Dihedral Group of Order $2^{n}$ where $n \geq 3$ As I was self studying finite group theory I noticed something intriguing and failed to provide proof for the claim. What I noticed was that the order of the Dihedral group of order $2^{n}$ where $n \geq 3$ is always equal to four times its largest conjugacy class. I was curious if this was true in general, I have a strong feeling it is, but can't seem to prove it. I researched it to see if there was any literature on the topic but can't seem to find any. I also tried to prove the claim myself by induction but I can't seem to make progress. Any helpful tips or direction would be greatly appreciated.   
 A: This can be accomplished by classifying the conjugacy classes of the dihedral groups. The key is to understand what conjugation does.


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*If you conjugate a flip across an axis $\ell$ by an element $g$, you get a flip across $g\cdot\ell$.

*Conjugating a rotation $R$ by a rotation yields $R$, and by a flip yields $R^{-1}$.
In the dihedral group of an $n$-gon, if $n$ is odd then any axis of reflection can be rotated to any other axis of reflection, so the reflections form a single conjugacy class. If $n$ is even, though, the axes of reflection have two orbits under rotations: the axes through edges, and the ones through vertices, so there are two largest conjugacy classes and they have half of the group's reflections, and half of the group's elements are reflections, so both conjugacy classes have a quarter of the group elements.
Conjugacy classes of rotations are smaller, since they include only rotations of a given order, which implies the largest conjugacy class of rotations has at most $\phi(n)\le n/2$ elements ($n$ is even).
This classification of conjugacy classes generalizes to the orthogonal group $O(2)$, the matrix Lie group of all rotations and reflections of the two-dimensional plane. One can go further and classify the conjugacy classes in $O(3)$ too, which have rotations and reflections of 3D space.
