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Just out of curiosity, I wonder whether there are non-amenable groups with arbitrarily large Tarski numbers. The Tarski number $\tau(G)$ of a discrete group $G$ is the smallest $n$ such that $G$ admits a paradoxical decomposition with $n$ pieces: $\exists A_1,\ldots,A_k,B_1\ldots,B_l\subset G$, $\exists g_1,\ldots,g_k,h_1\ldots,h_l\in G$ such that $k+l=n$ and $$G = \bigsqcup_{i=1}^k A_i\sqcup\bigsqcup_{j=1}^l B_j = \bigsqcup_{i=1}^k g_iA_i = \bigsqcup_{j=1}^l h_jB_j \quad\mbox{(disjoint unions)}.$$ Tarski's theorem says that $\tau(G)<\infty$ iff $G$ is non-amenable. It is known that $\tau(G)=4$ iff $G$ contains a non-abelian free subgroup. (See a survey paper by Ceccherini-Silberstein, Grigorchuck, and de la Harpe)

If $G$ is a non-amenable group such that every $m$ generated subgroup of it is amenable, then it satisfies $\tau(G)>m+2$ (because one may assume $g_1=e=h_1$ in the paradoxical decomposition). Such $G$ probably exists, but I do not know any examples even for $m=2$.

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I think this is still an open problem. – Misha Jul 25 '13 at 2:03
If I understand correctly, despite Mark's stellar answer, all the questions about Tarski numbers listed in the survey paper remain open, right? – Dan Sălăjan Jul 26 '13 at 13:16
up vote 136 down vote accepted

It is indeed an open problem, as Misha said. But here is a solution. In E. Golod, Some problems of Burnside type. 1968 Proc. Internat. Congr. Math. (Moscow, 1966) pp. 284-289. Izdat. ”Mir”, Moscow, Golod announced, for every $m$ an infinite finitely generated torsion group all of whose $m$-generated subgroups are finite. A proof can be found in Ershov, Mikhail Golod-Shafarevich groups: a survey. Internat. J. Algebra Comput. 22 (2012), no. 5, 1230001, 68 pp (Theorem 3.3). The proof starts with a Golod-Shafarevich group $G$. If you assume that $G$ has property (T) (such groups exist by Ershov, see the survey), the resulting group will have property (T). Thus there exists a finitely generated infinite property (T), hence non-amenable, group with arbitrary large Tarski number.

Correction. Misha Ershov sent me two corrections.

  1. It is easier to deduce the answer from Theorem 3.3 and Ershov's theorem that any GS group is non-amenable, it also can be found in the survey (this does follow from existence of property (T) GS group).

  2. There exists a generalized GS group with property (T) and all m-gen. subgroups finite (for every fixed m). Thus a property (T) group with arbitrary large Tarski number also exists.

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So the problem was open when you started writing the answer, and solved by the time you'd finished!? – HJRW Jul 25 '13 at 10:41
That is correct. – Mark Sapir Jul 25 '13 at 10:43
That seems eminently deserving of a +1! – HJRW Jul 25 '13 at 10:56
The reason it was open was that when the problem was formulated by Ceccherini-Silberstein, Grigorchuk, and de la Harpe, GS groups with property (T) (or even with property ($\tau$)) were not known. In fact many people believed that such groups do not exist. Also I am not sure that they considered GS groups. After Ershov found a GS group with property (T), nobody noticed that it solves the problem - till now. – Mark Sapir Jul 25 '13 at 11:28
This is great, indeed. I never imagined it could be solved so quickly! – Narutaka OZAWA Jul 26 '13 at 2:54

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