Theorem 2

Let G be a Lie group.

(i) The mapping H ↦ L(H) is a bijection of the set of integral subgroups of G onto the set of Lie subalgebras of L(G) admitting a topological supplement.

(ii) Let H be an integral subgroup of G. Every connected Lie subgroup germ of G with Lie algebra L(H) is an open submanifold of H which generates H.

(a) Let h be a Lie subalgebra of L(G) admitting a topological supplement. Let H_1 be a Lie subgroup germ of G such that L(H_1) = h (§ 4, Theorem 3). H_1 can be chosen so that it is connected. Let H be the subgroup of G generated by H_1. There exists (§ 1, Corollary to Proposition 22) a Lie group structure on H such that H_1 is an open submanifold of H and the canonical injection of H into G is an immersion. As H_1 is connected, H is connected and is hence an integral subgroup of G. Then L(H) = L(H_1) = h. This proves that the mapping considered in (i) is surjective.

(b) Let H be an integral subgroup of G and N_1 a connected Lie subgroup germ of G with Lie algebra L(H). As the canonical injection of H into G is an immersion, there exists an open subgroup germ H_1 of H which is at the same time a submanifold of G and hence a Lie subgroup germ of G with Lie algebra L(H). On the other hand, let N be the subgroup of G generated by N_1; by part (a) of the proof, it has the structure of an integral subgroup of G such that N_1 is an open submanifold of N. By § 4, Theorem 3, H_1 ∩ N_1 is open in H_1 and N_1. Hence the subgroup of G generated by H_1 ∩ N_1 is equal on the one hand to H and on the other to N. Therefore the Lie groups H and N are equal. This proves (ii) and also proves that the mapping considered in (i) is injective.