Fast way of finding a homomorphism 
Show that $D_{12}$, the dihedral group of order $12$, is isomorphic to the direct product $D_6\times C_2$.

I can do it by just checking that each element in one has an equivalent element in the other, but is there a quicker way of showing that there must exist an isomorphism?
 A: As has been discussed in the comments, there are two answers, each depending on which notation you mean. Does $D_n$ mean the dihedral group of order $n$, or of order $2n$? My original answer assumed that the OP meant $D_n$ has order $n$, but in the comments it is apparent that they mean it has order $2n$. So I'll first answer the question they meant, and then I'll leave my original answer.

Answer 1: If $D_6$ has order $6$ then you can show that $D_{12}\cong D_6\times C_2$ as follows. The idea is to show that it is the internal direct product of $D_6$, the dihedral group of order $6$, and $C_2$, so $D_{12}=MN$ where $M\cap N=1$, and $M$ and $N$ are both normal in $D_{12}$.
Take $N$ to be the natural copy of $D_6$ in $D_{12}$: remember that $D_{12}$ is acting on a hexagon, and we can make this look like a triangle by rotating twice. So $D_6$ embeds by rotating twice. Take $M$ to be the half-twist. Clearly, $M\cong C_2$ and $M\cap N=1$. Finally, note that $M$ is central in $D_{12}$, which implies that both $M$ and $N$ are normal in $D_{12}$ (because $mn=nm$ for all $n\in N$ so $mnm^{-1}=m$, and normality follows as $D_{12}=MN$). This proves the result.

Answer 2: If $D_6$ has order $12$ then there doesn't exist an isomorphism.
The centre of a group $G$ is the set $\{g\in G\mid gx=xg \:\forall\: x\in G\}$. Isomorphism preserve the centre of a group (heck, they preserve everything! But centre's are what are relevant here). So,
Hint: The centre of $D_{2n}$, the dihedral group of $4n$ has order $2$. (The non-trivial element is the half-twist: it commutes with everything.)
