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I know that by a result of Blyth and Robinson published in 1995, all finite simple groups that all of their proper non-abelian simple sections are isomorphic to one of the simple groups A5 or PSL(2,7) are characterized. I am looking for a characterization of all finite simple groups that all of their proper non-abelian simple sections are isomorphic to one of the simple groups A5, PSL(2,7) or PSL(2,9).

Any help is appreciated!

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    $\begingroup$ What might such a characterization look like, roughly? $\endgroup$
    – Derek Holt
    Oct 23 at 12:07
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    $\begingroup$ The Blyth-Robinson 1995 paper, published in Comm. Algebra, is here behind paywall (DOI link). It might be useful to describe the characterization. (I'd expect some kind of composition series of bounded length — with an explicit bound, independent of the group— in which blocks are either solvable or semisimple??) $\endgroup$
    – YCor
    Oct 23 at 12:39
  • $\begingroup$ semi-simple? $\endgroup$
    – markvs
    Oct 23 at 19:14
  • $\begingroup$ @markvs for a finite group I'd (somewhat abusively) use "semisimple" as "direct product of finite non-abelian simple groups". $\endgroup$
    – YCor
    Oct 23 at 22:20
  • $\begingroup$ But you can get examples with arbitrarily large semisimple length by taking iterated wreath products of $A_5$. I am still unsure of what sort of general description one might expect of this class of groups. $\endgroup$
    – Derek Holt
    Oct 24 at 7:46
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I think that this problem can be broken down to a relatively transparent residual case. If $G$ is such a group, you can only hope to control the structure of $G/{\rm Sol}(G)$, where ${\rm Sol}(G)$ is the largest solvable normal subgroup of $G$, and you can also only hope to control the structure of $G^{(\infty)}$, the terminal member of the derived series of $G$. Hence we only consider perfect groups $G$ with ${\rm Sol}(G) = 1.$

Then $F^{\ast}(G)$ is a direct product of simple groups, each isomorphic to one of ${\rm PSL}(2,7)$, ${\rm PSL}(2,9)$ or $A_{5}.$ Furthermore, $G/F^{\ast}(G)$ is isomorphic to a subgroup of ${\rm Out}(F^{\ast}(G)).$ Since the structure of this outer automorphism group is reasonably transparent, it should be possible to continue to pin down the structure of $G$.

Later edit: This answer applies to the question originally asked.

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  • $\begingroup$ Dear Prof. Robinson, thank you very much for your answer. Applogiz, I edited my question. I asked my question for finite "simple" groups. How do you think about simple groups with the above property? $\endgroup$
    – user53093
    Oct 23 at 18:42
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    $\begingroup$ This is a striking example where the failure of writing the question properly drastically changes its meaning. $\endgroup$
    – YCor
    Oct 23 at 18:47
  • $\begingroup$ @YCor, Yes you are right. Sorry for this, but it was a careless and a typo in my question. $\endgroup$
    – user53093
    Oct 23 at 18:57

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