Proposition 5

Let G be a finite-dimensional real or complex Lie group and A an integral subgroup of G. Then $DA = DA$. In particular, A is a normal subgroup of $\overline{A}$ and $\overline{A}/A$ is commutative.

Let $a = L(A)$. Let $G_1$ be the set of $g \in G$ such that

Then $G_1$ is a closed subgroup of $G$. If $y \in a$, then $\exp y \in G_1$, by § 6, no. 4, Corollary 3 (ii) to Proposition 10. Hence $G_1$ contains $A$ and therefore $\overline{A}$. Thus, for $g \in \overline{A}$, $L(\mathrm{Int}\, g)$ leaves $a$ stable and therefore $\mathrm{Int}\, g$ leaves $A$ stable; more precisely, $L(\mathrm{Int}\, g)$ defines the identity automorphism of $a/\mathscr{D}a$ and hence $\mathrm{Int}\, g$ defines the identity automorphism of $A/DA$. This proves that $(\overline{A}, A) \subset DA$. With the real Lie group structure on $G$, $\overline{A}$ is a Lie subgroup (§ 8, no. 2, Theorem 2); let $b$ be its Lie algebra. Let $G_2$ be the set of $g \in G$ such that

By the above, $G_2 \supseteq A$ and hence $G_2 \supseteq \overline{A}$. Therefore, for $g \in \overline{A}$, Int $g$ leaves DA stable and defines the identity automorphism of $\overline{A}/DA$. Hence $DA \supseteq D\overline{A}$.