Homomorphism from $S_n$ to an abelian group Any homomorphism from $S_n$ to an abelian group $G$ is given by   $\;f(\sigma) = e$, if $\sigma$ is an even permutation, and $f(\sigma)= a$, where order of $a =2$, if $\sigma$ is an odd permutation.
I can prove that this is a homomorphism, but what guarantees that there is no other homomorphism other than this, and why are all even permutations mapped onto the identity of $G$? 
I only know that under a homomorphism, identity of $G_1$ goes to identity of $G_2$ and kernels are normal subgroups ($A_n$ are normal), but why there are no possibilities in which kernel is not $A_n$?? I am confused.

Hint If $f : S_n \to G$ is a group homomorphism and $f((i,j))=a$ than $a^2=e$ in $G$.
Therefore, $f$ takes each transposition in some element of order 2.
Next, if $G$ is abelian, use the fact that 
$$(i,j)(1,i)(i,j)=(1,j)$$
And
$$(1,j)(1,i)(1,j)=(i,j)$$
to deduce that 
$$f((1,i))=f((i,j))=f((i,j)) \forall (i,j)$$
Finally, write each permutation as a product of transpositions.
