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 $xHy^{ -1} \subset AA^{ -1},$ for all $(x,y)\in A^2\setminus (aH)^2.$ \end{ itemize} where $G$ is the subgroup generated by ${ A^{ -1}A}.$ Assuming that $A^{ -1}A\neq G$ 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 $$A^{ -1}AK=KA^{ -1}A=A^{ -1}A\ \text{ and}\ 6|K|\ge |A^{ -1}A|=3|H|.$$
Categories: math.NT