I bumped into this relation between rotations and reflections of a regular n-gon: $$xa = a^{n-1}x$$ where $x$ is a reflection and $a$ is a rotation by $\frac{2\pi}{n}$. I couldn't find any rigorous proof, so I'm trying to do it by myself, in a purely algebraic fashion.
First I tried it with some simple practical examples of triangles, squares, and hexagons, and it seems to hold water. Then I started trying to work it out.
It took me longer than it should have to notice that, in fact, $(xa)^{-1} = a^{n-1}x$. This is because $x^2 = id$ and $a^n = id$:
$$(xa)^{-1} = a^{-1}x^{-1} =a^{n-1}x^{-1}=a^{n-1}x.$$
However, proving that $xa = a^{n-1}x$ (assuming it is actually true) seems beyond my abilities... If I could prove that $a^{n-1}x$ is its own inverse, it would follow easily, but after several attempts I got nowhere. Ergo, either it is not true, or I'm missing something ^^
I'll be grateful for any suggestions!
A rotation about the origin by an angle $2\theta$ is a combination of two reflections in the lines $L_1, L_2$ through the origin such that angle between $L_1, L_2$ is $\theta$. The lines $L_1, L_2$ are arbitrary as long as they pass through the center of rotation and make an angle $\theta$ between them. Let $x$ be the reflection in line $L$. Then the left hand side can be written as $R_L R_L R_{L_2}$ where $L_2$ makes an angle $\frac{\pi}{n}$ with $L$. Thus the result is just a reflection about $L_2$. The right hand side can similarly be written as $R_{L_1} R_L R_L$ where $L_1$ and $L$ make an angle $\frac{(n-1)\pi}{n}$ with $L$. Note that this is same as line $L_2$. Thus the two motions are the same.