Proving $r^i s = sr^{-i}$ in $D_{2n}$

Question

Let $D_{2n}$ be the group of symmetries of the regular $n$-gon. Define $r$ to be a counterclockwise rotation through $\frac{2\pi}{n}$ radians and $s$ a reflection in a fixed line of symmetry. I am trying to prove a result from Dummit and Foote's algebra book, namely that $r^i s = sr^{-i}$ for all $0 \leq i \leq n$. I'm having two main difficulties, but here is the attempt I wrote down:

We induct on $n$. If $i = 0$, the statement reduces to $s = s$. Assume the statement inductively for some $i \geq 0$. Then we have \begin{align*} r^{i+1} s & = r\left(r^i s\right) \\ & = r\left(sr^{-i} \right) & & \text{induction hypothesis} \\ & = (rs)r^{-i} \\ & = \left(sr^{-1}\right) r^{-i} & & \text{$i=1$ case} \\ & = s\left(r^{-1} r^{-i}\right) \\ & = sr^{-(i+1)} \end{align*}

There are two immediate problems I have with this proof:

(1) The book recommends induction on $i$, but I'm accustomed to proving such a statement for all $i$. I don't know how to prove it for all $0 \leq i \leq n$. I assume there is some kind of modular equivalence built into $D_{2n}$ so it will still hold for higher $i$, but with less convenient names, but how would I, in general, go about proving the statement only for a select $i$? Do I require that the $i$ in my induction hypothesis be strictly less than $n$?

(2) I didn't prove the $i=1$ case, but I assumed it in my induction step. I don't know how to prove that in general. I can prove it in, say, $D_{10}$, but only by drawing the Pentagon and performing the symmetries. Is there a better way to prove this in general without a picture?

Answer 1

In last paragraph of page 24, it mentions that after we label the vertices of a regular $n$-gon from $1$ to $n$ in a clockwise manner,

Since symmetries are rigid motions one sees that once the position of the ordered pair of vertices $1,2$ has been specified, the action of the symmetry on all remaining vertices is completely determined.

To prove $rs=sr^{-1}$, the authors give the hint that work out separately what each side in this equation does to vertices $1$ and $2.$

The authors define $r$ to be the rotation clockwise about the origin through $2\pi/n$ radian and $s$ to be the reflection about the line of symmetry through vertex $1$ and the origin.

First it is clear that $r(i)=i+1\bmod n$ and $r^{-1}(i)=i-1\bmod n$.

Next you can check that \begin{align*}rs(1)&=r(1)=2,\\ rs(2)&=r(n-1)=1.\end{align*} and \begin{align*}sr^{-1}(1)&=s(n-1)=2,\\ sr^{-1}(2)&=s(1)=1.\end{align*}

Therefore we conclude that $rs=sr^{-1}$.