Talk:Presentation of a group

message: Given two presentations, the problem of determining if they give the same group is undecidable, in the Godelian sense.  Viz there is no algorithm or Turing machine that can systematically explore presentations and determine whether they specify the same group in a finite amount of time (in general -- in any particular case, you just might be able to work it out). This is isomorphic to the halting problem. Do we have an article on this?  
(There is an intuitive way to look at this: for any given presentation, you can try to write out every possible string that it generates; the problem is one of trying to compare all possible strings to determine if the same set was generated. Finitely-presented groups are kind-of-like context-free grammars, so one has a similar problem there: do two different grammars generate the same set of sentences? The CFG article does briefly review the undecidability problem.) linas 14:30, 25 July 2006 (UTC)
thread_title:  Undecidability 

message: I'm looking for info on the following presentation: <x,y| z=xyx^{-1}y^{-1}, zx=xz> which is a lot like the Heisenberg group, except that the second relation is not there: zy\ne yz. I have in hand the matrix representations for this group (including an inf-dimensional one) but wanted to know if this thing has a name, is studied in general, etc. Thanks linas 17:31, 11 March 2007 (UTC)
thread_title: a particular presentation

message: Just a place to sign edits to the math rating.Dan Hoey 19:36, 11 March 2007 (UTC)
thread_title: Math rating

message: Maybe I'm doing something wrong, but if I define a group (of permutations) S by the following four generators:
s1 := (1,2)(3,4)(5,6), s2 := (1,5)(2,4)(3,6),
s3 := (1,2)(3,5)(4,6), s4 := (1,4)(2,5)(3,6),
then they are clearly different, of order two, and they satisfy:
s3 * s1 == s1 * s3, s2 * s4 == s4 * s2,
s1 * s2 * s1 == s2 * s1 * s2, s2 * s3 * s2 == s3 * s2 * s3,
s3 * s4 * s3 == s4 * s3 * s4, s1 * s4 * s1 == s4 * s1 * s4,
particularly, s1 * s2, s2 * s3, s3 * s4 and s4 * s1 are of order 3.
However, S is not a S5 group but merely a S4 group with only 24 elements?!
IMO, the problem is here that for the generators as above we have:
s1 * s2 * s1 == s3 * s4 * s3 and s1 * s4 * s1 == s2 * s3 * s2,
whereas in S5 we would have:
(1, 2)(2, 3)(1, 2) <> (3, 4)(4, 5)(3, 4) and (1, 2)(4, 1)(1, 2) <> (2, 3)(3, 4)(2, 3).
Unfortunately, the presentation for S5 as in the table nowhere requires that:
s1 * s2 * s1 must be different from s3 * s4 * s3, as well as that
s1 * s4 * s1 must be different from s2 * s3 * s2.  —Preceding unsigned comment added by IvoP (talk • contribs) 12:16, 1 April 2009 (UTC)
replies: text: Hopefully the presentation for S5 mentions that s4*s1 should have order 2, not order 3.  <a,b,c,d:aa,bb,cc,dd,(ab)^3,(bc)^3,(cd)^3,(da)^3,(ac)^2,(bd)^2> is an infinite group (that has S4 as a quotient).  It has derived series: 2, 3, 2^4, Z^3, ....  On the other hand, S5 = <a,b,c,d:aa,bb,cc,dd,(ab)^3,(bc)^3,(cd)^3,(ac)^2,(ad)^2,(bd)^2>. JackSchmidt (talk) 12:51, 1 April 2009 (UTC)
thread_title:  Coxeter presentation of S5 

message: Shouldn't this article be called "Presentation (mathematics)" or "Presentation (group theory)" by the usual Wikipedia naming conventions?  — Preceding unsigned comment added by 207.241.137.116 (talk) 02:00, 15 January 2012 (UTC)
replies: text: No: the parenthetical disambiguation is not used except when required for disambiguation.  A natural title, as is used here and would generally be the term used in the references, is preferred.  See Wikipedia:Naming_conventions#Precision and disambiguation. — Quondum☏✎ 18:48, 15 January 2012 (UTC)
thread_title:  Wikipedia naming conventions 

message: OK, dumb question: normally, in defining a group in terms of generators, one talks about the set of all finite but unbounded strings of group elements (the Kleene star of these).  Sometimes, one can take the limit and consider the set of infinitely-long strings of group elements.  The question: in the latter case, when such a limit is possible, is the result, by necessity, always a topological group?  Or are there completions that are not?  linas (talk) 21:16, 28 August 2012 (UTC)
replies: text: Oh silly me, never mind. The definition of topological groups requires the group action to be continuous, and clearly, one can imagine completions where the action would not be continuous.  Hmm. New question: can anything be said, in general, about the completions of groups to an infinite number of products? (other than the obvious ones e.g. finite groups, or the completeness of reals under addition, etc.) linas (talk) 21:35, 28 August 2012 (UTC)
replies: text: Just checking that it is clear: infinite strings of generators are not the same as infinite products.  (You can't concatenate infinite strings in a reasonable way.  So you don't get a group this way.)  Instead, infinite strings of generators should be thought of as infinite paths in the Cayley graph of the group. If you are very careful about how you do things (ie use geodesic rays in the Cayley graph, based at the identity) and you are lucky with your group (ie it is Gromov hyperbolic) then the set of rays (up to an equivalence relation) forms a boundary/space at infinity for the group in a natural _geometric_ way.  The geometry of the group and its boundary then informs the algebra of the original group.  As an easy example - the trivial group has no points at infinity.  The infinite cyclic group ZZ has two points at infinity.  Any other finitely generated free group has a Cantor set at infinity. best,Sam nead (talk) 19:03, 29 August 2012 (UTC)
thread_title:  Stupid question. 

message: Substituted at 02:30, 5 May 2016 (UTC)
thread_title: Assessment comment

message: The other day I saw a group "presentation" of the following form:  \langle x,y \, | \, g^n = 1 \; \forall g \rangle 
The idea being that this was a group with 2 generators, with the only relation being that *every* element to the power of n is equal to the identity.
This article doesn't discuss these types of presentations.
I guess you would call this a finitely generated but infinitely related presentation.
That is, there are just two generators, x and y, but the relations are actually the infinite set:
1^n = 1 ,\, x^n = 1,\, y^n = 1,\, (xx)^n = 1,\, (xy)^n = 1,\, (yx)^n = 1,\, (yy)^n = 1,\, \ldots 
ranging over every word made of x and y (and their inverses). 
Perhaps the article should address these types of presentations? --AndreRD (talk) 13:32, 13 May 2019 (UTC)
replies: text: See Burnside problem. A link to this page would be a good idea. jraimbau (talk) 09:15, 14 May 2019 (UTC)
thread_title:  ⟨x,y | g^n = 1 ∀g⟩