Fully invariant Lie subring

This article describes a Lie subring property: a property that can be evaluated for a subring of a Lie ring
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ANALOGY: This is an analogue in Lie ring of a property encountered in group. Specifically, it is a Lie subring property analogous to the subgroup property: fully invariant subgroup
View other analogues of fully invariant subgroup | View other analogues in Lie rings of subgroup properties (OR, View as a tabulated list)

Definition

A subring $S$ of a Lie ring $L$ is termed a fully invariant Lie subring if, for every endomorphism $\sigma$ of $L$, $\sigma (S)\subseteq S$.

Relation with properties in related groups

Lazard Lie ring

Suppose $G$ is a Lazard Lie group and $L$ is a Lazard Lie ring. Under the natural bijection between $L$ and $G$, fully invariant subrings of $L$ correspond to fully invariant subgroups of $G$.

Relation with other properties

Stronger properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
Fully invariant subgroup of additive group of a Lie ring		Fully invariant subgroup of additive group of Lie ring is derivation-invariant and fully invariant
Lie subring invariant under any additive endomorphism satisfying a comultiplication condition

Weaker properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
Characteristic subring of a Lie ring

Metaproperties

Metaproperty name	Satisfied?	Proof	Statement with symbols
transitive Lie subring property	Yes	full invariance is transitive for Lie rings	If $A\leq B\leq L$ are Lie rings, with $A$ fully invariant in $B$ and $B$ fully invariant in $L$, then $A$ is fully invariant in $L$.
Lie bracket-closed Lie subring property	Yes	full invariance is Lie bracket-closed for Lie rings	If $A,B$ are fully invariant Lie subrings of a Lie ring $L$, so is $[A,B]$.
strongly intersection-closed Lie subring property	Yes	full invariance is strongly intersection-closed for Lie rings	If $A_{i},i\in I$ are all fully invariant Lie subrings of a Lie ring $L$, then so is $\bigcap _{i\in I}A_{i}$.