Proposition 3

Suppose that K is ultrametric. Let G be a compact Lie group and $\alpha$ an infinitesimal automorphism of G. There exist an open subgroup I of K and a law of analytic operation $(\lambda, g) \mapsto \phi_\lambda(g)$ of I on G with the following properties: (1) if D is the associated law of infinitesimal operation, then $D(1) = \alpha$; (2) for all $\lambda \in I, \phi_\lambda \in \mathrm{Aut}\, G$. As G is compact, there exist an open subgroup I' of K and a law of analytic operation $(\lambda, g) \mapsto \phi_\lambda(g)$ of I' on G with property (1) of the proposition (§ 4, no. 7, Corollary 2 to Theorem 6). We write $\phi_\lambda(g) = f_g(\lambda)$ for $\lambda \in I'$ and $g \in G$. Then, for $g_1, g_2$ in $G$ and $\lambda \in I'$,

and $(f_{g_1}f_{g_2})(0) = g_1 g_2 = f_{g_1g_2}(0)$. Hence $f_{g_1'g_2'}(\lambda) = f_{g_1'}(\lambda) f_{g_2'}(\lambda)$ for $(g_1', g_2', \lambda)$

in a neighbourhood of $(g_1, g_2, 0)$ (Differentiable and Analytic Manifolds, R, 9.1.8). As $G$ is compact, there exists an open subgroup $I$ of $I'$ such that $f_{g_1g_2}(\lambda) = f_{g_1}(\lambda) f_{g_2}(\lambda)$ for all $g_1 \in G, g_2 \in G, \lambda \in I$. In other words, $\phi_\lambda \in \mathrm{Aut}\, G$ for $\lambda \in I$.