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## Main Question or Discussion Point

Hello,

Let's have a group

I consider the subgroup [itex]\left\langle A^B \right\rangle[/itex] which is called

It is clear that [itex]A\cap \left\langle A^B \right\rangle = A[/itex].

What about [itex]B\cap \left\langle A^B \right\rangle[/itex]?

Do

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 [itex]\left\langle A^B \right\rangle[/itex] which is called

*conjugate closure*of*A*with respect to*B*, and it is the subgroup generated by the set: [tex]A^B=\{ b^{-1}ab \;|\; a\in A,\; b\in B\}[/tex]It is clear that [itex]A\cap \left\langle A^B \right\rangle = A[/itex].

What about [itex]B\cap \left\langle A^B \right\rangle[/itex]?

Do

*B*and the conjugate closure [itex]\left\langle A^B \right\rangle[/itex] have trivial intersection?