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In the paper http://projecteuclid.org/download/pdf_1/euclid.ijm/1255632506 it is stated without proof that the commutator subgroup of the modular group is a free group of rank $2$.

Can anyone give a reference for the proof of this fact?

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  • $\begingroup$ Serre's book "Trees" is one place where you can find the necessary material $\endgroup$ – Geoff Robinson May 10 '14 at 18:23
  • $\begingroup$ This is not a research-level question, but a well-known statement. See the answer of user39082, or in-general, try to play ping-pong on the upper-half-plane with two inverse unipotents or so. $\endgroup$ – Asaf May 10 '14 at 18:24
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The commutator group is free because it acts freely on the Bass-Serre tree. See the answers to https://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz

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  • $\begingroup$ thank you for your answer! $\endgroup$ – Veeeo May 10 '14 at 18:39
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Another (essentially equivalent) way to see this is that ${\rm PSL}(2,\mathbb{Z})$ is the free product of a cyclic group of order $2$ and order $3,$ so it has Euler-Poincare -Wall characteristic $\frac{1}{2}+\frac{1}{3} - 1 = \frac{-1}{6}.$ Its commutator subgroup clearly has index $6$ (it has a homomorphic image of order $6$ by properties of the free product, while every Abelian homomorphic image clearly has order dividing $6$). Hence the commutator subgroup has E-P-W characteristic $-1$, so must be free of rank $2$ (it is a torsion free subgroup of an amalgam of two finite groups, so is free).

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If we regard the modular group as $\langle x\rangle \ast \langle y\rangle$, $x^2=y^3=1$, then its commutator group is freely generated by $a=xyxy^2$ and $b=xy^2xy$. So its a free group of rank $2$. In contrast, for example, the commutator group $G'$ of the Picard group $G=PSL_2(\mathbb{Z}[i])$ is not a free group.

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