solutions to exercises of representations and characters of groups by gordon james and martin...
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7/28/2019 Solutions to Exercises of Representations and Characters of Groups by Gordon James and Martin Liebeck
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Solutions to Exercises of Representations and Characters of Groups by Gordon
James and Martin Liebeck
Solutions by Jay Taylor† (University of Aberdeen)Last Updated 22/07/2011
This document contains solutions to the exercises in the book “Rep-
resentations and Characters of Groups” written by Gordon James and
Martin Liebeck. Throughout we will freely use the notation used in
this book. All non explicit references will refer to results from the
book. A table of contents is given on the next page and at the
head of each chapter/section all exercises from that chapter/section
which have a complete solution are listed. There may be solutions
to other exercises in this document, (which are not contained in this
list), but such exercises are considered to have only a partial solution.
† email [email protected].
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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
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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
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Chapter 3. Group Representations
Exercise 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercise 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercise 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercise 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Exercise 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Exercise 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Exercise 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Exercise 3.1. Assume ρ is a representation of G then because am = 1 we must have
In = 1ρ = amρ = (aρ)m = Am. Now if Am = In then we have (ai )ρ = Ai for all i ∈ Z.
Therefore
(ai a j )ρ = ai + j ρ = Ai + j = Ai A j = (ai ρ)(a j ρ).
Exercise 3.2. We can see that
A3 = I2 B3 =
1 0
0 e 2πi/3
3=
1 0
0 e 2πi
= I2,
C 3 =
0 1
−1 −1
3=
0 1
−1 −1
0 1
−1 −1
0 1
−1 −1
=
−1 −1
1 0
0 1
−1 −1
= I2.
Hence ρ1, ρ2 and ρ3 are representations by the above. Clearly ρ1 is not faithful as Gρ1 = I2.
However ρ2 and ρ3 are faithful. We can see in the calculation of C 3 that C 2 = I2, hence
ρ3(g ) = I2
⇒g = 1. Also B2
= I2 as e 4πi/3
= 1 and so ρ2 is faithful.
Exercise 3.3. Now we know that every element of G can be written ai b j for some 0 i
n − 1 and 0 j 1, using the relations. We certainly have that (1)n = (−1)2 = (1) and
(−1)(1)(−1) = (1), so defining the map ρ : G → GL(1, F ) by (ai b j )ρ = (−1) j we have ρ
is a representation by Example 1.4 and aρ = (1), bρ = (−1) as required.
Exercise 3.4. We are showing that the definition of equivalent representations is an equiv-
alence relation.
Reflexive clearly we have gρ = I−1n (gρ)In for all g ∈ G so ρ is equivalent to ρ.
1
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Chapter 3 2
Symmetric assume ρ is equivalent to σ then there exists T such that, for all g ∈ G,
we have gσ = T −1(gρ)T . However this gives us gρ = (T −1)−1(gσ)T −1 and so σ is
equivalent to ρ.
Transitive assume ρ is equivalent to σ and σ is equivalent to τ then there exists T, S
such that, for all g ∈ G, we have gσ = T −1(gρ)T and gτ = S−1(gσ)S. Then this
gives us that, for all g ∈ G, we have gτ = S−1(T −1(gρ)T )S = (T S)−1(gρ)(T S). In
other words ρ is equivalent to τ .
Exercise 3.5. We first check that the matrices satisfy the relations of G. So,
A6 = e iπ/3 0
0 e −iπ/3
6
= e 2iπ 0
0 e −2iπ = I2
B2 =
0 1
1 0
0 1
1 0
= I2,
B−1AB =
0 1
1 0
e iπ/3 0
0 e −iπ/3
0 1
1 0
,
=
0 e −iπ/3
e iπ/3 0
0 1
1 0
,
= e −iπ/3 0
0 e iπ/3 ,
= A−1
C 2 =
1/2
√ 3/2
−√ 3/2 1/2
1/2
√ 3/2
−√ 3/2 1/2
=
−1/2
√ 3/2
−√ 3/2 −1/2
C 4 =
−1/2
√ 3/2
−√ 3/2 −1/2
−1/2
√ 3/2
−√ 3/2 −1/2
=
−1/2 −√
3/2√ 3/2 −1/2
C 6 = −1/2 −√ 3/2
√ 3/2 −1/2 −1/2√
3/2
−√ 3/2 −1/2 = 1 0
0 1D2 =
1 0
0 −1
1 0
0 −1
=
1 0
0 1
D−1CD =
1 0
0 −1
1/2
√ 3/2
−√ 3/2 1/2
1 0
0 −1
=
1/2 −√
3/2√ 3/2 1/2
= C −1
By Example 1.4 the above calculations are enough to show that ρ1 and ρ4 are representations
of D12. Let ar b s , ap b q ∈ D12 then we have, (recalling that the calculations above show that
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Chapter 3 3
the matrices respect the rules of D12),
(ar
b s
ap
b q
)ρ2 = (ar
−p
b s +q
)ρ2 (ar
b s
ap
b q
)ρ3 = (ar
−p
b s +q
)ρ,= A3(r −p )(−B)s +q = (−A)r −p Bs +q ,
= A3r (−1)s A−3p Bs (−B)q = (−A)r (−1)−p A−p Bs Bq ,
= A3r (−B)s A3p (−B)q = (−A)r Bs (−A)p Bq ,
= (ar b s )ρ2(ap b q )ρ2 = (ar b s )ρ3(ap b q )ρ3.
Recall that (−1)−p = (−1)p (−1)−2p = (−1)p [(−1)2]−p = (−1)p . Therefore ρ2 and ρ3 are
representations of D12.
We first of all consider the other powers of A and C . We can see the following holdtrue
A3 =
e iπ 0
0 e −iπ
= −I2 C 3 =
−1/2
√ 3/2
−√ 3/2 −1/2
1/2
√ 3/2
−√ 3/2 1/2
= −I2,
which also gives us C 5 = C 3C 2 = −C 2. Now if ai b j ∈ ker(ρ1) then we have (ai b j )ρ1 = I2
but this gives us Ai B j = I2 ⇒ Ai = B− j . However no power of A, other than 6, is equal
to I2 or B, therefore we must have the kernel is trivial and the representation is faithful. A
similar argument shows that if ai b j ∈ ker(ρ4) then we must have C i = D− j but no power of
C , other than 6, is equal to I2 or D so the representation is faithful. Clearly ρ2 is not faithfulas a2ρ2 = A6 = I2. Finally we also have ρ3 is not faithful as a3ρ3 = (−A)3 = −A3 = I2.
We have that the representations ρ1 and ρ4 are in fact equivalent. This is a little
more enlightening if we write e iπ/3 = 12
+√ 32 i . It is then maybe easier to see that A is
the diagonalised form of C . What are the eigenvalues of C ? Well this has characteristic
polynomial λ2−λ+ 1, whose solutions are precisely λ1 = e iπ/3 = 12
+√ 32
i and λ2 = e −iπ/3 =12
−√ 32 i . What are the corresponding eigenvectors?
1/2√
3/2
−√ 3/2 1/2 v 1
v 2 = 1
2 +
√ 3
2 i v 1
v 2 ⇒ −
√ 3/2i
√ 3/2
−√ 3/2 −√ 3/2i v 1
v 2 = 0
0 ,
⇒
i −1
1 i
v 1
v 2
=
0
0
.
Now the second line is just −i times the first and so we have v 2 = i v 1. Similarly for the
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Chapter 3 4
second we get v 2 = −i v 1. Then let T be the matrix
T = 1 1−i i ⇒ T −1 = 1
2i i −1
i 1 .
Using this we see that
T −1(bρ4)T =1
2i
i −1
i 1
1 0
0 −1
1 1
−i i
,
=1
2i
i 1
i −1
1 1
−i i
,
= 12i
0 2i 2i 0
,
= bρ1.
Similarly we have
T −1(aρ4)T =1
2i
i −1
i 1
1/2
√ 3/2
−√ 3/2 1/2
1 1
−i i
,
= 12i 1/2i + √ 3/2 −1/2 + √ 3/2i
1/2i − √ 3/2 1/2 +
√ 3/2i
1 1−i i ,
=1
2i
i +
√ 3 0
0 i − √ 3
,
= aρ1.
As the matrices are equivalent on the generators it is enough to see that in fact ρ4 is
equivalent to ρ1. None of the other representations are equivalent. This can be seen by
checking the characteristic polynomial of aρ2 and aρ3 and seeing that they are different,
(recall equivalent matrices have the same characteristic polynomial).
Exercise 3.6. We define the following permutation matrices in GL(3, F )
A =
0 1 0
0 0 1
1 0 0
B =
0 1 0
1 0 0
0 0 1
.
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Chapter 3 5
We check that these satisfy the relations of D8, so
A3 =0 1 0
0 0 1
1 0 0
0 1 00 0 1
1 0 0
0 1 00 0 1
1 0 0
=0 0 1
1 0 0
0 1 0
0 1 00 0 1
1 0 0
=1 0 0
0 1 0
0 0 1
,
B2 =
0 1 0
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
=
1 0 0
0 1 0
0 0 1
,
B−1AB =
0 1 0
1 0 0
0 0 1
0 1 0
0 0 1
1 0 0
0 1 0
1 0 0
0 0 1
=
0 0 1
0 1 0
1 0 0
0 1 0
1 0 0
0 0 1
=
0 0 1
1 0 0
0 1 0
= A−1
By Example 1.4 this means that the map ρ : D8 → GL(3, F ) given by (ai b j )ρ = Ai B j , for
all 0 i 3 and 0 j 1 is a representation. Is it faithful? Well if ai b j ∈ ker(ρ) then we
have Ai B j = I2 ⇒ Ai = B− j but no power of A is equal to I3 or B, except for A3 and so
our representation is faithful.
Exercise 3.7. Assume ρ is a representation of degree 1 then it is a map from G to
GL(1, F ) ∼= F . Therefore G/ ker(ρ) ∼= im(ρ) ⊆ F and as F is a field we have G/ ker(ρ) is
abelian.
Exercise 3.8. If we have (gρ)(hρ) = (hρ)(gρ) then this gives us that (gh)ρ = (hg )ρ.
However this only implies gh = hg if ρ is an isomorphism. A classic counter-example
is the trivial representation ρ : G → GL(1, F ) given by gρ = I for all g ∈ G. Clearly
(gρ)(hρ) = (hρ)(gρ) for all g, h ∈ G but G isn’t necessarily abelian so we don’t necessarily
have gh = hg .
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Chapter 4. F G-modules
Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Exercise 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Exercise 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Exercise 4.1. Let u 1 = v 1 + v 2 + v 3, u 2 = v 1 − v 2 and u 3 = v 1 − v 3. Then the change of
basis matrix from B 1 = v 1, v 2, v 3 to B 2 = u 1, u 2, u 3 is given as followsv 1
v 2
v 3
=
1
3
1 1 1
1 −2 1
1 1 −2
u 1
u 2
u 3
Now the matrices of S3 = 1, (12), (13), (23), (123), (132) with respect to the basis B 1are given by the standard permutation matrices and so
[1]B1=
1 0 0
0 1 0
0 0 1
[(12)]B1 =
0 1 0
1 0 0
0 0 1
[(13)]B1
=
0 0 1
0 1 0
1 0 0
,
[(23)]B1=
1 0 0
0 0 1
0 1 0
[(123)]B1
=
0 1 0
0 0 1
1 0 0
[(132)]B1
=
0 0 1
1 0 0
0 1 0
.
So to calculate the matrices in the new basis B 2 we just multiply by the change of basis
matrix on the right
[1]B2 =
1
31 1 1
1 −1 01 0 −1
1 0 0
0 1 00 0 1
1 1 1
1 −2 11 1 −2
= 1 0 0
0 1 00 0 1
,
[(12)]B2=
1
3
1 1 1
1 −1 0
1 0 −1
0 1 0
1 0 0
0 0 1
1 1 1
1 −2 1
1 1 −2
=
1 0 0
0 −1 0
0 −1 1
,
[(13)]B2=
1
3
1 1 1
1 −1 0
1 0 −1
0 0 1
0 1 0
1 0 0
1 1 1
1 −2 1
1 1 −2
=
1 0 0
0 1 −1
0 0 −1
,
6
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Chapter 4 7
[(23)]B2=
1
3
1 1 1
1 −1 0
1 0 −1
1 0 0
0 0 1
0 1 0
1 1 1
1 −2 1
1 1 −2
=
1 0 0
0 0 1
0 1 0
,
[(123)]B2=
1
3
1 1 1
1 −1 0
1 0 −1
0 1 0
0 0 1
1 0 0
1 1 1
1 −2 1
1 1 −2
=
1 0 0
0 −1 1
0 −1 0
,
[(132)]B2=
1
3
1 1 1
1 −1 0
1 0 −1
0 0 1
1 0 0
0 1 0
1 1 1
1 −2 1
1 1 −2
=
1 0 0
0 0 −1
0 1 −1
.
What we notice about the matrices in the basisB 2 is that they fall into block form, namely
a 1 × 1 block and a 2 × 2 block. In Chapter 5 we will come to see that this means the
representation is reducible.
Exercise 4.2. We check the definition for V to be an F G module. The following statements
are for all v ∈ V .
(a) We clearly have v g ∈ V as ±v ∈ V because V is a vector space.
(b) Let g, h ∈ Sn then we have
v (gh) = v if gh is even,−v if gh is odd.
=
v if g is even and h is even,
−v if g is odd and h is even,
−v if g is even and h is odd,
v if g is odd and h is odd.
= v g if h is even,
−v g if h is odd.
= (v g )h.
(c) Recall 1 ∈ Sn is even so we clearly have v 1 = v .
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Chapter 4 8
(d) Let λ ∈ F then
(λv )g =λv if g is even,
−λv if g is odd.= λ
v if g is even,−v if g is odd.
= λ(v g ).
(e) For all u ∈ V we have
(u + v )g =
u + v if g is even,
−u − v if g is odd.
= u if g is even,
−u if g is odd.
+ v if g is even,
−v if g is odd.
= ug + v g.
Therefore V certainly is an F G-module.
Exercise 4.3. We check the conditions of Proposition 4.6 to show that this is indeed an
RQ8-module.
(a) Clearly we have v i g ∈ V for all 1 i 4 and g ∈ G.
(b) This is clear as we’re defining the action of of Q8 on V by acting one letter at a time,
i.e. v k (ab ) = (v k a)b .
(c) We have that 1 = a4 and so we can see that the following hold
v 11 = v 1a4 = −v 2a3 = v 1a2 = −v 2a = v 1,
v 21 = v 2a4 = −v 1a3 = v 2a2 = −v 1a = v 2,
v 31 = v 3a4 = −v 4a3 = −v 3a2 = v 4a = v 3,
v 41 = v 4a4 = v 3a3 = −v 4a2 = −v 3a = v 4.
(d) This is clear by how we’re defining the module.
Therefore, by Proposition 4.6, we have that V is an RQ8 module.
Exercise 4.4. Let E ij be the standard basis for the vector space of all matrices over F .
Then for some aij ∈ F and σ ∈ Sn we have
A =
ni,j =1
aij E ij B =
ni,j =1
aij E σ(i ) j .
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Chapter 4 9
Now consider the matrix P defined by P =n
k =1 E σ(k )k . We can see that this is a permu-
tation matrix, as there is only one non-zero entry in each row and column. Now, this gives
us that
P A =
nk =1
E σ(k )k
ni ,j =1
ai j E ij =
ni,j,k =1
aij E σ(k )k E ij =
ni,j,k =1
δik aij E σ(k ) j =
ni,j =1
aij E σ(i ) j = B.
Similarly if τ ∈ Sn such that B =n
i ,j =1 ai j E iτ ( j ), then we define a permutation matrix
P =n
k =1 E kτ (k ) which gives us
AP =
n
i,j =1aij E ij
n
k =1E kτ (k ) =
n
i,j,k =1aij E ij E kτ (k ) =
n
i,j,k =1δ jk aij E iτ (k ) =
n
i ,j =1ai j E iτ ( j ) = B.
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Chapter 5. F G-submodules and reducibiity
Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Exercise 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Exercise 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Exercise 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Exercise 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Exercise 5.1. It is clear that the properties of Proposition 4.6 hold and so V is an F G-
module. Now V is of dimension 2 so there is only one possible non-trivial F G submoduleof dimension 1. Consider U = span(α, β) ⊆ V , when is this an F G submodule of V ?
It’s clearly a 1-dimensional subspace of V as (0, 0) ∈ U and for all k 1, k 2 ∈ F we have
k 1(α, β) + k 2(α, β) = (k 1 + k 2)(α, β) ∈ U .
What about the action of G on U ? Well we have (α, β)1 ∈ U and (α, β)a = ( β, α) ∈U if ( β, α) = k (α, β) for some k ∈ F . We can see that letting k = ±1 turns U
into an F G module and so U is either span(1, 1) or span(1, −1). Hence U is an
F G submodule of V . Now as vector spaces are classified by their dimension we have
0
, span
(1, 1)
, span
(1,
−1)
and V are all the F G submodules of V .
Exercise 5.2. Assume ρ : G → GL(n, F ) is a reducible representation, then the corre-
sponding F G-module F n given by v g = v (gρ) for all v ∈ F n and g ∈ G is reducible. In
other words there exists a proper submodule W of V . Now assume σ : G → GL(n, F ) is
equivalent to ρ, then there exists a matrix T ∈ GL(n, F ) such that gσ = T −1(gρ)T for all
g ∈ G.
We claim that W T is also a proper submodule of F n under σ, in other words (w T )g =
(w T )(gσ) ∈ W T for all w T ∈ W T and g ∈ G. This is clear to see as for g ∈ G and
w ∈ W we have
(w T )g = (w T )(gσ) = (w T )(T −1(gρ)T ) = w (gρ)T ⊆ W T
because w (gρ) ∈ W . Therefore W T is a submodule of F n under σ. Note that because T
is invertible we must have the dimension of W T is the same as the dimension of W . In
other words W = 0 or F n implies W T = 0 or F n and we’re done.
Exercise 5.3. We break the four representations down individually. In each case we’re
looking for a 1-dimensional submodule of C2, lets call it W with basis element w = (α, β).
10
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Chapter 5 11
Recall that e ikπ/3 = e −ikπ/3 unless e ikπ/3 is real, in other words k = 3n for some n ∈ Z.
(a) Let H = b G the cyclic subgroup of order 2. Then C2 is also a CH -module.
Consider U to be a 1-dimensional submodule of C2 then by question 1 we haveeither (1, 1) ∈ U or (1, −1) ∈ U . However we have (1, 1) and (1, 1)a are linearly
independent, as are (1, −1) and (1, −1)a. Therefore dim U 2 and so U cannot be
a 1-dimensional submodule. Hence ρ1 is an irreducible representation.
(b) We have the actions of a and b are
w a =
α β −1 0
0 −1
=−α − β
= −w ∈ W,
w b = α β 0 11 0
= β α ∈ W ⇒ β = kβ,
for some k ∈ C. Therefore W = span(1, 1) and W = span(1, −1) are submod-
ules of C2, which means ρ2 is reducible.
(c) We can see that ρ3 is irreducible by a virtually identical argument to that of ρ1.
(d) We have that ρ4 is equivalent to ρ1, therefore ρ1 irreducible implies ρ4 irreducible.
Exercise 5.4. Let G = (123), (456), (23)(45).
(a) We can see that
a3 = (123)3 = (132)(123) = 1,
b 3 = (456)3 = (465)(456) = 1,
c 2 = (23)(45)(23)(45) = 1,
ab = (123)(456) = (456)(123) = ba,
c −1ac = (23)(45)(123)(23)(45) = (132) = a−1,
c −1bc = (23)(45)(456)(23)(45) = (465) = b −1.
In a similar fashion to the dihedral groups this gives us that any element of G can be
written as ai b j c k with 0 i , j 2 and 0 k 1. Therefore |G| 18. However
|a, b | = 9 and by Lagrange’s theorem we have 9 | |G| but a, b = G and so we
must have |G| = 18.
(b) Let ω be a primitive cube root of unity, i.e. ω3 = 1 and ω = 0, ω2 = 0. Then all the
primitive cubed roots of unity are 1, ω and ω2. Let ε = ω and η = ω2 then we have ρ
is a representation of G. As in previous cases, this is true if the matrices satisfy the
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Chapter 5 12
relations of the group. So,
(aρ)3 = ω 00 ω−1
3 = ω3
00 ω−3
= I2 (bρ)3 = ω2
00 ω−2
3 = ω6
00 ω−6
= I2,
(cρ)2 =
0 1
1 0
0 1
1 0
= I2,
(cρ)−1(aρ)(cρ) =
0 1
1 0
ω 0
0 ω−1
0 1
1 0
,
=
0 ω−1
ω 0
0 1
1 0
,
=
ω−1 0
0 ω
,
= (aρ)−1,
(cρ)−1(bρ)(cρ) =
0 1
1 0
ω2 0
0 ω−2
0 1
1 0
,
=
0 ω−2
ω2 0
0 1
1 0
,
=ω−2 0
0 ω2
,
= (aρ)−1.
Hence it is a representation.
(c) We must have that ε and η are cubed roots of unity because ε3 = η3 = 1. If these
are not primitive then, for example, we have ε = 1 or ε2 = 1 which gives us a or
a2 ∈ ker(ρ). Hence ρ is not faithful. Now assume that they are primitive cubed roots
of unity as above. Then we have (aρ)(bρ) = I2
, which means ab ∈
ker(ρ). Therefore
there are no values for which ρ is a faithful representation.
(d) Consider the action of ρ on C2. We want to find a non-trivial, in other words 1-
dimensional, submodule of C2 say U . Consider the cyclic subgroup H = c G of
order 2. Now if U is a CG submodule then it’s also a CH module and by question 1
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Chapter 5 13
we must have that U contains either u 1 = (1, 1) or u 2 = (1, −1).
u 1a = 1 1 ε 00 ε−1
= ε ε−1 ∈ U ⇒ ε = 1,
u 1b =
1 1 η 0
0 η−1
=
η η−1
∈ U ⇒ η = 1,
u 2a =
1 −1 ε 0
0 ε−1
=
ε −ε−1
∈ U ⇒ ε = 1,
u 2b =
1 −1
η 0
0 η−1
=
η −η−1
∈ U ⇒ η = 1.
Therefore we have ρ is irreducible as long as (ε, η) = (1, 1). If ε = η = 1 then the
subspace U = span(1, 1) is a submodule of C2.
Exercise 5.5. Let V = 0 and 0g = 0 for all g ∈ G. Then V is a CG-module which is
neither reducible nor irreducible.
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Chapter 6. Group algebras
Exercise 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Exercise 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Exercise 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Exercise 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Exercise 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Exercise 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Exercise 6.1. Let G = D8 = a, b | a4 = b 2 = 1, b −1ab = a−1.(a) We have x = a + 2a2 and y = b + ab − a2, which gives us
xy = (a + 2a2)(b + ab − a2),
= −2 − a3 + ab + 3a2b + 2a3b,
y x = (b + ab − a2)(a + 2a2),
= ba + 2ba2 + aba + 2aba2 − a3 − 2a4,
= a3b + 2a2b + b + 2a3b − a3 − 2,
= −2 − a3 + b + 2a2b + 3a3b,
x 2 = (a + 2a2)(a + 2a2),
= a2 + 2a3 + 2a3 + 4a4,
= 4 + a2 + 4a3.
(b) Recall that G = 1, a , a2, a3,b,ab,a2b, a3b . Then if z = b + a2b we have
z a = (b + a2b )a = ba + a2ba = a3b + ab = a(a2b + b ) = az ,
z b = (b + a
2
b )b = b
2
+ a
2
b
2
= b
2
+ ba
2
b = b (b + a
2
b ) = bz ,z a2 = (z a)a = a(z a) = a2z ,
za3 = az a2 = a2z a = a3z ,
z ab = az b = abz,
z a2b = azab = a2zb = a2bz ,
z a3b = az a2b = a2z ab = a3z b = a3bz .
Now as the elements of G form a basis for the group algebra, it is clear that z r = r z
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Chapter 6 15
for all r ∈ CG.
Exercise 6.2. Let G = C 2×
C 2 =
(x i , y j )|
0 i , j 1 and x 2 = y 2 = 1
. Consider the
action of C 2 × C 2 on the group algebra basis, then we have
(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(1, 1) = λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ),
(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(1, y ) = λ2(1, 1) + λ1(1, y ) + λ4(x, 1) + λ3(x, y ),
(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(x, 1) = λ3(1, 1) + λ4(1, y ) + λ1(x, 1) + λ2(x, y ),
(λ1(1, 1) + λ2(1, y ) + λ3(x, 1) + λ4(x, y ))(x, y ) = λ4(1, 1) + λ3(1, y ) + λ2(x, 1) + λ1(x, y ).
So with respect to the basis G of F G we can see that the matrices of the regular represen-
tation are going to be
[(1, 1)]G =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
[(1, y )]G =
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
,
[(x, 1)]G =
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
[(x, y )]G =
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
.
Exercise 6.3. In short, no. Take r = 1−a and s = 1+ a then we have r s = (1−a)(1+a) =
1 + a − a − a2 = 0. In other words, the group algebra is not in general an integral domain.
Exercise 6.4. We let G = g 1, . . . , g n be a finite group and c =n
i =1 g i .
(a) Let h ∈ G, then h = g j for some 1 j n. Recall that given any 1 i n there
exists a unique 1 k, n such that g i g j = g k and g j g i = g . Then we have
ch = n
i =1
g i g j =
n
i =1
g i g j =
n
i =1
= g k = c,
hc = g j
n
i =1
g i
=
ni =1
g j g i =
n=1
= g = c.
(b) Therefore we have
c 2 = (g 1 + · · · + g n)c = c + · · · + c n times
= |G|c.
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Chapter 6 16
(c) So let’s examine what’s happening to the basis G. We have g i ϑ = g i c = c for all
1 i n. Therefore the matrix of ϑ with respect to G is
[ϑ]G =
1 . . . 1
.... . .
...
1 . . . 1
.
Exercise 6.5. Recall that as a consequence of the definition of an F G module we have
0g = (u − u )g = ug − ug = 0 for all g ∈ G. Therefore, let r =n
i =1 αi g i ∈ F G then we
have
0r = 0n
i =1
αi g i =
n
i =1
0(αi g i ) =
n
i =1
(0αi )g i =
n
i =1
0g i = 0.
Also, for any v ∈ V we have v 0 = v (1 − 1) = v 1 − v 1 = v − v = 0.
Let G = 1, g 2, . . . , g n, with n > 1, and let V = span1 ⊆ F G be a 1-dimensional
vector subspace of the group algebra, such that 1g i = −1 for all 2 i n and 1.1 = 1.
Then this is certainly a submodule of F G. However we have 1(1+g i ) = 1.1+1g i = 1−1 = 0
but neither 1 nor 1 + g i are zero.
Exercise 6.6. We have to check that W is a sub module. Let w 1 = 1 + ω2a + ωa2 and
w 2 = b + ω2ab + ωa2b We have G =
1, a , a2,b,ab,a2b
and we can see that
w 1a = a + ω2a2 + ω w 2a = ba + ω2aba + ωa2ba,
= ω(1 + ω2a + ωa2) = ω2(b + ω2ab + ωa2b ),
w 1a2 = a2 + ω2 + ωa w 2a2 = ba2 + ω2aba2 + ωa2ba2,
= ω2(1 + ω2a + ωa2) = ω(b + ω2ab + ωa2b ),
w 1b = b + ω2ab + ωa2b w 2b = b 2 = ω2ab 2 + ωa2b 2,
= 1(b + ω2ab + ωa2b ) = 1(1 + ω2a + ωa2),
w 1
ab = (ab + ω2a2b + ωb ) w 2
ab = bab + ω2abab + ωa2bab,
= ω(b + ω2ab + ωa2b ) = ω2(1 + ω2a + ωa2),
w 1a2b = (a2b + ω2b + ωab ) w 2a2b = ba2b + ω2aba2b + ωa2ba2b,
= ω2(b + ω2ab + ωa2b ) = ω(1 + ω2a + ωa2).
Hence W is a 2-dimensional submodule of CG. We check its irreducibility, to show this
we just have to show that there is no 1-dimensional submodule of W . Let 0 = U =
spanαw 1 + βw 2 be a 1-dimensional vector subspace of W . How does G act on the basis
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Chapter 6 17
of U ? Well, recalling ω2 = ω,
(αw 1 + βw 2)a = αw 1a + βw 2a = αωw 1 + βω2
w 2 ∈ U ⇒ α = 0 or β = 0,(αw 1 + βw 2)b = αw 1b + βw 2b = αw 2 + βw 1 ∈ U ⇒ β = α,
for some ∈ C. However if both conditions are satisfied then α = β = 0 and U is the 0module. Hence W is an irreducible module.
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Chapter 7. F G-homomorphisms
Exercise 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Exercise 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Exercise 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Exercise 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Exercise 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Exercise 7.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Exercise 7.1. For all u 1, u 2 ∈ U, λ ∈ F and g ∈ G we have
(u 1 + u 2)(ϑφ) = ((u 1 + u 2)ϑ)φ = (u 1ϑ + u 2ϑ)φ = (u 1ϑ)φ + (u 2ϑ)φ = u 1(ϑφ) + u 2(ϑφ),
(λu 1)(ϑφ) = ((λu 1)ϑ)φ = (λ(u 1ϑ))φ = λ((u 1ϑ)φ) = λ(u 1(ϑφ)),
(ug )(ϑφ) = ((uϑ)g )φ = (uϑφ)g.
because uϑ ∈ V and φ is an F G-homomorphism. Therefore ϑφ is an F G homomorphism.
Exercise 7.2. Let V = spanv 1, . . . , v 5 be the permutation module defined by v i g = v ig
for all g ∈ (12345) S5. Let σ = (12345) and define a map ϑ : V → F G byv i ϑ = σi and extend linearly, we claim this is an F G-isomorphism. It is clear to see that
(v i σ)ϑ = v i +1ϑ = σi +1 = σi σ = (v i ϑ)σ. As σ generates G we have that ϑ is an F G-
homomorphism. It’s clear that it’s then a bijection.
Exercise 7.3. We first show that V 0 is a vector subspace of V . We clearly have 0 ∈ V 0
as 0g = 0 for all g ∈ G. Now if v , v ∈ V 0 and λ1, λ2 ∈ F then for all g ∈ G we
have (λ1v 1 + λ2v 2)g = (λ1v 1)g + (λ2v 2)g = λ1(v 1g ) + λ2(v 2g ) = λ1v 1 + λ2v 2 and so
λ1v 1 + λ2v 2 ∈ V 0, hence V 0 is a vector subspace of V . It’s clearly then closed under the
action of G by definition and so is an F G-submodule.Now let v 1, v 2 ∈ V , λ ∈ F and h ∈ H then we have
(v 1 + v 2)ϑ =g ∈G
(v 1 + v 2)g =g ∈G
v 1g +g ∈G
v 2g = v 1ϑ + v 2ϑ,
(λv 1)ϑ =g ∈G
(λv 1)g =g ∈G
λ(v 1g ) = λg ∈G
v 1g = λ(v 1ϑ),
(v 1h)θ =g ∈G
v 1gh =
g ∈G
v 1g
h = (v 1ϑ)h.
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Chapter 7 19
It’s clear that the image is contained in V 0 as summing over gh ∈ G is the same as summing
over g ∈ G. Let v 0 ∈ V 0 then we havev 0|G|
ϑ =
g ∈G
v 0g
|G| =g ∈G
v 0|G| = v 0,
which gives us the map is surjective.
Exercise 7.4. Let φ : V → W be the isomorphism between V and W . Now let w 0 ∈ W 0 then
there exists v ∈ V such that w 0 = v φ. We have, for all g ∈ G, that w 0g = w 0 ⇒ (v φ)g =
v φ ⇒ v g = v ⇒ v ∈ V 0 because φ is an isomorphism. So W 0 ⊆ V 0φ and for v 0 ∈ V 0 we
have, for all g ∈ G, that (v 0φ)g = (v 0g )φ = v 0φ. Therefore V 0φ ⊆ W 0 ⇒ W 0 = V 0φ, which
gives us the restriction map φ|V 0 : V 0 → W 0 is an isomorphism.
Exercise 7.5. Let v 1, v 2, v 3, v 4 be the standard basis for the permutation module V . Then
we have that
v 1(12) = v 2 v 2(12) = v 1 v 3(12) = v 3 v 4(12) = v 4,
v 1(34) = v 1 v 2(34) = v 2 v 3(34) = v 4 v 4(34) = v 3,
v 1(12)(34) = v 2 v 2(12)(34) = v 1 v 3(12)(34) = v 4 v 4(12)(34) = v 3.
