Conjugacy classes for rotations of $D_{2n}$

Question

It says in my notes that for a dihedral group $D_{2n}$, if $n$ is even, then each conjugacy class has at most $1$ element. It says that for the conjugacy class $C_{r^k}$ where $r^k$ is some reflection, the conjugacy class is $\{r^{\frac{n}{2} = k}\}$ (or just $\{r^{k}\}$).

I didn't follow this. Say we have $D_8$. Then $n = 4$, so $k = 2$. Then $r^2$ is just moving a vertex on a octagon over two vertices. But what about $r^1$, $r^3$, $r^5$, and $r^7$? $1$, $3$, $5$, and $7$ are not even numbers, so this says that they are not in any conjuacy classes since they are not of the form $r^2$, and the only conjuacy classes of $D_8$ is $\{r^{2}\}$. But we know that each element should belong to at least one conjugacy class, so how is this possible?

Answer 1

The dihedral group $D_{2n}$ is the symmetry group of the regular $n$-gon. In case you're wondering what the conjugacy classes actually look like, here it is laid out.

If $n>2$ is even the conjugacy classes are:

• $\{I\}$ where $I$ is the identity element.
• $\{S\}$ where $S$ is the $180^\circ$ rotation.
• Pairs $\{R,R^{-1}\}$ where $R$ is any other rotation.
• The set of all reflections across lines through two vertices of the $n$-gon.
• The set of all reflections across lines through midpoints of edges of the $n$-gon.

If $n>1$ is odd the conjugacy classes are:

• $\{I\}$ where $I$ is the identity element.
• Pairs $\{R,R^{-1}\}$ where $R$ is any nontrivial rotation.
• The set of all reflections in the symmetry group.

The above characterization can be derived from the following geometric facts:

• Conjugating a rotation by a rotation does nothing; they commute.
• Conjugating a rotation by a reflection yields the inverse rotation.
• Conjugating a flip across a line $\ell$ by a rotation yields a flip across ($\ell$ twice rotated).
• (I'll leave conjugating flips by flips as an exercise.)

Answer 2

Something is certainly incorrect in your notes: a group is abelian if and only if every conjugacy class has size one, and the dihedral group $D_{2n}$ is not abelian for $n>2$. Possibly what is meant is that the center of $D_{2n}$ consists of at most one non-trivial element; and for $n$ even, this element is $r^{n/2}$.