Theorem 1

Let $\mathfrak{g}$ be a compact Lie algebra. The Cartan subalgebras of $\mathfrak{g}$ (Chap. VII, §2, no. 1, Def. 1) are its maximal commutative subalgebras; in particular, $\mathfrak{g}$ is the union of its Cartan subalgebras. The group Int($\mathfrak{g}$) operates transitively on the set of Cartan subalgebras of $\mathfrak{g}$.

Since $\mathfrak{g}$ is reductive, its Cartan subalgebras are commutative (Chap. VII, §2, no. 4, Cor. 3 of Th. 2). Conversely, let $\mathfrak{t}$ be a commutative subalgebra of $\mathfrak{g}$. By §1, no. 3, Prop. 1, ad $x$ is semi-simple for all $x\in \mathfrak{t}$; by Chap. VII, §2, no. 3, Prop. 10, there exists a Cartan subalgebra of $\mathfrak{g}$ containing $\mathfrak{t}$. This proves the first assertion of the theorem.

Now let $\mathfrak{t}$ and $\mathfrak{t}'$ be two Cartan subalgebras of $\mathfrak{g}$. We prove that there exists $u\in$ Int($\mathfrak{g}$) such that $u(\mathfrak{t}) =\mathfrak{t}'$. By Prop. 1 of §1, no. 3, we can assume that $\mathfrak{g}$ is of the form L(G), where G is a connected compact Lie group, and can choose a separating invariant symmetric bilinear form F on $\mathfrak{g}$. Let $x$ (resp. $x'$) be a regular element of $\mathfrak{g}$ such that $\mathfrak{t}=\mathfrak{g}^0(x)$ (resp. $\mathfrak{t}'=\mathfrak{g}^0(x')$) (Chap. VII, §3, no. 3, Th. 2). Applying Lemma 1 with K = G, we see that there exists $k\in G$ such that [(Ad $k$)$(x), x']$ is orthogonal to $\mathfrak{g}$ with respect to F, and hence is zero; then (Ad $k$)$(x)\in \mathfrak{g}^0(x') =\mathfrak{t}'$, so $\mathfrak{g}^0$((Ad $k$)$(x)$) $=\mathfrak{t}'$ since (Ad $k$)$(x)$ is regular. We conclude that (Ad $k$)$(\mathfrak{t}) =\mathfrak{t}'$, hence the theorem.