Let $S_n$ denote the symmetric group on $n$ letters and $C_n$ denote the cyclic group of order $n$. Consider $(C_2 \times C_2 \times C_2) \rtimes S_3$ where $S_3$ acts on $(g_1, g_2, g_3) \in C_2 \times C_2 \times C_2$ as follows: Given $\sigma \in S_3$, $\sigma \cdot (g_1, g_2, g_3) = (g_{\sigma^{-1}(1)}, g_{\sigma^{-1}(2)}, g_{\sigma^{-1}(3)})$.
My question: Is $(C_2 \times C_2 \times C_2) \rtimes S_3 \cong S_4 \times C_2$.
Progress: Clearly they have the same order. I can show that they indeed have the same center. I have computed the number of elements of each order as follows:
\begin{array}{c | c | c}
\text{ order } & \text{ # of elements }\\
1 & 1 \\
2 & 19 \\
3 & 8 \\
4 & 12 \\
6 & 8
\end{array}
Both groups of the same number of elements of each order. I've also determined that proving this isomorphism is equivalent to $(C_2 \times C_2 \times C_2) \rtimes S_3$ having a subgroup isomorphic to $S_4$. The logic goes as follows:
Clearly if the two groups are isomorphic, then $(C_2 \times C_2 \times C_2) \rtimes S_3$ has a subgroup isomorphic to $S_4$. If $(C_2 \times C_2 \times C_2) \rtimes S_3$ has a subgroup isomorphic to $S_4$, then this subgroup must intersect $Z(G)$ trivially, as $Z(S_4)$ is trivial. Further, $(C_2 \times C_2 \times C_2) \rtimes S_3 = S_4Z(G)$. Then since the center is normal, $(C_2 \times C_2 \times C_2) \rtimes S_3 \cong S_4 \rtimes Z(G) \cong S_4 \rtimes C_2$ where $Z(G)$ acts by conjugation on $S_4$. Since $Z(G)$ is the center, we just have $(C_2 \times C_2 \times C_2) \rtimes S_3 \cong S_4 \times C_2$.
I'm fairly stuck at this point. I want to maybe try an find some elements that satisfy the Coexeter relations sitting in $(C_2 \times C_2 \times C_2) \rtimes S_3$.
Best Answer
Adding a non-GAP argument.
A key ingredient is the fact that $S_4=V_4\rtimes S_3$, where $V_4\unlhd S_4$ is the copy of Klein-4 whose non-trivial elements are the three products of two disjoint 2-cycles. Furthermore, conjugation by elements of $S_3$ gives all the six permutations of those three type $(2,2)$ permutations.
Let use view $N=C_2\times C_2\times C_2$ additively as a 3-dimensional vector space over $\Bbb{F}_2$: $$ N=\{(x_1,x_2,x_3)\mid x_1,x_2,x_3\in\Bbb{F}_2\}. $$ We can write $N$ as a direct sum $N=Z\oplus V$ where $$Z=\langle(1,1,1)\rangle\qquad \text{and}\qquad V=\{(x_1,x_2,x_3)\in N\mid x_1+x_2+x_3=0\}.$$ Both $Z$ and $V$ are stable under the action of $S_3$. Also, $S_3$ acts trivially on $Z$, and permutes the elements of $V$ the same way it permutes that copy of $V_4\le S_4$ - all the six ways of permuting the vectors $110,101,011$ are gotten by permuting the coordinates.
Putting all this together gives $$ N\rtimes S_3\simeq Z\times (V\rtimes S_3)\simeq C_2\times S_4. $$ The subgroup $Z$ is the center of this group.