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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?

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The group of motions of the plane is solvable of class two, so your two rotations can't generate a free group (and satisfy an explicit relation, e.g. $[[u,v],u[u,v]u^{-1}]=1$). But you have a variant of the paradox, using a free semigroup, implying, if I remember well that there is no finitely additive (unbounded!) measure on the bounded Borel sets, for which the unit disc has measure 1 (amenability of the group of motion provides an invariant mean on subset of the plane, which necessarily takes value 0 on the bounded subsets. –  YCor Oct 3 '12 at 14:44
    
Sorry, I don't understand your notation. If $u$ and $v$ are the two rotations, $v=tu$, with $t$ translation by the vector $OA$, with $A=v(O)$, $v^{-1}=t'u$ ($t'$ similarly translation by $OB$, with $B=v^{-1}(O)$) ; then $ut=Tu$ (the new translation vector being rotation of the first one by $alpha$). So how can $u$ and $v$ related, especially by a short relation ? –  Feldmann Denis Oct 3 '12 at 15:27
    
@FeldmannDenis Brackets usually denote the commutator $[g, h] = g^{−1}h^{−1}gh.$ –  Trevor Wilson Oct 3 '12 at 17:04
    
Oups... Sorry for my misunderstanding. And yes, the problem stems indeed from the existence of this relation. Next, I have to find the mistake in my "proof" that no such relation exists (even when the two angles of rotation are distinct), but this has no interest for MathOverfloxw, I am afraid :-( –  Feldmann Denis Oct 3 '12 at 17:46
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