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

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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.

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I have an answer, but you aren't going to like it.

There are 52 groups of order 48, up to isomorphism (I'll number them by the second part of their GAP ID). First, just typing your two groups into GAP, it says that they're isomorphic, so we know what we're trying to prove.

Note that our highest order element has order 6, so we can rule out 1, 2, 4, 5, 6, 7, 8, 9, 10, 13, 14, 18, 20, 23, 24, 25, 26, 27, 29, 36, 37, and 44, as all of these have elements of order at least 8. That leaves us with 3, 11, 12, 15, 16, 17, 19, 21, 22, 28, 30, 31, 32, 33, 34, 35, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, and 52 to worry about.

Neither of our groups is nilpotent, which rules out 21, 22, 45, 46, 47 and 52, leaving us with 3, 11, 12, 15, 16, 17, 19, 28, 30, 31, 32, 33, 34, 35, 38, 39 40, 41, 42, 43, 48, 49, 50, and 51.

Both of our groups have 10 conjugacy classes of elements, which rules out everything except for 30, and 48. Of those, 48 is exactly $S_4 \times C_2$, so we just need to show that $(C_2 \times C_2 \times C_2)\rtimes S_3$ is not the same as group 30, which is $A_4 \rtimes C_4$. But the Frattini Subgroup of $A_4 \rtimes C_4$ is $C_2$, while the Frattini subgroup of $(C_2 \times C_2 \times C_2)\rtimes S_3$ is trivial, so indeed these are not isomorphic, hence the result.

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