Find the number of group homomorphisms from $\mathbb{Z}_8$ to $S_3$. 
Find the number of group homomorphisms from $\mathbb{Z}_8$ to $S_3$.

My try:
I know $\mathbb{Z}_8$ is a cyclic group of order $8$, and that the number of group homomorphisms from a cyclic group to any other group is determined where the generators of the cyclic group get mapped.
It is easy to notice that the generators of $\mathbb{Z}_8$ are
$\langle 1\rangle,\langle 3\rangle,\langle 5\rangle $ and $\langle 7\rangle $.
I also know that the order of the image divides the order of $S_3$ (as it forms a subgroup of $S_3$).
Also if $f$ be the homomorphism, then $f(0)= (1)(2)(3)$.
So till now I got $f(0)=(1)(2)(3)$,
$|f(2)| \mid 6$, $|f(3)| \mid 6$, $|f(4)| \mid 6$, $|f(5)| \mid 6$, $|f(6)| \mid 6$, $|f(7)| \mid 6$, $|f(8)| \mid 6$. So the orders of $|f(i)|$ can be $1$ or $2$ or $3$ or $6$.
But how can I proceed from here?
 A: Let $\varphi\colon\Bbb Z_8\longrightarrow S_3$ be a homomorphism. Since $1$ has order $8$, the order of $\varphi(1)$ must divide $8$. That is, it must be $1$, $2$, $4$ or $8$. But $S_3$ has no element whose order is $4$ or $8$, and the only $\varphi$ such that $\varphi(1)=\operatorname{Id}$ is the trivial homomorphism. Finally, $S_3$ has exactly three elements whose order is $2$ (the three transpositions), and if $\tau$ is any of them you can define $\varphi(n)=\tau^n$; it is a group homomorphism and it maps $1$ into $\tau$.
Therefore, there are $4$ such homomorphisms.
A: Suppose that $\phi:\mathbb{Z}_8\to S_3$ is a homomorphism then $\phi(\mathbb{Z}_8)\leq S_3$ using properties of homomorphism $\phi(\mathbb{Z}_8)$ is cyclic group
$\phi(Z_8)\approx\mathbb{Z}_3 $ or $ \mathbb{Z}_2$ Since $|\phi(\mathbb{Z_8})| \;divides \;|\mathbb{Z}_8|=8$ $\Rightarrow \phi(\mathbb{Z}_8)\approx \mathbb{Z}_2$ there are 3 element of order 2 in $S_3$ so we have 3 homomorphism and one trivial homomorphism=4 homomorphism are there
A: The subgroups of $\Bbb Z_8=\{0,1,2,3,4,5,6,7\}$ are:
\begin{alignat}{1}
&H_1:=\{0\} \\
&H_2:=\{0,4\} \\
&H_3:=\{0,2,4,6\} \\
&H_4:=\Bbb Z_8 \\
\end{alignat}
and they are all normal, as $\Bbb Z_8$ is cyclic and hence Abelian. The subgroups $H_1$ and $H_2$ can't be kernels of homomorphisms $\Bbb Z_8\to S_3$, because the respective images would have order $8(>6)$ and $4(\nmid 6)$. The subgroup $H_3$ is the kernel of three distinct homomorphisms: they map the elements of $H_3$ to $()$ (by definition of kernel), and the other elements (all of order $8$) to one same $2$-cycle: in fact, they can't be mapped to $3$-cycles (because $3\nmid 8$), nor to different $2$-cycles (because the image set wouldn't be a subgroup of $S_3$). So, for $H_3$ as kernel we have the three homomorphisms:
$$0\mapsto (), 1\mapsto(ij), 2\mapsto (), 3\mapsto(ij), 4\mapsto (), 5\mapsto(ij), 6\mapsto (), 7\mapsto(ij)$$
The case of kernel equal to $H_4$, i.e. to the whole $\Bbb Z_8$, corresponds to the trivial homomorphism, where all the elements are mapped to $()$. So, there are overall four homomorphisms $\Bbb Z_8\to S_3$.
