Uniquely p-divisible implies subring annihilated by subring of Lie ring of derivations is also uniquely p-divisible

Statement
Suppose $$p$$ is a prime number. Suppose $$L$$ is a Lie ring that is uniquely $$p$$-divisible, i.e., for every $$a \in L$$, there is a unique $$b \in L$$ such that $$pb = a$$. Suppose $$D$$ is the fact about::Lie ring of derivations of $$L$$ and $$C$$ is a Lie subring of $$D$$. Consider the subring $$K$$ of $$L$$ given by:

$$K := \{ a \in L \mid d(a) = 0 \ \forall \ d \in C \}$$

In other words, $$K$$ is a fact about::subring annihilated by a subring of the Lie ring of derivations of the Lie ring $$L$$.

Then, $$K$$ is uniquely $$p$$-divisible, i.e., for any $$a \in K$$, there exists a unique $$b \in K$$ such that $$pb = a$$.

Analogous facts in other algebraic structures

 * Uniquely p-divisible implies fixed-point subgroup of a subgroup of the automorphism group is also uniquely p-divisible

Applications

 * Center of uniquely p-divisible Lie ring is uniquely p-divisible

Proof
Given: A prime number $$p$$. A Lie ring $$L$$ with the property that for every $$a \in L$$, there is a unique $$b \in L$$ such that $$pb = a$$. $$D$$ is the Lie ring of derivations of $$L$$ and $$C$$ is a Lie subring of $$D$$. Consider the subring $$K$$ of $$L$$ given by:

$$K := \{ a \in L \mid d(a) = 0 \ \forall \ d \in C \}$$

$$\! a$$ is an element of $$K$$.

To prove: There is a unique $$b \in K$$ such that $$pb = a$$.

Proof: The overall strategy is to first find $$b \in L$$ and then show that, in fact, $$b$$ must be in $$K$$.