Minimal sumsets in finite solvable groups. (Q960955)

Given a group \(G\) and positive integers \(r,s\leq |G|\), the authors denote by \(\mu_G(r,s)\) the least possible size of a product set \(AB=\{ab\mid a\in A,\;b\in B\}\), where \(A,B\) run over all subsets of \(G\) of size \(r,s\), respectively. In a previous paper, they have completely determined \(\mu_G\) when \(G\) is Abelian [J. Algebra 287, No. 2, 449-457 (2005; Zbl 1095.11012)], however the function \(\mu_G\) is largely unknown for \(G\) non-Abelian, in part because efficient tools for proving lower bounds on \(\mu_G\) are still lacking in that case. The main result of this paper is a lower bound on \(\mu_G\) for finite solvable groups, obtained by building it up from the Abelian case with suitable combinatorial arguments. The result may be summarized as follows: if \(G\) is finite solvable of order \(m\), then \(\mu_G(r,s)\geq\mu_{G'}(r,s)\), where \(G'\) is any Abelian group of the same order \(m\). Equivalently, \(\mu_G(r,s)\geq\min_{h\mid m}\{(\lceil\frac{r}{h}\rceil+\lceil\frac{s}{h}\rceil-1)h\}\). One nice application is the full determination of the function \(\mu_G\) for the dihedral group \(G=D_n\) and all \(n\geq 1\). The authors prove that \(D_n\) has the same \(\mu\)-function as an Abelian group of order \(2n\).