If $\mathfrak g$ has an invertible derivation, then $\mathfrak g$ is nilpotent.

Asked Sep 02 '18 at 18:59

Active Sep 02 '18 at 18:59

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Consider a finite dimensional Lie algebra $\mathfrak g$ and let $D$ be an invertible derivation of $\mathfrak g$. Prove that $\mathfrak g $ is nilpotent.

For any $X\in \mathfrak g$, we can write $ad (DX) = [D,ad(X)].$

Since exists $D^{-1}$, we see that $ad(X) = ad(D^{-1}DX) = [D^{-1},ad(DX)].$ So for any $X$, $ad(X)= [D^{-1},ad(DX)]$.

How can I proceed from here? I guess that I must eventually use Engel theorem.

asked Sep 02 '18 at 18:59

user2345678

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https://math.stackexchange.com/questions/2901344/if-d-is-a-derivation-of-a-lie-algebra-and-d-x-lambda-x-rightarrow-t – Tsemo Aristide Sep 02 '18 at 19:37

If $\mathfrak g$ has an invertible derivation, then $\mathfrak g$ is nilpotent.

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