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irreducibility criterion is very useful.

Proposition 4.17. If ρ : G ’’ GLC (V ) is a non-zero representation, then V is irreducible
if and only if (χρ |χρ ) = 1.

Proof. If V = n1 V1 • · · · • nr Vr , then by orthonormality of the χj ,

n2 .
(χρ |χρ ) = ( ni χi | nj χj ) = ni nj (χi |χj ) = j
i j i j j

if and only if n2 + · · · + n2 = 1. Remembering that the nj are non-negative
So (χρ |χρ ) = 1 r
1
integers we see that (χρ |χρ ) = 1 if and only if all but one of the nj is zero and for some k,
∼ Vk and so is irreducible.
nk = 1. Thus V =

Notice that for the character table of S3 we can check that the characters satisfy this criterion
and are also orthonormal. Provided we believe that the rows really do represent characters we
have found an orthonormal basis for the class functions C(S3 ). We will return to this problem
later.

Example 4.18. Let us assume that the above character table for S3 is correct and let
ρ = ρreg be the regular representation of S3 on the vector space V = C[S3 ]. Let us take as a
basis for V the elements of S3 . Then
ρσ „ = σ„,
hence the matrix [ρσ ] of ρσ relative to this basis has 0™s down its main diagonal, except when
σ = e for which it is the 6 — 6 identity matrix. The character is χ given by

6 if σ = e,
χ(σ) = Tr[ρσ ] =
0 otherwise.
Thus we obtain
1 1
(χρ |χ1 ) = χρ (σ)χ1 (σ) = 6 = 1,
6 6
σ∈S3
1 1
(χρ |χ2 ) = χρ (σ)χ2 (σ) = 6 = 1,
6 6
σ∈S3
1 1
(χρ |χ3 ) = χρ (σ)χ3 (σ) = (6 — 2) = 2.
6 6
σ∈S3

Hence we have
C[S3 ] ∼ V1 • V2 • V3 • V3 = V1 • V2 • 2V3 .
=
4. CHARACTER TABLES 43

In fact we have seen the representation V3 already in Problem Sheet 2, Qu. 5(b). It is easily
veri¬ed that the character of that representation is χ3 .
Of course, in order to use character tables, we ¬rst need to determine them! So far we do
not know much about this beyond the fact that the number of rows has to be the same as
the number of conjugacy classes of the group G and the existence of the 1-dimensional trivial
character which we will always denote by χ1 and whose value is χ1 (g) = 1 for g ∈ G. The
characters of the distinct complex irreducible representations of G are the irreducible characters
of G.

Theorem 4.19. Let G be a ¬nite group. Let χ1 , . . . , χr be the distinct complex irreducible
characters and ρreg the regular representation of G on C[G].
a) Every complex irreducible representation of G occurs in C[G]. Equivalently, for each
irreducible character χj , (χρreg |χj ) = 0.
b) The multiplicity nj of the irreducible Vj with character χj in C[G] is given by

nj = dimC Vj = χj (e).

So to ¬nd all the irreducible characters, we only have to decompose the regular representa-
tion!

Proof. Using the formual¦

|G| if g = e,
χρreg (g) =
0 if g = e,
we have
1 1
nj = (χρreg |χj ) = χρreg (g)χj (g) = χρ (e)χj (e) = χj (e).
|G| reg
|G|
g∈G

Corollary 4.20. We have
r r
n2 = (χρreg |χj )2 .
|G| = j
j=1 j=1

The following result also holds but the proof requires some Algebraic Number Theory.

Proposition 4.21. For each irreducible character χj , nj = (χρreg |χj ) divides the order of
G, i.e., nj | |G|.

The following row and column orthogonality results for the character table of a group G are
very important.

Theorem 4.22. Let χ1 , . . . , χr be the distinct complex irreducible characters of G and e =
g1 , . . . , gr be a collection of representatives for the conjugacy classes of G and for each k, let
CG (gk ) be the centralizer of gk .
a) Row orthogonality: For 1 i, j r,
r
χi (gk )χj (gk )
= (χi |χj ) = δij .
| CG (gk )|
k=1

b) Column orthogonality: For 1 i, j r,
r
χk (gi )χk (gj )
= δij .
| CG (gi )|
k=1
44 4. CHARACTER THEORY

Proof.
a) We have
1
δij = (χi |χj ) = χi (g)χj (g)
|G|
g∈G
r
1 |G|
= χi (gk )χj (gk )
|G| | CG (gk )|
k=1

(since the conjugacy class of gk contains |G|/| CG (gk )| elements)
r
χi (gk )χj (gk )
= .
| CG (gk )|
k=1

b) Let ψs : G ’’ C be the function given by

1 if g is conjugate to gs ,
ψs (g) =
0 if g is not conjugate to gs .
By Theorem 4.15, there are »k ∈ C such that
r
ψs = »k χk .
k=1

But then »j = (ψs |χj ). We also have
1
(ψs |χj ) = ψs (g)χj (g)
|G|
g∈G
r
ψs (gk )χj (gk )
=
| CG (gk )|
k=1

χj (gs )
= ,
| CG (gs )|

hence
r
χj (gs )
ψs = χj .
| CG (gs )|
j=1

Thus we have the required formula
r
χj (gt )χj (gs )
δst = ψs (gt ) = .
| CG (gs )|
j=1


5. Examples of character tables
Equipped with the results of the last section, we can proceed to ¬nd some character tables.
For abelian groups we have the following result which follows from what we have seen already
together with the fact that in an abelian group every conjugacy class has exactly one element.

Proposition 4.23. Let G be a ¬nite abelian group. Then there are |G| distinct complex
irreducible characters, each of which is 1-dimensional. Moreover, in the regular representation
each irreducible occurs with multiplicity 1, i.e.,

C[G] ∼ V1 • · · · • V|G| .
=
5. EXAMPLES OF CHARACTER TABLES 45


Example 4.24. Let G = g0 ∼ Z/n be cyclic of order n. Let ζn = e2πi/n , the ˜standard™
=
primitive nth root of unity. Then for each k = 0, 1, . . . , (n ’ 1) we may de¬ne a 1-dimensional
representation ρk : G ’’ C— by
r rk
ρk (g0 ) = ζn .
The character of ρk is χk given by
r rk
χk (g0 ) = ζn .
Clearly these are all irreducible and non-isomorphic.

Let us consider the orthogonality relations for these characters. We have
n’1
1 r r
(χk |χk ) = χk (g0 )χk (g0 )
n
r=0
n’1
1 kr kr
= ζn ζn
n
r=0
n’1
1 n
= 1= = 1.
n n
r=0

For 0 k< (n ’ 1) we have
n’1
1 r r
(χk |χ ) = χk (g0 )χ (g0 )
n
r=0
n’1
1 kr
ζn ζnr
=
n
r=0
n’1
1 (k’ )r
= ζn .
n
r=0
By row orthogonality this sum is 0. This is a special case of the following identity which is often
used in many parts of Mathematics.

Lemma 4.25. Let d ∈ N, m ∈ Z and ζd = e2πi/d . Then
d’1
d if d | m,
mr
ζd =
0 otherwise.
r=0

Proof. We give a proof which does not use character theory!
m
If d m, then ζd = 1. Then we have
d’1 d’1
m(r+1)
m mr
ζd ζd = ζd
r=0 r=0
d
ms
= ζd
s=1
d’1
mr
= ζd ,
r=0

hence
d’1
m mr
(ζd ’ 1) ζd = 0,
r=0
46 4. CHARACTER THEORY

and so

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