(Edit) Let $G$ be a group. Two subsets $A,B$ of $G$ are said to be equidecomposable if there exists a finite partition $A=\bigsqcup_{i=1}^nA_i$ and $a_i\in G$ such that $B=\bigsqcup_{i=1}^na_iA_i$.

Say that a group has Property (X) if it has a subset equidecomposable to a proper subset of itself. Clearly this implies being infinite.

The negation of (X) passes to subgroups. The group $\mathbb{Z}$ has (X) (for $n=1$, ans hence all non-torsion (=non-periodic) groups have (X). Using a paradoxical decomposition, all non-amenable groups have (X).

Does every infinite finitely generated amenable group have (X)?

The only remaining cases are periodic and amenable.

  • 1
    $\begingroup$ It is not hard to check that a group has (X) if and only if some finitely generated subgroup has (X). In particular, locally finite groups do not have (X), and a positive answer to the question is equivalent to say that groups with (X) are exactly the non-locally-finite groups. $\endgroup$ – YCor Apr 17 '18 at 9:24
  • $\begingroup$ @YCor Yes, You are exactly right! Thanks for the edition! $\endgroup$ – Meisam Soleimani Malekan Apr 17 '18 at 13:20
  • $\begingroup$ Additional remark: say that $G$ has (X)$_n$ if it has (X) with this given $n$ (this is weaker when $n$ grows). For instance, $G$ has (X)$_1$ iff $G$ is non-torsion. More generally, it is easy to show that if every $n$-generated subgroup of $G$ is finite, then $G$ does not have (X)$_n$. I don't know if the converse holds, but it is enough to show that there does not exist any uniform $n$ working for all non-locally-finite groups. $\endgroup$ – YCor Apr 17 '18 at 22:33
  • $\begingroup$ PS now the converse is obtained in Robin's answer! $\endgroup$ – YCor Apr 18 '18 at 19:21
  • $\begingroup$ Note (slightly in regards with the initial less precise formulation of the question): I do not know any example of a group for which such a decomposition is used to prove that it's infinite (and for which it's not completely trivial for any other reason). $\endgroup$ – YCor Apr 18 '18 at 19:24

The answer is yes: every infinite finitely generated group has a subset which is equidecomposable to a proper subset of itself.

This follows from a theorem of Brandon Seward: Every finitely generated infinite group $G$ admits a translation-like action by the group $\mathbb{Z}$ of integers. https://arxiv.org/abs/1104.1231

What this means is that there exists a free action of $\mathbb{Z}$ on $G$ such that each element of $\mathbb{Z}$ acts as an element of the wobbling group of $G$. That is, there exists a bijection $f:G\rightarrow G$ having no finite orbits, along with a finite partition $G=\bigsqcup_{i=1}^nX_i$ and finitely many group elements $a_1,\dots , a_n\in G$ such that $f(x)=a_ix$ for all $x\in X_i$. If we take $T$ to be a subset of $G$ containing exactly one point from each orbit, and define $A$ to be the set $A = \bigsqcup _{n\geq 0} f^n(T)$, then $f(A)$ is a proper subset of $A$, and the partition $A=\bigsqcup _{i=1}^n A_i$, where $A_i := A\cap X_i$, along with the group elements $a_1,\dots , a_n$, witness that $A$ is equidecomposable with $f(A)$.

Edit: I realized there is also a simple direct argument which also shows that $n$ is bounded above by the minimum size of a finite subset $S$ of $G$ which generates an infinite subsemigroup of $G$. Consider the (left) Cayley graph of this subsemigroup $H$ with respect to its generating set $S$, i.e., with vertex set $H$ and with a directed edge from $h$ to $sh$ for each $s\in S$ and $h\in H$. This graph is infinite and locally finite, so by Kőnig's lemma it contains an infinite geodesic ray, say with vertices $h_0,h_1,\dots$. Let $A := \{ h_0,h_1,h_2,\dots \}$ and let $B=A \setminus \{ h_0 \}$. For each $n\geq 0$ there is some $s_n\in S$ such that $h_{n+1}=s_nh_n$, so $A$ is partitioned into finitely many sets $A_s$, $s\in S$, where $A_s$ consists of those $h_n$ for which $s_n=s$. Then $B$ is partitioned by $sA_s$, $s\in S$, so $A$ and $B$ are equidecomposable.

  • $\begingroup$ Nice! Indeed, you're right (as regards your edit). Indeed what you really need is a bounded-displacement permutation of $G$ with an infinite orbit, which indeed follows from the existence of a geodesic. The more difficult thing achieved by Seward is to get a bounded-displacement permutation with only infinite orbits. $\endgroup$ – YCor Apr 18 '18 at 19:21

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