Commutator of subgroup with normalizing symmetric subset equals commutator with subgroup generated

Statement
Suppose $$G$$ is a group, $$H$$ is a subgroup of $$G$$, and $$A$$ is a symmetric subset of $$G$$ such that $$A \subseteq N_G(H)$$, i.e., $$A$$ is contained in the normalizer of $$H$$. Then, we have:

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

where $$[L,M]$$ denotes the commutator, namely, the subgroup:

$$\langle [l,m] \mid l \in L, m \in M \rangle$$.

Related facts

 * Commutator of Abelian subgroup with normalizing subgroup equals commutator of generating sets
 * Subgroup normalizes its commutator with any subset
 * Commutator of the whole group and a subset implies normal
 * Commutator of a normal subgroup and a subset implies 2-subnormal
 * Commutator of a 2-subnormal subgroup and a subset implies 3-subnormal
 * Commutator of two subgroups is normal in join

Proof in terms of the left-action definition of commutator
This proof follows the left-action definition of commutator:

$$[a,h] = aha^{-1}h^{-1}$$.

Given: A group $$G$$, a subgroup $$H$$ of $$G$$, a symmetric subset $$A$$ of $$N_G(H)$$.

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

Proof: Since $$A$$ is symmetric, every element of $$\langle A \rangle$$ can be written as a product of elements of $$A$$. Thus, $$[\langle A \rangle, H]$$ is generated by elements of the form:

$$[a_1a_2 \dots a_n, h]$$

it suffices to prove that every such element is in $$[A,H]$$. We do this by induction on $$n$$.

Suppose $$[a,h]$$ is in $$[A,H]$$ for all $$a$$ that can be written as products of length at most $$n - 1$$ among letters of $$A$$ and their inverses. Now, consider a commutator:

$$[a,h]$$

where $$a = a_1a_2 \dots a_n$$ with $$a_i \in A$$. Thus, we have:

$$a = a_1b$$,

where the induction hypothesis applies to $$b$$. Thus:

$$[a,h] = aha^{-1}h^{-1} = aha^{-1}(bhb^{-1})^{-1}(bhb^{-1})h^{-1} = a_1(bhb^{-1})a_1^{-1}(bhb^{-1})^{-1}(bhb^{-1})h^{-1} = [a_1,bhb^{-1}][b,h]$$.

Now, $$[b,h] \in [A,H]$$ by the induction hypothesis. Also, since $$A \subseteq N_G(H)$$, $$\langle A \rangle \le N_G(H)$$, so $$bhb^{-1} \in H$$. Thus, $$[a_1,bhb^{-1}] \in [A,H]$$. Thus, the product is in $$[A,H]$$.