TITLE: On what sets can $\mathfrak{S}_n$ act transitively?
QUESTION [6 upvotes]: I would like to know $\mathfrak{S}_n$ could act faithfully transitively on sets with $m$ elements, with $m > n$.
I know that it is not possible if $m = n+1$ except for $n = 5$.
Any ideas ?

REPLY [5 votes]: Here is a way to understand the faithful, transitive actions of $S_n$. We concentrate on small $|X|$. Clearly $\binom{n}{1} \leq X$, and we describe the $X$ that are “near” to $\binom{n}{k}$.
For $n$ sufficiently large compared to $k$, the answer is very simple: $S_n$ is (basically) acting on the subsets of size $k$ of $\{1,2,\ldots,n\}$. For small $k$ we can list the exceptions.

Claim (large $n$): For each $k$ there is some $N$ such that for $n>N$ the only faithful transitive actions of $S_n$ on $X$ with $\binom{n}{k} \leq |X| < \binom{n}{k+1}$ have a stabilizer $H$ satisfying $$\underbrace{S_1 \times \cdots \times S_1}_{k\text{ factors}} \times A_{n-k} \leq H \leq S_k \times S_{n-k},$$ where the left hand and right hand groups are the natural subgroups of $S_n$ acting on the sets $\{1\},\{2\}, \ldots, \{k\}, \{k+1, \ldots, n\}$ and $\{1,2,\ldots,k\}, \{k+1,\ldots,n\}$. This means that $$\binom{n}{k} \leq |X| \leq 2k! \binom{n}{k}=2n(n-1)\cdots(n-k+1).$$

Depending on what kind of relationship you want $|X|$ and $n$ to have, we can make the claim more specific. For instance, if you want $|X|$ to grow linearly with $n$, then for large enough $n$, you'll want $|X| < \binom{n}{2}$.

Claim ($k=1$):  If $S_n$ has a faithful, transitive action on a set $X$ with $|X| < \binom{n}{2}$, then the stabilizer $H$ of a point is conjugate to exactly one of:

the natural $S_1 \times S_{n-1}$ with $|X|= \phantom{2}\binom{n}{1}$, or
the natural $S_1 \times A_{n-1}$ with $|X|= 2\binom{n}{1}$

unless $H$ exists in an exceptional way:

$n=5$, $|X|=6$, $H$ is the natural affine $\operatorname{AGL}(1,5) = F_{20}$
$n=6$, $|X|=6$, $H$ is the natural almost simple $\operatorname{PGL}(2,5)$, 
  outer conjugate to $(H,|X|)=(S_1\times S_5,\binom{n}{1})$
$n=6$, $|X|=10$, $H$ is the natural wreath $S_3 \wr S_2$
$n=6$, $|X|=12$, $H$ is the natural almost simple $\operatorname{PSL}(2,5)$, 
  outer conjugate to $(H,|X|)=(S_1 \times A_5,2\binom{n}{1})$

or $n$ is so small, that some rows need to be ignored:

$n\leq 3$, in which case no such $X$ exists, since $\binom{n}{2} \leq \binom{n}{1}$
$n=4$, in which case the second row is too big, since $\binom{n}{2} \leq 2\binom{n}{1}$


This says that actions of linear size are basically the natural action on $n$ points. We can also handle quadratic easily enough. Again for $n$ large enough, you'll want $|X| < \binom{n}{3}$.

