7
$\begingroup$

It is known that the symmetric group $S_n$ can act transitively on $n+1$ elements if and only if $n=5$.

Are there similar classifications for $S_n$ acting transitively on $n+k$ elements, where $k$ is fixed? For example, is it known whether there are only finitely many such $n$ for every $k$?

$\endgroup$
13
  • 4
    $\begingroup$ I think that for $n >7$, there is a gap between $n-1$ and $\frac{(n-1)(n-2)}{2}$ for irreducible character degrees of $S_{n}$ . On the other hand $S_{n}$ does have a transtive permutation representation of degree $\frac{n(n-1)}{2}$ (on unordered pairs of distinct elements). Yoou may find more information in the books of Fulton and Harris, or James. $\endgroup$ Commented Jan 18, 2021 at 19:38
  • 1
    $\begingroup$ I think you'll find a good overview of what is known in Dixon-Mortimer's Permutation groups. $\endgroup$
    – abx
    Commented Jan 18, 2021 at 19:51
  • 2
    $\begingroup$ Of course I forgot that $A_{n-1}$ is a subgroup of index $2n$ in $S_{n}.$ $\endgroup$ Commented Jan 18, 2021 at 20:57
  • 1
    $\begingroup$ In fact for $n>5$, the smallest set of size greater than $n$ on which $S_n$ acts transitively has size $n(n-1)/2$. The corresponding result for $A_n$, with a lot more detail, is proved in Theorem 5.2A of Dixon & Mortimer - the proof is indeed based on the O'Nan-Scott Theorem. Almost simple maximal subgroups are not problematic, because they are generally much smaller than the intransitive and imprimitive maximal subgroups. $\endgroup$
    – Derek Holt
    Commented Jan 18, 2021 at 21:21
  • 1
    $\begingroup$ @DerekHolt Doesn't $S_n$ act transitively on $S_n/A_{n-1}$ which has size $2n$? $\endgroup$
    – Wojowu
    Commented Jan 18, 2021 at 21:23

1 Answer 1

9
$\begingroup$

Fix $k > 0$.

Suppose that $n > 6$ and $\frac{n(n-3)}{2} > k$. If $[S_n : H] \leq n+k$, then $H$ is one of the following: $S_n$, $A_n$, $S_{n-1}$, or $A_{n-1}$. So in particular if $[S_n : H] = n+k$, then $k = n$ and $H = A_{n-1}$.

See Theorem 5.2B in "Permutation Groups" by Dixon and Mortimer, and also this Math.SE question.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .