# 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)
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:

## 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 $G$.
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