Proposition 12

Let G be a finite-dimensional Lie group. For $L(G)$ to be nilpotent, it is necessary and sufficient that G have a nilpotent open subgroup. Suppose that G has a nilpotent open subgroup $G_0$. By the Corollaries to Propositions 4 and 6, no. 2, $C^iL(G_0) = \{0\}$ for sufficiently large $i$. Hence $L(G_0) = L(G)$ is nilpotent.

Suppose that $L(G)$ is nilpotent. If $K = \mathbf{R}$ or $\mathbf{C}$, the identity component $G_0$ of G is nilpotent by the Corollary to Proposition 4, no. 2, and $G_0$ is open in G. If K is ultrametric, the Corollary to Proposition 6, no. 2, proves that there exist an open subgroup $G_1$ of G, an integer $i > 0$ and a neighbourhood V of e in G such that $C^iG_1 \cap V = \{e\}$. Then, if $G_0$ is a sufficiently small subgroup of $G_1$, $C^iG_0 \subset V$, hence $C^iG_0 = \{e\}$ and $G_0$ is nilpotent.

Let $g$ be a nilpotent Lie algebra. The Hausdorff series $H(X, Y)$ corresponding to $g$ has only a finite number of non-zero terms and we know (Chapter II, § 6, no. 5, Remark 3) that the law of composition $(x, y) \mapsto H(x, y)$ defines a group structure on $g$. Suppose further that $g$ is complete and normable. Clearly the law H is a continuous-polynomial (Differentiable and Analytic Manifolds, R, Appendix). Hence $g$, together with the law H, is a Lie group G, said to be associated with $g$. By § 4, no. 2, Lemmas 2 and 3, $L(G) = g$. The identity mapping $\phi$ of $g$ into $G$ is an exponential mapping of $G$ such that $\phi(\lambda x)\phi(\lambda'x) = \phi((\lambda + \lambda')x)$ for all $x \in g, \lambda \in K, \lambda' \in K$. Every Lie subalgebra $\mathfrak{h}$ of $g$ admitting a topological supplement is a Lie subgroup $H$ of $G$ and $L(H) = \mathfrak{h}$.