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

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 $$G$$.

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) uses::Subgroup normalizes its commutator with any subset: If $$H \le G$$ 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 $$G$$, $$[G,H] \le H$$. Thus, $$[A,H] \le H$$.
 * 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 $$G$$.