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Let $f,g\colon \mathbb R \to \mathbb R$ be the permutations defined by $f\colon x \mapsto x+1$ and $g\colon x \mapsto x^3$, or maybe even have $g\colon x \mapsto x^p$, $p$ an odd prime.

In the book, by Pierre de la Harpe, Topics in Geometric Group Theory section $\textrm{II.B.40}$, as a research problem, it asks to find an appropriate "ping-pong" action to show that the group, under function composition, $G=\langle f,g \rangle$ is a free group of rank two.

Is there such a proof? That is, is there a proof where the key insight is having that group act in such a way to apply the ping-pong lemma(table-tennis lemma)? I have not been able to find such a proof either by working on it, or in the literature.

Maybe we don't have such a proof but do we have a proof that $G$ contains a free subgroup of rank two, akin to proofs for torsion-free hyperbolic group, or the Tits alternative. I am not sure how obvious it is that $G$ is hyperbolic, or linear. I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly.

Note that there are proofs of this theorem, but as far as I know, they do not use the ping-pong lemma.

The only proofs of the result(and more general things) I know of are in :

  • Free groups from fields by Stephen D. Cohen and A.M.W. Glass

  • The group generated by $x \mapsto x+1$ and $x \mapsto x^p$ is free. by Samuel White

  • Arithmetic permutations by S.A. Adeleke and A.M.W. Glass

DMcMor
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  • The book contains quite a lot of these "research problems". Could it be that the author hasn't worked them out and instead offers them as, say, research proposals for bachelor or master theses? Perhaps they are open problems, including this one here? Notice that the existing proofs are quite complicated. – Martin Brandenburg May 24 '15 at 21:53
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    @MartinBrandenburg I agree, the Andrews-Curtis conjecture, 3d Poincare conjecture, "is every hyperbolic group residually finite?", are only some of the research problems he has in the book. It has been 15 years or so since the book was published, so I am wondering is such a proof has been done, and I don't think at the time of the book being published there was such a proof. It is also really difficult to find papers on this, the only reason I found the ones sited above was because he sited them! (It is sometimes a little ambiguous as to "how solved" the research problems are) –  May 24 '15 at 22:15
  • Given that I found the idea of such a proof quite compelling, I figured others had the same though, and maybe someone had found one since the book was written. –  May 24 '15 at 22:17
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    "I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly": this is not quite true. To find free groups in linear groups is now easy, but to show that two given elements generate a free group is usually delicate. For instance I think the following is open (or known but hard): let $G$ be a Zariski-dense subgroup of $SL(n,\mathbf{C})$ be non-virtually-solvable, and let $x\in G$ be of infinite order: does there exist $y\in G$ such that $\langle x,y\rangle$ is free? Typically if $x$ is unipotent, it's hard. – YCor May 28 '15 at 19:33
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    I don't expect that linearity would be easier than free. For instance, the subgroup $H$ generated by $f,g$ and in addition $h:x\mapsto 2x$ is not linear because of the relations $hfh^{-1}=f^2$, $ghg^{-1}=h^3$. Indeed the last implies that $g^nhg^{-n}f^{\pm 1}g^nh^{-1}g^{-n}=f^{\pm 2^{(3^n)}}$. If these are complex matrices, the norm of the left hand is bounded exponentially, and hence the norm of $f^m$ partially growth logarithmically (both for $m\ge 0$ and $m\le 0$). This is possible only if $f$ has finite order, so every representation of $H$ maps $f$ to an element of finite order. – YCor May 28 '15 at 21:01

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