Finite Permutation Groups with Few Orbits Under the Action on the Power Set

TL;DR

This paper develops a general method to classify permutation groups with a small number of orbits on the power set, extending previous classifications for groups with exactly n+1 orbits to those with n+r orbits for 2 ≤ r ≤ 15.

Contribution

It introduces a new systematic approach to classify permutation groups with few orbits on the power set, applicable for small values of r, using computational tools.

Findings

Classified permutation groups with n+r orbits for 2 ≤ r ≤ 15.

Established a general method for such classifications.

Extended known classifications beyond the case of n+1 orbits.

Abstract

We study the orbits under the natural action of a permutation group $G \subseteq S_n$ on the powerset $\mathscr{P}(\{1, \dots , n\})$. The permutation groups having exactly $n+1$ orbits on the powerset can be characterized as set-transitive groups and were fully classified in \cite{BP55}. In this paper, we establish a general method that allows one to classify the permutation groups with $n+r$ set-orbits for a given $r$, and apply it to integers $2 \leq r \leq 15$ using the computer algebra system GAP.

Tables

$n$	$G$	$\left|G\right|$	GAP ID
4	$C_2$	2	4S2
6	${\text{A}}_4$	12	6S31
6	$C_5⋊C_4$	20	6S38
6	${\text{A}}_5$	60	6S50
6	${\text{S}}_5$	120	6S53

$n$	$G$	$\left|G\right|$	GAP ID
5	$C_4$	4	5S6
5	${\text{S}}_3$	6	5S10
5	$C_6$	6	5S11
5	$D_8$	8	5S12
5	$D_{12}$	12	5S15
6	$D_{12}$	12	6S33
7	${\text{S}}_5$	120	7S89
7	${\text{A}}_6$	360	7S93
7	${\text{S}}_6$	720	7S94
8	$\left(\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_2\right)⋊C_2\right)⋊C_3$	192	8T38
8	$\left(\left(\left(C_2\times C_2\times C_2\right)⋊\left(C_2\times C_2\right)\right)⋊C_3\right)⋊C_2$	192	8T40
8	$\left(\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_3\right)⋊C_2\right)⋊C_3$	288	8T42
8	$\left(\left(\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_2\right)⋊C_2\right)⋊C_3\right)⋊C_2$	384	8T44
8	$\left(\left({\text{A}}_4\times {\text{A}}_4\right)⋊C_2\right)⋊C_2$	576	8T45
8	$\left({\text{A}}_4\times {\text{A}}_4\right)⋊C_4$	576	8T46
8	$\left({\text{S}}_4\times {\text{S}}_4\right)⋊C_2$	1152	8T47
9	$\left(C_3\times C_3\right)⋊C_8$	72	9T15
9	$\left(C_3\times C_3\right)⋊Q{D}_{16}$	144	9T19
12	$M_{11}$	7920	12P1

$n$	$G$	$\left|G\right|$	GAP ID
4	$C_2$	2	4S3
6	$C_6$	6	6S17
8	$\left(\left(C_2\times C_2\times C_2\right)⋊\left(C_2\times C_2\right)\right)⋊C_3$	96	8S242
8	$\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_2\right)⋊C_3$	96	8S247
8	$\left(\left(\left(C_2\times C_2\times C_2\right)⋊\left(C_2\times C_2\right)\right)⋊C_3\right)⋊C_2$	192	8S268
8	$\left(\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_2\right)⋊C_3\right)⋊C_2$	192	8S270
8	${\text{A}}_7$	2520	8S293
8	${\text{S}}_7$	5040	8S294
12	$\mathrm{PSL}$$(2,11)⋊C_2$	1320	12T218

$n$	$G$	$\left|G\right|$	GAP ID
5	$C_2\times C_2$	4	5S5
6	$C_2\times {\text{A}}_4$	24	6S40
6	${\text{S}}_4$	24	6S41
6	$C_2\times {\text{S}}_4$	48	6S49
7	${\text{A}}_5$	60	7S81
8	$\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_3\right)⋊C_2$	96	8S240
9	$\left(C_3\times C_3\right)⋊Q_8$	72	9S370
9	${\text{A}}_8$	20160	9S551
9	${\text{S}}_8$	40320	9S552
10	${\text{S}}_6$	720	10T32

$n$	$G$	$\left|G\right|$	GAP ID
6	$C_5$	5	6S14
6	${\text{S}}_3$	6	6S16
6	$C_3\times C_3$	9	6S28
6	$D_{10}$	10	6S29
6	$\left(C_3\times C_3\right)⋊C_2$	18	6S35
6	$C_3\times {\text{S}}_3$	18	6S37
6	${\text{S}}_3\times {\text{S}}_3$	36	6S45
8	$\mathrm{GL}$$(2,3)$	48	8S216
10	${\text{A}}_6$	360	10S1396
10	$\mathrm{PSL}$$(2,8)$	504	10S1448
10	$\mathrm{PSL}$$(2,8)⋊C_3$	1512	10S1539
10	${\text{A}}_9$	181440	10S1590
10	${\text{S}}_9$	362880	10S1591
12	$\mathrm{PSL}$$(2,11)$	660	12T179

