Commutator of a group and a subset implies normal

This article describes a computation relating the result of the commutator operator on two known subgroup properties or properties of subsets of groups: (i.e., improper subgroup and subset of a group), to another known subgroup property (i.e., normal subgroup)
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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 $G$.

Related facts

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. Subgroup normalizes its commutator with any subset: If $H\leq G$ 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 $G$.