(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)?
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The only remaining cases are periodic and amenable.



Can infiniteness of finitely generated groups be read by a "paradoxical" decomposition?
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.


