Exercise 15, § 1

(a) Let $a$ be a Lie algebra, $b$ a subalgebra of $a$, $c$ an ideal of $b$ and $z$ the centralizer of $c$ in $a$. Then $[z, b] \subset z$. Deduce that $b + z$ is a subalgebra of $a$ in which $z$ is an ideal. (b) Let $g$ be a Lie algebra, $a$ a subinvariant subalgebra of $g$ and $b$ an ideal of $a$. If the centralizer of $b$ in $a$ and the centralizer of $a$ in $g$ are zero, the centralizer $\mathfrak{z}$ of $\mathfrak{b}$ in $\mathfrak{g}$ is zero. (Let $\mathfrak{h} = \mathfrak{a} + \mathfrak{z}$, which is a subalgebra by (a); $\mathfrak{a}$ is subinvariant in $\mathfrak{h}$; if $\mathfrak{z} \neq \{0\}$, then $\mathfrak{a} \neq \mathfrak{b}$, hence $\mathfrak{a}$ is an ideal in $\mathfrak{a}'$, with $\mathfrak{a} \neq \mathfrak{a}' \subset \mathfrak{h}$; let $a' \in \mathfrak{a}'$, $a' \notin \mathfrak{a}$ and $a' = a + z$, $a \in \mathfrak{a}$, $z \in \mathfrak{z}$, $z \neq 0$; show that $[z, a]$ is contained in $\mathfrak{a}$ and commutes with $\mathfrak{b}$, whence $[z, a] = \{0\}$ giving a contradiction.)

(c) Let $\mathfrak{g}$ be a Lie algebra with centre zero, $\mathfrak{D}_1$ the Lie algebra of derivations of $\mathfrak{g}$, $\mathfrak{D}_2$ the Lie algebra of derivations of $\mathfrak{D}_1$, ... $\mathfrak{g}$ is identified with an ideal of $\mathfrak{D}_1, \mathfrak{D}_1$ with an ideal of $\mathfrak{D}_2, \ldots$ (cf. Exercise 12 (b)). Using (b) and Exercise 12 (b), show that the centralizer of $\mathfrak{g}$ in $\mathfrak{D}_i$ is zero for all $i$. (For a sequel to this exercise, cf. § 4, Exercise 15.)