# Class two implies commutator map is endomorphism

## Statement

### Statement with left-action convention

Suppose $G$ is a nilpotent group whose nilpotency class is two. Then, for any element $x \in G$, the maps: $\! y \mapsto [x,y]$

and: $\! y \mapsto [y,x]$

are endomorphisms of $G$. Here: $\! [x,y] = xyx^{-1}y^{-1}$

These endomorphisms map to inverse elements, so their images coincide.

The image of these endomorphism lie in the derived subgroup of $G$, hence in the center of $G$, so it is abelian. The kernel of this endomorphism contains the center of the group, more specifically, it is the centralizer of $x$ in $G$.

### Statement with right-action convention

Suppose $G$ is a nilpotent group whose nilpotency class is two. Then, for any element $x \in G$, the maps: $\! y \mapsto [x,y]$

and: $\! y \mapsto [y,x]$

are endomorphisms of $G$. Here: $\! [x,y] = x^{-1}y^{-1}xy$

These endomorphisms map to inverse elements, so their images coincide.

The image of these endomorphism lie in the derived subgroup of $G$, hence in the center of $G$, so it is abelian. The kernel of this endomorphism contains the center of the group, more specifically, it is the centralizer of $x$ in $G$.

## Proof

### Proof with the left-action convention

CONVENTION WARNING: This article/section uses the left-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.

Given: A group $G$ of nilpotency class two, an element $x \in G$

To prove: The map $y \mapsto [x,y]$ is an endomorphism of $G$

Proof: It suffices to show that if $y_1,y_2 \in G$, then: $[x,y_1y_2] = [x,y_1][x,y_2]$

The crucial fact we use is that since $G$ has nilpotency class two, the commutator $[x,y_2]$ is in the center, and hence it commutes with $y_1$.

Thus: $[x,y_2] = y_1[x,y_2]y_1^{-1} = y_1xy_2x^{-1}y_2^{-1}y_1^{-1}$

Plugging this in, we get: $[x,y_1y_2] = xy_1x^{-1}y_1^{-1}y_1xy_2x^{-1}y_2^{-1}y_1^{-1} = xy_1y_2x^{-1}y_2^{-1}y_1^{-1} = [x,y_1y_2]$

(an analogous proof works with the other convention for commutators).