Prove that order of $D_{2n}$ is exactly $2n$ I have been reading Dummit & Foote, and I got stuck on exercise 1 in ch 1.1, where the authors asked to show that a group with general dihedral group presentation has order exactly $2n$.
To be specific, by general dihedral group presentation, it means
$$
D_{2n} = \langle r,s ~\vert~ r^n = s^2 = 1 ,~ rs = sr^{-1} \rangle
$$
and the exercise asks to show $\lvert D_{2n} \rvert = 2n$
It's clear that any element in $D_{2n}$ can be uniquely represented by $s^br^m$, where $b \in \{0, 1\}$ and $m \in [0,n-1] \cap \mathbb{N}$. I try to find contradiction. That assuming $s=1$ seems dumb - it doesn't seem to contradict anything, though - so I started by assuming $\lvert r \rvert = m < n$. Here's what I tried:
$$
\begin{align*}
  r^m &= 1 && \text{ Assumption} \\
  s^2 r^m &= 1 && \text{ Multiply }s^2=1\text{ from left} \\
  s r^{n-m} s &= 1 && \text{ Apply "commute" law} \\
  s r^{n-m} &= s && \text{ Multiply }s\text{ from right} \\
  r^{n-m} &= 1 &&\text{ Cancellation law}
\end{align*}
$$
I find no contradictions here; it seems to suggest $\lvert r \rvert$ can be any factor of $n$. For example, both assuming that $\lvert r \rvert = 5$ in $D_{20}$ or assuming that $\lvert r \rvert = 2$ in $D_{16}$ seem to violate nothing. But from these we have $\lvert D_{20} \rvert \leq 10$ and $\lvert D_{16} \rvert \leq 4$, which is absurd.
I've read this, this, and this, but they don't seem to answer my question. The first one, with my intro-level algebra knowledge, provided an alternative way to see the presentation of $D_{2n}$, besides the 2D polygon example in Dummit & Foote. The second one illustrated that why a presentation like this leads to a group with order $\leq 2n$ (which I agree with), but give no reason to why there's lower bound on order of such presentation. The third seem to be wrong, because it seems the case when the remainder is $0$ is not considered, which is why I'm stuck in the beginning.
 A: You won't be able to just manipulate expressions as you are doing, because the group $D_{2m}$, with $m|n$, satisfies all the relevant identities, but has order $2m$. For all you know, you are working in that group...
You already know that there are at most $2n$ elements. You want to show that there are exactly $2n$ elements; that is, that you have not "missed" any relations that follow from yours.
This is usually the "hard part" of looking at groups given by presentations. What you usually want to do is use von Dyck's Theorem in some way (the universal property of the presentation) to show that the group has at least $2n$ elements, by finding surjective maps from your group to one with $2n$ elements.
That means finding groups $K$ whose order you already know, and elements $\rho,\sigma\in K$ that satisfy the relations that $r$ and $s$ do, so that you get a morphism from $D_{2n}$ onto $\langle \rho,\sigma\rangle$. Then you know  that $|D_{2n}|\geq |\langle \rho,\sigma\rangle|$. If you can find a group of order $2n$ to do this with, that will give you the inequality you are missing.
One possibility is to use some of the other interpretations of the dihedral group: look at $S_n$, and consider the element $\rho=(1,2,\ldots,n)$, and $\sigma=(2,n-1)(3,n-2)\cdots$. Then verify that $\rho^n=\sigma^2=e$, $\rho\sigma = \sigma\rho^{-1}$, and that $\langle \rho,\sigma\rangle$ has exactly $2n$ elements.
