Let $G$ be a connected, reductive group with maximal torus $T$ and Borel subgroup $B$ containing $T$. Let $\Phi = \Phi(G,T)$ be the set of roots of $T$ in $G$, and let $\Phi^+ = \Phi(B,T)$. There is a unique base $\Delta$ of $\Phi$ contained in $\Phi^+$: it is the set of $\alpha \in \Phi^+$ which is not the sum of any two elements of $\Phi^+$.
For any subset $I \subseteq \Delta$, let $$T_I = (\bigcap\limits_{\alpha \in I} \textrm{Ker } \alpha)^0$$ and let $H = Z_G(T_I)$. Then $H$ is a connected, reductive group with maximal torus $T$. I am trying to understand why
$$\Phi(H,T) = [I]$$ where $[I]$ is the set of $\alpha \in \Phi$ which are linear combinations of elements in $I$.
The inclusion $\supseteq$ is easy: the Lie algebra $\mathfrak h$ is the set of $X \in \mathfrak g$ for which $\textrm{Ad } t X = X$ for all $t \in T_I$. Let $\alpha = c_1\beta_1 + \cdots + c_m\beta_m$, where $c_i \in \mathbb{Z}$ and $\beta_i \in I$. If $0 \neq X \in \mathfrak g_{\alpha}$, and $t \in T_I$, then
$$\textrm{Ad } tX = \alpha(t)X = \beta_1(t)^{c_1} \cdots \beta_m(t)^{c_m}X = X$$ This shows that $\mathfrak g_{\alpha} \subseteq \mathfrak h$, whence $\alpha$ is a root of $T$ in $H$.
I have not been able to show the other inclusion yet. I tried showing that if $\alpha \not\in [I]$, then $\mathfrak g_{\alpha} \cap \mathfrak h = 0$: for $\alpha \not\in [I]$, we can write
$$\alpha = c_1\beta_1 + \cdots + c_m\beta_m$$ for $c_i \neq 0$ and $\beta_i \in \Delta$, where say $\beta_1, ... , \beta_s \not\in I$, and $\beta_{s+1}, ... , \beta_m \in I$. If $X \in \mathfrak g_{\alpha} \cap \mathfrak h$, $t \in T_I$, then
$$X = \textrm{Ad } t X = \alpha(t)X = \beta_1(x)^{c_s} \cdots \beta_s(t)^{c_s}X$$ which shows that $T_I \subseteq \textrm{Ker } \beta$, where $\beta = c_1\beta_1 + \cdots + c_s\beta_s$. Is it obvious how to continue?