Small sets without unique products in torsion-free groups (Q6580178)

If \(A\) and \(B\) are subsets of a group \(G\) then an element \(g \in AB\) is called a unique product for the pair \((A,B)\) if there is exactly one way to write \(g = ab\) with \(a \in A\) and \(b\in B\). Besides, a group \(G\) is called a unique product group (or \(UP\) group) if for any two finite nonempty subsets \(A\) and \(B\) of \(G\), there is a unique product for the pair \((A,B)\).\N\NThe authors note the following Kaplansky's (strong) zero-divisor conjecture: if \(R\) is a domain and \(G\) is a torsion-free group, then the group ring \(R[G]\) of \(G\) over \(R\) is a domain.\N\NThe basic results of the paper are the following.\N\NTheorem 1.2. If \(A\) is a nonempty subset of a torsion-free group such that \(A^2\) has no unique product, then \(\vert A \vert \geq 8\), and this bound is sharp.\N\NCorollary 1.3. If \(R\) is a domain, \(G\) is a torsion-free group, and \(\alpha \in R[G] - {0}\) satisfies \({\alpha}^2 = 0\), then \(\alpha\) has at least eight elements of \(G\) in its support.\N\NTheorem 1.4. Let \(A\) and \(B\) be subsets of a torsion-free group. Assuming \(AB\) has no unique product, then \(\vert A \vert + \vert B \vert \geq \) 16. More specifically, the following five bounds hold:\N\NIf \(\vert A \vert \) = 3, then \(\vert B \vert \geq \) 19.\N\NIf \(\vert A \vert \) = 4, then \(\vert B \vert \geq \) 14.\N\NIf \(\vert A \vert \) = 5, then \(\vert B \vert \geq \) 11.\N\NIf \(\vert A \vert \) = 6, then \(\vert B \vert \geq \) 10.\N\NIf \(\vert A \vert \) = 7, then \(\vert B \vert \geq \) 9.\N\NWhen \(\vert A \vert \) is small, the authors find lower bounds on \(\vert B \vert \).\N\NAs a consequence, the authors obtain that any counterexample to Kaplansky's zero-divisor conjecture must be quite complicated, if it exists at all. They give new examples of torsion-free groups that are not unique product groups.