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 $G$.

Related facts

Applications

• Product with commutator equals join with conjugate: If $H\leq G$ and $g\in G$, $H[g,H]=\langle H,H^{g}\rangle$.
• Commutator of two subgroups is normal in join

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.

With the left-action convention

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.