These calculations tell us that V 0 = spanv 1 + v 2, v 3 + v 4. Alternatively, by exercise 3, wehave that (F G)0 = spang ∈G g . By exercise 4, we have that if V and F G are isomorphic
then V 0 and (F G)0 are isomorphic. However V has dimension 2 and F G has dimension 1,
so they can’t be isomorphic.
Exercise 7.6. Let G = x | x 2 = 1.
(a) We check the properties of an F G-homomorphism. For all (α1+ βx ), ( γ 1+δx ) ∈ F G,
λ ∈ F we have
((α1 + βx ) + ( γ 1 + δx ))ϑ = ((α + γ )1 + ( β + δ)x )ϑ,
= (α + γ − β − δ)(1 − x ),
= (α − β)(1 − x ) + ( γ − δ)(1 − x ),
= (α1 + βx )ϑ + ( γ 1 + δx )ϑ.
(λ(α1 + βx ))ϑ = ((λα)1 + (λβ)x )ϑ ((α1 + βx )x )ϑ = ( β1 + αx )ϑ,
= (λα − λβ)(1 − x ) = ( β − α)(1 − x ),
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Chapter 7 20
= λ(α − β)(1 − x ) = (α − β)(x − 1),
= λ((α1 + βx )ϑ) = (α
− β)(1
−x )x,
= (α1 + βx )ϑx.
Therefore ϑ is an F G-homomorphism.
(b) Now for any α1 + βx ∈ F G we have
(α1 + βx )ϑ2 = ((α − β)(1 − x ))ϑ,
= (α − β)((1 − x )ϑ),
= (α − β)(1 − (−1))(1 − x ),
= 2(α − β)(1 − x ),
= 2((α1 + βx )ϑ).
Therefore ϑ2 = 2ϑ.
(c) What is [ϑ] with respect to the standard basis G? This is
[ϑ]G =
1 −1
−1 1
.
What are the eigenvalues of this matrix? There the solutions of the equation (1 −λ2) − 1 = λ(λ − 2) = 0, which are λ = 0 and λ = 2. Calculating the eigenvector
with respect to λ = 2 gives us
v 1 v 2
1 −1
−1 1
= 2
v 1 v 2
⇒
v 1 v 2
−1 −1
−1 −1
=
0 0
⇒ v 2 = −v 1.
Clearly any vector is an eigenvector of λ = 0. Therefore we have that B = 1−x , 1 +
x form a basis such that the matrix [ϑ]B has the required form. Note that we can
see that by setting
T =
1 −1
1 1
and T −1 =
1
2
1 1
−1 1
we have
T [ϑ]GT −1 =1
2
−1 1
1 1
1 −1
−1 1
1 1
−1 1
,
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Chapter 7 21
=1
2
2 −2
0 0
1 1
−1 1
,
=
2 0
0 0
,
= [ϑ]B
Then T is our change of basis matrix and we have
1 0
1 −1
1 1
=
1 −1
,
0 1 1 −1
1 1
=
1 1
.
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Chapter 8. Maschke’s Theorem
Exercise 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Exercise 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Exercise 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Exercise 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Exercise 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Exercise 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Exercise 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Exercise 8.1. Let B = v 1, v 2 then we can see that the following is true
[x ]B =
0 1
−1 −1
.
We now try and diagonalise this matrix. What are the eigenvalues? Well these are the
solutions of the characteristic polynomial −λ(−1 − λ) + 1 = λ2 + λ + 1 = 0, which gives
us solutions λ1 = −e iπ/3 or λ2 = −e −iπ/3. Therefore the eigenvectors for λ1 and λ2 have
the form
v 1 v 2
0 1
−1 −1
= λi
v 1 v 2
⇒
v 1 v 2
−λi 1
−1 −1 − λi
=
0 0
,
⇒−λi v 1 = v 2,
v 1 + (−1 − λi )v 2 = 0,
⇒ (1 + λ + λ2)v 1 = 0,
which means v 1 can be chosen arbitrarily in each case. Therefore we have a direct sum
V = spanv 1 − λ1v 2 ⊕ spanv 1 − λ2v 2.
Exercise 8.2. Now G = 1, x , y , x y ∼= C 2 × C 2 such that x 2 = y 2 = 1 and xy = y x . We
want to find all 1-dimensional submodules of RG, say U . Now taking G as the standard
basis we must have that the basis element of a 1-dimensional submodule has the form
α1 + βx + γy + δx y for some α, β , γ, δ ∈ R. We consider the action of the group on this
basis element
(α1 + βx + γy + δx y )1 = α1 + βx + γy + δxy,
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Chapter 8 23
(α1 + βx + γy + δx y )x = β1 + αx + δy + γxy,
(α1 + βx + γy + δx y ) y = γ 1 + δx + αy + βxy,
(α1 + βx + γy + δx y )xy = δ1 + γx + βy + αxy.
Clearly the first element will always lie in U . For the rest of the elements to lie in U we
must have there exists r 1, r 2 and r 3 ∈ R such that
α = r 1 β = r 2 γ = r 3δ
β = r 1α = r 2δ = r 3 γ
γ = r 1δ = r 2α = r 3 β
δ = r 1 γ = r 2 β = r 3α
⇒ r 21 = r 22 = r 23 = 1.
Now the only linearly independent solutions to these equations are the quadruples (1, 1, 1, 1),
(1, 1, −1, −1), (1, −1, 1, −1) and (1, −1, −1, 1). Therefore we have
RG = span1 + x + y + x y ⊕ span1 + x − y − x y ⊕ span1 − x + y − xy ⊕ span1 − x + y − xy .
Note that we could also have got this information by finding all the eigenvectors of x in the
regular representation.
Exercise 8.3. Let G be any non-trivial finite group and consider a 2-dimensional vector
subspace V = spanv 1, v 2 ⊆ CG. By defining v g = v for all v ∈ V and g ∈ G we have V
is a submodule of CG. Then define a homomorphism ϑ : V → V by (αv 1 + βv 2)ϑ = αv 2.
Now this is a CG homomorphism because
(α1v 1 + β1v 2 + α2v 1 + β2v 2)ϑ = α1v 2 + α2v 2 = (α1v 1 + β1v 2)ϑ + (α2v 1 + β1)ϑ,
(λ(αv 1 + βv 2))ϑ = λ(αv 2) = λ((αv 1 + βv 2)ϑ),
((αv 1 + βv 2)g )ϑ = (αv 1 + βv 2)ϑ = αv 2 = (αv 2)g = (αv 1 + βv 2)ϑg.
However we can see that ker(ϑ) = im(ϑ) = spanv 2.
Exercise 8.4. Assume ρ : G → GL(2,C) is reducible. Then by the matrix form of
Maschke’s theorem we have, for all g ∈ G
gρ =
λg 0
0 µg
,
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Chapter 8 24
for some suitable λg , µg ∈ C determined by g . Now if this is true then (gρ)(hρ) = (hρ)(gρ)
for all g, h ∈ G because diagonal matrices commute but this implies G is abelian. Therefore
we must have ρ is irreducible.
Exercise 8.5. Consider the subspace U = span(1, 0) ⊆C2. Now clearly we have
1 0
1 0
n 1
=
1 0
,
which means U is a 1-dimensional submodule of C2. Let’s try and find a submodule U such
that C2 = U ⊕ U . Now any vector (v 1, v 2) ∈ C2 will be linearly independent of (1, 0) if
v 2
= 0, so consider U = span
(v 1, v 2)
. This gives us
v 1 v 2
1 0
n 1
=
v 1 + nv 2 v 2
.
If this is in U then there exists k ∈ C with v 1 = k (v 1 + nv 2) and v 2 = kv 2. The second
condition gives k = 1 but if this is the case then the first condition gives us nv 2 = 0 ⇒ n = 0.
Hence there can be no submodule U and so C2 is not completely reducible.
Exercise 8.6. Let ( , ) be a complex inner product on U × V .
(a) We first check that [ , ] is also a complex inner product on U
×V . By definition 14.2
this involves checking the following, for all u 1, u 2 ∈ U , v ∈ V and λ1, λ2 ∈ C
[u 1, v ] =x ∈G
(u 1x , v x ) =x ∈G
(v x , u 1x ) =x ∈G
(v x , u 1x ) = [v x , u 1x ],
[λ1u 1 + λ2u 2, v ] =x ∈G
((λ1u 1 + λ2u 2)x , v x ) = λ1[u 1, v ] + λ2[u 2, v ],
[u, u ] =x ∈G
(ux,ux ) >0
> 0 if u = 0.
(b) We first show a property of the inner product. For any g ∈ G we have
[ug,vg ] =x ∈G
(uxg,vxg ) =
x =xg ∈G(ux , v x ) = [u, v ].
Therefore, for every g ∈ G and v ∈ U ⊥ we have
[u , v g ] = [ug −1, v g g −1] = [ug −1, v ] = 0
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Chapter 8 25
because v ∈ U ⊥ and ug −1 ∈ U because U is a submodule. Hence U ⊥ is indeed a
submodule of V .
(c) Clearly Maschke’s thereom holds, as given any submodule U of V we have U ⊥ is also
a submodule such that V = U ⊕ U ⊥.
Exercise 8.7. By Maschke’s theorem we have the group algebra is completely reducible
and so CG = V 1 ⊕ · · · ⊕ V k for some irreducible CG modules. Now CG is faithful as a CG
module, therefore we have v x = v for some v ∈ V i . Consider the following set
K i = x ∈ G | v x = v for all v ∈ V i ,
we claim this is a normal subgroup of G. Clearly 1 ∈ K as v 1 = v for all v ∈ V i , also forall x, y ∈ K i we have v (xy ) = (v x ) y = v y = v and so x y ∈ K i . Finally if x ∈ K i then
v x = v ⇒ v = v x −1 and so x −1 ∈ K i , which gives us K is a subgroup. If x ∈ K i and
g ∈ G then v (g −1x g ) = (v g −1)xg = v (g −1g ) = v and so K i is a normal subgroup of G.
However we clearly have K i = G and so, as G is simple, we have K i = 1 which means V i
is a faithful irreducible CG submodule.
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Chapter 9. Schur’s Lemma
Exercise 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Exercise 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Exercise 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Exercise 9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Exercise 9.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Exercise 9.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Exercise 9.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Exercise 9.1. For C 2 = x | x 2 = 1 we have the irreducible representations over C are
given by ρi : C 2 → GL(1,C) such that
x ρ1 = (1) x ρ2 = (−1).
Let ω = e 2πi/3. For C 3 = y | y 3 = 1, the irreducible representations over C are given by
ρi : C 3 → GL(1,C) such that
y ρ1 = (1) y ρ2 = (ω) y ρ3 = (ω2).
For C 2 × C 2 ∼= a, b | a2 = b 2 = 1 and ab = ba we have the irreducible representations
over C are given by ρi : C 2 × C 2 → GL(1,C) such that
xρ1 = (1) xρ2 = (1) xρ3 = (−1) xρ4 = (−1),
y ρ1 = (1) y ρ2 = (−1) y ρ3 = (1) y ρ4 = (−1).
Exercise 9.2. We have G = C 4 × C 4 ∼= a4 = b 4 = 1 and ab = ba.
(a) Recall that every irreducible representation of C 4 = x | x 4 = 1 over C is given byρi : C 4 → GL(1,C) such that x ρ1 = (1), xρ2 = (i ), xρ3 = (−1) and xρ4 = (−i ).
Therefore we have ρ : G → GL(1,C) given by aρ = (−1) and bρ = (−1) is an
irreducible representation of G over C. Now every element in G is of the form ai b j
and so clearly
(ai b j )2ρ = (a2)i (b 2) j ρ = (a2ρ)i (b 2ρ) j = (1)i (1) j = (1).
26
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Chapter 9 27
(b) The elements of order 2 in G are 1, a2, b 2 and a2b 2. If σ existed then a2σ = b 2σ =
(−1) but (a2b 2)σ = (a2σ)(b 2σ) = (−1)(−1) = (1). Hence σ cannot exist.
Exercise 9.3. Let g 1, . . . , g r generate G and let ξ1, . . . , ξr be distinct primitive ni th roots
of unity. Then the following map ρ : G → GL(r,C) given by
(g i 11 . . . g i r r )ρ =
ξi 11 . . . 0
.... . .
...
0 . . . ξi r r
is a faithful irreducible representation of G.
Yes. Let r = 2 and n1 = n2. Then let ξ1 and ξ2 be primitive n1th and n2th roots of unity
respectively. Then if g 1, g 2 generate G = C n1 × C n2 we have the map τ : G → GL(1,C)given by (g i 11 g i 22 )τ = (ξi 11 ξi 22 ) is a faithful irreducible representation of degree 1 < r = 2.
Exercise 9.4. We have this is a representation if the matrices satisfy the relations of the
group. It is quick to check that
(aρ)4 =
−7 10
−5 7
4=
−1 0
0 −1
2= I2,
(bρ
)2 = −5 6
−4 5−5 6
−4 5 =I2,
(bρ)−1(aρ)(bρ) =
−5 6
−4 5
−7 10
−5 7
−5 6
−4 5
,
=
5 −8
3 −5
−5 6
−4 5
,
=
7 −10
5 −7
,
= (aρ)−1
.
Therefore ρ is indeed a representation of D8. Now consider a matrix which commutes with
gρ for all g ∈ D8, then in particular we havea b
c d
−7 10
−5 7
=
−7 10
−5 7
a b
c d
⇒−7a − 5b 10a + 7b
−7c − 5d 10c + 7d
=
−7a + 10c −7b + 10d
−5a + 7c −5b + 7d
,
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Chapter 9 28
a b
c d
−5 6
−4 5
=
−5 6
−4 5
a b
c d
⇒−5a − 4b 6a + 5b
−5c − 4d 6c + 5d
=
−5a + 6c −5b + 6d
−4a + 5c −4b + 5d
.
This tells us that −5b = 10c and −4b = 6c , which means b = c = 0. Using this we have
10a = 10d ⇒ a = d and so the matrix M is just a scalar multiple of the identity. In other
words, ρ is an irreducible representation of D8.
We now do exactly the same for the representation σ of D8. We first check that this is
a representation by checking the relations.
(aσ)4 =5 −6
4 −5
4=1 0
0 1
2= I2,
(bρ)2 =
−5 6
−4 5
−5 6
−4 5
= I2,
(bρ)−1(aρ)(bρ) =
−5 6
−4 5
5 −6
4 −5
−5 6
−4 5
,
= −1 0
0−
1−5 6
−4 5 ,
=
5 −6
4 −5
,
= (aρ)−1.
We comment on how many unique matrices there are. Now a2σ = I2 and a3σ = aρ. Also
we can see that bρ = −aρ and a−1ρ = aρ. Therefore we only have one condition on a
matrix which commutes with gσ for every g ∈ G. In particular this is
a b c d
5 −64 −5
=5 −6
4 −5
a b c d
⇒ 5a + 4b −6a − 5b 5c + 4d −6c − 5d
=5a − 6c 5b − 6d
4a − 5c 4b − 5d
.
This means the only conditions on our matrix are that 4b = −6c and 10b = 6(d − a).
Therefore the following matrix commutes with all (gσ)10 −6
4 0
,
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Chapter 9 29
which means σ is not an irreducible representation.
Exercise 9.5. Clearly z = g ∈Gg is in the centre of CG. By Proposition 9.14 we have
there exists λ ∈ C such that v z = λv for all v ∈ V .
Exercise 9.6. Let G = D6 = a, b | a3 = b 2 = 1 and bab = a−1.
(a) We want to show that g (a + a−1) = (a + a−1)g for g = a or b , then it follows that
a + a−1 ∈ Z (CG). We have that
a(a+a−1) = a2+1 = (a+a−1)a b (a+a−1) = ba+ba−1 = a−1b +ab = (a+a−1)b.
(b) Consider u 1 = 1 + ω
2
a + ωa
2
and u 2 = b + ω
2
ab + ωa
2
b then we have
u 1(a + a−1) = (a + a−1) + ω2a(a + a−1) + ωa2(a + a−1),
= (ω + ω2)1 + (1 + ω)a + (1 + ω2)a2,
= (ω + ω2)u 1,
u 2(a + a−1) = b (a + a−1) + ω2ab (a + a−1) + ωa2b (a + a−1),
= (ω + ω2)b + (1 + ω)ab + (1 + ω2)a2b,
= (ω + ω2)u 2.
Therefore, given any w ∈ W we have w (a + a−1) = (ω + ω2)w .
Exercise 9.7. We consider the centres of the following groups and use Proposition 9.16.
(a) Consider C n = x | x n = 1. Let ω = e 2πi/n then ω is a primitive nth root of
unity. We know that the map ρ : C n → GL(1,C) given by x ρ = (ω) is an irreducible
representation of C n and is clearly faithful as all powers of ω are distinct.
(b) Let D8 = a, b | a4 = b 2 = 1 and bab = a3 = 1, a , a2, a3,b,ab,a2b, a3b . Now it’s
quick to check that Z (D8) = 1, a2 ∼= C 2. Consider the representation σ : D8 →
GL(2,C
) given byaσ =
i 0
0 −i
bσ =
0 1
1 0
then because i is a primitive fourth root of unity we have this is a faithful irreducible
representation of D8.
(c) It’s clear to see that the Z (G×H ) = Z (G)×Z (H ) and therefore Z (C 2×D8) ∼= C 2×C 2,
which is not cyclic. Therefore there cannot exist a faithful irreducible C(C 2 × D8)
module by Proposition 9.16.
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Chapter 9 30
(d) Now Z (C 3 × D8) ∼= C 3 × C 2 ∼= C 6 and so is cyclic. Let ξ = e 2πi/3 then we have the
representation π : C 3 × D8 → GL(2,C) given by
(x , 1)π =
ξ 0
0 ξ−1
(1, a)π =
i 0
0 −i
(1, b )π =
0 1
1 0
is a faithful irreducible representation of C 3 × D8.
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Chapter 10. Irreducible modules and the group algebra
Exercise 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Exercise 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Exercise 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Exercise 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Exercise 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Exercise 10.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Exercise 10.1. Let V = spang ∈G g ∈ G be the trivial submodule of CG. We now wantto see if there exists a 1-dimensional submodule, say U , of CG such that U ∼= V . Let
U = spanu , then for U to be the trivial submodule we need ug = u for all g ∈ G. In
other words |G|u = u (
g ∈G g ) = (
g ∈G g )u ⇒ u ∈ V and so U = V . Therefore there is
only one unique trivial CG module.
Exercise 10.2. Let G = x | x 4 = 1. We consider i as a fourth root of unity and define
the following elements of CG,
v 0 = 1 + x + x 2
+ x 3
,v 1 = 1 + i x − x 2 − i x 3,
v 2 = 1 − x + x 2 − x 3,
v 3 = 1 − i x − x 2 + i x 3.
Then V i = spanv i are irreducible CG modules because v 0x = v 0, v 1x = i v 1, v 2x = −v 2
and v 3x = −iv 3. Therefore we have CG = V 0 ⊕ V 1 ⊕ V 2 ⊕ V 3.
Exercise 10.3. Consider u 1 = 1 + a + a2 + a3 − b − ab − a2b − a3b ∈ CG. Then we have
the following
u 1a = a + a2 + a3 + 1 − a3b − b − ab − a2b = u 1,
u 1b = b + ab + a2b + a3b − 1 − a − a2 − a3 = −u 1.
Hence U 1 = spanu 1 is the required 1-dimensional CG submodule. Alternatively let u 2 =
1 − a + a2 − a3 + b − ab + a2b − a3b and u 3 = 1 − a + a2 − a3 − b + ab − a2b + a3b then
31
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Chapter 10 32
we have that
u 2a = a − a2
+ a3
− 1 + a3
b − a + ab − a2
b = −u 2,u 2b = b − ab + a2b − a3b + 1 − a + a2 − a3 = u 2,
u 3a = a − a2 + a3 − 1 − a3b + b − ab + a2b = −u 3,
u 3b = b − ab + a2b − a3b − 1 + a − a2 + a3 = −u 3.
Therefore U 2 = spanu 2 and U 3 = spanu 3 are the required 1-dimensional CG modules.
Exercise 10.4. Let G = D8 = a, b | a4 = b 2 = 1 and bab = a3 then define the following
elements of CG
v 0 = 1 + a + a2 + a3 w 0 = bv 0 = b + ba + ba2 + ba3,
v 1 = 1 + i a − a2 − ia3 w 1 = bv 1 = b + i ba − ba2 − iba3,
v 2 = 1 − a + a2 − a3 w 2 = bv 2 = b − ba + ba2 − ba3,
v 3 = 1 − i a − a2 + ia3 w 3 = bv 3 = b − i ba − ba2 + iba3.
Now as in question 2 we have that spanv i are Ca modules and it’s clear to see that
spanw i are also Ca modules. Now b acts on these elements in the following way
v 0b = b + ba3 + ba2 + ba = w 0 w 0b = 1 + a3 + a2 + a = v 0,
v 1b = b + iba3 − ba2 − iba = w 3 w 1b = 1 + i a3 − a2 − i a = v 3,
v 2b = b − ba3 + ba2 − ba = w 2 w 2b = 1 − a3 + a2 − a = v 2,
v 3b = b − iba3 − ba2 + iba = w 1 w 3b = 1 − i a3 − a2 + i a = v 1.
Therefore this gives us that spanv 0, w 0, spanv 1, w 3, spanv 2, w 2 and spanv 3, w 1 are
Cb modules.
Now we have that spanv 0, w 0 is reducible as U 0 = spanv 0 + w 0, (the trivial module),
and U 1 = spanv 0 − w 0 are CG submodules. Also, by question 3, we have spanv 2, w 2 isreducible as U 2 = spanv 2 + w 2 and U 3 = spanv 2 − w 2 are CG submodules. We quickly
check that U 4 = spanv 1, w 3 and U 5 = spanv 3, w 1 are irreducible. Now a acts on a linear
combination of the basis in the following way
(αv 1 + βw 3)a = αiv 1 − βi w 3 ( γv 3 + δw 1)a = − γi v 3 + δi w 1.
So if these are bases of 1-dimensional submodules then there exists k and such that
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Chapter 10 33
α = ki α, β = −kiβ and γ = −i γ , δ = iδ. However no such k and exist, therefore the
modules are irreducible and we have
CG = U 0 ⊕ U 1 ⊕ U 2 ⊕ U 3 1-dim
⊕ U 4 ⊕ U 5 2-dim
.
because v 0 + w 0, v 0 − w 0, v 2 + w 2, v 2 − w 2, v 1, v 3, w 1, w 3 form a basis for CG. Note that
non of the 1-dimensional modules are isomorphic but U 4 ∼= U 5 by the isomorphism sending
v 1 → w 1 and w 3 → v 3.
Now by Theorem 10.5 we have that every 1-dimensional irreducible representation of
D8 is equivalent to one of the following
ρ0 : a → (1) ρ1 : a → (1) ρ2 : a → (−1) ρ3 : a → (−1)
b → (1) b → (−1) b → (1) b → (−1)
and any 2-dimensional irreducible representation of D8 is equivalent to
ρ4 : a →
i 0
0 −i
b →
0 1
1 0
.
Exercise 10.5. Let ψ : U 1
→U 2 be the CG isomorphism between these modules. Let
0 = λ ∈ C then we define a map ϑ : U 1 → V such that uϑ = u + λ(uψ). We claim this
is a CG homomorphism. We check the following properties for all u 1, u 2 ∈ U 1, µ ∈ C and
x ∈ G
(u 1 + u 2)ϑ = (u 1 + u 2) + λ((u 1 + u 2)ψ) = (u 1 + λ(u 1ψ)) + (u 2 + λ(u 2ψ)) = u 1ϑ + u 2ϑ,
(µu 1)ϑ = µu 1 + λ((µu 1)ψ) = µ(u 1 + λ(u 1ψ)) = µ(u 1ϑ),
(u 1x )ϑ = u 1x + λ(u 1xψ) = u 1x + (λ(u 1ψ))x = u 1ϑx.
What is the kernel of ϑ? Well we require that
u ∈ ker(ϑ) ⇔ uϑ = 0 ⇔ u + λuψ = 0 ⇔ u = 0
as the sum is direct. Now im(ϑ) is a submodule of V and U 1 ∼= im(ϑ). This argument
works equally well for U 2 and we’re done.
Exercise 10.6. We first show that this is irreducible. Let U = spanαv 1 + βv 2 ⊆ V be a
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Chapter 10 34
1-dimensional submodule, then we have
(αv 1 + βv 2)a = i αv 1 − iβv 2 ∈ U ⇒ α = k 1iα,β = −k 1i β,(αv 1 + βv 2)b = αv 2 − βv 1 ∈ U ⇒ α = k 2α, β = −k 2 β,
for some k 1, k 2 ∈ C but these only exist if α = β = 0 and so U is the 0 module. Therefore
V is irreducible.
Now consider a 2-dimensional CG submodule, say U , with basis u 1 = 1 − i a − a2 + i a3 +
ib − ab − ia2b + a3b and u 2 = −i + a + ia2 − a3 + b − i ab − a2b + ia3b . Then we have
u 1a = a − i a2 − a3 + i + i a3b − b − iab + a2b = iu 1,
u 1b = b − i ab − a2b + ia3b + i a2 − a3 − i + a = u 2,
u 2a = −ia + a2 + i a3 − 1 + a3b − i b − ab + i a2b = −i u 2,
u 2b = −ib + ab + ia2b − a3b + a2 − ia3 − 1 + ia = −u 1.
Therefore U is an irreducible CG submodule which is isomorphic to V .
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Chapter 11. More on the group algebra
Exercise 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Exercise 11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Exercise 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Exercise 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Exercise 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Exercise 11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Exercise 11.1. If G is non-abelian then not all CG modules have dimension 1, however weknow we always have the trivial module with dimension 1. Therefore we have 1+
5i =2 d 2i =
6 then we must have one module of dimension 2 and two modules of dimension 1.
Exercise 11.2. If G is of order 12 then there are four possibilities for the degrees of the
irreducible representations. Either 112, 1102, 1422 or 133.
By exercise 3.5 we have that ρ1 and ρ3 are irreducible representations of D12 with degree
2. We recall that these representations are not isomorphic because ρ1 is faithful but ρ3 is
not. Therefore we must have the degrees of the irreducible representations of D12 are 1422.
Exercise 11.3. We have that G = g 1, . . . , g n forms a basis for the group algebra CG.
We define a family of CG homomorphisms by r φi = g i r for all r ∈ CG and 1 i n. Let
φ : CG → CG be any CG-homomorphism then
1φ = λ1g 1 + · · · + λng n,
for some λi ∈ C. Therefore, for any r ∈ CG we have
r φ = (1r )φ = (1φ)r,
= (λ1g 1 + · · · + λng n)r,
= λ1g 1r + · · · + λng nr,
= r (λ1φ1 + · · · + λnφn)
and so φi span HomCG(CG,CG). We check that they’re linearly independent by evaluating
both sides of the following equation at 1
0 = 1(λ1φ1 + · · · + λnφn),
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Chapter 11 36
= λ1g 1 + · · · + λng n.
Now the g i ’s are linearly independent and so we must have λ1 = · · · = λn = 0 and we’redone.
Exercise 11.4. Let v 1, . . . , v n be the natural basis for the permutation module V . Then
spanv 1 + · · · + v n is the unique trivial submodule module of CG. Therefore by Corollary
11.6 we have dim(HomCG(V, U )) = 1.
Exercise 11.5. We can construct a basis for HomCG(CG, U 3) in roughly the same way as
in the proof of Proposition 11.8. We have v 1, w 2 form a basis for U 3. Now define two maps
φ1 : CG
→U 3 and φ2 : CG
→U 3 by r φ1 = v 1r and r φ2 = w 2r . Then consider a CG
homomorphism φ : CG → U 3. For some λ1, λ2 ∈ C we have
1φ = λ1v 1 + λ2w 2.
Hence for any r ∈ CG we have
r φ = (1r )φ = (1φ)r = λ1v 1r + λ2w 2r = r (λ1φ1 + λ2φ2).
So φ1, φ2 span HomCG(CG, U 3) and they’re a basis because if 0 = µ1φ1 + µ2φ2 then
0 = µ1(1φ1) + µ2(1φ2) = µ1v 1 + µ2w 2,
which gives us µ1 = µ2 = 0.
We wish to now find a basis for the vector space HomCG(U 3,CG). Recall from the proof
of Proposition 11.4 that
HomCG(U 3,CG) ∼= HomCG(U 3, U 1) ⊕ HomCG(U 3, U 2)
⊕HomCG(U 3, U 3)
⊕HomCG(U 3, U 4),
∼= HomCG(U 3, U 3) ⊕ HomCG(U 3, U 4),
as vector spaces because CG = U 1 ⊕ U 2 ⊕ U 3 ⊕ U 4. Now ϑ1 : U 3 → U 3 and ϑ2 : U 3 → U 4
given by uϑ1 = u and uϑ2 = bu for all u ∈ U 3 are basis elements for HomCG(U 3, U 3) and
HomCG(U 3, U 4) respectively. Therefore they form a basis for HomCG(U 3,CG).
Exercise 11.6. This means that we can express V as a direct sum V ∼= k i =1 V d i i , where
V d i i = V i ⊕ · · · ⊕ V i for d i copies. Also we can express W as a direct sum W ∼=
k i =1 V e i i .
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Chapter 11 37
Therefore we have
dim(HomCG(V, W )) = dim(HomCG(
k i =1
V d i i ,
k i =1
V e i i )),
=
k i =1
d i e i dim(HomCG(V i , V i )),
=
k i =1
d i e i .
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Chapter 12. Conjugacy classes
Exercise 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Exercise 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Exercise 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Exercise 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Exercise 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Exercise 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Exercise 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Exercise 12.1. Clearly we have 1 ∈ C G(x ) as 1x = x 1. Now if g ∈ C G(x ) then gx = x g ⇒xg −1 = g −1x ⇒ g −1 ∈ C G(x ). Finally, if g, h ∈ C G(x ) then (gh)x = gx h = x (gh) which
gives us gh ∈ C G(x ) and so C G(x ) is a subgroup of G. If g ∈ Z (G) then gy = y g for all
y ∈ G, in particular for x ∈ G and so Z (G) ⊆ C G(x ).
Exercise 12.2. Now g ∈ C G(x ) ⇔ gx = x g ⇔ g (xz ) = (xz )g ⇔ g ∈ C G(xz ). Therefore
we have |g G| = |G|/|C G(x )| = |G|/|C G(xz )| = |(gz )G|.
Exercise 12.3. Let G = Sn.
(a) Well (12) is conjugate to σ ∈ Sn if and only if they have the same cycle type. Now
there aren2
2-cycles in Sn which gives us |(12)Sn | =
n2
. Now if g ∈ C G((12)) then
g −1(12)g = (12), which gives us (1g 2g ) = (12). Therefore we have g = y or (12) y ,
where y is an element which fixes 1 and 2. There are (n −2)! such elements y and so
|C G((12))| = 2 · (n − 2)! which gives us |(12)Sn | = |G|/|C G((12))| = n!/2!(n − 2)! =n2
as required.
(b) Choose a 3 element subset from 1, . . . , n, say i , j , k . Then (i jk ) and (ikj ) are
the only distinct 3-cycles from this set. In otherwords there are 2n3 3-cylces in
Sn and so |(123)Sn | = 2n3. Now choose a 4 element subset from 1, . . . , n, say
i , j , k , . Then (i j )(k), (ik )( j) and (i)( jk ) are the only distinct 2-2-cycles you
can make from this set. Hence there are 3n4
distinct 2-2-cycles in Sn, which gives
us |(12)(34)Sn | = 3n4
.
(c) We consider the sizes of the conjugacy classes in S6. We express this in table 12.1.
Exercise 12.4. Now the cycle types appearing in A6 are: (1), (2,2), (2,4), (3), (3,3) and
(5). Now (12)(34) commutes with (56), which is odd. We have (12)(3456) commutes
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Chapter 12 39
Cycle Shape Representative |g G| Reason
(1) 1 1 –(2) (12) 15
62
(2,2) (12)(34) 45
62
42
/2
(2,2,2) (12)(34)(56) 1562
42
/3 × 2
(2,3) (12)(345) 120 262
43
(2,4) (12)(3456) 90
62
× 3!
