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Let $S$ be a finite nonabelian simple group such that the exact sequence

$$1 \to S \to {\rm Aut}(S) \to {\rm Out}(S) \to 1$$

is nonsplit, where $S$ is identified with ${\rm Inn}(S)$. Then there is always a (not necessarily unique) minimal subgroup $A$ in ${\rm Out}(S)$ with respect to the condition that

$$1 \to S \to S.A \to A \to 1$$

is nonsplit. Where can I find a classification of all such pairs $(S,A)$? In particular, is it always true that $|A|=2$? I checked it with the ATLAS that $S$ cannot be sporadic and, if it is alternating, then $S\cong A_6$ and $S.A\cong M_{10}$.

Update: (inspired by Derek Holt's example below)

Can there be nonisomorphic minimal nonsplit extensions $S.A_1$ and $S.A_2$ for a given simple group $S$ with $A_1\cong A_2$?

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It is not always true that $|A|=2$. The outer automorphism of $A_6 \cong {\rm PSL}_2(9)$ induced in $M_{10}$ is the product of a field automorphism and a diagonal automorphism of ${\rm PSL}_2(9)$ and I believe that it is true in general that the product of a field and a diagonal automorphism of the same order of a finite simple group of Lie type gives rise to a nonsplit extension. I checked this by computer in the case of $S = {\rm PSL}_3(64)$, where we get a nonsplit extension $S.A$ with $|A|=3$.

I am guessing that these are essentially the only instances of nonsplit extensions of finite simple groups by their automorphism groups, but I am not certain, and I am afraid that I not aware of any definitive results of this type. It is an interesting question.

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  • $\begingroup$ @Derek thank you. That is an interesting example! If I understand it right, it may be generalized to construct a minimal nonsplit extension with $|A|=r$ for an arbitrary prime $r$. If we take $S={\rm PSL}_r(q^r)$, where $q\equiv 1 (r)$, and define $\phi$ and $\delta$ to be a field and diagonal automorphisms of $S$, both of order $r$, then $S.A =\langle S, \phi\delta \rangle$ should be nonsplit. $\endgroup$
    – Anvita
    Commented Jun 3, 2013 at 1:00
  • $\begingroup$ @Derek However, there are several ways to find a subgroup in $\langle\phi,\delta\rangle$ of order $r$ other than $\langle\phi\rangle$ and $\langle\delta\rangle$; namely, $r-1$ ways, and $\langle \phi\delta \rangle$ is only one of them. Do the other choices result in a nonsplit extension as well? Could there be nonisomorphic such minimal nonsplit extensions? $\endgroup$
    – Anvita
    Commented Jun 3, 2013 at 1:04
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    $\begingroup$ @Anvita: Yes, they are all nonsplit extensions, and so the answer to your new question is yes, there are such nonisomorphic nonpslit extensions. I have convinced myself that the extensions really are nonsplit in the case of a product of diagonal and field automorphisms of order $r$ of ${\rm PSL}_r(q^r)$, where $r|q-1$, but the calculations are a bit messy. $\endgroup$
    – Derek Holt
    Commented Jun 3, 2013 at 14:34

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