Classifying groups such that $G \simeq \mathbb{Z}_3 \rtimes (\mathbb{Z}_2 \times \mathbb{Z}_2)$

Question

I am thinking G will be $D_{12}$; however, I am not sure how to prove that all semi-direct product are isomorphic or explicitly get $D_{12}$. We have $\mathbb{Z}_2\times \mathbb{Z}_2$ has two generators $(1,0)$ and $(0,1)$ so we have three choices of nontrivial homomorphism $\psi : \mathbb{Z}_2 \times \mathbb{Z}_2 \rightarrow Aut(\mathbb{Z}_3) = \mathbb{Z}_2 = \{1,-1\}$ is just mapping $(1,0),(0,1)$ to $\pm 1$ and both don't go to 1. Why is all semi-direct producted achieved in this way isomorphic though ?

Answer 1

This is true when $\psi$ is nontrivial. Notice that $Ker \psi$ has order $2$. Hence, $\mathbb Z_3 Ker\psi$ isomorphic to $\mathbb Z_6$.(it acts trivially on $\mathbb Z_3$.) Set $H=\mathbb Z_3 Ker\psi$

Now let $x\in G-H$. Then $x=ab$ where $a \in \mathbb Z_3$ and $b\in \mathbb Z_2\times \mathbb Z_2-Ker\psi$.

Notice that order of $b$ is two and $b^{-1}ab=a^{-1}$. Thus,

$$x^2=abab=ab^{-1}ab=aa^{-1}=1$$.

order of $x$ is also $2$.

As, index of $H$ is $2$, $H$ is normal in $G$ and $G=H<x>$

Thus, $x$ acts on $H$ by conjugation. If $x$ acts trivially on $H$ then $G$ is abelian. But this is not the case as $\psi$ is nontrivial.

$Aut(H)\cong Z_2$. We must have $h^x=h^{-1}$ for $H=<h>$.

Then $G=<x,h|x^2=h^6=1 \ \ and \ \ h^x=h^{-1} >\cong D_{12}$.