Existence of Cartan subalgebras

If $\mathfrak{g}=0$, the zero subalgebra is nilpotent and equals its own normalizer, hence is a Cartan subalgebra. Assume now $\mathfrak{g}\mathrm{\ne }0$. There is a nonzero semisimple element: if every element of $\mathfrak{g}$ had nilpotent adjoint operator, then [L2] would make $\mathfrak{g}$ nilpotent, hence solvable, so its radical would be $\mathfrak{g}\mathrm{\ne }0$, contradicting semisimplicity [L6]. Choose an element $x$ whose adjoint operator is not nilpotent and let $x=x_s+x_n$ be its decomposition from [L1]; if $x_s=0$ then ${\mathrm{ad}}_x={\mathrm{ad}}_{x_n}$ is nilpotent, contrary to the choice of $x$, so $x_s\mathrm{\ne }0$ is semisimple.

A toral subalgebra of maximal dimension exists and is maximal by inclusion. Let $\mathfrak{t}$ be any maximal toral subalgebra; it is nonzero since the zero subalgebra is contained in $\mathbb{C}{x}_s$ from step 1.1. By [L7] the subalgebra $\mathfrak{t}$ is abelian, so the operators ${\mathrm{ad}}_h$, $h\in \mathfrak{t}$, are pairwise commuting, and each is semisimple by [L7]; hence by [L4] and [L8] they are simultaneously diagonalisable and $\mathfrak{g}={⨁}_{\lambda \in {\mathfrak{t}}^{\ast }}{\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\lambda }=\{x\in \mathfrak{g}:[h,x]=\lambda (h)x\text{ for all }h\in \mathfrak{t}\}\mathrm{.}$ Put $\mathfrak{l}={\mathfrak{g}}_0=C_{\mathfrak{g}}(\mathfrak{t})$.

For $x\in \mathfrak{l}$ the Jordan parts lie in $\mathfrak{l}$: by [L1] and [L9] there is a polynomial $p$ with ${\mathrm{ad}}_{x_s}=p({\mathrm{ad}}_x)$, and $[{\mathrm{ad}}_x,{\mathrm{ad}}_h]={\mathrm{ad}}_{[x,h]}=0$ for $h\in \mathfrak{t}$ [L6, L8], so $[{\mathrm{ad}}_{x_s},{\mathrm{ad}}_h]=0$ and therefore ${\mathrm{ad}}_{[x_s,h]}=0$, that is, $[x_s,h]\in Z(\mathfrak{g})=0$; thus $x_s\in \mathfrak{l}$, and $x_n=x-x_s\in \mathfrak{l}$ as well. Moreover $\mathfrak{t}+\mathbb{C}{x}_s$ is toral: it is a subalgebra because $[x_s,\mathfrak{t}]=0$, it is abelian, and each of its elements has semisimple adjoint operator by [L4] since ${\mathrm{ad}}_{x_s}$ and the commuting operators ${\mathrm{ad}}_h$ are semisimple. By the maximality of $\mathfrak{t}$ we get $x_s\in \mathfrak{t}$.

Invariance of $B$ gives $(\lambda (h)+\mu (h))B(y,z)=B([h,y],z)+B(y,[h,z])=0$ for $y\in {\mathfrak{g}}_{\lambda }$, $z\in {\mathfrak{g}}_{\mu }$ and $h\in \mathfrak{t}$; if $\lambda +\mu \mathrm{\ne }0$ some $h$ has $(\lambda +\mu )(h)\mathrm{\ne }0$, so $B({\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\mu })=0$.

For $x\in \mathfrak{l}$ we have ${\mathrm{ad}}_{\mathfrak{l}}(x)={\mathrm{ad}}_{\mathfrak{l}}(x_n)$, because $x_s\in \mathfrak{t}$ centralises $\mathfrak{l}$; by [L1] the operator ${\mathrm{ad}}_{x_n}$ is nilpotent, so every adjoint operator of $\mathfrak{l}$ is nilpotent on $\mathfrak{l}$ and [L2] makes $\mathfrak{l}$ nilpotent.

The algebra $\mathfrak{l}$ is abelian. It is nilpotent by step 4.1, hence solvable, so [L3] supplies a common eigenvector in the nonzero module $\mathfrak{g}$. Its line is invariant. Apply [L3] to the quotient by that line and then to successive nonzero quotients; the dimension drops at each step, and lifting the resulting invariant flag gives a basis of $\mathfrak{g}$ in which all ${\mathrm{ad}}_x$, $x\in \mathfrak{l}$, are upper triangular. For $x\in [\mathfrak{l},\mathfrak{l}]$ the operator ${\mathrm{ad}}_x$ is a sum of commutators of upper triangular operators, hence strictly upper triangular, hence nilpotent; consequently $B(x,y)=\mathrm{tr}({\mathrm{ad}}_x{\mathrm{ad}}_y)=0$ for every $y\in \mathfrak{l}$, since a strictly upper triangular operator times an upper triangular operator stays strictly upper triangular.

The restriction $B{\mathrm{\mid }}_{\mathfrak{l}}$ is nondegenerate: if $x\in \mathfrak{l}$ satisfies $B(x,\mathfrak{l})=0$, then for every nonzero weight $\lambda$ we have $B(x,{\mathfrak{g}}_{\lambda })=0$ by step 3.2, and $B(x,\mathfrak{l})=0$ by hypothesis, so $B(x,\mathfrak{g})=0$ and [L5] gives $x=0$. Applying this to step 5.1 yields $[\mathfrak{l},\mathfrak{l}]=0$.

Every element of $\mathfrak{l}$ is semisimple: for $x\in \mathfrak{l}$ we have $x_n\in \mathfrak{l}$ by step 3.1, and $[x_n,y]=[x,y]-[x_s,y]=0$ for every $y\in \mathfrak{l}$ because $x\in \mathfrak{l}$ and $x_s\in \mathfrak{t}$; hence ${\mathrm{ad}}_{x_n}$ commutes with ${\mathrm{ad}}_y$ and the product ${\mathrm{ad}}_{x_n}{\mathrm{ad}}_y$ is nilpotent, so $B(x_n,y)=0$ for all $y\in \mathfrak{l}$. Nondegeneracy from step 6.1 forces $x_n=0$. Thus $\mathfrak{l}$ is abelian and consists of semisimple elements, i.e. $\mathfrak{l}$ is toral; since $\mathfrak{t}\subseteq \mathfrak{l}$, maximality gives $\mathfrak{l}=\mathfrak{t}$.

Finally $N_{\mathfrak{g}}(\mathfrak{t})=\mathfrak{t}$: if $x\in N_{\mathfrak{g}}(\mathfrak{t})$ and $x={\sum }_{\lambda }{x}_{\lambda }$ is its decomposition from step 2.1, then for every $h\in \mathfrak{t}$ we have $[h,x]={\sum }_{\lambda }\lambda (h)x_{\lambda }\in \mathfrak{t}={\mathfrak{g}}_0\mathrm{.}$ Uniqueness of the direct weight-space decomposition forces every nonzero-weight component $\lambda (h)x_{\lambda }$ of this sum to vanish. For each $\lambda \mathrm{\ne }0$, choose $h\in \mathfrak{t}$ with $\lambda (h)\mathrm{\ne }0$; then $x_{\lambda }=0$. Therefore $x\in {\mathfrak{g}}_0=\mathfrak{l}=\mathfrak{t}$. Since $\mathfrak{t}$ is abelian and hence nilpotent and equals its normalizer, [L7] makes it a Cartan subalgebra. The Axiom of Choice was inherited through [L1] and [L9].