(3) (123) 40 263
(3,3) (123)(456) 40 263(4) (1234) 90 66
4
(5) (12345) 144 4 × 3 × 265
(6) (123456) 120 5!
Table 12.1: Conjugacy Classes of S6
with (12), which is odd. We have (123) commutes with (45), which is odd. We have
(123)(456) commutes with (14)(25)(36), which is odd. However there is no odd element
which commutes with a cycle of shape (5). Therefore (12345)A6
= (12345)S6.
Exercise 12.5. Any normal subgroup of A5 must be a union of conjugacy classes and
the order must divide the order of the group, namely 60. Now the divisors of 60 are
2,3,4,5,6,10,12,15,20 and 30. Now the conjugacy class sizes of A5 are 1,12,15 and 20.
However as we must always include the identity it’s clearly impossible to construct a normal
subgroup of order 15, 20 or 30. Therefore A5 is simple.
Exercise 12.6. We have Q8 = a, b | a4 = 1, b 2 = a2, b −1ab = a−1. Now straight away we
have b −1ab = a3 and so a, a3 are conjugate. We have a2 is the only element of order 2 so
it’s in a conjugacy class on its own. We have a3ba = a2b and b −1(ab )b = a3b . Thereforethe conjugacy classes are 1, a2, a, a3, ab,a3b and b, a2b . Therefore a basis for
Z (CQ8) is
1, a2, a + a3, ab + a3b, b + a2b.
Exercise 12.7. Now we have from the class formula that |G| = |Z (G)| +
x i ∈Z (G) |x Gi |.Now p | |G| and p | |x Gi | for each x i ∈ Z (G) so we must have p | |Z (G)| and so Z (G) = 1.
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Chapter 12 40
Now assume G does not have a conjugacy class of size p . Then we have p 2 | |x Gi | for
each x i ∈ Z (G) and clearly p 2 | |G| = p n. If n 3 this gives us p 2 | |Z (G)| but |Z (G)| = p
and so we must have G has a conjugacy class of size p .
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Chapter 13. Characters
Exercise 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Exercise 13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Exercise 13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Exercise 13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Exercise 13.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Exercise 13.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Exercise 13.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Exercise 13.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Exercise 13.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Exercise 13.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Exercise 13.1. Now we have G = 1, a , a2, a3, a4, a5,b,ab,a2b, a3b, a4b, a5b . We first
see that the conjugacy classes of G are
1 a3 a, a5 a2, a4 b, a2b, a4b ab,a3b, a5b
by (12.12). Now the matrices of the representations are as follows
a2ρ1 =
ω2 0
0 ω−2
a3ρ1 =
1 0
0 1
a4ρ1 =
ω 0
0 ω−1
a5ρ1 =
ω2 0
0 ω−2
,
abρ1 =
0 ω
ω−1 0
a2bρ1 =
0 ω2
ω−2 0
a3bρ1 =
0 1
1 0
a4bρ1 =
0 ω
ω−1 0
,
a5bρ1 =
0 ω2
ω−2 0
a2ρ2 =
1 0
0 1
a3ρ2 =
−1 0
0 1
a4ρ2 =
1 0
0 1
,
a5ρ2 =−1 0
0 1
abρ2 =
−1 0
0 −1
a2bρ2 =
1 0
0 −1
a3bρ2 =
−1 0
0 −1
,
a4bρ2 =
1 0
0 −1
a5bρ2 =
−1 0
0 −1
.
Recall that X 3 = 1 can be written as (X − 1)(X 2 + X + 1) = 0. So if ω is a root of
unity we have ω2 + ω = −1 ⇒ ω + ω−1 = −1. Therefore the characters associated
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Chapter 13 42
are the ones given in table 13.1. It’s easy to see that ker(ρ1) = ker(χ1) = 1, a3 and
ker(ρ2) = ker(χ2) = 1, a2, a4.
g 1 a3 a a2 b ab
χ1(g ) 2 2 −1 −1 0 0
χ2(g ) 2 0 0 2 0 −2
Table 13.1: Two characters of D12
Exercise 13.2. Well we know that there are four irreducible characters of C 4 = x | x 4 = 1which are all linear. Consider i as a primitive fourth root of unity then the irreduciblerepresentations are xρ j = i j −1 for 1 j 4. Therefore the irreducible characters are given
in table 13.2. From the table we can see χreg(g ) = χ1(g ) + χ2(g ) + χ3(g ) + χ4(g ) is
4 = |G| if g = 1 and 0 otherwise.
g 1 x x 2 x 3
χ1(g ) 1 1 1 1
χ2(g ) 1 i −1 −i
χ3(g ) 1 −1 1 −1χ4(g ) 1 −i −1 i
Table 13.2: The irreducible characters of C 4
Exercise 13.3. Well χ(g ) = | fix(g )|, in particular we can easily see that the character
values are then given by χ((12)) = 5 and χ((16)(235)) = 2.
Exercise 13.4. Let χ : G → C be a non-zero character. If χ is a homomorphism then
χ(1) = χ(12
) = χ(1)2
⇒ χ(1)(χ(1) − 1) = 0 ⇒ χ(1) = 1, (note χ(1) = 0 because it’sthe dimension of the module), in other words χ is a linear character.
Exercise 13.5. Let ρ : G → GL(n,C) be the corresponding irreducible represenation of χ.
If z ∈ Z (G) then z ρ ∈ Z (GL(n,C)), which means z ρ = λIn for some λ ∈ C. We have
z m = 1 ⇒ λmIn = In ⇒ λm = 1 and so λ is an mth root of unity. Now for all g ∈ G we
have
(z g )ρ = (zρ)(gρ) = λIn(gρ) = λ(gρ).
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Chapter 13 43
Therefore taking traces gives us χ(zg ) = λχ(g ) for all g ∈ G.
We comment that irreducibility is required to say z ρ is a scalar of λn. If the represen-
tation is reducible then you could have different scalars of different block identities.
Exercise 13.6. Now using Theorem 13.11 we have for some λ ∈ C and all h ∈ G we have
|χ(g )| = χ(1) ⇔ gρ = λIn for some λ ∈ C,
⇔ (gρ)(hρ) = (hρ)(gρ) for all h ∈ G,
⇔ (gh)ρ = (hg )ρ,
⇔ gh = hg because ρ is faithful,
⇔g
∈Z (G).
Therefore Z (G) = g ∈ G | |χ(g )| = χ(1). Note that if χ is reducible then (gρ)(hρ) =
(hρ)(gρ) ⇒ gρ = λIn.
Exercise 13.7. Let ρ : G → GL(n,C) be a representation of a finite group G.
(a) Define a map ψ : G → GL(1,C) by gψ = (det(gρ)). We claim this is a homo-
morphism. For all g, h ∈ G we have (gh)ψ = (det((gh)ρ)) = (det((gρ)(hρ))) =
(det(gρ))(det(hρ)) = (gψ)(hψ). Clearly δ is then a character as δ(g ) = tr(gψ) and
it’s linear because δ(1) = tr(1ψ) = tr(1) = 1.
(b) Recall that matrices A, B ∈ GL(n,C) we have
det(A−1BA) = det(A)−1 det(B) det(A) = det(B).
Now ker(δ) = g ∈ G | δ(g ) = δ(1) = g ∈ G | det(gρ) = 1. By the above matrix
fact we have that similar matrices are in the same left coset of G/ ker(ρ). Let g, h ∈ G
then we have the left cosets (g −1hg ) ker(δ) = h ker(δ) ⇔ (hg ) ker(δ) = (gh) ker(δ),
which means G/ ker(δ) is abelian.
(c) Let N = g ∈ G | δ(g ) > 0, we claim this is a normal subgroup of G. Now 1 ∈ N because δ(1) = 1 > 0. Let x , y ∈ N then δ(xy ) = δ(x )δ( y ) > 0 and so xy ∈ N .
Finally if x ∈ N then δ(x −1) = 1δ(x )
> 0 and so x −1 ∈ N , which means N is a subgroup.
Let x ∈ N and g ∈ G then δ(g −1xg ) = δ(g −1)δ(x )δ(g ) = δ(x ) > 0, which gives us
N is a normal subgroup of G. If g, h ∈ G then gN = hN ⇔ gh−1 ∈ N ⇔ δ(gh−1) >
0 ⇔ δ(g )δ(h)
> 0 ⇔ δ(g ), δ(h) > 0 or δ(g ), δ(h) < 0. Therefore as there exists an
element h ∈ G with δ(h) = −1 we have N is a normal subgroup of index 2 in G.
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Chapter 13 44
Exercise 13.8. Let G be the standard basis of CG, with dim(CG) = |G| = n, and
ρreg : G → CG be the regular representation. We know that [gρreg]G are permutation ma-
trices associated with some permutation σg ∈ Sn. Now we have det([gρreg ]G) = sgn(σg ).Consider, as in question 7, the linear character δ : g → det([gρreg ]G). We have 2 | |G| and
so there exists an element of order 2 in G. In other words there exists an element g ∈ G
such that σg is a transposition and so δ(g ) = −1. Therefore by part c of question 7 we
have there exists a normal subgroup of index 2 in G.
Exercise 13.9. Let g ∈ G be an element of order 2. Let ρ : G → GL(n,C) be the
corresponding representation to χ. We have from position 9.11 that there exists a basis
B such that [gρ]B is a diagonal matrix, whose entries are square roots of unity. Therefore
χ(g ) = r − s for some r, s ∈ N where r is the number of +1’s and s the number of −1’s onthe diagonal of gρ. Now χ(1) = r + s and so we have r − s = r + s − 2s ≡ r + s (mod 2).
If s is even then s = 2s for some s ∈ N and r − s = r + s − 4s ≡ r + s (mod 4),
in other words χ(g ) ≡ χ(1) (mod 4). If s is not even then det([gρ]B) = (−1)s = −1.
Therefore considering the linear character δ : g → det([gρ]B) from question 7, we have
δ(g ) = −1 which gives us there exists a normal subgroup of index 2 by part c.
Exercise 13.10. Consider the regular character χreg = χ1+· · ·+χn where χi are irreducible
characters. Now χreg(1) = |G| but χreg(x ) = 0 in other words
χ1(1) + · · · + χn(1) = |G| χ1(x ) + · · · + χn(x ) = 0.
Therefore we cannot have χi (x ) = χi (1) for all i because |G| > 0.
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Chapter 14. Inner products of characters
Exercise 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Exercise 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Exercise 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Exercise 14.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercise 14.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercise 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercise 14.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercise 14.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercise 14.1. We have that
χ, χ =32
24+
(−1)2
4+
02
3+
32
8+
(−1)2
4= 2,
χ, ψ =3 · 3
24+
(−1) · 1
4+
0 · 0
3+
(−1) · 3
8+
(−1) · (−1)
4=
32
24− 32
24= 0,
ψ, ψ =32
24+
12
4+
0
3+
(−1)2
8+
(−1)2
4= 1.
Therefore ψ is an irreducible character but χ is not.
Exercise 14.2. We calculate the characters χi corresponding to the representations ρi .
Using the following matrix calculations
a2ρ1 =
−1 0
0 −1
abρ1 =
0 i
i 0
a2ρ2 =
−1 0
0 −1
abρ2 =
i 0
0 −i
,
a2ρ3 = 1 0
0 1 abρ3 = −1 0
0−
1 ,
we give the values of the characters in table 14.1.
As χ1 = χ2 we have ρ1 and ρ2 are equivalent representations. Also χ3 = χ2 and χ3 = χ1
means ρ3 is not equivalent to either χ1 or χ2.
Exercise 14.3. Now we have that tr(gσ) = tr(T −1g (gρ)T g ) = tr(gρ) for all g ∈ G. There-
fore σ and ρ have the same character and by Theorem 14.21 this gives us σ and ρ are
equivalent representations.
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Chapter 14 46
g 1 a a2 ab b
χ1(g ) 2 0 −2 0 0χ2(g ) 2 0 −2 0 0
χ3(g ) 2 0 2 −2 0
Table 14.1: The characters of ρ1, ρ2 and ρ3
Exercise 14.4. We consider the inner product of χ with itself,
χ, χ =1
|G
| χ(g )χ(g ) =
1
|G
| g ∈G|χ(g )| =
1
|G
| g ∈Gχ(g ) >
|G|
|G
|= 1.
Therefore χ must be reducible as we cannot have χ, χ = 1.
Exercise 14.5. Expanding the inner product of χreg, χ we see
χreg, χ =1
|G|g ∈G
χreg(g )χ(g ) =1
|G| |G|χ(1) = χ(1),
because χ(1) ∈ N \ 0.
Exercise 14.6. In Exercise 11.4 we showed that if V is the permutation module and U is
the trivial module then dim(HomCG(V, U )) = 1. Now by Theorem 14.24 we have
1 = dim(HomCG(V, U )) = π, 1Sn.
Alternatively, a more direct approach is to use the Orbit-Stabiliser theorem to show thatg ∈G | fix(g )| = |G|.
Exercise 14.7. Now we know that
ψ, ψ = d 21 + · · · + d 2k .
Now if ψ, ψ = 1 then we must have that d i = 1 for just one i and so ψ is an irreducible
character. If ψ, ψ = 2 then we must have two d i = 1 and the rest zero, therefore
ψ = χi 1 + χi 2. If ψ, ψ = 3 then we must have ψ = χi 1 + χi 2 + χi 3. Finally if ψ, ψ = 4
then either ψ = χi 1 + χi 2 + χi 3 + χi 4 or ψ = 2χi 1.
Exercise 14.8. No. For example consider the trivial character and the sign character of
Sn. Then we have χ(g ) = 1Sn(g ) + χsgn(g ) = 2 or 0. Now the trivial and sign characters
are irreducible so χ = 2φ.
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Chapter 15. The number of irreducible characters
Exercise 15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Exercise 15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Exercise 15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Exercise 15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Exercise 15.1. We consider the inner product of χ with the irreducibles
χ, χ1 =
1
6 (19 − 3 + 2 · (−2)) =
12
6 = 2,
χ, χ2 =1
6(19 + 3 + 2 · (−2)) =
18
6= 3,
χ, χ3 =1
6(2 · 19 + 0 + 2 · (−2) · (−1)) =
42
6= 7.
Therefore we have χ = 2χ1 + 3χ2 + 7χ3. All the coefficients are positive integers and so
χ is a character of S3.
Exercise 15.2. We check the appropriate inner products
ψ1, χ1 =1
6ψ2, χ1 =
3
6=
1
2ψ3, χ1 =
2
6=
1
3,
ψ1, χ2 =1
6ψ2, χ1 =
−1 · 3
6= −1
2ψ3, χ1 =
2
6=
1
3,
ψ1, χ3 =2
6=
1
3ψ2, χ1 = 0 ψ3, χ1 =
−2
6= −1
3.
Therefore we have
ψ1 =1
6χ1 +
1
6χ2 +
1
3χ3 ψ2 =
1
2χ1 − 1
2χ2 ψ3 =
1
3χ1 +
1
3χ2 − 1
3χ3.
Exercise 15.3. Examining the inner products we have
ψ, χ1 =6
12+
0
12+
3
6− 3
6− 1 + i
4+
−1 + i
4= 0,
ψ, χ2 =6
12+
0
12− 3
6− 3
6+
−i + 1
4+
−i − 1
4= 0,
ψ, χ3 =6
12+
0
12+
3
6− 3
6+
1 + i
4+
1 − i
4= 1,
47
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Chapter 15 48
ψ, χ4 =6
12+
0
12− 3
6− 3
6+
i − 1
4+
−1 − i
4= −1,
ψ, χ5 =12
12 +0
12 −3
6 +3
6 +0
4 +0
4 = 1,
ψ, χ6 =12
12+
0
12+
3
6+
3
6+
0
4+
0
4= 2.
Therefore ψ = χ3− χ4 + χ5 + 2χ6. So ψ is not a character of G because χ4 has a negative
coefficient.
Exercise 15.4. Let G be a group with |G| = 12.
(a) We show that the centre of a group never has index 2. If the centre does have index
2 then G = Z (G)
x Z (G) for some x
∈G. However consider xz
∈xZ (G) then
for g ∈ Z (G) we clearly have g (x z ) = (xz )g . Now consider x z ∈ x Z (G) then
(xz )(xz ) = (xz )(x z ) and so xz ∈ Z (G) means xZ (G) = Z (G) ⇒ G = Z (G).
Therefore Z (G) cannot have index 2 in the group. If G is abelian then there are
12 conjugacy classes as |G| = 12. Assume G is non-abelian then |Z (G)| | |G| but
we’ve just shown |Z (G)| = 6 therefore |Z (G)| 4. So there are at most 4 conjugacy
classes with order 1. All the other conjugacy classes have order at least 2 which leaves
us with |G| 4 · 1 + 5 · 2 = 14, which isn’t possible. Therefore G cannot have exactly
9 conjugacy classes.
(b) By Exercise 11.2 we have the dimensions of the irreducible modules of a group of order 12 are either 112, 1102, 1422, 133. Therefore the group has either 12, 6 or 4
conjugacy classes. Now the cyclic group C 12 has 12 conjugacy classes. The dihedral
group D12 has precisely 6 conjugacy classes by (12.12). Finally by Example 13.25 we
have A4 has precisely 4 conjugacy classes.
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Chapter 16. Character tables and orthogonality relations
Exercise 16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Exercise 16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Exercise 16.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Exercise 16.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Exercise 16.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Exercise 16.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Exercise 16.1. Recall that C 2 × C 2 = (1, 1), (1, y ), (x, 1), (x, y ) such that x 2 = y 2 = 1then the character table is as in table 16.1.
g (1, 1) (1, y ) (x, 1) (x , y )
χ1(g ) 1 1 1 1
χ2(g ) 1 −1 1 −1
χ3(g ) 1 1 −1 −1
χ4(g ) 1
−1
−1 1
Table 16.1: The character table of C 2 × C 2
Exercise 16.2. Using the column orthogonality relations we get the character table of G
to be as in table 16.2.
g g 1 g 2 g 3 g 4 g 5
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 1 −1 −1
χ3(g ) 1 1 −1 1 −1
χ4(g ) 1 1 −1 −1 1
χ5(g ) 2 −2 0 0 0
Table 16.2: The character table of G from exercise 16.2
49
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Chapter 16 50
Exercise 16.3. We first use the column othogonality relations to find χ3(1) and χ4(1).
We can see that g 4 is an element of order 2. This is because g 4 ∈ C G(g 4) and o (g 4) ||C G(g 4)| ⇒ o (g 4) = 2. The column relation on g 4 says 12 + 02 + |χ3(g 4)|2 + |χ4(g 4)|2 = 2,which gives us |χ3(g 4)|2 + |χ4(g 4)|2 = 1. We know from Corollary 13.10 that χ4(g 4) ≡ 0
(mod 2) so χ4(g 4) = 0. Finally the column relations on g 1 and g 4 give us 1 + χ3(g 4) = 0
so the character table so far is as in table 16.3
g g 1 g 2 g 3 g 4
χ1(g ) 1 1 1 1
χ2(g ) 2 α β 0
χ3(g ) 1 x 1 x 2 −1
χ4(g ) 2 y 1 y 2 0
Table 16.3: Partial character table of G, |G| = 10
We now use the column orthogonality relations of g 2, g 4 and g 3, g 4 to get
1 − x 1 = 0 ⇒ x 1 = 1 1 − x 2 = 0 ⇒ x 2 = 1.
Finally we use the column orthogonality of g 1, g 2 and g 1, g 3 to get the values
1 + 2α + 1 + 2 y 1 = 0 ⇒ y 1 = −1 − α = β,
1 + 2 β + 1 + 2 y 2 = 0 ⇒ y 2 = −1 − β = α.
Therefore the final character table of G is as in table 16.4
g g 1 g 2 g 3 g 4
χ1(g ) 1 1 1 1
χ2(g ) 2 α β 0
χ3(g ) 1 1 1 −1
χ4(g ) 2 β α 0
Table 16.4: The character table of G from exercise 16.3
Exercise 16.4. We have the character table as in the question.
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Chapter 16 51
(a) From the column orthogonality of g 2 we have
3 + |ζ |2
+ |ζ |2
= 7 ⇒ 2|ζ |2
= 4 ⇒ |ζ |2
= 2.
Now using the orthogonality of g 1 and g 2 we have
3 + 3ζ + 3ζ = 0 ⇒ ζ + ζ = −1 ⇒ 2Re(ζ ) = −1 ⇒ Re(ζ ) = −1
2.
Using this and combining it with the first equation we have
−
1
22
+ Im(ζ )2 = 2
⇒
1
4
+ Im(ζ )2 = 2,
⇒ 1 + 4 Im(ζ )2 = 8,
⇒ 4Im(ζ )2 = 7,
⇒ Im(ζ )2 =7
4,
⇒ Im(ζ ) =
√ 7
2.
Therefore ζ = −1+√ 7i
2.
(b) Recall that we have χ(g ) is real valued if g is conjugate to g −1
. Now the fact thatχ4(g 2) and χ5(g 2) are complex, means that g −12 ∈ g G2 . Now we know χi (g −12 ) = χi (g )
and so the column of the character table reads 1, 1, 1, ζ , ζ . Also recall that for all
x ∈ G we have xg 2 = g 2x ⇔ g −12 x = xg −12 . Therefore |C G(g −12 )| = |C G(g 2)| = 7.
This leaves us with the third column of the character table, which is given in table
16.5.
g g 1 g 2 g −12
χ1(g ) 1 1 1χ2(g ) 1 1 1
χ3(g ) 1 1 1
χ4(g ) 3 ζ ζ
χ5(g ) 3 ζ ζ
Table 16.5: Three columns of the character table of G from exercise 16.4
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Chapter 16 52
Exercise 16.5. Let X = g ∈ G | k i =1 χi (g )χi (g ) = |G|. Recall from the column
orthogonality relation thatk
i =1
χi (z )χi (z ) = |C G(z )|.
Then we have
x ∈ X ⇔k
i =1
χi (z )χi (z ) = |G| ⇔ |C G(z )| = |G| ⇔ z ∈ Z (G),
in other words Z (G) = X as required.
Exercise 16.6. Now let C be the character table of G. Now we know that χ(g ) = χ(g −1)
and so C is obtained from C be rearranging the columns. If we swap two columns in a
matrix then we multiply the determinant by −1. Therefore det C = det C = ± det(C ). If
det C = det C then det C is real and if det C = − det(C ) then det C is purely imaginary.
We consider the matrix C T C . Now we can see that (C T C )ij =k
=1 χ(g i )χ(g j ) =
δij |C G(g i )| by the column orthogonality relations. Therefore we have
| det C |2 = (det C )(det C ) = det(C T C ) =
k i =1
|C G(g i )|.
Finally for G = C 3, recall that ω = 1+√ 3i 2 and ω2 = 1−√ 3i
2 , then we have the determinant
of the character table to be1 1 1
1 ω ω2
1 ω2 ω
= (ω2 − ω) − (ω − ω2) + (ω2 − ω),
= 3(ω2 − ω),
= −3√
3i .
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Chapter 17. Normal subgroups and lifted characters
Exercise 17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
Exercise 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Exercise 17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Exercise 17.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
Exercise 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Exercise 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Exercise 17.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Exercise 17.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Exercise 17.1. Let G = Q8 = a, b | a4 = 1, b 2 = a2, b −1ab = a−1.
(a) From Exercise 12.6 we have the conjugacy classes are 1, a2, a, a3, ab,a3b and b, a2b .
(b) We can see that a−1b −1ab = a−2 = a2. Therefore we can see that G = a2 ∼= C 2.
Now, let H = a2 then we have G/H = H,aH,bH,abH and (aH )2 = (bH )2 = 1.
Therefore it’s easy to see that G/H ∼= C 2 × C 2. Using table 16.1 we can easily see
the character table of G/H is given in table 17.1.
gH H aH bH abH
χ1(g ) 1 1 1 1
χ2(g ) 1 −1 1 −1
χ3(g ) 1 1 −1 −1
χ4(g ) 1 −1 −1 1
Table 17.1: The character table of G/H
We can now use this to obtain the linear characters of Q8. We see that a2H = H
and so the linear characters are given in table 17.2.
(c) Clearly there is one more character to find and it has degree 2. Therefore using the
column orthogonality relations it’s easy to see the character table of Q8 is as in table
17.3. Comparing the two we can see that Q8 has the same character table as D8 but
Q8 ∼= D8. Therefore two groups with the same character table are not necessarily
isomorphic.
53
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Chapter 17 54
g 1 a2 a ab b
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 −1 1 −1
χ3(g ) 1 1 1 −1 −1
χ4(g ) 1 1 −1 −1 1
Table 17.2: The linear characters of Q8
g 1 a2 a ab b
|C G(g )| 8 8 4 4 4
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 −1 1 −1
χ3(g ) 1 1 1 −1 −1
χ4(g ) 1 1 −1 −1 1
χ5(g ) 2 −2 0 0 0
Table 17.3: The character table of Q8
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Chapter 17 55
Exercise 17.2. It’s clear that a7 = b 3 = 1 by the elements cycle shape. Now,
b −1
ab = (1b 2b 3b 4b 5b 6b 7b ) = (1357246) = (1234567)(1234567) = a2
.
Therefore the relations hold and G = a, b | a7 = b 3 = 1, b −1ab = a2.
(a) It’s clear from the relations that every element can be written in the form b i a j for
some 0 i 6 and 0 j 2. So we can see that |G| 21. However G contains
the cyclic subgroups a and b , which have orders 7 and 3 respectively. Therefore
7 | |G| and 3 | |G|, which means |G| = 21.
(b) First we note that the conjugacy classes either have size 3 or 7. Now clearly a
and a
2
are conjugate by the relation b −1
ab = a
2
. Using this relation we get theconjugacy classes a, a2, a4 and a3, a5, a6. Now using this relation again we get
b −1(ba)b = ab = ba2 and so we potentially get the conjugacy classes ba,ba2, ba4and ba3, ba5, ba6. Again using the relation we get b −1(b 2a)b = bab = b 2a2 and
so we potentially get the conjugacy classes b 2a, b 2a2, b 2a4 and b 2a3, b 2a5, b 2a6.
What about b and b 2? Recall from the relations that we have a−1b = ba−2 and
a−1b 2 = b 2a4. Then a−1ba = ba5a = ba6 and a−3ba3 = ba−6a3 = ba4. Also
a−1b 2a = b 2a5 and a−3b 2a3 = b 2a12a3 = b 2a. Therefore the conjugacy classes of G
are
1, a, a2, a4, a3, a5, a6,
b,ba,ba2, ba3, ba4, ba5, ba6, b 2, b 2a, b 2a2, b 2a3, b 2a4, b 2a5, b 2a6.
(c) What’s the derived subgroup of G? Well [a, b ] = a−1b −1ab = a and so G = a.
Therefore G/G = G , bG , b 2G ∼= C 3 because (bG )3 = G. Therefore the charac-
ter table of G/G is as in table 17.4. Note that in table 17.4 ω = e 2πi/3 = 1+√ 3i
2is a
3rd root of unity.
gG G bG b 2G
χ1(g ) 1 1 1
χ2(g ) 1 ω ω2
χ3(g ) 1 ω2 ω
Table 17.4: The character table of G/G from exercise 17.2
We can see that aG = a3G = G and so we can lift the above character table
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Chapter 17 57
g 1 a a3 b b 2
|C G(g )
|21 7 7 3 3
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 1 ω ω
χ3(g ) 1 1 1 ω ω
χ4(g ) 3 α β 0 0
χ5(g ) 3 α β 0 0
Table 17.6: The partial character table of G from exercise 17.2
From the row orthogonality we get
3
21+
α
7+
β
7= 0 ⇒ 3 + 3α + 2 β = 0 ⇒ β = −1 − α.
If α = 1+√ 7i
2 then β = −1 − −1+√ 7i
2= −1−√
7i 2
= α. In other words the character
table for G is as in table 17.7.
g 1 a a3 b b 2
|C G(g )| 21 7 7 3 3
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 1 ω ω
χ3(g ) 1 1 1 ω ω
χ4(g ) 3 α α 0 0
χ5(g ) 3 α α 0 0
Table 17.7: The character table of G from exercise 17.2
Exercise 17.3. Recall that |G/G | = |G|/|G| counts the number of linear characters. Now
|G | | |G| and so |G/G | = 1, 2, 3, 4, 6 or 12. From Exercise 15.4(b) we have that a group
of order 12 has either 4, 6 or 12 conjugacy classes, in other words it has either 4, 6 or 12
irreducible characters. Assume that G has 12 conjugacy classes then G is abelian and so
we must have G has 12 linear characters.
Assume G is not abelian and has 6 irreducible characters, say χ1, χ2, χ3, χ4, χ5 and χ6
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Chapter 17 58
then
χ1(1)2 + χ2(1)2 + χ3(1)2 + χ4(1)2 + χ5(1)2 + χ6(1)2 = 12.
In this case we cannot have 6, 3, 2 or 1 linear characters. However we can have 4 linear
characters with χ5(1) = χ6(1) = 2.
Assume G is not abelian and has 4 irreducible characters, then we must have
χ1(1)2 + χ2(1)2 + χ3(1)2 + χ4(1)2 = 12.
In this case we cannot have 4, 3 or 2 linear characters. However we can have 3 linear
characters by letting χ4(1) = 3.
Therefore we have the following options for G. Either G is abelian and so has 12
conjugacy classes and 12 linear characters, G is not abelian and has 6 conjugacy classes
with 4 linear characters or G has 4 conjugacy classes and 3 linear characters. Note that
this shows there is no perfect group, (i.e. a group G such that G = G), of order 12. Also
clearly there is no simple group of order 12 as G is always a non-trivial normal subgroup in
this case.
Exercise 17.4. We use proposition 17.14 to construct a new character from the linear
character given. From the above Exercise we have that because G is of order 12 and G
has 6 conjugacy classes then G has four linear characters. Let χ4 = χ and χ5 = φ thenχ6 = χ5χ4. We have χ3 = χ4 because χ4 isn’t real, which means χ4 is another irreducible
character. Finally we get χ2 by using the column orthogonality relations with χ1. Hence
we have the character table is given as in table 17.8.
g 1 g 2 g 3 g 4 g 5 g 6
χ1(g ) 1 1 1 1 1 1
χ2(g ) 1 −1 −1 1 1 1
χ3(g ) 1 i −i 1 −1 −1χ4(g ) 1 −i i 1 −1 −1
χ5(g ) 2 0 0 −1 −1 2
χ6(g ) 2 0 0 −1 1 −2
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,
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Chapter 17 59
χ3 = χ3, χ4 = χ, χ5 = φ, χ6 = χφ. We can now determine the order of the centralisers
and hence the conjugacy classes using the column orthogonality relations and the orbit
stabiliser theorem. We give these in table 17.9.
g 1 g 2 g 3 g 4 g 5 g 6
|C G(g )| 12 4 4 6 6 12
|g G| 1 3 3 2 2 1
Table 17.9: The conjugacy classes of G from exercise 17.4
Exercise 17.5. We can read off from the character table what the kernels are. Thereforethe normal subgroups of D8 are
ker(χ1) = D8 ker(χ2) = 1, a , a2, a3 ker(χ3) = 1, a2, b , a2b ,
ker(χ4) = 1, a2,ab,a3b ker(χ5) = 1 ker(χ2) ∩ ker(χ3) = 1, a2.
Exercise 17.6. Let G = T 4n = a, b | a2n = 1, an = b 2, b −1ab = a−1.
(a) Let ρ : T 4n → GL(2,C) be the map defined in the question. To show that this is a
representation it is enough to show that the relations hold.
(aρ)2n =
ε 0
0 ε−1
2n=
ε2n 0
0 ε−2n
= I2,
(aρ)n =
εn 0
0 ε−n
=
εn 0
0 εn
=
0 1
εn 0
0 1
εn 0
= (bρ)2,
(bρ)−1(aρ)(bρ) =
0 εn
1 0
ε 0
0 ε−1
0 1
εn 0
,
= 0 εn−1
ε 0 0 1
εn 0 ,
=
ε2n−1 0
0 ε
,
= (aρ)−1.
Therefore this is indeed a representation of T 4n.
(b) We start by considering the conjugacy classes for T 4n. We claim that, for 1 r
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Chapter 17 60
n − 1, the conjugacy classes are as follows
1 |C G(1)| = |G| = 4n,an |C G(an)| = |G| = 4n,
ar , a−r |C G(a)| = |a| = 2n,
ba2 j | 0 j n − 1 |C G(b )| = |b | = 4,
ba2 j +1 | 0 j n − 1 |C G(ab )| = |b | = 4.
