A nonempty subset $S$ of a group $G$ is called *small* if there is an infinite sequence of elements $g\_n$ in $G$ such that the translated sets $g\_nS$ are pairwise disjoint. 


Question: Is there a group which is a (disjoint) union of three small subsets, but it is not a union of two small subsets? 


Remark: Such a group must be non-amenable (clear) and must not contain a copy of the non-abelian free group (in fact it is an exercise to see that the groups which are a union of two small sets are exactly those containing the free group).


Bonus question: Is every non-amenable group a finite union of small subsets?



Partition of a group into small subsets
I believe the answer to the bonus question is negative, that is, there exist non-amenable (finitely generated) groups which are not representable as a union of finitely many small subsets. This is directly related to the problem of constructing non-amenable groups
with arbitrarily large Tarski numbers discussed [here](https://mathoverflow.net/questions/137678/non-amenable-groups-with-arbitrarily-large-tarski-number). Here is a sketch of the
argument.


Using a variation of the method that Mark Sapir has described in the above post, one can construct a finitely generated non-amenable group G with the following property:


For every n there is a finite index subgroup G(n) of G such that every n-generated subgroup of G(n) is finite.


Now suppose such G is a union of k small subsets S\_1,..., S\_k. Then each finite index subgroup H of G is also a union of k subsets, namely the intersections of S\_1,...,S\_k
with H, which are still small (as subsets of H). This easily implies that H must have a non-amenable subgroup generated by at most k^2 elements, which is a contradiction.


