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χ(g) =
2 if g = e,




0 if g ∈ H.
/
Taking inner products with the irreducible characters χj we obtain the following.
1
(χ|χ1 )D8 = (2 ’ 2 + 0 + 0 + 0) = 0,
8
1
(χ|χ2 )D8 = (2 ’ 2 + 0 + 0 + 0) = 0,
8
1
(χ|χ3 )D8 = (2 ’ 2 + 0 + 0 + 0) = 0,
8
1
(χ|χ4 )D8 = (2 ’ 2 + 0 + 0 + 0) = 0,
8
1
(χ|χ5 )D8 = (4 + 4 + 0 + 0 + 0) = 1.
8
Hence we must have χ = χ5 , giving another derivation of the representation ρ5 .
Theorem 4.32 (Frobenius Reciprocity). There is a linear isomorphism
HomG (W ‘G , V ) ∼ HomH (W, V “G ).
=
H H

Equivalently on characters we have
(χσ ‘G |χρ )G = (χσ |χρ “G )H .
H H

Proof. See §16 of [A&B].
Example 4.33. Let σ be the irreducible representation of S3 with character χ3 and un-
derlying vector space W . Decompose the induced representation W ‘S4 into its irreducible
S3
summands over the group S4 .
Proof. Let
W ‘S4 ∼ n1 V1 • n2 V2 • n3 V3 • n4 V4 • n5 V5 .
S3 =
Then
nj = (χj |χσ ‘S4 )S4 = (χj “S4 |χσ )S3 .
S3 S3
7. REPRESENTATIONS OF SEMI-DIRECT PRODUCTS 51


To evaluate the restriction χj “S4 we take only elements of S4 lying in S3 . Hence we have
S3
1
n1 = (χ1 “S4 |χσ )S3 = (2 + 0 ’ 2) = 0,
S3
6
1
n2 = (χ2 “S4 |χσ )S3 = (1 · 2 + 0 + 1 · ’2) = 0,
S3
6
1
n3 = (χ3 “S4 |χσ )S3 = (3 · 2 + 0 + 0 · ’2) = 1,
S3
6
1
n4 = (χ4 “S4 |χσ )S3 = (3 · 2 + 0 + 0 · ’2) = 1,
S3
6
1 6
n5 = (χ5 “S4 |χσ )S3 = (2 · 2 + 0 + ’2 · ’1) = = 1.
S3
6 6
Hence we have
W ‘S4 ∼ V3 • V4 • V5 .
S3 =

7. Representations of semi-direct products
Recall the notion of a semi-direct product group G = N H; this has N G, H G,
H © N = {e} and HN = N H = G. We will describe a way to produce the irreducible
characters of G from those of the groups N G and H G.
Proposition 4.34. Let • : Q ’’ G be a homomorphism and ρ : G ’’ GLC (V ) be a
representation of G. Then the composite •— ρ = ρ —¦ • is a representation of Q on V . Moreover,
if •— ρ is irreducible over Q, then ρ is irreducible over G.
Proof. The ¬rst part is clear.
For the second, suppose that W ⊆ V is a G-subspace. Then for h ∈ Q and w ∈ W we have
(•— ρ)h w = ρ•(h) w ∈ W.
Hence W is a Q-subspace. By irreducibility of •— ρ, W = {0} or W = V , hence V is irreducible
over G.
The representation •— ρ is called the representation on V induced by • and we often denote
the underlying Q-module by •— V . If j : Q ’’ G is the inclusion of a subgroup, then j — ρ = ρ “G ,
Q
the restriction of ρ to Q.
In the case of G = N H, there is a surjection π : G ’’ H given by
π(nh) = h (n ∈ N, h ∈ H),
as well as the inclusions i : N ’’ G and j : H ’’ G. We can apply the above to each of these
homomorphisms.
Now let ρ : G ’’ GLC (V ) be an irreducible representation of the semi-direct product G =
N H. Then i— V decomposes as
i— V = W1 • · · · • Wm
where Wk is a non-zero irreducible N -subspace. For each g ∈ G, notice that if x ∈ N and
w ∈ W1 , then
ρx (ρg w) = ρxg w = ρg ρg’1 xg w = ρg w
for w = ρg’1 xg w. Since g ’1 xg ∈ g ’1 N g = N ,
gW1 = {ρg w : w ∈ W1 }
is an N -subspace of i— V . If we take
± 
 
˜
W1 = ρg wg : wg ∈ W1 ,
 
g∈G
52 4. CHARACTER THEORY

˜
then we can verify that W1 is a non-zero N -subspace of i— V and in fact is also a G-subspace of
˜
V . Since V is irreducible, this shows that V = W1 .
Now let
H1 = {h ∈ H : hW1 = W1 } ⊆ H.
Then we can verify that H1 H G. The semidirect product
G1 = N H1 = {nh ∈ G : n ∈ N, h ∈ H1 } G
also acts on W1 since for nh ∈ G1 and w ∈ W1 ,
ρnh w = ρn ρh w = ρn w ∈ W1
where w = ρh w; hence W1 is a G1 -subspace of V “G1 . Notice that by the second part of
G
Proposition 4.34, W1 is irreducible over G1 .
Lemma 4.35. There is a G-isomorphism
W1 ‘G1 ∼ V.
G=

