Prove that $S_3\times\mathbb{Z}_2$ is isomorphic to $D_6$ Prove that $S_3\times \mathbb{Z}_2$ is isomorphic to $D_6$.
My attempt:
I tried to decompose $D_6$ as the internal direct product of subgroups $H$ and $K$ isomorphic to $S_3$ and $\mathbb{Z}_2$ respectively:
Let $H=\{1, r^2, r^4, s, r^2s, r^4s\}$
Let $K=\{1, r^3s\}$
$HK=D_6$
$H\cap K=\{1\}$
However, taking $h=r^2$ and $k=r^3s$,
\begin{equation*}
\begin{split}
hk&=r^2\cdot r^3s \\
&=r^5s
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
kh&=r^3s\cdot r^2 \\
&=r^3sr^2ss \\
&=r^3r^{-2}s \\
&= rs
\end{split}
\end{equation*}
$hk\neq kh$
So, this doesn't work. How do I solve this problem?

Note that $Z(D_6) = \langle r^3\rangle$. This is your candidate for the copy of $C_2$. 
