Commutator of a normal subgroup and a subset implies 2-subnormal

This article describes a computation relating the result of the commutator operator on two known subgroup properties or properties of subsets of groups: (i.e., normal subgroup and subset of a group), to another known subgroup property (i.e., 2-subnormal subgroup)
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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup realizable as the commutator of a normal subgroup and a subset) must also satisfy the second subgroup property (i.e., 2-subnormal subgroup)
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Statement

Suppose $G$ is a group, $H$ is a normal subgroup of $G$, and $A$ is any subset of $G$. Then, the subgroup:

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

is a 2-subnormal subgroup of $G$.

Related facts

Commutators between things of the same type

• Normality is commutator-closed: The commutator of two normal subgroups is normal.
• Characteristicity is commutator-closed
• Commutator of subnormal subgroups is subnormal iff their join is subnormal

Similar facts

• Commutator of a group and a subset implies normal
• Commutator of a 2-subnormal subgroup and a subset implies 3-subnormal
• Commutator of two subgroups is normal in join
• Commutator of a 3-subnormal subgroup and a finite subset implies subnormal

Opposite facts

• Commutator of a normal subgroup and a subgroup not implies normal
• Commutator of a 3-subnormal subgroup and a subset not implies subnormal
• Commutator of a 3-subnormal subgroup and a finite subset not implies 4-subnormal

Facts used

1. Subgroup normalizes its commutator with any subset: If $H\leq G$ and $A$ is a subset of $G$, $H$ normalizes $[A,H]$.

Proof

Given: A group $G$, a normal subgroup $H$ of $G$, a subset $A$ of $G$.

To prove: $[A,H]$ is 2-subnormal in $G$.

Proof:

1. (Given data used: $H$ is normal in $G$): Since $H$ is normal in $G$, $[G,H]\leq H$. Thus, $[A,H]\leq H$.
2. (Fact used: fact (1)): By fact (1), $H$ normalizes $[A,H]$. Combining this with step (1), we get that $[A,H]$ is normal in $H$.
3. We thus have that $[A,H]$ is normal in $H$, and $H$ is normal in $G$. Thus, $[A,H]$ is 2-subnormal in $G$.