I'm reading Humphrey's Linear Algebraic Groups (GTM 21). I don't understand the proof of Corollary 25.3(d).
Corollary. Let $G$ be reductive, $rank_{ss}G = 1$, $T$ a maximal torus of $G$, $Z = Z(G)^\circ$. Then:
(a) $(G,G)$ is semisimple, of dimension 3.
(b) $G = (G,G)\cdot Z$, the intersection of $(G,G)$ with $Z$ being finite.
(c) $C_G(T) = T$, and in particular, $Z(G) \subset T$.
(d) If $\varphi : G \to \mathrm{PGL}(2,\mathsf{K})$ is an epimorphism, $\operatorname{Ker}\varphi = Z(G)$.
You don't need to read all the proof. I can understand the proof of (a) (b) (c). I just don't understand the last sentence: "So (c) and (d) follows."
Proof. We get from part (f) of the theorem an epimorphism $\varphi : G \to \mathrm{PGL}(2,\mathsf{K})$, with $(\operatorname{Ker}\varphi)^\circ = R(G)$. Since $G$ is reductive, $R(G) = Z$ and is a torus. On the other hand, $\mathrm{PGL}(2, \mathsf{K})$ is its own derived group: for example, otherwise the derived group would be solvable, contrary to the semisimplicity of $\mathrm{PGL}(2,\mathsf{K})$. It follows that $\varphi$ maps $(G,G)$ onto $\mathrm{PGL}(2,\mathsf{K})$. Combined with the fact that $(G,G)$ is connected and the fact that $(G,G)\cap Z$ is finite, this proves both (a) and (b). It is easy to see that a maximal torus in $\mathrm{PGL}(2,\mathsf{K})$ is its own centralizer, by computing in $\mathrm{SL}(2,\mathsf{K})$, so $C_G(T)$ and $T$ have the same image under $\varphi$, forcing $T \subset C_G(T) \subset T\cdot \operatorname{Ker}\varphi$. But $T$ has finite index in the right side, while $C_G(T)$ is connected. So (c) and (d) follows.
It seems that the proof doesn't say anything about $\operatorname{Ker}\varphi$ except that $(\operatorname{Ker}\varphi)^\circ = Z$. we also know that $Z(G) \subset \operatorname{Ker}\varphi$, because the center of $\mathrm{PGL}(2,\mathsf{K})$ is trivial. So $Z(G)$ has finite index in $\operatorname{Ker}\varphi$.
How can I prove that $\operatorname{Ker}\varphi \subset Z(G)$? Am i missing something?