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I'm trying to understand the proof given in this answer of the fact that finite-dimensional nilpotent Lie algebras are closed under extensions. The only step I don't understand in the argument is when the text claims that $\operatorname{ad}_x|_K$ is nilpotent. Why is so? I assume here one is using the hypothesis that says that $K$ is nilpotent, but how?

Elías Guisado Villalgordo
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    It is not true that finite-dimensional nilpotent Lie algebras are closed under extensions. In the post you link to, the hypothesis "all $ad_x\vert _K$ are nilpotent" is explicitly stated in the question. – Torsten Schoeneberg May 02 '22 at 14:24

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We need the assumption that ${\rm ad}(x)|_K$ is nilpotent for all $x \in L$. Otherwise the claim is false. In fact, nilpotent Lie algebras are not closed under extensions in general. Consider the extension of the abelian (hence nilpotent) $1$-dimensional Lie algebra $K$ by the abelian (nilpotent) Lie algebra $K$ given by the short exact sequence $$ 0\rightarrow K \rightarrow \mathfrak{r}_2(K)\rightarrow K\rightarrow 0 $$ where $\mathfrak{r}_2(K)$ is the non-abelian Lie algebra of dimension $2$ with basis $(x,y)$ and $[x,y]=x$, which is not nilpotent.

Dietrich Burde
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