On products of sets in groups (Q1334936)

Let \(G\) be a group, \(B\) a finite subset of \(G\) containing 1, and \(A\) a non-empty finite subset of \(G\). The authors prove the following result: Either \(AB= A\langle B\rangle\) or \(| AB|\geq | A|+ \lfloor{2\over 3}(| B|+ 1)\rfloor\) if \(B= B^{-1}\) or \(B\cap B^{-1}=\{1\}\). The proof is based on properties of Cayley graphs.