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 $(C_2 \times C_2 \times C_2) \rtimes S_3$.