Lie algebra is nilpotent iff all two dimensional subalgebras are abelian Math Grip Substitute

I’m nerve-wracking to demonstrate that if $\mathfrak$ is a Lie algebra terminated an algebraically shut bailiwick and every 2-d subalgebra is abelian so $\mathfrak$ is nilpotent.

By an generalisation all I pauperization to prove is that $[\mathfrak,\mathfrak]$ is of rigorously glare assign than $\mathfrak$, and I opinion the easiest way to do this would be by contradiction but I can’t see how to go.

Can anyone leap me a resolving to this job?

This is for alteration, not cooking.

Cipher that understand nearly $g$ is nilpotent, let $h$ be a Matted subalgebra of $g$, for every $x\in h$, let $ad_x:h\rightarrow h$ outlined by $ad_x(y)=[x,y]$. Since $h$ is outlined complete an algebraical close discipline, $ad_x$ has an eigenvalue. $[x,y]=cy$. If $c\neq 0$, it implies for apiece $n(ad_x)^n(y)\neq 0$. This is impossible since $g$ is nilpotent. Hypothesis that $h=Vect(x,y)$, $[x,y]=cx+dy$, $(ad_x)^2=[x,cx+dy]=d(cx+dy)$ you conjecture recursively that $(ad_x)^(y)=d^n(cx+dy)$ since $g$ is nilpotent, $d=0$. This implies that $ad_y(x)=-cx$. Since $ad_y$ is nilpotent, $c=0$. So $h$ is commutative.

On the former discard, speculation that every subalgebra of attribute 2 is commutative, let $y$ be an eigenvector of $ad_x$, $Vect(x,y)$ is a Matte subalgebra, so it is commutative, hence $[x,y]=0$. This implies that $ad_x$ is nilpotent an the theorem of Engel implies the closing.

answered Apr 19 ’16 at 18:22

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