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Proposition 2.18 from "Elementary abelian $p$-subgroups of algebraic groups" by R. Griess. used the term "simply connected almost simple linear algebraic group $G$" without defining it.

In the abstract, "nearly simple linear algebraic group" is defined as: the connected component of the identity $G^{\circ}$ is perfect, $C_G(G^{\circ}) = Z(G^{\circ})$ and $G^{\circ}/Z(G^{\circ})$ is simple.

I don't suppose they are the same... What is the definition for "almost simple" here?

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  • $\begingroup$ The connected component of the identity of a nearly simple linear algebraic group is almost simple. $\endgroup$
    – Wilberd van der Kallen
    Commented Nov 21 at 8:49
  • $\begingroup$ So you mean like $G$ is almost simple if G is connected, perfect and $G/Z(G)$ is simple? Thanks for your input! $\endgroup$
    – scsnm
    Commented Nov 21 at 9:04
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    $\begingroup$ The standard definition of "almost simple" is that a group $G$ is almost simple if $S \le G \le {\rm Aut}(S)$ for some nonabelian simple group $S$. I would guess that the authors are using the standard meaning. $\endgroup$
    – Derek Holt
    Commented Nov 21 at 9:26
  • $\begingroup$ Sorry this may sound stupid. But isn't your definition for finite groups?... $\endgroup$
    – scsnm
    Commented Nov 21 at 9:33
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    $\begingroup$ It's not my definition! But it appears that the one I gave is not the definition of an almost simple algebraic group that is being used here, so ignore my comment. $\endgroup$
    – Derek Holt
    Commented Nov 22 at 10:34

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