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What is the asymptotics of the number of the maximal subgroups of $S_n$ (as a function of $n$)? This must be written down somewhere...

EDIT I am actually more interested in the number of conjugacy classes of maximal subgroups (the difference is graphically illustrated by Derek's and Gerry's comments.)

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    $\begingroup$ Are you interested in oeis.org/A066115 "Number of conjugacy classes of maximal subgroups of the symmetric group $S_n$? $\endgroup$ Nov 24, 2015 at 12:36
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    $\begingroup$ Why do you expect it to be written down somewhere? People are more often interested in the conjugacy classes rather than the subgroups themselves. $\endgroup$
    – Derek Holt
    Nov 24, 2015 at 12:54
  • $\begingroup$ I would guess that the number of maximal subgroups is dominated by the $(p-2)!$ conjugates of the maximal subgroup of $S_p$ of order $p(p-1)$, when $p$ is prime. The number of conjugacy classes of maximal subgroups is almost certainly something between $n/2$ and $n$, and closer to $n/2$, because most of them are intransitive. But to prove that, you would need to bound the number of primitive maximals, which may have been done already. $\endgroup$
    – Derek Holt
    Nov 24, 2015 at 13:35
  • $\begingroup$ @DerekHolt In fact I, too, am more interested in the number of conjugacy classes. I will look up Gerry's reference, but if you have another one, do tell... $\endgroup$
    – Igor Rivin
    Nov 24, 2015 at 13:43
  • $\begingroup$ @GerryMyerson Sadly, the oeis seems to have no comments on the asymptotic growth of this quantity... $\endgroup$
    – Igor Rivin
    Nov 24, 2015 at 13:44

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With regard to the conjugacy class question, you should refer to this:

Liebeck, Martin W.; Shalev, Aner Maximal subgroups of symmetric groups. J. Comb. Theory, Ser. A 75, No.2, 341-352 (1996).

The following is a quote from the ZBMath review by W. Knapp:

The purpose of this paper is to give estimates on the number of conjugacy classes of maximal subgroups of the finite symmetric groups $S_n$, on $n$ letters in terms of $n$. It is shown that this number is of the form $(\frac12+o(1))n$. The main work has to be done in establishing that $S_n$ has at most $n^{6/11+o(1)}$ conjugacy classes of primitive maximal subgroups. Of course, the O’Nan-Scott Theorem and the classification of finite simple groups are used.

The same paper is also relevant to the original question of the OP (about the number of maximal subgroups). A further quote from the review:

In the course of the proof, it is shown that any finite almost simple group has at most $n^{17/11+o(1)}$ maximal subgroups of index $n$.

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