# Subgroup normalizes its commutator with any subset

## Statement

### Statement with the right-action convention

Suppose $G$ is a group, $H$ is a subgroup of $G$, and $A$ is a subset of $G$. Then, $H$ normalizes the following subgroup of $G$:

$[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle$.

Here, $[a,h] := a^{-1}h^{-1}ah$ denotes the commutator of the two elements.

Note that since $[a,h]^{-1} = [h,a]$, the subgroup $[A,H]$ equals the subgroup $[H,A]$.

### Statement with the left-action convention

Suppose $G$ is a group, $H$ is a subgroup of $G$, and $A$ is a subset of $G$. Then, $H$ normalizes the following subgroup of $G$:

$[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle$.

Here, $[a,h] := aha^{-1}h^{-1}$ denotes the commutator of the two elements.

Note that since $[a,h]^{-1} = [h,a]$, the subgroup $[A,H]$ equals the subgroup $[H,A]$.

### Difference between the two statements

Note also that the subgroup $[A,H]$ defined in the right-action convention is the subgroup $[A^{-1},H]$ as per the left-action convention. However, the truth of the two statements is still equivalent since the quantification is over all subsets of $G$.

## Facts used

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

$\! [x,yz] = [x,z][x,y]^z$.

where $a^g = g^{-1}ag$. and $[a,b] = a^{-1}b^{-1}ab$.

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

$\! [x,yz] = [x,y]c_y([x,z])$

where $[g,h] := ghg^{-1}h^{-1}$ and $c_g(h) := ghg^{-1}$.

## Proof

### With the right-action convention

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

To prove: $H$ normalizes $[A,H]$.

Proof: Since $[a,h]$ (with $a \in A, h \in H$) form a generating set for $H$, it suffices to show that $[a,h]^k \in [A,H]$ for any $k \in H$. Let's do this. By fact (1), we have:

$\! [a,hk] = [a,k][a,h]^k$.

This rearranges to give:

$\! [a,h]^k = [a,k]^{-1}[a,hk]$.

Note that since $k,h \in H$, $hk \in H$. Thus, the right side is a left quotient of two elements in $[A,H]$, hence it is in $[A,H]$, completing the proof.