2011
DOI: 10.2140/involve.2011.4.293
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Jacobson’s refinement of Engel’s theorem for Leibniz algebras

Abstract: We develop Jacobson's refinement of Engel's Theorem for Leibniz algebras. We then note some consequences of the result.Since Leibniz algebras were introduced in [Loday 1993] as a noncommutative generalization of Lie algebras, one theme has been to extend Lie algebra results to Leibniz algebras. In particular, Engel's theorem has been extended in [Ayupov and Omirov 1998;Barnes 2011;Patsourakos 2007]. In the second of these works, the classical Engel's theorem is used to give a short proof of the result for Leib… Show more

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Cited by 23 publications
(18 citation statements)
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“…The second method greatly shortens the proof. The short proof given here shows the result using Lie sets and follows that in [12]. A Lie subset of a Leibniz algebra A is a subset that is closed under multiplication.…”
Section: For a Leibniz Algebra A The Series Of Idealssupporting
confidence: 53%
See 1 more Smart Citation
“…The second method greatly shortens the proof. The short proof given here shows the result using Lie sets and follows that in [12]. A Lie subset of a Leibniz algebra A is a subset that is closed under multiplication.…”
Section: For a Leibniz Algebra A The Series Of Idealssupporting
confidence: 53%
“…As an immediate consequence of Proposition 4.2, we have Several authors have given various forms of Engel's theorem for Leibniz algebras. The Leibniz versions were proven without using the standard Lie algebra result in [4] and [27], and using the standard result in [8] and [12]. The second method greatly shortens the proof.…”
Section: For a Leibniz Algebra A The Series Of Idealsmentioning
confidence: 99%
“…Thus, that nilpotency is 2-recognizeable in Leibniz algebras follows from all left multiplications being nilpotent, Engel's theorem. This result, shown in several places, can be cast as in Jacobson's refinement to Engel's theorem for Lie algebras, see [6], a result that we use.…”
mentioning
confidence: 81%
“…Proof If L is nilpotent of length less than three, then the result holds by the previous theorem, so let L be nilpotent of As D in nilpotent, D ⊆ K, we show K is a Cartan subalgebra of L. Assume y ∈ N L (K), but y / ∈ K. Then xy ∈ K, but y / ∈ S x , so xy / ∈ S x , contradiction. So y ∈ K thus K is its own normalizer in L. K = K ∩ (D + N (L)) = D + (K ∩ N (L)), and since D is nilpotent by definition, and (K ∩ N (L)) is a nilpotent ideal of K, K is nilpotent by [4]. Now assume D is contained in two Cartan subalgebras of L, E 1 and E 2 .…”
mentioning
confidence: 99%