Claim ($k=2$): If $S_n$ has a faithful, transitive action on a set $X$ with $|X|< \binom{n}{3}$, then the stabilizer $H$ of a point is conjugate to exactly one of:

the natural $S_1 \times S_{n-1}$ with $|X|= \phantom{2}\binom{n}{1}$,
the natural $S_1 \times A_{n-1}$ with $|X|= 2\binom{n}{1}$,
the natural $S_2 \times S_{n-2}$ with $|X|= \phantom{2}\binom{n}{2}$,
the natural $S_2 \times A_{n-2}$ with $|X|= 2\binom{n}{2}$,
the natural $A_2 \times S_{n-2}$ with $|X|= 2\binom{n}{2}$,
the natural $2(A_2 \times A_{n-2})$ with $|X|= 2\binom{n}{2}$,
the natural $A_2 \times A_{n-2}$ with $|X|= 4\binom{n}{2}$,

unless $H$ is a new group:

$n=5$, $|X|=6$, $H$ is the natural affine $\operatorname{AGL}(1,5) = F_{20}$
$n=6$, $|X|=6$, $H$ is the natural almost simple $\operatorname{PGL}(2,5)$, 
  outer conjugate to $(H,|X|)=(S_1\times S_5,\binom{n}{1})$
$n=6$, $|X|=10$, $H$ is the natural wreath $S_3 \wr S_2$
$n=6$, $|X|=12$, $H$ is the natural almost simple $\operatorname{PSL}(2,5)$, 
  outer conjugate to $(H,|X|)=(S_1 \times A_5,2\binom{n}{1})$
$n=6$, $|X|=15$, $H$ is an outer conjugate of $(H,|X|) = (S_2 \times S_4,\binom{n}{2})$
$n=7$, $|X|=30$, $H$ is the natural almost simple $\operatorname{PSL}(3,2)$
$n=8$, $|X|=30$, $H$ is the natural affine $\operatorname{ASL}(3,2)$
$n=8$, $|X|=35$, $H$ is the natural wreath $S_4 \wr S_2$

or $H$ does not exist / should be ignored:

$n\leq 3$, in which case no such $X$ exists, since $\binom{n}{3} \leq \binom{n}{1}$
$n=4,5$, in which case the fourth through the seventh row are too big, since $\binom{n}{3} \leq \binom{n}{2}$
$n=6,7,8$, in which case the fifth through seventh row are too big, since $\binom{n}{3} \leq 2\binom{n}{2}$
$n=9,10$, in which case the seventh row is too big, since $\binom{n}{3} \leq 4\binom{n}{2}$


This says the actions of quadratic size are basically the natural actions on pairs of points. The exceptions are either due to stabilizers not existing / having the right index for tiny $n$ (for example, $A_n$ has order $n!/2$ unless $n=0,1$), or honest to goodness other families from the O'Nan Scott classification: acting on block systems to give wreath product stabilizers, affine subgroups acting naturally on $p^n$ points, almost simple groups (either acting in in their “natural” way, or in other primitive ways), etc.
We could even continue to cubic:

Claim $(k=3)$: For $|X| < \binom{n}{4}$, the only new things happening are $$S_1 \times S_1 \times S_1 \times A_{n-3} \leq H \leq S_3 \times S_{n-3}$$ in general (that is $i\binom{n}{3}$ for $i=1,2,2,2,3,4,6,6,6,12$ up to conjugacy). The new exceptions are:

$n=4$, $|X|=6$, an exceptional $H=C_4$
$n=10$, $|X|=126$, the natural wreath $S_5 \wr S_2$
$n=12$, $|X|=462$, the natural wreath $S_6 \wr S_2$

The orders with problematic rows that may need to be ignored stop at $n=52$.

This says the actions of cubic size are basically the natural action on triples of points.
Caveat: Personally, I'm only interested in small $n$. For small $n$ there are very few exceptions, so these claims seem very useful. Also the proofs are tedious but easy. For general $n$, I get very weird inequalities that are basically ridiculous (find $n$ such that $n > n!$) but they have enough complication that I'm not sure how to prove them. I would not be surprised if some of them are still open. Some of the exceptional cases were only checked by order and multiplicity, so for $n < 52$ there could be an error where on of the general rows does not exist ($A_1$ vs $S_1$ type thing), while some other $H$ does exist. One could check this more carefully, but I think my lists give the right feel.