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.


  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.


[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.

  • $\begingroup$ $SL_p$ does not have trivial center when viewed as an scheme. That might be the issue here. $\endgroup$ – Tobias Kildetoft Apr 25 '17 at 17:57
  • $\begingroup$ @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. $\endgroup$ – Alexander Bors Apr 25 '17 at 18:31
  • $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Apr 26 '17 at 8:37
  • $\begingroup$ 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). $\endgroup$ – Tobias Kildetoft Apr 26 '17 at 8:55
  • $\begingroup$ @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 $G_1/\ker{\varphi}\rightarrow G_2$. $\endgroup$ – Alexander Bors Apr 26 '17 at 11:41

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