Proposition 4

Let G be a finite-dimensional real or complex Lie group. Let A, B, C be integral subgroups of G such that [L(A), L(C)] ⊂ L(C) and

If [L(A), L(B)] ⊂ L(C), then (A, B) ⊂ C. If [L(A), L(B)] = L(C), then (A, B) = C.

Suppose that $[L(A), L(B)] \subset L(C)$. The sum $L(A) + L(B) + L(C)$ is a Lie subalgebra of $L(G)$. By considering an integral subgroup of $G$ with Lie algebra $L(A) + L(B) + L(C)$, the problem is reduced to the case where

and $G$ is connected. Then $L(C)$ is an ideal of $L(G)$. Suppose first that $G$ is simply connected. Then $C$ is a normal Lie subgroup of $G$ (§ 6, no. 6, Proposition 14). Let $\phi$ be the canonical morphism of $G$ onto $G/C$. Then

and hence $\phi(A)$ and $\phi(B)$ commute by the Hausdorff formula; therefore $(A, B) \subset C$. In the general case, let $G'$ be the universal covering of $G$ and $A', B', C'$ the integral subgroups of $G'$ such that $L(A') = L(A), L(B') = L(B), L(C') = L(C)$. Then $(A', B') \subset C'$ and $A, B, C$ are the canonical images of $A', B', C'$ in $G$, whence $(A, B) \subset C$. On the other hand, $(A, B)$ is the underlying set of an integral subgroup of $G$ (§ 6, no. 2, Corollary to Proposition 4) and its Lie algebra contains $[L(A), L(B)]$ (Proposition 3). If

then $(A, B) \supset C$, whence $(A, B) = C$.