Proposition 23

Let $G$ be a finite-dimensional real or complex Lie group, $r$ the radical of $L(G)$ (Chapter I, § 5, Definition 2) and $n$ the largest nilpotent ideal of $L(G)$ (Chapter I, § 4, no. 4). *Let $R$ (resp. $N$) be the integral subgroup of $G$ with Lie algebra $r$ (resp. $n$). Then $R$ (resp. $N$) is a solvable (resp. nilpotent) Lie subgroup of $G$, which is invariant under every continuous automorphism of $G$. Every solvable (resp. nilpotent) connected normal subgroup of $G$ is contained in $R$ (resp. $N$).

The group $R$ is solvable ($§ 6$, Proposition 19). Suppose that $K = R$. Let $G'$ be a connected solvable normal subgroup of $G$. Then $\overline{G'}$ is a connected solvable (no. 1, Corollary 2 to Proposition 1) normal Lie subgroup of $G$ ($§ 8$, no. 2, Theorem 2). Hence $L(\overline{G'})$ is a solvable ideal of $L(G)$, whence $L(\overline{G'}) \subset r$ and $\overline{G'} \subset R$. In particular, $\overline{R} \subset R$, hence $R$ is closed and therefore is a Lie subgroup of $G$. Suppose that $K = C$. Let $H$ be the underlying real Lie subgroup of $G$. If $r'$ is the radical of $L(H)$, $ir'$ is a solvable ideal of $L(H)$, whence $r' = ir'$; hence $r \subset r' \subset r$ and, by the above, $R$ is closed in $H$ and therefore in $G$; thus $R$ is a Lie subgroup of $G$. Every connected solvable normal subgroup of $G$ is a connected solvable normal subgroup of $H$ and therefore contained in $R$. Hence we have proved for $K = C$ and for $K = R$ that $R$ is the largest connected solvable normal subgroup of $G$; therefore $R$ is invariant under every continuous automorphism of $G$. The proof for $N$ is completely analogous.