Commutator of a 3-subnormal subgroup and a finite subset implies subnormal

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

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

is a 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 2-subnormal subgroup and a subset implies 3-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) uses::3-subnormal implies finite-conjugate-join-closed subnormal
 * 2) uses::Product with commutator equals join with conjugate
 * 3) uses::Subgroup normalizes its commutator with any subset

Proof
Given: A group $$G$$, a 3-subnormal subgroup $$H$$, a finite subset $$A$$ of $$G$$.

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

Proof: