Corollary

Suppose that K is ultrametric. Let G be a finite-dimensional Lie group with Lie algebra g. There exists an open subgroup $G_0$ of G such that, for all $i$, $D^iG_0$ (resp. $C^iG_0$) is a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$ (resp. $\mathscr{C}^ig$).

(a) By Proposition 3 applied inductively, for every open subgroup $G_1$ of G and for all $i$, $D^iG_1$ contains a Lie subgroup germ of G with Lie algebra $\mathscr{D}^ig$.

(b) Let $G'$ be an open subgroup of G such that, for $i \leq n$, $D^iG'$ is a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$. By Proposition 6, there exist open subgroups $H_1, H_2$ of $D^nG'$, such that $(H_1, H_2)$ is a Lie subgroup with Lie algebra $\mathscr{D}^{n+1}g$. Let $G''$ be an open subgroup of $G'$ small enough for $D^nG'' \subset H_1 \cap H_2$. Then $D^{n+1}G'' \subset (H_1, H_2)$. The relations

prove, using (a), that $D^iG''$ is, for $i \leq n + 1$, a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$.

(c) There exists an integer $p$ such that $\mathscr{D}^pg = \mathscr{D}^{p+1}g = \cdots$. By the above, there exists an open subgroup $G_0$ of G such that $D^iG_0$ is, for $i \leq p$, a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$. But, by (a), the same assertion remains true for $i > p$ since $D^pG_0 \supset D^iG_0$ for $i > p$.

(d) The argument is similar for the $C^i$.