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Consider the conjugation action of $D_n$ on $D_n$. Prove that the number of conjugacy classes of the reflections are

$\begin{cases} 1 &\text{ if } n=\text{odd} \\ 2 &\text{ if } n=\text{even} \end{cases} $

I tried this: Let $σ$ be a reflection. And $ρ$ be the standard rotation of $D_n$.

$$ρ^l⋅σρ^k⋅ρ^{-l}=σρ^{k-2l}$$ $$σρ^l⋅σr^k⋅ρ^{-l}σ=σρ^{-k+2l}$$

If $n$ is even, it depends on $k$ if $-k+2l$ will stay even. But if $n$ is odd, then at some point $-k+2l=|D_n|$ and therefore you will also get the even elements. So independent of $k$ you will get all the elements. Is this the idea ?

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What have you tried? –  user47805 Apr 12 '13 at 15:08
I'm a little lost. Let $\sigma$ be a reflection. I know that $D_n \sigma$ contains only reflection, and I think I need to prove that it contains all the reflection when $n$ is uneven. And only half(?) of the reflection when $n$ is even. I tried the multiplication rules of $D_n$, but it didn't help. –  Kasper Apr 12 '13 at 15:14
Just to clarify, we're looking for the conjugacy classes from taking $\sigma D_n\sigma^{-1}$ correct? –  user47805 Apr 12 '13 at 15:34
@user47805 I don't know that notation, my book defines the conjugacy class of $x$ as $Gx=\{gxg^{-1} :g∈G \}$ but I think that is the same. –  Kasper Apr 12 '13 at 15:38
Figure out what you get when you conjugate by $\sigma$ and what you get when you conjugate by $\rho$. At that point you should be able to work out what happens when you conjugate by an arbitrary element of G. –  Noah Snyder Apr 12 '13 at 15:40

2 Answers 2

Hint. The Orbit-Stabilizer theorem gives you that $[G:C_G(g)]$ is the size of the conjugacy class containing $g$. When $n$ is odd, a reflection $g$ commutes only with itself (why?), so $g$ has $[G:C_G(g)]=|G|/2$ elements, which are easily identified as the other reflections. Now, use this same technique to figure out the answer for the case of even $n$, keeping in mind that dihedral groups $D_{2n}$ have nontrivial centers when $n$ is even (why?).

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I find it very difficult to prove the statments you are making here. Something like this ? Let $sr^ks=ssr^k$. Then $r^{-k}=r^k$. Then $-k≡k \pmod{n}$. Then $2k≡n \pmod{n}$. Then $n$ is even. –  Kasper Apr 13 '13 at 16:14
@Kasper Yes this is exactly the right path! For example, I would write, "assume $s$ commutes with an nontrivial element of the form $r^k$. Then $$r^ks=sr^k\Leftrightarrow s^{-1}r^ks=r^k\Leftrightarrow r^{-k}=r^k\Leftrightarrow r^{2k}=1,$$thus $o(r^k)=2$ since we assumed that $r^k$ was nontrivial. Thus $n$ must be even." (Clearly this is very close to what you had.) The other cases are treated quite similarly. –  Alexander Gruber Apr 13 '13 at 16:51

A different approach: Draw a picture!

If $n$ is even, some reflection in $D_n$ are about lines connecting two opposite vertices, some reflections are bout lines connecting the midpoints of two opposite edges. The two types aren't related by conjugacy (see below). OTOH if $n$ is odd, all the symmetry axes connect a vertex to a midpoint of an edge.

If $\sigma$ is the reflection about line $L$, and $\tau$ is any rotation of the plane around the origin, then $\tau\sigma\tau^{-1}$ is the reflection about the line $\tau(L)$: it maps points on the line $\tau(L)$ to themselves, preserves angles and distances, and has eigenvalue $-1$, so it is a reflection.

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This is not the recommended solution, when you are first encountering groups. But it is something to keep in mind. –  Jyrki Lahtonen Apr 13 '13 at 10:58

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