Find all of the homomorphisms $ \varphi: S_3 \to \mathbb{Z}_{4} $. I'd like to find all of the homomorphisms $ \varphi: S_3 \to \mathbb{Z}_{4} $.
What I've tried so far:
I tried to do $ \varphi(Id) = \bar{0} = \bar{4} $ (as somebody used here). But then I realized that it was because the identity element coincide with the generator of domain group $ \mathbb{Z}_{15} $.
And, as far as I know, $ S_3 $ doesn't have a generator.
I really have no idea of how to do it.

** Corrected: $ S_3 $ has two generators: $ d_1 $ and $ d_2 $. Even though, I can't use the same trick that in the other post.
$$ S_3 = \left \{ Id=\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}, d_1=\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}, d_2=\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}, t_1=\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}, t_2=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}, t_3=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix} \right \} $$
 A: The group $S_3$ has only $\{\mathit{Id}\}$, $A_3$ (the alternating subgroup) and $S_3$ as normal subgroups. If $\ker\varphi=S_3$, then we have the trivial homomorphism. We can rule out $\ker\varphi=\{\mathit{Id}\}$, because $S_3$ is not abelian.
Let's assume $\ker\varphi=A_3$. Then $\varphi$ induces an injective homomorphism $\hat\varphi\colon S_3/A_3\to\mathbb{Z}_4$ and there's only one of them, because $\mathbb{Z}_4$ has just one (cyclic) subgroup of order two, namely $\{\bar{0},\bar{2}\}$.
Finish up proving that indeed there is only one homomorphism $\varphi$ such that $\ker\varphi=A_3$.
A: Hint: Every 3-cycle must map to some $a\in\mathbb Z_4$ with the property $a+a+a=0$, and there is only one possibility for that.
Since the product of two different transpositions is a 3-cycle, this leaves very few possibilities for what the image of each of the three transpositions can be.
A: Suppose you already know that:
$S_3 \cong \langle a,b: a^3 = b^2 = e;ba = a^2b\rangle$.
Then any homomorphism $\phi:S_3 \to \Bbb Z_4$ is completely determined by $\phi(a)$ and $\phi(b)$.
Now the order of $\phi(a)$ divides the order of $a$, which is $3$ (a prime). So $\phi(a)$ must have order $1$ or $3$. However, $\Bbb Z_4$ has no elements of order $3$ ($3$ does not divide $4$), so we know $\phi(a) = \overline{0}$.
Thus $\phi$ is completely determined by $\phi(b)$, which must have order $1$, or order $2$ (again, $2$ is a prime number, which is rather convenient for us).
If $\phi(b)$ has order $1$, $\phi$ must send everything to $\overline{0}$. This is the trivial (zero) homomorphism.
On the other hand, if $\phi(b)$ has order $2$, it must be that $\phi(b) = \overline{2}$, as this is the only element of $\Bbb Z_4$ of order 2. This allows us to write the sole non-trivial homomorphism explicitly as:
$\phi(e) = \overline{0}\\ \phi(a) = \overline{0}\\ \phi(a^2) = \overline{0}\\ \phi(b) = \overline{2}\\ \phi(ab) = \overline{2}\\ \phi(a^2b) = \overline{2}$
(If you have trouble following this, substitute $(1\ 2\ 3)$ for $a$, and $(1\ 2)$ for $b$).
