Counting all homomorphism from $S_4$ to $\mathbb{Z_6}$

Question

I have to count number of homomorphism from $S_4$ to $\mathbb{Z_6}$.

One approach that I know is by finding the possible kernel of homomorphism from $S_4$ to $\mathbb{Z_6}$.

I am using another approach by finding the generators of $S_4$. We know that $S_4$ can be generated by a $2$-cycle, say $\sigma$ and a $3$-cycle, say $\tau$: Thus any homomorphism $f\colon S_4\to \mathbb{Z_6}$ can be determined by finding $f(\tau)$ and $f(\sigma)$ completely. It is clear that $f(\sigma)^2=1$ since $\sigma^2=1$ and $f(\tau)^3=1$ since $\tau^3=1$. Now we have to search for possible number of elements in $Z_6$ whose order divide $2$ and $3$ respectively.

My confusion: We know that $S_4$ can also be generated by a $2$-cycle and a $4$-cycle. Perhaps, above written procedure may give wrong results in such case since order of four cycle is $4$. I am not able to understand where I am going wrong.

I would be very much grateful if anybody could clear my doubt. Thank you very much for your time.

Answer 1

This is very similar to another recent question that asks to find all homomorphisms from $S_4$ to $\Bbb Z_2$. Certainly one of the simpler ways to answer your question is to determine the possible images of a transposition $\tau$. Once you know $\phi(\tau)$ for some transposition $\tau$, you have completely described the homomorphism $\phi$ since the transpositions generate $S_n$ and are all conjugate (so that their image is the same in the abelian target group). The only constraint on $\phi(\tau)$ is that be of order $2$. Hence,

If $G$ is an abelian group, there is one non-trivial morphism $S_n\to G$ for every element $g\in G$ or order $2$ (and the trivial morphism of course).

There is only one element of order $2$ in $\Bbb Z_6$, the element $3$, so that there is precisely one non trivial morphism $S_4\to\Bbb Z_6$ (plus the trivial morphism of course).