Determine all possible $\phi$ Let $\phi: S_6 \to \mathbb{Z}/6\mathbb{Z}$ be a homomorphism. Explain why $[S_6,S_6]$, the commutator subgroup of $S_6$, is a subset of ker($\phi$) and after that determine all possible $\phi$. 

For the first part I picked a random element in the commutator subgroup $a^{-1}b^{-1}ab$. From the definition of homomorphisms it follows that:
$\phi(a^{-1}b^{-1}ab)=\phi(a^{-1})\phi(b^{-1})\phi(a)\phi(b)$
Since $\mathbb{Z}/6\mathbb{Z}$ is abelian we can rearrange the above as follows:
$\phi(a^{-1})\phi(a)\phi(b)\phi(b^{-1})=\phi(a^{-1}abb^{-1})=\phi(1_G)=1_H$
That proves the inclusion.
When it comes to the second part I get totally stuck. I'm not even sure I fully understand the question. How do I do this?
 A: 1) Because $$S_6/\ker(\varphi)\cong \text{Rang}(\varphi)\leq \mathbb Z/6\mathbb Z,$$
and thus $S_6/\ker(\varphi)$ is abelian. Therefore $[S_6,S_6]\leq \ker(\varphi)$.
Notice that $[G,G]$ is the smallest group s.t. $G/H$ is abelian. Therefore, $G/H$ is abelian if and only if $[G,G]$ is a subgroup of $H$. 
2) By the fondamental theorem of quotient group, $$\hat \varphi:S_6/\ker(\varphi)\longrightarrow \mathbb Z/6\mathbb Z$$ 
is well defined and the unique homomorphism s.t. $$\hat \varphi(x\ker\varphi)=\varphi(x).$$
Since $S_6/\ker(\varphi)=\left<(123456)\ker\varphi\right>$ (why ?), you'll get that $\hat\varphi$ is only depending on $\hat\varphi((123456)\ker\varphi)$. Now, you finally get the result (and thus that $\varphi$ only depending on $\varphi((123456))$). At the end, you should find 2 homomorphism that are $$\varphi_1:(123456)\longmapsto 1$$
and
$$\varphi_2:(123456)\longmapsto 5.$$
Moreover, you have the trivial homomorphism (i.e. that send everything to $0$), and thus, you have 3 homomorphisms. 
