3
$\begingroup$

Let $G$ be a connected reductive affine algebraic group over an algebraically closed field $k$ whose semisimple rank (i.e., dimension of a maximal torus of $G/R\left(G\right)$ , where $R\left(G\right)$ is the identity component of the intersection of all Borel subgroups in G ) is equal to $1$. Consider $\varphi:G\twoheadrightarrow\mathsf{\mathrm{PGL}_{2}}\left(k\right)$ epimorphism of algebraic groups. I want to prove that $ker\left(\varphi\right)=Z\left(G\right)$ . I found a detailed proof here https://mathoverflow.net/questions/158674/why-is-the-semisimple-quotient-of-a-reductive-group-with-semisimple-rank-1-equal but I don't get a point. Namely, I don't understand why the fact that $H$ is abelian implies that it is a group of multiplicative type, hence central in G (actually, I can't understand both the implications). Thanks in advance for any comments, helps or any suggestions.

$\endgroup$
1
  • $\begingroup$ An obvious example of a group like $G$ could be $\mathrm{SL}_{2}$. I don't have any further examples, sorry. $\endgroup$
    – Bimbumbam
    Sep 13, 2016 at 17:20

1 Answer 1

0
$\begingroup$

The point is that $H$, being a kernel, is a normal subgroup of $G$ and thus has reductive component group and thus is reductive. But, this implies that $H^\circ$ is a torus and thus one has an extension

$$1\to H^\circ\to H\to \pi_0(H)\to 1$$

where the latter is a multiplicative group by classical theory. This then implies that $H$ itself is multiplicative quite simply.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.