Commutator of a transitively normal subgroup and a subset implies normal

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., transitively normal subgroup) must also satisfy the second subgroup property (i.e., subgroup whose commutator with any subset is normal)
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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:

$[H,S]=⟨[h,s]\mid h\in H,s\in S⟩$

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. 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 $G$.