Which of the following group is isomorphic to the group $S_3\times \mathbb Z_2$?

Question

The group $S_3\times \mathbb Z_2$ is isomorphic to one of the following groups:

$\mathbb Z_{12}$, $\mathbb Z_6\times \mathbb Z_2$, $A_4$, $D_6$…?

I know that $\mathbb Z_{12}$, $\mathbb Z_6\times \mathbb Z_2$ will not be the answer. So, which one between $A_4$, $D_6$…?

Answer 1

$$G=S_3\times\mathbb{Z_2}$$ is not abelian and not cyclic hence it's not isomorphic to $\mathbb{Z_{12}}$ and $\mathbb{Z_6}\times\mathbb{Z_2}$. Also the group $A_4$ doesn't have an element of order $6$ but $G$ has, say $\{(1,2,3),1\}$, so correct option is $D_6$

Answer 2

$D_6$ contains normal subgroups $G, H$ such that $G \cong S_3$ and $H \cong \mathbb{Z}_2$. Since $G \cap H=\{1\}$ you can conclude $D_6 \cong G \times H$. I don't know if this helps, depends on your mathematical background.