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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 $\lim_{r \to \infty}r/\iota(r)=2$?

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Sebastien Palcoux
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1 Answers1

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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

Mike Pierce
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verret
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  • Does it uses the classification? – Lior Bary-Soroker Jul 24 '19 at 09:09
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    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. – verret Jul 24 '19 at 23:17