= {1}, so by Exercise 7 we have |G ∩ Z(G)| = 2, so either G 0 ∩ Z(G) = G 0 or G 0 ∩ Z(G) = Z(G). We know that G/G 0 is abelian so G/G 0 ∼ 6 Q8 and G/Z(G) ∼ 6 Q8 by part (a). Therefore G/(G 0 ∩ Z(G)) ∼ 6 Q8 . = = =

Chapter 27.

Exercise Exercise Exercise Exercise

27.1 27.2 27.3 27.4

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Character table of the simple group of order 168 . . . .

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143 143 148 149

Exercise 27.1. Let us fix a matrix "

# x y X= ∈ SL(2, p) z w and assume that X ∈ Z(SL(2, p)). Then " #" # " #" # " # " # x y 1 1 1 1 x y x x +y x +z y +w = ⇒ = , z w 0 1 0 1 z w z z +w z w ⇒ z = 0 and x = w . Choosing another appropriate matrix we find "

x y 0 x

#" # " #" # " # " # 1 0 1 0 x y x +y y x y = ⇒ = , 1 1 1 1 0 x x x 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 g1 is correct. Now g2 = 2I2 = −I2 ∈ Z(SL(2, 3)), therefore the information for g2 is clear.

143

Chapter 27

144



g1 g2 g3 g4 g5 g6 g7

 1 0 = 0 1   2 0 = 0 2   0 1 = 2 0   1 1 = 0 1   1 2 = 0 1   2 1 = 0 2   2 2 = 0 2

Order

|CG (gi )|

|giG |

1

24

1

2

24

1

4

4

6

3

6

4

3

6

4

6

6

4

6

6

4

Table 27.1: The conjugacy classes of SL(2, 3) Start by considering X ∈ CG (g3 ) then we have # # " # " #" # " #" " z w 2y x 0 1 x y 0 1 x y , = ⇒ = 2x 2y 2w z 2 0 z w z w 2 0 # " x y . ⇒X= 2y x Now det(X) = x 2 − 2y 2 = x 2 + y 2 = 1. Our only choices are (x, y ) = (0, 1), (0, 2), (1, 0) or (2, 0) therefore |CG (g3 )| = 4, which means |g3G | = 24/3 = 6. Now as for the order of g3 we have " #" #!2 " #2 " # 0 1 0 1 2 0 4 0 g34 = = = = I2 . 2 0 2 0 0 2 0 4 Consider an element X ∈ CG (g4 ) then we have " #" # " #" # " # " # x y 1 1 1 1 x y x x +y x +z y +w = ⇒ = , z w 0 1 0 1 z w z z +w z w

Chapter 27

145 # x y . ⇒X= 0 x "

Now det(X) = x 2 = 1 then our options are y = 0, 1, 2 and x = 1 or 2, hence |CG (g4 )| = 6, which means |g4G | = 24/6 = 4. Now as for the order of g4 we have # # " #" # " #" #" " 1 0 1 1 1 2 1 1 1 1 1 1 = I2 . = = g43 = 0 1 0 1 0 1 0 1 0 1 0 1 Consider an element X ∈ CG (g5 ) then we have # # " # " #" # " #" " x + 2z y + 2w x 2x + y 1 2 x y 1 2 x y = , = ⇒ z w z 2z + w 0 1 z w z w 0 1 " # x y . ⇒X= 0 x It’s clear from the previous calculation for the order of g5 we have #" #" " 1 1 2 1 2 g53 = 0 1 0 1 0

that |CG (g5 )| = 6, which means |g5G | = 4. Now as # # " #" # " 1 0 1 1 1 2 2 = I2 . = = 0 1 0 1 0 1 1

Consider an element X ∈ CG (g6 ) then we have # # " # " #" # " #" " 2x + z 2y + w 2x x + 2y 2 1 x y 2 1 x y , = ⇒ = 2z 2w 2z z + 2w 0 2 z w z w 0 2 " # x y . ⇒X= 0 x Hence again it’s clear from the previous calculation that we have |CG (g6 )| = 6, which means |g6G | = 4. Now as for the order of g6 we have g66 =

" #" #" #!2 2 1 2 1 2 1 = 0 2 0 2 0 2

"

1 1 0 1

#" #!2 " #2 2 1 2 0 = = I2 . 0 2 0 2

Finally consider an element X ∈ CG (g7 ) then we have " #" # " #" # " # " # x y 2 2 2 2 x y 2x 2x + 2y 2x + 2z 2y + 2w = ⇒ = , z w 0 2 0 2 z w 2z 2z + 2w 2z 2w

Chapter 27

146 # x y . ⇒X= 0 x "

