If is a homomorphism then is completely determined by its value on a generating set.

We will use the generating set .

Let and .

Then .

Thus, it isnecessarythat divides .

Since it means .

Note that thus .

Since it means (by above) .

Thus, .

If then .

This is the trivial homomorphism i.e. everything is mapped to .

The other possibility is and then .

We will prove this defines a homomorphism.

Begin with .

Conjugating* by powers of gives and .

Therefore, .

Note that .

It follows that .

This means if iseventhen it is a product of even number of transpositions and so since even number of times is . If isoddthen is a product of odd number of transpositions and so since odd number of times is . Therefore, is simply the mapping which sends even permutations into and odd permutations into . It is trivial to proof now that it is indeed a homomorphism.

Thus, .

And like you guessed it does indeed have only two elements.

*)Definition, conjugated by means the product .