Derivation equals endomorphism for Lie ring iff it is abelian

Statement

The following are equivalent for a Lie ring $L$:

1. $L$ is an Abelian Lie ring (?), i.e., the Lie bracket on $L$ is identically zero.
2. A function from $L$ to itself is a derivation of $L$ if and only if it is an endomorphism of $L$.
3. Every endomorphism of $L$ is a derivation of $L$.
4. Every automorphism of $L$ is a derivation of $L$.
5. The identity map is a derivation of $L$.

Related facts

• Inner derivation implies endomorphism for class two Lie ring
• Fully invariant implies ideal for class two Lie ring

Proof

(1) implies (2)

If $L$ is an abelian Lie ring, then the Lie bracket on $L$ is identically zero. Thus, the Leibniz rule for derivations as well as the Lie bracket-preservation condition for endomorphisms is satisfied by all functions from $L$ to itself. Thus, the following are equivalent for a function from $L$ to itself:

• It is a derivation of $L$.
• It is an endomorphism of the underlying abelian group of $L$.
• It is an endomorphism of $L$ as a Lie ring.

This shows that (2) holds.

(2) implies (3) implies (4) implies (5)

This is obvious.

(5) implies (1)

Suppose the identity map is a derivation of $L$. Then, for any $x,y\in L$, we have, by the Leibniz rule:

$[x,y]=[x,y]+[x,y]$.

This simplifies to $[x,y]=0$, so $L$ is abelian.