Commutator of a group and a subset implies normal

Statement
Suppose $$G$$ is a group and $$A$$ is any subset of $$G$$. Consider the commutator:

$$[A,G] := \langle [a,g] \mid a \in A, g \in G \rangle$$.

Here, $$[a,g]$$ denotes the commutator:

$$[a,g] = a^{-1}g^{-1}ag$$.

The subgroup $$[A,G]$$ is a normal subgroup of $$G$$.

Properties we can prove about the subgroup obtained as the commutator
When both subgroups are of the same kind:


 * Normality is commutator-closed
 * Characteristicity is commutator-closed
 * Commutator of subnormal subgroups is subnormal iff their join is subnormal

When one is a particularly nice subgroup and the other is arbitrary:


 * Commutator of a group and a subgroup implies normal
 * Commutator of a group and a subgroup of its automorphism group 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 3-subnormal subgroup and a finite subset implies subnormal
 * Commutator of two subgroups is normal in join

Properties we cannot prove about the subgroup obtained as the commutator

 * Commutator of a normal subgroup and a subset 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

Relation with commutators

 * Subgroup normalizes its commutator with any subset
 * Commutator of a group and a subgroup of its automorphism group is normal
 * Commutator of subgroup with normalizing symmetric subset equals commutator with subgroup generated

Facts used

 * 1) uses::Subgroup normalizes its commutator with any subset: If $$H \le G$$ and $$A \subseteq G$$, then $$H$$ normalizes the commutator $$[A,H]$$.

Proof
Given: A group $$G$$, a subset $$A \subseteq G$$.

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

Proof: By fact (1), setting $$H = G$$, we obtain that $$G$$ normalizes $$[A,G]$$. Thus, $$[A,G]$$ is normal in $$G$$.