The number of subgroups of ${\frak S}_n$ 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.
 A: 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:


*

*Derek Holt, Enumerating subgroups of the symmetric group
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.
A: 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.
