# Commutator of a group and a subset implies normal

From Groupprops

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

## Contents

## Statement

Suppose is a group and is any subset of . Consider the commutator:

.

Here, denotes the commutator:

.

The subgroup is a normal subgroup of .

## 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

- Subgroup normalizes its commutator with any subset: If and , then normalizes the commutator .

## Proof

**Given**: A group , a subset .

**To prove**: is normal in .

**Proof**: By fact (1), setting , we obtain that normalizes . Thus, is normal in .