Subgroup normalizes its commutator with any subset

Statement with the right-action convention
Suppose $$G$$ is a group, $$H$$ is a subgroup of $$G$$, and $$A$$ is a subset of $$G$$. Then, $$H$$ normalizes the following subgroup of $$G$$:

$$[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle$$.

Here, $$[a,h] := a^{-1}h^{-1}ah$$ denotes the commutator of the two elements.

Note that since $$[a,h]^{-1} = [h,a]$$, the subgroup $$[A,H]$$ equals the subgroup $$[H,A]$$.

Statement with the left-action convention
Suppose $$G$$ is a group, $$H$$ is a subgroup of $$G$$, and $$A$$ is a subset of $$G$$. Then, $$H$$ normalizes the following subgroup of $$G$$:

$$[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle$$.

Here, $$[a,h] := aha^{-1}h^{-1}$$ denotes the commutator of the two elements.

Note that since $$[a,h]^{-1} = [h,a]$$, the subgroup $$[A,H]$$ equals the subgroup $$[H,A]$$.

Difference between the two statements
Note also that the subgroup $$[A,H]$$ defined in the right-action convention is the subgroup $$[A^{-1},H]$$ as per the left-action convention. However, the truth of the two statements is still equivalent since the quantification is over all subsets of $$G$$.

Applications

 * Product with commutator equals join with conjugate: If $$H \le G$$ and $$g \in G$$, $$H[g,H] = \langle H, H^g \rangle$$.
 * Commutator of two subgroups is normal in join

Facts used

 * 1) uses::Formula for commutator of element and product of two elements: With the right-action convention, this is expressed as:

$$\! [x,yz] = [x,z][x,y]^z$$.

where $$a^g = g^{-1}ag$$. and $$[a,b] = a^{-1}b^{-1}ab$$.

With the left-action convention, it is expressed as:

$$\! [x,yz] = [x,y]c_y([x,z])$$

where $$[g,h] := ghg^{-1}h^{-1}$$ and $$c_g(h) := ghg^{-1}$$.

With the right-action convention
Given: A group $$G$$, a subgroup $$H$$, a subset $$A$$ of $$G$$.

To prove: $$H$$ normalizes $$[A,H]$$.

Proof: Since $$[a,h]$$ (with $$a \in A, h \in H$$) form a generating set for $$H$$, it suffices to show that $$[a,h]^k \in [A,H]$$ for any $$k \in H$$. Let's do this. By fact (1), we have:

$$\! [a,hk] = [a,k][a,h]^k$$.

This rearranges to give:

$$\! [a,h]^k = [a,k]^{-1}[a,hk]$$.

Note that since $$k,h \in H$$, $$hk \in H$$. Thus, the right side is a left quotient of two elements in $$[A,H]$$, hence it is in $$[A,H]$$, completing the proof.