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Fearless Symmetry by Robert Gross, Avner Ash

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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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