Let $A$ be a non-empty subset of the symmetric group $S_3$ and put $A^{-1}=\{ a^{-1}:a\in A\}$.
Prove that:
(1) If $|A|=3$ and $A$ is not a subgroup then $|A^{-1}A|=|AA^{-1}|=5$.
(2) If $|A|\neq 3$ then $A^{-1}A=AA^{-1}\leq S_3$.
(Note that if $|A|\geq 4$ then $A^{-1}A=AA^{-1}= S_3$, see Is it true that if $|A|>\frac{|G|}{2}$ then $A^{-1}A=AA^{-1}=G$?).