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.

  • $\begingroup$ 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


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.


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