Let ‎$‎L‎$‎ be a Lie algebra. why if ‎$‎L‎$ ‎be‎ supersolvable then ‎$‎L'=[L,L]‎$ ‎ is nilpotent.‎

Question

Let ‎$‎L‎$‎ be a Lie algebra. why if ‎$‎L‎$ ‎be‎ supersolvable then ‎$‎L'=[L,L]‎$ ‎(derived algebra) is nilpotent.‎

Answer 1

Every supersolvable Lie algebra is solvable. For a solvable Lie algebra $L$, over an algebraically closed field of characteristic zero, Lie's theorem implies that $[L,L]$ is nilpotent - the adjoint operators $ad(x)$ can be simultaneously put into upper-triangular form, and then the adjoint operators of $[L,L]$ are strictly upper-triangular, hence nilpotent. By Engel's theorem, $[L,L]$ is nilpotent.

Note that $[L,L]$ need not be nilpotent for solvable Lie algebras over arbitrary fields (there are counter examples).