Theorem 2

Let $G$ be a finite-dimensional Lie group over $\mathbf{R}$ or $\mathbf{Q}_p$. Every closed subgroup of $G$ is a Lie subgroup of $G$. More generally, let $U$ be a symmetric open neighbourhood of $e$ in $G$ and $H$ a non-empty closed subspace of $U$ such that the conditions $x \in H$, $y \in H$ and $xy^{-1} \in U$ imply $xy^{-1} \in H$. Then $H$ is a Lie subgroup germ of $G$.

Let $\mathfrak{h}$ be the Lie subalgebra tangent to $H$ at $e$ (§ 4, no. 5, Definition 2). There exists a Lie subgroup germ $H_0$ of $G$ with Lie algebra $\mathfrak{h}$ and contained in $H$. We prove that $H_0$ is open in $H$ with the topology induced by that on $G$. This will prove that $H$ is an analytic submanifold of $G$ and will establish the theorem.

There exist a vector subspace $\mathfrak{k}$ supplementary to $\mathfrak{h}$ in $L(G)$, symmetric open neighbourhoods $V_1, V_2$ of zero in $\mathfrak{h}$ and $\mathfrak{k}$ respectively and an exponential mapping $\phi$ of $G$ defined on $V_1 + V_2$ and possessing the following properties:

(a) the mapping $(a_1, a_2) \mapsto \phi(a_1)\phi(a_2)$ is an analytic isomorphism of $V_1 \times V_2$ onto an open subset $V$ of $G$; (b) $\phi(V_1) \subset H_0$; (c) $V^2 \subset U$.

We shall prove (and this will achieve the proof) that there exists an open neighbourhood $V'_2$ of 0 in $V_2$ such that $H \cap (\phi(V_1)\phi(V'_2)) = \phi(V_1)$.

Suppose that the assertion is false. Then we can find a sequence $(x_n)$ in $V_1$ and a sequence $(y_n)$ in $V_2 - \{0\}$ tending to 0, such that $\phi(x_n)\phi(y_n) \in H$ for all $n$. Then $\phi(y_n) \in H$ by (c).

If $K = \mathbf{Q}_p$, it can be assumed that $V_2$ is an additive subgroup of $\mathfrak{k}$ and that $\phi(pa) = \phi(a)^p$ for all $a \in V_2$ and all $p \in \mathbf{Z}$. Then $\phi(\lambda y_1) \in H$ for all $\lambda \in \mathbf{Z}$ and hence by continuity for all $\lambda \in \mathbf{Z}_p$. The mapping $f : \lambda \mapsto \phi(\lambda y_1)$ of $\mathbf{Z}_p$ into $G$ is analytic and takes its values in $H$ and $(T_0f)(1) = y_1$. Hence $y_1 \in \mathfrak{h}$, which is absurd. The theorem is therefore established in the case of $\mathbf{Q}_p$.

If $K = \mathbf{R}$, it can be assumed that $V_2$ is connected and that $y_n$ belongs to $\frac{1}{2}V_2 - \{0\}$. By taking a subsequence of $(y_n)$ if necessary, we can find a sequence $(\lambda_n)$ of non-zero scalars such that $\lambda_n^{-1}y_n$ tends to an element $y$ of $V_2 - \{0\}$. The sequence $(\lambda_n)$ tends to 0. Let $\lambda \in \mathbf{R}$ be such that $\lambda y \in \frac{1}{2}V_2$; we prove that $\exp(\lambda y) \in H$. It can be assumed that $\lambda \lambda_n^{-1}y_n \in \frac{1}{2}V_2$ for all $n$. Let $k_n \in \mathbf{Z}$ be such that $|\lambda - k_n \lambda_n|$ tends to 0. For $n$ sufficiently large, $(\lambda - k_n \lambda_n)\lambda_n^{-1}y_n \in \frac{1}{2}V_2$ and hence $k_n y_n \in \frac{1}{2}V_2$. Hence $\exp(h y_n) \in H$ for $h$ an integer and $0 \leq |h| \leq |k_n|$ (as is seen by induction on $|h|$). Then

Hence the mapping $f : \lambda \mapsto \exp \lambda y$, where $\lambda y \in \frac{1}{4}V_2$, takes its values in $H$ and $(T_0f)(1) = y$. Hence $y \in \mathfrak{h}$, which is absurd. The theorem is thus established in the case of $\mathbf{R}$.

Theorem 2 is no longer true if $G$ is not assumed to be finite-dimensional (Exercise 12).