Question on conjugate closure of subgroups

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  • #1
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Main Question or Discussion Point

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 [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?
 

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  • #2
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Well, let's say we have something like [itex]b^{-1}ab=b' \in B[/itex]. Then [itex]a=bb'b^{-1}[/itex]. Now, clearly this implies that [itex]a \in B[/itex]. But, [itex]A\cap B = 1[/itex]. So...
 
  • #3
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Hi Robert,
thanks for the answer. I believe that with your argument you have essentially proved the base case of induction. The elements of [itex]\left\langle A^B \right\rangle[/itex] have the form: [itex](b_1^{-1} a_1 b_1) (b_2^{-1} a_2 b_2) \ldots (b_n^{-1} a_n b_n)[/itex].

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.
 

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