Proof. See the books [J&L] and [J-PS].
Thus every irreducible of G = N H arises from an irreducible of N which extends to a
representation (actually irreducible) of such a subgroup N K N H = G for K H but
to no larger subgroup.
Example 4.36. Let D2n be the dihedral group of order 2n. Then every irreducible repre-
sentation of D2n has dimension 1 or 2.
Proof. We have D2n = N H where N = ± ∼ Z/n and H = {e, β}. The n distinct
=
irreducibles ρk of N are all 1-dimensional by Example 4.24. Hence for each of these we have a
subgroup Hk H such that the action of N extends to N Hk and so the corresponding induced
representation Vk ‘D2 n is irreducible over D2n with dimension |D2n /(N Hk )| = 2/|Hk |. Every
Hk
irreducible of D2n occurs this way.
For n = 4, it is a useful exercise to identify all the irreducibles in the character table in this
way.
CHAPTER 5


Some applications to group theory

In this chapter we will see some applications of representation theory to Group Theory.

1. Characters and the structure of groups
In this section we will give some results relating the character table of a ¬nite group to its
subgroup structure.
Let ρ : G ’’ GLC (V ) be a representation for which dimC V = n. De¬ne the subset
ker χρ = {g ∈ G : χρ (g) = χρ (e)}

Proposition 5.1. ker χρ = ker ρ and hence ker χρ is a normal subgroup of G.

Proof. For g ∈ ker χρ , let v = {v1 , . . . , vn } be a basis of V consisting of eigenvectors of ρg ,
so ρg vk = »k vk for suitable »k ∈ C, and indeed each »k is a root of unity and so has the form
»k = etk i for tk ∈ R. Then
n
χρ (g) = »k .
k=1
Recall that for t ∈ R, eti = cos t + i sin t. Hence
n n
χρ (g) = cos tk + i sin tk .
k=1 k=1

Since χρ (e) = n,
n
cos tk = n,
k=1
which can only happen if each cos tk = 1, but then sin tk = 0. So we have all »k = 1 which
implies that ρg = IdV . Thus ker χρ = ker ρ as claimed.
Now let χ1 , . . . , χr be the distinct irreducible characters of G and r = {1, . . . , r}.
r
Proposition 5.2. k=1 ker χk = {e}.
r
Proof. Set K = k=1 ker χk G. By Proposition 5.1, for each k, ker χk = ker ρk , hence
ker ρk there is a factorisation of ρk : G ’’ GLC (Vk ),
N G. Indeed, since N
ρ
p
G ’ G/K ’k GLC (Vk ),
’ ’
where p : G ’’ G/K is the quotient homomorphism. As p is surjective, it is easy to check
that ρk is an irreducible representation of G/K, with character χk . Clearly the χk are distinct
irreducible characters of G/K and nk = χk (e) = χk (eK) are the dimensions of the corresponding
irreducible representations.
By Corollary 4.20, we have
n2 + · · · + n2 = |G|
1 r
since the χk are the distinct irreducible characters of G. But we also have
n2 + · · · + n2 |G/K|
1 r
53
54 5. SOME APPLICATIONS TO GROUP THEORY

since the χk are some of the distinct irreducible characters of G/K. Combining these we have
|G| |G/K| which can only happen if |G/K| = |G|, i.e., if K = {e}. So in fact
r
ker χk = {e}.
k=1

Proposition 5.3. A subgroup N G is normal if and only it has the form

N= ker χk
k∈S

for some subset S ⊆ r.

Proof. Let N G and suppose the quotient group G/N has s distinct irreducible repre-
sentations σk : G/N ’’ GLC (Wk ) (k = 1, . . . , s) with characters χk . Each of these gives rise to
˜
a composite representation of G
q σ
σk : G ’ G/N ’k GLC (Wk )
’ ’

and again this is irreducible because the quotient homomorphism q : G ’’ G/N is surjective.
This gives s distinct irreducible characters of G, so each χσk is actually one of the χj .
By Proposition 5.2 applied to the quotient group G/N ,
s s
ker σk = ker χk = {eN },
˜
k=1 k=1

hence since ker σk = q ’1 ker σk , we have
s s
ker χσk = ker σk = N.
k=1 k=1

Conversely, for any S ⊆ r, ker χk G since for each k, ker χk G.
k∈S

Corollary 5.4. G is simple if and only if for every irreducible character χk = χ1 and
e = g ∈ G, χk (g) = χk (e). Hence the character table can be used to decide whether G is simple.

Corollary 5.5. The character table can be used to decide whether G is solvable.

Proof. G is solvable if and only if there is a sequence of subgroups

{e} = G G · · · G1 G0 = G
’1

for which the quotient groups Gs /Gs+1 are abelian. This can be seen from the character table.
For a solvable group we can take the subgroups to be the lower central series given by G(0) = G,
and in general G(s+1) = [G(s) , G(s) ]. It is easily veri¬ed that G(s) G and G(s) /G(s+1) is abelian.
By Proposition 5.3 we can now check whether such a sequence of normal subgroups exists using
the character table.

We can also de¬ne the subset

ker |χρ | = {g ∈ G : |χρ (g)| = χρ (e)}.

Proposition 5.6. ker |χρ | is a normal subgroup of G.
2. A RESULT ON REPRESENTATIONS OF SIMPLE GROUPS 55

Proof. If g ∈ ker |χρ |, then using the notation of the proof of Proposition 5.1, we ¬nd that
n n
2
sin tk |2
|χρ (g)| = | cos tk + i
k=1 k=1
2 2
n n
= cos tk + sin tk
k=1 k=1
n n

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