Question on conjugate closure of subgroups

In summary, the conversation discussed the concept of subgroup and conjugate closure within a larger group. The subgroup is generated by the set of elements that are the product of an element in the first subgroup and an element in the second subgroup. It was observed that the intersection of the original subgroup and its conjugate closure is equal to the original subgroup. The question was then raised about whether the intersection of the second subgroup and its conjugate closure is trivial. The conversation then delved into a possible proof by induction and the need to prove that certain elements within the conjugate closure must be equal to the identity.
  • #1
mnb96
715
5
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
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
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.
 

1. What is the definition of conjugate closure of subgroups?

The conjugate closure of subgroups is the smallest subgroup that contains all conjugates of a given subgroup. In other words, it is the subgroup that is obtained by taking the union of a subgroup and all its conjugates.

2. How is the conjugate closure of subgroups related to group theory?

The concept of conjugate closure of subgroups is an important concept in group theory. It helps in understanding the structure and properties of groups, and is used in proving theorems and solving problems related to groups.

3. Can you give an example of conjugate closure of subgroups?

One example of conjugate closure of subgroups is the subgroup of even permutations in the symmetric group Sn. This subgroup is closed under conjugation, and its conjugate closure is the entire symmetric group Sn.

4. How is the conjugate closure of subgroups different from the normal closure of subgroups?

The conjugate closure of subgroups is the smallest subgroup that contains all conjugates of a given subgroup, while the normal closure of subgroups is the smallest normal subgroup that contains a given subgroup. In other words, the normal closure is a subset of the conjugate closure.

5. Why is the concept of conjugate closure of subgroups important?

The conjugate closure of subgroups is important because it helps in identifying and studying the various subgroups of a given group. It also plays a crucial role in the study of symmetry and in proving theorems related to groups and their properties.

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