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**mathematicalbagpiper** Show that $\displaystyle D_{2n} \cong D_n \times \mathbb{Z}_2$

I'm using $\displaystyle x, y$ to represent the elements in $\displaystyle D_{2n}$ and $\displaystyle a, b$ for those elements in $\displaystyle D_n$. My first idea was to do this:

Map $\displaystyle x^{2i} \in D_{2n}$ to $\displaystyle (a^i, 0) \in D_n \times \mathbb{Z}_2$ since that forms a cyclic subgroup of the correct order.

Now I can't figure out where to go from here, or if my initial idea is even remotely in the ballpark.