Ideal property is centralizer-closed

This article gives the statement, and possibly proof, of a Lie subring property (i.e., ideal of a Lie ring) satisfying a Lie subring metaproperty (i.e., centralizer-closed Lie subring property)
View all Lie subring metaproperty satisfactions | View all Lie subring metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for Lie subring properties
Get more facts about ideal of a Lie ring |Get facts that use property satisfaction of ideal of a Lie ring | Get facts that use property satisfaction of ideal of a Lie ring|Get more facts about centralizer-closed Lie subring property

Statement

Suppose $L$ is a Lie ring, $S$ is an ideal of $S$, and $C=C_{L}(S)$ is the centralizer of $S$ in $L$. Then, $C$ is also an ideal of $S$.

Related facts

Analogue for group theory

• Normality is centralizer-closed

Generalizations

The general version of this result is: invariance under any set of derivations is centralizer-closed, which reduces to this result when we take the inner derivations (see Lie ring acts as derivations by adjoint action). Another special case of this general result is:

• Derivation-invariance is centralizer-closed

Proof

Hands-on proof using Jacobi identity

Given: A Lie ring $L$, an ideal $S$ of $L$. $C=C_{L}(S)$ is the centralizer of $S$ in $L$.

To prove: $C$ is an ideal of $L$, i.e., for any $x\in L$ and $c\in C$, $[x,c]\in C$.

Proof: To show $[x,c]\in C$ it suffices to show that $[[x,c],s]=0$ for all $s\in S$.

By the Jacobi identity:

$[[x,c],s]+[[c,s],x]+[[s,x],c]=0$.

Since $C$ centralizes $S$, $[c,s]=0$, so the second term is zero. Further, since $S$ is an ideal, $[s,x]\in S$, and since $C$ centralizes $S$, $[[s,x],c]=0$. Thus, both the second and third term on the left side are zero, forcing $[[x,c],s]=0$.