# 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.

• @AdamZalcman: That's a rather bold assertion; in general you can't just ignore other identities in the presentation. In fact, that's the whole issue: showing that the other identities do not somehow imply a relation $r^m=1$ with some $m\lt n$. How do you deduce that this is the presentation of the subgroup generated by $r$? Commented Oct 7, 2021 at 18:18
• @AdamZalcman How large is the subgroup generated by $r$ in $$\langle r, s | r^4 = s^3 = 1, rs = r^2s^2\rangle?$$ (I don't see how your argument is sufficient to find orders) Commented Oct 7, 2021 at 18:38
• @AdamZalcman: "We see that all relators that lack an $s$ have the form $r^{kn}$ for $k\in\mathbb{Z}$." That is where I thnk you are jumping. I don't think it is anywhere near as "obvious" as you seem to think it is. So please do not elide that assertion; prove it. Commented Oct 7, 2021 at 18:39
• @AdamZalcman: Or to vary Brian's example a bit so that the element doesn't trivially cancel, let's take $\langle r,s\mid r^4=s^2=1, sr=r^2s\rangle$. The relation subgroup is the normal closure of the subgroup generated by $r^4$, $s^2$, and $s^{-1}r^{-2}sr$, all reduced words in the free group. "We see that all relators that lack an $s$ are of the form $r^{4k}$ for $k\in\mathbb{Z}$" ? So we conclude that the order of $r$ in that case is $4$? But $s^{-1}rs = srs = r^2s^2 = r^2$, so $1 = r^4 = (s^{-1}rs)^2 = s^{-1}r^2s$, hence $s^{-1}r^2s=1$, so $r^2=1$. The same "argument" fails here. Commented Oct 7, 2021 at 18:49
• @ArturoMagidin I was really confused when you said "trivially cancel", but I just now looked at it and realized I mistyped the relations. Ooops! I meant $$\langle r, s | r^4 = s^3 = 1, rs = s^2r^2\rangle$$ The point is still that the group is trivial, but it's just "easy" rather than "trivial" to see. Commented Oct 7, 2021 at 22:15

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.