Characteristic not implies derivation-invariant: Difference between revisions
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{{Lie subring property non-implication| | |||
stronger = characteristic subring of a Lie ring| | |||
weaker = derivation-invariant Lie subring}} | |||
==Statement== | ==Statement== | ||
A [[fact about::characteristic subring of a Lie ring]] need not be a [[fact about::derivation-invariant Lie subring|derivation-invariant subring]]. | A [[fact about::characteristic subring of a Lie ring;2| ]][[characteristic subring of a Lie ring]] need not be a [[fact about::derivation-invariant Lie subring;2| ]][[derivation-invariant Lie subring]]. | ||
==Related facts== | ==Related facts== | ||
Latest revision as of 15:12, 30 June 2013
This article gives the statement and possibly, proof, of a non-implication relation between two Lie subring properties. That is, it states that every Lie subring satisfying the first Lie subring property (i.e., characteristic subring of a Lie ring) need not satisfy the second Lie subring property (i.e., derivation-invariant Lie subring)
View a complete list of Lie subring property non-implications | View a complete list of Lie subring property implications
Get more facts about characteristic subring of a Lie ring|Get more facts about derivation-invariant Lie subring
Statement
A characteristic subring of a Lie ring need not be a derivation-invariant Lie subring.
Related facts
Similar facts
Opposite facts
- Fully invariant subgroup of additive group of Lie ring is derivation-invariant and fully invariant
- Characteristic subgroup of additive group of odd-order Lie ring is derivation-invariant and fully invariant
- Derivation equals endomorphism for Lie ring iff it is abelian
- Inner derivation implies endomorphism for class two Lie ring
Facts used
Proof
By fact (1), there exists a finite abelian group with a characteristic subgroup that is not fully invariant in . Consider as an abelian Lie ring, with trivial Lie bracket. Then, the automorphisms of as a Lie ring are the same as the automorphisms as a group, so is a characteristic subring of . Further, the derivations of as a Lie ring are precisely its endomorphisms as a group (because the Leibniz rule condition is vacuous). Thus, since is not fully invariant by assumption, is not derivation-invariant.