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Let $A$ be a finite subset of a torsion free group $G$ with $|A|\geq3$. Does the set $AA^{-1}$ have the "unique product" property (i.e. there exist an element $c\in AA^{-1}$ that is uniquely written as a product $ab$ with $a\in A$, $b\in A^{-1}$)? In particular, what can we say if we consider the Passman four group: $$G:=\langle x,y: (x^2)^y=x^{-2}, (y^2)^x=y^{-2}\rangle$$

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  • $\begingroup$ I suppose you already know that, but the answer is positive for abelian groups. $\endgroup$
    – Wojowu
    Commented Jun 17, 2019 at 10:05
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    $\begingroup$ @Wojowu Yes, the answer is positive for all right orderable groups containing Abelian ones. $\endgroup$
    – MSMalekan
    Commented Jun 17, 2019 at 10:15
  • $\begingroup$ Examples of torsion free groups without the UP property have been constrcuted by Rips and Segev. I would be suprised if they are not counter example to your question.sciencedirect.com/science/article/pii/0021869387901256 $\endgroup$
    – Thomas
    Commented Jun 27, 2019 at 14:17
  • $\begingroup$ @Thomas, no, Rips-Segev example won't work. The UP propery is about the product of two different sets $ AB$ In the paper $B$ is of size 4, a concrete set. $\endgroup$
    – user6976
    Commented Nov 18, 2019 at 0:05

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