Product with commutator equals join with conjugate

Statement

Suppose $H\le G$ is a subgroup and $A\subseteq G$ is a subset. Define:

$[A,H]=⟨a^{-1}{h}^{-1}ah\mid a\in A,h\in H⟩$

and:

$H^A=⟨a^{-1}Ha\mid a\in A⟩$.

Then, we have:

$⟨H,H^A⟩=⟨H,[A,H]⟩$.

Further, since $H$ normalizes $[A,H]$, we have:

$⟨H,H^A⟩=H[A,H]$.

Related facts

Applications

• Strongly paranormal implies paranormal
• Commutator subgroup satisfies ascending chain condition on subnormal subgroups implies subnormal join property

Facts used

1. Subgroup normalizes its commutator with any subset

Proof

Given: A subgroup $H\le G$, a subset $A\subseteq G$.

To prove: $⟨H,H^A⟩=⟨H,[A,H]⟩$ = $H[A,H]$.

Proof:

1. $a^{-1}ha\in ⟨H,[A,H]⟩$ for all $a\in A,h\in H$, and this: Note that $a^{-1}ha=(a^{-1}ha{h}^{-1})h$. We have that $a^{-1}ha{h}^{-1}\in [A,H]$ and $h\in H$, giving the required result.
2. $H^A\le ⟨H,[A,H]⟩$: This is an immediate consequence of step (1).
3. $a^{-1}{h}^{-1}ah\in H^A$ for all $a\in A,h\in H$: Note that $a^{-1}{h}^{-1}ah=(a^{-1}{h}^{-1}a)h$. We have that $a^{-1}{h}^{-1}a\in H^A$ and $h\in H$, giving the required result.
4. $[A,H]\le H^A$:This is an immediate consequence of step (3).
5. $H$ normalizes $[A,H]$, and thus, $⟨H,H^A⟩=⟨H,[A,H]⟩=H[A,H]$: This follows from fact (1).