# When are $((C_2 \times C_2) \rtimes C_3) \rtimes C_2$ and $((C_2 \times C_2) \rtimes C_2) \rtimes C_3$ isomorphic?

Let's consider $\mathfrak{G}:=((C_2 \times C_2) \rtimes_{\phi} C_3) \rtimes_{\nu} C_2$ (which I do believe is $\mathcal{S}_4$, please confirm or argue against) and $G:=((C_2 \times C_2) \rtimes_{\mu} C_2) \rtimes_{\xi} C_3$.

Under which hypothesis does there exist an isomorphism $\chi: \mathfrak{G} \rightarrow G$?

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It's hard to argue for or against when we don't know what $\phi$ and $\nu$ are. –  JSchlather Dec 23 '12 at 13:32
They may also do not exist. That's what I want to know. –  Ivan Dec 23 '12 at 14:30
The first group could be $S_4$ for an appropriate choice of $\phi$ and $\nu$. There would be an isomorphism if all of the maps $\phi$, $\nu$, $\mu$ and $\xi$ were trivial, in which case both groups would be isomorphic to $C_2 \times C_2 \times C_2 \times C_3$ (but that's rather obvious isn't it?). There are other possible choices which would make the groups isomorphic. –  Derek Holt Dec 23 '12 at 17:53

First, let's take a look at $\mathfrak{G}$ and determine under which $\phi$ and $\nu$ we get $\mathfrak{G}\cong S_4$.

The automorphism group of the Klein $V$ group is $S_3$, since $(1,0)$, $(0,1)$, and $(1,1)$ are interchangable. If we assume that $\phi$ is nontrivial, then $C_3$ embeds into $\text{Aut}(V)$ by acting as a $3$-cycle on these elements. The resulting group is $A_4$.

The automorphism group of $A_4$ has nine elements of order $2$. Three of them will yield $A_4\rtimes C_2\cong A_4 \times C_2$; these are the inner automorphisms induced from the $V$ group. Every other nontrivial automorphism of order $2$ will produce $A_4\rtimes C_2\cong S_4$.

So you see that if we assume that

1. $\phi$ is nontrivial, and
2. $\nu$ embeds $C_2$ into $\text{Out}((C_2\times C_2)\rtimes_\phi C_3 )$,

then $\mathfrak{G}=((C_2\times C_2) \rtimes_\phi C_3 )\rtimes C_2\cong S_4$.

Now let's look at $G$. Every automorphism of $V$ of order $2$ produces $D_4$ when we take $(C_2\times C_2)\rtimes C_2$, which is clear from the geometric interpretation of $D_4$ (consider elements of $C_2\times C_2$ as coordinates). However, $\text{Aut}(D_4)$ is isomorphic to $D_4$, and thus has no elements of order $3$. So the only possibility for $\xi$ is the trivial homomorphism. Thus $G$ can never be isomorphic to $S_4$.

So let's go back and think. How could we have chosen $\phi$ and $\nu$ so that $\mathfrak{G}\cong G$? Aside from choosing $\phi,\nu,$ and $\mu$ to be trivial, the only way for this to happen would be to choose $\phi$ to be trivial and $\nu$ so that $C_2$ centralizes $C_3$ and acts faithfully on $V$, or to choose $\mu,\nu$ trivial and $\phi,\xi$ nontrivial.

So $\mathfrak{G}\cong G$ if and only if $\mathfrak{G}\cong G \cong (C_2)^3\times C_3$, $\mathfrak{G}\cong G \cong D_4\times C_3$, or $\mathfrak{G}\cong G \cong A_4\times C_2$.

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Lovely, thanks, especially for last sentence. –  Ivan Dec 23 '12 at 21:05
It is also possible to have them both isomorphic to $A_4 \times C_2$ (i.e. $\mu$ and $\nu$ trivial and $\phi$, $\xi$ nontrivial). –  Derek Holt Dec 24 '12 at 15:21
Oh yeah, I forgot about that in the last line. Thanks Derek. –  Alexander Gruber Dec 24 '12 at 22:29