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$T$ is a maximal torus of $G$, and $P$ is the set of characters $\beta$ of $T$ for which the weight space $$\mathfrak g_{\beta} = \{ X \in \mathfrak g : \textrm{Ad } t(X) = \beta(t)X, \textrm{ for all } t \in T\}$$ is nonzero. So $\mathfrak g$ is the direct sum of $\mathscr L(Z_G(T))$ and the $\mathfrak g_{\beta} : \beta \in P$.

$P'$ is the set of $\beta \in P$ for which $G_{\beta} := Z_G(\textrm{Ker } \beta)^0)$ is not solvable. I had a few questions about the following paragraph (Springer, Linear Algebraic Groups) which I was hoping anyone intimately familiar with root data might be able to easily answer.

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1 . Why is $(\textrm{Ker } \alpha)^0 = (\textrm{Ker } \beta)^0$?

2 . Why does $\alpha$ being a rational multiple of $\beta$ imply that $\alpha \in P'$?

Not all rational multiples of an element of $P$ should still be in $P$. After all, $P$ is finite.

3 . Why does the axiom (RD 2) hold? In other words, if $\alpha, \beta \in R$, and $s_{\alpha}(x) = x - \langle x, \alpha^{\wedge} \rangle \alpha$, why is $s_{\alpha}(\beta)$ still a member of $R$?

Since $\alpha \in R \subseteq P' \subseteq P$, I know that $s_{\alpha}$ stabilizes $P$ and $P'$. But it's not clear at all to me why $s_{\alpha}$ stabilizes $R$.

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