Product with commutator equals join with conjugate

Statement

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

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

and:

$H^{A}=\langle a^{-1}Ha\mid a\in A\rangle$.

Then, we have:

$\langle H,H^{A}\rangle =\langle H,[A,H]\rangle$.

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

$\langle H,H^{A}\rangle =H[A,H]$.

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 $H\leq G$, a subset $A\subseteq G$.

To prove: $\langle H,H^{A}\rangle =\langle H,[A,H]\rangle$ = $H[A,H]$.

Proof:

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