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.

The group homomorphism provides an isomorphism $S_n/\text{ker}(f)\cong \text{im}(f)$ whereby $\text{ker}(f)$ is a normal subgroup. In this case you have  $\ker(f)\in\{1,A_n,S_n\}$. G abelian implies $\ker(f)\neq 1$ and therefore $S_n/\text{ker}(f)\cong 1$ or $S_n/\text{ker}(f)\cong\mathbb{Z}/ 2\mathbb{Z}$
For n=4 there is the addition normal subgroup $V_4$.
