Exercise 39, § 9

Let G be a finite-dimensional connected real Lie Group, H an integral subgroup of G and $\overline{H}$ its closure in G. Suppose that H has property (F) (Exercise 38).

(a) Let $x \in \overline{H}$. Then $\mathrm{Ad}_{L(G)}x$ leaves $L(H)$ stable (no. 2, Proposition. 5); let $u(x)$ be its restriction to $L(H)$. Then $u(\overline{H}) = \mathrm{Ad}_{L(H)}(H)$. (Observe that $\mathrm{Ad}_{L(H)}(H)$ is dense in $u(\overline{H})$ and apply property (F).)

(b) Let C be the centre of $\overline{H}$. Then $\overline{H} = C.H$ and C is the closure of the centre of H. (By (a), if $x \in \overline{H}$, there exists $y \in H$ such that $\mathrm{Ad}_{L(H)}x = \mathrm{Ad}_{L(H)}y$. Then $\mathrm{Ad}(x^{-1}y)$ is the identity on $L(H)$, hence $x^{-1}y \in Z_{\overline{H}}(H)$ and $x \in Z_{\overline{H}}(H).H$. By continuity, $Z_{\overline{H}}(H)$ is equal to C. The group C is closed in $\overline{H}$ and hence in G; therefore $\overline{C} \cap \overline{H} \subset C$. Let $x \in C$. There exists a sequence $(x_n)$ in H such that $x_n$ tends to x. Then $\mathrm{Ad}_{L(H)}x_n$ tends to 1 and hence there exists a sequence $(y_n)$ in H such that $y_n$ tends to e and $\mathrm{Ad}_{L(H)}y_n = \mathrm{Ad}_{L(H)}x_n$. Then $y_n^{-1}x_n \in C \cap H$ and $y_n^{-1}x_n$ tends to x.)

(c) $H = \overline{H}$ if and only if the centre of H is closed in G. (Use (b).)