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The group $S_3\times \mathbb Z_2$ has order $12$. I know four groups of order $12$: $$\mathbb Z_{12},\mathbb Z_{2}\times \mathbb Z_{6},A_4,D_{12}.$$

But it seems that none of them is isomorphic to $S_3\times \mathbb Z_2$. So what group is $S_3\times \mathbb Z_2$ isomorphic to? Is there a fifth group of order $12$?

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    $\begingroup$ no, it's $D_{12}$ $\endgroup$ – graeme Jul 3 at 3:00
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    $\begingroup$ there is a 5th group of order 12 though, it is this one $\endgroup$ – graeme Jul 3 at 3:02
  • $\begingroup$ Cf. this question $\endgroup$ – J. W. Tanner Jul 3 at 3:14
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It is $D_{12}$. Define a homomorphism $\phi$ such that $\left(\begin{pmatrix}1&2&3\end{pmatrix},1\right)\mapsto r_6$ $\left(\text{where }r_6=\begin{pmatrix}1&2&3&4&5&6\end{pmatrix}\right)$, and let $\left(\begin{pmatrix}2&3\end{pmatrix},0\right)\mapsto s$ $\left(\text{where } s=\begin{pmatrix}1&6\end{pmatrix}\begin{pmatrix}2&5\end{pmatrix}\begin{pmatrix}3&4\end{pmatrix}\right)$. This should show that the groups are indeed isomorphic.

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It's not abelian, so it's $A_4$ or $D_{12}$.

But $A_4$ has no subgroup of order six. And $\rho=((123),1)\in S_3×\Bbb Z_2$ has order six.

That leaves $D_{12}$.

$\sigma =((12),0)\in S_3×\Bbb Z_2$ has order two. Now we check that $(\rho\sigma)^2=((13),0))^2=(e,0)=e\in S_3×\Bbb Z_2 $.

Those are the relations for $D_{12}$.

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