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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.

Arturo Magidin
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usb_naming_is_confusing
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  • Welcome to MSE! If you already know that every element of $D_{2n}$ can be uniquely represented as $s^br^m$ with $s\in{0, 1}$ and $m\in{0, \dots, n-1}$ then you have a bijection from $D_{2n}$ to $A={(b, m)|s\in{0, 1}, m\in{0, \dots, n-1}}$ and since $|A|=2n$ then you're done. – Adam Zalcman Oct 07 '21 at 18:04
  • I think that's based on the assumption $\lvert r \rvert = n$, but this don't appear in the presentation. So I tried to deduce $\lvert r \rvert = n$ with only the presentation, by first assuming it's wrong and try to find contradiction. But I got lost. – usb_naming_is_confusing Oct 07 '21 at 18:08
  • That first link you've given deals directly with giving a lower bound on the order. – Brian Moehring Oct 07 '21 at 18:15
  • The subgroup generated by $r$ has presentation $\langle r|r^n=1\rangle$ which is the cyclic group of order $n$. Therefore, the order $|r|$ of $r$ in $D_{2n}$ is $n$. Adding in the observation that the index of the subgroup is $2$, we obtain the result. – Adam Zalcman Oct 07 '21 at 18:16
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    @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 $r$? – Arturo Magidin Oct 07 '21 at 18:18
  • @Brian Moehring Sorry I didn't phrase the problem properly, but I don't think finding a set modulo operator satisfying the presentation provides lower bound. Similar argument is also made in Dummit & Foote, where the authors stated that the 2D n-polygons satisfy $D_{2n}$ means there's at least $2n$ elements, but I'm still not convinced, e.g. the $D_{20}$ with $\lvert r \rvert = 5$ above. – usb_naming_is_confusing Oct 07 '21 at 18:20
  • Do you understand the universal property of the presentation? The group with this presentation has the following property: given any group $G$, and any elements $\rho,\sigma\in G$ such that $\rho^n=1$, $\sigma^2=1$, and $\rho\sigma=\sigma\rho^{-1}$ hold in $G$, there exists a unique group homomorphism $\phi\colon D_{2n}\to G$ such that $\phi(r)=\rho$ and $\phi(s)=\sigma$. The rigid motions of the 2-dimensional $n$-gon satisfy these properties and has $2n$ elements. Since there must a morphism $D_{2n}\to\text{rigid motions}$, which is surjective, that tells you $|D_{2n}|\geq 2n$. – Arturo Magidin Oct 07 '21 at 18:23
  • So what is it that you are still "not convinced" about? If $|r|=5$, then the rigid motions of the $10$-gon do not satisfy the relations, because the single-vertex rotation $\rho$ has $\rho^5\neq I$. – Arturo Magidin Oct 07 '21 at 18:24
  • @陳矩翰 If you give some group that satisfies the conditions of the presentation (two generators which satisfy the given equations) with at least $2n$ elements, then the group presentation gives a group with at least $2n$ elements. All your declaration about $D_{20}$ and $|r|=5$ means is that $D_{10}$ is a quotient of $D_{20}$, which is true but unrelated to determining the order of $D_{20}$. – Brian Moehring Oct 07 '21 at 18:25
  • @ArturoMagidin $\langle G|R\rangle = F(G)/N(R)$ where $F(G)$ is the free group generated by the set $G$ and $N(R)$ is the normal closure of relators $R$ in $F(G)$. In the present case $N(R)$ is generated by $r^n$, $s^2$ and $srsr$. So in this case we see that all relators that lack an $s$ have the form $r^{kn}$ for $k\in\mathbb{Z}$. – Adam Zalcman Oct 07 '21 at 18:33
  • Oh I think I understand now. If I find a group satisfying the presentation, of which order is $g$, then I'm can be sure the presentation meant a group with lower bound $g$ - but that don't prevent me to find another group satisfying the presentation, of which order is $h > g$, and now I have a better lower bound on the group order behind the presentation. Is that right? – usb_naming_is_confusing Oct 07 '21 at 18:33
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    @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) – Brian Moehring Oct 07 '21 at 18:38
  • And based on this I know why I'm stuck. At first I thought a presentation may mean different groups, in particular with different order. But it's universal, which means it represents exactly one group up to homomorphism. So things like I did above is merely finding quotients of the group, and setting lower bounds on the order. – usb_naming_is_confusing Oct 07 '21 at 18:38
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    @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. – Arturo Magidin Oct 07 '21 at 18:39
  • Proving every element has a presentation in a finite set gives an upper bound. Finding quotients give a lower bound. If you are lucky enough and find a lower bound via quotients that agrees with your upper bound via representatives, then you're golden: you know exactly what the order is in that case. Which is what happens here. Note that in the abstract it's a hard problem to decide, given a presentation, whether a particular expression represents the trivial element in the group. – Arturo Magidin Oct 07 '21 at 18:42
  • I think that's exactly what I was missing. Thx ArturoMagidin @BrianMoehring – usb_naming_is_confusing Oct 07 '21 at 18:43
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    @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. – Arturo Magidin Oct 07 '21 at 18:49
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    @AdamZalcman: I think your error is saying "$N(R)$ is generated by $r^n$, $s^2$, and $srsr$." It is not generated by those elements, it is the normal closure of the subgroup generated by those elements, as you had noted earlier. Note that by the Nielsen-Schreier formula, $N(R)$ is free of rank $2n+1$, so the three elements you give certainly do not suffice to generate it without taking normal closure. – Arturo Magidin Oct 07 '21 at 18:55
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    @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. – Brian Moehring Oct 07 '21 at 22:15
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    @BrianMoehring: Yes, that one doesn't suffer from the issue of the mis-typed one. In any case, we are saying the same thing: the explanation given for why $r$ "must" have order $n$ does not seem sufficient, and seems to miss precisely the problems that one encounters in situations like these. – Arturo Magidin Oct 08 '21 at 00:54
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    @ArturoMagidin You are right. Thank you for pinpointing the error in my reasoning and for the helpful counterexamples! $N(R)$ is indeed not generated by the relators, but by their conjugates e.g. $gr^ng^{-1}$ for $g\in F(G)$. This makes it substantially more involved to analyze the structure of $F(G)/N(R)$. Having thought a little bit more about it I now completely agree that this is the "hard part" of the problem. Thanks again! – Adam Zalcman Oct 08 '21 at 16:34
  • @BrianMoehring Thank you for the helpful counterexamples above! – Adam Zalcman Oct 08 '21 at 16:34

1 Answers1

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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.

Shaun
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Arturo Magidin
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