Determine the number of homomorphisms from $S_{3} \rightarrow \Bbb Z_{2} \times \Bbb Z_{4}$.

Question

Determine the number of homomorphism from $S_{3} \rightarrow \Bbb Z_{2} \times \Bbb Z_{4}$.

My attempt: A homomorphism from $S_{3} \rightarrow \Bbb Z_{2} \times \Bbb Z_{4}$ is a homomorphism into an abelian group. Therefore,

${\rm hom}(S_{3},\Bbb Z_{2} \times \Bbb Z_{4})= {\rm hom}\left(\frac{S_{3}}{[S_{3},S_{3}]},\Bbb Z_{2} \times \Bbb Z_{4}\right)$, where $[S_{3},S_{3}]$ is the normal subgroup of $S_{3}$ generated by the elements of the form $aba^{-1}b^{-1}$ and $[S_{3},S_{3}]=A_{3}$.

${\rm hom}(S_{3},\Bbb Z_{2} \times \Bbb Z_{4})= {\rm hom}(\Bbb Z_{2},\Bbb Z_{2} \times \Bbb Z_{4})$.

Next my idea is to calculate the number of elements in $\Bbb Z_{2} \times \Bbb Z_{4}$ whose order is divisible by 2. I get 4 elements of order 4, 3 elements of order 2, and one element of order 1.

Anyone can please suggest to me, is this direction correct to think this question?

Answer 1

Alternatively, note that all homomorphisms $S_{3} \to \mathbb Z_{2} \times \mathbb Z_{4}$ are of the form $f \times g$, for homomorphisms $f : S_{3} \to \mathbb Z_{2}$ and $g : S_{3} \to \mathbb Z_{4}$. Since both $\mathbb Z_{2}$ and $\mathbb Z_{4}$ embed into $\mathbb C^\times$, you can start by considering all homomorphisms $S_{3} \to \mathbb C^\times$. For that, see this answer.

Answer 2

What you have done so far is correct. Let me continue. We have\begin{align*} \text{Hom}(S_3,\mathbf{Z}/2\mathbf{Z}\times \mathbf{Z}/4\mathbf{Z}) &\cong \text{Hom}(\mathbf{Z}/2\mathbf{Z},\mathbf{Z}/2\mathbf{Z}\times \mathbf{Z}/4\mathbf{Z}) \\ & \cong\text{Hom}(\mathbf{Z}/2\mathbf{Z},\mathbf{Z}/2\mathbf{Z})\times \text{Hom}(\mathbf{Z}/2\mathbf{Z}, \mathbf{Z}/4\mathbf{Z}) \end{align*} where the last line is the universal property of the product. Now use that $$\text{Hom}(\mathbf{Z}/2\mathbf{Z},G)\cong \{x\in G\mid x^2=1 \}$$ (noting $G$ multiplicatively).

Answer 3

The possibilities for the kernel are $1$, $A_3$ and $S_3$. The first case is not allowed, as it would give rise to an embedding $S_3\hookrightarrow\Bbb Z_2\times\Bbb Z_4$, which is not possibile because $6\nmid 8$. The second case corresponds to $\operatorname{im}\phi\cong C_2$ (first homomorphism theorem), namely to $(12)$, $(13)$ and $(23)$ being all sent to one same element of order $2$. In the codomain there are three elements of order $2$, and any of them as image of the three $2$-cycles gives rise to a homomorphism. So there are three nontrivial homomorphisms, plus the trivial one (corresponding to the whole $S_3$ as kernel), so four overall.