Fully invariant subgroup of additive group of Lie ring is derivation-invariant and fully invariant

Statement
Suppose $$L$$ is a Lie ring. Suppose $$S$$ is a fact about::fully invariant subgroup of the additive group of $$L$$. Then:


 * 1) $$S$$ is a Lie subring of $$L$$, i.e., $$S$$ is closed under the Lie ring operations of $$L$$.
 * 2) $$S$$ is a fact about::derivation-invariant Lie subring of $$L$$, i.e., any derivation of $$L$$ sends $$S$$ to itself.
 * 3) $$S$$ is a fact about::fully invariant Lie subring of $$L$$: any endomorphism of $$L$$ as a Lie ring sends $$S$$ to itself.

Similar facts

 * Homomorph-containing subgroup of additive group of Lie ring is self-derivation-invariant and homomorph-containing

Applications

 * Characteristic subgroup of additive group of odd-order Lie ring is derivation-invariant and fully invariant

Proof
(Using notation as in the statement above).

Since $$S$$ is a subgroup of $$L$$ by definition, proving (1) and (2) only requires us to show that $$S$$ is closed under arbitrary derivations (this would imply both (1) and (2) since the Lie bracket with an element of $$S$$ is itself an inner derivation). This, in turn, follows from the fact that any derivation is an endomorphism of the underlying abelian group structure, and $$S$$, being fully invariant, is thus sent to itself by this endomorphism.

For (3), note that any endomorphism of $$L$$ as a Lie ring is also an endomorphism of the underlying abelian group structure of $$L$$. Since $$S$$ is fully invariant, the endomorphism must send $$S$$ to within itself.