Recall that the dihedral group $D_{6} \cong \mathbb{Z}_{6} \rtimes _{\phi} \mathbb{Z}_{2},$ where the reflection $\mu$ acts on the rotation $\rho$ by \begin{equation*}
\prescript{\mu}{}{\rho} = \rho^{-1}.
\end{equation*} Show that we also have: $$D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$$ Hint: Label the vertices of a regular hexagon $1,2,3,4,5,6$ consecutively, and consider the stabilizer of the set $\{ 2,4,6 \}$ by the action of $D_{6}$ on the set of vertices.
My question is:
I do not know how considering the stabilizer of the set $\{ 2,4,6 \}$ by the action of $D_{6}$ on the set of vertices of the regular hexagon will prove that $D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$ Could anyone explain this to me, please?
Best Answer
So $D_6$ has a subgroup isomorphic to $S_3$, namely the stabilizer of $\{2,4,6\}$.
It also has a two element subgroup, isomorphic to $\Bbb Z_2$, whose intersection with $S_3$ is trivial (consider a reflection that doesn't stabilise $\{2,4,6\}$.)
By counting $S_3\Bbb Z_2=D_6$.
Thus $D_6\cong S_3×\Bbb Z_2$.