Problem
Find all the semidirect products $G=\mathbb Z_3 \rtimes_{\phi} \mathbb Z_4$. Show that one of these is non abelian and is not isomorphic to $\mathbb A_4$
I am a bit lost with this problem. First I've tried to find all the possible morphisms $$\phi:\mathbb Z_4 \to Aut(\mathbb Z_3)$$
The only two possible automorphisms in $\mathbb Z_3$ are $Id$ and $-Id$. $\phi$ is determine by the value it takes at $1$, so I have $\phi_1(1)=Id$ and $\phi_2(1)=-Id$.
For the case $\phi_1=Id$, we can think the external semidirect product as a cartesian product $\mathbb Z_4 \oplus \mathbb Z_3$ with $(z_1,z_2)+(w_1,w_2)=(z_1+\phi_1(z_2)(w_1),z_2+w_2)=(z_1+w_1,z_2+w_2)$. I don't know to which group $\mathbb Z_4 \oplus \mathbb Z_3$ is isomorphic to (it is clear it has to be an abelian group of order $12$).
For the case $\phi_2=-Id$, I don't know what the external semidirect product looks like and to which "known" group is isomorphic to.
Any suggestions would be appreciated.
$\mathbf Z_4\oplus\mathbf Z_3$ can be given another name. For a hint, what's the order of $(1,1)$? You aren't supposed to show that the other semidirect product is isomorphic to some known group, just that it's not isomorphic to $A_4$. The easiest way to do this is to remember that elements of $A_4$ all have order $2$ and $3$ and find an element of the semidirect product of higher order.