Derivation-invariance is transitive

This article gives the statement, and possibly proof, of a Lie subring property (i.e., derivation-invariant Lie subring) satisfying a Lie subring metaproperty (i.e., transitive Lie subring property)
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Get more facts about derivation-invariant Lie subring |Get facts that use property satisfaction of derivation-invariant Lie subring | Get facts that use property satisfaction of derivation-invariant Lie subring|Get more facts about transitive Lie subring property

ANALOGY: This is an analogue in Lie rings of a fact encountered in group. The old fact is: characteristicity is transitive.
Another analogue to the same fact, in the same new context, is: characteristicity is transitive for Lie rings
View other analogues of characteristicity is transitive|View other analogues from group to Lie ring (OR, View as a tabulated list)

Statement

A derivation-invariant Lie subring of a derivation-invariant Lie subring is a derivation-invariant Lie subring. In symbols, if $L$ is a Lie ring with subrings $A,B$ such that $B$is a derivation-invariant Lie subring of $L$ and $A$ is a derivation-invariant Lie subring of $B$, then $A$ is a derivation-invariant Lie subring of $L$.

Related facts

• Derivation-invariant subring of ideal implies ideal

Proof

Given: A Lie ring $L$ with Lie subrings $A\le B\le L$. $B$ is a derivation-invariant Lie subring of $L$ and $A$ is a derivation-invariant subring of $B$.

To prove: $A$ is a derivation-invariant subring of $L$.

Proof: Suppose $d$ is a derivation of $L$.

Since $B$ is a derivation-invariant subring of $L$, $d$ restricts to a map from $B$ to itself. Let $d^{\prime }$ be the restriction of $d$ to $B$. Clearly, $d^{\prime }$ is a derivation of $B$.

Since $A$ is derivation-invariant in $B$, $d^{\prime }$ restricts to a map from $A$ to itself. Thus, $d$ restricts to a map from $A$ to $A$.