$n$	$G$	$\left|G\right|$	GAP ID
5	$C_3$	3	5S4
5	${\text{S}}_3$	6	5S9
7	$D_{14}$	14	7S48
7	$C_2\times \left(C_5⋊C_4\right)$	40	7S75
7	${\text{S}}_5$	120	7S87
7	$C_2\times {\text{A}}_5$	120	7S88
7	$C_2\times {\text{S}}_5$	240	7S92
8	$\mathrm{SL}$$(2,3)$	24	8S154
9	$\left(C_2\times C_2\times C_2\right)⋊C_7$	56	9S355
9	$\left(\left(C_3\times C_3\times C_3\right)⋊C_3\right)⋊C_2$	162	9S457
9	$\left(\left(C_3\times C_3\times C_3\right)⋊C_3\right)⋊C_2$	162	9S458
9	$\left(C_2\times C_2\times C_2\right)⋊\left(C_7⋊C_3\right)$	168	9S462
9	$\left(\left(C_3\times C_3\times C_3\right)⋊C_3\right)⋊\left(C_2\times C_2\right)$	324	9S497
9	$\mathrm{PSL}$$(3,2)⋊C_2$	336	9S499
9	$\left(\left(\left(C_3\times C_3\times C_3\right)⋊\left(C_2\times C_2\right)\right)⋊C_3\right)⋊C_2$	648	9S522
9	$\left(\left(\left(C_3\times C_3\times C_3\right)⋊\left(C_2\times C_2\right)\right)⋊C_3\right)⋊C_2$	648	9S524
9	$\left(\left(\left(\left(C_3\times C_3\times C_3\right)⋊\left(C_2\times C_2\right)\right)⋊C_3\right)⋊C_2\right)⋊C_2$	1296	9S534
9	$\left(C_2\times C_2\times C_2\right)⋊\mathrm{PSL}(3,2)$	1344	9S535
10	$\left(C_5\times C_5\right)⋊\left(\left(C_4\times C_4\right)⋊C_2\right)$	800	10S1496
10	$\left(C_2\times C_2\times C_2\times C_2\right)⋊{\text{S}}_5$	1920	10S1542
10	$C_2\times \left(\left(C_2\times C_2\times C_2\times C_2\right)⋊{\text{A}}_5\right)$	1920	10S1543
10	$C_2\times \left(\left(C_2\times C_2\times C_2\times C_2\right)⋊{\text{S}}_5\right)$	3840	10S1561
10	$\left({\text{A}}_5\times {\text{A}}_5\right)⋊C_2$	7200	10S1569
10	$\left({\text{A}}_5\times {\text{A}}_5\right)⋊\left(C_2\times C_2\right)$	14400	10S1576
10	$\left({\text{A}}_5\times {\text{A}}_5\right)⋊C_4$	14400	10S1577
10	$\left({\text{A}}_5\times {\text{A}}_5\right)⋊D_8$	28800	10S1584
11	${\text{A}}_{10}$	1814400	11S3091
11	${\text{S}}_{10}$	3628800	11S3092

$n$	$G$	$\left|G\right|$	GAP ID
4	1	1	4S1
6	$C_4\times C_2$	8	6S24
6	$D_8$	8	6S27
6	$D_8\times C_2$	16	6S34
8	$C_7⋊C_6$	42	8S196
8	$\mathrm{PSL}(3,2)$	168	8S264
10	$\left(C_5\times C_5\right)⋊C_8$	200	10S1311
10	$\left(C_5\times C_5\right)⋊\left(C_8⋊C_2\right)$	400	10S1418
10	$\left(C_2\times C_2\times C_2\times C_2\right)⋊{\text{A}}_5$	960	10S1504
10	$\left(C_2\times C_2\times C_2\times C_2\right)⋊{\text{S}}_5$	1920	10S1541
12	${\text{A}}_{11}$	11!/2
12	${\text{S}}_{11}$	11!

$n$	$G$	$\left|G\right|$	GAP ID
5	$C_2\times C_2$	4	5S7
6	$D_8$	8	6S26
7	$C_7$	7	7S23
7	$C_3\times {\text{S}}_3$	18	7S51
7	$C_5⋊C_4$	20	7S53
7	${\text{S}}_4$	24	7S63
7	$C_3\times {\text{A}}_4$	36	7S71
7	$\left(C_3\times C_3\right)⋊C_4$	36	7S73
7	${\text{S}}_3\times {\text{S}}_3$	36	7S74
7	$C_2\times {\text{S}}_4$	48	7S78
7	${\text{A}}_4\times {\text{S}}_3$	72	7S82
7	$\left(C_3\times {\text{A}}_4\right)⋊C_2$	72	7S83
7	$C_3\times {\text{S}}_4$	72	7S84
7	$\left({\text{S}}_3\times {\text{S}}_3\right)⋊C_2$	72	7S85
7	${\text{S}}_4\times {\text{S}}_3$	144	7S90
8	$\left(C_4\times C_4\right)⋊C_2$	32	8S181
8	$\left(\left(C_4\times C_4\right)⋊C_2\right)⋊C_2$	64	8S226
8	$\left(\left(C_4\times C_4\times C_4\right)⋊C_4\right)⋊C_2$	64	8S227
8	$\left(\left(C_4\times C_4\times C_4\right)⋊C_4\right)⋊C_2$	64	8S228
8	$\left(D_8\times D_8\right)⋊C_2$	128	8S259
8	$C_2\times {\text{S}}_5$	240	8S272
8	$C_2\times {\text{A}}_6$	720	8S286
8	${\text{S}}_6$	720	8S287
8	$C_2\times {\text{S}}_6$	1440	8S292
9	$C_9⋊C_6$	54	9S354
9	$\mathrm{PSL}(3,2)$	168	9S460
11	$\mathrm{PSL}(2,11)$	660	11S2754
13	${\text{A}}_{12}$	12!/2
13	${\text{S}}_{12}$	12!

