Proposition 5

Let G be a Lie group whose Lie algebra is compact.

a) Assume that G is connected. Then G has a largest compact subgroup K; it is connected. There exists a closed central vector subgroup (no. 2) N of G such that G is the direct product $N\times K$.

b) Assume that the group of connected components of G is finite. Then:

(i) Every compact subgroup of G is contained in a maximal compact subgroup.

(ii) If $K_1$ and $K_2$ are two maximal compact subgroups of G, there exists $g\in G$ such that $K_2=gK_1g^{-1}$.

(iii) Let K be a maximal compact subgroup of G. Then $K\cap G_0$ is equal to $K_0$; it is the largest compact subgroup of $G_0$.

(iv) There exists a closed central vector subgroup N of $G_0$, normal in G, such that, for any maximal compact subgroup K of G, $G_0$ is the direct product of $K_0$ by N, and G is the semi-direct product of K by N.

a) We retain the notations of Prop. 4. The projection of Ker $f$ onto V is a finite subgroup of the vector group V, hence is reduced to the identity element. It follows that Ker $f$ is contained in $T\times S$, hence that G is the direct product of the vector group $N =f(V)$ and the compact group $K =f(T\times S)$. Every compact subgroup of G has a projection onto N that is reduced to the identity element, hence is contained in K. This proves a).

b) Assume now that $G/G_0$ is finite. By $a), G_0$ is the direct product of its largest compact subgroup M and a vector subgroup P; the subgroup M of G is clearly normal. Let $\mathfrak{n}$ be a vector subspace complement of L(M) in L(G), stable under the adjoint representation of G (no. 1 and no. 3, Prop. 3); this is an ideal of L(G) and we have $L(G) = L(M)\times \mathfrak{n}$. Let N be the integral subgroup of G with Lie algebra $\mathfrak{n}$; by Chap. III, §6, no. 6, Prop. 14, it is normal in G. The projection of L(G) onto L(P) with kernel L(M) induces an isomorphism from $\mathfrak{n}$ to L(P); it follows that the projection of $G_0$ onto P induces an étale morphism from N to P; since P is simply-connected, this is an isomorphism, and N is a vector group. The morphism $(x, y) \rightarrow xy$ from $M\times N$ to $G_0$ is an injective étale morphism (since $M\cap N$ is reduced to the identity element), hence an isomorphism. It follows that N is a closed subgroup of G and that the quotient $G/N$ is compact, since $G_0/N$ is compact and $G/G_0$ is finite (General Topology, Chap. III, §4, no. 1, Cor. 2 of Prop. 2).

By Integration, Chap. VII, §3, no. 2, Prop. 3, every compact subgroup of G is contained in a maximal compact subgroup, these are conjugate, and for any maximal compact subgroup K of G, G is the semi-direct product of K by N. Since $G_0$ contains N, it is the semi-direct product of N by $G_0\cap K$; it follows that $G_0\cap K$ is connected, hence equal to $K_0$, since $K/(G_0\cap K)$ is isomorphic to $G/G_0$, hence finite; finally, $K_0$ is clearly the largest compact subgroup of $G_0$ by a).