Commutator of a 2-subnormal subgroup and a subset implies 3-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., 2-subnormal subgroup and subset of a group), to another known subgroup property (i.e., 3-subnormal subgroup)
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Statement

Suppose $G$ is a group, $H$ is a 2-subnormal subgroup and $A$ is a subset of $G$. Then, the commutator:

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

is a 3-subnormal subgroup of $G$.

Related facts

• Commutator of a group and a subset implies normal
• Commutator of a normal subgroup and a subset implies 2-subnormal
• Commutator of a 3-subnormal subgroup and a finite subset implies subnormal
• Commutator of a group and a subgroup implies normal
• Normality is commutator-closed
• Characteristicity is commutator-closed
• Subgroup normalizes its commutator with any subset
• Product with commutator equals join with conjugate

Facts used

1. 2-subnormality is conjugate-join-closed
2. Product with commutator equals join with conjugate
3. Subgroup normalizes its commutator with any subset

Proof

Given: A group $G$, a 2-subnormal subgroup $H$, a subset $A$ of $G$.

To prove: $[A,H]$ is 3-subnormal in $G$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$\langle H,H^{A}\rangle =H[A,H]$	Fact (2)	$H$ is a subgroup of $G$.
2	$H$ normalizes $[A,H]$.	Fact (3)
3	$[A,H]$ is a normal subgroup of the subgroup $\langle H,H^{A}\rangle$.			Steps (1), (2)
4	$\langle H,H^{A}$ is 2-subnormal in $G$.	Fact (1)	$H$ is 2-subnormal in $G$.		$\langle H,H^{A}\rangle$ is a join of conjugates of $H$, so the given and fact apply.
5	$[A,H]$ is a normal subgroup of $\langle H,H^{A}\rangle$, which in turn is a 2-subnormal subgroup of $G$. Thus, $[A,H]$ is a 3-subnormal subgroup of $G$.			Steps (3), (4)	Step-combination direct.