What is a conjugacy class of reflection? I have a problem to do, asking to show that $D_{2n}$ has two conjugacy classes of reflections if n is even, but only one if n is odd.
My question is, what is a conjugacy class of reflection?
I have seen that there is a solution to this question on this site, but I don't want to lose the value of the question by looking up the solution.
Thanks
 A: A reflection of the plane is a linear transformation (ok, it could be affine, but assume that the center of the regular $n$-gon is at the origin) such that it has two orthogonal eigenspaces, one belonging to eigenvalue $+1$ and the other to eigenvalue $-1$. These properties are invariant under conjugation by orthogonal transformations, so any conjugate of a reflection is another reflection.
If the eigenspace of $\lambda=+1$ (=the line that you reflect w.r.t) of a reflection $s$ intersects the polygon at the points $P$ and $Q$, and $g$ is any symmetry of the $n$-gon, then
$gsg^{-1}$ is the reflection w.r.t. the line connecting the points $g(P)$ and $g(Q)$.
If $n$ is odd, then for any reflection $r\in D_{2n}$ one of the points $P,Q$ is a vertex and the other is the midpoint of an edge opposite to that vertex. Because $D_{2n}$ acts transitively on the set of vertices, all the vertices (and hence also all the corresponding midpoints) occur as $PQ$-pairs. Thus there is only a single conjugacy class of reflections.
OTOH if $n$ is even, then both $P$ and $Q$ are either vertices or both are midpoints of an edge. The group $D_{2n}$ does not mix these two kinds of points. For example if $P$ is a vertex, then so is $g(P)$ for all $g\in D_{2n}$. But again $D_{2n}$ acts transitively on both vertices and midpoints, so we get two conjugacy classes of reflections.

As an extra let me point out that by the same line of reasoning all the reflections of $D_{2n}$ become conjugates in $D_{4n}$. See this answer for a discussion of this phenomenon in the case $n=4$.

And a very different way of looking at it:
The group $D_{2n}$ is defined by generators and relations as 
$$D_{2n}=\langle a, b\mid a^2=b^2=(ab)^n=1\rangle.$$
If $n$ is even this implies that we get four homomorphisms from $D_{2n}\to C_2=\{\pm1\}$ by arbitrarily mapping $a\mapsto \pm1$, $b\mapsto \pm1$. Because $C_2$ is abelian all the conjugates are mapped to same elements in all those homomorphisms. Therefore $a^ib^j$ and
$a^kb^\ell$ can be conjugates only if $i\equiv k$ and $j\equiv\ell\pmod2$.
OTOH if $n$ is odd, then any homomorphism $D_{2n}\to C_2$ must map $ab\mapsto1$, so the images of $a$ and $b$ are equal. Thus the determinant (sending both $a,b$ to $-1$ is the only non-trivial homomorphism from $D_{2n}$ to $C_2$ in this case.
Combining this with the bit about conjugates of reflections being reflections also allows you to identify the conjugacy classes.
A: The elements of $D_{2n}$ are geometric transformations of the plane, leaving an $n$-gon fixed. So it makes sense for an element of $D_{2n}$ to be a reflection.
A conjugacy class of a reflection is a conjugacy class of an element which is a reflection. :-)
A: This is a vague attempt to inspire geometric intuition, rather than a rigorous demonstration of the point.
Think about the conjugacy classes of the dihedral group as symmetries of a polygon. Thinking geometrically, you might see that conjugating a reflection with a rotation leaves the reflection altered by an even number of vertices.
If there are $2n$ vertices conjugation keeps the parity the same. If there are $2n+1$ vertices, you go twice round, hitting all the vertices.
