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Let be $\mathcal{L}$ a finite-dimensional Lie algebra. How I can prove that if every $2-$dimensional Lie subalgebra of $\mathcal{L}$ is abelian, then $\mathcal{L}$ is nilpotent?

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  • $\begingroup$ You mean complex Lie algebra. This result is false, say, for real Lie algebras (e.g., the Lie algebra of the group of motions of the Euclidean plane has only abelian 2-dimensional subalgebras, but is not nilpotent). $\endgroup$ – YCor Jun 2 '16 at 21:31
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Let $x\in L$ and consider the adjoint operator $ad(x)$ for $x\in L$. Its eigenvectors $y\neq 0$ are given by $[x,y]=\lambda y$. Suppose that $\lambda\neq 0$. Then $\langle x,y\rangle $ is a $2$-dimensional non-abelian subalgebra, a contradiction to the assumption. Hence $\lambda=0$ and all adjoint operators have only $\lambda=0$ as eigenvalue. Hence they are all nilpotent. By Engel's theorem, $L$ is nilpotent.

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    $\begingroup$ You are welcome. The other direction is also true. If $L$ is nilpotent, then all $2$-dimensional subalgebras are abelian. $\endgroup$ – Dietrich Burde May 24 '16 at 19:34
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    $\begingroup$ I have just proved the other direction. My problem was this direction. $\endgroup$ – Vincenzo Zaccaro May 24 '16 at 19:35

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