Theorem 1

Let $\alpha \in R(G,T)$. The centralizer $Z_{\alpha}$ of the kernel of $\alpha$ is a connected closed subgroup of G; its centre is Ker $\alpha$; its derived group $D(Z_{\alpha}) = S_{\alpha}$ is a connected closed semi-simple subgroup of G of rank 1. We have $R(Z_{\alpha},T) =\{\alpha ,-\alpha \}$ and dim $Z_{\alpha}=$ dim T + 2.

Let $Z'_{\alpha}$ be the centralizer of (Ker $\alpha$ )$_0$. By Lemma 2, this is a connected closed subgroup of G, and $R(Z'_{\alpha},T)$ is the set of $\beta \in R(G,T)$ such that $\beta$((Ker$\alpha$ )$_0$) $=\{1\}$. Clearly, $\{\alpha ,-\alpha \} \subset R(Z'_{\alpha},T)$. Conversely, let $\beta \in R(Z'_{\alpha},T)$; since (Ker $\alpha$ )$_0$ is of finite index in Ker $\alpha$, there exists an integer $r\not= 0$ such that $t^{r\beta}= 1$ for $t\in$ Ker $\alpha$. From the exactness of the sequence

$0\longrightarrow \mathbf{Z}\longrightarrow X(T)\longrightarrow$ X(Ker $\alpha$ )$\longrightarrow 0$

corresponding by duality to the exact sequence

0 $\longrightarrow$ Ker$\alpha \longrightarrow$ T $\longrightarrow^{\alpha}\mathbf{U}\longrightarrow 0$,

it follows that $r\beta$ is a multiple of $\alpha$; by Chap. VIII, §2, no. 2, Th. 2 (i), this implies that $\beta \in \{\alpha ,-\alpha \}$. Thus, $R(Z'_{\alpha},T) =\{\alpha ,-\alpha \}$. It follows (Lemma 2) that the centre of $Z'_{\alpha}$ is Ker$\alpha$, so $Z'_{\alpha}= Z_{\alpha}$. Finally, by Cor. 1 of Prop. 4 (§1, no. 4), $D(Z_\alpha)$ is a connected closed semi-simple subgroup of $G$; it is of rank $1$ because $\mathscr{L}(Z_\alpha)(\mathbf C)=\mathfrak{g}^\alpha+\mathfrak{g}^{-\alpha}+[\mathfrak{g}^\alpha,\mathfrak{g}^{-\alpha}]$.