Corollary 4

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$, and $\mathfrak{a}$ a subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$.

(i) $\mathfrak{a}$ is equal to its own normalizer in $\mathfrak{g}$.

(ii) Assume that $k=\mathbf{R}$ or $\mathbf{C}$; let G be a Lie group with Lie algebra $\mathfrak{g}$, A the integral subgroup of G with Lie algebra $\mathfrak{a}$. Then A is a Lie subgroup of G, and it is the identity component of the normalizer of A in G.

Let $\mathfrak{n}$ be the normalizer of $\mathfrak{a}$ in $\mathfrak{g}$. Since $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{n}$ (Example $3$)$,\{0\}$ is a Cartan subalgebra of $\mathfrak{n}/\mathfrak{a}$ (Cor. 2), hence is equal to its normalizer in $\mathfrak{n}/\mathfrak{a}$; in other words, $\mathfrak{n}=\mathfrak{a}$. Assertion (ii) follows from (i) and Chap. III, §9, no. 4, Cor. of Prop. 11.