Proposition 5

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{c}$ its centre, $\mathfrak{h}$ a vector subspace of $\mathfrak{g}$. Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ if and only if $\mathfrak{h}$ contains $\mathfrak{c}$ and $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$.

Assume that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$. Since $[\mathfrak{c},\mathfrak{g}]\subset \mathfrak{h}$, we have $\mathfrak{c}\subset \mathfrak{h}$. On the other hand, $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$ by Cor. 2 of Prop. 4.

Assume that $\mathfrak{h}\supset \mathfrak{c}$ and that $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$. Let $f$ be the canonical morphism from $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{c}$. The algebra $\mathfrak{h}$, which is a central extension of $\mathfrak{h}/\mathfrak{c}$, is nilpotent. Let $\mathfrak{n}$ be the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$. If $x\in \mathfrak{n}$, $[f(x),\mathfrak{h}/\mathfrak{c}]\subset \mathfrak{h}/\mathfrak{c}$, hence $f(x)\in \mathfrak{h}/\mathfrak{c}$, and so $x\in \mathfrak{h}$. This proves that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$.