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Let $G$ be a reductive algebraic group, $T$ one of its maximal tori, and $\Phi$ the root system relative to $T$. Let $X(T)$ denote the character group of $T$. Then,

The rank of the subgroup $R$ of $X(T)$ generated by $\Phi$ is equal to the semisimple rank of $G$.

Here, semisimple rank of $G$ is defined to be the rank of the semisimple group gotten by factoring the radical of $G$ from $G$.

In order to prove the assertion, the author said:

We need only prove this when $G$ is semisimple. Let $R'$ be the annihilator of $R$ in the 1-psg group $Y(T)$, relative to the dual pairing $X(T) \times Y(T) \rightarrow \mathbb{Z}$. It just has to be shown that $R'=0$. Now $< \alpha, \lambda> =0$ for all $\alpha \in \Phi$ means that $ \lambda (\mathbb{G}_m) \subseteq \mathrm{Ker} \alpha$, hence that $\lambda (\mathbb{G}_m) \subseteq T_{\alpha}$, for all $\alpha$. $\lambda (\mathbb{G}_m) \subseteq Z(G)^{\circ}$, which is trivial because $G$ is semisimple. Therefore $\lambda =0$.

(Page 160 of James Humphrey's Linear Algebraic Groups, GTM21)

I don't know how is the rank of a subgroup of $X(T)$ defined. Is it defined to be the dimension of it over the base field? Also, why does only the semisimple case need to be considered? Why does the triviality of $R'$ imply the result?

Many thanks!

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