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Some believe that if $L$ is a nilpotent Leibniz algebra and $N$ is a nilpotent ideal such that $N\subset Z^l(L)$ and $L/N$ is nilpotent then $L$ is nilpotent. In this theorem 3.1 I read a proof of this and I think is not true. I want a counterexample to show that this is not true.

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    Why do you not think it is true? Is there some problem with the proof? – 2017-02-08
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    @TobiasKildetoft:yes there is a problem in proof.we have $L^{m+n}\supseteq [L^m,L^n]$ but in this proof use converse – 2017-02-08
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    http://link.springer.com/content/pdf/10.1007/978-94-011-5072-9_1.pdf you can read this – 2017-02-08
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    No, I cannot read that as it is behind a paywall and I am not at a university with access right now. Anyway, you should include the specific issues you have with the proof in the question itself. – 2017-02-08
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    @TobiasKildetoft:llemma:for any $i$,$j\in \mathbb Z [L^i,L^j]\subseteq L^{i+j}$ this article prove this – 2017-02-08
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    But having inclusion in one direction does not rule out equality in special cases. – 2017-02-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53244/discussion-between-parisa-and-tobias-kildetoft). – 2017-02-08
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    Theorem 3.1 seems to say that L is nilpotent if and only if L/N^2 is nilpotent. – 2017-02-08
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    But that isn't what the Theorem is saying. It talks about L factored by the derived subalgebra of N (which is the Frattini subalgebra of N). – 2017-02-08
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    @DavidTowers:I want to prove theorem ,but I have problem where it use equation $L^{m+n}=[L^m,L^n]$ – 2017-02-08
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    @DavidTowers:the proof of this theorem is same to theorem 2 of "Some characterizations of nilpotent lie algebras cHAo".I think the proof of this theorem have problem. – 2017-02-08

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This result follows from Theorem 5.5 of Barnes, D. (2011). Some theorems on Leibniz algebras. Comm. Algebra 39(7):2463–2472.