Exercise 6, § 10

a) Show the equivalence of the properties:

(i) $\mathfrak{g}$ has no nilpotent element $\not= 0$.

(ii) $\mathfrak{g}$ has no parabolic subalgebra $\not=\mathfrak{g}$.

(Use Cor. 2 of Th. 1.)

Such an algebra is called anisotropic.

b) Let $\mathfrak{p}$ be a minimal parabolic subalgebra of $\mathfrak{g},\mathfrak{r}$ the radical of $\mathfrak{p}$, and $\mathfrak{s}=\mathfrak{p}/\mathfrak{r}$. Show that $\mathfrak{s}$ is anisotropic. (Remark that, if $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{s}$, the inverse image of $\mathfrak{q}$ in $\mathfrak{p}$ is a parabolic subalgebra of $\mathfrak{g}$, cf. §3, Exerc. $5a$)$.)$