Corollary 1

Let G be a Lie group. There exists a neighbourhood of e which contains no finite subgroup distinct from {e}. If K = R or C, there exists an open neighbourhood of e which contains no subgroup distinct from {e}.

We write L(G) = L. Choose a norm on L defining the topology on L and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all x, y in L.

Suppose that K = R or C. Let G' be the Lie group germ defined by L. There exist an open ball U' of centre 0 in G' and an isomorphism $\phi$ of the Lie group germ U' onto an open neighbourhood U of e in G. Let V' = $\frac{1}{2} U'$, $V = \phi(V')$, H be a subgroup of G contained in V and $h \in H$. Let $x = \phi^{-1}(h) \in V'$. If $x \neq 0$, there exists an integer $n > 0$ such that $x, 2x, \ldots, nx$ are in V', $(n+1)x \in U'$, $(n+1)x \notin V'$. Then $h, h^2, \ldots, h^n$ are in V,

which is absurd. Hence H = {e}.

Suppose that K is ultrametric. It suffices to prove the corollary when G is the Lie group associated with L. If $g \in G$, the powers of g evaluated in G are the elements of $\mathbf{Z}g$ evaluated in L. These are distinct if $g \neq e$. Hence G contains no finite subgroup distinct from {e}.