$n$	$G$	$\left|G\right|$	GAP ID
6	${\text{S}}_3$	6	6S19
6	${\text{A}}_4$	12	6S30
6	${\text{S}}_4$	24	6S44
8	$C_2\times {\text{A}}_4$	24	8S157
8	$\left(C_2\times C_2\times C_2\right)⋊C_4$	32	8S176
8	$C_2\times {\text{S}}_4$	48	8S214
8	$\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_2\right)⋊C_2$	64	8S229
8	${\text{S}}_5$	120	8S257
10	$\left(C_5\times C_5\right)⋊D_8$	200	10S1305
10	$\left(C_5\times C_5\right)⋊Q_8$	200	10S1307
10	$\left(C_5\times C_5\right)⋊\left(C_4\times C_2\right)$	200	10S1309
10	$\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_5\right)⋊C_4$	320	10S1386
10	$\left(C_5\times C_5\right)⋊\left(\left(C_4\times C_2\right)⋊C_2\right)$	400	10S1419
10	$C_2\times \left(\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_5\right)⋊C_4\right)$	640	10S1472
14	${\text{A}}_{13}$	$13!/2$
14	${\text{S}}_{13}$	$13!$

$n$	$G$	$\left|G\right|$	GAP ID
5	$C_2$	2	5S3
6	$C_2\times C_2\times C_2$	8	6S22
6	$D_8$	8	6S23
7	$C_2\times {\text{A}}_4$	24	7S60
7	${\text{S}}_4$	24	7S67
7	${\text{S}}_4$	24	7S70
8	${\text{S}}_4$	24	8S158
9	$\left(C_3\times C_3\right)⋊C_6$	54	9S352
9	$\left(C_3\times C_3\times C_3\right)⋊C_3$	81	9S401
9	$\left(\left(C_3\times C_3\right)⋊C_3\right)⋊\left(C_2\times C_2\right)$	108	9S425
9	$\left(\left(C_3\times C_3\times C_3\right)⋊C_3\right)⋊C_2$	162	9S459
9	$((C_3\times C_3\times C_3)⋊(C_2\times C_2)⋊C_3$	324	9S496
9	$\left({\text{S}}_3\times {\text{S}}_3\times {\text{S}}_3\right)⋊C_3$	648	9S523
9	$C_2\times {\text{A}}_7$	5040	9S548
9	${\text{S}}_7$	5040	9S549
9	$C_2\times {\text{S}}_7$	10080	9S550
10	$\left(\left(C_2\times C_2\times C_2\times C_2\right)⋊C_5\right)⋊C_4$	320	10S1385
13	$M_{12}$	95040	$\star$
15	${\text{A}}_{14}$	$14!/2$
15	${\text{S}}_{14}$	$14!$

Equations

$$s(G)\ge s(G_1)\times \cdots \times s(G_m).$$

$$s(G)\ge \left(\genfrac{}{}{0ex}{}{s+m-1}{s-1}\right).$$

$$n=2,4,6,8,10,12,14,16,24.$$

$$n=2,4,6,8,12.$$

$$n=3,4,\dots ,16,19,23,24,25,43.$$

$$n=3,4,5,6,7,8,9,11,12.$$

$$n=4,6,8,10,12,14,16,18,22,24,34,42.$$

$$n=4,6,8,10,12.$$

$$n=5,7,9,11,13,15,17,19,21,23,25,33,41,43.$$

$$n=4,5,6,7,8,9,10,11,12.$$

Full text

1. Introduction

Throughout the paper, we let $n\geq 2$ denote a positive integer and let $N=\{1,\dots,n\}$.111We exclude case $n=1$ as the only group action on the set {1} is the trivial action. By a permutation group on n letters we mean a subgroup $G$ of $\textup{{S}}_{n}$ endowed with the natural action $(g,x)\mapsto gx\coloneqq g(x):G\times N\to N$. We call $n$ the degree of the permutation group $G$. The action of $G$ on $N$ induces an action of $G$ on $\mathscr{P}(N)$, given by $(g,X)\mapsto gX=\{gx:x\in X\}:G\times\mathscr{P}(N)\to\mathscr{P}(N)$. In this case, the elements being acted on are subsets of $N$. Accordingly, we shall call the orbits under this action set-orbits. Note that for all $g\in G$ and $X\subseteq N$, $|gX|=|X|$. Since there are $n+1$ distinct sizes of subsets of $N$, it follows that there are at least $n+1$ distinct set-orbits. Additionally, this shows that all sets in the same set-orbit will have the same cardinality. A set-orbit containing sets of size $t$ will be called a $t$-set-orbit. For a given permutation group $G$ on $n$ letters, the number of distinct $t$-set-orbits under the action of $G$ on $\mathscr{P}(N)$ will be denoted by $s_{t}(G)$ and the total number of set-orbits will be denoted by $s(G)$. Clearly $s(G)=\sum_{s=0}^{n}s_{t}(G)$.

Definition 1.1** (Beaumont and Peterson, [2]).**

Given an integer $0\leq t\leq n$, a permutation group $G$ on $n$ letters is called t-set-transitive if for all $t$-element subsets $S,T\subseteq N$, there exists $g\in G$ such that $gS=T$.

