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 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 Derivation-invariant Lie subring (?) of $L$, i.e., any derivation of $L$ sends $S$ to itself.
3. $S$ is a Fully invariant Lie subring (?) of $L$: any endomorphism of $L$ as a Lie ring sends $S$ to itself.

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.