Question on conjugate closure of subgroups

Hello,
Let's have a group G and two subgroups A<G and B<G such that the intersection of A and B is trivial.
I consider the subgroup $\left\langle A^B \right\rangle$ which is called conjugate closure of A with respect to B, and it is the subgroup generated by the set: $$A^B=\{ b^{-1}ab \;|\; a\in A,\; b\in B\}$$

It is clear that $A\cap \left\langle A^B \right\rangle = A$.

What about $B\cap \left\langle A^B \right\rangle$?
Do B and the conjugate closure $\left\langle A^B \right\rangle$ have trivial intersection?

Well, let's say we have something like $b^{-1}ab=b' \in B$. Then $a=bb'b^{-1}$. Now, clearly this implies that $a \in B$. But, $A\cap B = 1$. So...
thanks for the answer. I believe that with your argument you have essentially proved the base case of induction. The elements of $\left\langle A^B \right\rangle$ have the form: $(b_1^{-1} a_1 b_1) (b_2^{-1} a_2 b_2) \ldots (b_n^{-1} a_n b_n)$.