most recent 30 from http://mathoverflow.net 2013-05-23T14:00:02Z http://mathoverflow.net/feeds/question/39990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39990/reference-request-for-two-generator-subgroups-of-a-free-group Yemon Choi 2010-09-26T02:53:35Z 2010-09-26T03:47:02Z

<p>According to B. Fine, G. Rosenberger, <em>On restricted Gromov groups</em>, Comm. Algebra 20 (1992) 2171--2181, Gromov proved the following in his long article introducing word-hyperbolic groups:</p> <blockquote> <p>Let $x$ and $y$ be elements of a torsion-free word-hyperbolic group. Either the subgroup generated by $x$ and $y$ is cyclic, or there exists $n$ such that the subgroup generated by $x^n$ and $y^n$ is free of rank 2.</p> </blockquote> <p>As a very special case of this, we get the following corollary</p> <blockquote> <p>Let $x$ and $y$ be non-commuting elements in the free group on two generators. Then the subgroup generated by $x$ and $y$ contains a copy of the free group on two generators.</p> </blockquote> <p>I am trying to cite this corollary as efficiently as possible, for background motivation in something I'm writing. Does anyone know of something slightly, erm, more accessible for the non-specialist than Gromov's original article? I don't really know any geometric group theory beyond some of the terminology and Nielsen-Schreier, but the result seems like it shouldn't be too hard to prove directly, modulo some standard results on free groups. Unfortunately, I don't really have space to sketch any proof in what I'm writing.</p> <p>(So, to clarify, what I'm really hoping for is an answer saying that the result is easily deduced from material in, say, Section Z of Lyndon &amp; Schupp or similar.)</p> <p><strong>EDIT/UPDATE:</strong> my thanks to John Stillwell and Ian Agol for pointing out what should have been blindingly obvious, namely that the result is a trivial consequence of Nielsen-Schreier, and for politely <em>not</em> pointing out what is just as obvious, that I should think harder before asking questions.</p>

http://mathoverflow.net/questions/39990/reference-request-for-two-generator-subgroups-of-a-free-group/39993#39993 John Stillwell 2010-09-26T03:08:15Z 2010-09-26T03:08:15Z

<p>By Nielsen-Schreier, the subgroup $F$ of $F_2$ generated by $x$ and $y$ is free. Since $x$ and $y$ do not commute, $F$ is not the free group of rank 1, so it must contain a free group of rank 2</p>

http://mathoverflow.net/questions/39990/reference-request-for-two-generator-subgroups-of-a-free-group/39995#39995 Agol 2010-09-26T03:15:46Z 2010-09-26T03:15:46Z

<p>This is deduced from <a href="http://books.google.com/books?id=aiPVBygHi_oC&amp;lpg=PP1&amp;dq=lyndon%2520schupp&amp;pg=PA8#v=onepage&amp;q&amp;f=false" rel="nofollow">Proposition 2.11</a> of Lyndon-Schupp, which says that a subgroup of a free group is free. If the subgroup generated by $x$ and $y$ is a free group of rank one, then $x$ and $y$ commute. So the subgroup they generate must be a free group of rank 2. </p>