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In an introduction to abstract algebra, I was recently introduced to the idea of presenting a group - minimally, a group is just a set of generators along with a set of relations amongst the generators. I believe that I have, at least, a rather basic understanding of this idea. On the other hand, I don't quite understand when one knows that they have a sufficient amount of relations to uniquely characterize the group at hand. For example, a common example for generators and relations is the Dihedral group $ D_n = \{ \rho, \tau : \;\rho^n = 1, \tau^2 =1, \tau\rho\tau^{-1}=\rho^{-1} \}$. Clearly there are two generators here: a rotation $ \rho $ by an angle $ 2\pi/n$ and a reflection $ \tau $. What I don't understand is exactly how one knows that these three relations as listed are sufficient to characterize the group. When listing the relations, I see that each of these properties are true, but how does one know that they cannot stop with just $ \rho^n = 1$ and $ \tau^2= 1 $, the most basic properties of $D_n$? A small bit of clarification here would be greatly appreciated as I feel as though I am missing something obvious.

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The relation $\tau \rho \tau ^{-1} = \rho ^{-1}$ tells you about how $\tau$ and $\rho$ interact. Without it, you would just have a free group on two letters modulo those order relations.

You can consider $D_{2*4} = D_8$ the rotations of the square and convince yourself that composing rotations and reflections does indeed satisfy such a relation.

An informal explanation now on why these three relations should be sufficient:

By the geometric argument I used in $D_8$, we should be able to convince ourselves that there is a group of order $2n$ with the relations as described in the presentation. So any such presentation gives rise to a group of order $2n$ or larger. Now I claim that the relation between $\rho$ and $\tau$ will restrict the order to at most $2n$.

The relation $\tau \rho \tau ^{-1} = \rho ^{-1}$ is usually called a commuting relation, because it tells you how to move the $\tau$ and $\rho$ between each other. With this relation, we can present every word in $D_{2n}$ as $\rho ^i \tau ^j$ where $0 \le i < n$ and $j = 0,1$. Now it is clear that indeed there can be at most $2n$.

This still goes back to what I hoped to convey originally which was that we need a third relation that tells us how the rotations and reflections behave together. I hope this clarifies things a bit.

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    $\begingroup$ The group with reations just $r^n=t^2=1$ is not abelian, and not even finite. $\endgroup$ – Mariano Suárez-Álvarez Dec 19 '15 at 7:16
  • $\begingroup$ I think that should be the free product $\mathbb{Z}/n\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}$. $\endgroup$ – André 3000 Dec 19 '15 at 15:08
  • $\begingroup$ So in general, there must be a relation between each generator? I don't think you addressed the problem in general of how one knows that they have sufficient relations amongst the generators $\endgroup$ – P7E Dec 19 '15 at 17:40
  • $\begingroup$ Yes sorry, I primarily work with free abelian groups and made that assumption here. $\endgroup$ – Future Dec 19 '15 at 20:43
  • $\begingroup$ I added additional arguments attempting to address why the three conditions should be both sufficient and necessary by expanding on what I said originally. Let me know if you would like me to clarify further. $\endgroup$ – Future Dec 19 '15 at 20:57
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This won't address your specific question here, but more of my general feeling about presentations.

Key idea: Presentations make it easy to communicate the particular group you're working with, but are generally hard to come up with, or work with!


For example, there are lots and lots of groups of order $96$ -- 231 of them, to be precise. But if you've found an interesting one (say, this guy), how in the world would you describe it to someone, especially if it doesn't belong to a fairly well-known family, or have a nice description as (semi)direct products?

That's where a presentation comes into play. Supposing you have such a presentation, you just write it down, tell your friend, and that's that. Your job is done!

This is ignoring the fact that it's really nontrivial to determine a set of relations that pins down your group. I've never even thought of doing this, but I'd wager it's not a pleasant task. Why would I be willing to wager that?

Let's go back to your friend, when she receives receives the compact presentation you sent earlier. She has her work cut out for her! See this answer of mine for an idea of the kind of work required just to list elements, for a group of order only $8$. Long story short, it's completely nontrivial to actually unpack a presentation, in general. This is without even mentioning the word problem, which in a sense makes precise how difficult it is.

So in summation, group presentations are nice as exactly that -- presentations. If you have any other description of the group to work with, chances are, it'll be easier than working with the presentation.

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