In terms of the action of $G$ on $\mathscr{P}(N)$, we see that $G$ is $t$-set-transitive if and only if $s_{t}(G)=1$. In other words, a $t$-set-transitive group is a permutation group with exactly one $t$-set-orbit. Clearly, all permutation groups are [math]-set-transitive and $n$-set-transitive. A $1$-set-transitive permutation group is simply transitive on $N$.

Definition 1.2** (Beaumont and Peterson, [2]).**

A permutation group $G$ on $n$ letters is called set-transitive if $G$ is $t$-set-transitive for all integers $0\leq t\leq n$.

Set-transitive groups were studied as early as 1944 by Neumann and Morgenstern [12]. In [2], Beaumont and Peterson proved that a set-transitive permutation group on $n$ letters, with $n\notin\{5,6,9\}$, always contains the alternating group $\textup{{A}}_{n}$. There has also been significant research devoted to bounding the number of set-orbits $s(G)$ of a degree $n$ permutation group $G$. A trivial lower bound is $s(G)\geq 2^{n}/|G|$. In [5], Cameron proved that if $G$ has order $\exp(o(n^{1/2}))$, then $s(G)=(2^{n}/|G|)(1+o(1))$. In [1], Babai and Pyber showed that if $G$ does not contain $\textup{{A}}_{l}$ ($l>t\geq 4$) as a composition factor, then $\frac{\log_{2}s(G)}{n}\geq\frac{c}{t}$ for some positive constant $c$. In the same paper, they raised the following question: what is $\inf(\frac{\log_{2}s(G)}{n})$ over all solvable degree $n$ permutation groups $G$? This question was answered in a recent paper by Yang in [13].

On the other hand, there has been relatively little work done on the problem of classifying groups in terms of their number of set-orbits. Beaumont and Peterson successfully classified all set-transitive permutation groups in [2], and Kantor classified $2,3,4$-set-transitive groups which are not $2,3,4$-transitive in [9]. These classifications lend itself to a natural generalization in the following sense. Viewed in terms on the action of $G$ on $\mathscr{P}(N)$, we see that $G$ is set-transitive if and only if $s(G)=n+1$. Our paper seeks to completely classify the permutation groups $G$ on $n$ letters satisfying $s(G)=n+r$, for small positive integers $r$. The paper is laid out in the following manner. In Sections 1 and 2, we give necessary definitions and useful facts. In Section 3, we develop a general method mimicking the strategy in [2]. In Section 4, we exemplify the method by classifying groups with $n+2$, $n+3$, $n+4$, and $n+5$ set-orbits. Then we use GAP  [6] to calculate all such groups for $r\leq 15$. Before continuing, we state the following useful facts from [2] and [10].

Given a permutation group $G$ on $n$ letters:

1. (1)

If $G$ contains a $t$-set-transitive subgroup $H$, then $G$ is $t$-set-transitive. 2. (2)

If $G$ is $t$-transitive, then $G$ is $s$-set-transitive for all positive integers $s\leq t$. 3. (3)

The symmetric group $\textup{{S}}_{n}$ is set-transitive. 4. (4)

The alternating group $\textup{{A}}_{n}$ is set-transitive for all $n\geq 3$. 5. (5)

If $s_{t}(G)=1$ for some $2\leq t\leq\left\lfloor\frac{n}{2}\right\rfloor$, then $G$ is primitive.222Throughout this paper, we let $\lfloor x\rfloor$ be the floor function.

2. Main Theorems

Theorem 3 in [2] states that if $G$ is $t$-set-transitive, then $G$ is $(n-t)$-set-transitive for any integer $1\leq t\leq n-1$. It is easy to generalize this to the following observation:

Lemma 2.1**.**

Let $G$ be a permutation group on $n$ letters and let $0\leq t\leq n$ be an integer. Then $s_{t}(G)=s_{n-t}(G)$.

Note that there are $n+1$ distinct sizes of sets in $\mathscr{P}(N)$. Hence, if $n$ is odd, there are an even number of distinct set sizes, and so Lemma  2.1 implies that a permutation group on $n$ letters must have an even number of set-orbits. Therefore, if $n$ is odd and a permutation group $G$ on $n$ letters has $s(G)=n+r$, then $r$ will have to be odd as well. Now consider the situation where $n$ is even and a permutation group $G$ on $n$ letters has an odd number of set-orbits. In this case, since $s_{t}(G)=s_{n-t}(G)$ for all $0\leq t\leq n$, it follows that $s_{n/2}(G)$ must be odd. We summarize these results in the following lemma.

Lemma 2.2**.**

Let $G$ be a permutation group on $n$ letters and suppose $s(G)=n+r$.

1. (1)

If $r$ is even, then $n$ is even. 2. (2)

If $r$ is odd and $n$ is even, then $s_{n/2}(G)$ is odd.

Theorem 2.3**.**

[Livingstone and Wagner, [10]] Given a permutation group $G$ on $n$ letters and an integer $1\leq t\leq\frac{n}{2}$, we have $s_{t-1}(G)\leq s_{t}(G)$.

Remark*.*

This theorem implies that if $1\leq t<\left\lfloor\frac{n}{2}\right\rfloor-1$ then $s_{t}(G)>1$ only if $s_{t-1}(G)>1$. We will make frequent use of this fact in our classification.

Lemma 2.4**.**

Let $G$ be $\left(\left\lfloor\frac{n}{2}\right\rfloor+k\right)$-set-transitive for a positive integer $k$. If there exists a prime $p$ such that $\left\lfloor\frac{n}{2}\right\rfloor+k<p\leq n$, then $G$ is $(n-p+1)$-transitive.

