On Orbit Equivalent Permutation Groups

Home , System of imprimitivity

ON ORBIT EQUIVALENT PERMUTATION GROUPS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of the Ohio State University

Keyan Yang

*****

The Ohio State University 2008

Dissertation Committee: Approved by Prof. Akos´ Seress, Advisor

Prof. Ronald Solomon Advisor Prof. Michael Davis Graduate Program in Mathematics

ABSTRACT

Two permutation groups G, H ≤ Sym(Ω) are called orbit equivalent if they have the same orbits on the power set of Ω. Primitive orbit equivalent permutation groups were determined by Seress. In this thesis we prove results toward the classiﬁcation of two-step imprimitive, orbit equivalent permutation groups, which is the next natural step in the program of classifying all transitive, orbit equivalent pairs.

Along the way, we also prove that with a short explicit list of exceptions, all primitive groups have at least four regular orbits on the power set of the underlying set.

ii Dedicated to Binyi

iii ACKNOWLEDGMENTS

I would like to thank my advisor Akos´ Seress for his patience, guidance and constant encouragement which made this work possible. I am grateful for the countless enlightening discussions and suggestions as well as the intellectual support he has provided.

I am grateful to Professor Cai-heng Li for his helpful discussions. I have appre- ciated the many conversations we have had. I would like to thank Professor Ron

Solomon, who has always been ready to share his knowledge with me. I have appreci- ated his enthusiasm and generosity in helping me. I would like to thank the member of my dissertation committee, Professor Michael Davis for his invaluable time.

I am thankful for the Department of Mathematics for their support. I especially thank Cindy Bernlohr for her patience.

Last but not the least I wish to thank my family for their constant support and encouragement.

iv VITA

1978 ...... Born in Hubei, China

2000 ...... B.Sc. in Mathematics, China University of Geosciences, China

2003 ...... MS. in Mathematics, Fudan University, China

2003-Present ...... Graduate Teaching Associate, The Ohio State University

PUBLICATIONS

1. A.´ Seress and K. Yang, On orbit equivalent, two-step imprimitive permutation groups. To appear in ”Computational Group Theory and the Theory of Groups” Contemporary Mathematics vol. 470, American Mathematical Society, 2008.

2. K. Yang and F. Li, Classiﬁcation of torsion-free sheaves on a rational curve with a triple point, Journal of Fudan University. 2004 Vol.43 No.3 P.366-370.

v FIELDS OF STUDY

Major Field: Mathematics

Specialization: Combinatorics

vi TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... ii

Acknowledgments ...... iv

Vita...... v

List of Tables ...... ix

CHAPTER PAGE

1 Introduction ...... 1

1.1 Introduction ...... 1 1.2 Structure of the thesis and main results ...... 2

2 Preliminaries ...... 7

2.1 Permutation groups ...... 7 2.2 Primitive groups ...... 7 2.3 Wreath products ...... 8 2.4 The O’Nan-Scott Theorem ...... 9 2.5 Previous results ...... 11

3 Regular orbits of primitive groups ...... 13

3.1 Product action ...... 13 3.2 An estimate ...... 15 3.3 Proof of Theorem 1.1 ...... 16

4 General results on orbit equivalent groups ...... 22

4.1 Basic results ...... 22 4.2 Groups that are orbit equivalent to a regular group . . . . 23

vii 4.3 Maximal blocks ...... 26 4.4 Orbit equivalent pairs of wreath products ...... 31

5 A ∈ (γ) and B ∈ (γ) ...... 33

5.1 Clean subgroups isomorphic to An ...... 33 5.2 Proof of Theorem 1.2 ...... 37 5.3 Sm o Sn ...... 41 5.4 Sn × Sm ...... 43 5.4.1 Regular orbits of Sn × Sm ...... 44 5.4.2 Groups that are orbit equivalent to Sn × Sm ...... 47 6 Miscellaneous orbit equivalent 2-step imprimitive permutation groups . 55

6.1 A ∈ (α) and B ∈ (β) ...... 55 6.2 A ∈ (α) and B ∈ (γ)...... 67 6.3 A ∈ (β) and B ∈ (α) ...... 70 6.4 A ∈ (β) and B ∈ (γ) ...... 78

7 Orbit equivalent permutation groups up to degree 15 ...... 82

8 GAPcode...... 84

Bibliography ...... 89

viii LIST OF TABLES

TABLE PAGE

6.1 Groups with no regular orbit ...... 71

ix CHAPTER 1

INTRODUCTION

1.1 Introduction

The action of a permutation group G on Ω naturally induces an action on the power set of Ω. Two permutation groups G, H ≤ Sym(Ω) are called orbit equivalent, in notation G ≡ H, if they have the same orbits on the power set. In general, two orbit equivalent permutation groups are not necessarily equal. An open problem posed about 30 years ago was to determine all transitive orbit equivalent pairs.

