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?

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 13 '14 at 13:47
  • $\begingroup$ 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 $t\in X$. $\endgroup$ Nov 13 '14 at 15:03
  • $\begingroup$ @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? $\endgroup$
    – Bartek
    Nov 14 '14 at 11:13
  • $\begingroup$ 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 $\mathbb Z$. $\endgroup$ Nov 14 '14 at 11:27
  • 3
    $\begingroup$ 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 $\mathbb Z^2$. $\endgroup$
    – fedja
    Nov 28 '14 at 1:08

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