# $G/Z\left(G\right)$ cannot be isomorphic to $\mathrm{SL}_{2}\left(k\right)$

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.

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

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.