Showing the Dihedral Group is Generated by Two Elements

Question

I am trying to show the Dihedral group $D_{2n} = \langle r,s ~|~ r^n s^2 = 1, rs = sr^{-1} \rangle$ can be generated by $s$ and $sr$.

I found a relatively brief proof here, but if I am not mistaken it presupposes the following theorem:

Let $G = \langle x_1,x_2,...,x_n ~|~ R_1 \rangle$, where $R_1$ is a collection of relations on the generators $x_1,..,x_n$. If $x_i \in \langle y_1,y_2,...,y_k ~|~ R_2 \rangle$ for all $i=1,2,...,n$, then $G = \langle y_1,y_2,...,y_k ~|~ R_2 \rangle$.

Is this right? Although it seems obvious, I am having a little trouble proving it. Here is my shot at part of the proof:

If $g \in G$, then $g = f(x_1,...,x_n)$, where $f(x_1,...x_n)$ is some expression involving the generators. But the generators $x_1,...x_n$ can be written in terms of $y_1,...,y_k$. Therefore $g = f(y_1,...,y_k) \in \langle y_1,y_2,...,y_k ~|~ R_2 \rangle$.

However, this seems a little flaky and non-rigorous. So, my questions are, is this the theorem the Project Crazy Project is presupposing; and how would one prove it rigorously?

Answer 1

Perhaps easier and simpler:

$$r=s\cdot sr\implies \langle s,r\rangle\le\langle s,sr\rangle$$

and since ther other inclusion is trivial we're done.

Answer 2

Fix two elements $r,s\in D_{2n}$ where $r^n=1$ and $s^2=1$. Note that if $sr^u=sr^v$, then $$r^u=s^2r^u=s(sr^u)=s(sr^v)=s^2r^v=r^v,$$ but $r^u,r^v\in\langle r\rangle$ and so $u=v$. Also note that if $r^u=sr^v$ then$$r^u=(sr)r^{v-1}=(r^{-1}s)r^{v-1}=r^{-1}(sr)r^{v-2}=r^{-2}sr^{v-2}=...=r^{-v}s$$ and so $r^{u+v}=s\in\langle r\rangle\cap\langle s\rangle$, a contradiction. Therefore, we show that the set $$\{1,r,r^2,...,r^{n-1},s,sr,sr^2,...,sr^{n-1}\}$$ has $2n$ elements. Thus $$D_{2n}=\{1,r,r^2,...,r^{n-1},s,sr,sr^2,...,sr^{n-1}\}=\langle r,s\rangle.$$

To complete the proof just see that $\langle r,s\rangle=\langle s,sr\rangle$ (see @DonAntonio answer.)