Banach-Tarski vs von Neumann While not so well known, the von Neumann paradox is built among the same lines, in dimension 2 and with transforms within the special linear group. But what is wrong  with the following "proof" : it is not very hard to show that the group generated by the two  rotations of angle $\alpha$ around $O$ and $C$, with $OC<1/100$, say, and $\alpha/\pi$ irrational and small (say $<1/100$ too), is non-abelian and free. The von Neumann construction applied to two disjoint unit discs then split them in four sets (plus a few fixed points) and sends those sets injectively, by rotations and  translations, to disjoint sets  included in the union of four copies of the unit disc, this  union having something like 1.04 area of the disc. Obviously, this is very wrong, as the Banach-Tarski construction of a full additive measure invariant by isometries of the plane preclude such a thing. Where am I mistaken?
 A: The mistake appears at the very begining of the "proof". In order to obtain the paradoxical decomposition we need a free group of rank $2$. Denote one of the above rotations by $\sigma$ and the other by $\tau$. One can easily check that the isometry $\sigma^2 \tau^2 \sigma^{-2} \tau^{-2}$ is a translation and the same about $\sigma^{-2} \tau^{-2} \sigma^{2} \tau^{2}$. This implies that the isometries commute and therefore the group generated by $\sigma$ and $\tau$ is not a free group of rank $2$.
It should be noted that the von Neumann paradox uses elements from the special affine group $SA_2(\mathbb{R})$ and not the linear transformations alone (linear transformations fix the origin so it cannot be duplicated). For a variant of Banach-Tarski paradox on the punctured unit disk (the origin removed) you may consult the paper "A free group of piecewise linear transformations", Coll. Math. 125 (2011), 141-146.
Also, the paradox mentioned by YCor above (so called Sierpinski-Mazurkiewicz paradox) uses the free group generated by the translation $f(z) = z + 1$ and the rotation $g(z) = e^{iz}$, but it does not exclude the existence of Borel measure as suggested above. Note that there is a Banach measure (finitely additive, isometry invariant measure defined on all subsets of $\mathbb{R}^2$ that normalizes the unit square and extends the Lebesgue measure) which restriction to Borel sets has the desired properties. All that we can say in this case is that Sierpinski-Mazurkiewicz paradox excludes the possibility of existence of any finitely additive, isometry invariant measure defined on all subsets of $\mathbb{R}^2$ that normalizes the paradoxical set constructed by Sierpinski and Mazurkiewicz.
