# Thread: D2n Isomorphic To Dn X Z2

1. ## D2n Isomorphic To Dn X Z2

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

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

Map $x^{2i} \in D_{2n}$ to $(a^i, 0) \in D_n \times \mathbb{Z}_2$ 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 $D_{2n} \cong D_n \times \mathbb{Z}_2$

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

Map $x^{2i} \in D_{2n}$ to $(a^i, 0) \in D_n \times \mathbb{Z}_2$ 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 $x^{2i} \in D_{2n}$ to $(a^i, 0) \in D_n \times \mathbb{Z}_2$. The correspondence is completed by mapping $x^n \in D_{2n}$ to $(e,1) \in D_n \times \mathbb{Z}_2$, and $y \in D_{2n}$ to $(b,0) \in D_n \times \mathbb{Z}_2$ (where $e$ is the identity element of $D_n$).