Homomorphism Between Non-Abelian Group and Abelian Group

I know that when finding homomorphisms between groups, for a cyclic group to any other group, then the homomorphism is completely determined by where you send the generator. However, I have two questions regarding homomorphisms between non-abelian groups and abelian groups.

For instance, I know that for a homomorphism between $S_3$ and $C_4$ (cyclic group of order 4), you map the commutator subgroup, $A_3$ to the unit element of $C_4$. However, what do you do with the even permutations, and why must you map the commutator subgroup to the unit element of $C_4$?

Also, how should I approach finding the homomorphisms between $C_2 \times C_2$ (direct product of two cyclic groups of order 2) and $S_3$?

• Regarding the first question: If $\phi$ is a homomorphism from $S_3$ to $C_4$, then its kernel and image must satisfy $|S_3| = |\text{ker}(\phi)||\text{im}(\phi)|$, i.e. $6 = |\text{ker}(\phi)||\text{im}(\phi)|$. Since $|\text{im}(\phi)|$ must divide $|C_4| = 4$ and (as the previous equation shows) it must also divide $6$, the only two possibilities are $|\text{im}(\phi)| = 1$ or $2$. In the former case we have the trivial homomorphism; in the latter case, we must have $|\text{ker}(\phi)| = 3$. Since $A_3$ is the unique subgroup of $S_3$ with this order, the result follows. – Bungo May 5 '17 at 3:53
• So, $\phi$ must map $A_3$ to the unit element of $C_4$. Everything else is therefore mapped to the unique generator of the unique subgroup of order $2$ contained in $C_4$. – Bungo May 5 '17 at 3:58

$C_2$ is isomorphic to a subgroup of $S_3$ so that should help with the second question. As to the first, the commutator is generated by products of the form $xyx^{-1}y^{-1}$ so any homomorphism into an abelian group would allow these elements to commute, and this cancel. That is $$\phi(xyx^{-1}y^{-1})=\phi(x)\phi(y)\phi(x^{-1})\phi(y^{-1})=\phi(x)\phi(x^{-1})\phi(y)\phi(y^{-1})=1$$ so the generators of the commutator are in the kernel, so the commutator subgroup is too.