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I'm studying about Lie algebras using J.E. Humphreys' book ("Introduction to Lie Algebras and Representation Theory"). On page 19 he says:

It is obvious that $L$ will be solvable if $[LL]$ is nilpotent.

But it is not obvious to me. Why is it true?

Silvia Ghinassi
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Combalge
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1 Answers1

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Suppose that $[L,L]$ is nilpotent. Now $L/[L,L]$ is abelian, hence solvable. It follows that both $[L,L]$ and $L/[L,L]$ are solvable. Then $L$ is solvable, too, because every extension of a solvable Lie algebra by a solvable Lie algebra is itself solvable (we have the extension $0\rightarrow [L,L]\rightarrow L\rightarrow L/[L,L]\rightarrow 0$). The argument is also valid over fields of characteristic $p>0$, whereas the converse statement, i.e., that $L$ solvable implies $[L,L]$ nilpotent, need not be true in that case.

Dietrich Burde
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