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Dec 10, 2018 at 19:41 history edited Derek Holt CC BY-SA 4.0
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Dec 10, 2018 at 15:03 comment added Derek Holt No, you are not being dense, I am using a result about maximal subgroups of subgroups of wreath products here. I will add an explanation to my answer, but that will have to wait until this evening.
Dec 10, 2018 at 14:30 comment added Nick Gill Derek, great comments. Thanks. One thing though: it's not completely clear to me that whenever, say, $A_n$, is a section of a finite group $G$, then its intransitive maximals will yield distinct maximals for $G$... Am I just being dense?
Dec 10, 2018 at 11:52 comment added Derek Holt I think that the answer to your second question is yes if you replace composition factor by chief factor.
Dec 10, 2018 at 10:51 comment added Derek Holt The answer to your first question is yes, essentially because the number of parabolic maximal subgroups of groups of Lie type increases with their Lie rank. For example ${\rm PSL}_n(q)$ has $n-1$ of these. And $A_n$ has about $n/2$ intransitive maximal subgroups with two orbits. But $A_5 \wr C_p$ has $5$ classes of maximal subgroups for all primes $p$, so the answer to the second question is no.
Dec 10, 2018 at 10:32 comment added Nick Gill Thanks Derek, this is really interesting. Let me idly speculate on the basis of 2 minutes of thought: do you think there could be a function $f(c)$ such that if $G$ is a group with at most $c$ conjugacy classes of maximal subgroups, then all simple groups involved with $G$ are of Lie type of rank at most $f(c)$, or alternating of size at most $f(c)$ (or sporadic)? I wonder if one could also bound the number of non-abelian compostion factors by a function of $c$? (I realise this is asymptotics which is not in the spirit of the original post...)
Dec 10, 2018 at 8:41 history edited Derek Holt CC BY-SA 4.0
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Dec 10, 2018 at 8:05 history edited Derek Holt CC BY-SA 4.0
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Dec 9, 2018 at 22:03 history answered Derek Holt CC BY-SA 4.0