Proof.

By Theorem 7, Corollary 1 in [2], it suffices to show that $p>\max\left(\left\lfloor\frac{n}{2}\right\rfloor+k,n-\left\lfloor\frac{n}{2}\right\rfloor-k\right)=\left\lfloor\frac{n}{2}\right\rfloor+k$. This is true by assumption, so we are done. \hbox to0.0pt{\sqcap\hss}\sqcup

Lemma 2.5**.**

Let $k$ be a positive integer and let $G$ be a permutation group on $n$ letters that does not contain $\textup{{A}}_{n}$. If there exists a prime $p$ such that $\left\lfloor\frac{n}{2}\right\rfloor+k<p<\frac{2n}{3}$, then $G$ is not $\left(\left\lfloor\frac{n}{2}\right\rfloor+k\right)$-set-transitive.

Proof.

Assume for contradiction that $G$ is $\left(\left\lfloor\frac{n}{2}\right\rfloor+k\right)$-set-transitive, then by Lemma 2.4 such a group $G$ is $\left(n-p+1\right)$-transitive. Since $G$ does not contain $\textup{{A}}_{n}$, $G$ is at most $(\frac{n}{3}+1)$-transitive (see [4, page 152]). Now notice that $n-p+1>n-\frac{2n}{3}+1$ since $p<\frac{2n}{3}$ and since $n-\frac{2n}{3}+1=\frac{n}{3}+1$, we have reached a contradiction and thus are done. \hbox to0.0pt{\sqcap\hss}\sqcup

Note that both Lemmas 2.4 and 2.5 hold for $k=0$ when $n$ is even. If we find a maximum $k_{0}$ for which there exists a prime $p$ such that $\left\lfloor\frac{n}{2}\right\rfloor+k_{0}<p<\frac{2}{3}n$, then $G$ cannot be $\left\lfloor\frac{n}{2}\right\rfloor+k$ set-transitive for any $k\leq k_{0}$. If $n$ is odd and such a $k_{0}$ exists, then $G$ is not $\left\lfloor\frac{n}{2}\right\rfloor+1$ set-transitive. Thus it is not $\left\lfloor\frac{n}{2}\right\rfloor$ set-transitive either. These results are summarized in the following corollary.

Corollary 2.6**.**

Let $G$ be a permutation group on $n$ letters not containing $\textup{{A}}_{n}$, where $n$ is even (odd). Let $k_{0}$ be the greatest nonnegative (positive) integer such that there exists a prime $p$ with $\left\lfloor\frac{n}{2}\right\rfloor+k_{0}<p<\frac{2n}{3}$. Then for all $0\leq k\leq k_{0}$, $G$ is not $\left\lfloor\frac{n}{2}\right\rfloor+k$ set-transitive.

Proof.

If $G$ is even, then the existence of such a nonnegative $k_{0}$ shows that there exists a prime $p$ such that $\frac{1}{2}+k\leq\frac{n}{2}+k_{0}<p<\frac{2n}{3}$. By Lemma 2.5, $G$ is not $\frac{n}{2}+k$ set-transitive. For the case of $G$ being odd, a positive $k_{0}$ shows that there exists a prime $p$ such that $\left\lfloor\frac{n}{2}\right\rfloor+k\leq\left\lfloor\frac{n}{2}\right\rfloor+k_{0}<\frac{2n}{3}$. Since this holds at least for $k=1$, it holds for $k=0$ since $n-\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)=\left\lfloor\frac{n}{2}\right\rfloor$. \hbox to0.0pt{\sqcap\hss}\sqcup

Theorem 2.7**.**

A permutation group $G$ on $n$ letters not containing $\textup{{A}}_{n}$ is not $\left\lfloor\frac{n}{2}\right\rfloor+k$ set-transitive for any positive integer values of $k\leq k_{0}$ where $k_{0}$ is the largest integer such that $48-\frac{n+1}{2}\leq k_{0}\leq\frac{5}{54}n-\frac{1}{2}$.

Proof.

First note that $48\leq\frac{n+1}{2}+k_{0}$. Due to a result by Breusch [3], which states that there exists a prime between $x$ and $\frac{9x}{8}$ for $x\geq 48$, there exists a prime $p$ between $\frac{n+1}{2}+k_{0}$ and $\frac{9}{16}n+\frac{9}{8}k_{0}+\frac{9}{16}$. Note that $p>\left\lfloor\frac{n}{2}\right\rfloor+k$ for all $k\leq k_{0}$. Since $k_{0}\leq\frac{5}{54}n-\frac{1}{2}$, we have $p<\frac{9}{16}n+\frac{9}{8}k_{0}+\frac{9}{16}\leq\frac{9}{16}n+\frac{9}{8}\left(\frac{5}{54}n-\frac{1}{2}\right)+\frac{9}{16}=\frac{2}{3}n.$ Thus by Lemma 2.5, $G$ cannot be $\left\lfloor\frac{n}{2}\right\rfloor+k$ set-transitive for any $k\leq k_{0}$. \hbox to0.0pt{\sqcap\hss}\sqcup

Remark*.*

For $k=0$ the theorem still holds true for even $n$. For odd $n$, $G$ is not $\left\lfloor\frac{n}{2}\right\rfloor$ set-transitive when a positive $k_{0}$ exists because $G$ will not be $\left\lfloor\frac{n}{2}\right\rfloor+1$ set-transitive. This theorem is powerful as it shows that often sets of the same size lie in different orbits. The larger $n$ is, the more set-orbits $G$ will have. The theorem can be applied to make an upper bound on the amount of letters $G$ can permute and have exactly $n+r$ set-orbits. For example, the first value of $n$ for which we get an applicable $k_{0}$ is $n=81$, which gives $k_{0}=7$. This implies that a permutation group $G$ on $n$ letters not containing $\textup{{A}}_{n}$ has at least $14$ additional set-orbits, which leads to a corollary.