Therefore again we confirm that |CG (g7 )| = 6, which means |g7G | = 4. Now as for the order of g7 we have " g76 =

#!2 #" #" 2 2 2 2 2 2 = 0 2 0 2 0 2

#2 #!2 " #" " 2 0 1 2 2 2 = I2 . = 0 2 0 1 0 2

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 g4 is conjugate to g5 and g6 is conjugate to g7 . Well " #" # " #" # " # " # x y 1 1 1 2 x y x x +y x + 2z y + 2w = ⇒ = , z w 0 1 0 1 z w z z +w z w " # x y ⇒X= . 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 g6 and g7 we find "

x y z w

#" # " #" # " # " # 2 1 2 2 x y 2x x + 2y 2x + 2z 2y + 2w = ⇒ = , 0 2 0 2 z w 2z z + 2w 2z 2w " # x y ⇒X= 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) and SL(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 are (" # " # " # ) (" # " # " # " # ) 0 1 2 1 2 2 1 1 1 0 0 1 2 1 Z, Z, Z Z, Z, Z, Z 2 0 1 1 2 1 0 1 2 1 2 2 2 0 (" # " # " # " # ) (" # ) 1 2 1 0 0 1 1 1 1 0 Z, Z, Z, Z Z . 0 1 1 1 2 1 2 0 0 1

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 in A4 . 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+2 3i then in table 27.2 we write down the lifted characters from PSL(2, 3) to obtain four irreducible characters of SL(2, 3). g

g1

g2

g3

g4

g5

g6

g7

|CG (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 d5 , d6 and d7 be the remaining degrees then we know that di | |G| = 24 and di2 6 24 ⇒ di 6 4. Therefore the degrees are either 1, 2, 3 or 4. We also know that 24 = 12 + 12 + 12 + 32 + d52 + d62 + d72 ⇒ 12 = d52 + d62 + d72 . The only integer solution to this is d5 = d6 = d7 = 2. Therefore the remaining three characters are all of degree 2. Before carrying on we note that g2−1 ∈ g2G , g3−1 ∈ g3G , g5 = g4−1 and g7 = g6−1 . Therefore the columns of g2 and g3 are real and χ(g5 ) = χ(g4 ), χ(g7 ) = χ(g6 ) for all irreducible characters χ of G. Indeed because of the orders of the elements we will have that the columns of g2 and g3 will be integer valued. Also, we know that |CG (g3 )| = 4 which means the remaining entries in the column of g3 are zeros. Also because g2 is an involution we know χ(g2 ) is an integer for all irreducible characters of G and χ(g2 ) ≡ χ(1) (mod 2). Using this information and the column orthogonality relations between g1 and g2 we can determine the column of g2 . 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

g1

g2

g3

g4

g5

g6

g7

|CG (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 g5 = g4−1 and g7 = g6−1 this fills in the remaining details. Taking the inner product of χ1 with χ4 we find hχ1 , χ4 i =

2 a+a+b+b 2 − +0+ ⇒ 0 = 2a + 2b ⇒ b = −a. 24 24 6

Also considering the inner product of χ4 with itself gives us hχ4 , χ4 i =

22 22 4a2 + +0+ ⇒ 24 = 8 + 16a2 ⇒ a2 = 1 ⇒ a = ±1. 24 24 6

Now g4 is an element of order 3 and a is an integer so we must have χ(g4 ) ≡ χ(1) (mod 3) for all irreducible characters, which give us a = −1. Recall that χ4 χ2 and χ4 χ3 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

g1

g2

g3

g4

g5

g6

g7

|CG (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

χ5 (g)

2

−2

0

χ6 (g)

2

−2

χ7 (g)

3

3

1

1

−ω

−ω

2

ω

ω2

0

−ω 2

−ω

ω2

ω

−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.

t1 t2 t3 t4 t5

 1 = 0  2 = 0  3 = 0  1 = 0  1 = 0

 0 Z 1  0 Z 4  0 Z 5  1 Z 1  6 Z 1

Order

|CG (gi )|

|giG |

1

21

1

3

3

7

3

3

7

7

7

3

7

7

3

Table 27.5: The conjugacy classes of the subgroup T 6 PSL(2, 7) Clearly the information about t1 is correct. Let X ∈ CT (t2 ) be a generic element then " #" # " #" # " # " # x y 2 0 2 0 x y 2x 4y 2x 2y Z= Z⇒ Z= Z, 0 x −1 0 4 0 4 0 x −1 0 4x −1 0 4x −1 ⇒ y = 0.

