On the cardinality of sumsets in torsion-free groups. (Q2918794)

Let \(G\) be a torsion-free group and \(A,B\) its finite subsets. It was shown by \textit{J. H. B. Kemperman} [Nederl. Akad. Wet., Proc., Ser. A 59, 247-254 (1956; Zbl 0072.25605)] that if \(|A|,|B|\geq 2\), then NEWLINE\[NEWLINE|AB|\geq |A|+|B|-1,NEWLINE\]NEWLINE and \textit{L. V. Brailovsky} and \textit{G. A. Freiman} [J. Algebra 130, No. 2, 462-476 (1990; Zbl 0697.20019)] described the case in which one has equality here.NEWLINENEWLINE The main result of the paper shows that for any positive integer \(k\) there exists a constant \(c(k)\) (one can take \(c(k)=32(k+3)^6\)) with the following property: if \(A,B\) are finite subsets of \(G\), \(A\) is not a subset of a left coset of a cyclic subgroup of \(G\) and \(|B|\geq c(k)\), then NEWLINE\[NEWLINE|AB|>|A|+|B|+k.NEWLINE\]NEWLINE In the case \(k=1\) (with \(c(1)=4\)) this was earlier obtained by \textit{Y. O. Hamidoune, A. S. Lladó} and \textit{O. Serra} [Combinatorica 18, No. 4, 529-540 (1998; Zbl 0930.20034)].NEWLINENEWLINE If \(G\) is a unique product group, then it is shown (Lemma 12) that the main result holds with \(c(k)=4(2k+3)^3\), and the example of the Klein bottle group implies that in this case \(c(k)\) must grow at least as a quadratic polynomial. The authors conjecture that in the case \(A=B\) one has \(c(k)=ak+b\) with suitable \(a,b\).