# Free subgroup of Diff([0,1])?

It is well known that the group of diffeomorphisms of the circle contains free non-Abelian subgroups. Is it true (known) that the group of diffeomorphisms of the interval $[0,1]$ contains free subgroups? One approach to get a positive answer can be the following. Consider all functions $f_a=\frac{\exp(ax)-1}{\exp(a)-1}$, $a>1$, which are smooth diffeomorphisms of $[0,1]$. First prove that the group generated by these functions does not satisfy any non-trivial law. Now for any non-trivial word $w$ in two variables consider the function $w(f_a,f_b)$ (for every $a,b$). The set of pairs $(a,b)\in \mathbb{R}^2$ for which this function is the identity function has dimension at most 1. Then by Baire category theorem, since the set of words $w$ is countable, there are $a,b$ such that $f_a,f_b$ freely generate a free subgroup. Does this argument actually work?

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This kind of question (zoology of subgroups of diffeomorphisms of intervals) it can be important to mention the differentiability class ($Diff^\infty$, $Diff^2$, $Diff^1$). Examples by Navas show that they are very different. I assume you mean $Diff^\infty$? – YCor Sep 20 '12 at 7:24
@Yves Cornulier : The reference in my answer shows that the differentiability class does not matter for this particular question. – Andy Putman Sep 21 '12 at 3:16