This as a very enjoyable post. Thanks for posting it.

Regarding remark 3, I don’t know if this qualifies as a more “direct” proof but corollary 8 can be proved via Lie’s theorem and vice-versa. In one direction, corollary 8 can be proved using Lie’s theorem by first showing that any derivation maps the radical of into itself and second by using Lie’s theorem on the solvable ideal , where is the image of in under the ad map, to show that is nilpotent.

Conversely, corollary 8 can be used to prove Lie’s theorem inductively by taking a solvable Lie algebra , a codimension 1 ideal , and then showing that the space generated by 1-dimensional ideals of is both a non-trivial subspace (by induction) and an ideal in Showing that is an ideal requires cor. 8 and this produces a common eigenvector of the adjoint representation, showing that every solvable ideal has a complete flag consisting of ideals. From there, we can work out Lie’s theorem for any representation by considering the direct sum , where is considered as an abelian ideal.

As a bonus, theorem 7 also follows from cor. 8 and Lie’s theorem.

Best Regards,

Alejandro Ginory

*[Corrected, thanks – T.]*

I was wondering why one need to take the ideal A to be of codimension 1? Already in the nilpotent case for example, i don’t see in which step we need A to satisfy that (also do we need H to be abelian?). I get the feeling that it is enough to be able to decompose the lie algebra into A \oplus H with A an ideal of the lie algebra.

Thanks for your answer!

Best regards

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