# Thread: D2n Isomorphic To Dn X Z2

1. ## D2n Isomorphic To Dn X Z2

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.

2. Originally Posted by 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.
This result is only true when n is odd. In that case, it is correct to map $\displaystyle x^{2i} \in D_{2n}$ to $\displaystyle (a^i, 0) \in D_n \times \mathbb{Z}_2$. The correspondence is completed by mapping $\displaystyle x^n \in D_{2n}$ to $\displaystyle (e,1) \in D_n \times \mathbb{Z}_2$, and $\displaystyle y \in D_{2n}$ to $\displaystyle (b,0) \in D_n \times \mathbb{Z}_2$ (where $\displaystyle e$ is the identity element of $\displaystyle D_n$).