What is a conjugacy class of reflection?

Question

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

Answer 1

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.

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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 $n=4$.

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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.

Answer 2

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. :-)

Answer 3

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.