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On a question of permutation groups acting on the power set

  • Jinbao Li , Yong Yang EMAIL logo and Mengxi You
Published/Copyright: February 4, 2025

Abstract

In this work, we settle a conjecture about the number of set-orbits of permutation group acting on the power set.

MSC 2010: 20B05

1 Introduction

Let G be a subgroup of the symmetric group S n and let Ω = { 1 , 2 , , n } where n 2 . The natural action of G on Ω is G × Ω Ω where ( g , m ) g m , g G , and m Ω . This action induces the action of the permutation group G on the power set P ( Ω ) by the mapping G × P ( Ω ) P ( Ω ) where ( g , X ) g X { g x : x X } . It is easy to see that G takes an element X of P ( Ω ) to another element Y of P ( Ω ) where X and Y are both subsets of Ω that contain the same number of elements. We call the orbits under this action set-orbits, and for some positive integer 0 t n , we denote the number of t -set-orbits as s t ( G ) where t -set-orbit is a set-orbit containing sets of size t . Let s ( G ) = t = 0 n s t ( G ) , and we call s ( G ) to be the number of set-orbits under the action of G on P ( Ω ) . We are interested in the following classification question.

Question. Given some integer r 1 , what are all the permutation groups G such that s ( G ) = n + r , where n is the degree of the permutation group?

Set-transitive groups were studied as early as 1944 by von Neumann and Morgenstern [1]. The case for set-transitive groups when r = 1 was studied by Beaumont and Peterson [2]. Their main result was that a group G that does not contain the alternating group A n cannot be n 2 set-transitive, and thus not set-transitive, with exceptions only when n = 5 , 6 , 9 , and the authors proceeded to classify all such exceptions. In [3,4], the cases where 2 r 33 were classified.

Gintz et al. [4] proposed a conjecture that there is no permutation group of degree n > r 11 that will have n + r set-orbits. In this article, we confirm this conjecture.

2 Lemmas

In this section, some lemmas are stated that are used in the later part of the study.

Lemma 2.1

If G is a permutation group acting on n elements and t is an integer such that 0 t n , then s t ( G ) = s n t ( G ) .

Proof

This is [3, Lemma 2.1].

Lemma 2.2

Suppose that G is a permutation group acting on n elements with s ( G ) = n + r and n > r 11 , G must be a primitive group.

Proof

This is [4, Proposition 2.8].□

Lemma 2.3

Let G be a primitive subgroup of S n .

  1. If G is not 3-transitive, then G < n n .

  2. If G does not contain A n , then G < 50 n n .

Proof

This is [5, Corollary 1.1].□

3 Proof of the main result

In this section, we provide two algorithms that are used later.

Algorithm 1
Input: n : The degree of the permutation group
1: Grplist All the primitive permutation groups with degree n not containing A n .
2: M O The largest order of all the groups in G r p l i s t .
3: value ← Binary Variable(1/0)
4: if G r p l i s t is none, then
5: value ← 0
6: else if 2 n M O 2 n 0 , then
7: value ← 0
8: else
9: value ← 1
10: end if
Output value

If the value obtained by Algorithm 1 is equal to 0, then s ( G ) 2 n for all the primitive permutation group G with degree n except S n and A n .

Algorithm 2
Input: n : The degree of the permutation group
1: G r p l i s t All the primitive permutation groups with degree n not containing A n .
2: for G in G r p l i s t do
3: e s t i m a t e = 2 s 0 ( G ) + 2 s 1 ( G ) = 4 ( s 1 ( G ) = 1 since G is primitive)
4: if 2 n
5: s = n 2
6: for k = 2 to s
7: if k s
8: n o r b = 2 s k ( G )
9: else
10: n o r b = s k ( G )
11: end if
12: e s t i m a t e = e s t i m a t e + n o r b
13: if e s t i m a t e 2 n
14: Break Loop
15: end if
16: end for
17: else
18: s = [ n 2 ]
19: for k = 2 to s
20: n o r b = 2 s k ( G )
21: e s t i m a t e = e s t i m a t e + n o r b
22: if e s t i m a t e 2 n
23: Break Loop
24: end if
25: end for
26: end if
27: end for
Output: n , G , e s t i m a t e 2 n

