How does one show that $r^{n/2} \in D_{2n}$ commutes with $sr^i$? Here's what I've tried:
Showed that $r^{n/2}$ commutes with $s$.
For $sr^i$, for integer $i \in [1, n-1]$, I tried this: $sr^i r^{n/2} = r^{-(i+{n/2})}s=r^{-(i-n/2)}s=r^{-i}r^{n/2} s$.
But here's where I get stuck. Does anybody know what to do next?
Note that we have to be a little bit careful here: we need to assume $n$ is even for $r^{n/2}$ to have any meaning. Then, it is easy to check that $|r^{n/2}|=2$, so $r^{n/2}=r^{-n/2}$. Then, using the relation: $r^ks=sr^{-k}$, we have:
$r^{n/2}sr^i=sr^{-n/2}r^i=sr^{n/2}r^i=sr^{n/2+i}=sr^ir^{n/2}$, so $r^{n/2}$ commutes with $sr^i$.