Corollary 2.8**.**

If a permutation group $G$ on $n$ letters that does not contain $\textup{{A}}_{n}$ has less than $n+16$ set-orbits, then $n\leq 81$.

The following Lemmas are in  [1].

Lemma 2.9**.**

If $L\leq G\leq{\operatorname{Sym}}(\Omega)$, then $s(G)\leq s(L)\leq s(G)\cdot|G:L|$.

Lemma 2.10**.**

Assume $G$ is intransitive on $\Omega$ and has orbits $\Omega_{1},\dots,\Omega_{m}$. Let $G_{i}$ be the restriction of $G$ to $\Omega_{i}$. Then

[TABLE]

Proof.

Since $G\leq G_{1}\times\cdots\times G_{m}$ we can apply Lemma  2.9. Clearly $s(G_{1}\times\cdots\times G_{m})=s(G_{1})\times\cdots\times s(G_{m})$. \hbox to0.0pt{\sqcap\hss}\sqcup

Lemma 2.11**.**

Let $G$ be a transitive permutation group acting on a set $\Omega$ where $|\Omega|=n$. Let $(\Omega_{1},\dots,\Omega_{m})$ denote a system of imprimitivity of $G$ with maximal block-size $b$ $(1\leq b<n;b=1$ if and only if $G$ is primitive; $bm=n$). Let $N$ denote the normal subgroup of $G$ stabilizing each of the blocks $\Omega_{i}$. Let $G_{i}={\operatorname{Stab}}_{G}(\Omega_{i})$, and denote $s=s(G_{1})$. Then

[TABLE]

3. Outline of Methods

In this section we will outline a step-by-step method on how we fully classify groups with $n+r$ set-orbits for $2\leq r\leq 5$ which can also be applied to classify groups with $n+r$ set-orbits for even greater $r$. To outline our method we first reduce the amount of letters $n$ on which $G$ could act, then once we have a reasonable sized list we can test a number of specific permutation groups for the remaining $n$ values. Throughout the whole method we assume that any permutation group we consider does not contain $\textup{{A}}_{n}$ because if it did then $s(G)=n+1$. We will exemplify how to do some of the steps in the calculation sections of this paper.

Step 0

Choose the $r$ value for which you want to classify all groups with $s(G)=n+r$. Let $k_{0}=\left\lfloor\frac{r-1}{2}\right\rfloor$ and find the smallest $n$ such that $k_{0}$ fits in the bounds specified by Theorem 2.7. The smallest $n$ for which any $k_{0}$ appears is $n=81$ with $k_{0}=7$. This is “Step [math]” because for any reasonable $r$, say $r<16$, we know that $n\leq 81$.

Step 1

Now that we have an upper bound on $n$ we can start eliminating some of the possible $n$. Right away, if $r$ is even then we can eliminate all the odd $n$ by Lemma 2.2. Since we need $n+r$ set-orbits, we know that $G$ cannot be $s$ set-transitive for at most $r-1$ different set sizes $s$. This is where we can use Corollary 2.6. If $n$ is odd then we need a $k_{0}$ value such that there is a prime $\frac{n-1}{2}+k_{0}<p<\frac{2n}{3}$ and $2k_{0}>r-1$. If $n$ is even then we need a $k_{0}$ value such that there is a prime $\frac{n}{2}+k_{0}<p<\frac{2n}{3}$ and $2k_{0}+1>r-1$. This is because then we would know that $s_{t}(G)>1$ for $2k_{0}$ different $s$ values in the odd case, and $2k_{0}+1$ different $s$ values in the even case (due to the fact that $s_{\frac{n}{2}}(G)=s_{n-\frac{n}{2}}(G)$). If we look at a table of primes and find any such $p$ values for the necessary $k_{0}$ for a given $n$, then we can remove that $n$ from the list of candidates.

Step 2: Miller’s Method

Now we look at our remaining $n$ values and apply a theorem of Miller [11, vol. III, p. 439] which states that if $n=mp_{0}+r$, $p_{0}$ is prime, $m\in\mathbb{N}$, $p_{0}>m$, $r>m$, then a group $G$ on $n$ symbols, not containing $\textup{{A}}_{n}$, cannot be more than $r$-transitive. We decompose $n$ so we have values of $m$, $p_{0}$, and $r$ that fit the conditions and we try to find a sufficiently small $r$. Using Lemmas 2.4 and 2.5, we see if we can find a small enough $p$ such that $\left\lfloor\frac{n}{2}\right\rfloor+k_{1}<p\leq n$, then $n-p+1>r$ will contradict that $G$ is $\left\lfloor\frac{n}{2}\right\rfloor+k_{1}$ set-transitive. For this, $k_{1}$ is the same as $k_{0}$ in the last step but it is not necessary that $\left\lfloor\frac{n}{2}\right\rfloor+k_{1}<\frac{2n}{3}$. If we reach a contradiction, then we can remove that $n$ from the list.

