Proposition 10

Suppose that $p \neq 0$. Let $G$ be a finite-dimensional Lie group. Let $G_f$ be the set of $x \in G$ for which there exists a strictly increasing sequence $(n_i)$ of integers such that $x^{n_i}$ tends to $e$ as $i$ tends to $+\infty$. (i) $G_f$ is open in $G$. (ii) There exists one and only one mapping $\psi$ of $G_f$ into $L(G)$ with the following properties: (a) $\psi(x^n) = n \psi(x)$ for all $x \in G_f$ and all $n \in \mathbf{Z}$; (b) there exists an open neighbourhood $V$ of $e$ in $G$ such that $\psi|V$ is the inverse mapping of an injective exponential mapping. (iii) The mapping $\psi$ is analytic.

There exists an open subgroup of $G$ which is isomorphic to a standard group (no. 3, Theorem 4). Assertion (i) then follows from Lemma 1.

Let $U$ be an open subgroup of $L(G)$ and $\phi : U \to \phi(U)$ an exponential mapping of $G$ with the properties of Proposition 3 of no. 2. $U$ can be assumed to be small enough for $\phi(U) \subset G_f$. Let $x \in G_f$. There exists $m \in \mathbf{Z} - \{0\}$ such that $x^m \in \phi(U)$. The element $\frac{1}{m} \phi^{-1}(x^m)$ does not depend on the choice of $m$. For let $m' \in \mathbf{Z}$ be such that $x^{m'} \in \phi(U)$. Then $x^{m m'} \in \phi(U)$ and

whence our assertion. Let $\psi(x) = \frac{1}{m} \phi^{-1}(x^m)$. Then $\psi|\phi(U) = \phi^{-1}$. On the other hand, if $n \in \mathbf{Z}$, then

Hence $\psi$ has properties (a) and (b) of the proposition. In a neighbourhood of $x$, $\psi$ is composed of the mappings $x \mapsto x^m$, $y \mapsto \phi^{-1}(y)$ and $z \mapsto \frac{1}{m} z$; hence $\psi$ is analytic on $G_f$.

Finally, let $\psi'$ be a mapping of $G_f$ into $L(G)$ and $V'$ a neighbourhood of $e$ in $G_f$ such that $\psi'(x^n) = n \psi'(x)$ for $x \in G_f$ and $n \in \mathbf{Z}$ and such that $\psi'|V'$ is the inverse mapping of an injective exponential mapping. Then $\psi$ and $\psi'$ coincide on a neighbourhood $W$ of $e$. If $x \in G_f$, there exists $n \in \mathbf{Z}$ such that $x^n \in W$. Then

and hence $\psi = \psi'$.