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Because of my interest in this question, I listed the subgroups of ${\frak S}_n$ for $1\le n\le4$. I found that the number of subgroups are, respectively, $1,2,6,24$. It might be a coincidence, or it could reveal a deep connection.

Is it always the case that ${\frak S}_n$ has exactly $n!$ (non-proper) subgroups ?

Edit. Because of my erroneous claim about ${\frak S}_4$, I tried to delete the question. But this cannot be done once an answer has been posted. So I asked the help of the front desk. Decision is pending.

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    $\begingroup$ Are you sure of your computations? According to oeis.org/A005432, the group $S_4$ has $30$ distinct subgroups, not $24$. $\endgroup$
    – Francesco Polizzi
    Commented Oct 14, 2015 at 12:00
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    $\begingroup$ Taken from an answer on MSE :math.stackexchange.com/q/76176 The number of distinct subgroups of the symmetric group on n points are given for n ≤ 13 in oeis:A005432 (and I checked by hand that the number of subgroups of $S_{4}$ is indeed $30$). $\endgroup$
    – Geoff Robinson
    Commented Oct 14, 2015 at 12:00
  • $\begingroup$ Yeah, questions really should not be deleted with upvoted/accepted answers, except under extraordinary circumstances. Errors in the post don't by themselves merit a deletion; I'm very sorry. $\endgroup$
    – Todd Trimble
    Commented Oct 14, 2015 at 14:48

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As Francesco Polizzi mentions, the answer is no alredy for ${\frak S}_4$: there are $30$ subgroups, but $4!=24$.

Here are some more (very small) calculations:

\begin{array}{|c|c|c|c|} \hline \mathrm{group}& \mathrm{\# subgroups} & n! \\ \hline {\frak S}_1 & 1 &1\\ \hline {\frak S}_2 & 2 &2\\ \hline {\frak S}_3 & 6 &6\\ \hline {\frak S}_4 & 30 &24\\ \hline {\frak S}_5 & 156 &120\\ \hline {\frak S}_6 & 1455 & 720\\ \hline {\frak S}_7 & 11300 & 5040\\ \hline {\frak S}_8 & 151221 & 40320\\ \hline {\frak S}_9 & 1694723 & 362880\\ \hline \end{array}

You can find the number of subgroups up to ${\frak S}_{18}$ in here:

Also, according to this math.stackexchange, the asymptotics for the number of subgroups is $\log(\#\text{sub}) = \Theta(n^2)$.

As for the number of subgroups for arbitrary ${\frak S}_n$, the problem seems wide open.

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If you look at my answer to this question, you will see that the number of proper subgroups grows at least as fast as $2^{n^2/6},$ so much faster than $n!.$ This is a result of Laci Pyber.

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  • $\begingroup$ The MSE answer I mentioned in my comment to the original question also mentions this asymptotic behaviour. $\endgroup$
    – Geoff Robinson
    Commented Oct 14, 2015 at 16:00

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