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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! enter image description here

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

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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.

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Corresponding to the fact that, if a polygon has an even number of sides, the axis of reflections are of one kind (the lines going through a vertex and the middle of an edge) but if the polygon has an even number of sides, they are of two kinds (the ones going though opposite vertices and the ones going through opposite edges' middles). –  PseudoNeo Apr 24 '13 at 16:38
    
Of course, the first “even” in the last comment should be an “odd”. –  PseudoNeo Apr 24 '13 at 17:13
    
Ahh thank you, this makes a lot of sense now. –  user51327 Apr 24 '13 at 18:03

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