In Algorithm 2, we use Lemma 2.1 to simplify the calculation. For a given integer n , we calculate the value of e s t i m a t e 2 n of all primitive permutation groups with degree n except S n and A n . If e s t i m a t e 2 n obtained by Algorithm 2 is less than 0, then e s t i m a t e 2 n is equal to s ( G ) 2 n . If e s t i m a t e 2 n obtained by Algorithm 2 is greater than or equal to 0, then e s t i m a t e 2 n is less than or equal to s ( G ) 2 n . The GAP [6] code of Algorithms 1 and 2 is available in [7].

Theorem 3.1

There is no permutation group of degree n > r 11 that will have n + r set-orbits.

Proof

We prove the result by contradiction. Suppose G is a permutation group of degree n > r 11 that has n + r set-orbits, then G has to be primitive by Lemma 2.2.

Suppose G contains A n . Then, G is S n or A n . The alternating group A n is set-transitive except for n = 2 . The symmetric group S n is set-transitive. Since s ( G ) = n + 1 , G does not contain A n .

By Lemma 2.3, we have G < 50 n n . Thus, s ( G ) > 2 n 50 n n . We define a function

f ( x ) = 2 x 50 x x 2 x ( x > 0 ) .

Let

g ( x ) = x ln 2 ( x + 1 ) ln x ln 2 ln 50 ( x > 0 ) .

Then,

g ( x ) = ln 2 ln x 2 x 1 x 1 x .

Let

h ( x ) = ln x 2 x .

Then,

h ( x ) = 2 ln x 4 x x .

When x > e 2 , h ( x ) is monotonically increasing, and thus g ( x ) is monotonically increasing. We note that g ( 56 ) > 0 , and therefore, g ( x ) > 0 when x 56 . Thus,

x ln 2 ( x + 1 ) ln x ln 2 ln 50 > 0 ( x 56 ) , ln 2 x 50 x x > ln 2 x , 2 x 50 x x > 2 x .

We have shown that f ( x ) > 0 when x 56 . Thus, when n 56 , we have

s ( G ) 2 n > f ( n ) 2 n > 0 , s ( G ) > 2 n > n + r .

Therefore, we know that the degree of G is less than 56.

Next we use Algorithm 1 to calculate the cases where 12 n < 56 . The calculation shows that s ( G ) 2 n > n + r except when n = 12 , 13 , 14 , 15 , 16 , 21 , 22 , 23 , 24 , 32 .

We then use Algorithm 2 to calculate the cases where n = 12 , 13 , 14 , 15 , 16 , 21 , 22 , 23 , 24 , 32 , and the results are in Table 1.

From Table 1, we see that when n = 12 , s ( G ) 2 n = e s t i m a t e 2 n = 5 , 10 , 2 , or 4 . We can calculate explicitly that in these cases, s ( G ) = 19 , 14, 22, or 20, but r is equal to 11, and n + r = 23 s ( G ) .

In all the other cases, s ( G ) 2 n 0 , and thus s ( G ) 2 n > n + r . Therefore, n is not equal to any integer greater than 12.□

