Find all group homomorphism from $(\Bbb Z_4, +_4)$ to $(S_3,\circ)$.
Attempt: Let $e_G$ denote the identity element in a group $G$. The map $f(a)=e_{S_3}$ is the trivial group homomorphism.
Now, notice that $\Bbb Z_4$ is cyclic, i.e., $\Bbb Z_4 = \langle 1,3 \rangle$. Let $|a|$ be the order of an element $a$ of a group $G$. Then $|1|=|3|=4$. Next, for any $a \in G$ with $|a|=n$, we must have $|f(a)| \mid n$. Hence, $|f(a)| \mid 4$ for all $a \in \Bbb Z_4$. Now, the non-trivial elements in $S_3$, which the order divide $4$, are $(1 \ 2), (1 \ 3),$ and $(2 \ 3)$. Hence, we found three mappings, namely, $f(a)=(1 \ 2)^a, f(a)=(1 \ 3)^a$, and $f(a)=(2 \ 3)^a$, for any $a \in \Bbb Z_4$, where $(x \ y)^k$ denotes the composition of $(x \ y)$ with itself $k$ times. It's not difficult to show that indeed these mappings are a group homomorphism.
Therefore, in total, there are $4$ group homomorphism from $\Bbb Z_4$ to $S_3$, where $1$ is the trivial group homomorphism, and the other three are non-trivial homomorphism.
Is it work? Thanks in advanced.
Once you solve such a problem, it's worth trying to extend it to a more general task: for an integer $n \geq 2$ and a group $G$, count the number of homomorphisms $\mathbf Z/n\mathbf Z \to G$. Show this number equals the number of $g \in G$ such that $g^n = 1$ (i.e., the order of $g$ divides $n$).
When $n = 4$ and $G = S_3$, you want to count solutions to $g^4 = 1$ in $S_3$, which amounts to counting elements of order $1$ or $2$ (other elements have order $3$), and there are $4$ of those elements in $S_3$.