166 CHAPTER 15

Examples of Characters

To illustrate these concepts further, it will be helpful to have a

concrete example. Suppose H =

{1,2,3}

, the permutation group of

{1, 2, 3}. We call it

3

for short.

We need to refer to the elements of

3

in terms of just two of

those elements. We let σ be the following permutation:

1 → 2

2 → 3

3 → 1.

Let τ be this permutation:

1 → 2

2 → 1

3 → 3.

It is a fact that all of the elements in

3

can be written in terms

of σ and τ (and the identity permutation e). We usually use power

notation for the composition of a permutation with itself, and so σ

2

is another way to write σ ◦ σ , which is the permutation

1 → 3

2 → 1

3 → 2.

The six elements of

3

are {e, σ , σ

2

, τ , τ ◦ σ , τ ◦ σ

2

}.

EXERCISE: Show that σ

3

= τ

2

= e, and that σ ◦ τ = τ ◦ σ

2

.

EXERCISE: Show that

3

has three conjugacy classes: {e},

{σ , σ

2

}, and {τ, τ ◦ σ , τ ◦ σ

2

}.

SOLUTION: Recall that if x and y are in the same conjugacy

class, then there is some element g in the group so that

y = g ◦ x ◦ g

−1

. Now, it is not too difﬁcult to check that

THE GREEKS HAD A NAME FOR IT 167

g ◦ e ◦ g

−1

will always be e,soe is the only element in its

conjugacy class. If you do a bit of computation, you can check

that τ ◦ σ ◦ τ

−1

= σ

2

,soσ and σ

2

are in the same conjugacy

class. A bit more work gives the two equations

σ ◦ τ ◦ σ

−1

= τ ◦ σ and σ

2

◦ τ ◦ (σ

2

)

−1

= τ ◦ σ

2

,soτ , τ ◦ σ , and

τ ◦ σ

2

are in the same conjugacy class. And then more

trial-and-error shows that there is no element g in

3

such

that g ◦ σ ◦ g

−1

= τ, and similarly there is no way to conjugate

any element in the second conjugacy class and get something

in the third one.

In order to deﬁne a representation r, we need to say what r(σ )

and r(τ ) are. This will totally determine r, because every element of

3

can be written in terms of σ and τ . Because r is a representation,

we must have r(τ ◦ σ ) = r(τ )r(σ ), and r(σ

2

) = r(σ)

2

, and so on.

One representation of

3

in GL(3, Z) is given by

r(σ ) =

⎡

⎢

⎣

010

001

100

⎤

⎥

⎦

, r(τ ) =

⎡

⎢

⎣

010

100

001

⎤

⎥

⎦

.

But just choosing matrices for r(σ ) and r(τ ) does not guarantee

that r really extends to a representation. Remember that r(e) must

be the identity matrix, I, and this means that there are a few things

to check before we know that this is indeed a representation:

EXERCISE: Check that r(σ )

3

= I and r(τ )

2

= I.

In fact, there is yet another relationship that these matrices

must satisfy. Because σ ◦ τ = τ ◦ σ

2

,ifr is going to be a representa-

tion, we must have r(σ )r(τ ) = r(τ )r(σ )

2

.

EXERCISE: Check that the matrix equation r(σ )r(τ ) =

r(τ )r(σ )

2

is true.

It can be proven that this is enough checking for this group

3

and in fact r does extend to a bona ﬁde representation from

3

to

GL(3, Z).

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