mr

ζd = 0.

r=0

If d | m then

d’1 d’1

mr

ζd = 1 = d.

r=0 r=0

the case where n = 3 and G = g0 ∼ Z/3. The

As a special case of Exercise 4.24, consider =

character table of G is

2

e g0 g0

χ1 1 11

2

χ2 1 ζ3 ζ3

2

χ3 1 ζ3 ζ3

order 4, so G ∼ Z/2 — Z/2. The character

Example 4.26. Let G = a0 , b0 be abelian of =

table of G is as follows.

e a0 b0 a0 b0

χ1 = χ00 1 1 1 1

χ10 1 ’1 1 ’1

χ01 11 ’1 ’1

χ11 1 ’1 ’1 1

Example 4.27. The character table of the quaternion group of order 8, Q8 is as follows.

1 ’1 i j k

χ1 1 1 1 1 1

χi 1 1 1 ’1 ’1

χj 1 1 ’1 1 ’1

χk 1 1 ’1 ’1 1

χ2 2 ’2 0 0 0

Proof. There are 5 conjugacy classes:

{1}, {’1}, {i, ’i}, {j, ’j}, {k, ’k}.

As always we have the trivial character χ1 . There are 3 homomorphisms Q8 ’’ C— given by

ρi (ir ) = 1 and ρi (j) = ρi (k) = ’1,

ρj (jr ) = 1 and ρj (i) = ρi (k) = ’1,

ρk (kr ) = 1 and ρk (i) = ρk (j) = ’1.

These provide three 1-dimensional representations with characters χi , χj , χk taking values

χi (ir ) = 1 and χi (j) = χi (k) = ’1,

χj (jr ) = 1 and χj (i) = χi (k) = ’1,

χk (kr ) = 1 and χk (i) = χk (j) = ’1.

Since |Q8 | = 8, we might try looking for a 2-dimensional complex representation. But the de¬ni-

tion of Q8 provides us with the inclusion homomorphism j : Q8 ’’ GLC (C2 ), where we interpret

the matrices as taken in terms of the standard basis. The character of this representation is χ2

given by

χ2 (1) = 2, χ2 (’1) = ’2, χ2 (±i) = χ2 (±j) = χ2 (±k) = 0.

This completes the determination of the character table of Q8 .

5. EXAMPLES OF CHARACTER TABLES 47

Example 4.28. The character table of the dihedral group of order 8, D8 , is as follows.

e ±2 ± β ±β

χ1 1 1 1 1 1

χ2 1 1 1 ’1 ’1

χ3 1 1 ’1 1 ’1

χ4 1 1 ’1 ’1 1

χ5 2 ’2 0 0 0

Proof. The elements of D8 are

e, ±, ±2 , ±3 , β, ±β, ±2 β, ±3 β

and these satisfy the relations

±4 = e = β 2 , β±β = ±’1 .

The conjugacy classes are the sets

{e}, {±2 }, {±, ±3 }, {β, ±2 β}, {±β, ±3 β}.

There are two obvious 1-dimensional representations, namely the trivial one ρ1 and also ρ2 ,

where

ρ2 (±) = 1, ρ2 (β) = ’1.

The character of ρ2 is determined by

χ2 (±r ) = 1, χ2 (β±r ) = ’1.

A third 1-dimensional representation comes from the homomorphism ρ3 : D8 ’’ C— given by

ρ3 (±) = ’1, ρ3 (β) = 1.

The fourth 1-dimensional representation comes from the homomorphism ρ4 : D8 ’’ C— for

which

ρ4 (±) = ’1, ρ4 (β) = ’1.

The characters χ1 , χ2 , χ3 , χ4 are clearly distinct and thus orthonormal.

Before describing χ5 as the character of a 2-dimensional representation, we will determine

it up to a scalar factor. Suppose that

χ5 (e) = a, χ5 (±2 ) = b, χ5 (±) = c, χ5 (β) = d, χ5 (β±) = e

for a, b, c, d, e ∈ C. The orthonormality conditions give (χ5 |χj ) = δj 5 . For j = 1, 2, 3, 4, we

obtain the following linear system:

®

® a ®

11 2 2 2 0

b

1 1 2 ’2 ’2

c = 0

(5.1) °1 1 ’2 2 ’2» °0»

°d»

1 1 ’2 ’2 2 0

e

which has solutions

b = ’a, c = d = e = 0.

If χ5 is an irreducible character we must also have (χ5 |χ5 ) = 1, giving

a2

12 2

1= a +a = ,

8 4

and so a = ±2. So we must have the stated bottom row. The corresponding representation is

that of Example 3.5, viewed as a complex representation. This is easily seen to have χ5 as its

character.

48 4. CHARACTER THEORY

Remark: The groups Q8 and D8 have identical character tables even though they are non-

isomorphic! Thus character tables cannot always tell non-isomorphic groups apart.

Example 4.29. The character table of the symmetric group S4 , is as follows.

e (12) (12)(34) (123) (1234)

[1] [6] [3] [8] [6]

χ1 1 1 1 1 1

χ2 1 ’1 1 1 ’1

χ3 3 1 ’1 0 ’1

χ4 3 ’1 ’1 0 1

χ5 2 0 2 ’1 0

Proof. Recall that the conjugacy classes correspond to the di¬erent cycle types of which

are represented by the following list of elements where the numbers in brackets give the sizes of

the conjugacy classes:

e [1], (12) [6], (12)(34) [3], (123) [8], (1234) [6].

