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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

Mohsen Shahriari
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Lazare
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2 Answers2

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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 $S_{18}$.

Dietrich Burde
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  • Thanks a lot for your references. I known the case n=p, i also read something about $p$-subgroup of $S_n$. Do you think there exist results like that for particularly $n$ ? I mean for exemple for odd numbers ? – Lazare Aug 01 '20 at 21:14
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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.

Derek Holt
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  • When you said "elementary abelian two subgroups" you mean isomorphic to the trivial group and $\mathbb{Z}/2 \mathbb{Z}$ ? Thanks a lot for this conjecture, have you got recent references (articles) for your upper bound? – Lazare Aug 01 '20 at 21:06
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    For a prime $p$, an elementary abelian $p$-group is an abelian group $G$ with $g^p=1$ for all $g \in G$. It is isomorphic to a direct product of copies of ${\mathbb Z}/p{\mathbb Z}$. As I said, I am not sure exactly what the best known upper bound is - I will need to hunt around for references. – Derek Holt Aug 01 '20 at 22:02
  • Thanks for your answer. If you re-find your references, don't hesitate to make them here. I'm very interested on this upper bound. – Lazare Aug 02 '20 at 13:51
  • Another question about this number: There exists a lower-bound (untrivial) for the number of subgroups of $S_n$ ? – Lazare Aug 04 '20 at 13:34
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    Yes I answered that question in my answer! It would be a good exercise for you to try and prove that, just to try and get a feeel for the problem. – Derek Holt Aug 04 '20 at 15:02
  • Sorry, i was focused on your upper-bound! Do you have the reference of the Pyber's work ? (Not necessary if the method is only to use abelian p-group, but maybe there are comments on the upper bound) – Lazare Aug 04 '20 at 15:37