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?

  • $\begingroup$ Did you look in the literature on "Tarski numbers"? This is not quite what you are asking about, but close. In particular, take a look at the proof that every non-amenable f.g. group is paradoxical, see if you can use the proof to settle your bonus question. $\endgroup$
    – Misha
    Jul 18, 2013 at 14:00
  • $\begingroup$ Thanks Misha. Indeed one motivation is to quantify the Von Neumann problem, like the Tarski number do. In particular the first question is about low complexity counterexamples. Since the Tarski number are not understood (to put it mildly) I was hoping to find an easier way to measure non-amenability. One can show quite easily that if a group is paradoxical with one of the "halves" puzzle equivalent with the whole group by cutting in two pieces only, then the group is indeed a finite union of small sets... but no such example is known, except if the group contains the free subgroup. $\endgroup$ Jul 18, 2013 at 14:24
  • $\begingroup$ @EricTressler: Thanks for reply, but I don t understand what is the group in your example... $\endgroup$ Jul 18, 2013 at 14:26
  • $\begingroup$ Dan: I see, but did you try to extract an answer to the bonus question from the proof of paradoxality for nonamenable groups rather than merely from its statement? (If you did not, it might be worth trying.) Also, maybe you should directly ask Grigorchuk, he might know some interesting examples. $\endgroup$
    – Misha
    Jul 18, 2013 at 16:55
  • $\begingroup$ @Misha: thanks for the suggestion, it looks like worth trying indeed. Btw, the only proof I know is via Folner and Marriage but this cannot be Tarski´s original proof, I guess... $\endgroup$ Jul 19, 2013 at 14:38

1 Answer 1


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. 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.

  • $\begingroup$ I accept this very interesting answer. The main question is unsolved for the moment, but this might change. $\endgroup$ Aug 17, 2013 at 8:39

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