5
$\begingroup$

A number $n$ is called insipid if the groups having a core-free maximal subgroup of index $n$ are exactly $A_n$ and $S_n$. There is an OEIS enter for these numbers: A102842. There are exactly $486$ insipid numbers less than $1000$.

Question: Are there infinitely many insipid numbers?

Let $\iota(r)$ be the number of insipid numbers less than $r$. The following plot (from OEIS) leads to:

Bonus question: Is it true that $\lim_{r \to \infty}r/\iota(r)=2$?

enter image description here

$\endgroup$
16
$\begingroup$

Almost all $n$ are insipid. In fact, the number of non-insipid numbers at most $n$ grows like $2n/\log n$. See the paper

Cameron, Peter J.; Neumann, Peter M.; Teague, David N. On the degrees of primitive permutation groups. Math. Z. 180 (1982), 141–149. doi.org/10.1007/BF01318900

$\endgroup$
  • $\begingroup$ Does it uses the classification? $\endgroup$ – Lior Bary-Soroker Jul 24 at 9:09
  • 1
    $\begingroup$ Very much so. The first sentence of the paper is "The recently announced classification of the finite simple groups has made possible some striking results about permutation groups which had previously defied all attempts at proof. " I think even today, not much can be said without the classification. In the mathscinet review, Praeger writes that pre-classification results only "suggested" that the number of insipid numbers is "probably infinite", so even this much weaker result wasn't known. $\endgroup$ – verret Jul 24 at 23:17

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.