Step 3

After applying Miller’s Method we have reduced the number of possible groups on $n$ letters that can have $n+r$ set-orbits. We will now reduce the number of groups even further using the following argument. For a given $n$ and $r$ where $r<n-4$, if $s_{2}(G)>1$ then it follows that $s(G)>n+r$, in which case $G$ does not have $n+r$ set-orbits. Thus, we can assume $s_{2}(G)=1$ which implies that $G$ is primitive by Theorem 6 in [2]. Similarly, if $r<n-2$, $s_{1}(G)>1$ implies that $s(G)>n+r$, in which case $G$ does not have $n+r$ set-orbits. By assuming $s_{1}(G)=1$ in this case, we know that $G$ is transitive by Theorem 5 in [2]. In these two cases, we will use [7] to find the structure of the transitive groups. To reduce even further, we will introduce a fact from [2] which follows simply from the orbit stabilizer theorem.

Fact 3.1**.**

If a permutation group $G$ on $n$ letters is $s$ set-transitive, then $n\choose s$ divides $|G|$.

Since we know $s_{k_{1}}(G)=1$, then $n\choose{k_{1}}$ must divide $|G|$. Under the above assumption, $G$ is primitive or transitive, in which case one only needs to check a few groups. If the above restriction on $r$ and $n$ does not hold, then we cannot use the fact that $s_{t}(G)=1$ for any $t$. Under this situation, one must check all non-trivial subgroups of $\textup{{S}}_{n}$. We can use GAP to compute these cases.

Step 4: Computation

At this point, we have a list of possible $n$ for the degree of $G$, and a list of possible groups for each $n$. Now we can simply run a GAP program to calculate $s(G)$ for each candidate, and list out the ones that have $s(G)=n+r$ as desired.

Remark*.*

We will do this process by hand for $r=2,3,4,5$ to exemplify the process and then use a GAP program to go up to $r=15$. Since we are going in a linear order for $r=2,3,4,5$, we will often run into the same group twice. An example of this is when we consider all primitive groups on eight letters such that ${8\choose 3}=56$ divides $|G|$, and in a later section we consider all primitive groups such that ${8\choose 2}=28$ divides $|G|$. We will not consider the same groups twice if we have already calculated $s(G)$, but rather just list the groups we know have $s(G)=n+r$ from previous sections and then only consider the groups whose order is divisible by $28$ but not $56$. There will be several different instances where we can use previous knowledge to reduce the possible number of groups with $n+r$ set-orbits.

4. Groups With Few Set-Orbits

Groups with $n+2$ set-orbits Assume a group $G$ on the set $N=\{1,\ldots,n\}$ has $n+2$ set-orbits.

Looking at only even $n$ and applying Step 1 with $k_{0}=1$, we are left with the following possibilities:

[TABLE]

To exemplify Step 2, we will show one of the calculations done. Since $24=1\times 19+5$, then $G$ is at most $5$-transitive. Using $k_{1}=1$ we need to find the smallest prime $p$ such that $13<p\leq 24$, so $p=17$. Thus, if $G$ was $13$ set-transitive, then it would be $n-p+1=8$-transitive, which would contradict that it is at most $5$-transitive. Thus, we know a permutation group on $24$ letters cannot have $s(G)=n+2$.

We spare the reader from having to see any more of these calculations. At the end of this method, we are left with

[TABLE]

Before continuing to Step 3, we take care of the trivial case of $n=2$. The only permutation groups on $2$ letters are the trivial group and $\textup{{S}}_{2}$. It happens that the trivial group has $s(G)=4$, so we must include it. We will no longer consider $n=2$ for any $r$.

Now we move on to Step 3. We consider transitive groups on $4$ letters and primitive groups on $6,8,12$ letters such that $15,56,792$ divides the group orders, respectively.

Now we move on to Step 4 (computation). We show all of the groups for which $s(G)=n+2$ in the following table. For the GAP ID we let nTr be TransitiveGroup(n,r), nPr be PrimitiveGroup(n,r). We let nSr be ConjugacyClassesSubgroups($S_{n}$)$[\mathrm{r}]$.

[TABLE]

Groups with $n+3$ set-orbits From the previous computations we know that $C_{2}\times C_{2}$ on $4$ letters and $\rm{PSL}(2,7)$ on eight letters have $s(G)=n+3$.

For Step 1, we use a $k_{0}$ value of 2 for odd $n$ and 1 for even $n$. We find primes in the necessary range and are left with

[TABLE]

Applying Step 2, we are left with

[TABLE]

For $n=3$, we check all subgroups of $\textup{{S}}_{3}$. For $n=4,5$ we check the transitive groups. For $n=6,7,8,9,11,12$ we check primitive groups with order divisible by $15,21,56,84,330,792$, respectively. Note that for the even $n$ we are already done from the previous section. We show the results for all groups with $n+3$ set-orbits in a table, using the same GAP identification key as in the previous section.

[TABLE]

Groups with $n+4$ set-orbits We know that the same numbers we were unable to remove in the earlier sections will reappear. Since for even numbers we need $k_{0}=2$, the following $n$’s can no longer be removed using Step 1: $n=18,22,34,42$. So we must continue with

[TABLE]

After Step 2 we have only

[TABLE]

So we must check all subgroups of $\textup{{S}}_{4}$, the transitive groups on $6$ letters, and the primitive groups on $8,10,12$ letters that whose order is divisible by $28,120,495$. We do not recheck the primitive groups on $8$ letters divisible by 56.

