# Nilpotent Lie Algebras and 2-dimensional Lie Subalgebras

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?

• 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). – YCor Jun 2 '16 at 21:31

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.
• You are welcome. The other direction is also true. If $L$ is nilpotent, then all $2$-dimensional subalgebras are abelian. – Dietrich Burde May 24 '16 at 19:34