Product with commutator equals join with conjugate

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

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

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:

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

Applications

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

Facts used

 * 1) uses::Subgroup normalizes its commutator with any subset

Proof
Given: A subgroup $$H \le G$$, 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) $$H^A \le \langle H, [A,H] \rangle$$: 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) $$[A,H] \le H^A$$: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).