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Consider the following statement about a connected, reductive group $G$ over a field $k$:

Every finite, normal subgroup $N$ of $G$ is central.

In characteristic $0$, this is true, and the proof is easy: since $N$ is smooth, it suffices to show that $N(\overline k)$ is contained in $Z(G)(\overline k)$; so we can consider, for $n \in N(\overline k)$, the orbit map $g \mapsto g n g^{-1}$ on $G_{\overline k}$, whose image is a connected subscheme of $N_{\overline k}$, hence equals the singleton $\{n\}$.

One cannot reason in the same way in positive characteristic, since not every subgroup scheme of $G$ is smooth. Is the statement nonetheless still true?

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Oops, it turns out that I already knew the answer to this, in a different context.

This can fail in characteristic $2$: the kernel of the exceptional isogeny $\operatorname{SO}_{2n + 1} \to \operatorname{Sp}_{2n}$ is not central. I originally supposed that this sort of exceptional behavior might be peculiar to characteristic $2$ (a supposition to which @YCor responded in a comment), but @FriedrichKnop points out in a comment that it is not so: in any non-$0$ characteristic, the kernel of the Frobenius isogeny is not central.

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  • $\begingroup$ Where does the fact in odd characteristic come from? $\endgroup$
    – YCor
    Commented Dec 23, 2021 at 0:40
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    $\begingroup$ The Frobenius kernel is not central either. $\endgroup$ Commented Dec 23, 2021 at 9:25
  • $\begingroup$ The problem is solved by Chevalley's classification of isogenies. $\endgroup$ Commented Dec 23, 2021 at 9:32
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    $\begingroup$ Springer, T. A. Linear algebraic groups. Second edition. Theorem 9.6.5. $\endgroup$ Commented Dec 26, 2021 at 8:53
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    $\begingroup$ Yes that's true if you include the exceptional isogeny of $G_2$ in characteristic $3$. $\endgroup$ Commented Dec 26, 2021 at 18:02

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