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This question concerns the classification of semisimple linear algebraic groups $G$ (over some fixed algebraically closed field $k$), which I am currently learning from Malle and Testerman's book [1].

Below, I list some facts from [1, Section 9.2]. My problem is that I think these facts are contradictory in conjunction, and I would be grateful for any clarification as to what I am misunderstanding.

Facts:

  1. For indecomposable "abstract" root systems $\Phi$ not of type $\operatorname{D}_{2n}$, the (semi)simple algebraic groups $G$ whose root system is isomorphic to $\Phi$ are in bijection with subgroups of the fundamental group of $\Phi$, $\Lambda(\Phi)$, a finite abelian group depending only on $\Phi$ (and not, e.g., on the characteristic of $k$) [1, Theorem 9.13 and text passage afterward]. The $G$ corresponding to the whole fundamental group $\Lambda(\Phi)$ under this correspondence is called simply connected [1, Definition 9.14].
  2. The simply connected semisimple algebraic group with root system of type $\operatorname{A}_{n-1}$ is $\operatorname{SL}_n$, and the fundamental group $\Lambda(\operatorname{A_{n-1}})$ is cyclic of order $n$ [1, Table 9.2, p. 72].
  3. If $G$ is any semisimple group with root system $\Phi$, then there is an isogeny (surjective algebraic group homomorphism with finite kernel) from $G_{\operatorname{sc}}$, the simply connected semisimple group with root system isomorphic to $\Phi$, onto $G$ [1, Proposition 9.15]. Moreover, the kernel of that isogeny is a finite central subgroup of $G_{\operatorname{sc}}$ [1, paragraph before Proposition 9.15].

Said contradiction arises when combining these facts in positive characteristic. Let $p$ be a prime, $k=\overline{\mathbb{F}_p}$, and consider simple linear algebraic groups $G$ over $k$ with root system isomorphic to $\operatorname{A}_{p-1}$. By Facts 1 and 2, there should be precisely two such $G$ (as $\mathbb{Z}/p\mathbb{Z}$ has precisely two subgroups). And by Facts 2 and 3, each such $G$ is of the form $\operatorname{SL}_p(k)/N$, where $N$ is some subgroup of the center $\zeta\operatorname{SL}_p(k)$. But this center is trivial, so $\operatorname{SL}_p(k)$ is the only such quotient group, a contradiction.

Reference:

[1] G. Malle and D. Testerman, Linear Algebraic Groups and Finite Groups of Lie Type, Cambridge University Press (Cambridge studies in advanced mathematics, 133), 2011.

Alexander Bors
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  • $SL_p$ does not have trivial center when viewed as an scheme. That might be the issue here. – Tobias Kildetoft Apr 25 '17 at 17:57
  • @TobiasKildetoft: Thank you for your comment. "Center" definitely means "center of the underlying abstract group" in this context, and the center in this sense of $\operatorname{SL}_p$ should be trivial, if I am not completely mistaken. – Alexander Bors Apr 25 '17 at 18:31
  • I took a quick look through this and indeed, I can also not figure out what is going on, as both Malle-Testerman and Springer (which they refer to for the proof of the isogeny theorem for the given case) work in the context of varieties rather than schemes. – Tobias Kildetoft Apr 26 '17 at 08:37
  • Actually, some further thoughts: You should check that there is an isogeny given as the restriction of the projection $GL_n\to PGL_n$ to $SL_n$. When $n = p$ is the characteristic of the field, this is indeed an isomorphism of abstract groups, but I have a feeling it might not be one of varieties, as the inverse might not be a morphism of varieties (but I did not actually write everything up nicely enough to check this). – Tobias Kildetoft Apr 26 '17 at 08:55
  • @TobiasKildetoft: Thank you. You are right, this isogeny can't be an isomorphism of algebraic groups, since $\operatorname{SL}_n$ and $\operatorname{PGL}_n$ have nonisomorphic root data, as Malle and Testerman point out in their example 9.16(1). So my mistake was the assumption that a surjective homomorphism of algebraic groups $\varphi: G_1\rightarrow G_2$ induces an isomorphism of algebraic groups $G_1/\ker{\varphi}\rightarrow G_2$. – Alexander Bors Apr 26 '17 at 11:41
  • Yes, something like that. But also worth noting is that in fact usually it does, the conditions for when this is are just quite technical (and whether they can even be formulated nicely in terms of varieties I am not sure. Nor could I formulate them myself without looking up a bunch of stuff). – Tobias Kildetoft Apr 26 '17 at 12:07

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