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 $n$?

  • $\begingroup$ 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

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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.