Proposition 1

Let G be a Lie group germ and H and $H'$ two Lie subgroup germs. In order that $L(H) \supset L(H')$, it is necessary and sufficient that $H \cap H'$ be open in $H'$.

If $H \cap H'$ is open in $H'$, then $L(H') = L(H \cap H') \subset L(H)$. Suppose that $L(H) \supset L(H')$. Let i, $i'$ be the canonical injections of H, $H'$ into G. By shrinking $H'$ if necessary, it can be assumed that there exists a morphism $\psi$ of $H'$ into H such that $L(\psi)$ is the canonical injection of $L(H')$ into $L(H)$ (no. 1, Theorem 1). Then $L(i \circ \psi) = L(i')$ and hence there exists a neighbourhood V of $e_{H'}$ in $H'$ such that $i \circ \psi$ and $i'$ coincide on V (Theorem 1). Therefore $V \subset H$, hence $V \subset H \cap H'$ and $H \cap H'$ is open in $H'$ (§ 1, no. 10).