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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 ?

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  • $\begingroup$ Any two of your three homomorphisms $\Bbb Z_2\times\Bbb Z_2\to{\rm Aut}(\Bbb Z_3)$ are related by an automorphism of $\Bbb Z_2\times\Bbb Z_2$, which induces an isomorphism between the two versions of $\Bbb Z_3\rtimes(\Bbb Z_2\times\Bbb Z_2)$. Recall our discussion here. $\endgroup$ – arctic tern Nov 13 '16 at 20:08
  • $\begingroup$ It's not true. The direct product $Z_3 \times Z_2 \times Z_2$ is abelian and is not isomorphic to $D_{12}$, $\endgroup$ – Derek Holt Nov 13 '16 at 20:19
  • $\begingroup$ @DerekHolt I think Adeek is talking about ones that are "achieved in this way" via one of the "three choices of nontrivial homomorphism[s]." $\endgroup$ – arctic tern Nov 13 '16 at 23:28
  • $\begingroup$ @arctictern But the first sentence in the post is incorrect. $\endgroup$ – Derek Holt Nov 14 '16 at 9:05
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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 $G=<x,h|x^2=h^6=1 \ \ and \ \ h^x=h^{-1} >\cong D_{12}$.

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