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

amin
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    Welcome to MSE! It helps to add your thoughts/efforts so users can see where it is you're having trouble with your question. – Vincent Aug 04 '14 at 08:54

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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).

Dietrich Burde
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