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!