Derivation equals endomorphism for Lie ring iff it is abelian

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


 * 1) $$L$$ is an fact about::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

(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.