# Subgroup normalizes its commutator with any subset

## Contents

## Statement

### Statement with the right-action convention

Suppose is a group, is a subgroup of , and is a subset of . Then, normalizes the following subgroup of :

.

Here, denotes the commutator of the two elements.

Note that since , the subgroup equals the subgroup .

### Statement with the left-action convention

Suppose is a group, is a subgroup of , and is a subset of . Then, normalizes the following subgroup of :

.

Here, denotes the commutator of the two elements.

Note that since , the subgroup equals the subgroup .

### Difference between the two statements

Note also that the subgroup defined in the right-action convention is the subgroup as per the left-action convention. However, the truth of the two statements is still equivalent since the quantification is over *all* subsets of .

## Related facts

### Applications

- Product with commutator equals join with conjugate: If and , .
- Commutator of two subgroups is normal in join

## Facts used

- Formula for commutator of element and product of two elements: With the right-action convention, this is expressed as:

.

where . and .

With the left-action convention, it is expressed as:

where and .

## Proof

### With the right-action convention

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

**To prove**: normalizes .

**Proof**: Since (with ) form a generating set for , it suffices to show that for any . Let's do this. By fact (1), we have:

.

This rearranges to give:

.

Note that since , . Thus, the right side is a left quotient of two elements in , hence it is in , completing the proof.

### With the left-action convention

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