Suppose L is a finite dimensional Lie Algebra over an algebraically closed field of characteristic zero. I want to show that $[L,rad(L)]$ is nilpotent.
I was given a hint that all operators of the form $ad_{[L,rad(L)]}(x)$ are nilpotent which proves this result via Engel's Theorem. I am, however, unclear on how one shows this. I assume Lie's Theorem is used, but I can't figure out how.
There is an elementary proof as follows. Denote by $R=rad(L)$ and set $L_1:=R+\langle y\rangle$ for an element $y\in L$. Then we have $[L_1,L_1]\subseteq [R,R]+[R,\langle y\rangle]\subseteq R$, so that $[L_1,L_1]$ is a solvable ideal of $L$, and hence $L_1$ is solvable, too. It follows that $[L_1,L_1]$ is nilpotent, i.e., $ad(x)$ is nilpotent for all $x\in [L_1,L_1]$. Now let $x=[a,b]\in[R,L]$ a pure commutator. Then there is a $y\in L$ with $b\in R+\langle y\rangle$, i.e., we have $x\in [L_1,L_1]$, so that $ad(x)$ is nilpotent. Now since $y$ is arbitrary, $x=[a,b]$ runs through whole $[R,L]$, which yields that $ad(x)$ is nilpotent for all $x\in [R,L]$. By Engel's theorem, $[R,L]$ is nilpotent.
More generally one can show that $D(rad(L))\subseteq nil(L)$ for every derivation $D$ of $L$. This uses the Killing form and Cartan's solvability criterion. For an inner derivation $D=ad(x)$ it follows that $ad(L)(rad(L))=[L,R]$ is a nilpotent ideal in $L$.
