Commutator of a group and a subgroup 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 subgroup), to another known subgroup property (i.e., normal subgroup)
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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., subgroup realizable as the commutator of the whole group and a subgroup) must also satisfy the second subgroup property (i.e., normal subgroup)
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Statement

Statement with symbols

Suppose $G$ is a group and $H$ is any subgroup. Then, the commutator $[G,H]$, defined as:

$[G,H] := \langle [g,h] \mid g \in G, h \in H \rangle$

is a normal subgroup of $G$.

Related facts

Stronger facts

• Commutator of a group and a subset implies normal
• Subgroup normalizes its commutator with any subset
• Commutator of a group and a subgroup of its automorphism group is normal

Other related facts

• Commutator of two subgroups is normal in join
• Commutator of a normal subgroup and a subset implies 2-subnormal
• Commutator of a 2-subnormal subgroup and a subset implies 3-subnormal
• Normality is commutator-closed
• Characteristicity is commutator-closed

Breakdown for Lie rings

• Lie bracket of a Lie ring and a subring need not be an ideal

Facts used

1. Subgroup normalizes its commutator with any subset: If $K \le G$ and $A$ is a subset of $G$, then $K$ normalizes the commutator $[A,K] = [K,A]$.

Proof

Given: A group $G$, a subgroup $H$.

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

Proof: Apply fact (1) to $K = G$ and $A = H$. We get that $G$ normalizes $[G,H]$. Hence, $[G,H]$ is normal in $G$.