Inner derivation implies endomorphism for class two Lie ring

ANALOGY: This is an analogue in Lie rings of a fact encountered in group. The old fact is: class two implies commutator map is endomorphism.
View other analogues of class two implies commutator map is endomorphism|View other analogues from group to Lie ring (OR, View as a tabulated list)

Statement

Suppose $L$ is a Lie ring of nilpotency class two (?). In other words, $[[x,y],z]=0$ for all $x,y,z\in L$. Then, any inner derivation of $L$ is an endomorphism of $L$ as a Lie ring.

Related facts

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.