As a follow-up to my previous question, the next two exercises state:
Use the given generators and relations to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then:
- a) $rx = xr^{-1}$, and
- b) $x$ has order 2, and hence $D_{2n}$ is generated by $s$ and $sr$, each with order 2.
Here, $D_{2n}$ has the "usual" presentation $D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.$
For part a), since $x$ is not a power of $r$, I have written $ x = sr^k$ for some $0 \leq k \leq n - 1$.
So then $rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}$, but this doesn't seem to be moving me in the right direction.
Should I be using induction on the power of $r$ in the element $x$ ?
Then for part b), I simply wrote $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1$, but I doubt that I have sufficiently justified my work.
From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble.
Thanks in advance for your help!