Derivation equals endomorphism for Lie ring iff it is abelian
The following are equivalent for a Lie ring :
- is an Abelian Lie ring (?), i.e., the Lie bracket on is identically zero.
- A function from to itself is a derivation of if and only if it is an endomorphism of .
- Every endomorphism of is a derivation of .
- Every automorphism of is a derivation of .
- The identity map is a derivation of .
- Inner derivation implies endomorphism for class two Lie ring
- Fully invariant implies ideal for class two Lie ring
(1) implies (2)
If is an abelian Lie ring, then the Lie bracket on 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 to itself. Thus, the following are equivalent for a function from to itself:
- It is a derivation of .
- It is an endomorphism of the underlying abelian group of .
- It is an endomorphism of 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 . Then, for any , we have, by the Leibniz rule:
This simplifies to , so is abelian.