If A, B are subgroups of G such that $b^{-1}Ab\subset A$ for all $b\in B$, show that AB is a subgroup of G My attemptation: First I wanted to prove that the set AB is closed under multiplication. That means if $a,a'\in A, b,b'\in B$, then $aba'b'\in AB$. This is hard to show . So then I considered $b^{-1}a^{-1}$, Since$b^{-1}a^{-1}=(b^{-1}a^{-1}e)e$ where $b^{-1}a^{-1}e \in A, e\in B$, so it's closed under inversion. Now, if $b=e$, then $aba'b'\in AB$, if $b\neq e$ and $b'=b^{-1}$, then the same is true. However, I didn't know what I can do next. I'm still sure that I have to utilize the condition that $b^{-1}Ab\subset A$ since if it's not true then the statement we're proving is not necessarily true.

Note that $aba'b'= a(ba'b^{-1})bb'$ and $(ab)^{-1}=b^{-1}a^{-1}= (b^{-1}a^{-1}b)b^{-1}$.