# About nilpotency of Leibniz algebras

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.

• Why do you not think it is true? Is there some problem with the proof? – Tobias Kildetoft Feb 8 '17 at 11:55
• @TobiasKildetoft:yes there is a problem in proof.we have $L^{m+n}\supseteq [L^m,L^n]$ but in this proof use converse – pink floyd Feb 8 '17 at 12:40
• link.springer.com/content/pdf/10.1007/978-94-011-5072-9_1.pdf you can read this – pink floyd Feb 8 '17 at 12:43
• 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. – Tobias Kildetoft Feb 8 '17 at 12:44
• @TobiasKildetoft:llemma:for any $i$,$j\in \mathbb Z [L^i,L^j]\subseteq L^{i+j}$ this article prove this – pink floyd Feb 8 '17 at 12:52