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?
Best Answer
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 $\hat 4, \hat 8$.