Determine how many group homomorphisms from $\Bbb{Z}_3$ (with addition as the group operation) to $S_4$ are there

Question

First of all, is $\Bbb{Z}_3$ with addition a set that contains ${0,1,2}$ with $\bmod(3)$? Then, $h(1)=$ identity element is a possible homomorphism. If $h(1)$ is not the identity element, are we trying to find how many element in S4 has a order of $3$? If so, given $S_4$ is isomorphic to the symmetry group of a cube, we have the elements included rotation $270^\circ$ are of order $3$. If that's the case the answer equal to $13$ since there are 13 rotational symmetry axis in a cube?

Answer 1

As the order of $\mathbb Z_3$ is equal to $3$, if $h$ is a non trivial requested homomorphism, $h(1)$ must be an element of order $3$, i.e. a $3$-cycle.

A $3$-cycle of $S_4$ fixes one element. And that element being chosen, the are $2$ $3$-cycles. Therefore in total $8$ $3$-cycles. Adding the trivial homomorphism gives in total $9$ homomorphisms.

Answer 2

Let $G$ and $H$ be groups, and $h\colon G\to H$ a homomorphism. Then, if $g\in G$ has finite order $o(g)$, we get:

\begin{alignat}{1} e_H &= h(e_G) \\ &=h(g^{o(g)}) \\ &=(h(g))^{o(g)} \\ \end{alignat}

whence $o(h(g))\mid o(g)$. In our case, $G=\Bbb Z_3=\{0,1,2\}$ and $o(1)=o(2)=3$. Therefore, there are at most as many nontrivial such homomorphisms as the number of $h$'s which send $0$ to $()$, $1$ to some $3$-cycle and $2$ to some $3$-cycle either. Now, $h(1)$ and $h(2)$ can't be the same $3$-cycle, say $\sigma$, because $h(\Bbb Z_3)=\{(),\sigma\}$ is not a subgroup of $S_4$. Hence, there are actually at most as many nontrivial such homomorphisms as the number of $h$'s which send $0$ to $()$, $1$ to some $3$-cycle, say $\sigma$, and $2$ to a different $3$-cycle, say $\tau$ ($\ne\sigma$). Finally, $\{(),\sigma,\tau\}\le S_4\iff \sigma\tau=()$ (closure requirement), which cuts the number of combinations down to $8$, namely:

\begin{alignat}{1} &h(\Bbb Z_3)=\{\space(),\space h(1)=(1)(234),\space h(2)=(1)(243)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(1)(243),\space h(2)=(1)(234)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(2)(134),\space h(2)=(2)(143)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(2)(143),\space h(2)=(2)(134)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(3)(124),\space h(2)=(3)(142)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(3)(142),\space h(2)=(3)(124)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(4)(123),\space h(2)=(4)(132)\} \\ &h(\Bbb Z_3)=\{\space(),\space h(1)=(4)(132),\space h(2)=(4)(123)\} \\ \end{alignat}

So, with the trivial one ($h(\Bbb Z_3)=\{()\}$), there are $9$ homomorphisms from $\Bbb Z_3$ to $S_4$ altogether.