Prove the radical $N$ is nilpotent.

Question

I have been reading the called "Structure of Real Lie Algebras" by P. Turkowski. I have got hard time to prove the following theorem THEOREM 2.2. Let $R(x) \in Der(N)$ for all x in $S$, and let it define the semidirect sum $L = N\oplus S$. For a reducible representation $R$ of the Levi factor $S$ (completely reducible since $S$ is semisimple) that does not possess a zero representation in its decomposition into the direct sum of irreducible representations, the radical $N$ is nilpotent.

please refer to page 200 of following paper: http://www.sciencedirect.com/science/article/pii/002437959290259D

Answer 1

It's quite hard to read this paper as it's written in a very non-standard language.

In more standard language:

Let $\mathfrak{g}$ be a Lie algebra (over a field of characteristic zero of the form $\mathfrak{r}\rtimes\mathfrak{s}$, with $\mathfrak{r}$ solvable and $\mathfrak{s}$ semisimple. Suppose that the representation of $\mathfrak{s}$ on the abelianization of $\mathfrak{r}$, induced by the adjoint action, does not contain the trivial 1-dimensional representation. Then $\mathfrak{r}$ is nilpotent.

Proof: we can suppose the field algebraically closed. Let's prove that for every finite-dimensional representation, the image of $\mathfrak{r}$ is conjugate to strictly upper triangular matrices. We argue by induction on the dimension $d$ of the rep. For $d=1$, this is clear since $\mathfrak{r}$ is contained (by the assumption) in the derived subalgebra. For $d>1$, if the rep is reducible, we have a triangular block decomposition and we can argue using the inductive hypothesis. So it remains to deal with the case the rep is irreducible. The image of $\mathfrak{r}$ can be made triangular; in particular $[\mathfrak{r},\mathfrak{r}]$ has a common eigenvector (for the eigenvalue 0); the set of such vectors and zero form an invariant subspace, and by irreducibility we deduce that $[\mathfrak{r},\mathfrak{r}]$ acts trivially. So the action of $\mathfrak{r}$, since it factors through its abelianization, decomposes into characteristic subspaces and since $\mathfrak{r}$ is an ideal, they are preserved by $\mathfrak{s}$. Thus, by irreducibility, there is a single characteristic subspace, for some eigenvalue $t$. Since $\mathfrak{r}$ is contained in the derived subalgebra (by the main assumption), it acts with trace zero and hence $t=0$. This finishes the induction step.

The induction being proved, we apply it to any representation, with the conclusion that the image of $\mathfrak{r}$ is nilpotent. Applying this to the adjoint representation, which has central kernel, we deduce that $\mathfrak{r}$ is nilpotent.