Two Inverse results

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 $$A^{-1}AK=KA^{-1}A=A^{-1}A\ \text{and}\ 6|K|\ge |A^{-1}A|=3|H|.$$