Commutator of a transitively normal subgroup and a subset implies normal

Statement
Suppose $$G$$ is a group, $$H$$ is a transitively normal subgroup of $$G$$ (i.e., every normal subgroup of $$H$$ is normal in $$G$$), and $$S$$ is a subset of $$G$$. Then, the commutator:

$$[H,S] = \langle [h,s] \mid h \in H, s \in S \rangle$$

is a normal 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

Facts used

 * 1) uses::Subgroup normalizes its commutator with any subset

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

To prove: $$[H,S]$$ is normal in $$G$$.

Proof:


 * 1) $$H$$ is normal in $$G$$: This follows because $$H$$ is given to be transitively normal in $$G$$.
 * 2) $$[H,S]$$ is a normal subgroup of $$H$$: Since $$H$$ is normal in $$G$$, $$[H,S]$$ is a subgroup of $$H$$. By fact (1), $$H$$ normalizes $$[H,S]$$. Thus, $$[H,S]$$ is normal in $$H$$.
 * 3) $$[H,S]$$ is a normal subgroup of $$G$$: This follows from the previous step and the fact that $$H$$ is transitively normal in $$G$$.