Find all group homomorphisms $f: \Bbb Z_6 \to S_3$.
I was asked this on a quiz and at the time I did not remember that the only normal subgroups of $S_3$ are $\{e\}, A_3$ and $S_3$ itself. This becomes quite straightforward from here as $\Bbb Z_6/\{e\} \cong f(\Bbb Z_6) = S_3$ but $\Bbb Z_6$ is abelian and $S_3$ is not so $\ker(f) =\{e\}$ cannot work.
The last two options work so one can conclude that there are two possible homomorpihsms.
I wanted to know if this can be done by noting that $(1)=f(0)=f(1+1+1+1+1+1)=6f(1)$ which implies that $f(1)$ needs to be an permutation $\tau \in S_3$ such that $6\tau = (1)$ i.e. multiplied six times by itself gives the identity. Is it possible to conclude this without computing $6\tau$ for every $\tau \in S_3$?
Are you looking for morphisms $f:\mathbb Z_n\to S_3$ for $n=3$ or $n=6$? Whatsoever, $f$ is determined by $f(1)$, which can be any element $\tau$ of $S_3$ such that $\tau^n=id$ (and knowing the normal subgroups of $S_3$ is useless).
If $n=6$, any element of $S_3$ will do (by Lagrange theorem).
If $n=3$, $f(1)$ is either $id$ or one of the two $3$-cycles. There are indeed two possibilities for $\ker(f)$, but three possibilities for $f$.