Show that there is no subgroup of $S_n$ of order $(n-1)!/n$. I am trying to show that $S_n$ does not have a subgroup of order $(n-1)!/n$ for any $n$ other than $6$. I have checked it to be true up to $S_{13}$. Any ideas? 
Of course, if $n$ is prime then that order isn't an integer, so obviously there can't be a subgroup of that order. But what about composite $n$?
 A: This conjecture is true, here is a sketch of a proof.
You are looking for a group $G$ of index $n^2$ in $S_n$.
$G$ is either 
(1) intransitive, 
(2) transitive but imprimitive,
(3) primitive.
In the first case, we must have $G\leq S_a \times S_{n-a}$ for some $1\leq a\leq n/2$. But this implies that ${n}\choose{a}$ divides $n^2$. It is not hard to see that this implies that $a=1$ hence $G$ is contained in $S_{n-1}$. We are now asking about a subgroup of index $n$ in $S_{n-1}$, and we can repeat the whole argument here to find that the only option is $F_{20}$ in $S_5$.
Suppose now that $G$ is transitive but imprimitive, so $G\leq S_{n/a}^{a} \rtimes S_a$, for some divisor $1<a<n$ of $n$. As before, this implies that $\frac{n!}{a!(n/a)!^a}$ divides $n^2$ which, with a bit of work, can be shown not to happen. (The left side is typically much bigger than $n^2$.)
Finally, if $G$ is primitive, then it is quite small, in fact $|G|\leq 4^n$, for example by 
Praeger, Cheryl E.; Saxl, Jan
On the orders of primitive permutation groups. 
Bull. London Math. Soc. 12 (1980), no. 4, 303–307. 
and, as in the imprimitive case, this is too small to have index $n^2$. (Except for small $n$ which need to be checked by hand.)
