# Sets which are unions of translates of each other but aren't single translates

I'm a hobbyist mathematician so any question I ask here might be at risk of closure. I hope this one is good enough, but I'm not sure. This is a continuation of two questions I asked on math.stackexchange.com: About translating subsets of $\Bbb R^2$ and About translating subsets of $\Bbb Z$. The question is

For which abelian groups $M$ can we find $A,B\subseteq M$ such that

• $A$ is a union of translated (only translations are allowed) copies of $B$;
• $B$ is a union of translated copies of $A$;
• $A$ is not a a single translated copy of $B$ (and the other way around, which follows)?

David Moews showed in an answer to the first question that such sets exist in $\Bbb Z^\omega$ and therefore in any group in which $\Bbb Z^\omega$ embeds. Certainly no such sets can be found in finite groups. Can we describe all groups that have such two sets?

• For those who might not follow the MSE link: $\mathbb Z^\omega$ here denotes direct sum, not Cartesian product. That is, such sets exist in any abelian group of infinite rank. Nov 13 '14 at 13:47
• Generalizing the finite case, there are no such sets in torsion groups, and they satisfy the stronger condition that $A=A+X$ implies $A=A+t$ for every $t\in X$. Nov 13 '14 at 15:03
• @Emil so this means that if the property of not having such sets is closed under finite products, then this property is just the finiteness of rank, right? Nov 14 '14 at 11:13
• I don't know whether it is realistic to expect the property to be closed under products, but anyway, then you'd still need to prove it holds for $\mathbb Q$, it's not enough to have it for $\mathbb Z$. Nov 14 '14 at 11:27
• Anyway, if you don't mind countable and highly overlapping unions, $y>\sqrt 2 x$ and $y>\sqrt 2 x+\frac 12$ is a simple explicit pair in $\mathbb Z^2$. Nov 28 '14 at 1:08