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?

  • $\begingroup$ I don't remember all these details when I learned linear algebraic groups. But I do remember constantly switching between Borel, Springer, and Humphreys when I was stuck on one book's explanation of something. $\endgroup$
    – D_S
    Dec 31, 2017 at 16:01

1 Answer 1


I'm not sure what Humphreys had in mind, but I think the following argument works. If $N$ is a finite normal subgroup of a connected group, then $N$ is contained in the center of that group (Exercise 7.11). We have $(\operatorname{Ker}\phi)^0 = Z(G)^0 = Z$, the radical of $G$. And we want to show that in fact $\operatorname{Ker}\phi = Z(G)$.

In the quotient group $G/(\operatorname{Ker}\phi)^0$, the subgroup $\operatorname{Ker}\phi/(\operatorname{Ker}\phi)^0$ is finite, hence contained in the center of $G/(\operatorname{Ker}\phi)^0$. Let $x \in \operatorname{Ker}\phi$, so that $x(\operatorname{Ker}\phi)^0$ is in the center of $G/(\operatorname{Ker}\phi)^0$. Then for all $y \in G$, $xyx^{-1}y^{-1} \in (\operatorname{Ker}\phi)^0 = Z$. But also $xyxy^{-1}$ is in the derived group of $G$, which has finite intersection with $Z$ (Lemma 19.5). So the image of the morphism

$$G \rightarrow G, y \mapsto xyx^{-1}y^{-1}$$

is finite and irreducible, hence a point. Therefore, $x \in Z(G)$.

On the other hand, if $x \in Z(G)$, then as you say, the image of $x$ in $G/\operatorname{Ker}\phi \cong \operatorname{PGL}_2$ is trivial, because $\operatorname{PGL}_2$ has trivial center.


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