Table 1

Results of Algorithm 2

n G e s t i m a t e 2 n
12 M 11 5
12 M 12 10
12 PSL ( 2 , 11 ) 2
12 PGL ( 2 , 11 ) 4
13 C 13 34
13 D 26 18
13 C 13 C 3 2
13 C 13 C 4 36
13 C 13 C 6 20
13 AGL ( 1 , 13 ) 20
13 PSL ( 3 , 3 ) 4
14 PSL ( 2 , 13 ) 14
14 PGL ( 2 , 13 ) 2
15 A 7 14
15 A 6 8
15 S 6 6
15 PSL ( 4 , 2 ) 4
16 C 16 C 5 48
16 C 16 D 10 36
16 AGL ( 1 , 16 ) 38
16 ( A 4 × A 4 ) C 2 12
16 ( C 16 C 5 ) . C 4 8
16 AGL ( 1 , 16 ) C 2 22
16 C 16 . S 3 × S ( 3 ) 10
16 C 16 . C 9 C 4 4
16 A Γ L ( 1 , 16 ) 4
16 ( S 4 × S 4 ) C 2 4
16 C 16 . PSL ( 4 , 2 ) 0
16 A Γ L ( 2 , 4 ) 12
16 A S L ( 2 , 4 ) 2 8
16 AGL ( 2 , 4 ) 14
16 A S L ( 2 , 4 ) 20
16 C 16 . S 6 12
16 C 16 . A 6 16
16 C 16 S 5 20
16 C 16 A 5 24
16 C 16 A 7 2
17 C 17 66
17 D 34 34
17 C 17 C 4 2
17 C 17 C 8 30
17 AGL ( 1 , 17 ) 50
17 PSL ( 2 , 2 4 ) 4
17 PSL ( 2 , 2 4 ) C 2 16
17 PSL ( 2 , 2 4 ) C 4 = P Γ L ( 2 , 2 4 ) 2
21 PGL ( 2 , 7 ) 52
21 A 7 40
21 S 7 38
21 PSL ( 3 , 4 ) = M 21 24
21 P Σ L ( 3 , 4 ) 12
21 PGL ( 3 , 4 ) 20
21 P Γ L ( 3 , 4 ) 20
22 M 22 4
22 M 22 C 2 12
23 C 23 134
23 D 46 68
23 C 23 C 11 44
23 AGL ( 1 , 23 ) 8
23 M 23 2
24 M 24 1
24 PSL ( 2 , 23 ) 52
24 PGL ( 2 , 23 ) 24
32 AGL ( 1 , 32 ) 32
32 A Γ L ( 1 , 32 ) 46
32 A S L ( 5 , 2 ) 4
32 PSL ( 2 , 31 ) 156
32 PGL ( 2 , 31 ) 80

Acknowledgments

The authors are grateful to the referees for the valuable suggestions which greatly improved the manuscript.

  1. Funding information: This work was partially supported by a grant from the Simons Foundation (No. 918096, to YY), the Scientific Research Foundation for Advanced Talents of Suqian University (2022XRC069), and the Natural Science Foundation of Chongqing, China (cstc2021jcyj-msxmX0511).

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. All authors have contributed equally.

  3. Conflict of interest: The authors state no conflicts of interest.

  4. Data availability statement: The GAP code associated with this study is in [7].

References

[1] J. von Neumann and O. Morgenstern, Theory of Games and Economic Behavior, 2nd rev. edn, Princeton University Press, Princeton, NJ, 1947. Search in Google Scholar

[2] R. A. Beaumont and R. P. Peterson, Set-transitive permutation groups, Canad. J. Math. 7 (1955), 35–42. 10.4153/CJM-1955-005-xSearch in Google Scholar

[3] A. Betz, M. Chao-Haft, T. Gong, T. M. Keller, A. Ter-Saakov, and Y. Yang, Finite permutation groups with few orbits under the action on the power set, Rocky Mountain J. Math. 51 (2021), no. 5, 1553–1565. 10.1216/rmj.2021.51.1553Search in Google Scholar

[4] M. Gintz, M. Kortje, M. Laurence, Z. Wang, and Y. Yang, Permutation groups with few orbits on the power set. II, Publ. Math. Debrecen 102 (2023), no. 1–2, 237–254. 10.5486/PMD.2023.9392Search in Google Scholar

[5] A. Maróti, On the orders of primitive groups, J. Algebra 258 (2002), no. 2, 631–640. 10.1016/S0021-8693(02)00646-4Search in Google Scholar

[6] The GAP Group, GAP - Groups, Algorithms, and Programming, Version 4.13.0; 2024. https://www.gap-system.org. Search in Google Scholar

[7] https://github.com/youmx111/ON-A-QUESTION-OF-PERMUTATION-GROUP-ACTING-ON-POWER-SET. Search in Google Scholar

Received: 2024-08-15
Revised: 2024-11-27
Accepted: 2024-11-28
Published Online: 2025-02-04

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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