Semisimple linear algebraic groups in positive characteristic

Question

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.