Note that there are n + 3 conjugacy classes. We now argue why these are indeed the
conjugacy classes of T 4n. However before doing this we make a passing comment on
the multiplication in T 4n. We have b −1
ab = a−1
⇔ ab = ba−1
⇔ b = a−1
ba−1
⇔ba = a−1b .
• 1. This is clear.
• an. It’s clear that a C G(an) but because an = b 2 we also have b C G(an). In other words |C G(an)| = |G| = 4n, which means |(an)G| = [G :
C G(an)] = 1.
• ar , a−r . We have b −1ar b = a−r from the relations and so ar , a−r ⊆ (ar )G.
What about the centraliser? Well a C G(ar ) and so we have
2 |(ar )G| = [G : C G(ar )] [G : a] = 2.
• ba2 j | 0 j n − 1. We have a−1(ba2 j )a = (a−1b )a2 j +1 = ba2 j +2 = ba2( j +1)
from the relations, so certainly ba2 j | 0 j n − 1 ⊆ b G. What about the
centraliser of b ? Well b C G(b ) which gives us
n |b G| = [G : C G(b )] [G : b ] =4n
4= n.
• ba2 j +1
| 0 j n − 1. We have a−1
(ba2 j +1
)a = (a−1
b )a2 j +2
= ba2 j +3
=ba2( j +1)+1 from the relations, so certainly ba2 j +1 | 0 j n − 1 ⊆ (ba)G. So
far we’ve used 1 + 1 + 2(n − 1) + n = 3n elements and there are n elements in
this set. Therefore this accounts for all the elements and we’re done.
We claim that the degree 2 representation from part (a) is irreducible. This in
fact follows from Exercise 8.4, if ε = ±1, but we will show the inner product of the
character is equal to 1 for verification. Recall that ε2n = 1 ⇒ εn = ±1. We choose
ε = ±1 then the character is as in table 17.10.
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Chapter 17 61
g 1 an ar b ba
|C G(g )
|4n 4n 2n 4 4
χ(g ) 2 −2 εr + ε−r 0 0
Table 17.10: The character of the representation from Exercise 17.6(a)
We now check the inner product of this character with itself to verify that ρ is an
irreducible representation.
χ, χ
=
22
4n
+(−2)2
4n
+
n−1
r =1(εr + ε−r )(εr + ε−r )
2n
,
=1
n+
1
n+
n−1r =1
|εr |2 + εr ε−r + ε−r εr + |εr |22n
.
Recall that εr = ε−r then the summand simplifies to
=2
n+
n−1r =1
2 + ε2r + ε−2r
2n,
= 2n
+ n − 1n
+
n−1r =1
ε2r
n,
= 1 +1
n− 1
n,
= 1.
Therefore the character is irreducible. However note that this is not true if ε = ±1
becausen−1
j =1 ε2 j = n − 1 in this case.
We briefly explain why
n−1 j =1 = ε2 j = −1. Recall ε is a solution of the polynomial
X 2n
− 1 = 0. Now we have
X 2n − 1 = 0 ⇔ (X n + 1)(X n − 1) = 0,
⇔ X n + 1 = 0 or X n − 1 = 0,
⇔ X n + 1 = 0 or (X − 1)(X n−1 + · · · + X 2 + X + 1) = 0.
Consider ε2, then this is a solution of X n − 1 = 0 because (ε2)n = ε2n = 1. If
ε2 = 1 ⇔ ε = ±1 then we have ε2 is a solution of X n−1 + · · · + X + 1 = 0. This
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Chapter 17 62
gives us the required identity for the character calculation. Note, this explains why if
ε = ±1 we have the character is not irreducible
The linear characters are the lifts from the derived subgroup. Using the rela-tions we have b −1(a−1b )a = b −1(ba)a = a−2, so G = a2. This means G/G =
G, aG , bG ,baG and so there are |G/G | = 4 linear characters. If n is even we
have G/G ∼= C 2 × C 2 because b 2 = an ∈ G . If this is the case then the character
table for G is as in table 17.11.
g 1 an ar 1r n−1
b ba
|C G(g )| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 1 1 −1 −1
φ2 1 1 (−1)r 1 −1
φ3 1 1 (−1)r −1 1
χk (g )0k n−1
2 2(−1)k εkr + ε−kr 0 0
Table 17.11: The character of T 4n, when n is even
If n is odd then G/G ∼= C 4 because b 2G = anG = aG and b 3G = bb 2G =banG = baG . Therefore we lift the four linear characters of C 4 to the group G and
we have the character table is as in table 17.12.
Notice that we get a character χk for each 1 k n − 1 when ε is a primitive
2nth root of unity, for example ε = e 2πi/2n. Note that taking n k 2n−1 wouldn’t
give us any more unique characters. This gives us (n − 1) + 4 = n + 3 irreducible
characters and so these are all the irreducible characters of G.
Exercise 17.7. Let G = U 6n = a, b | a2n = b 3 = 1, a−1ba = b −1.
(a) Let ρ : U 6n → GL(2,C) be the map defined in the question. To show this is arepresentation it is enough to show that the relations hold
(aρ)2n =
0 ε
ε 0
2n=
ε2 0
0 ε2
n=
ε2n 0
0 ε2n
= I2,
(bρ)3 =
ω 0
0 ω2
3=
ω3 0
0 ω6
= I2,
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Chapter 17 63
g 1 an ar 1r n−1
b ba
|C G(g )| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 −1 (−1)r i −i
φ2 1 1 1 −1 −1
φ3 1 −1 (−1)r −i i
χk (g )1k n−1
2 2(−1)k εkr + ε−kr 0 0
Table 17.12: The character of T 4n, when n is odd
(aρ)−1(bρ)(aρ) =
0 ε−1
ε−1 0
ω 0
0 ω2
0 ε
ε 0
,
=
0 ε−1ω2
ε−1ω 0
0 ε
ε 0
,
=
ω2 0
0 ω
,
= (bρ)−1.
Therefore ρ certainly is a representation of U 6n.
(b) We start by considering the conjugacy classes of U 6n. We claim that, for 0 j n−1,
the conjugacy classes are as follows
a2 j |C G(1)| = |G| = 6n,
a2 j b, a2 j b 2 |C G(b )| = |b, a2| = 3n,
a2 j +1, a2 j +1b, a2 j +1b 2 |C G(a)| = |a| = 2n.
Note that there are n + n + n = 3n conjugacy classes. We now argue why these are
indeed the conjugacy classes of U 6n. However before doing this we make a passing
comment on the multiplication in U 6n. We have a−1ba = b −1 ⇔ a−1b −1a = b ⇔b −1a = ab and a−1ba = b −1 ⇔ ba = ab −1. Therefore ba j = a j b if j is even and a j b −1
if j is odd.
• a2 j . We have b −1(a2 j )b = a2 j b −1b = a2 j from the relations. This means
b C G(a2 j ) and so C G(a2 j ) = G, which gives us |(a2 j )G| = [G : C G(a j )] = 1.
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Chapter 17 64
• a2 j b, a2 j b 2. We have a−1(a2 j b )a = a2 j −1(ab −1) = a2 j b 2, so certainly we
have a2 j b, a2 j b 2 ⊆ (a2 j b )G. What about the centraliser? We have b C G(a2 j b ) because b C G(a j ) and C G(b ). However we also have a−2(a2 j b )a2 =a−1(a2 j b 2)a = a2 j −1(ab −2) = a2 j b . Therefore a2 C G(a2 j b ), which leaves us
with
2 |(a2 j b )G| = [G : C G(a2 j b )] [G : a2, b ] =6n
3n= 2.
• a2 j +1, a2 j +1b, a2 j +1b 2. We have
b −1a2 j +1b = a2 j +1b 2 and b −2a2 j +1b 2 = a2 j +1b 4 = a2 j +1b
from the relations, so certainly we have a2 j +1, a2 j +1b, a2 j +1b 2 ⊆ (a2 j +1)G.
What about the centraliser? Clearly a C G(a2 j +1), which gives us
3 |(a2 j +1)G| = [G : C G(a2 j +1)] [G : a] =6n
2n= 3.
We claim that the representation defined in part (a) is irreducible. This is true by
Exercise 8.4 for any 2nth root of unity ε. We will also show the inner product is equal
to 1 for verification. The character of the representation is given in table 17.13.
g a2 j a2 j +1 a2 j b
|C G(g )| 6n 2n 3n
χ(g ) 2ε2 j 0 −ε2 j
Table 17.13: The character of the representation from Exercise 17.7(a)
These values come from the following matrix calculations
(a2 j ρ) =
ε2 j 0
0 ε2 j
(a2 j +1ρ) =
0 ε2 j +1
ε2 j +1 0
(a2 j bρ) =
ωε2 j 0
0 ω2ε2 j
and the fact that ω2 + ω = −1. The inner product of the character is
χ, χ =
n−1 j =0
2ε2 j 2ε2 j
6n+
n−1 j =0
−ε2 j −ε2 j
3n,
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Chapter 17 65
=
n−1
j =02|ε2 j |2
3n+
n−1
j =0|ε2 j |2
3n,
=2
3n· n +
1
3n· n,
= 1.
We now consider the derived subgroup of G. We know from the relations that
b −1a−1ba = b −2 = b , which gives us that G = b . Therefore |G/G | = 6n/3 =
2n and so there are 2n linear characters lifted from G. We have that G/G =
G, aG ∼= C 2n because (aG ) = a2nG = G. It’s easy to see that a2 j G = a2 j G ,
a2 j +1G = a2 j +1G and a2 j bG = a2 j G. Therefore the character table of G is as in
table 17.14.
g a2 j 0 j n−1
a2 j +10 j n−1
a2 j b 0 j n−1
|C G(g )| 6n 2n 3n
φk (g )0k 2n−1
ε2kj ε2k ( j +1) ε2kj
χ(g )0n−1
2ε2j 0 −ε2j
Table 17.14: The character table of U 6n
for ε some primitive 2nth root of unity, for example ε = e 2πi/2n. This gives us
2n + n = 3n irreducible characters, which is the number of conjugacy classes and
hence all the irreducible characters.
Exercise 17.8. Let G = V 8n = a, b | a2n = b 4 = 1, ba = a−1b −1, b −1a = a−1b , for n odd.
(a) Let ρ : V 8n → GL(2C) be the map defined in the question. To show this is a
representation it is enough to show that the relations hold
(aρ)2n =
ε 0
0 −ε−1
2n=
ε2n 0
0 (−)2nε−2n
= I2,
(bρ)4 =
0 1
−1 0
4=
−1 0
0 −1
2= I2,
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Chapter 17 66
(bρ)(aρ) =
0 1
−1 0
ε 0
0 −ε−1
,
=
0 −ε−1
−ε 0
,
=
ε−1 0
0 −ε
0 −1
1 0
,
= (aρ)−1(bρ)−1,
(bρ)−1(aρ) =
0 −1
1 0
ε 0
0 −ε−1
,
=0 ε−1
ε 0
,
=
ε−1 0
0 −ε
0 1
−1 0
,
= (aρ)−1(bρ).
Therefore this is indeed a representation of V 8n.
(b) We start by considering the conjugacy classes for V 8n. We claim, for 1 r n−12 and
0 s n − 1, the conjugacy classes are as follows
1 |C G(1)| = |G| = 8n,
b 2 |C G(b 2)| = |G| = 8n,
a2r , a−2r |C G(a2)| = |a, b 2| = 4n,
a2r b 2, a−2r b 2 |C G(a2b 2)| = |a, b 2| = 4n,
a2s +1, a−2s −1b 2 |C G(a)| = |a, b 2| = 4n,
a2b, a2b 3 | 0 n − 1 |C G(b )| = |b | = 4,
a2+1b, a2+1b 3 | 0 n − 1 |C G(ab )| = |b | = 4.
Note that 2 | n−1 because n is odd. We now argue that these are indeed the conjugacy
classes of V 8n. Before doing this we note some incredibly useful multiplication formulae
for the group
ba = a−1b −1 b −1a = a−1b ba−1 = ab −1 b −1a−1 = ab,
ba2r = a−2r b b −1a2r = a−2r b −1 ba2s +1 = a−2s −1b −1 b −1a2s +1 = a−2s −1b.
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Chapter 17 67
• 1. This is clear.
•
b 2
. We have ab 2a−1 = aa−1b −2 = b 2, so
a
C G(b 2). Clearly we have
b C G(b 2), which gives us C G(b 2) = G so |(b 2)G| = 1.
• a2r , a−2r . We have ba2r b −1 = a−2r bb −1 = a−2r , so certainly a2r , a−2r ⊆(a2r )G . What about the centraliser? Clearly a C G(a2r ) but we also have
b 2(a2r )b −2 = ba−2r bb −2 = a2r bb −1 = a2r so b 2 C G(a2r ). This gives us
2 |(a2r )G| = [G : C G(a2r )] [G : a, b 2] =8n
4n= 2.
• a2r b 2, a−2r b 2. We have b (a2r b 2)b −1 = a−2r bb = a−2r b 2, so certainly we have
a2r
b 2
, a−2r
b 2
⊆ (a2r
b 2
)G
. What about the centraliser? By the previous pointwe have b 2 C G(a2r ) so b 2 C G(a2r b 2). Now a(a2r b 2)a−1 = a2r +1b 2a−1 =
a2r +1a−1b −2 = a2r b 2. Therefore a C G(a2r b 2) and so we have
2 |(a2r b 2)G| = [G : C G(a2r b 2)] [G : a, b 2] = 2.
• a2s +1, a−2s −1b 2. We have ba2s +1b −1 = a−2s −1b −1b −1 = a−2s −1b 2, so cer-
tainly a2s +1, a−2s −1b 2 ⊆ (a2s +1)G. What about the centraliser? Clearly a C G(a2s +1) but we also have b 2(a2s +1)b −2 = ba−2s −1b −3 = a2s +1b −4 = a2s +1 so
b 2 C G(a2s +1). This gives us
2 |(a2r )G| = [G : C G(a2r )] [G : a, b 2] = 2.
• a2b, a2b 3 | 0 n − 1. We have aba−1 = aab −1 = a2b 3 and ab 3a−1 =
ab −1a−1 = a2b . Recall that n is odd so when we conjugate by an we get
anba−n = ananb −1 = b 3 and anb 3a−n = anb −1a−n = ananb = b . So certainly we
have a2b, a2b 3 | 0 n − 1 ⊆ b G. What about the centraliser? Clearly
b
C G(b ) so we have
2n |b G| = [G : C G(b )] [G : b ] =8n
4= 2n.
• a2+1b, a2+1b 3 | 0 n − 1. We have a(ab )a−1 = a2ab −1 = a3b 3 and
a(ab 3)a−1 = a2b −1a−1 = a3b . Recall that n is odd so when we conjugate
by an we get an(ab )a−n = an+1ba−n = an+1anb −1 = ab 3 and an(ab 3)a−n =
an+1b −1a−n = an+1anb = ab . So certainly we have a2+1b, a2+1b 3 | 0
n − 1 ⊆ (ab )G. There are 2n elements in this set and we’ve used 1 + 1 +
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Chapter 17 68
(n − 1) + (n − 1) + 2n + 2n = 6n elements so far. Therefore we must have
|(ab )G| = 2n and we’re done.
We know that the representation defined in part (a) is irreducible by Exercise8.4. However we also check the inner product of the character for verification. In
table 17.15 we give the character of this representation with respect to the conjugacy
classes defined above.
g 1 b 2 a2r a2r b 2 a2s +1 b ab
|C G(g )| 8n 8n 4n 4n 4n 4 4
χ(g ) 2 −2 ε2r + ε−2r −ε2r − ε−2r ε2s +1 − ε−2s −1 0 0
Table 17.15: The character of the representation from Exercise 17.8(a)
These values come from the following matrix calculations
(b 2)ρ =
−1 0
0 −1
(a2r )ρ =
ε2r 0
0 ε−2r
(a2r b 2)ρ =
−ε2r 0
0 −ε−2r
,
(a2s +1)ρ = ε2s +1 0
0 −ε−2s −1 (ab )ρ =
0 ε
ε−1
0 .
Recall that for an nth root of unity εk = ε−k . Then the inner product of the
character with itself is as follows
χ, χ =22
8n+
(−2)2
8n+
n−12
r =1
(ε2r + ε−2r )(ε2r + ε−2r )4n
+
n−12
r =1
(ε2r + ε−2r )(ε2r + ε−2r )4n
+
n−1
s =0(ε2s +1 − ε−2s −1)(ε2s +1 − ε−2s )
4n,
=8
8n+ 2
n−12
r =1
2|ε2r |2 + ε2r ε2r + ε−2r ε−2r
4n
+
n−1s =0
2|ε2s +1|2 − ε2s +1ε2s +1 − ε−2s −1ε−2s −1
4n,
=2
2n+
2
2n· n − 1
2+
1
2n· n +
n−12
r =1
ε4r + ε−4r
2n−
n−1s =0
ε4s +2 + ε−4s −2
4n,
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Chapter 17 69
=2 + n − 1 + n
2n+
n−12
r =1ε4r + ε−4r
2n−
n−1
s =0ε4s +2 + ε−4s −2
4n,
=2n + 1
2n+
n−12
r =1
ε4r + ε−4r
2n−
n−1s =0
ε4s +2 + ε−4s −2
4n.
We investigate these sums a little further to make them more enlightening. Now
ε−4r = εn−4r = ε2n−4r . We have 1 r n−12
⇒ 4 4r 2n − 2. From this we can
see that 2n − (2n − 2) 2n − 4r 2n − 4 ⇒ 2 2n − 4r 2n − 4. So for the first
sum we have
n−12
r =1
ε4r + ε−4r =
n−12
r =1
ε4r +
n−32
r =1
ε4r +2,
= ε4 + ε8 + · · · + ε2n−2 + ε2 + ε6 + · · · + ε2n−4,
=
n−1r =1
ε2r .
Considering the second sum we have ε−4s −2 = εn−4s −2 = ε4n−4s −2 = ε4(n−s )−2. So
if 0 s n − 1 then 0 4s 4n − 4 ⇒ 2 4s + 2 4n − 2. This gives us that
4n−
(4n−
2) 4n−
(4s + 2) 4n−
2⇒
2 4(n−
s )−
2 4n−
2. Using this in
the second sum we see
n−1s =0
ε4s +2 + ε−4s −2 =
n−1s =0
ε4s +2 + ε4(n−s )−2,
= 2
n−1s =0
ε4s +2,
= 2(ε2 + ε6 + · · · + ε2n−4 + ε2n + ε2n+4 + · · · + ε4n−2),
= 2(ε2 + ε6 +
· · ·+ ε2n−4 + 1 + ε4 +
· · ·+ ε2n−2),
= 2
n−1s =0
ε2s .
Recall that ε is an nth root of unity and so is a solution to the polynomial X n−1 =
0. We have that 1 is a solution of this polynomial so we get a factorisation
X n − 1 = (X − 1)(X n−1 + · · · + X 2 + X + 1) = 0.
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Chapter 17 70
If ε is an nth root of unity then so is ε2 because (ε2)n = (εn)2 = 12 = 1. Now
ε2 = 1 ⇒ ε = ±1 but ε = −1 because n is odd therefore εn = (−1)n = −1.
Assume ε = 1 then ε2 is a solution of X n − 1 = 0 but not a solution of X − 1 = 0.Therefore ε2 is a solution of X n−1 + · · · + X + 1 = 0. In other words
n−1s =0 ε2s = 0
andn−1
r =1 ε2r = −1. Using all this information in the calculation of our inner product
we have
χ, χ =2n + 1
2n+
n−1r =1
ε2r
2n− 2
n−1s =0
ε2s
4n=
2n + 1 − 1
2n= 1.
Finally, to show this we had to make the assumption that ε = 1. Assume ε = 1
then we have the inner product to be
χ, χ =2n + 1
2n+
n−1r =1
ε2r
2n− 2
n−1s =0
ε2s
4n=
(2n + 1) + (n − 1) − n
2n=
2n
2n= 1.
Therefore χ is an irreducible character of V 8n for any nth root of unit ε.
We now consider the linear characters of G. The derived group is generated by
[a, b ] = a−1b −1ab = a−2b 2 and hence also its inverse b −2a2 = b 2a2. Now (b 2a2)2 =
b 2a2b 2a2 = a4, so a4 G . Recall n = 2v + 1 is odd so a2n = 1 ⇒ a4v +2 = 1 ⇒a4v = a−2, which means a2 a4 G . Using the fact that a−2 ∈ G we have
(b 2a2)a−2 = b 2
∈G . Therefore G =
b 2, a2
, which gives us
|G/G
|= 8n/2n = 4.
In other words G/G = G, aG , bG ,abG with (aG )2 = a2G = G and (bG )2 =
b 2G = G . Clearly G/G ∼= C 2 × C 2 so we can lift the character table of C 2 × C 2
to the whole of G. Checking the cosets of G we can see a2r G = 22r b 2G = G and
a2s +1G = aG . Therefore the character table so far is given in table 17.16. Note
that in table 17.16 ε a primitive nth root of unity and 0 k n − 1.
g 1 b 2 a2r a2r b 2 a2s +1 b ab
|C G(g )| 8n 8n 4n 4n 4n 4 4
λ1(g ) 1 1 1 1 1 1 1
λ2(g ) 1 1 1 1 1 −1 −1
λ3(g ) 1 1 1 1 −1 1 −1
λ4(g ) 1 1 1 1 −1 −1 1
χk (g ) 2 −2 ε2kr + ε−2kr −ε2kr − ε−2kr ε(2s +1)k − ε−(2s +1)k 0 0
Table 17.16: The partial character table of V 8n
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Chapter 17 71
There are 2n + 3 conjugacy classes but in the above table we only have n + 4
irreducible characters. Therefore there are n − 1 irreducible characters left. We could
try and combine a linear character with the χk but this in fact will not give us anew character. So, there must be more irreducible representations. Consider the
representation σ : G → GL(2,C) given by
aσ =
ω 0
0 ω−1
bσ =
0 1
1 0
,
for some 2nth root of unity ω. We check that this indeed a representation of V 8n by
checking the relations.
(aσ)2n =
ω 0
0 ω−1
2n=
ω2n 0
0 ω−2n
= I2,
(bσ)4 =
0 1
1 0
4=
1 0
0 1
2= I2,
(bσ)(aσ) =
0 1
1 0
ω 0
0 ω−1
=
0 ω−1
ω 0
=
ω−1 0
0 ω
0 1
1 0
= (aσ)−1(bσ)−1,
(bσ)−1(aσ) = 0 1
1 0ω 0
0 ω−1 = 0 ω−1
ω 0 = ω−1 0
0 ω0 1
1 0 = (aσ)−1(bσ).
This representation is irreducible, as long as ω = ±1, by Exercise 8.4. We also
check the inner product of the character for verification. The character for σ is given
in table 17.17.
g 1 b 2 a2r a2r b 2 a2s +1 b ab
|C G(g )| 8n 8n 4n 4n 4n 4 4
φ(g ) 2 2 ω2r + ω−2r ω2r + ω−2r ω2s +1 + ω−2s −1 0 0
Table 17.17: The character of σ for V 8n
Assume ω = ±1 then we have the inner product to be
φ, φ =22
8n+
22
8n+ 2
n−12
r =1
(ω2r + ω−2r )(ω2r + ω−2r )4n
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Chapter 17 72
+
n−1
s =0(ω2s +1 + ω−2s −1)(ω2s +1 + ω−2s −1)
4n,
=2
2n+
n−12
r =1
2 + ω4r + ω−4r
4n+
n−1s =0
2 + ω4s +2 + ω−4s −2
4n,
=2
2n+
2
2n· n − 1
2+
2n
4n+
n−1r =1
ω2r
2n+ 2
n−1s =0
ω2s
4n,
=2 + (n − 1) + n − 1
2n,
=2n
2n,
= 1.
Most of the arguments for the above calculation are identical to the previous inner
product calculation, however the arguments for the sums are slightly different. Recall
ω is a 2nth root of unity, hence it is a solution of the polynomial X 2n − 1 = 0. We
have a factorisation X 2n − 1 = (X n +1)(X n − 1) = 0. Now ω2 is an nth root of unity
because (ω2)n = ω2n = 1, hence it is a solution of the polynomial X n − 1 = 0. We
have a factorisation
X n
− 1 = (X − 1)(X n−1
+ · · · + X 2
+ X + 1) = 0.
If ω2 = 1 ⇔ ω = ±1 then ω2 is a solution of the polynomial X n−1 + · · · + X + 1 = 0
and this gives us the desired results as before.
Assume ω = ±1 then we have the inner product calculation gives
φ, φ =2n + 1
2n+
n−1r =1
ω2r
2n+ 2
n−1s =0
ω2s
4n=
(2n + 1) + (n − 1) + n
2n=
4n
2n= 2
and so the character is not irreducible in this case. This confirms what we’re told by
Exercise 8.4 that the representation is irreducible as long as ω = ±1.
Let η be a primitive 2nth root of unity then η2 is a primitive nth root of unity.
Now letting ω = η and ε = η2 we have the character table of V 8n is as in table 17.18.
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Chapter 17 73
g 1 b 2 a2r 1r n−1
2
a2r b 21r n−1
2
a2s +10s n−1
b ab
|C G(g )| 8n 8n 4n 4n 4n 4 4
λ1(g ) 1 1 1 1 1 1 1
λ2(g ) 1 1 1 1 1 −1 −1
λ3(g ) 1 1 1 1 −1 1 −1
λ4(g ) 1 1 1 1
−1
−1 1
χk (g )0k n−1
2 −2 η4kr + η−4kr −η4kr − η−4kr η2k (2s +1) − η−2k (2s +1) 0 0
φ(g )1 n−1
2
2 2 η2r + η−2r η2r + η−2r η(2s +1) + η−(2s +1) 0 0
Table 17.18: The character table of V 8n
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Chapter 18. Some elementary character tables
Exercise 18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Exercise 18.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Exercise 18.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Exercise 18.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Exercise 18.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Exercise 18.1. We need one rotation symmetry of order 4, so we use the obvious a =
(1234). We also need a reflection of order 2 such that b −1(1234)b = (1234)−1 = (1432).Letting b = (14)(23) we have b −1(1234)b = (1b 2b 3b 4b ) = (4321) = (1432). Therefore
D8∼= H = (1234), (14)(23) S4.
Explicitly we have the elements of the subgroup and the conjugacy classes are
H = 1, (13), (24), (12)(34), (13)(24), (14)(23), (1234), (1432),
1
(13)(24)
(1234), (1432)
(24), (13)
(14)(23), (12)(34)
.
Using this information and the table in Example 16.3(3), (page 161), we have the character
table of D8 is as in table 18.1.
g 1 (13)(24) (13) (12)(34) (1234)
|C G(g )| 8 8 4 4 4
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 −1 −1 1
χ3(g ) 1 1 −1 1 −1
χ4(g ) 1 1 1 −1 −1
χ5(g ) 2 −2 0 0 0
π(g ) 4 0 2 0 0
Table 18.1: The character table of D8
Clearly π is not the sum of the four linear characters as the values on (13) don’t agree.
74
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Chapter 18 75
Therefore it is the sum of χ5 and two linear characters. The dimension of the trivial module
in the permutation module is always one, so we know one of the linear characters is the
trivial character. Therefore it’s clear to see that π = χ1 + χ4 + χ5.Note: the decomposition of the permutation character is dependent upon the choice
of D8 as a subgroup of S4. Choosing b to be a reflection in a diagonal of the square will
result in π = χ1 + χ3 + χ5, (see solutions - page 419).
Exercise 18.2. By section 18.3 we have the character tables for all dihedral groups. Ex-
plicitly the character table of D12 is as in table 18.2. Recall ε = e 2πi/6 is a primitive 6th
root of unity. Explicitly we have ε = 1+√ 3i
2 , ε2 = −1+√ 3i
2 and ε4 = −ε = −1−√ 3i
2 .
g 1 a3
a a2
b ab |C G(g )| 12 12 6 6 2 2
χ1(g ) 1 1 1 1 1 1
χ2(g ) 1 1 1 1 −1 −1
χ3(g ) 1 −1 −1 1 1 −1
χ4(g ) 1 −1 −1 1 −1 1
ψ1(g ) 2 −2 1 −1 0 0
ψ2(g ) 2 2−
1−
1 0 0
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(χ1) = G ker(χ4) = a2, ab ,
ker(χ2) = a ker(ψ1) = 1,
ker(χ3) =
a2, b
ker(ψ2) =
a3
.
Using Proposition 17.5 we have that these are nearly all the normal subgroups of D12. In
fact the only subgroup that is missing is ker(χ3) ∩ ker(χ4) = a2. Therefore the seven
distinct normal subgroups of D12 are 1, a3, a2, a, a2, b , a2, ab and G.
Exercise 18.3. See Exercise 17.6
Exercise 18.4. See Exercise 17.7
Exercise 18.5. See Exercise 17.8
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Chapter 19. Tensor products
Exercise 19.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
Exercise 19.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
Exercise 19.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
Exercise 19.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Exercise 19.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
Exercise 19.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
Exercise 19.1. Examining the inner products we see
χψ,φ =1
|G|g ∈G
χψ(g )φ(g ) χψ,φ =1
|G|g ∈G
χψ(g )φ(g ),
=1
|G|g ∈G
χ(g )ψ(g )φ(g ) =1
|G|g ∈G
χ(g )ψ(g )φ(g ),
=1
|G|g ∈G
χ(g )ψ(g )φ(g ) =1
|G|g ∈G
ψ(g )χ(g )φ(g ),
= g ∈G χ(g )ψφ(g ) =
1
|G| g ∈G ψ(g )χφ(g ),
= χ,ψφ = ψ,χφ.
Therefore χψ,φ = χ,ψφ = ψ,χφ as required.
Exercise 19.2. Let χ and ψ be irreducible characters of G then, using Exercise 1, we have
χψ, 1G = χ, ψ1G = χ, ψ =
1 if χ = ψ,
0 if χ
= ψ.
Recall that if ψ is an irreducible character then so is ψ, by Proposition 13.15. Therefore
the final equality comes from Theorem 14.12.
Exercise 19.3. Let U be a CG module which affords the unfaithful character χ. As χ is
not faithful we know there exists a non-identity element g ∈ ker(χ) such that
ug = u for all u ∈ U.
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Chapter 19 77
The element g is not conjugate to 1, therefore there exists an irreducible character ψ such
that ψ(g ) = ψ(1), (by Proposition 15.5).
Let n 0 be an integer and consider the CG module constructed as the n-fold tensorproduct of U , say W = U ⊗ · · · ⊗ U . Clearly
w g = (u 1 ⊗ · · · ⊗ u n)g = u 1g ⊗ · · · ⊗ u ng = u 1 ⊗ · · · ⊗ u n = w for all w ∈ W.
Consider any submodule W ⊆ W and let the character afforded by this submodule be φ.
Then it’s clear that φ(g ) = φ(1) because w g = w for all w ∈ W . So if ψ(g ) = ψ(1) we
must have χn, ψ = 0 because ψ cannot correspond to a submodule of W .
Exercise 19.4. We express the characters χS, χA, φS and φA in table 19.1. We also calcu-late what g 2 is for each representative of the conjugacy classes. Then we use the formulae
for χS and χA given in the summary of Chapter 19.
g 1 (123) (12)(34) (12345) (13452)
g 2 1 (132) 1 (13524) (14235)
χ(g ) 5 −1 1 0 0
χS(g ) 15 0 3 0 0
χA(g ) 10 1 −2 0 0
φ(g ) 3 0 −1 1+√ 5
21−√
52
φS(g ) 6 0 2 1 1
φA(g ) 3 0 −1 1+√ 5
21−√
52
Table 19.1: Decomposing squares of characters of A5
It is worthwhile commenting that (12345) is conjugate to (13524) by (2354) ∈ A5
and (13452) is conjugate to (14235) by (3425)∈A5
. Therefore the squared five cycles
lie in the opposite conjugacy class to where they started. We explain in more depth the
calculations in the above table. For a start we have1 +
√ 5
2
2
=1 + 2
√ 5 + 5
4=
3 +√
5
2
1 − √
5
2
2
=1 − 2
√ 5 + 5
4=
3 − √ 5
2
Using this information we can then show the explicit calculations of φS and φA on the five
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Chapter 19 78
cycles.
