Inner derivation implies endomorphism for class two Lie ring

Statement
Suppose $$L$$ is a fact about::Lie ring of nilpotency class two. In other words, $$inner derivation of $$L$$ is an endomorphism of $$L$$ as a Lie ring.

Applications

 * Fully invariant implies ideal for class two Lie ring

Opposite facts

 * Derivation not implies endomorphism for class two Lie ring

Other related facts

 * Derivation equals endomorphism for Lie ring iff it is abelian
 * Fully invariant subgroup of additive group of Lie ring is derivation-invariant and fully invariant

Proof
Given: A Lie ring $$L$$ of nilpotency class two, an element $$x \in L$$.

To prove: The map $$y \mapsto [x,y]$$ is an endomorphism of $$L$$ as a Lie ring.

Proof: The map is clearly an endomorphism of the additive group of $$L$$, so it suffices to show that it preserves the Lie bracket. In other words, we need to show that for $$y, z \in L$$, we have:

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

But since $$L$$ has nilpotency class two, both sides are zero, so the result holds.