Two Inverse results

Abstract

Let $A$ be a subset of group $G_0$ with $|A^{-1}A|\le 2|A|-2.$ We show that there are an element $a\in A$ and a non-null proper subgroup $H$ of $G$ such that one of the following holds: \begin{itemize} \item $x^{-1}Hy\subset A^{-1}A,$ for all $(x,y)\in A^2\setminus (Ha{)}^2,$ \item $xH{y}^{-1}\subset A{A}^{-1},$ for all $(x,y)\in A^2\setminus (aH{)}^2.$ \end{itemize} where $G$ is the subgroup generated by $A^{-1}A.$ Assuming that $A^{-1}A\ne G$ and that $|A^{-1}A|<\frac{5|A|}{3},$ we show that there are a normal subgroup $K$ of $G$ and a subgroup $H$ with $K\subset H\subset A^{-1}A$ and $2|K|\ge |H|$ such that $$A^{-1}AK=K{A}^{-1}A=A^{-1}A\text{and}6|K|\ge |A^{-1}A|=3|H|.$$

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• An analytic approach to a weak non-Abelian Kneser-type theorem