Show that $D_{12}$ is isomorphic to $D_6\times C_2$ Show that $D_{12}$ is isomorphic to $D_6 \times C_2$, where $D_{2n}$ is the dihedral group of order $2n$ and $C_2$ is the cyclic group of order $2$. 
I'm somewhat in over my head with my first year groups course. This is a question from an example sheet which I think if someone answered for me could illuminate a few things about isomorphisms to me. 
In this situation, does one use some lemma (that the direct product of two subgroups being isomorphic to their supergroup(?) if certain conditions are satisfied)? Does $D_{12}$ have to be abelian for this? Do we just go right ahead and search for a fitting bijection? Can we show the isomorphism is there without doing any of the above?
If someone could please answer the problem in the title and talk their way through, they would be being very helpful.
Thank You.

Assuming $D_{n}$ is the dihedral group of order $n$, I would proceed as follows. Note that $D_{6} \cong S_{3}$, and $S_{3}$ is generated by $(12)$ and $(123)$. Therefore $D_{6} \times C_{2} = \langle ((12),[0]),((123),[1]) \rangle$. Next note that $D_{12} = \langle r,s | \, r^{6}=s^{2}=e , s^{-1}rs=r^{-1} \rangle$, map the generators of $D_{6} \times C_{2}$ to the generators of $D_{12}$, and show this extends to a bijection.
