Proposition 3

Let G be a Lie group. There exists an exponential mapping $\phi$ of G with the following properties: (i) $\phi$ is defined on an open subgroup U of the additive group $L(G)$; (ii) $\phi(U)$ is an open subgroup of G and $\phi$ is an isomorphism of the analytic manifold U onto the analytic manifold $\phi(U)$; (iii) $\phi(nx) = \phi(x)^n$ for all $x \in U$ and all $n \in \mathbf{Z}$.

Let $L(G)$ be given a norm compatible with its topology and such that $\| [x, y] \| \leq \| x \| \| y \|$ for $x, y$ in $L(G)$. Let $G_1$ be the Lie group defined by $L(G)$. Let $\psi = \mathrm{Id}_{G_1}$, which is an exponential mapping of $G_1$. For all $\mu > 0$, let $L_\mu$ be the set of $x \in L(G)$ such that $\| x \| < \mu$. Then, for $\mu$ sufficiently small, $L_\mu$ is an open subgroup of the additive group $L(G)$, $\psi(L_\mu)$ is an open subgroup of $G_1$ (§ 4, no. 2, Lemma 3), $\psi|L_\mu$ is an isomorphism of analytic manifolds of $L_\mu$ onto $\psi(L_\mu)$ and $\psi(nx) = \psi(x)^n$ for all $x \in L_\mu$ and all $n \in \mathbf{Z}$. The $L_\mu$ form a fundamental system of neighbourhoods of 0 in $L(G)$. By Theorem 1, there exist $\mu$ and an open subgroup $G'$ of G such that $\psi(L_\mu)$ and $G'$ are isomorphic, whence the proposition.