Non-amenable groups with arbitrarily large Tarski number? 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 $m=2$.
 A: 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.

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*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).


*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.
