I wonder does the following statement holds: 1. For any sufficiently large $n$ the symmetric group $S_n$ contains at least one 2-transitive subgroup other than $S_n$ itself and $A_n$?

  1. Actually I'm interested in whether every symmetric group $S_n$, for sufficiently large $n$, contains a transitive subgroup whose order is divisible by $n(n-1)$. If the first statement holds then it obviously implies the second.

I would appreciate any comments on the above two questions.

  • 3
    I'm pretty sure the 2-transitive question should be answered with a NO. Given that you want it to hold for all $n$, to show a NO answer you can just study those $n$ which are not prime powers -- this rules out the affine groups. So you only have to look at almost simple groups. The list of 2-transitive actions of almost simple groups is well-known... I would have thought you could go through this list and cook up an infinite increasing sequence of positive non-prime-power integers that don't occur as degrees of 2-transitive actions... – Nick Gill Sep 13 at 11:03
  • 2
    The 2-transitive groups have been classified assuming the classification of finite simple groups. Some references are given in the question mathoverflow.net/questions/32351/doubly-transitive-groups. From this it seems to me you probably don't get examples for ALL sufficiently large n but I am no expert. – Benjamin Steinberg Sep 13 at 11:04
  • @BenjaminSteinberg You get examples for infinitely many $n$, including all prime powers $n$ (the affine groups), but the proportion of integers for which there are examples tends to $0$. – Derek Holt Sep 13 at 16:14
  • @DerekHolt, this is why I put all in caps. – Benjamin Steinberg Sep 13 at 16:52
up vote 14 down vote accepted

It has been pointed out in comments that the answer to Question 1 is no. The answer to Question 2 is also no.

Let $p$ be a prime with $p \equiv 1 \bmod 72$, and let $n = 2p+1$. Then $n \equiv 3 \bmod 9$, so $n$ is not prime and is not a proper power. Also, $n \equiv 3 \bmod 8$, so $n+1$ is not a power of $2$.

I claim that no transitive subgroup $G$ of $S_n$ apart from $A_n$ and $S_n$ has order divisible by $n(n-1)$.

Such a $G$ contains an element $g$ of order $p$, and it would have to be primitive because the maximal imprimitive subgroups of $S_n$ are wreath products of the form $S_a \wr S_b$ with $ab=n$, and their orders are not divisible by $p$ (note that $n$ odd implies $a,b \ge 3$). Now by an old theorem of Jordan, if $g$ was a single $p$-cycle then we would have $G=A_n$ or $S_n$, so we can assume that $g$ is a product of two disjoint $p$-cycles.

So $G$ is either 2-transitive, or its point stabilizer has two orbits of the same length $p$, and then $G$ is $3/2$-transitive.

$3/2$-transitive groups are classified in Corollary 3 of a paper by Liebeck, Praeger and Saxl. The list consists of $2$-transitive groups, Frobenius groups, affine groups, and almost simple groups of even degree or degree $21$. A Frobenius group with an element of order $p$ would have to be affine, and is ruled out by our choice of $n$, which is odd, not prime, and not a proper power.

So we are just left with the $2$-transitive groups. For sufficiently large $n$, the only such groups of odd degree (apart from $A_n$ and $S_n$) are the Suzuki groups ${\rm Sz}(2^k)$, which are of degree $2^{2k}+1$, and so cannot occur in this context, the unitary groups $U_3(2^k)$ of degree $2^{3k}+1$, ditto, and the groups containing $L_k(q)$ of degree $n=(q^k-1)/(q-1)$.

Groups in the last class have order dividing $$q^{k(k-1)/2}(q^{k}-1)(q^{k-1}-1)\cdots(q-1)e,$$ where $q=r^e$ with $r$ prime. Our choice of $n$ ensures that $n+1$ is not a power of $2$, which rules out $q=2$, and for $q>2$ , we see that all prime divisors of $|G|$ are less than $p = (n-1)/2$.

  • I don't see offhand the reasoning behind your 4th paragraph ("Since such a $G$..."). Could you give a short hint ? – BS. Sep 16 at 13:30
  • Yes I missed out a step there. I have added more explanation. – Derek Holt Sep 16 at 16:18

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