# 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 $m=2$.

• I think this is still an open problem. Jul 25, 2013 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? Jul 26, 2013 at 13:16

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.

• So the problem was open when you started writing the answer, and solved by the time you'd finished!?
– HJRW
Jul 25, 2013 at 10:41
• That is correct.
– user6976
Jul 25, 2013 at 10:43
• That seems eminently deserving of a +1!
– HJRW
Jul 25, 2013 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.
– user6976
Jul 25, 2013 at 11:28
• This is great, indeed. I never imagined it could be solved so quickly! Jul 26, 2013 at 2:54