φS((12345)) = 123 +
√ 52 + 1 −
√ 52 = 12 · 42 = 1,
φA((12345)) =1
2
3 +
√ 5
2− 1 − √
5
2
=
1
2
2 + 2
√ 5
2
=
1 +√
5
2,
φS((13452)) =1
2
3 − √
5
2+
1 +√
5
2
=
1
2· 4
2= 1,
φA((13452)) =1
2
3 − √
5
2− 1 +
√ 5
2
=
1
2
2 − 2
√ 5
2
=
1 − √ 5
2.
Using the character table for A5 from Example 20.14, page 221, it’s easy to spot that
φA = ψ4 and φS = ψ1 + ψ3. Now χS and χA are a little more difficult to spot. We first
consider the inner products with themselves, this gives us
χS, χS =152
60+
32
4=
225 + 135
60=
360
60= 6,
χA, χA =102
60+
12
3+
(−2)2
4=
100 + 20 + 60
60=
180
60= 3.
Therefore we have d 21 + d 22 + d 23 + d 24 + d 25 = 6 or 3. In the first case our only option is 2, 1, 1
and in the second case our only option is 1, 1, 1. Using this information and working on the
values of χS(1) and χA(1) it’s easy to see that χA = ψ2+ ψ4+ ψ5 and χS = ψ1+ ψ2+ 2ψ3.
Exercise 19.5. We express all the information in the question in table 19.2 and use the
formulas for χS and χA to complete the table.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
g 2i g 1 g 1 g 2 g 5 g 4 g 4 g 5
|C G(g i )| 24 24 4 6 6 6 6
χ(g ) 2 −2 0 −ω2 −ω ω ω2
χS(g ) 3 3 −1 0 0 0 0
χA(g ) 1 1 1 ω ω2 −1 ω
Table 19.2: Symmetric and alternating components of χ2 in Exercise 19.5
To show that χS and χA are irreducible characters we consider the inner products with
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Chapter 19 79
themselves. This gives us
χS, χS =32
24 +32
24 +(−
1)2
4 =9 + 9 + 6
24 =24
24 = 1,
χA, χA =12
24+
12
24+
12
4+
ωω
6+
ω2ω2
6+
(−1)2
6+
ωω
6,
=1 + 1 + 6 + 4 + 4 + 4 + 4
24,
=24
24,
= 1.
Recalling that ωω = ω
2
ω2
= |ω|2
= 1. Therefore the characters are irreducible. Howeverit is not clear that χ itself is an irreducible character. So we check the inner product
χ, χ =22
24+
(−2)2
24+
ω2ω2
6+
ωω
6+
ωω
6+
ω2ω2
6=
4 + 4 + 4 + 4 + 4 + 4
24=
24
24= 1.
Therefore we have three irreducible characters of G and the trivial character gives us a
fourth. Now χA is a linear character, so we know straight away that χχA and χ2A will be
irreducible characters and as we can see below in the table, they are distinct from the other
characters. Finally χ2A is another irreducible linear character so we have χχ2
A is another
irreducible character. Therefore the character table for G is as in table 19.3.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
|C G(g i )| 24 24 4 6 6 6 6
1G 1 1 1 1 1 1 1
χA(g ) 1 1 1 ω ω2 −1 ω
χ2A(g ) 1 1 1 ω2 ω 1 ω2
χ(g ) 2
−2 0
−ω2
−ω ω ω2
χχA(g ) 2 −2 0 −1 −1 −ω 1
χχ2A(g ) 2 −2 0 −ω −ω2 ω ω
χS(g ) 3 3 −1 0 0 0 0
Table 19.3: The character table of G from Exercise 19.5
Exercise 19.6. Recall the character table of D6 from Example 16.3(1) on page 160. There
are 9 conjugacy classes in D6 × D6, represented by pairs of conjugacy classes from each
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Chapter 19 80
copy of D6. Therefore the character table is as in table 19.4.
g i (1, 1) (1, a) (1, b ) (a, 1) (a, a) (a, b ) (b , 1) (b , a) (b , b )
χ1 × χ1(g ) 1 1 1 1 1 1 1 1 1
χ1 × χ2(g ) 1 1 −1 1 1 −1 1 1 −1
χ1 × χ3(g ) 2 −1 0 2 −1 0 2 −1 0
χ2 × χ1(g ) 1 1 1 1 1 1 −1 −1 −1
χ2 × χ2(g ) 1 1 −1 1 1 −1 −1 −1 1
χ2 × χ3(g ) 2 −1 0 2 −1 0 −2 1 0
χ3 × χ1(g ) 2 2 2 −1 −1 −1 0 0 0χ3 × χ2(g ) 2 2 −2 −1 −1 1 0 0 0
χ3 × χ3(g ) 4 −2 0 −2 1 0 0 0 0
Table 19.4: The character table of D6 × D6
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Chapter 20. Restriction to a subgroup
Exercise 20.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Exercise 20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
Exercise 20.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
Exercise 20.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Exercise 20.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
Exercise 20.1. We have G = S4 and H = (1234), (13) G.
(a) Let ϕ : H → D8 = a, b | a4 = b 2 = 1,bab = a−1 be the map defined by (1234)ϕ =a and (13)ϕ = b . Clearly (1234)4 = 1 and (13)2 = 1. Also (13)(1234)(13) =
(3214) = (1432) = (1234)−1. Therefore ϕ is certainly a homomorphism. It’s clear
that ker(ϕ) = 1 and Im(ϕ) = D8 therefore ϕ is an isomorphism.
(b) Recall the conjugacy classes of H are given by 1, (13)(24), (1234), (1432),
(13), (24), (14)(23), (12)(34). The character tables of S4 and D8, (taken from
Section 18.1 on page 180 and Example 16.3(3) on page 162 respectively), are given
in tables 20.1 and 20.2.
g 1 (12) (123) (12)(34) (1234)
|C G(g )| 24 4 3 8 4
χ1(g ) 1 1 1 1 1
χ2(g ) 1 −1 1 1 −1
χ3(g ) 2 0 −1 2 0
χ4(g ) 3 1 0 −1 −1
χ5(g ) 3 −1 0 −1 1
Table 20.1: The character table of S4
Note that [G : H ] = 248
= 3 and H is not normal as it contains only 2 of the 6 four
cycles in S4, so it is not a union of conjugacy classes. Now the restricted character
table of S4 to H is as in table 20.3.
Just by inspection it’s easy to see that
χ1 ↓ H = ψ1 χ4 ↓ H = ψ3 + ψ5,
81
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Chapter 20 82
g 1 a2 a b ab
|C G(g )
|8 8 4 4 4
ψ1(g ) 1 1 1 1 1
ψ2(g ) 1 1 1 −1 −1
ψ3(g ) 1 1 −1 1 −1
ψ4(g ) 1 1 −1 −1 1
ψ5(g ) 2 −2 0 0 0
Table 20.2: The character table of D8
h 1 (13)(24) (1234) (13) (14)(23)
|C H(h)| 8 8 4 4 4
(χ1 ↓ H )(h) 1 1 1 1 1
(χ2 ↓ H )(h) 1 1 −1 −1 1
(χ3 ↓ H )(h) 2 2 0 0 2
(χ4 ↓ H )(h) 3 −1 −1 1 −1
(χ5 ↓ H )(h) 3 −1 1 −1 −1
Table 20.3: The characters of S4 restricted to H
χ2 ↓ H = ψ4 χ5 ↓ H = ψ2 + ψ5,
χ3 ↓ H = ψ1 + ψ4.
Therefore ψ1, ψ2, ψ3, ψ4 and ψ5 all occur as components of the restricted characters.
Exercise 20.2. From Exercise 12.4 we know that the cycle types appearing in A6 are (1),
(2, 2), (4, 2), (3), (3, 3) and (5). We also know that the only class that splits are the (5)
cycles. From Exercise 12.3 we know that these conjugacy classes have respective orders 1,
45, 90, 40, 40, 72 and 72. The character table of S6, (taken from Example 19.17 on page
205), is given in table 20.4. In the table we have shaded the columns which correspond to
cycle types in A6.
It’s easy to see from the shaded columns of the character table that
χ1 ↓ A5 = χ2 ↓ A5 χ3 ↓ A5 = χ4 ↓ A5 χ5 ↓ A5 = χ6 ↓ A5,
χ7 ↓ A5 = χ8 ↓ A5 χ9 ↓ A5 = χ10 ↓ A5.
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Chapter 20 83
Class (1) (2) (3) (2, 2) (4) (3, 2) (5) (2, 2, 2) (3, 3) (4, 2) (6)
|C G(g )
|720 48 18 16 8 6 5 48 18 8 6
χ1(g ) 1 1 1 1 1 1 1 1 1 1 1
χ2(g ) 1 −1 1 1 −1 −1 1 −1 1 1 −1
χ3(g ) 5 3 2 1 1 0 0 −1 −1 −1 −1
χ4(g ) 5 −3 2 1 −1 0 0 1 −1 −1 1
χ5(g ) 10 2 1 −2 0 −1 0 −2 1 0 1
χ6(g ) 10 −2 1 −2 0 1 0 2 1 0 −1
χ7(g ) 9 3 0 1 −1 0 −1 −3 0 1 0
χ8(g ) 9
−3 0 1 1 0
−1
−3 0 1 0
χ9(g ) 5 1 −1 1 −1 1 0 −3 2 −1 0
χ10(g ) 5 −1 −1 1 1 −1 0 3 2 −1 0
χ11(g ) 16 0 −2 0 0 0 1 0 −2 0 0
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)
|C H(h)| 360 9 8 5 5 9 4
ψ1(h) 1 1 1 1 1 1 1
ψ2(h) 5 2 1 0 0 −1 −1
ψ3(h) 10 1 −2 0 0 1 0
ψ4(h) 9 0 1 −1 −1 0 1ψ5(h) 5 −1 1 0 0 2 −1
ψ6(h) 8 α1 α2 α3 α4 α5 α6
ψ7(h) 8 β1 β2 β3 β4 β5 β6
Table 20.5: The partial character table of A6
Using the column orthogonality relations with the first column and each of the remaining
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Chapter 20 84
columns we get a system of equations
1 + 10 + 10 + 0 − 5 + 8α1 + 8 β1 = 01 + 5 − 20 + 9 + 5 + 8α2 + 8 β2 = 0
1 + 0 + 0 − 9 + 0 + 8α3 + 8 β3 = 0
1 + 0 + 0 − 9 + 0 + 8α4 + 8 β4 = 0
1 − 5 + 10 + 0 + 10 + 8α5 + 8 β5 = 0
1 − 5 + 0 + 9 − 5 + 8α6 + 8 β6 = 0
⇒
α1 + β1 = −2,α2 + β2 = 0,
α3 + β3 = 1,
α4 + β4 = 1,
α5 + β5 = −2,
α6 + β6 = 0.
Using the column orthogonality relations of the columns with themselves we get a system
of equations
1 + 4 + 1 + 0 + 1 + α21 + β21 = 9
1 + 1 + 4 + 1 + 1 + α22 + β2
2 = 8
1 + 0 + 0 + 1 + 0 + α23 + β2
3 = 5
1 + 0 + 0 + 1 + 0 + α24 + β2
4 = 5
1 + 1 + 1 + 0 + 4 + α25 + β2
5 = 9
1 + 1 + 0 + 1 + 1 + α26 + β2
6 = 4
⇒
α21 + β21 = 2,
α22 + β2
2 = 0,
α23 + β2
3 = 3,
α24 + β2
4 = 3,
α25 + β2
5 = 2,
α26 + β2
6 = 0.
Recall that the inverse of any element in A6 is of the same cycle shape as the original
element. Therefore every element inA6 is conjugate to its inverse. This tells us that all
the entries in the character table of A6 are real. Therefore we can straight away say that
α2 = β2 = α6 = β6 = 0 and α1 = β1 = α5 = β5 = −1. Now β3 = 1 − α3 and
α23 + β2
3 = 3 ⇒ α23 + (1 − α3)2 = 3,
⇒ 2α23 − 2α3 + 1 = 3,
⇒ α23 − α3 − 1 = 0,
⇒ α3 =1 ± √
5
2.
Now choose α3 = 1+√ 5
2then β3 = 1−√
52
. It’s clear that α4 and β4 are solutions of the
same equations as α3 and β3. We also know that ψ6 and ψ7 are different characters so
must have α4 = β3 and β4 = α3. This gives us the character table of A6 is as in table 20.6.
Exercise 20.3. If H is an abelian subgroup of G then every irreducible character, say ψi ,
of H is linear. Certainly, as H is abelian, we have H is a normal subgroup of G. Therefore
by Clifford’s Thereom we have χ ↓ H = e (ψ1 + · · · + ψm) for some positive integer e . Now
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Chapter 20 85
Class (1) (3) (2, 2) (5) (5) (3, 3) (4, 2)
|C H(h)
|360 9 8 5 5 9 4
ψ1(h) 1 1 1 1 1 1 1
ψ2(h) 5 2 1 0 0 −1 −1
ψ3(h) 10 1 −2 0 0 1 0
ψ4(h) 9 0 1 −1 −1 0 1
ψ5(h) 5 −1 1 0 0 2 −1
ψ6(h) 8 −1 0 1+√ 5
21−√
52
−1 0
ψ7(h) 8
−1 0 1−√
52
1+√ 5
2
−1 0
Table 20.6: The character table of A6
it’s clear that
χ(1) = (χ ↓ H )(1) = e (ψ1(1) + · · · + ψm(1)) = em,
where m = |H |.From Proposition 20.5 we have that
mi =1 e 2 = e 2m [G : H ] = n. However, as e and
m are positive integers we clearly have
em e 2m [G : H ] = n ⇒ χ(1) n.
Exercise 20.4. Let χ be an irreducible character of G. Let ψ1, . . . , ψm be the irreducible
characters of H . Then χ ↓ H = e 1ψ1 + · · · + e mψm. If H is a subgroup of index 3 then by
Propotisition 20.5 we have
e 21 + · · · + e 2m 3.
Therefore the only possibilities are that one of the e i
is 1, two of the e i
are 1 or three of
the e i are 1. In other words
χ ↓ H, χ ↓ H H = e 21 + · · · + e 2m = 1, 2 or 3.
We want to find examples of each of these cases occurring. Consider Exercise 1 of
this chapter. We have D8∼= (1234), (13) G is a subgroup of index 3 because [G :
(1234), (13)] = |G||D8| = 24
8= 3. Keeping the notation of that exercise we have χ1 ↓ H =
ψ1 and χ3 ↓ H = ψ1 + ψ4 which shows the possibilities of χ ↓ H, χ ↓ H = 1 or 2.
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Chapter 20 86
For the final example consider the group A4. The character table of A4, (taken from
18.2 on page 181), is given in table 20.7.
g (1) (12)(34) (123) (132)
|C G(g )| 12 4 3 3
χ1(g ) 1 1 1 1
χ2(g ) 1 1 ω ω2
χ3(g ) 1 1 ω2 ω
χ4(g ) 3 −1 0 0
Table 20.7: The character table of A4
Now the linear characters are lifts from A4/V 4 where V 4 = 1, (12)(34), (13)(24), (14)(23)is the derived subgroup of A4. Note that |A4/V 4| = 3, so V 4 is a subgroup of index 3 in A4.
Also we have (12)(34)2 = (13)(24)2 = 1 and (12)(34)(13)(24) = (14)(23) so V 4 ∼= C 2×C 2.
Therefore we know the character table of V 4, which we give in table 20.8 along with the
restricted character χ4 ↓ V 4.
h (1) (123) (132)
|C V 4(h)| 12 4 3 3
ψ1(h) 1 1 1 1
ψ2(h) 1 1 −1 −1
ψ3(h) 1 −1 1 −1
ψ4(h) 1 −1 −1 1
χ4 ↓ V 4 3 −1 −1 −1
Table 20.8: The character table of V 4
By inspection we can see that χ4 ↓ V 4 = ψ2 + ψ3 + ψ4.
Exercise 20.5. If an irreducible representation, say χ, of S7 restricts to an irreducible
representation of A7 then there is another irreducible representation, say χ of S7 such
that χ ↓ A7 = χ ↓ A7. Also if χ restricts to a sum of two irreducible representations of A7
then they have the same degree and so we must have 2 | χ(1).
Using this information we can immediately see that 1, 15, 21 and 35 are four of the
degrees of the irreducible characters of A7. As 20 occurs on its own we must have that
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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.
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Chapter 21. Induced modules and characters
Exercise 21.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Exercise 21.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Exercise 21.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
Exercise 21.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Exercise 21.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Exercise 21.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
Exercise 21.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
Exercise 21.1. We have G = D8 = a, b | a4 = b 2 = 1, b −1ab = a−1 and H = a2, b G.
Note that H ∼= C 4, which means |H | = 4.
(a) Let u = 1 − a2 + b − a2b then
ua2 = (1 − a2 + b − a2b )a2 = a2 − 1 + ba2 − a2ba2 = −1 + a2 − b + a2b = −u ∈ U,
ub = (1 − a2 + b − a2b )b = b − a2b + b 2 − a2b 2 = 1 − a2 + b − a2b = u ∈ U.
Therefore U is a submodule of CH .
(b) We have the induced module to be
U ↑ G = span1 − a2 + b − a2b, a − a3 + ba − a2ba,b − a2b + b 2 − a2b 2,
= span1 − a2 + b − a2b, a − a3 − ab + a3b .
Therefore u and ua are a basis for U ↑ G.
(c) Now the character of the CH module U is given in table 21.1.
h 1 a2
b a2
b |C H(h)| 4 4 4 4
χ(h) 1 −1 1 −1
Table 21.1: The character of the CH module U
We have the conjugacy classes of D8 to be 1, a2, a, a3, b, a2b and ab,a3b .
Only 1, a2 and b, a2b lie in H but b, a2b splits in two. Therefore the values
88
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Chapter 21 89
of the induced character are
(χ ↑ G)(1) = |G
||H |χ(1) = 2
(χ ↑ G)(a2) =|C G(a2)||C H(a2)|χ(a2) = −2
(χ ↑ G)(b ) = |C G(b )|
ψ(b )
|C H(b )| +ψ(a2b )
|C H(a2b )|
= 4 · 1 − 1
4= 0
In particular the induced character is as in table 21.2. We know from the character
table of D8 that χ ↑ G is irreducible but we also have χ ↑ G, χ ↑ G = 22
8+ (−2)2
8= 1.
g 1 a2 a b ab |C G(h)| 4 4 4 4
(χ ↑ G)(g ) 2 −2 0 0 0
Table 21.2: The induced character of the CH module U
Exercise 21.2. Let G = S4 and H = (123) ∼= C 3.
(a) We have the character tables of S4 and C 3 = a | a3 = 1, (taken from 18.1 on
page 180 and Example 16.3(2) on page 160 respectively), are given in tables 21.3
and 21.4. In the tables ω = e 2πi/3 is a primitive 3rd root of unity.
g 1 (12) (123) (12)(34) (1234)
|C G(g )| 24 4 3 8 4
χ1(g ) 1 1 1 1 1
χ2(g ) 1 −1 1 1 −1
χ3(g ) 2 0 −1 2 0
χ4(g ) 3 1 0 −1 −1χ5(g ) 3 −1 0 −1 1
Table 21.3: The character table of S4
The restricted character table of S4 to H ∼= C 3 is then given in table 21.5. Using the
character table of C 3 it is easy to see that
χ1 ↓ H = χ2 ↓ H = ψ1 χ3 ↓ H = ψ2 + ψ3 χ4 ↓ H = χ5 ↓ H = ψ1 + ψ2 + ψ3.
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Chapter 21 90
g 1 a a2
|C G(g )
|3 3 3
ψ1(g ) 1 1 1
ψ2(g ) 1 ω ω2
ψ3(g ) 1 ω2 ω
Table 21.4: The character table of C 3
g 1 (123) (132)
|C G(g )
|24 4 3
(χ1 ↓ H )(g ) 1 1 1
(χ2 ↓ H )(g ) 1 1 1
(χ3 ↓ H )(g ) 2 −1 −1
(χ4 ↓ H )(g ) 3 0 0
(χ5 ↓ H )(g ) 3 0 0
Table 21.5: The characters of S4 restricted to H
(b) Using Frobenius reciprocity it is easy to see that
ψ1 ↑ G = χ1 + χ2 + χ4 + χ5 ψ2 ↑ G = ψ3 ↑ G = χ3 + χ4 + χ5.
Exercise 21.3. Let ψ be a character of H G and let U be the corresponding CH module
that affords ψ as a character. Now U ∼= U 1 ⊕· · ·⊕ U m for some irreducible CH -submodules
U i of CH . Let ψi be the irreducible character of U afforded by the submodule U i , then
ψ = ψ1
+· · ·
+ ψm
. In definition 21.9 we define
U ↑ G = (U 1 ↑ G) ⊕ · · · ⊕ (U m ↑ G) ⇒ ψ ↑ G = ψ1 ↑ G + · · · + ψm ↑ G.
Therefore we have (ψ ↑ G)(1) = (ψ1 ↑ G)(1) + · · · + (ψm ↑ G)(1).
Consider one of the irreducible CH submodules, say U k . Then we have that U k has
a basis v 1, . . . , v , for some v 1, . . . , v ∈ CH . We define the induced module to be
U k (CG) = spanxg | x ∈ U k , g ∈ G. Let v ∈ U then we have v g ∈ U if and only if g ∈ H .
So, let S be a right transversal for H in G. Then v j s | 1 j , s ∈ S is a basis for the
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Chapter 21 91
induced module U k (CG). In other words
(ψk ↑ G)(1) = dim(U k ↑ G) = dim(U k (CG)) = |S| dim(U k ) = |S|ψk (1) = |G
||H |ψk (1).
Now extending this to the whole of ψ we get
(ψ ↑ G)(1) = (ψ1 ↑ G)(1) + · · · + (ψm ↑ G)(1),
=|G||H |ψ1(1) + · · · +
|G||H |ψm(1),
=|G||H |(ψ(1) + · · · + ψ(m)),
= |G||H |ψ(1),
as required.
Exercise 21.4. Let φ be any irreducible character of G then taking the inner product we
have
(ψ(χ ↓ H )) ↑ G, φG = ψ(χ ↓ H ), φ ↓ H H,
=
ψ, χ
↓Hφ
↓H
H,
= ψ,χφ ↓ H H,
= ψ ↑ G,χφG,
= (ψ ↑ G)χ, φG.
Therefore as φ was an arbitrary irreducible character of G we must have that ψ(χ ↓ H ) ↑G = (ψ ↑ G)χ. Note the results on inner products used above come from Exercise 19.1.
Exercise 21.5. For a start we have the elements of H have the form
a = (1234567) ab = (137)(254) ab 2 = (157)(364),
a2 = (1234567) a2b = (156)(273) a2b 2 = (126)(475),
a3 = (1357246) a3b = (175)(346) a3b 2 = (165)(237),
a4 = (1473625) a4b = (124)(365) a4b 2 = (134)(276),
a5 = (1526374) a5b = (143)(267) a5b 2 = (173)(245),
a6 = (1765432) a6b = (162)(457) a6b 2 = (142)(356),
a7 = 1 b = (235)(476) b 2 = (253)(467).
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Chapter 21 92
Therefore every element is either the identity a 7 cycle or a 3-3 cycle. Recall the conjugacy
classes of S7 are determined by their cycle type. Then using the formula in Proposition
21.23 we have
(ψ ↑ S7)(1) =|G||H |ψ(1),
(ψ ↑ S7)((1234567)) = |C G(a)|
ψ(a)
|C H(a)| +ψ(a3)
|C H(a3)|
,
(ψ ↑ S7)((123)(456)) = |C G(b )|
ψ(b )
|C H(b )| +ψ(b 2)
|C H(b 2)|
If σ ∈ S7 is not the identity, a 7 cycle or a 3-3 cycle then we will have (ψ ↑ S7)(σ) = 0.
Therefore this gives us that the induced character is as in table 21.6.
Class (1) (3, 3) (7)
(φ ↑ S7)(g ) 240 12 2
(ψ ↑ S7)(g ) 720 0 −1
Table 21.6: The characters φ ↑ S7 and ψ ↑ S7 of S7
Exercise 21.6. By Corollary 21.20 or Exercise 21.3, we have (ψ ↑ G)(1) = [G : H ]ψ(1)or, in other words, [G : H ]ψ(1) = d 1χ1(1)+ · · · + d k χk (1). Using Frobenius Reciprocity we
have
d i = ψ ↑ G, χi G = ψ, χi ↓ H H.
Now χi ↓ H = d i ψ +φ where φ is a character of H or 0. Therefore, as a restricted character
as the same degree as the original character, χi (1) = (χi ↓ H )(1) = d i ψ(1) + φ(1)
d i ψ(1). Using this we get
d 21ψ(1) + · · · + d 2k ψ(1) d 1χ1(1) + · · · + d k χk (1) ⇒ (d 21 + · · · + d 2k )ψ(1) [G : H ]ψ(1),
⇒k
i =1
d 2i [G : H ]
because ψ(1) > 0.
Exercise 21.7. Supose that H is a normal subgroup of index 2 in G and let ψ be an
irreducible character of H .
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Chapter 21 93
Proposition 21.1 (20.9). We have that either
(a) ψ ↑ G is an irreducible character with degree 2ψ(1), or
(b) ψ ↑ G is the sum of two distinct irreducible characters of G, which both have degree
ψ(1).
Proof. Let χ1, . . . , χk be the irreducible characters of G then there exists d 1, . . . , d k such
that ψ ↑ G = d 1χ1 + · · · + d k χk . By the above theorem we havek
i =1 d 2i [G : H ] = 2.
Now the d i are positive integers and so we have only two options. Either ψ ↑ G = χi or
ψ ↑ G = χi + χ j for some i , j with i = j . In the first case you know, by Proposition 21.20
or Exercise 3, that (ψ ↑ G)(1) = [G : H ]ψ(1) = 2ψ(1). Therefore χi (1) = 2ψ(1).
For the second case we have by Frobenius Reciprocity that
1 = ψ ↑ G, χi G = ψ, χi ↓ H H,
1 = ψ ↑ G, χ j G = ψ, χ j ↓ H H.
Therefore ψ is a constituent of χi ↓ H and χ j ↓ H . This tells us that χi (1) = (χi ↓ H )(1)
ψ(1) and χ j (1) = (χ j ↓ H )(1) ψ(1). However 2ψ(1) = χi (1) + χ j (1), so we must have
χi (1) = χ j (1) = ψ(1).
Proposition 21.2 (20.11). Assume ψ ↑ G is irreducible. Aside from ψ there is only one
other irreducible character of H , say φ, such that φ↑
G = ψ↑
G.
Proof. If ψ ↑ G is irreducible then ψ ↑ G = χ for some irreducible character χ of G. By
Frobenius Reciprocity we know that χ ↓ H = ψ + φ where φ is an irreducible character of
H or φ = 0. Assume φ = 0. We know that a restriction of an irreducible character has the
same degree as the original character. Therefore,
2ψ(1) = (ψ ↑ G)(1) = ((ψ ↑ G) ↓ H )(1) = (χ ↓ H )(1) = ψ(1).
However this is a contradiction as ψ(1) = 0.
Therefore we have φ is an irreducible character of H . Using the Frobenius Reciprocitytheorem we have
1 = φ, χ ↓ H H = φ ↑ G, χG.
In other words φ ↑ G = χ = ψ ↑ G and we’re done.
Proposition 21.3 (20.12). Assume ψ ↑ G = χ1+χ2 for some irreducible characters χ1, χ2
of G. If φ is an irreducible character of H such that φ ↑ G has χ1 or χ2 as a constituent
then φ = ψ.
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Chapter 21 94
Proof. If χ1 is a constituent of φ ↑ G then χ1 has coefficient 1 by the above Proposition
20.9 for induction. Therefore, using Frobenius Reciprocity, we have
1 = ψ ↑ G, χ1G = ψ, χ1 ↓ H H 1 = φ ↑ G, χ1G = φ, χ1 ↓ H H.
Also ψ(1) = χ1(1) and φ(1) = χ1(1), so ψ = χ1 = φ and we’re done.
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Chapter 22. Algebraic integers
Exercise 22.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Exercise 22.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Exercise 22.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Exercise 22.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Exercise 22.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Exercise 22.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Exercise 22.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Exercise 22.1. Let χi be the irreducible characters of G. Then we know that
|G| = |G/G | +
χ(1)>1
χ(1)2
where the sum is over irreducible characters of G. We note that |G| | |G|, so our only
possibilities are that |G/G | = 1, 3, 5 or 15. Now 52 = 25 so no character can have
χ(1) = 5. So,
•
|G/G
| = 1 because 1 + 32 = 10 <
|G
|and 1 + 32 + 32 = 19 >
|G
|.
• |G/G | = 3 because 3 + 32 = 12 < |G| and 3 + 32 + 32 = 21 > |G|.• |G/G | = 5 because 5 + 32 = 14 < |G| and 5 + 32 + 32 = 23 > |G|.
Therefore |G/G | = 15, which means there are 15 linear characters and so G is abelian.
Exercise 22.2. Every irreducible character of G has degree which is a divisor of 16, i.e.
1, 2, 4, 8 or 16. However 82 = 64 > |G| and 162 = 256 > |G|, also 42 = 16 = |G| but we
always have at least one linear character, so our only possibilities are 1 or 2.
Recall that the number of linear characters is |G/G | and the only possibilities for this
are 1, 2, 4, 8 or 16. Now as χ(1)2
= |G| we have the only possibilities for the degrees of the irreducible characters are
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 2, 2
1, 1, 1, 1, 2, 2, 2.
In other words the number of conjugacy classes in G is either 7, 10 or 16.
95
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Chapter 22 96
Exercise 22.3. Note that G is non-abelian so |G/G | = pq . Now the only possibilities for
|G/G | are 1, p or q .
(a) We have χ(1) | |G| for all irreducible characters χ of G, so either χ is linear orχ(1) = p or q .
(b) Recall that q < p so pq < p 2, which means no character can have degree p . Therefore
we have
|G| = |G/G | +
χ(1)>1
χ(1) ⇒ |G| = |G/G | + q 2n.
Now q | |G| ⇒ q | |G/G | but the only possibilities for |G/G | are 1, p or q so we
must have |G/G | = q . In other words pq = |G| = q |G| ⇒ |G| = p .
(c) We now know that pq = q + q 2n ⇒ p = 1 + qn ⇒ n = p −1q . Now we know n must bean intger, so q | p −1 as required. Finally we can see the group has q linear characters
and p −1q
irreducible characters of order q . In other words G has q + p −1q
conjugacy
classes.
Exercise 22.4. Let φ be a character of a group G such that φ(g ) = φ(h) for all non-identity
elements g, h ∈ G.
(a) We check the multiplicity of the trivial module in φ. Now taking inner products we
see
φ, 1G =1
|G|g ∈G
φ(g )1G(g ) =1
|G|g ∈G
φ(g ) =1
|G|(φ(1) + (|G| − 1)φ(h)),
= φ(h) +φ(1) − φ(h)
|G| ,
for some non-identity element h ∈ G. Now assume φ = a1G + bχreg and recall that
χreg, 1G = 1. Therefore the above inner product calculation has taken into account
the multiplicity of 1G inside χreg. So
φ = φ(h) + φ(1) − φ(h)|G| 1G + b (χreg − 1G).
Now evaluating at the identity we get
φ(1) =
φ(h) +
φ(1) − φ(h)
|G|
+ b (|G| − 1) ⇒ (φ(1) − φ(h))|G| − 1
|G| = b (|G| − 1),
⇒ b =φ(1) − φ(h)
|G| .
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Chapter 22 97
Therefore φ = a1G + bχreg with a = φ(h) and b = φ(1)−φ(h)|G| for some non-identity
element h ∈ G.
(b) We know that φ is a character and hence the multiplicity of any irreducible character
in φ is an integer. We notice from above that a + b is the multiplicity of the trivial
character and hence is an integer. Also φ(1) is an integer, which means a1g (1) +
bχreg(1) = a + b |G| is an integer.
(c) Let χ = 1G be an irreducible character of G then
φ, χ = a1G + bχreg, χ = a1G, χ + b χreg, χ = bχ(1)
because the multiplicity of any irreducible character in χreg is the degree of the char-acter. Now φ is a character and hence the multiplicity of any irreducible character in
φ is an intger, which means bχ(1) is an integer.
(d) We have that bχ(1) is an integer but so is χ(1) so we must have b is an integer. We
have a + b is an integer but we’ve just determined b is an integer, so we must have
a is an integer as well. In otherwords the character φ is integer valued.
Exercise 22.5. Let G be a group of odd order.
