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Question: What is the most frequent order of subgroups of $S_n$? More precisely: Let $a_k$ be the number of subgroups of $S_n$ with order $k$. What is the maximum of $a_k$?

This question came up during a discussion of the open problem of determining the size of the largest antichain in the subgroup lattice of $S_n$.

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  • $\begingroup$ "Most frequent" in terms of number of conjugacy classes of subgroups, or really in terms of number of subgroups? $\endgroup$
    – Stefan Kohl ♦
    Commented Jul 10, 2015 at 22:59
  • $\begingroup$ In terms of number of subgroups. I will edit the question to make this obvious. $\endgroup$
    – Daniel Soltész
    Commented Jul 10, 2015 at 23:00
  • $\begingroup$ I would guess k=2. Do you have any data or OEIS sequences? Gerhard "Yet Another Use For Mode" Paseman, 2015.07.10 $\endgroup$
    – Gerhard Paseman
    Commented Jul 10, 2015 at 23:03
  • $\begingroup$ @GerhardPaseman Well I don't have any OEIS sequence for a_k, but I have one for the total number of subgroups (oeis.org/A005432). But k can not be small by a counting argument. The number of $2$ element subgroups is trivially at most $n!$, but the total number of subgroups has order of magnitude $c^{n^2}$, so the most frequent order by the pidgeon-hole must be at least $c^{n^2}/n!$ which is larger. $\endgroup$
    – Daniel Soltész
    Commented Jul 10, 2015 at 23:25
  • $\begingroup$ How do you prove there are $c^{n^2}$ subgroups? It seems likely to me that by inspecting such a proof it would not be too hard to find the $k$ that maximizes $a_k$. $\endgroup$
    – Eric Wofsey
    Commented Jul 10, 2015 at 23:44

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This is already sort of in the comments, but just to say it clearly:

In the paper Enumerating Finite Groups of Given Order Author(s): L. Pyber Source: Annals of Mathematics, Second Series, Vol. 137, No. 1 (Jan., 1993), pp. 203-220

Pyber proves that the number of subgroups of $S_n$ is bounded below by $2^{(1/16 + o(1))n^2}.$ He shows this by noting that $S_n$ contains $C_2^{[\frac{n}2]},$ and the latter has the specified number of subgroups. He then conjectures that the lower bound is, in fact, sharp. In other words, modulo this conjecture, at least morally, the maximally present order of a subgroup is, indeed, a power of two, and the actual power of two is presumably $2^{[[\frac{n}2]/2]}.$ It seems that Pyber's conjecture is still open.

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  • $\begingroup$ @GeoffRobinson Of course, I meant "maximally abundant order of subgroup", I have fixed now, thanks! $\endgroup$
    – Igor Rivin
    Commented Jul 11, 2015 at 14:28

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