Remark

Preserving the notation of Proposition 22 and its corollary, suppose that K is of characteristic 0, that H is finite-dimensional and that the topology on U admits a countable base. Even with all these hypotheses it is possible that G is not closed in H (Exercise 3). But, if G is closed, G is a Lie subgroup of H. For the mapping $(g, h) \mapsto gh$ is a law of analytic left operation of G on H. The orbit of e is G. Our assertion then follows from Propositions 2 (iv) and 14 (iii).