I want to prove that for subgroups $A,B \leq G$ where $G$ has finite order and $A$ has the property $\forall x \in G, a \in A\;\exists n \in\mathbb{Z}: x(ax^{-1})^n \in A$, $AB$ and $BA$ are equal, even though they need not be groups in general.
It is clear that we only need to show that one of the products is a subset of the other; if we know that $AB \subseteq BA$, it follows that $AB \subseteq BA = ((BA)^{-1})^{-1} = (A^{-1}B^{-1})^{-1} = (AB)^{-1}$, thus $AB \subseteq (AB)^{-1}$. Since clearly, $|AB| = |(AB)^{-1}|$, it would follow that $AB = BA$ because $G$ is finite (the case $BA \subseteq AB$ can be proved similarly). I also know that if either $A$ or $B$ is a normal subgroup, then $AB$ is a group and $AB = BA$ follows trivially. In this case, though, $AB$ need not be a group and I don't know how to prove the statement from here.
Choose $x=ba$, giving $bab^{-n} \in A$ for some $n$, so $ba \in AB$, and hence $BA \subseteq AB$. Since $G$ is finite, this implies that they are equal.