Number of homomorphisms from $A_4$ to $\mathbb{Z}_{12}$

abstract-algebrafinite-groupsgroup-homomorphismgroup-theorypermutations

I'm trying to calculate a number of homomorphisms from alternating group $$A_4$$ to $$\mathbb{Z}_{12}$$.

We can generate entire $$A_4 = \langle a , b \rangle$$ with two elements: $$a =(12)(34), b = (123)$$ where $$a^2=(), b^3=()$$ and $$(ab)^3=()$$. To my understanding it is enough to define homomorphism on the set of generators respecting these relations.

$$f:A_4 \rightarrow \mathbb{Z}_{12}$$ is homomorphism if for every $$x,y \in A_4$$: $$f(xy) = f(x)+f(y)$$. So because of $$(ab)^3=()$$ we have $$0=f((ab)^3) = 3f(a) + 3f(b) = f(a)$$. And since order of $$f(x)$$ has to divide order of $$x$$ the only possibilities are: $$f(a) = 0$$ and $$f(b) \in \{0, 4, 8\}$$.

So including the trivial one it seems that there are only 3 such homomorphisms. Is my reasoning correct?

In this answer I have explained how to use $$\ker f$$ and the normal subgroups of $$A_4$$ in order to find all possible morphisms.
Clearly $$\ker f$$ can't be trivial, since then $$f$$ is an isomorphism which is false.
If $$\ker f=A_4$$ you get the trivial morphism.
The third case is $$\ker f=\{(1),(14)(23),(13)(24),(12)(34)\}$$.
If $$\sigma\in A_4$$ is a $$3$$-cycle, say $$\sigma=(1\ 2 \ 3)$$, then $$A_4=K\cup\sigma K\cup\sigma^2 K$$ (disjoint union), where $$K=\ker f$$. This shows that knowing $$f(\sigma)$$ is enough to determine $$f$$ completely. But $$3f(\sigma)=\hat 0$$, and thus $$f(\sigma)$$ can only be $$\hat 4, \hat 8$$.