Proof that $rs = sr^{-1}$ in $D_{2n}$

Question

In the group $D_{2n}$, we define $r$ to be a counterclockwise rotation by $\frac{2\pi}{n}$ and $s$ a reflection through a fixed line of symmetry. I'm trying to prove that $$rs = sr^{-1}.$$

I can prove smaller cases, like $n=3,4,5$ by drawing the figures and performing the rotations or reflections in succession, but I cannot figure out how to prove this for an arbitrary $n$.

I'd appreciate any hints or direction on how to proceed.

Answer 1

Hint

The relation is the same as $(sr)^2=1$. So it suffices to check that $sr$ is a reflection.

Drawing a picture makes it rather trivial: rotating a reflection gives another reflection (through the rotated axis).

Answer 2

Let $a_1,a_2,\dots,a_n$ denote the vertices of the $n$-gon ordered clockwise. If $a_1,a_2,a_3,\dots,a_n$ denotes the $n$-gon before performing any symmetries, then we see that

$$ a_1,a_2,a_3,\dots,a_n\mathop{\Rightarrow}^r a_n,a_1,a_2,\dots,a_{n-1}\tag1 $$ $$ a_n,a_1,a_2,\dots,a_{n-1}\mathop{\Rightarrow}^s a_n,a_{n-1},a_{n-2},\dots,a_1\tag2 $$ $$ a_n,a_{n-1},a_{n-2},\dots,a_1\mathop{\Rightarrow}^r a_1,a_n,a_{n-1},\dots,a_2\tag3 $$

By comparing the left hand side of (1) and the right hand side of (3), we see that the $rsr=s$, which gives the desired result.