Homomorphism from $S_n$ to an abelian group

Question

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.

Answer 1

If such a homomorphism $S_n \to G$ exists, then it induces a surjective homomorphism $S_n \to \langle a \rangle \cong C_2$. Therefore, the kernel has index $2$ and so is $A_n$ because it contains $A_n$.

Answer 2

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.

Answer 3

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 $V_4$.