Take the 2-minute tour ×  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 $m=2$.  share|