Chapter 27

150

Therefore |CT (t2 )| = 3 ⇒ |t2T | = 7. As for the order of t2 we have "

2 0 t23 = 0 4

#" #" # " #" # " # 2 0 2 0 4 0 2 0 1 0 Z= Z= Z = Z. 0 4 0 4 0 2 0 4 0 1

Let X ∈ CT (t3 ) then we have " #" # " #" # " # " # x y 3 0 3 0 x y 3x 5y 3x 3y Z= Z⇒ Z= Z, 0 x −1 0 5 0 5 0 x −1 0 5x −1 0 5x −1 ⇒ y = 0. Therefore |CT (t3 )| = 3 ⇒ |t3T | = 7. As for the order of t3 we have "

3 0 t33 = 0 5

#" #" # " #" # " # 3 0 3 0 2 0 3 0 6 0 Z= Z= Z = Z. 0 5 0 5 0 4 0 5 0 6

Let X ∈ CT (t4 ) then we have # # " # " #" # " #" " x x −1 + y x x +y 1 1 x y 1 1 x y Z, Z= Z⇒ Z= 0 x −1 0 x −1 0 1 0 x −1 0 x −1 0 1 ⇒ x = x −1 . So x = 1, (or 6 but this case is equivalent modulo the centre), which means |CT (t4 )| = 7 and |t4T | = 3. As for the order of t4 we have " #7 " # " # 1 1 1 7 × 1 1 0 t47 = Z= Z= Z = Z. 0 1 0 1 0 1 Let X ∈ CT (t5 ) then we " #" # " x y 1 6 1 Z= −1 0 x 0 1 0

have #" # " # " # −1 6 x y x 6x + y x 6x + y Z⇒ Z= Z, −1 −1 1 0 x 0 x 0 x −1 ⇒ x = x −1 .

Again this gives us |CT (t5 )| = 7 and |t5T | = 7. As for the order of t5 we have "

1 6 t57 = 0 1

#7

"

# " # 1 7×6 1 0 Z= Z= Z = Z. 0 1 0 1

Chapter 27

151

Therefore the information in the table is correct. All that is left is to make sure that t3 6∈ t2T and t5 6∈ t4T . Let X ∈ T then #" # " # " # # " " #" 2x 4y 3x 3y 3 0 x y x y 2 0 Z⇒ Z= Z. Z= 0 4x −1 0 5x −1 0 5 0 x −1 0 x −1 0 4 No such element exists in T therefore these elements are not conjugate. Similarly we have #" # " # " # # " " #" x x +y x 6x −1 + y 1 6 x y x y 1 1 Z⇒ Z= Z, Z= 0 x −1 0 x −1 0 1 0 x −1 0 x −1 0 1 ⇒ 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 ∼ = F7,3 . 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 restriction to T is g4G . It’s clear that [G : T ] = 8 so we have 1T ↑ G is as in table 27.6. g

g1

g2

g3

g4

g5

g6

168

8

4

3

7

7

(1T ↑ G)(g)

8

0

0

2

1

1

χ

7

1

0

0

|CG (g)|

−1 −1

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 character by Frobenius Reciprocity. Let χ = 1T ↑ G − 1G then taking the inner product we see hχ, χi =

72 1 1 1 49 + 21 + 42 + 56 168 + + + = = = 1. 168 8 4 3 168 168

Note we could also have seen this directly from the character table of PSL(2, 7) on

Chapter 27

152

page 318. (b) We now induce the non-trivial linear characters of T . By reading the values off the character table on page 240, we easily see the characters are as in table 27.7. Again we know this is an irreducible character of G from the character table but taking the inner product we see hλ1 ↑ G, λ1 ↑ Gi =

82 1 1 1 64 + 56 + 2 × 24 168 + + + = = = 1. 168 3 7 7 168 168

g

g1

g2

g3

g4

g5

g6

168

8

4

3

7

7

(λ1 ↑ G)(g)

8

0

0

−1

1

1

(λ2 ↑ G)(g)

8

0

0

−1

1

1

|CG (g)|

Table 27.7: The values of λ1 ↑ G and λ2 ↑ G (c) We now decompose the character χ into its symmetric and anti-symmetric parts. First we need to know what conjugacy classes the squares of the elements lie in. Now calculating the squares of the conjugacy class representatives we have " #" # " # " #" # " # 0 1 0 1 6 0 2 5 2 5 0 6 g22 = Z= Z g32 = Z= Z 6 0 6 0 0 6 2 2 2 2 1 0 " #" # " # " #" # " # 2 0 2 0 4 0 1 1 1 1 1 2 g42 = Z= Z g52 = Z= Z 0 4 0 4 0 2 0 1 0 1 0 1 " #" # " # 1 6 1 6 1 5 g62 = Z= Z. 0 1 0 1 0 1 Now it’s clear that g22 = 6g1 = g1 ∈ g1G , g32 = 6g2 = g2 ∈ g2G and g42 = g4−1 ∈ g4G . However g52 and g62 are not so clear. We claim g52 ∈ g5G , so we have "

a b c d

#" # " #" # " # " # 1 1 1 2 a b a a+b a + 2c b + 2d Z= ⇒ = , 0 1 0 1 c d c c +d c d ⇒ c = 0 and d = 4a.

