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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 $n$?

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  • $\begingroup$ Is this just a conjecture of yours, or is it part of an exercise or otherwise known to be true? $\endgroup$ – Thomas Andrews Mar 8 '16 at 2:20
  • $\begingroup$ It is a conjecture. I am unsure whether or not it is known, although I haven't found anything about this anywhere. I was hoping that if it is known, someone would let me know, and otherwise they might have a good idea. $\endgroup$ – user320832 Mar 8 '16 at 2:35
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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.)

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  • $\begingroup$ Thank you! I still need to check some of the details, but that makes sense overall. Do you know if this was known before, and if so where I might find it? Or did you just figure it out on your own? This may be a lemma in a paper I'm writing, and I need to give credit where credit is due. $\endgroup$ – user320832 Mar 8 '16 at 15:42
  • $\begingroup$ I don't think I'd ever seen the particular case $n^2$, but "large" subgroups of $S_n$ are very well-studied, and the general approach I outlined above is well known. If you want references, search for things like "Maximal subgroups of S_n" and so on. $\endgroup$ – verret Mar 9 '16 at 0:54

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