# Conjugacy classes of D2n? [duplicate]

I'm trying to understand this proof and I can follow it up the the second last paragraph where it states what happens if n is odd/even. I don't understand why there is just one conjugacy class when n is odd and two when n is even.

Help much appreciated, thanks!

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## marked as duplicate by Alexander Gruber♦, Davide Giraudo, Hagen von Eitzen, Arkamis, MicahApr 24 '13 at 17:42

If $n$ is odd, then $k-2i\bmod{n}$ takes every value in $\mathbb{Z}/n\mathbb{Z}$ as we vary $i$. However, if $n$ is even, it only takes values with the same parity as $k$. (You can try this on some small examples and try to come up with a proof if you don't find this clear). So in the case that $n$ is odd, $\{r\rho^{k-2i}:i\in\mathbb{Z}\}$ is the set of all reflections, whereas if $n$ is even it is only half of them, and the set of reflections breaks into two conjugacy classes.