Conjectures around number of subgroups of symmetric group

Question

I am asking if you know unsolved or recently solved conjectures around numbers of subgroups in symmetric or alternating groups. In fact, is there a formula depending of $n$ to count subgroups of order $k$ in the symmetric group $S_n$ ? Particularly, how subgroups $S_n$ contains ?

I know it is possible to use GAP to find this on the cases for $n=1,\dots,15$, but i don't know if formulas around these questions have already been discovered. If you have references on the topic, don't hesitate.

Thanks

Answer 1

In general, it is difficult to find a formula for the number of subgroups of $S_n$. However, for $n=p$ the answer is easy. If $n=p$ is prime then $S_p$ contains $(p-1)!/(p-1) = (p-2)!$ subgroups of order $p$.

For the general case, see the answers at MSE so far:

Enumerating all subgroups of the symmetric group

There are upper bounds for the number by Pyber and Shalev.

See also the paper by Derek Holt for a list of representatives of the conjugacy classes of subgroups of $S_n$ for $n ≤ 18$, including the $7274651$ classes of subgroups of $S_{18}$.

Answer 2

There is a conjecture of Pyber that the number of subgroups (or conjugacy classes of subgroups if you prefer!) of $S_n$ is $2^{\frac{n^2}{16}+o(n^2)}$.

It is easy to prove that this is a lower bound: you can do this by looking at elementary abelian two subgroups in which all orbits have length 1 or 2.

I think the best upper bound proven so far is something like $2^{\frac{n^2}{4}+o(n^2)}$ but I would need to check.