Cartan subalgebras are exactly maximal toral subalgebras

The implication "maximal toral $\Rightarrow$ Cartan" is [L1]. For the converse, let $\mathfrak{h}$ be a Cartan subalgebra; by [L3] it is nilpotent and $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Let $\mathfrak{g}={⨁}_{\alpha }{\mathfrak{g}}_{\alpha }$ be its generalized weight decomposition from [L2], so that ${\mathfrak{g}}_0=\{x\in \mathfrak{g}:({\mathrm{ad}}_H{)}^{k}x=0\text{ for all }H\in \mathfrak{h}\text{ and large }k\}$.

We have ${\mathfrak{g}}_0=\mathfrak{h}$: [L2] gives $\mathfrak{h}\subseteq {\mathfrak{g}}_0$; conversely every $X\in N_{\mathfrak{g}}(\mathfrak{h})$ lies in ${\mathfrak{g}}_0$, because $({\mathrm{ad}}_H{)}^{k}X=({\mathrm{ad}}_H{)}^{k-1}[H,X]$ with $[H,X]\in \mathfrak{h}$ and ${\mathrm{ad}}_H$ nilpotent on $\mathfrak{h}$ by [L5]; so $N_{\mathfrak{g}}(\mathfrak{h})\subseteq {\mathfrak{g}}_0$. If ${\mathfrak{g}}_0\mathrm{\ne }\mathfrak{h}$, then ${\mathfrak{g}}_0\mathrm{/}\mathfrak{h}$ is a nonzero module for the solvable algebra $\mathfrak{h}$, so by [L5] there is $X\mathrm{\notin }\mathfrak{h}$ whose image in ${\mathfrak{g}}_0\mathrm{/}\mathfrak{h}$ is a common eigenvector for all ${\mathrm{ad}}_H$; the diagonal functional has value $0$ on every $H$, because ${\mathrm{ad}}_H$ is nilpotent on ${\mathfrak{g}}_0$, so $[H,X]\in \mathfrak{h}$ for all $H$ and $X\in N_{\mathfrak{g}}(\mathfrak{h})\setminus \mathfrak{h}$, contradicting $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Hence ${\mathfrak{g}}_0=\mathfrak{h}$.

For $x\in \mathfrak{h}$ we have $x_s,x_n\in \mathfrak{h}$ and ${\mathrm{ad}}_{x_s}$ acts on each ${\mathfrak{g}}_{\alpha }$ as the scalar $\alpha (x)$: by [L4] write ${\mathrm{ad}}_{x_s}=p({\mathrm{ad}}_x)$; on ${\mathfrak{g}}_{\alpha }$ the operator ${\mathrm{ad}}_x$ is $\alpha (x)$ plus a commuting nilpotent operator, so $p({\mathrm{ad}}_x)=p(\alpha (x))$ plus a commuting nilpotent operator, while ${\mathrm{ad}}_{x_s}$ restricts semisimply to the invariant subspace ${\mathfrak{g}}_{\alpha }$; hence ${\mathrm{ad}}_{x_s}{\mathrm{\mid }}_{{\mathfrak{g}}_{\alpha }}=p(\alpha (x))\cdot 1$, and since $x_s=x-x_n$ with ${\mathrm{ad}}_{x_n}$ nilpotent on ${\mathfrak{g}}_{\alpha }$, comparison of scalar parts gives $p(\alpha (x))=\alpha (x)$. In particular ${\mathrm{ad}}_{x_s}$ commutes with every ${\mathrm{ad}}_H$, $H\in \mathfrak{h}$, so by [L7] $[x_s,H]=0$ and, $\mathfrak{g}$ being centerless [L6], $x_s\in C_{\mathfrak{g}}(\mathfrak{h})\subseteq N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$; then $x_n=x-x_s\in \mathfrak{h}$ as well.

$\mathfrak{h}$ is abelian: apply the common-eigenvector assertion in [L5] first to $\mathfrak{g}$ and then to each successive nonzero quotient by the invariant subspaces already obtained. Each quotient has smaller dimension, so this constructs a full invariant flag in finitely many steps. In a basis adapted to the flag, the solvable algebra $\mathfrak{h}$ acts triangularly, and for upper triangular matrices $A,B,C$ one has $\mathrm{tr}(ABC)=\mathrm{tr}(BAC)$ since both equal the sum of diagonal products; hence $B([H_1,H_2],H)=\mathrm{tr}({\mathrm{ad}}_{[H_1,H_2]}{\mathrm{ad}}_H)=0$ for all $H_1,H_2,H\in \mathfrak{h}$. For $X\in {\mathfrak{g}}_{\alpha }$ with $\alpha \mathrm{\ne }0$, the operator ${\mathrm{ad}}_H{\mathrm{ad}}_X$ maps ${\mathfrak{g}}_{\beta }$ into ${\mathfrak{g}}_{\beta +\alpha }$ by [L2], hence has zero trace because it has no diagonal blocks; therefore $B(H,X)=0$ for all $H\in \mathfrak{h}$, $\alpha \mathrm{\ne }0$, $X\in {\mathfrak{g}}_{\alpha }$. Combining the two orthogonality statements with ${\mathfrak{g}}_0=\mathfrak{h}$ from step 1.2 gives $B([H_1,H_2],\mathfrak{g})=0$, and nondegeneracy of $B$ [L6] forces $[H_1,H_2]=0$.

No nonzero element of $\mathfrak{h}$ has nilpotent adjoint operator: if $x\in \mathfrak{h}$ has ${\mathrm{ad}}_x$ nilpotent, then for $y\in \mathfrak{h}$ the operators ${\mathrm{ad}}_x,{\mathrm{ad}}_y$ commute by step 2.1, so ${\mathrm{ad}}_x{\mathrm{ad}}_y$ is nilpotent and $B(x,y)=0$; and for $X\in {\mathfrak{g}}_{\alpha }$ with $\alpha \mathrm{\ne }0$ we have $B(x,X)=0$ by the trace argument of step 2.1 with $H=x$. Hence $B(x,\mathfrak{g})=0$ and [L6] gives $x=0$.

By steps 1.3 and 3.1 every $x\in \mathfrak{h}$ has $x_n=0$, that is, $x=x_s$ is semisimple; with step 2.1 this makes $\mathfrak{h}$ a toral subalgebra by [L8]. It is maximal: if $\mathfrak{t}\supseteq \mathfrak{h}$ is toral, then $\mathfrak{t}$ is abelian with $[\mathfrak{t},\mathfrak{h}]=0$, so $\mathfrak{t}\subseteq C_{\mathfrak{g}}(\mathfrak{h})\subseteq N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$ and $\mathfrak{t}=\mathfrak{h}$. Hence $\mathfrak{h}$ is maximal toral, which is the converse implication. The zero algebra is covered by the convention that its zero subalgebra is both Cartan and maximal toral. The Axiom of Choice was inherited through [L1] and [L4].