(a) Assume g ∈ G such that g = g −1 then g 2 = 1. If g is not the identity then g has
order 2 and 2 | |G| but this cannot happen as G has odd order. Therefore g is theidentity.
(b) If χ = χ then we have χ(g ) = χ(g −1) for all g ∈ G. Recall |G| is odd, then 2 | |G|−1
and from part (a) we know there is no non-identity element of order 2. Let k = |G|−12
and define a set Ω = g 1, . . . , g k such that 1 ∈ Ω and if g i ∈ Ω then g −1i ∈ Ω. Then
the set Ω−1 = g −11 , . . . , g −1k is such that Ω ∩ Ω−1 = ∅. Then
χ, 1g =1
|G| g ∈Gχ(g )1G(g ) =
1
|G| g ∈Gχ(g ) =
1
|G|(χ(1) + 2
g i ∈Ωχ(g i ))
Now χ(g i ) is an algebraic integer for each g i ∈ Ω and so α =
g i ∈Ω χ(g i ) is an
algebraic integer.
(c) Assume χ, 1g = 0, then χ(1) + 2α = 0 ⇒ χ(1) = −2α. However χ(1) | |G| and
2 | χ(1) ⇒ 2 | |G| but G has odd order so this cannot happen. Therefore χ, 1G = 1
and so χ = 1G.
Exercise 22.6. Let G be a group such that |G| = 120 and G has exactly seven conjugacy
classes. Assume further that g contains an element of order 5 such that |C G(g )| = 5 and
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Chapter 22 98
g, g 2, g 3 and g 4 are all conjugate in G. Also, let χ1, . . . , χ7 be the irreducible characters of
G and, without loss of generality, assume χ1 = 1G.
(a) For a start we know that, by Theorem 22.16, that χ(g ) is an integer. So, using thecolumn orthogonality relations we have
χ2(g )2 + χ3(g )2 + χ4(g )2 + χ5(g )2 + χ6(g )2 + χ7(g )2 = 4.
By Corollary 22.27 we have χ(g ) ≡ χ(1) (mod 5) for any irreducible character χ, so
−2 χi (g ) 2. Assume χi (g ) = ±2 for some i and χ j (g ) = 0 for all j = i . Then
we have
|G
| 12 + 22 + 52
×5 = 130,
which is a contradiction. Therefore our only possibility is that for every irreducible
character χ either χ(g ) = 0, 1 or −1.
(b) Using Corollary 22.27 we know that χ(g ) ≡ χ(1) (mod 5) and χ(g ) = 0 for exactly
two characters. We know that χ(1) | |G| = 120 and χ(1)2 < |G| ⇒ χ(1) < 11,
which means our only options for χ(1) in this case is 5 or 10. However we can’t have
χ(1) = 10 because then we’d have |G| 102 + 52 = 125, which is a contradiction.
Therefore we have exactly two characters of degree 5.
(c) All the divisors of 120 which are less than 10 are 1, 2, 3, 4, 5, 6 and 8. Now for anyirreducible character χ, we have from part (a) that χ(1) ≡ χ(g ) ≡ 0, 1, 4 (mod 5)
and so we can’t have χ(1) = 2, 3 or 8. So our only options are 1, 4, 5 or 6. We know
there are exactly two characters of degree 5 by part (b) and so
χ4(1)2 + χ5(1)2 + χ6(1)2 + χ7(1)2 = 69.
Now 43 = 64 < 69, so we must have at least one character of degree 6. This
leaves us with χ5(1)2 + χ6(1)2 + χ7(1)2 = 33. Finally the only solution to this is
χ5(1) = χ6(1) = 4 and χ7(1) = 1. Reordering the characters according the degree
we have
χ1(1) = 1 χ1(g ) = 1,
χ2(1) = 1 χ2(g ) = 1,
χ3(1) = 4 χ3(g ) = −1,
χ4(1) = 4 χ4(g ) = −1,
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Chapter 22 99
χ5(1) = 5 χ5(g ) = 0,
χ6(1) = 5 χ6(g ) = 0,
χ7(1) = 6 χ7(g ) = 1.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2
(g ) 1 α1
α2
α3
α4
α5
1
χ3(g ) 4 β1 β2 β3 β4 β5 −1
χ4(g ) 4 γ 1 γ 2 γ 3 γ 4 γ 5 −1
χ5(g ) 5 δ1 δ2 δ3 δ4 δ5 0
χ6(g ) 5 ε1 ε2 ε3 ε4 ε5 0
χ7(g ) 6 κ1 κ2 κ3 κ4 κ5 1
Table 22.1: The partial character table of G from Exercise 22.6
So far we have the character table for G is as in table 22.1. Consider the columnof g 4. We have χ(g 4) ≡ χ(1) (mod 3) for each irreducible character and using the
column orthogonality relations we have
α23 + β2
3 + γ 23 + δ23 + ε23 + κ23 = 5.
Therefore each entry in this column is either 0, ±1 or ±2. This immediately tells
us that κ3 = 0. Now α3, β3, γ 3 are either 1 or −2 and δ3, ε3 are either −1 or 2.
Therefore no entry in the column is ±2 and we must have α3 = β3 = γ 3 = 1 and
δ3 = ε3 = −1. So this gives us that the character table is as in table 22.2.Consider the column of g 5. We have the order of g 5 is 22 and so χ(g 5) ≡ χ(1)
(mod 2) for all irreducible characters χ. Also using the column orthogonality on g 5
we have
α24 + β2
4 + γ 24 + δ24 + ε24 + κ24 = 3.
It’s clear that all these entries must be 1. Therefore α4, δ4 and ε4 are equal to
±1 and β4 = γ 4 = κ4 = 0. Using the column orthogonality relation with g 7 we get
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Chapter 22 100
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 α1 α2 1 α4 α5 1
χ3(g ) 4 β1 β2 1 β4 β5 −1
χ4(g ) 4 γ 1 γ 2 1 γ 4 γ 5 −1
χ5(g ) 5 δ1 δ2 −1 δ4 δ5 0
χ6(g ) 5 ε1 ε2
−1 ε4 ε5 0
χ7(g ) 6 κ1 κ2 0 κ4 κ5 1
Table 22.2: The partial character table of G from Exercise 22.6
1 + α4 = 0 ⇒ α4 = −1. Using the column orthogonality relation with g 4 we get
−δ4 − ε4 = 0 ⇒ ε4 = −δ4. Therefore we have the character table so far to be as in
table 22.3.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 α1 α2 1 −1 α5 1
χ3(g ) 4 β1 β2 1 0 β5 −1
χ4(g ) 4 γ 1 γ 2 1 0 γ 5 −1
χ5(g ) 5 δ1 δ2 −1 1 δ5 0
χ6(g ) 5 ε1 ε2 −1 −1 ε5 0χ7(g ) 6 κ1 κ2 0 0 κ5 1
Table 22.3: The partial character table of G from Exercise 22.6
Consider the column of g 3. We have χ(g 3) ≡ χ(1) (mod 2) for every irreducible
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Chapter 22 101
character χ. Now using the column orthogonality relation on g 3 we get
α2
2 + β2
2 + γ 2
2 + δ2
2 + ε2
2 + κ2
2 = 7.
Now every entry in the column must be 2. Now three of the entries are ±1, which
gives us β22 + γ 22 + κ2
2 = 4. Therefore two of these entries are zero and one is ±2.
Using the column orthogonality relations between g 3 and g 7 we have
1 + α2 − β2 − γ 2 + κ2 = 0.
We can’t have α2 = −1 as only one of the remaining values is non-zero, therefore
α2 = 1. Using the column orthogonality relations between g 3 and g 5 we have
1 − 1 + δ2 + −ε2 = 0 ⇒ δ2 = ε2.
Using the column orthogonality of g 3 with g 4 we get
2 + β2 + γ 2 − 2δ2 = 0.
Using this equation together with the relation from g 3 and g 7 we can see that 4 −2δ2 + κ2 = 0 ⇒ κ2 = 2δ2 − 4. Now δ2 = ±1, which gives us κ2 = −2 or −8. Clearlyκ2 can’t be −8 so this gives us δ2 = ε2 = 1 and κ2 = −2. Therefore the remaining
entries must be zero and we have the character table so far to be as in table 22.4.
Consider the column of g 2. We have g 2 is an element of order 2 and so χ(g 2) ≡χ(1) (mod 2) for all irreducible characters χ. Using the column orthogonality relation
on g 2 we see
α21 + β2
1 + γ 21 + δ21 + ε21 + κ21 = 11.
Therefore it’s clear that every entry in the column is 3. Say one entry is
±3
then that means there are three of the remaining entries equal to ±1 and the rest 0.
However this would imply 4 entries are congruent to 1 (mod 2) but there are clearly
only 3. Therefore every entry in the column is either 0, ±1 or ±2. In fact knowing
that α21 = δ21 = ε21 = 1 means that
β21 + γ 21 + κ2
1 = 8.
Therefore only one entry is 0 and the other two are ±2.
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Chapter 22 102
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 α1 1 1 −1 α5 1
χ3(g ) 4 β1 0 1 0 β5 −1
χ4(g ) 4 γ 1 0 1 0 γ 5 −1
χ5(g ) 5 δ1 1 −1 1 δ5 0
χ6(g ) 5 ε1 1
−1
−1 ε5 0
χ7(g ) 6 κ1 −2 0 0 κ5 1
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 + γ 1) + 5(δ1 + ε1) + 6κ1 = 0,
1 + α1 + δ1 + ε1
−2κ1 = 0,
1 + α1 + β1 + γ 1 − δ1 − ε1 = 0,
1 − α1 + δ1 − ε1 = 0,
1 + α1 − β1 − γ 1 + κ1 = 0.
Adding the second and third relations gives us 2 + 2δ1 − 2κ1 = 0 ⇒ κ1 = 1 + δ1.
Therefore we can’t have κ1 = −2 as δ1 = ±1. Using this in the relations we reduce
our system of equations to
1 + α1 + 4( β1 + γ 1) + 5(δ1 + ε1) + 6κ1 = 0,1 + α1 + β1 + γ 1 − δ1 − ε1 = 0,
1 − α1 + δ1 − ε1 = 0,
2 + α1 − β1 − γ 1 + δ1 = 0.
Adding the second and fourth relations again we get 3 + 2α1 − ε1 = 0 ⇒ ε1 =
2α1 + 3 but this gives us ε1 = 5 or 1. However ε1 = 5 by the congruence relations
and so we have ε = 1 and α = −1. Putting this into our relations, (and removing
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Chapter 22 103
any redundant equations), we get
5 + 4( β1 + γ 1) + 5δ1 + 6κ1 = 0,1 − β1 − γ 1 + δ1 = 0,
1 + δ1 = 0.
Clearly δ1 = −1 ⇒ κ1 = 0. Then the remaining relations tell us γ 1 = − β1. Now the
characters χ3 and χ4 are currently identical so there is no harm in choosing β1 = 2
and γ 1 = −2. This gives us the character table to be as in table 22.5.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7
Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 −1 1 1 −1 α5 1
χ3(g ) 4 2 0 1 0 β5 −1
χ4(g ) 4 −2 0 1 0 γ 5 −1
χ5(g ) 5 −1 1 −1 1 δ5 0
χ6(g ) 5 1 1 −1 −1 ε5 0
χ7(g ) 6 0 −2 0 0 κ5 1
Table 22.5: The partial character table of G from Exercise 22.6
To get the final column of g 6 it is easiest to use the row orthogonality relations
with the trivial representation. This gives us
1 − 10 + 15 + 20 − 30 + 20α5 + 24 = 0
4 + 20 + 20 + 20 β5 − 24 = 04 − 20 + 20 + 20 γ 5 − 24 = 0
5 − 10 + 15 − 20 + 30 + 20δ5 = 0
5 + 10 + 15 − 20 − 30 + 20ε5 = 0
6 − 30 + 20κ5 + 24 = 0
⇒
20 + 20α5 = 0,
20 + 20 β5 = 0,−20 + 20 γ 5 = 0,
20 + 20δ5 = 0,
−20 + 20ε5 = 0,
20κ5 = 0.
From this it is now easy to see that the complete character table for G is as follows in
table 22.6. Comparing this to the character table of S5 on page 201 we can indeed
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Chapter 22 104
see that these are the same.
g i g 1 g 2 g 3 g 4 g 5 g 6 g 7Order 1 2 2 3 4 6 5
|C G(g i )| 120 12 8 6 4 6 5
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 −1 1 1 −1 −1 1
χ3(g ) 4 2 0 1 0 −1 −1
χ4(g ) 4 −2 0 1 0 1 −1
χ5(g ) 5−
1 1−
1 1−
1 0
χ6(g ) 5 1 1 −1 −1 1 0
χ7(g ) 6 0 −2 0 0 0 1
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
+ c n−1X n
−1
+ · · · + c 1X + c 0,
with c 0 = (−1)n det(A) and c n−1 = − tr(A). All c i ∈ Z and so λ is a solution of such a
polynomial.
(⇐) Let λ be the solution of the polynomial
X n + an−1X n−1 + · · · + a1X + a0 ∈ Z[X ].
We now wish to construct an integer valued matrix A such that λ is a solution of det(A −XIn) = X n + an−1X n−1 + · · · + a1X + a0. Let A be the matrix
0 1 . . . 0...
. . . 1...
.... . .
...
−a0 −a1 . . . −an−1
.
We now want to discover what the determinant of A − XIn is. We proceed by induction.
We claim that det(A − XIn) = (−1)n(X n + an−1X n−1 + · · · + a1X + a0). We check that
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Chapter 22 105
this is the case for A a 2 × 2 matrix
det−X 1−a0 −a1 − X = −X (−a1 − X ) + a0 = X 2 + a1X + a0.
Therefore this is certainly true for n = 2. Now assume that this is true for n = k then the
k + 1th case is
det
−X 1 . . . 0...
. . . 1...
.... . . 1
−a0
−a1 . . .
−ak
−X
= (−X )det
−X 1 . . . 0...
. . . 1...
.... . . 1
−a1 −a2 . . . −ak − X
− det
0 1 . . . 0... −X 1
......
. . . 1
−a0 −a2 . . . −ak − X
,
= (−X )(−1)k (X k + ak X k −1 + · · · + a2X + a1) − (−1)k a0,
= (−1)k +1(X k +1 + ak X k + · · · + a2X 2 + a1X + a0).
Therefore the result holds by induction. Now det(A−
λIn
) = 0, so λ is an eigenvalue of A
and hence an algebraic integer.
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Chapter 23 107
g i 1 ar 1r n−1
2
b
|C G(g i )| 2n n 2
χ1(g ) 1 1 1
χ2(g ) 1 1 −1
ψ j (g )1 j n−1
2
2 ε j r + ε− j r 0
Table 23.1: The character table of D2n, for n odd
g i 1 am ar
1r m
−1b ab
|C G(g i )| 2n 2n n 4 4
χ1(g ) 1 1 1 1 1
χ2(g ) 1 1 1 −1 −1
χ3(g ) 1 (−1)m (−1)r 1 −1
χ4(g ) 1 (−1)m (−1)r −1 1
ψ j (g )1 j m−1
2 2(−1) j ε j r + ε− j r 0 0
Table 23.2: The character table of D2n, for n even
Clearly the χi are real characters in both tables but what about the ψ j ? Well for any
complex number z we have z + z = 2 Re(z ). Also it’s clear that ε = e −2πi/n = ε−1 and
so ψ j is also a real character of D2n. Therefore all the conjugacy classes of D2n are real
conjugacy classes. This means that every element of D2n is real and so there are 2n real
elements. Clearly every irreducible character is real, which gives us ιχ = 1 for all irreducible
characters χ.
Exercise 23.4. Let V be the 2-dimensional irreducible CG module associated to the repre-sentation ρ. Let B = v 1, v 2 be the basis of V then we have B = v 1 ⊗ v 2 − v 2 ⊗ v 1 is a
basis of A(V ⊗ V ). Let the matrix of g with respect to B be
gρB =
a b
c d
.
Therefore we have v 1g = av 1 + bv 2 and v 2g = cv 1 + dv 2.
We now consider the action of g on the antisymmetric tensor product module A(V ⊗V ).
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Chapter 23 108
We have
(v 1 ⊗ v 2 − v 2 ⊗ v 1)g = v 1g ⊗ v 2g − v 2g ⊗ v 1g,= (av 1 + bv 2) ⊗ (cv 1 + dv 2) − (cv 1 + dv 2) ⊗ (av 1 + bv 2),
= ad (v 1 ⊗ v 2) + bc (v 2 ⊗ v 1) − cb (v 1 ⊗ v 2) − ad (v 2 ⊗ v 1),
= (ad − bc )(v 1 ⊗ v 2 − v 2 ⊗ v 1).
Therefore χA(g ) = ad − bc = det(gρ). Now ιχ = −1 if and only if 1G is a constituent
of χA but this happens if and only if χA = 1G, (as χA is a linear character and hence
irreducible). In other words det(g ) = 1 for all g ∈ G.
Exercise 23.5. We let G = T 4n = a, b | a2n = 1, an = b 2, b −1ab = a−1 and ε = ±1 a2nth root of unity.
(a) Let ρ : G → GL(2,C) be the irreducible matrix represenation associated to V . It’s
clear that for a and b we have
aρ =
ε 0
0 ε−1
bρ =
0 1
εn 0
Now clearly det(aρ) = εε−1 = 1 and det(bρ) = −εn. Recall the determinant is
multiplicative and every element in T 4n can be written in the form ai b j with 1 i 2n
and 0 j 1. Now ιχ = −1 if and only if det(x ρ) = 1 for all x ∈ T 4n. Therefore
this only happens if −εn = 1 ⇒ εn = −1. So we have ιχ = 1 if εn = 1 and ιχ = −1
if εn = −1.
(b) We check the action of a on the bilinear form
β(v 1a, v 1a) = β(εv 1, εv 1) = ε2 β(v 1, v 1) = 0 = β(v 1, v 1),
β(v 2a, v 2a) = β(ε−1v 2, ε−1v 2) = ε−2 β(v 2, v 2) = 0 = β(v 2, v 2),
β(v 1a, v 2a) = β(εv 1, ε−1v 2) = εε−1 β(v 1, v 2) = β(v 1, v 2),
β(v 2a, v 1a) = β(ε−1v 2, εv 1) = ε−1εβ(v 2, v 1) = β(v 2, v 1).
Therefore β is invariant under the action of a. Now considering the action of b on
the bilinear form
β(v 1b, v 1b ) = β(v 2, v 2) = 0 = β(v 1, v 1),
β(v 2b, v 2b ) = β(εnv 1, εnv 1) = ε2n β(v 1, v 1) = 0 = β(v 2, v 2),
β(v 1b, v 2b ) = β(v 2, εnv 1) = εn β(v 2, v 1) = ε2n = 1 = β(v 1, v 2),
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Chapter 23 109
β(v 2b, v 1b ) = β(εnv 1, v 2) = εn β(v 1, v 2) = εn = β(v 2, v 1).
Therefore β is invariant under the action of b , hence invariant under the action of allelements x ∈ T 4n.
Using Theorem 23.16 we have ιχ = 1 if there exists a symmetric bilinear form on
V , which happens if β(v 2, v 1) = β(v 1, v 2) = 1, in other words εn = 1. Alternatively
the theorem tells us that ιχ = −1 if there exists a skew-symmetric bilinear form on
V , which happens if β(v 2, v 1) = − β(v 1, v 2), in other words εn = −1.
(c) Every element in T 4n has the form ai b j for some 1 i 2n and 0 j 1. Now
assume j = 0 then (ai )2 = a2i = 1 ⇔ i = n or 2n, however if i = 2n then ai = 1, which
is not an element of order 2. Assume j = 1 then (ai b )2 = ai bai b = ai a−i b 2 = an
= 1.
Therefore the only element of order 2 in T 4n is an.
(d) Now the character table for T 4n with n even is as in table 23.3.
g 1 an ar 1r n−1
b ba
g 2 1 1 a2r an an
|C G(g )| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 1 1 −1 −1
φ2 1 1 (−1)r 1 −1
φ3 1 1 (−1)r −1 1
χk (g )0k n−1
2 2(−1)k εkr + ε−kr 0 0
χ2k (g ) 4 4 2 + ε2kr + ε−2kr 0 0
(χk )S(g ) 3 3 1 + ε2kr + ε−2kr (−1)k (−1)k
(χk )A(g ) 1 1 1 (−1)k +1 (−1)k +1
Table 23.3: The character table of T 4n, with n even
It’s obvious that ι1G = ιφ1 = ιφ2 = ιφ3 = 1 because 12G = φ2
1 = φ22 = φ2
3 = 1G and
(1G)S = 1G, (1G)A = 0. Now all that’s left to consider is the indicator of the χk . It’s
clear that χk is real because εkr + ε−kr = εkr + εkr = 2Re(εkr ).
Now in the table above we have considered χ2k and decomposed it into its sym-
metric and alternating parts. If k is odd then k + 1 is even and (χk )A = 1G, which
gives us ιχk = −1. If k is even then k + 1 is odd and (χk )A = φ1. Therefore 1G must
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Chapter 23 110
be a constituent of (χk )S and so (ιχk ) = 1. So we have the Frobenius-Schur Count
of Involutions is
χ
(ιχ)χ(1) = 1 + 1 + 1 + 1 +
n−1k =1
2(−1)k = 4 − 2 = 2,
because n is even so the sum equals −2.
If n is odd then we have the character table of T 4n to be as in table 23.4.
g 1 an ar 1r n−1
b ba
g 2 1 1 a2r an an
|C G(g )| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 −1 (−1)r i −i
φ2 1 1 1 −1 −1
φ3 1 −1 (−1)r −i i
χk (g )1k n−1
2 2(−1)k εkr + ε−kr 0 0
χ2k (g ) 4 4 2 + ε2kr + ε−2kr 0 0
(χk )S(g ) 3 3 1 + ε2kr + ε−2kr (−1)k (−1)k
(χk )A(g ) 1 1 1 (−1)k +1 (−1)k +1
Table 23.4: The character table of T 4n, with n odd
It’s clear that 12G = φ22 = 1G so ι1G = ιφ2 = 1 and φ1, φ3 aren’t real so ιφ1 = ιφ3 = 0.
Now all that’s left to consider is the indicator of the χk . However the situation here
is clearly identical to the situation for when n is even. Therefore the Frobenius-Schur
Count of Involutions is
χ
(ιχ)χ(1) = 1 + 0 + 1 + 0 +
n−1k =1
2(−1)k = 2 + 0 = 2,
because n is odd so the sum equals 0. This confirms that there is only one involution
in T 4n.
Exercise 23.6. Let χ be an irreducible character of G such that ιχ = −1. Let V be the
irreducibleCG module which affords χ as a character. By Theorem 23.16 we have ιχ = −1
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Chapter 23 111
if and only if there exists a non-zero G-invariant skew-symmetric bilinear form on V . Let’s
call this bilinear form β. Now let v 1, . . . , v m be a basis for the CG module V . We consider
the vector subspaceU = u ∈ V | β(u, v ) = 0 for all v ∈ V .
It’s clear that this is a vector subspace but it is also a CG module. Recall that V G = V
and so given u ∈ U, v ∈ V and g ∈ G we have β(ug,vg ) = β(u, v ) = 0 and so UG = U .
Now V is an irreducible CG submodule and β is non-zero so we must have U = 0. Now
let β(v i , v j ) = x ij for all 1 i j m then we can construct a matrix X with
X =
0 x 1,2 . . . x 1,m−1 x 1,m
−x 1,2 . . . x 2,m...
. . ....
−x 1,m−1. . . x m−1,m
−x 1,m −x 2,m . . . −x m−1,m 0
.
We can see that the matrix X is skew-symmetric, i.e. X T = −X , because it’s coming
from a skew-symmetric bilinear form. By the fact that U = 0 we know that non of the
off diagonal entries in X are 0, hence the determinant of X is non-zero. We consider the
determinant of X T in two ways. We can see that we can get from X T to X by applying
m row and column swaps and multiplying by −1. Every time we make a row or column
swap we multiply the determinant by −1 and so det(X T ) = (−1)m+1 det(X ). Alterantively
det(X T ) = det(−X ) = − det(X ), so this gives us det(X ) = (−1)m det(X ). Therefore m is
even and we’re done.
Exercise 23.7. Let V be a vector space over R and let v 1, . . . , v n be any basis of V .
Recall that given a vector subspace U ⊆ V we define U ⊥ with respect to β1 by
U ⊥ = v ∈ V | β1(u, v ) = 0 for all u ∈ U .
Also, as vector spaces, we have V = U ⊕U ⊥. We now wish to construct a basis e 1, . . . , e nof V which is orthonormal with respect to β1.
We start by letting f 1 = v 1. Consider the vector subspace U 1 = spanv 1 and the
standard projection map π1 : V → U 1. Define f 2 = v 2 − π1(v 2) ∈ U ⊥1 . Note that f 2 = 0
because if f 2 = 0 then v 2 = π1(v 2) ⇒ v 2 ∈ spanv 1. However this cannot happen because
v 1, . . . , v n is a basis for V .
In a similar fashion we consider the vector subspace U 2 = spanv 1, v 2 and the standard
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Chapter 23 112
projection map π2 : V → U 2. Define f 3 = v 3 − π2(v 3) ∈ U ⊥2 . Again we have f 3 = 0 because
if f 3 = 0 then v 3 = π2(v 3) ⇒ v 3 ∈ spanv 1, v 2. However this cannot happen because
v 1, . . . , v n is a basis for V .Carrying on with this process gives us a set f 1, . . . , f n such that β(f i , f j ) = 0 for all
i = j . Now we define f i = f i /
β1(f i , f i ), then e 1, . . . , e n is an orthonormal basis of V
with respect to β1. Note that
β(f i , f i ) ∈ R because β(w , w ) > 0 for all non-zero w ∈ V .
Note that in general this process is known as the Gram Schmidt algorithm.
We now want to show that f 1, . . . , f n can be transformed into a basis which is also
orthogonal on β. Let X be an n × n matrix such that X i j = β(f i , f j ) then X is a real
symmetric matrix. By general matrix theory there exists a real orthogonal matrix Q, (i.e.
QT = Q−1), such that QXQ−1 = D where D is a diagonal matrix. Note that this a special
case of the usual diagonalisation process for complex valued matrices.
The above process corresponds to a change of basis for V . Let Q = (q i j ) then we define
a new basis
e i =
n j =1
q ij f j .
Now because QXQ−1 is diagonal we have β(e i , e j ) = 0 for any i = j . Also the ma-
trix representing β1(f i , f j ) is the identity so conjugating by Q does nothing. This means
β1(e i , e j ) = δi j as required.
Exercise 23.8. Let V and W be RG-modules.
(a) Let ϑ : V → W be an RG-homomorphism. Now im(ϑ) is an RG submodule of W
and ker(ϑ) is an RG submodule of V . We have V and W are irreducible RG modules
so either Im(ϑ) = W and ker(ϑ) = 0 or Im(ϑ) = 0 and ker(ϑ) = V . This gives
you either ϑ is an RG isomorphism or ϑ is the zero map.
(b) Now if ϑ : V → V is an RG isomorphism then ϑ is also a CG isomorphism. If V
is an irreducible CG module then by Schur’s Lemma there exists λ ∈ C such that
ϑ = λ1V . However ϑ is also an RG homomorphism so we must have λ
∈R otherwise
im(ϑ) = V is not an RG-module.
(c) Consider G = C 3 = x | x 3 = 1. We have a 2-dimensional irreducible RG submodule
V = span1 − a, 1 − a2. Let v 1 = 1 − a and v 2 = 1 − a2 then we have the action of
a on these basis elements to be
v 1a = (1 − a)a = a − a2 = v 2 − v 1 v 2a = (1 − a2)a = a − 1 = −v 1.
We can define a map ϑ : V → V by v ϑ = av . Note that multiplication on the left
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Chapter 23 113
is well defined as V is a submodule of the group algebra. Also, as G is commutative,
we will have the action of the group on the left to be the same as the action of the
group on the right.So, with respect to the basis B = v 1, v 2 we have
[a]B =
−1 −1
1 0
.
Now considering the eigenvalues of this matrix we have the characteristic polynomial
to be
det([a]B − x I2) = −1
−x
−1
1 −x = −x (−1 − x ) + 1 = x
2
+ x + 1.
This polynomial has eigenvalues −1±3i 2
∈ R. Therefore the map ϑ cannot be a real
scalar multiple of the identity map.
Exercise 23.9. Let Ω = Hx 1, . . . , H x n. The map ρg : Ω → Ω is a permutation of Ω if it
is a bijection. It’s clearly 1-1 because
(Hx i )ρg = (Hx j )ρg ⇒ Hx i g = Hx j g ⇒ Hx i = Hx j ⇒ x i = x j
because the right cosets of H partition the group G. It is also surjective as given Hx i ∈ Ω
we have Hx i g −1 ∈ Ω and (Hx i g −1)ρg = Hx i g −1g = Hx i . Therefore ρg is a permutation of
Ω.
We need to show that for all g, h ∈ G we have (gh)ρ = (gρ)(hρ) or in other words
ρgh = ρg ρh. For any Hx i ∈ Ω we have
(Hx i )ρgh = Hx i gh = (Hx i g )ρh = (Hx i )ρg ρh.
Now our choice of Hx i
was arbitrary and so ρgh
= ρg
ρh
, which gives us ρ is a homomorphism
from G to Sym(Ω).
Now we have ker(ρ) = g ∈ G | ρg = idΩ. If ρg is such a map then for all Hx i ∈ Ω we
have
(Hx i )ρg = (Hx i ) idΩ ⇔ Hx i g = Hx i ,
⇔ H = Hx i gx −1i ,
⇔ x i gx −1i ∈ H,
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Chapter 23 114
⇔ g ∈ x −1i Hx i .
Therefore ker(ρ) = ∩x ∈Gx −1
Hx as required.We have a bijection ϕ : Ω → 1, . . . , n given by (Hx i )ϕ = i . We have a map from
ψ : Sym(Ω) → Sn given by σ → ϕ−1σϕ. Now this is in fact an isomorphism, we can see
it’s a homomorphism as for any σ, τ ∈ Sym(Ω) we have
(στ )ψ = ϕ−1(στ )ϕ = (ϕ−1σϕ)(ϕ−1τϕ) = (σψ)(τψ).
Also we can define an inverse map ψ−1 : Sn → Sym(Ω) by πψ−1 = ϕπϕ−1. Note that the
composition of homomorphisms is again a homomorphism. Therefore if H is a subgroup
of index n in G then we have a homomorphism ρ : G → Sym(Ω), hence a homomorphismρψ : G → Sn. It’s clear that ker(ρψ) = ker(ρ) = ∩x ∈Gx −1Hx ⊆ H .
Exercise 23.10. Let G be a finite group with an involution t ∈ G such that C G(t ) ∼= C 2.
Now let χ1, . . . , χk be the irreducible characters of G and ψ1, ψ2 be the irreducible characters
of C 2. Recall that all the irreducible characters of C 2 are linear.
As t 2 = 1 we have t −1 ∈ t G, which means χi (t ) is an integer for each 1 i k .
Furthermore we know that χi (t ) ≡ χi (1) (mod 2) for each 1 i k . From the column
orthogonality relations we have
k i =1
χi (1)2 = |G|k
i =1
χi (t )2 = 2.
As the χi (t ) are integers we must have χi (t ) = ±1 for two i , j ∈ 1, . . . , k with i = j
and χi (t ) = 0 for all other irreducible characters. As χi (1) ≡ χi (t ) (mod 2) for every
irreducible character this means that only two of the character degrees are odd and the rest
are even.
We now consider the restriction of the irreducible characters to the centraliser C G(t ).
We know that χi ↓
C G
(t ) = d 1
ψ1
+ d 2
ψ2
for each 1 i k and that d 21
+ d 22 [G :
C G(t )] = 2. Therefore the only possibilities are that d 1, d 2 are 0 or 1. We have that
χi (1) = (χi ↓ C G(t ))(1) = d 1ψ1(1) + d 2ψ2(1) = d 1 + d 2.
So the degree of every irreducible character of G is either 1 or 2. However we know there
are two characters of odd degree and all the other characters have even degree. This means
G has two linear characters and all other irreducible characters are of degree 2. In other
words [G : G ] = 2.
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Chapter 23 115
Assume G is a simple group. We have G G is a normal subgroup of G and clearly
G = G so G = 1. This means G is abelian, which means there are only two characters
and they’re both linear. So G has the character table of C G(t ) ∼= C 2 which means G ∼= C 2.
