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d’1
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 ,

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