Proposition 6

Suppose that K is ultrametric. Let G be a finite-dimensional Lie group. Let A, B, C be Lie subgroups of G such that $[L(A), L(C)] \subset L(C)$, $[L(B), L(C)] \subset L(C)$. If $[L(A), L(B)] \subset L(C)$, there exist open subgroups A', B' of A, B such that $(A', B') \subset C$. If $[L(A), L(B)] \subset L(C)$, there exist open subgroups A', B', C' of A, B, C such that $(A', B') = C'$.

Suppose that $[L(A), L(B)] \subset L(C)$. As in the proof of Proposition 4, the problem reduces to the case where $L(C)$ is an ideal of $L(G)$. Then, by replacing G by an open subgroup, it reduces to the case where C is normal in G ($§ 7$, no. 1, Proposition 2). Let $\phi$ be the canonical morphism of G onto $G/C$. Then

By the Hausdorff formula, there exist open subgroups A', B' of A, B such that $\phi(A')$ and $\phi(B')$ commute, whence $(A', B') \subset C$. Suppose further that By Proposition 3, the Lie subalgebra tangent to $(A', B')$ at $e$ contains $L(C)$. Hence $(A', B')$ contains a Lie subgroup germ of G with Lie algebra $L(C)$. Therefore $(A', B')$ is an open subgroup of C.