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Chapter 25. Characters of groups of order pq
Exercise 25.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
Exercise 25.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
Exercise 25.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Exercise 25.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
Exercise 25.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
Exercise 25.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Exercise 25.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Exercise 25.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Exercise 25.1. We only have to check that the four group axioms hold. Let G be the
group specified in the question. Clearly I2 ∈ G as 0 ∈ Zp and 1 ∈ Z∗p . Associativity is given
to us by the fact that matrix multiplication is associative. We now check closure,1 y
0 x
1 y
0 x
=
1 y + y x
0 xx
.
Now y + y x ∈Zp because x
∈Z∗p ⇒
y x ∈Zp and Zp is a group under addition. Clearly
x, x ∈ Z∗p ⇒ x x ∈ Z∗p and so the product is in G.
Finally we show there exists an inverse element. Recall that Z∗p and Zp are groups under
multiplication and addition respectively. Therefore if x ∈ Z∗p then x −1 ∈ Z∗p and y ∈ Zp
then − y ∈ Zp . This means that1 − y x
0 x −1
∈ G and
1 y
0 x
1 − y x −1
0 x −1
=
1 − y x −1 + y x −1
0 xx −1
=
1 0
0 1
.
Exercise 25.2. We have F 11,5 = a, b | a11
= b 5
= 1 and b −1
ab = a3
, note that 35
=243 = 22 × 11 + 1 ≡ 1 (mod 11). Also the remaining powers of 3 are
32 = 9 ≡ 9 (mod 11) 33 = 27 ≡ 5 (mod 11) 34 = 81 ≡ 4 (mod 11).
Therefore 3 is a primitive root modulo 11. We have S = 1, 3, 4, 5, 9 and so |Z∗p /S| = 2,
so a set of left coset representatives for S in Z∗p is given by S, 2S.
Now G/G ∼= C 5 so let ε = e 2πi/5 be a primitive fifth root of unity. By Theorem 25.10
there are q = 5 linear characters corresponding to the lifts of C 5 and |Z∗p /S| = 2 irreducible
116
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Chapter 25 118
are such that u q ≡ v q ≡ 1 (mod p ). Therefore setting b = b k we get
G1 = a, b | ap
= b q
= 1 and b −1
ab = av
= G2.
Hence G1∼= G2.
Exercise 25.4. Let p = 2, q = (p − 1)/2 and G = F p,q .
(a) Now u is a generator for Z∗p as Z∗p is cyclic and u is a non-identity element. Note that
−1 is the only element of order 2 in Z∗p therefore we have there exists an integer m
such that
u m
≡ −1 (mod p )
⇔u 2m
≡u q
≡1 (mod p ),
⇔ 2 | q,
⇔ 2 | p − 1
2,
⇔ p − 1 ≡ 0 (mod 4),
⇔ p ≡ 1 (mod 4).
(b) Now a−1 ∈ aG ⇔ a−1 = as ⇔ s ≡ −1 (mod p ) for some s ∈ S, (the set of all powers
of u ). Therefore by the above result we have −1 ∈ S if and only if p ≡ 1 (mod 4).
(c) Note that q = (p − 1)/2 ⇒ 2 = (p − 1)/q , which means |Z∗p /S| = 2. Hence there
are two irreducible characters of G of degree q which we label φ1 and φ2. Consider
1, v to be a set of left coset representatives for S in Z∗p then we have two conjugacy
classes aG and (av )G. We now use the orthogonality relations to investigate the values
of φ1(a) and φ2(a). For 1 i 2 we have
0 = φi , 1G = j
φi (g j )1G(g j )
|C G(g j )| =1 + φi (a) + φi (av )
p ⇒ φi (a) + φi (av ) = −1.
Using the inner product of φi with itself we have
1 = φi , φi = j
φi (g j )φi (g j )
|C G(g j )| =q + φi (a)φi (a) + φi (av )φi (av )
p ,
⇒ p − q = φi (a)φi (a) + φi (av )φi (av ),
⇒ p − p − 1
2= φi (a)φi (a) + φi (av )φi (av ),
⇒ p + 1
2= φi (a)φi (a) + φi (av )φi (av ).
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Chapter 25 119
We now proceed to break this down in to cases. Assume p ≡ 1 (mod 4) then a is
conjugate to a−1, which means φi = φi . Recall that from above we have φi (av ) =
−1 − φi (a) so we have
φi (a)2 + φi (av )2 =p + 1
2⇒ φi (a)2 + (φi (a) + 1)2 =
p + 1
2,
⇒ 2φi (a)2 + 2φi (a) + 1 =p + 1
2,
⇒ 4φi (a)2 + 4φi (a) + 1 − p = 0,
⇒ φi (a) =−4 ±
42 − 4 · 4 · (1 − p )
2 · 4,
⇒ φi (a) = −4
± 42p
2 · 4 ,
⇒ φi (a) =−1 ± √
p
2.
Now assume p ≡ −1 (mod 4) then a is not conjugate to a−1, which means
a−1 ∈ (av )G. This means that φi (av ) = φi (a−1) = φi (a). Therefore we have
φi (a) + φi (av ) = −1 ⇒ φi (a) + φi (a) = −1,
⇒ 2Re(φi (a)) = −1,
⇒ Re(φi (a)) = −12
.
Using the second equation we have that
φi (a)φi (a) + φi (av )φi (av ) =p + 1
2⇒ 2φi (a)φi (a) =
p + 1
2,
⇒ Re(φi (a))2 + Im(φi (a))2 =p + 1
4,
⇒ Im(φi (a))2 =p + 1
4− 1
4,
⇒ Im(φi (a))2 =p
4 ,
⇒ Im(φi (a)) = ±√
p
4,
⇒ φi (a) =−1 ± √
pi
2.
Comparing with the question this settles the two possible cases.
(d) Let ε = e 2πi/p be a primitive p th root of unity. Consider, as in the question, δ = ±1
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Chapter 25 120
dependent upon p then we know that
−1 ± √ δp 2
= φ1(a) =
p −12
k =1
εu k
.
Now Z∗p is a cyclic group of order p − 1 and u is a generator for Z∗
p , therefore every
element in Z∗p can be expressed as a power of u .
We can choose u in the following way. Let n be the integer described in Theorem
25.1. Now p − 1 = 2q therefore we have
np −1 ≡ n2q ≡ (n2)q ≡ 1 (mod p ).
Also (n2)r ≡ n2r ≡ 1 (mod p ) for all 0 < r < q because if there was such an r then
we would have 0 < 2r < 2q = p − 1, which would invalidate the property of n. Hence
n2 is has order q modulo p . Therefore we can set u = n2, which gives us that u is a
quadratic residue modulo p . Indeed this means all powers of u are quadratic residues
modulo p .
In other words we have the desired result that
s ∈Q εs =−1 ± √
δp
2,
where Q is the set of all quadratic residues modulo p .
Exercise 25.5. First let us consider the conjugacy classes of E . Now it’s clear to see that
a, b C E (ai b j ), for any 0 i , j 2 but not i = j = 0. This means that |C E (ai b j )| 9,
so either the conjugacy class has size 1 or 2. However we have
c −1(ai b j )c = (c −1ac )i (c −1bc ) j = a−i b − j ,
so the conjugacy classes have order 2. Note that as a and b have odd order, no powers of a and b are involutions. This gives us the conjugacy classes
1 a, a2 b, b 2 ab,a2b 2 ab 2, a2b .
This only deals with half the elements of E . We have c C E (c ), which means
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Chapter 25 121
|c E | 9. However, keeping notation as before, we have
a−i
b − j
cb j
a j
= a−i
b −2 j
cai
= a−i
b −2 j
a−i
c = a−2i
b −2 j
c.
Therefore it’s clear that the final conjugacy class has order 9 and explicitly is
c , a c , bc , a2c, b 2c,abc,ab 2c, a2bc,a2b 2c .
Now a, b ∼= C 3×C 3 is a normal index 2 subgroup in E , hence we can use Clifford Theory
to give us more information about the induced characters. Recall that if η = e 2πi/3 = −1+i √ 3
2
then every irreducible character of C 3 is given by χk (ai ) = ηki , (with 0 k 2), and every
irreducible character of C 3 × C 3 is given by the products of these characters. Note thatthere are 6 conjugacy classes of E and hence 6 irreducible characters.
A generic irreducible character of a, b is given by χk χ(ai b j ) = ηki ηj = ηki +j with
0 k, 2. Notice that the conjugacy classes of ai b j split in to two conjugacy classes in
the subgroup a, b . Therefore we have the values of the induced characters to be
(χk χ ↑ E )(c ) = 0,
(χk χ ↑ E )(ai b j ) = |C E (ai b j )|
χk χ(ai b j )
|C a,b (ai b j )
|
+χk χ(a−i b − j )
|C a,b (a−i b − j )
|,
= 9 · ηki +j + η−ki −j
9,
= ηki +j + ηki +j ,
=
2 if ki + j ≡ 0 (mod 3),
−1 otherwise.
Let (k, ) denote the pair of indices coming from the character χk χ. Using the above
information we can see the following values of the induced characters
(0, 1) ⇒ j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(a) = 2,
(0, 2) ⇒ 2 j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(a) = 2,
(1, 0) ⇒ i ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(b ) = 2,
(1, 1) ⇒ i + j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(ab 2) = 2,
(1, 2) ⇒ i + 2 j ≡ 0 (mod 3) ⇒ (χ1χ2 ↑ E )(ab ) = 2
(2, 0) ⇒ 2i ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(b ) = 2,
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Chapter 25 122
(2, 1) ⇒ 2i + j ≡ 0 (mod 3) ⇒ (χ0χ1 ↑ E )(ab ) = 2,
(2, 2)
⇒2i + 2 j
≡0 (mod 3)
⇒(χ0χ1
↑E )(ab 2) = 2.
Therefore the following characters are equal
χ0χ1 ↑ E = χ0χ2 ↑ E χ1χ0 ↑ E = χ2χ0 ↑ E χ1χ1 ↑ E = χ2χ2 ↑ E,
χ1χ2 ↑ E = χ2χ1 ↑ E.
Recall that by Clifford theory we know that each induced character is either irreducible
or a sum of two irreducible characters each with multiplicity 1. The induced trivial character
always contains a copy of the trivial character, hence it is a sum of two linear characters.
The non-isomorphic induced characters are then as in table 25.2.
g 1 a b ab ab 2 c
|C E (g )| 18 9 9 9 9 2
χ0χ0 ↑ E 2 2 2 2 2 0
χ0χ1 ↑ E 2 2 −1 −1 −1 0
χ1χ0 ↑ E 2 −1 2 −1 −1 0
χ1χ2
↑E 2
−1
−1 2
−1 0
χ2χ1 ↑ E 2 −1 −1 −1 2 0
Table 25.2: Some characters induced from a, b to E
It’s easy to see that E = [a, c ], [b, c ] = a−2, b −2 = a, b , hence there are only
|G/G | = 18/9 = 2 linear characters of E . These both occur has constituents of χ0χ0 ↑ E ,
hence the remaining four characters induced from E must be irreducible. Therefore the
character table of E is as in table 25.3.
Exercise 25.6. Clearly we have Z (E ) = C E (a)∩C E (b )∩C E (c ) = a, b ∩c = 1, whichis obviously cyclic. However from the character table above we can see there is no faithful
irreducible character of E because there is no character such that ψi (x ) = ψi (1) for all
x ∈ E .
Exercise 25.7. We aim the find groups with the respective character degrees.
(a) Clearly if G exists then we have |G| = 3 ·12 + 4 ·32 = 3 + 36 = 39 = 3×13. Clearly G
cannot be abelian and so by Proposition 25.7 we have G ∼= F 13,3. To verify we clearly
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Chapter 25 123
g 1 a b ab ab 2 c
|C E (g )
|18 9 9 9 9 2
1E 1 1 1 1 1 1
ψ1 1 1 1 1 1 −1
ψ2 2 2 −1 −1 −1 0
ψ3 2 −1 2 −1 −1 0
ψ4 2 −1 −1 2 −1 0
ψ5 2 −1 −1 −1 2 0
Table 25.3: The character table of E
have 3 | 13 − 1 = 12, now F 13,3 has q = 3 linear characters and |Z∗p /S| = 12/3 = 4
characters of degree q = 3. Therefore F 13,3 is indeed the group we’re looking for.
(b) Clearly if G exists then we have |G| = 6 · 12 + 8 · 32 = 78 = 2 × 3 × 13. The group
F 13,3×C 2 has the character degrees that we’re looking for. Recall that the irreducible
characters of F 13,3×C 2 are the products of the irreducible characters of F 13,3 together
with the irreducible characters of C 2. Every irreducible character of C 2 is linear which
means the degree of the product is the same as the degree of the irreducible character
of F 13,3. Therefore there are 14 irreducible characters of F 13,3 × C 2, six of which are
linear and the remaining 8 have degree 3.
(c) Clearly if G exists then we have |G| = 6·12+3·22+8·32+4·62 = 234 = 2×32×13 =
6 × 39. The group S3 × F 13,3, (or alternatively D6 × F 13,3 as D6∼= S3), has the
character degrees that we’re looking for. Recall that S3 has two linear characters
and one irreducible character of degree 2. Therefore multiplying these degrees with
the character degrees from part (a) we get the desired result.
Exercise 25.8. First we aim to determine the conjugacy classes of G. Now
a
C G(a),
which means |C G(a)| 9 ⇒ |aG| 6. However for any 1 j 5 we have b − j ab j = a2 j ,
which gives us aG = a, a2, a4, a8, a16, a32 = a, a2, a4, a5, a7, a8.
Now consider the remaining powers of a. We have
b −2a3b 2 = (a3)4 = a12 = a3,
so a, b 2 C G(a3), which means |C G(a3)| 27 ⇒ |(a3)G| 2. However b −1a3b = a6, so
we must have (a3)G = a3, a6.
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Chapter 25 125
are |G/G | = 54/9 = 6 linear characters, which are the lifts from G/G ∼= b ∼= C 6. Let
ε = e 2πi/6 = 1+i √ 3
2then the linear characters of G are as in table 25.4.
g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5
|C G(g )| 54 9 27 9 9 6 18 6 18 6
ψ0(g ) 1 1 1 1 1 1 1 1 1 1
ψ1(g ) 1 1 1 1 1 ε ε2 −1 −ε −ε2
ψ2(g ) 1 1 1 1 1 ε2 −ε 1 ε2 −ε
ψ3(g ) 1 1 1 1 1 −1 1 −1 1 −1
ψ4
(g ) 1 1 1 1 1−
ε ε2 1−
ε ε2
ψ5(g ) 1 1 1 1 1 −ε2 −ε −1 ε2 ε
Table 25.4: The linear characters of G from Exercise 25.8
From this we can work out the remaining character degrees of G. We know that
ψi (1) | |G| so our only options are 2, 3, 6 or 9. Therefore we have
ψ6(1)2 + ψ7(1)2 + ψ8(1)2 + ψ9(1)2 = 48.
We can see that no character has degree 9 as 92 = 81 > 48. Also 4 · 32 = 36 < 48 sowe must have a character of degree 6. Now 48 − 62 = 12 = 3 · 22, which means our only
option is to have one character of degree 6 and three characters of degree 2.
We now want to find a suitable subgroup to try and obtain the remaining irreducible
characters from. We can see from the group relations that (a3)3 = (b 2)3 = 1 and b −2a3b 2 =
(b −2ab 2)3 = (a4)3 = a3. Therefore N = a3, b 2 ∼= C 3 × C 3 is a subgroup of G. We
can see that N is a union of conjugacy classes and is therefore a normal subgroup of
G. Now |G/N | = 54/9 = 6 and clearly G/N is not abelian therefore the factor group
G/N =
Na,Nb
∼= D6
∼= S3.
As this subgroup is normal we can lift the characters of N to G. We lift the irreducible
character of degree 2 from the character table of D6 on page 125. First we note that
N = Nb 2 = Nb 4 = Na3,
Nb = Nb 3 = Nb 5,
Na = Nab 2 = Nab 4.
Therefore we have an irreducible character of degree 2 whose values are in table 25.5.
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Chapter 25 126
g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5
|C G(g )
|54 9 27 9 9 6 18 6 18 6
ψ6(g ) 2 −1 2 −1 −1 0 2 0 2 0
Table 25.5: A degree 2 character of G from Exercise 25.8
We can combine this irreducible character with the linear characters we already to have
to obtain new irreducible characters. It’s clear to see using the character table above that
ψ7 = ψ6ψ1 and ψ8 = ψ6ψ2 are two distinct irreducible characters of degree 2. There is
only one irreducible character of degree 6 left to find. However a simple argument using
the column orthogonality relations determines the final character. Therefore the charactertable for G is as in table 25.6.
g 1 a a3 ab 2 ab 4 b b 2 b 3 b 4 b 5
|C G(g )| 54 9 27 9 9 6 18 6 18 6
ψ0(g ) 1 1 1 1 1 1 1 1 1 1
ψ1(g ) 1 1 1 1 1 ε ε2 −1 −ε −ε2
ψ2(g ) 1 1 1 1 1 ε2
−ε 1 ε2
−ε
ψ3(g ) 1 1 1 1 1 −1 1 −1 1 −1
ψ4(g ) 1 1 1 1 1 −ε ε2 1 −ε ε2
ψ5(g ) 1 1 1 1 1 −ε2 −ε −1 ε2 ε
ψ6(g ) 2 −1 2 −1 −1 0 2 0 2 0
ψ7(g ) 2 −1 2 −1 −1 0 2ε2 0 −2ε 0
ψ8(g ) 2 −1 2 −1 −1 0 −2ε 0 2ε2 0
ψ9(g ) 6 0 −3 0 0 0 0 0 0 0
Table 25.6: The character table of G from Exercise 25.8
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Chapter 26. Characters of some p -groups
Exercise 26.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Exercise 26.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Exercise 26.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
Exercise 26.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
Exercise 26.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
Exercise 26.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
Exercise 26.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Exercise 26.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
Exercise 26.1. Firstly, assume G is abelian then G has p n linear characters and p n−2 −p n−2 = 0 irreducible characters of degree p . Therefore this is certainly true for such an
abelian group.
Now assume G is non-abelian. Let χ be an irreducible character of G then there exists
an irreducible character ψ of H such that χ ↓ H, ψH = 0 ⇒ χ, ψ ↑ H G = 0. This just
uses the fact that every irreducible character of H is a constituent of the restriction of some
irreducible character of G and Frobenius reciprocity. Now we know that (ψ
↑G)(1) = [G :
H ]ψ(1) = [G : H ] = p because H is abelian hence every character is linear. Therefore as χ
is a constituent of (ψ ↑ G) we have
1 χ(1) (ψ ↑ G)(1) = p ⇒ χ(1) = 1 or p,
because χ(1) | |G|.So every irreducible character of G is either linear or of degree p . Now suppose there
are x linear characters and y characters of degree p then
x + p 2 y = p n ⇒ x = p n − p 2 y ⇒ x = p 2(p n−2 − y ) p 2
because x is non-negative and non-zero because we always have the trivial character as
a linear character. Therefore we have x = p m for some m 2, which leaves us with
y = p n−2 − p m−2.
Exercise 26.2. It’s clear to see from the relations that G C G(z i ) for 0 i 2, hence
we have three conjugacy classes of size 1. Now from the last defining relation of the group
127
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Chapter 26 128
we get
b −1ab = az ⇒ b = a−1baz ⇒ a−1ba = bz 2.
Let 0 i , j , k 2 but such that we don’t have i = j = 0. Then in general we have that
b −k (ai b j )b k = (b −k ab k )i b j = (az k )i b j = ai b j z ik ,
a−k (ai b j )ak = ai (a−k bak ) j = ai (bz k ) j = ai b j z jk .
Therefore we have the conjugacy classes of G to be
1 a,az,az 2 b 2, b 2z , b 2z 2 ab 2, ab 2z,ab 2z 2,
z a2, a2z , a2z 2 ab,abz,abz 2 a2b 2, a2b 2z , a2b 2z 2,
z 2 b,bz,bz 2 a2b, a2bz , a2bz 2.
So there are 11 conjugacy classes of G, which means there are 11 irreducible characters.
Note that a, z G is an abelian subgroup of order 32 = 9 and so [G : a, z ] = 3. Let
K = G = z in Theorem 26.6. Now |G/G | = 9 and the relation b −1ab = az tells us
(b −1ab )G = az G ⇒ (b −1ab )G = aG ⇒ (ab )G = (ba)G .
So G/G is abelian and contains two elements of order 3, so G/G ∼= C 3 × C 3.
Let η = e 2πi/3 = −1+√ 3i
2 then we have the character table for G is as in table 26.1.
Now the final two characters are induced linear characters of a, z ∼= C 3×C 3. We note that
the conjugacy classes 1, z and z 2 do not split upon restriction to a, z . However,
for 1 j 2, we have a j , a j z , a j z 2 splits into three conjugacy classes. Therefore if ψ is
an irreducible character of a, z then
(ψ ↑ G)(z j ) = [G : H ]ψ(z j ) = 3ψ(z j ),
(ψ ↑ G)(a j ) = |C G(a j )| ψ(a j )
|C a,z (a j )| +ψ(a j z )
|C a,z (a j z )| +ψ(a j z 2)
|C a,z (a j z 2)| ,
= ψ(a j ) + ψ(a j z ) + ψ(a j z 2),
= ψ(a j )(1 + ψ(z ) + ψ(z 2)).
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Chapter 26 129
g 1 z z 2 a a2 b b 2 ab a2b ab 2 a2b 2
|C G(g )
|27 27 27 9 9 9 9 9 9 9 9
χ1(g ) 1 1 1 1 1 1 1 1 1 1 1
χ2(g ) 1 1 1 1 1 η η2 η η η2 η2
χ3(g ) 1 1 1 1 1 η2 η η2 η2 η η
χ4(g ) 1 1 1 η η2 1 1 η η2 η η2
χ5(g ) 1 1 1 η η2 η η2 η2 1 1 η
χ6(g ) 1 1 1 η η2 η2 η 1 η η2 1
χ7(g ) 1 1 1 η2 η 1 1 η2 η η2 η
χ8(g ) 1 1 1 η2 η η η2 1 η2 η 1
χ9(g ) 1 1 1 η2 η η2 η η 1 1 η2
χ10(g ) 3 3η 3η2 0 0 0 0 0 0 0 0
χ11(g ) 3 3η2 3η 0 0 0 0 0 0 0 0
Table 26.1: The character table of G from Exercise 26.2
Now we can see that
ψ ↑ G, ψ ↑ G 32ψ(1)ψ(1)27
+ 32ψ(z )ψ(z )27
+ 32ψ(z 2)ψ(z 2)27
= 3 · 32
27= 1.
Hence we only get irreducible characters when ψ(z ) = 1 which confirms what we have in
the character table for G.
Exercise 26.3. We first need to determine the conjugacy classes of G. Now it’s clear that
for any 1 j 15 we have a C G(a j ), which means |(a j )G| 2. From the relations it’s
easy to see that b −1a j b = (b −1ab ) j = a− j . Therefore for 1 k 7 we have the following
conjugacy classes;
1
,
a8
,
ak , a−k
.
Now b C G(b ) which means |b G| 8. Now using the relations of the group we have
b −1ab = a−1 ⇒ ab = ba−1 ⇒ b = a−1ba−1 ⇒ a−1ba = a−2b.
Therefore a− j ba j = a−2 j b , which means b G = a2k b | 0 k 7. Likewise we get
a− j (ab )a j = a1−2 j b , which gives us (ab )G = a2k +1b | 0 k 7. In summary, with
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Chapter 26 130
1 7, we have the conjugacy classes of G to be
1 a2k
b | 0 k 7,a8 a2k +1b | 0 k 7,
a, a−.
Now a is an abelian subgroup of order 16 in G and hence of index 2. So take H = aand K = Z (G) = a8 in Theorem 26.6. Also note that a−1b −1ab = a−2, which means
G = a2. Now G/K ∼= a, b | a8 = b 2 = 1, b −1ab = a−1 = D16. Now we know the
character table of D16 from page 305, therefore we have the lifts from G/K to be as in
table 26.2.
g 1 a8 a a2 a3 a4 a5 a6 a7 b ab
|C G(g )| 32 32 16 16 16 16 16 16 16 4 4
χ1(g ) 1 1 1 1 1 1 1 1 1 1 1
χ2(g ) 1 1 1 1 1 1 1 1 1 −1 −1
χ3(g ) 1 1 −1 1 −1 1 −1 1 −1 1 −1
χ4(g ) 1 1 −1 1 −1 1 −1 1 −1 −1 1
χ5(g ) 2 2 0 −2 0 2 0 −2 0 0 0χ6(g ) 2 2
√ 2 0 −√
2 −2 −√ 2 0
√ 2 0 0
χ7(g ) 2 2 −√ 2 0
√ 2 −2
√ 2 0 −√
2 0 0
Table 26.2: The lifted characters from G/K to G of Exercise 26.3
The remaining four characters are induced from the abelian subgroup H = a. Let
ε = e 2πi/16 be a primitive 16th root of unity. Then every irreducible representation of H
takes the form ψi (a j ) = εij for 0 i , j 15. Now, for 1 k 7, we calculate the induced
character to be
(ψi ↑ G)(a8) = [G : H ]ψi (a8) = 2ε8i = 2(−1)i
(ψi ↑ G)(a j ) = |C G(a j )|
ψi (a j )
|C H(a j )| +ψi (a− j )
|C H(a− j )|
,
= ψi (a j ) + ψi (a− j ),
= εij + ε−ij ,
= εij + εi j ,
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Chapter 26 131
= 2 Re(εi j ).
Now it’s clear that Re(εk
) = Re(ε16−k
) = Re(εk
) for 1 k 7, which means ψk ↑ G =ψ16−k ↑ G for 1 k 7. So it’s sufficient to focus on the first eight characters to find
the remaining four irreducible characters of G. Recall that ε2 = e 2πi/8 =√ 2+i
√ 2
2. Straight
away we can see that ψ0 ↑ G = 2χ1 and in fact ψ8 ↑ G = χ3 + χ4 because ε8 = −1, in
other words they are reducible. Indeed, noticing that 2Re(ε2) =√
2, 2Re(ε4) = 0 and
2Re(ε6) = −√ 2, we can see that ψ2 ↑ G = χ6, ψ4 ↑ G = χ5 and ψ6 ↑ G = χ7.
Therefore we only get new distinct irreducible characters of G for i = 1, 3, 5 or 7.
Letting α = 2 Re(ε) = 2 cos(π/8) and β = 2 Re(ε3) = 2cos(3π/8), we get the character
table of G is as in table 26.3.
g 1 a8 a a2 a3 a4 a5 a6 a7 b ab
|C G(g )| 32 32 16 16 16 16 16 16 16 4 4
χ1(g ) 1 1 1 1 1 1 1 1 1 1 1
χ2(g ) 1 1 1 1 1 1 1 1 1 −1 −1
χ3(g ) 1 1 −1 1 −1 1 −1 1 −1 1 −1
χ4(g ) 1 1 −1 1 −1 1 −1 1 −1 −1 1
χ5
(g ) 2 2 0−
2 0 2 0−
2 0 0 0
χ6(g ) 2 2√
2 0 −√ 2 −2 −√
2 0√
2 0 0
χ7(g ) 2 2 −√ 2 0
√ 2 −2
√ 2 0 −√
2 0 0
χ8(g ) 2 −2 α√
2 β 0 − β −√ 2 −α 0 0
χ9(g ) 2 −2 β −√ 2 −α 0 α
√ 2 − β 0 0
χ10(g ) 2 −2 − β −√ 2 α 0 −α
√ 2 β 0 0
χ11(g ) 2 −2 −α√
2 − β 0 β −√ 2 α 0 0
Table 26.3: The character table of G from Exercise 26.3
Exercise 26.4. Let G = A,B,C,D where A,B,C,D are the 4×4 complex valued matrices
defined in the question.
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Chapter 26 132
(a) We now check that all pairs of generators commute modulo Z .
AB =
−1 0 0 00 1 0 0
0 0 1 0
0 0 0 −1
0 0 i 00 0 0 −i
i 0 0 0
0 −i 0 0
=
0 0 −i 00 0 0 −i
i 0 0 0
0 i 0 0
= −BA,
AC =
−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
0 i 0 0
i 0 0 0
0 0 0 −i
0 0 −i 0
=
0 −i 0 0
i 0 0 0
0 0 0 −i
0 0 i 0
= −CA,
AD =
−
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
=
0 0 0
−1
0 0 1 0
0 1 0 0
−1 0 0 0
= −DA,
BC =
0 0 i 0
0 0 0 −i
i 0 0 0
0 −i 0 0
0 i 0 0
i 0 0 0
0 0 0 −i
0 0 −i 0
=
0 0 0 1
0 0 −1 0
0 −1 0 0
1 0 0 0
= −CB,
BD =
0 0 i 0
0 0 0 −i
i 0 0 0
0 −i 0 0
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
=
0 i 0 0
−i 0 0 0
0 0 0 i
0 0 −i 0
= −DB,
CD =
0 i 0 0
i 0 0 0
0 0 0 −i
0 0 −i 0
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
=
0 0 i 0
0 0 0 i
−i 0 0 0
0 −i 0 0
= −DC.
Therefore every pair of generators commutes modulo Z . Clearly G is not abelianbut G/Z is abelian. We recall that if a quotient G/N is abelian then we must have
G N . In other words I = G Z ⇒ G = Z .
(b) We check the squares of the generating elements. It’s obvious that A2 = D2 = 1, so
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Chapter 26 134
Now checking the commutation with D gives us
x 11 0 0 x 140 x 22 x 23 0
0 x 32 x 33 0
x 41 0 0 x 44
0 0 0 10 0 1 0
0 1 0 0
1 0 0 0
=
0 0 0 10 0 1 0
0 1 0 0
1 0 0 0
x 11 0 0 x 140 x 22 x 23 0
0 x 32 x 33 0
x 41 0 0 x 44
,
⇒
x 14 0 0 x 11
0 x 23 x 22 0
0 x 33 x 32 0
x 44 0 0 x 41
=
x 41 0 0 x 44
0 x 32 x 33 0
0 x 22 x 23 0
x 11 0 0 x 14
,
⇒ X =
x 11 0 0 x 14
0 x 22 x 23 0
0 x 23 x 22 0
x 14 0 0 x 11
.
Now checking the commutation with C gives us
x 11 0 0 x 14
0 x 22 x 23 0
0 x 23 x 22 0
x 14 0 0 x 11
0 i 0 0
i 0 0 0
0 0 0
−i
0 0 −i 0
=
0 i 0 0
i 0 0 0
0 0 0
−i
0 0 −i 0
x 11 0 0 x 14
0 x 22 x 23 0
0 x 23 x 22 0
x 14 0 0 x 11
,
0 i x 11 −i x 14 0
ix 22 0 0 −i x 23
ix 23 0 0 −i x 22
0 i x 14 −i x 11 0
=
0 ix 22 ix 23 0
ix 11 0 0 i x 14
−i x 14 0 0 −ix 11
0 −ix 23 −ix 22 0
,
⇒ X =
x 11 0 0 0
0 x 11 0 0
0 0 x 11 0
0 0 0 x 11
.
Therefore X is a scalar multiple of the identity and we’re done.
(d) Now we have an irreducible representation of degree 4 which means that 12 + 42 =
17 |G| 32 but G is a 2 group, which means |G| = 2x but 24 = 16 therefore we
must have |G| = 25 = 32.
Recall from part (a) that G = Z , which means |G/G | = |G|/|G| = 32/2 = 16.
So the remaining representations of G are all 1-dimensional. By part (a) we know that
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Chapter 26 135
all generators commute in G/G , which means G/G ∼= C 2 × C 2 × C 2 × C 2. Now for
example an irreducible representation of C 2 is given by Aρ = −1 or Aρ = 1 = (−1)0.
So for 0 x , y , z , w 1 we get a linear representation of G given by
Ai B j C k DZ m → (−1)xr + y j +zk +w.
This gives us all the irreducible representations of G.
Exercise 26.5. Let G1, . . . , G9 be the non-abelian groups of order 16 with presentations as
given.
(a) We start by looking for a faithful irreducible representation of degree 2 for G1 = D16.
Examining the character table for G1 we can see that such an irreducible representationwould need to have trace 0 on b and trace ±√
2 on a. Recall ω = e 2πi/8 =√ 2+i
√ 2
2.
