Let $G$ be a multiplicative group and $A\subseteq G$ such that
1) $\forall a\in A, a^{-1}\in A$
2) $1\in A$
3) $AA \subseteq gA$ for some $g\in G$
Can we say that $A$ is a subgroup?
One can immediately show that $g\in A$ and that we have $g^{-1}A \subseteq A \subseteq gA$ but I believe you must use the symmetry property of $A$ (1) again to conclude that it is a subgroup (if it is).