The table of all groups with $s(G)=n+4$ is listed below.

[TABLE]

Groups with $n+5$ set-orbits From previous sections we have found five groups with $n+5$ set-orbits. For even $n$ we continue to use $k_{0}=2$, and for odd $n$ we take $k_{0}=3$. After Step 1 $\&$ 2 we have the same even numbers as in the above section, so $n=4,6,8,10,12$. The odd numbers after Step 1 will be the same as in the $n+3$ case, but now we must add $n=17,21,33,41$. We must apply Step 2 to the odd numbers

[TABLE]

After applying Step 2 we have

[TABLE]

Note that once again, we handled all the necessary computations for the even $n$ in the previous section. So we must check all subgroups of $\textup{{S}}_{5}$, the transitive subgroups of $\textup{{S}}_{7}$, and the primitive groups on $9,11$ letters whose order is divisible by $36,165$ but not $84,330$, respectively. We list all groups with $n+5$ set-orbits in the table below.

[TABLE]

Remaining Computations We indeed classify all the cases till $r=15$ using the same method.

The cases of $12\leq r\leq 15$ require some additional work. In each of these cases we could check all subgroups of $\textup{{S}}_{n}(n\leq 11)$ using GAP. Since our computers cannot use GAP to compute all subgroups of $\textup{{S}}_{n}(n\geq 12)$, we will use Lemmas 2.10 and 2.11 to eliminate non-transitive or imprimitive subgroups of $\textup{{S}}_{n}$. It’s important to note that for transitive and primitive subgroups of $\textup{{S}}_{n}$ we can still use GAP since GAP’s built in library has all primitive groups of degree less than or equal to $4096$. We shall discuss the case when $r=12$ in detail and then list the results for $13\leq r\leq 15$ as the process will be very similar.

In the case $r=12$, note that we may check the number of set-orbits of all the subgroups of $\textup{{S}}_{n}$ ($n\leq 11$) using GAP, and we can also check all the transitive and primitive subgroups of $\textup{{S}}_{n}$ for $n$ not too large. We will discuss how to handle the case when $n=12$, and the other possible $n$ values can be checked in a similar way.

Assume the action of group $G$ is not transitive, then the $1$-set-orbits will be partitioned into at least 2 orbits. Assume it is partitioned into at least 3 orbits, then $s_{1}\geq 3$ and by Theorem 2.3, we have $s_{i}\geq 3$ for $1\leq i\leq 11$. Thus $s(G)\geq 3\cdot 11+2=35=12+23$, and this is impossible. Then assume the $1$-set-orbits is partitioned in two, then $G$ can be a subgroup of $\textup{{S}}_{1}\times\textup{{S}}_{11},\textup{{S}}_{2}\times\textup{{S}}_{10},\textup{{S}}_{3}\times\textup{{S}}_{9},\textup{{S}}_{4}\times\textup{{S}}_{8},\textup{{S}}_{5}\times\textup{{S}}_{7},\textup{{S}}_{6}\times\textup{{S}}_{6}$. By Lemma 2.10 we check that in the corresponding cases, the set-orbits will be at least $24,33,40,45,48,49$. In which case the only possibility is a subgroup of $\textup{{S}}_{1}\times\textup{{S}}_{11}$, and in this case, $G$ is either $\textup{{S}}_{11}$ or $\textup{{A}}_{11}$.

Assume the action of group $G$ is transitive but not primitive, note that $12=2\cdot 6$, $3\cdot 4$, $4\cdot 3$, or $6\cdot 2$, and by applying Lemma  2.11, we see that all the cases will lead to more set-orbits.

Assume the action of group $G$ is transitive and primitive, we use GAP to run through all the possible primitive groups of degree $12$ and we see none of them satisfy the requirement.

We list all the results in the next few tables.

Groups with $n+6$ set-orbits:

Groups with $n+7$ set-orbits:

Groups with $n+8$ set-orbits:

Groups with $n+9$ set-orbits:

Groups with $n+10$ set-orbits:

Groups with $n+11$ set-orbits:

Groups with $n+12$ set-orbits:

Groups with $n+13$ set-orbits:

Groups with $n+14$ set-orbits:

Groups with $n+15$ set-orbits

$\star$ Here $M_{12}$ acts transitively on $12$ elements (with $14$ set-orbits in this action) and trivially on the other element.

5. Closing Remarks

Now that a general method is developed for calculating all groups with $n+r$ orbits where $r$ is not too large. The GAP code used for the calculation is available at  [8]. We have successfully classified all the cases for $r\leq 15$. We remark that by far the most computationally taxing step is finding all the subgroups of $\textup{{S}}_{n}$. If one could come up with a method to circumvent this, the classification could go much farther. Another computationally taxing step is calculating how many set-orbits a large group has. If both of these steps can be improved upon, the classification could go further.

Acknowledgements

This research was conducted under NSF-REU grant DMS-1757233 by Betz, Chao-Haft, Gong, and Ter-Saakov during the Summer of 2019 under the supervision of Keller and Yang. The authors gratefully acknowledge the financial support of NSF and also thank Texas State University for providing a great working environment and support. Keller and Yang were also partially supported by grants from the Simons Foundation (#280770 to Thomas M. Keller, #499532 to Yong Yang). The authors are grateful to the referee for the valuable suggestions which greatly improved the manuscript.