So there are 5 rows and columns in the character table. The sign representation sign : S4 ’’ C—

is 1-dimensional and its character is

χ2 (e) = χ2 ((12)(34)) = χ2 (123) = 1 and χ2 (12) = χ2 (1234) = ’1.

The 4-dimensional permutation representation ρ4 corresponding to the action on 4 = {1, 2, 3, 4}

˜

has character χρ4 given by

˜

χρ4 (σ) = number of ¬xed points of σ.

˜

So we have

χρ4 (e) = 4, χρ4 ((12)(34)) = χρ4 (1234) = 0, χρ4 (123) = 1, χρ4 (12) = 2.

˜ ˜ ˜ ˜ ˜

We know that this representation has the form

C[4] = C[4]S4 • W

where W is a S4 -subspace of dimension 3, for which the character χ3 is determined by

χ1 + χ3 = χρ4 ,

˜

and hence

χ3 = χρ4 ’ χ1 .

˜

So we obtain the following values for χ3

χ3 (e) = 3, χ3 ((12)(34)) = χ3 (1234) = ’1, χ3 (123) = 0, χ3 (12) = 1.

Calculating the inner product of this with itself gives

1

(χ3 |χ3 ) = (9 + 6 + 3 + 0 + 6) = 1,

24

and so χ3 is the character of an irreducible representation.

From this information we can deduce that the two remaining irreducibles must have dimen-

sions n4 , n5 satisfying

n2 + n2 = 24 ’ 1 ’ 1 ’ 9 = 13,

4 5

and thus we can take n4 = 3 and n5 = 2, since these are the only possible values up to order.

If we form the tensor product ρ2 — ρ3 we get a character χ4 given by

χ4 (g) = χ2 (g)χ3 (g),

hence the 4th line in the table. Then (χ4 |χ4 ) = 1 and so χ4 really is an irreducible character.

6. RECIPROCITY FORMULÆ 49

For χ5 , recall that the regular representation ρreg has character χρreg decomposing as

χρreg = χ1 + χ2 + 3χ3 + 3χ4 + 2χ5 ,

hence we have

1

χ5 = χρreg ’ χ1 ’ χ2 ’ 3χ3 ’ 3χ4 ,

2

which gives the last row of the table.

Notice that in this example, the tensor product ρ3 —ρ5 which is a representation of dimension

6 cannot be irreducible. Its character χρ3 —ρ5 must be a linear combination of the irreducibles,

5

χρ3 —ρ5 = (χρ3 —ρ5 |χj )χj .

j=1

Recall that for g ∈ S4 ,

χρ3 —ρ5 (g) = χρ3 (g)χρ5 (g).

For the values of the coe¬cients we have

1

(χρ3 —ρ5 |χ1 ) = (6 + 0 ’ 6 + 0 + 0) = 0,

24

1

(χρ3 —ρ5 |χ2 ) = (6 + 0 ’ 6 + 0 + 0) = 0,

24

1

(χρ3 —ρ5 |χ3 ) = (18 + 0 + 6 + 0 + 0) = 1,

24

1

(χρ3 —ρ5 |χ4 ) = (18 + 0 + 6 + 0 + 0) = 1,

24

1

(χρ3 —ρ5 |χ5 ) = (12 + 0 ’ 12 + 0 + 0) = 0.

24

Thus we have

χρ3 —ρ5 = χ3 + χ4 .

In general it is hard to predict how the tensor product of two representations decomposes in

terms of irreducibles.

6. Reciprocity formul¦

Let H G, ρ : G ’’ GLC (V ) a representation of G and σ : H ’’ GLC (W ) a representation

of H. Recall that the induced representation σ ‘G is of dimension |G/H| dimC W , while the

H

C

G has dimension dim V . We will write χ “G and χ ‘G for the characters of

restriction ρ “H ρH σH

these representations. First we show how to calculate the character of an induced representation.

Lemma 4.30. The character of the induced representation σ ‘G is given by

H

1

χσ ‘G (g) = χσ (x’1 gx).

H

|H|

x∈G

g∈xHx’1

Proof. See §16 of [A&B].

Example 4.31. Let H = {e, ±, ±2 , ±3 } D8 . Let σ : H ’’ C— be the 1-dimensional

representation of H given by

σ(±k ) = ik .

Decompose the induced representation σ ‘D8 into its irreducible summands over the group D8 .

H

50 4. CHARACTER THEORY

Proof. We will use the character table of D8 given in Example 4.28. Notice that H D8 ,

hence for x ∈ D8 we have xHx’1 = H. Let χ = χσ ‘D8 be the character of this induced

H

representation. We have

1

χσ (x’1 gx)

χ(g) =

4

x∈D8

g∈xHx’1

±

1 χσ (x’1 gx)

if g ∈ H,

= 4 x∈D8

0 if g ∈ H.

/

Thus if g ∈ H we ¬nd that

±

1 4χ (±) + 4χ (±3 ) if g = ±, ±3 ,

4 σ σ

1

8χσ (±2 ) if g = ±2 ,

χ(g) =

4

1

(8χσ (e)) if g = e.

4

Hence we have ±

i + i3 = 0 if g = ±, ±3 ,

’2 if g = ±2 ,