Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to a build a better understanding of presentations. I get that a a group has a presentation $\langle S \mid R \rangle$ if it is the "freest" group subject to the relations $R$.

But, for instance, how is it immediately clear that $\langle x, y \mid x^2 = y^2 \rangle$ is the same group as $\langle x, y | xyx^{-1}y \rangle$? Also, more generally, what are some ways to determine if two different presentations yield the same group?

share|improve this question
4  
One thing you should note is that presentations can be arbitrarily complicated to understand.. it's undecidable whether or not a presentation defines the identity group or not.. –  user58512 Mar 9 '13 at 19:15

3 Answers 3

up vote 5 down vote accepted

As user58512 points out in the comments, it is not always possible to determine whether two presentations define isomorphic groups. It has been shown that the general problem of determining whether a given presentation defines the trivial group is undecidable.

However, in many cases we can show (or even easily see) that two presentations represent the same group. One way to do this formally is using Tietze transformations. I suggest you try transforming your first presentation into the second using Tietze tranformations. As a hint, try starting by adding $xy$ as a new generator in the second presentation.

share|improve this answer
1  
@John: I'd recommend waiting a bit longer before accepting an answer. Often people don't look at questions that already have an accepted answer, and you might miss out on some useful extra enlightenment by accepting an answer so quickly. I'm glad you found this helpful, though! –  Tara B Mar 9 '13 at 19:48
    
In fact, Tietze transformations always work, it is just that there is not necessarily a way to do them algorithmically. –  user1729 Oct 18 '13 at 7:41

Sometimes the laws of the presentation imply eachother. But in general, you can think of them as quotient groups of the free groups, and the relations as congruence relations, or with manipulation, generators of the kernel (which is why the default relation is to the identity). And then it isn't usually immediately apparent what things are and aren't isomorphic, and the presentation may not all-knowing. So you prove the isomorphisms theorems.

share|improve this answer
    
I don't know what you mean by 'the presentation may not [be] all-knowing'. The presentation uniquely defines a group, so in that sense it is 'all-knowing'. But you may not be able to figure out everything you want to know about the group from the presentation, which is perhaps what you meant. –  Tara B Mar 9 '13 at 19:55
    
It's not all-knowing in the sense that it doesn't have the information to derive all other possible presentations. So it's not just looking within a single algebra, where if two equations give the same element, you can always find a path between them in the algebra. You have to step back, look at the meta-algebra - the category of groups. –  Loki Clock Mar 9 '13 at 20:05
    
Ah, but even the problem of deciding whether two words represent the same element of a group is in general undecidable! –  Tara B Mar 9 '13 at 20:07
1  
Summary of chat: Yes, if the groups are finitely presented (which is probably the context worth asking such a question in). –  Tara B Mar 9 '13 at 22:10
1  
Actually I'm fairly sure I was wrong in my argument. The two presentations won't necessarily have the 'same' generators (for example in John's example in this question, we don't have $x^2 = y^2$ in the second presentation - you have to transform the generating set in order to get from one presentation to the other), and so what I proposed won't always work. –  Tara B Mar 9 '13 at 22:32

It is a little difficult to try to explain this without drawing stuff, but if you draw a rectangle with one pair I of opposite sides identified and the other pair II identified in contrary directions, you can see you have two Moebius bands: one as a strip "in the middle" of the rectangle with the sides II identified, and the other one taking the strips above and before this first Möbius band (I'm sure there must be some site where this thing's done!).

Now you can apply the Seifert- van Kampen Theorem by joining the two bands on the middle of the rectangle, one above I and the other one below it, and this renders you the group $\,\Bbb Z *_{\Bbb Z}\Bbb Z\,$ , when the amalgation gives us just the relation $\,x^2=y^2\,$.

If you succeed in grabbing the geometric picture above you have your isomorphism at once, otherwise it is going to take way more time and sweat. If I can come up with something later I'll add it here.

Added: Massey's IV.5 seems to work on this but it is too complicated for me to understand it by merely reading it quickly. If you have some time try it.

Further added: We have $\,x^2=y^2\Longrightarrow x^2y^{-2}=1\,$ , and thus we can put

$$r:=xyx^{-1}\;,\;\;s:=xy^{-1}\Longrightarrow$$

$$ rsr^{-1}s=xyx^{-1}xy^{-1}xy^{-1}x^{-1}xy^{-1}=x^2y^{-2}=1$$

and this way we can pass from your first presentation to the second one...

share|improve this answer
    
I don't see how this answer would help with anything in the question other than the part about the two specific presentations, which I think was just an example rather than the main point. –  Tara B Mar 9 '13 at 19:57
    
Surely you meant something different rather than $z:=y$? –  Tara B Mar 9 '13 at 20:03
    
Can you see it now, @TaraB ? Even from the beginning I was trying to give some algebraic topology approach to see that the presentation with $\,x^2=y^2\,$ is Klein bottle's, and if the OP already knew the other presentation was also Klein bottle's then we're done. –  DonAntonio Mar 9 '13 at 20:03
    
I had another idea in mind with that substitution $\,y=z\,$ thing which, of course, is completely unnecessary. –  DonAntonio Mar 9 '13 at 20:05
    
I'm confused by your latest edit. Now what is $z$? –  Tara B Mar 9 '13 at 20:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.