# What group is $S_3\times \mathbb Z_2$ isomorphic to?

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 $$12$$?

• no, it's $D_{12}$ – graeme Jul 3 at 3:00
• there is a 5th group of order 12 though, it is this one – graeme Jul 3 at 3:02
• – J. W. Tanner Jul 3 at 3:14

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.
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 $$D_{12}$$.