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According to B. Fine, G. Rosenberger, On restricted Gromov groups, Comm. Algebra 20 (1992) 2171--2181, Gromov proved the following in his long article introducing word-hyperbolic groups:

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.

As a very special case of this, we get the following corollary

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.

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.

(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 & Schupp or similar.)

EDIT/UPDATE: 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 not pointing out what is just as obvious, that I should think harder before asking questions.

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  • $\begingroup$ I don't seem to be able to downvote my own question ... ;-) $\endgroup$
    – Yemon Choi
    Commented Sep 26, 2010 at 3:44
  • $\begingroup$ Yemon, I've always appreciated your polite and helpful contributions to MO, so I'm glad to return the favor. $\endgroup$
    – John Stillwell
    Commented Sep 26, 2010 at 18:14
  • $\begingroup$ Though you admit that maybe the question is too easy, the style of write up and concise nature deserves at least a +1 $\endgroup$
    – Owen Sizemore
    Commented Sep 27, 2010 at 3:41
  • $\begingroup$ Owen: It's not just that the question was too easy, it's that I mentioned in the question the very result which immediately answers the question! But thanks for the kind remarks anyway. $\endgroup$
    – Yemon Choi
    Commented Sep 27, 2010 at 4:14

2 Answers 2

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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

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  • $\begingroup$ Thanks! sorry for the dumb question. I added Nielsen-Schreier to the question without thinking properly about the actual statement of the theorem. $\endgroup$
    – Yemon Choi
    Commented Sep 26, 2010 at 3:44
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This is deduced from Proposition 2.11 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.

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  • $\begingroup$ Thanks! sorry for the dumb question. The software doesn't allow me to accept more than one answer, but if I could I'd accept this too. $\endgroup$
    – Yemon Choi
    Commented Sep 26, 2010 at 3:44
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    $\begingroup$ And in fact $x$ and $y$ are a basis for the free group they generate, since finitely generated free groups are Hopfian. $\endgroup$
    – Allen Hatcher
    Commented Sep 26, 2010 at 8:27
  • $\begingroup$ This is proved in Prop. 2.7 of the above reference. $\endgroup$
    – Ian Agol
    Commented Sep 26, 2010 at 18:46

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