Centralizer-free ideal implies automorphism-faithful

This article gives the statement and possibly, proof, of an implication relation between two Lie subring properties. That is, it states that every Lie subring satisfying the first Lie subring property (i.e., centralizer-free ideal) must also satisfy the second Lie subring property (i.e., automorphism-faithful Lie subring)
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ANALOGY: This is an analogue in Lie rings of a fact encountered in group. The old fact is: normal and centralizer-free implies automorphism-faithful.
Another analogue to the same fact, in the same new context, is: centralizer-free ideal implies derivation-faithful
View other analogues of normal and centralizer-free implies automorphism-faithful|View other analogues from group to Lie ring (OR, View as a tabulated list)

Statement

Suppose $L$ is a Lie ring and $I$ is a centralizer-free ideal of $L$, i.e., $I$ is an ideal of $L$ and its centralizer in $L$ is zero. Then, $I$ is an automorphism-faithful Lie subring (and hence an automorphism-faithful ideal) of $L$: Any non-identity automorphism of $L$ that restricts to an automorphism of $I$ restricts to a non-identity automorphism of $I$.

Related facts

Similar facts for Lie rings

• Centralizer-free ideal implies derivation-faithful

Analogues in groups

• Normal and centralizer-free implies automorphism-faithful
• Normal and self-centralizing implies coprime automorphism-faithful

Proof

Given: A Lie ring $L$, a centralizer-free ideal $I$ of $L$. An automorphism $\sigma$ of $L$ such that the restriction of $\sigma$ to $I$ is the identity map.

To prove: $\sigma (l)=l$ for all $l\in L$.

Proof: By the definition of automorphism, we have, for every $a\in A$:

$\!\sigma ([l,a])=[\sigma (l),\sigma (a)]$.

Since $A$ is an ideal and $a\in A$, $[l,a]\in A$. Thus, $\sigma ([l,a])=[l,a]$. Also, $\sigma (a)=a$. We thus have:

$\![l,a]=[\sigma (l),a]$.

By the biadditivity of the Lie bracket, this gives:

$\![\sigma (l)-l,a]=0$.

In other words, $\sigma (l)-l\in C_{L}(A)$. By assumption, $C_{L}(A)=0$, so $\sigma (l)=l$, completing the proof.