Now ω−1 = e 2πi/8 =√ 2−i √
22 , therefore we get such an irreducible representation
ρ : G1 → GL(2,C) by defining
aρ =
ω 0
0 ω−1
bρ =
0 1
1 0
.
It’s clear that (aρ)8 = (bρ)2 = I2 and so all that’s left to check is the remaining
relation.
(bρ)−1(aρ)(bρ) =
0 1
1 0
ω 0
0 ω−1
0 1
1 0
,
=
0 ω−1
ω 0
0 1
1 0
,
=
ω−1 0
0 ω
,
= (aρ)−1.
Now consider such an irreducible representation for G2. Recall that ω4 = −1 and
so if we define aρ as for G1 we have a4ρ = −I2. Therefore we must define bρ such
that (bρ)2 = (aρ)4 = −I. We define our representation ρ : G2 → GL(2,C) by
aρ =
ω 0
0 ω−1
bρ =
0 −1
1 0
.
It’s clear to see that (aρ)8 = I2 and (bρ)2 = (aρ)4 = −I2, so we only need to check
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Chapter 26 136
the remaining defining relation holds.
(bρ)−1(aρ)(bρ) = 0 1−1 0ω 0
0 ω−10 −1
1 0 ,
=
0 ω−1
−ω 0
0 −1
1 0
,
=
ω−1 0
0 ω
,
= (aρ)−1.
Now consider such an irreducible representation for G3. We notice that for G3 thetrace on a is not ±√
2 but ±i √
2. Now we can see that ω − ω−1 = i √
2 and also that
−ω−1 = ω3. So, we define our representation ρ : G3 → GL(2,C) by
aρ =
ω 0
0 ω3
bρ =
0 1
1 0
It’s clear to see that (aρ)8 = (bρ)2 = I2, so we only need to check the remaining
defining relation holds.
(bρ)−1(aρ)(bρ) =0 1
1 0
ω 00 ω3
0 11 0
,
=
0 ω3
ω 0
0 1
1 0
,
=
ω3 0
0 ω
,
= (aρ)−1.
Note that (ω3)3 = ω9 = ω.
Now consider such an irreducible representation for G4. Note that the presentation
of G4 written in the book is redundant so we rewrite this as G4 = a, b | a8 = b 2 =
1, b −1ab = a5. Note from the character table of G4 that we require the trace on a
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Chapter 26 137
to be 0 and that −ω = ω5. So, we define our representation ρ : G4 → GL(2,C) by
aρ = ω 00 ω5
bρ = 0 11 0
.
It’s clear to see that (aρ)8 = (bρ)2 = I2, so we only need to check the remaining
defining relation holds.
(bρ)−1(aρ)(bρ) =
0 1
1 0
ω 0
0 ω5
0 1
1 0
,
= 0 ω5
ω 0 0 1
1 0 ,
=
ω5 0
0 ω
,
= (aρ)−1.
Note that (ω5)5 = ω25 = ω.
Finally consider such an irreducible representation for G9. Note that there is no
redundancy in the presentation of G9. Now z is an element of order 4 and recall that
i is a fourth root of unity. From the character table we know that the trace on a
should be 2i and 0 on a and b . So, we define our representation ρ : G9 → GL(2,C)
by
aρ =
1 0
0 −1
bρ =
0 1
1 0
i 0
0 i
Now it’s clear that (aρ)2 = (bρ)2 = (z ρ)4 = I2 so we only need to check that the
remaining relations hold.
(aρ)(z ρ) = 1 0
0 −1i 0
0 i = i 0
0 −i = i 0
0 i 1 0
0 −1 = (z ρ)(aρ),
(bρ)(zρ) =
0 1
1 0
i 0
0 i
=
0 i
i 0
=
i 0
0 i
0 1
1 0
= (zρ)(bρ),
(bρ)−1(aρ)(bρ) =
0 1
1 0
1 0
0 −1
0 1
1 0
=
0 −1
1 0
0 1
1 0
=
−1 0
0 1
= (aρ)(z ρ)2.
(b) Recall from Proposition 9.16 that if there exists a faithful irreducible CG module
then Z (G), (the centre of G), is cyclic. Now the centre of G is the union of all
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Chapter 26 138
conjugacy classes of order 1 in the group. Reading off from the character tables we
get Z (G5) = Z (G6) = Z (G7) = Z (G8) = C 1 ∪ C 2 ∪ C 3 ∪ C 4 = 1, z , a2, a2z .
Now z = b 2 in G5 so Z (G5) = 1, b 2, a2, a2b 2 ∼= C 2 × C 2 which is not cyclic.Note that it is easier to see this is C 2 × C 2 once you know ba = ab −1 and ab = b −1a.
In G6, G7 and G8 we have az = z a so Z (G6) ∼= Z (G7) ∼= Z (G8) ∼= C 2 × C 2, which is
not cyclic.
(c) Let the morphism in the question be denoted by ρ : G5 → GL(2,C), then this gives
a representation of G5 if it satisfies the defining relations of G5. We note first that
clearly z 2 = 1. Now
(bρ)2 = i 2 0 0
0 (−i )2 0
0 0 12
= −1 0 0
0 −1 0
0 0 1
= (z ρ),
(aρ)4 =
0 1 0
1 0 0
0 0 i
0 1 0
1 0 0
0 0 i
2
=
1 0 0
0 1 0
0 0 −1
2
=
12 0 0
0 12 0
0 0 (−1)2
= I3.
Now we check the final relation
(bρ)−1(aρ)(bρ) = −i 0 0
0 i 0
0 0 1
0 1 0
1 0 0
0 0 i
i 0 0
0 −i 0
0 0 1
,
=
0 −i 0
i 0 0
0 0 i
i 0 0
0 −i 0
0 0 1
,
=
0 −1 0
−1 0 0
0 0 i
,
=
0 1 0
1 0 0
0 0 i
−1 0 0
0 −1 0
0 0 1
,
= (aρ)(z ρ).
Therefore this is indeed a representation of G5. Is it faithful? Well from the above it’s
easy to see that 1ρ,aρ,a2ρ, a3ρ,bρ,zρ,azρ,b −1aρ,abρ are all distinct elements of the
group generated by the matrices. Therefore there are more than 8 distinct elements
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Chapter 26 139
in this group which means it must have order 16, hence ρ is a faithful representation.
Now let φ : G6 → GL(2,C) be the morphism of G6 defined in the question. It’s
clear that (aφ)4 = (bφ)2 = (z φ)2 = 1. We first of all check the commuting relations
(aρ)(z ρ) =
i 0 0
0 −i 0
0 0 i
−1 0 0
0 −1 0
0 0 1
=
−i 0 0
0 i 0
0 0 i
=−
1 0 0
0 −1 0
0 0 1
i 0 0
0 −i 0
0 0 i
= (z ρ)(aρ)
(bρ)(z ρ) =
0 1 0
1 0 0
0 0 1
−1 0 0
0 −1 0
0 0 1
,
=
0 −1 0
−1 0 0
0 0 1
,
=−
1 0 0
0 −1 0
0 0 1
0 1 0
1 0 0
0 0 1
,
= (z ρ)(bρ)
Now we check that the final relation holds
(bρ)−1(aρ)(bρ) =
0 1 0
1 0 0
0 0 1
i 0 0
0 −i 0
0 0 i
0 1 0
1 0 0
0 0 1
,
=
0 −i 0
i 0 0
0 0 i
0 1 0
1 0 0
0 0 1
,
=
−i 0 0
0 i 0
0 0 i
,
= i 0 0
0
−i 0
0 0 i
−1 0 0
0
−1 0
0 0 1 ,
= (aρ)(z ρ).
Therefore this indeed a representation of G6 but is it faithful? Well again we see that
we have generated more than 8 distinct elements, hence the group generated by the
matrices must have order 16 so it is a faithful representation.
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Chapter 26 140
(d) Note that we get an irreducible representation of D8 by the morphism
a → i 00 −i
b → 0 11 0
.
Therefore our plan is to extend this by defining a morphism ρ : G7 → GL(3,C) by
aρ =
i 0 0
0 −i 0
0 0 1
bρ =
0 1 0
1 0 0
0 0 1
z ρ =
1 0 0
0 1 0
0 0 −1
.
It’s clear that the relations of G7 hold for these matrices. Notice that the the subgroup
generated by aρ and bρ gives us 8 distinct elements and multiplying anyone of theseby z ρ will give us another distinct element, hence the representation of faithful.
Likewise for G8 we define a morphism φ : G8 → GL(3,C) by
aρ =
i 0 0
0 −i 0
0 0 1
bρ =
0 1 0
−1 0 0
0 0 1
z ρ =
1 0 0
0 1 0
0 0 −1
.
Notice again that we have extended this idea from a 2-dimensional irreducible repre-
sentation of Q8. So it’s clear to see that this will indeed be a 3-dimensional faithful
irreducible representation of G8.
Exercise 26.6. Recall that if two groups have distinct character tables then they cannot
possibly be isomorphic. So the only possible isomorphisms are G1∼= G2, G5
∼= G6 and
G7∼= G8. Notice that we have G7
∼= D8 × C 2 and G8∼= Q8 × C 2 therefore we cannot have
G7∼= G8 otherwise this would give us D8
∼= Q8.
Now for G5 and G6 we list, in table 26.4, the representatives of the conjugacy classes
and the orders of those elements. Note that in G5 we have ab = b −aa and ba = ab −1 and
in G6 we have ab = baz and ba = abz .
g 1 z a2 a2z a a3 b a2b ab a3b
Order in G5 1 2 2 2 4 4 4 4 4 4
Order in G6 1 2 2 2 4 4 2 2 4 4
Table 26.4: The orders of elements of G5 and G6 from Exercise 26.6
Therefore it’s clear that the number of elements of order 2 and order 4 are different in G5
and G6, which means G5 ∼= G6.
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Chapter 26 141
In table 26.5 we now do the same thing for G1 and G2. Note that in both G1 and G2
we have ab = ba−1 and ba = a−1b .
g 1 a4 a2 a a3 b ab
Order in G1 1 2 4 8 8 2 2
Order in G2 1 2 4 8 8 4 4
Table 26.5: The orders of elements of G1 and G2 from Exercise 26.6
Therefore the number of elements of order 2 and order 4 in G1 and G2 are different so we
cannot have G1
∼= G2.
Exercise 26.7. Let G be a non-abelian group of order p 4.
(a) Now G is a p -group so this means Z (G) = 1 so |Z (G)| = 1 and G is not abelian
so |Z (G)| = p 4. Recall from Lemma 26.1 that if K Z (G) and G/K is cyclic,
then G is abelian. Well if |Z (G)| = p 3 then |G/Z (G)| = p 4/p 3 = p , which means
G/Z (G) ∼= C p . However this would imply G abelian but this is not the case, so
|Z (G)| = p or p 2.
If |Z (G)| = p 2 then we clearly have p 2 conjugacy classes of order 1. We want
to then show that this leaves us with p 3
−p conjugacy classes. Let x 1, . . . , x k be
representatives of the conjugacy classes of G then by the class equation we have
p 4 = p 2 +
x i ∈Z (G)|x Gi | ⇒
x i ∈Z (G)
|x Gi | = p (p 3 − p ) = p 2(p 2 − 1).
Therefore there are either p 3 − p conjugacy classes of order p or p 2 − 1 conjugacy
classes of order p 2. Assume x ∈ G but x ∈ Z (G), then Z (G) C G(x ) but Z (G) =C G(x ) because x ∈ Z (G). Therefore |C G(x )| = p 3 ⇒ |x G| = p , which gives us the
desired result.
(b) Recall that we have the number of linear characters of G is [G : G] = |G/G | and wehave
|G| = [G : G ] +
χ(1)>1
χ(1)2 ⇒ p 4 = [G : G] +
χ(1)>1
χ(1)2.
Now if χ(1) > 1 then χ(1) = p because χ(1) | |G| and if χ(1) p 2 otherwise the
above equation is not satisfied as [G : G ] 1. Therefore p 2 | p 4 and p 2 divides the
sum so we must have p 2 | [G : G]. Note that |G | = 1 because G is not abelian, so
we have [G : G ] = p 2 or p 3. In other words |G| = p or p 2.
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Chapter 26 142
Assume that |G | = p 2 then we have [G : G ] = p 2. Recall that any non-linear
character of G has order p so using the above equation we have
p 4 = p 2 +
χ(1)>1
p 2 ⇒
χ(1)>1
p 2 = p 2(p 2 − 1).
Therefore this tells us there are p 2 − 1 irreducible characters of order p and p 2 linear
characters. Or put another way we have p 2 + (p 2 − 1) = 2p 2 − 1 conjugacy classes
of G.
(c) Recall that G is a normal subgroup of G and that if G is a p -group then any normal
subgroup cannot intersect the centre trivially. This means p |G ∩ Z (G)| p 2
because G
∩Z (G) is a subgroup of G and Z (G). If
|G
∩Z (G)
|= p 2 then
|G
|=
|Z (G)| = p 2 but this means that the number of conjugacy classes is p 3 + p 2 − p and
2p 2 − 1. So this means p 3 + p 2 − p = 2p 2 − 1 ⇒ p 3 − p 2 − p + 1 = 0 but the only
solutions to the polynomial X 3 − X 2 − X + 1 = 0 are ±1. Therefore we must have
|G ∩ Z (G)| = p .
Exercise 26.8.
(a) Let G be any group and let Z = Z (G). Assume G/Z (G) ∼= Q8, then we can assume
there exists the following presentation for G/Z (G)
G/Z (G) = aZ,bZ | a4Z = Z, a2Z = b 2Z, (b −1ab )Z = a−1Z .
Now we have from the defining relations of Q8 that abZ = ba−1Z and baZ = a−1bZ .
Therefore
(a2b )Z = (ba−2)Z = ba2Z.
Note that the centre of G/Z must be trivial but we clearly have a2Z in the centre
of G/Z so we must have a2Z = Z , or in other words a2 ∈ Z . So we cannot have
G/Z ∼= Q8.(b) Certainly if G is an abelian group of order 16 then G = 1, so G/G ∩Z (G) ∼= Q8. Now
suppose G is non-abelian then G = 1, so by Exercise 7 we have |G ∩ Z (G)| = 2,
so either G ∩ Z (G) = G or G ∩ Z (G) = Z (G). We know that G/G is abelian so
G/G ∼= Q8 and G/Z (G) ∼= Q8 by part (a). Therefore G/(G ∩ Z (G)) ∼= Q8.
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Chapter 27. Character table of the simple group of order
168
Exercise 27.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Exercise 27.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Exercise 27.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
Exercise 27.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Exercise 27.1. Let us fix a matrix
X =
x y
z w
∈ SL(2, p )
and assume that X ∈ Z (SL(2, p )). Thenx y
z w
1 1
0 1
=
1 1
0 1
x y
z w
⇒
x x + y
z z + w
=
x + z y + w
z w
,
⇒ z = 0 and x = w .
Choosing another appropriate matrix we find
x y
0 x
1 0
1 1
=
1 0
1 1
x y
0 x
⇒
x + y y
x x
=
x y
x x + y
,
⇒ y = 0.
Therefore X = xI2 but X ∈ SL(2, p ) means det(X ) = x 2 = 1 ⇒ x = ±1 and we’re done.
Exercise 27.2. We start by noting that|
SL(2, 3)|
= 3·
(32
−1) = 3
·8 = 24. We first
start by calculating the conjugacy classes of SL(2, 3). Much as for the conjugacy classes
of PSL(2, 7) we claim that the information in table 27.1 covers all the conjugacy classes of
SL(2, 3).
We now aim to confirm that the information in this table is correct. Clearly the information
about g 1 is correct. Now g 2 = 2I2 = −I2 ∈ Z (SL(2, 3)), therefore the information for g 2 is
clear.
143
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Chapter 27 144
Order |C G(g i )| |g Gi |
g 1 = 1 0
0 1 1 24 1
g 2 =
2 0
0 2
2 24 1
g 3 =
0 1
2 0
4 4 6
g 4 =
1 1
0 1
3 6 4
g 5 = 1 2
0 1 3 6 4
g 6 =
2 1
0 2
6 6 4
g 7 =
2 2
0 2
6 6 4
Table 27.1: The conjugacy classes of SL(2, 3)
Start by considering X ∈ C G(g 3) then we havex y
z w
0 1
2 0
=
0 1
2 0
x y
z w
⇒
2 y x
2w z
=
z w
2x 2 y
,
⇒ X =
x y
2 y x
.
Now det(X ) = x 2 − 2 y 2 = x 2 + y 2 = 1. Our only choices are (x , y ) = (0, 1), (0, 2), (1, 0)
or (2, 0) therefore |C G(g 3)| = 4, which means |g G3 | = 24/3 = 6. Now as for the order of g 3
we have
g 43 =
0 1
2 0
0 1
2 0
2
=
2 0
0 2
2=
4 0
0 4
= I2.
Consider an element X ∈ C G(g 4) then we havex y
z w
1 1
0 1
=
1 1
0 1
x y
z w
⇒
x x + y
z z + w
=
x + z y + w
z w
,
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Chapter 27 145
⇒ X =
x y
0 x
.
Now det(X ) = x 2 = 1 then our options are y = 0, 1, 2 and x = 1 or 2, hence |C G(g 4)| = 6,
which means |g G4 | = 24/6 = 4. Now as for the order of g 4 we have
g 34 =
1 1
0 1
1 1
0 1
1 1
0 1
=
1 2
0 1
1 1
0 1
=
1 0
0 1
= I2.
Consider an element X ∈ C G(g 5) then we have
x y
z w 1 2
0 1 = 1 2
0 1x y
z w ⇒ x 2x + y
z 2z + w = x + 2z y + 2w
z w ,
⇒ X =
x y
0 x
.
It’s clear from the previous calculation that |C G(g 5)| = 6, which means |g G5 | = 4. Now as
for the order of g 5 we have
g 35 =
1 2
0 1
1 2
0 1
1 2
0 1
=
1 1
0 1
1 2
0 1
=
1 0
0 1
= I2.
Consider an element X ∈ C G(g 6) then we havex y
z w
2 1
0 2
=
2 1
0 2
x y
z w
⇒
2x x + 2 y
2z z + 2w
=
2x + z 2 y + w
2z 2w
,
⇒ X =
x y
0 x
.
Hence again it’s clear from the previous calculation that we have |C G(g 6)| = 6, which means
|g G6 | = 4. Now as for the order of g 6 we have
g 66 =
2 1
0 2
2 1
0 2
2 1
0 2
2
=
1 1
0 1
2 1
0 2
2
=
2 0
0 2
2= I2.
Finally consider an element X ∈ C G(g 7) then we havex y
z w
2 2
0 2
=
2 2
0 2
x y
z w
⇒
2x 2x + 2 y
2z 2z + 2w
=
2x + 2z 2 y + 2w
2z 2w
,
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Chapter 27 146
⇒ X =
x y
0 x
.
Therefore again we confirm that |C G(g 7)| = 6, which means |g G7 | = 4. Now as for the order
of g 7 we have
g 67 =
2 2
0 2
2 2
0 2
2 2
0 2
2
=
1 2
0 1
2 2
0 2
2
=
2 0
0 2
2= I2.
So the information in the table is correct. However we need to make sure that none of
the elements are conjugate. By the orders of the elements we only have to worry about
whether g 4 is conjugate to g 5 and g 6 is conjugate to g 7. Wellx y
z w
1 1
0 1
=
1 2
0 1
x y
z w
⇒
x x + y
z z + w
=
x + 2z y + 2w
z w
,
⇒ X =
x y
0 2x
.
However det(X ) = 2x 2 = 1 but x 2 = 1 for all x ∈ Z3 so X cannot be in SL(2, 3). Similarly
for g 6 and g 7 we find
x y
z w
2 1
0 2
=
2 2
0 2
x y
z w
⇒
2x x + 2 y
2z z + 2w
=
2x + 2z 2 y + 2w
2z 2w
,
⇒ X =
x y
0 2x
and such a matrix cannot lie in SL(2, 3). In other words the conjugacy classes in the table
are indeed correct.
Recall that Z = Z (SL(2, 3)) = I2, 2I2 is a non-trivial normal subgroup of SL(2, 3) andSL(2, 3)/Z = PSL(2, 3), which we know is isomorphic to A4 from the text. The elements
of PSL(2, 3) split up by conjugacy classes are0 1
2 0
Z,
2 1
1 1
Z,
2 2
2 1
Z
1 1
0 1
Z,
1 0
2 1
Z,
0 1
2 2
Z,
2 1
2 0
Z
1 2
0 1
Z,
1 0
1 1
Z,
0 1
2 1
Z,
1 1
2 0
Z
1 0
0 1
Z
.
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Chapter 27 147
Now we can see that the conjugacy classes of order 4 here correspond to the 3-cycles in A4
and the conjugacy class of order 3 correspond to the 2-2-cycles in A4. It’s not important
to know exactly how the conjugacy classes match up to the conjugacy classes of 3-cycles inA4. This is because interchanging the conjugacy classes in the character table just reorders
the irreducible characters. Therefore we can lift the character table of A4, (on page 181),
to SL(2, 3).
Let ω = e 2πi/3 = 1+√ 3i
2then in table 27.2 we write down the lifted characters from
PSL(2, 3) to obtain four irreducible characters of SL(2, 3).
g g 1 g 2 g 3 g 4 g 5 g 6 g 7
|C G(g )
|24 24 4 6 6 6 6
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 1 1 ω ω2 ω ω2
χ3(g ) 1 1 1 ω2 ω ω2 ω
χ4(g ) 3 3 −1 0 0 0 0
Table 27.2: Four irreducible characters of SL(2, 3)
There are 7 conjugacy classes, which means there are 7 irreducible characters in total. We
start by working out the degrees of the remaining characters. Let d 5, d 6 and d 7 be the
remaining degrees then we know that d i | |G| = 24 and d 2i 24 ⇒ d i 4. Therefore the
degrees are either 1, 2, 3 or 4. We also know that
24 = 12 + 12 + 12 + 32 + d 25 + d 26 + d 27 ⇒ 12 = d 25 + d 26 + d 27 .
The only integer solution to this is d 5 = d 6 = d 7 = 2. Therefore the remaining three
characters are all of degree 2. Before carrying on we note that g −12 ∈ g G2 , g −13 ∈ g G3 ,
g 5 = g −14 and g 7 = g −16 . Therefore the columns of g 2 and g 3 are real and χ(g 5) = χ(g 4),
χ(g 7) = χ(g 6) for all irreducible characters χ of G.
Indeed because of the orders of the elements we will have that the columns of g 2 and g 3
will be integer valued. Also, we know that |C G(g 3)| = 4 which means the remaining entries
in the column of g 3 are zeros. Also because g 2 is an involution we know χ(g 2) is an integer
for all irreducible characters of G and χ(g 2) ≡ χ(1) (mod 2). Using this information and
the column orthogonality relations between g 1 and g 2 we can determine the column of g 2.
Reordering the characters slightly we have the character table so far to be as in table 27.3,
where η, ε are complex numbers and a, b are real numbers.
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Chapter 27 148
g g 1 g 2 g 3 g 4 g 5 g 6 g 7
|C G(g )
|24 24 4 6 6 6 6
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 1 1 ω ω2 ω ω2
χ3(g ) 1 1 1 ω2 ω ω2 ω
χ4(g ) 2 −2 0 a a b b
χ5(g ) 2 −2 0 η η ε ε
χ6(g ) 2 −2 0 η η ε ε
χ7(g ) 3 3
−1 0 0 0 0
Table 27.3: The partial character table of SL(2, 3)
We explain why we have the entries for χ4, χ5 and χ6. Recall that if χ is an irreducible
character then so is χ. If all three characters were real then they would all be the same
and this cannot happen. Therefore at least one must be complex and hence comes with its
complex conjugate, we assume this to be χ5 and χ6 without loss of generality. Therefore
the final character must be real and recalling that g 5 = g −14 and g 7 = g −16 this fills in the
remaining details.
Taking the inner product of χ1 with χ4 we find
χ1, χ4 =2
24− 2
24+ 0 +
a + a + b + b
6⇒ 0 = 2a + 2b ⇒ b = −a.
Also considering the inner product of χ4 with itself gives us
χ4, χ4 =22
24+
22
24+ 0 +
4a2
6⇒ 24 = 8 + 16a2 ⇒ a2 = 1 ⇒ a = ±1.
Now g 4 is an element of order 3 and a is an integer so we must have χ(g 4)≡
χ(1) (mod 3)
for all irreducible characters, which give us a = −1. Recall that χ4χ2 and χ4χ3 must also
be irreducible characters of degree 2 therefore these must be χ4 and χ5. Hence we have
the character table of SL(2, 3) to be as in table 27.4.
Exercise 27.3. We can see from the character table that there is no non-identity element
g ∈ PSL(2, 7) such that χ(g ) = χ(1) for some irreducible character χ of G. Hence the
kernel of every character is trivial, which means there are no non-trivial normal subgroups.
Hence the group is simple.
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Chapter 27 149
g g 1 g 2 g 3 g 4 g 5 g 6 g 7
|C G(g )
|24 24 4 6 6 6 6
χ1(g ) 1 1 1 1 1 1 1
χ2(g ) 1 1 1 ω ω2 ω ω2
χ3(g ) 1 1 1 ω2 ω ω2 ω
χ4(g ) 2 −2 0 −1 −1 1 1
χ5(g ) 2 −2 0 −ω −ω2 ω ω2
χ6(g ) 2 −2 0 −ω2 −ω ω2 ω
χ7(g ) 3 3
−1 0 0 0 0
Table 27.4: The character table of SL(2, 3)
Exercise 27.4.
(a) We start by working out the conjugacy classes of the subgroup. We claim that they
conjugacy classes are as in table 27.5.
Order |C G(g i )| |g Gi |
t 1
= 1 0
0 1Z 1 21 1
t 2 =
2 0
0 4
Z 3 3 7
t 3 =
3 0
0 5
Z 3 3 7
t 4 =
1 1
0 1
Z 7 7 3
t 5 = 1 6
0 1Z 7 7 3
Table 27.5: The conjugacy classes of the subgroup T PSL(2, 7)
Clearly the information about t 1 is correct. Let X ∈ C T (t 2) be a generic element thenx y
0 x −1
2 0
0 4
Z =
2 0
0 4
x y
0 x −1
Z ⇒
2x 4 y
0 4x −1
Z =
2x 2 y
0 4x −1
Z,
⇒ y = 0.
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Chapter 27 150
Therefore |C T (t 2)| = 3 ⇒ |t T 2 | = 7. As for the order of t 2 we have
t 32 = 2 00 4
2 00 4
2 00 4
Z = 4 00 2
2 00 4
Z = 1 00 1
Z = Z.
Let X ∈ C T (t 3) then we havex y
0 x −1
3 0
0 5
Z =
3 0
0 5
x y
0 x −1
Z ⇒
3x 5 y
0 5x −1
Z =
3x 3 y
0 5x −1
Z,
⇒ y = 0.
Therefore|C T (t 3)
|= 3
⇒ |t T 3
|= 7. As for the order of t 3 we have
t 33 =
3 0
0 5
3 0
0 5
3 0
0 5
Z =
2 0
0 4
3 0
0 5
Z =
6 0
0 6
Z = Z.
Let X ∈ C T (t 4) then we havex y
0 x −1
1 1
0 1
Z =
1 1
0 1
x y
0 x −1
Z ⇒
x x + y
0 x −1
Z =
x x −1 + y
0 x −1
Z,
⇒ x = x −1.
So x = 1, (or 6 but this case is equivalent modulo the centre), which means |C T (t 4)| =
7 and |t T 4 | = 3. As for the order of t 4 we have
t 74 =
1 1
0 1
7Z =
1 7 × 1
0 1
Z =
1 0
0 1
Z = Z.
Let X ∈ C T (t 5) then we have
x y
0 x −11 6
0 1Z = 1 6
0 1x y
0 x −1Z ⇒ x 6x + y
0 x −1Z = x 6x −1 + y
0 x −1Z,
⇒ x = x −1.
Again this gives us |C T (t 5)| = 7 and |t T 5 | = 7. As for the order of t 5 we have
t 75 =
1 6
0 1
7Z =
1 7 × 6
0 1
Z =
1 0
0 1
Z = Z.
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Chapter 27 151
Therefore the information in the table is correct. All that is left is to make sure that
t 3 ∈ t T 2 and t 5 ∈ t T 4 . Let X ∈ T thenx y
0 x −1
2 0
0 4
Z =
3 0
0 5
x y
0 x −1
Z ⇒
2x 4 y
0 4x −1
Z =
3x 3 y
0 5x −1
Z.
No such element exists in T therefore these elements are not conjugate. Similarly we
havex y
0 x −1
1 1
0 1
Z =
1 6
0 1
x y
0 x −1
Z ⇒
x x + y
0 x −1
Z =
x 6x −1 + y
0 x −1
Z,
⇒x = 6x −1.
However no such element exists in T therefore these elements are not conjugate so
the table of conjugacy classes is correct.
Note that |T | = 21 = 3 × 7 and T is certainly not abelian. Then by Proposition
25.7 we have T ∼= F 7,3. The character table for this group was calculated on page 240.
It’s clear that it’s not necessary to know exactly how these groups are isomorphic to
use the character table as swapping the conjugacy classes just permutes the irreducible
characters.
We can see that the only conjugacy class of PSL(2, 7) that splits upon restrictionto T is g G4 . It’s clear that [G : T ] = 8 so we have 1T ↑ G is as in table 27.6.
g g 1 g 2 g 3 g 4 g 5 g 6
|C G(g )| 168 8 4 3 7 7
(1T ↑ G)(g ) 8 0 0 2 1 1
χ 7 −1 −1 1 0 0
Table 27.6: The values of 1T
↑G and χ = 1T
↑G
−1G
Recall that we always have a copy of the trivial character in the induced trivial char-
acter by Frobenius Reciprocity. Let χ = 1T ↑ G − 1G then taking the inner product
we see
χ, χ =72
168+
1
8+
1
4+
1
3=
49 + 21 + 42 + 56
168=
168
168= 1.
Note we could also have seen this directly from the character table of PSL(2, 7) on
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Chapter 27 153
so g 25 ∈ g G5 . Similarly we have
a b c d
1 60 1
Z = 1 50 1
a b c d
⇒ a 6a + b c 6c + d
= a + 5c b + 5d c d
,
⇒ c = 0 and d = 4a.
So again this element exists in G so g 26 ∈ g G6 . Using this information and the formula
on page 198, we see that χS and χA are as in table 27.8.
g g 1 g 2 g 3 g 4 g 5 g 6
g 2 g 1 g 1 g 2 g 4 g 5 g 6
|C G(g )| 168 8 4 3 7 7
χ 7 −1 −1 1 0 0
χS 28 4 0 1 0 0
χA 21 −3 0 0 0 0
Table 27.8: The decomposition of χ2
We inspect which of the characters we already know appear in χS. Now taking
inner products we see
χS, χ =7 × 28
168− 4
8+
1
3=
196 − 84 + 56
168=
168
168= 1,
χS, λ ↑ G =8 × 28
168− 1
3=
224 − 56
168=
168
168= 1,
χS, 1G =28
168+
4
8+
1
3=
28 + 84 + 56
168=
168
168= 1.
Therefore χ = ψ + χ + λ ↑ G + 1G for some character ψ of G. This character ψ has
values as in table 27.9.
g g 1 g 2 g 3 g 4 g 5 g 6
|C G(g )| 168 8 4 3 7 7
ψ 12 4 0 0 −2 −2
Table 27.9: The character ψ
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Chapter 27 155
χ
χ(g 1)χ(g 3) = 1 − 7 ± 3 ± 3 = 0.
Recall from the calculation in the chapter that χ(g 5) is complex for some irre-
ducible character χ. We assume this is χ5 and as χ5 is non-real we have χ6 = χ5 is
another irreducible character. Also g 6 = g −15 ⇒ χi (g 6) = χi (g 5) for each i = 5, 6.
This gives us the current character table of G. We now just have determine the value
of η.
Considering the column orthogonality relation of g 5 with g 3 we obtain
0 = 1 + η + η ⇒ 2Re(η) = −1 ⇒ Re(η) = −1
2.
Using the column orthogonality of g 5 with itself we get
7 = 1 + 1 + 1 + 2|η|2 ⇒ 2|η|2 = 4 ⇒ |η|2 = 2.
In other words this tells us that
Re(η)2 + Im(η)2 = 2 ⇒ 1
4+ Im(η)2 = 2 ⇒ Im(η)2 =
7
4⇒ Im(η) = ±
√ 7
2.