Theorem 3

(i) If L is a finite-dimensional Lie algebra, there exists a simply connected Lie group G such that L(G) is isomorphic to L.

(ii) Let G_1 and G_2 be two connected Lie groups, with G_1 simply connected. Let f be an isomorphism of L(G_1) onto L(G_2), φ the morphism of G_1 into G_2 such that L(φ) = f and N the kernel of φ. Then N is a discrete subgroup of the centre of G_1 and the morphism of G_1/N into G_2 derived from φ is a Lie group isomorphism. If G_2 is simply connected, φ is an isomorphism.

Let L be a finite-dimensional Lie algebra. There exists a finite-dimensional vector space E such that L can be identified with a Lie subalgebra of End(E) (Chapter I, § 7, Theorem 2). Let H be the integral subgroup of GL(E) with

Lie algebra L. Let $\hat{H}$ be its universal covering ($§ 1$, no. 9, Remark). Then $L(\hat{H})$ is isomorphic to L, whence (i).

Let $G_1, G_2, f, \phi, N$ be as in (ii). Then $\phi$ is étale, hence $\phi(G_1)$ is an open subgroup of $G_2$ and hence $\phi(G_1) = G_2$. On the other hand, N is discrete and hence contained in the centre of $G_1$ (Integration, Chapter VII, $§ 3$, Lemma 4). Clearly the morphism of $G_1/N$ onto $G_2$ derived from $\phi$ is a Lie group isomorphism. If $G_2$ is simply connected, every étale mapping of $G_1$ onto $G_2$ is injective and hence $N = \{e\}$.