Find the number of homomorphisms from $A_4\to\mathbb Z_3$ and $A_4\to S_3$?

Question

$\alpha :G\to G'$ is a homomorphism. Then $G/\ker(\alpha)\cong\alpha(G)$ according to the first isomorphism theorem. This implies

$\vert G\vert = \vert \ker(\alpha )\vert \vert \alpha ( G)\vert$.

We also know $\alpha(G )\le G'$.

So we need to identify number of possibilities of subgroups $H$ for different cases of $\ker(\alpha)$.

Also $\ker(\alpha) \unlhd G$ . I have borrowed the normal subgroups of $A_4$ from an assignment solution here.

Case (i). $\ker(\alpha)=A_4$

$\vert\alpha ( G)\vert = \frac{12}{12}=1$. One subgroup is possible.

Case (ii). $\ker(\alpha)=\{(1)\}$

$\vert\alpha ( G)\vert = \frac{12}{1}=12$. No subgroups are possible.

Case (iii). $\ker(\alpha)=\{(1),(14)(23),(13)(24),(12)(34)\}$

$\vert\alpha(G)\vert = \frac{12}{4}=3$. One subgroup is possible.

So the total number of homomorphisms should be $1+1=2$. But the answer is $3$. What is the error in my solution?

What does it mean when we say "there are $n$ homomorphisms from $G\to G'$"? Does it mean

a) the number of different subgroups in $G'$ that are homomorphic to $G$.

b) the number of elements that are involved in homomorphisms from $G\to G'$.

c) the number of different ways $\alpha(G)$ can be defined to generate homomorphisms from $G\to G'$.

Follow up after the hint by kabenyuk:

The multiplication table for the group $A_4$ is as follows.

$A_4$" />

The subgroups are as follows:

$H_0=\{(1)\}$,

$H_1$={(1),(123)(132)},

$H_2$={(1),(234),(243)},

$H_3$={(1),(134),(143)},

$H_4$={(1),(124),(142)},

$H_5$={(1),(12)(34)},

$H_6$={(1),(13)(24)},

$H_7$={(1),(14)(23)},

$H_8$={(1),(12)(34),(13)(24),(14)(23)},

I could find the following homomorphisms from subgroups ($H_1,H_2,H_3,H_4$) of $A_4$ to $\mathbb Z_3$ by constructing their multiplication tables.

$A_4$ to $\mathbb Z_3$" />

So, the two homomorphisms are

1. $\alpha(G)=e$
2. $\alpha(H)=f(\mathbb Z_3)$ for $H=\{H_1,H_2,H_3,H_4\}$.

Is my way of defining the homomorphisms correct? If it is correct what is third homomorphism that we could define?

Answer 1

Case (iii). $\ker(\alpha)=\{(1),(14)(23),(13)(24),(12)(34)\}$

If $\sigma\in A_4$ is a $3$-cycle, then $A_4=K\cup\sigma K\cup\sigma^2 K$ (disjoint union), where $K=\ker\alpha$. This shows that knowing $\alpha(\sigma)$ is enough to determine $\alpha$ completely. But $\alpha(\sigma)^3=e$, and thus $\alpha(\sigma)$ can only be a $3$-cycle of $S_3$. There are only two possibilities: $\alpha(\sigma)=(1\ 2\ 3)$ or $\alpha(\sigma)=(1\ 3\ 2)$.

Edit. Meanwhile the OP has edited the question and it seems that they wanted to find out the homomorphisms from $A_4$ to $\mathbb Z_3$. The answer is actually the same if one replaces "$\alpha(\sigma)$ can only be a $3$-cycle" by "$\alpha(\sigma)$ can be $\hat 1$ or $\hat 2$".