Question on conjugate closure of subgroups

[mnb96]

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?

 

[Robert1986]

Well, let's say we have something like b^{-1}ab=b&#039; \in B. Then a=bb&#039;b^{-1}. Now, clearly this implies that a \in B. But, A\cap B = 1. So...

 

[mnb96]

Hi Robert,
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).

I would like to prove/disprove that if such elements are contained in B, then they must be equal to the identity. And this is exactly where I get stuck.