Theorem 3

Let G be a Lie group germ and $\mathfrak{h}$ a Lie subalgebra of $L(G)$ admitting a topological supplement. There exists a Lie subgroup germ H of G such that $L(H) = \mathfrak{h}$. If $H_1$ and $H_2$ are Lie subgroup germs of G such that

then $H_1 \cap H_2$ is open in $H_1$ and $H_2$.

There exists a Lie group germ $H'$ with Lie algebra isomorphic to $\mathfrak{h}$ (Theorem 2). Shrinking $H'$ if necessary, it can be assumed that there exists a morphism $\phi$ of $H'$ into G such that $L(\phi)$ is an isomorphism of $L(H')$ onto $\mathfrak{h}$ (no. 1, Theorem 1). As $\mathfrak{h}$ admits a topological supplement, $\phi$ is an immersion at e. Hence, shrinking $H'$ further, it can be assumed that $\phi$ is an isomorphism of the manifold $H'$ onto a submanifold of G. This proves the existence of H. The second assertion follows from the following proposition: