# Lie algebra is nilpotent iff all two dimensional subalgebras are abelian?

I'm trying to prove that if $\mathfrak{g}$ is a Lie algebra over an algebraically closed field and every 2-d subalgebra is abelian then $\mathfrak{g}$ is nilpotent.

By an induction all I need to show is that $[\mathfrak{g},\mathfrak{g}]$ is of strictly lower dimension than $\mathfrak{g}$, and I thought the easiest way to do this would be by contradiction but I can't see how to proceed.

Can anyone give me a solution to this problem?

This is for revision, not homework.

Suppose that $g$ is nilpotent, let $h$ be a 2-dimensional subalgebra of $g$, for every $x\in h$, let $ad_x:h\rightarrow h$ defined by $ad_x(y)=[x,y]$. Since $h$ is defined over an algebraic closed field, $ad_x$ has an eigenvalue. $[x,y]=cy$. If $c\neq 0$, it implies for each $n(ad_x)^n(y)\neq 0$. This is impossible since $g$ is nilpotent. Suppose that $h=Vect(x,y)$, $[x,y]=cx+dy$, $(ad_x)^2=[x,cx+dy]=d(cx+dy)$ you deduce recursively that $(ad_x)^{n+1}(y)=d^n(cx+dy)$ since $g$ is nilpotent, $d=0$. This implies that $ad_y(x)=-cx$. Since $ad_y$ is nilpotent, $c=0$. Thus $h$ is commutative.
On the other side, suppose that every subalgebra of dimension 2 is commutative, let $y$ be an eigenvector of $ad_x$, $Vect(x,y)$ is a 2-dimensional subalgebra, thus it is commutative, thus $[x,y]=0$. This implies that $ad_x$ is nilpotent an the theorem of Engel implies the result.