Therefore det(X) = 4a2 = 1 ⇒ a2 = 2 ⇒ a = 3 or 4, hence this element exists in G

Chapter 27

153

so g52 ∈ g5G . Similarly we have "

a b c d

#" # " #" # " # " # 1 6 1 5 a b a 6a + b a + 5c b + 5d Z= ⇒ = , 0 1 0 1 c d c 6c + d c d ⇒ c = 0 and d = 4a.

So again this element exists in G so g62 ∈ g6G . Using this information and the formula on page 198, we see that χS and χA are as in table 27.8. g

g1

g2

g3

g4

g5

g6

g2

g1

g1

g2

g4

g5

g6

168

8

4

3

7

7

1

0

0

|CG (g)| χ

7

χS

28

4

0

1

0

0

χA

21

−3

0

0

0

0

−1 −1

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 7 × 28 4 1 196 − 84 + 56 168 − + = = = 1, 168 8 3 168 168 8 × 28 1 224 − 56 168 hχS , λ ↑ Gi = − = = = 1, 168 3 168 168 4 1 28 + 84 + 56 168 28 hχS , 1G i = + + = = = 1. 168 8 3 168 168 hχS , χi =

Therefore χ = ψ + χ + λ ↑ G + 1G for some character ψ of G. This character ψ has values as in table 27.9. g

g1

g2

g3

g4

g5

g6

|CG (g)|

168

8

4

3

7

7

ψ

12

4

0

0

−2

−2

Table 27.9: The character ψ

Chapter 27

154

Considering the inner product of ψ with itself we have hψ, ψi =

122 42 22 22 144 + 21 × 16 + 24 × 2 × 4 672 + + + = = = 4. 168 8 7 7 168 168

So either ψ is the sum of four irreducible characters or is 2 copies of one irreducible character. If ψ was a sum of four distinct irreducible characters then this would give us 7 irreducible characters overall but there are only 6. Therefore ψ = 2φ where φ is an irreducible character of G. (d) We have four irreducible characters so far of degrees 1, 6, 7 and 8. There are two irreducible characters of G remaining so we must have 168 = 12 + 62 + 72 + 82 + χ5 (1)2 + χ6 (1)2 ⇒ 18 = χ5 (1)2 + χ6 (1)2 . Now χi (1)2 6 18 ⇒ χi (1) 6 4 for i = 5, 6. The only integer solution to this problem is χ5 (1) = χ6 (1) = 3. Therefore the character table of G is as in table 27.10. g

g1

g2

g3

g4

g5

g6

168

8

4

3

7

7

1G

1

1

1

1

1

1

φ

6

2

0

0

χ

7

λ↑G

8

0

χ5

3

χ6

3

|CG (g)|

−1 −1

1

0

0

0

−1

1

1

−1

1

0

η

η

−1

1

0

η

η

−1 −1

Table 27.10: The character table of PSL(2, 7) We explain how we obtained the current values for χ5 and χ6 . Using the column orthogonality relation of g4 with itself we obtain that the remaining values for g4 must be 0, 0. Similarly using the column orthogonality of g2 and g3 with themselves we get the remaining values have to be ±1. The sign is determined by taking the column orthogonality with g1 , i.e. X χ

χ(g1 )χ(g2 ) = 1 + 12 − 7 ± 3 ± 3 = 0,

Chapter 27

155 X

χ(g1 )χ(g3 ) = 1 − 7 ± 3 ± 3 = 0.

χ

Recall from the calculation in the chapter that χ(g5 ) is complex for some irreducible character χ. We assume this is χ5 and as χ5 is non-real we have χ6 = χ5 is another irreducible character. Also g6 = g5−1 ⇒ χi (g6 ) = χi (g5 ) 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 g5 with g3 we obtain 1 0 = 1 + η + η ⇒ 2 Re(η) = −1 ⇒ Re(η) = − . 2 Using the column orthogonality of g5 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 √ 7 1 7 2 2 . Re(η) + Im(η) = 2 ⇒ + Im(η) = 2 ⇒ Im(η) = ⇒ Im(η) = ± 4 4 2 2

2

Therefore we can take η =

√ −1+ 7i 2

and we’re done.