Commutator of a group and a subset implies normal

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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)
View a complete list of commutator computations

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:

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

Properties we cannot prove about the subgroup obtained as the commutator

Relation with commutators

Facts used

  1. 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.