We will use the generating set .
Let and .
Thus, it is necessary that divides .
Since it means .
Note that thus .
Since it means (by above) .
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 .
Note that .
It follows that .
This means if is even then it is a product of even number of transpositions and so since even number of times is . If is odd then 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.
And like you guessed it does indeed have only two elements.
*)Definition, conjugated by means the product .