Proposition 11

Let G be a finite-dimensional real or complex Lie group, g its Lie algebra, A an integral subgroup of G and $a = L(A)$. Then $N_G(A) = N_G(a)$ and $N_G(A)$ is a Lie subgroup of G containing $\overline{A}$, with Lie algebra $n_g(a)$. The equality $N_G(A) = N_G(a)$ follows from § 6, no. 2, Corollary 2 to Proposition 3. By Proposition 10, $N_G(A)$ is then a Lie subgroup of G with Lie algebra $n_g(a)$. Hence $N_G(A)$ is closed. As $N_G(A) \supset A$, $N_G(A) \supset \overline{A}$.