A trivial observation is that if G ≡ H, then the group generated by G and H is also orbit equivalent to G and H. Hence, it is enough to consider the pairs (G, H) such that H < G. Clearly, if G ≤ Sym(Ω) has a regular orbit, i.e an orbit of size |G| on P(Ω), then no proper subgroup H < G is orbit equivalent to G.

In [18], Siemons proved that two orbit equivalent permutation groups must share the block systems of imprimitivity. Much research work [4] [17] [20] was done on regular orbits of primitive groups and primitive orbit equivalent pairs. Cameron,

Neumann, and Saxl [4] proved that with the exception of ﬁnitely many primitive groups, a primitive group with no regular orbits on the power set of the permutation domain must contain An, but they did not give estimates on the number of exceptional

1 groups. Seress [17] determined all primitive groups with no regular orbits and all primitive orbit equivalent pairs of groups .

Although the primitive case was completely solved, little was known about transitive orbit equivalent pairs. Regarding the orbit structure of a transitive group, in [19],

Siemons proved the orbits of G acting on the power set were completely determined by the orbits of G on the n∗−subsets of Ω, where (1/2)(n − 1) ≤ n∗ ≤ (1/2)(n + 1).

In this dissertation, we start to investigate the group structure of transitive orbit equivalent pairs. We take the natural step and focus on two-step imprimitive groups.

This is the ﬁrst step towards transitive groups from primitive groups.

1.2 Structure of the thesis and main results

In Chapter 2, we provide basic theory and notation for permutation groups. We also list the previous results concerning orbit equivalent permutation groups.

In Chapter 3, we prove the following result concerning regular orbits of primitive groups.

Theorem 1.1. Suppose that a primitive group G ≤ Sym(Ω) has a regular orbit, but no more than three regular orbits on P(Ω). Let n = |Ω| and let k be the number of regular orbits on P(Ω). Then (G, n, k) must be one of the following: (C2, 2, 1), (C3, 3, 2),

(D14, 7, 2), (AGL(1, 8), 8, 3), (PSL(2, 11), 12, 2), (PSL(2, 13), 14, 2), (A7, 15, 2), or

4 (C2 .S5, 16, 2).

We divide primitive permutation groups into the following three classes.

2 (α) Primitive groups with at least one regular orbit on the power set. By Theo-

rem 1.1, with eight exceptions, any primitive group with a regular orbit has at

least four regular orbits.

(β) Primitive groups G ≤ Sn with no regular orbits but (G, n) 6= (An, n) ,(Sn, n).

(γ)(An, n) and (Sn, n).

In chapter 4, we prove some basic properties of orbit equivalent permutation groups.

An imprimitive group G with block system Σ and ∆ as a block is called two-step

Σ ∆ imprimitive, if both the induced group G and the restriction G∆ on the block ∆ are primitive.

Starting in Chapter 5, we investigate two-step imprimitive orbit equivalent groups

Σ ∆ based on the structure of their primitive constituents G and G∆. In Chapter 5, we give the pairs H < G with G ≡ H when the primitive constituents of the two-step imprimitive group G are of type (γ) according to our clas- siﬁcation. The following results were proved.

Σ ∆ Theorem 1.2. Let H < G and H ≡ G. Suppose that (G , n) and (G∆, |∆|) are of type (γ). Suppose b := |∆| ≥ 2 and n := |Σ| ≥ 9 and let K and L be the kernels of the action on Σ of G and H , respectively.

Then one of the following holds.

n−1 n 1. b = 2 and C2 ≤ L ≤ K ≤ C2 or 1 ≤ L ≤ K ≤ C2.

n−1 n−1 2. b = 3 and C3 ≤ L ≤ K ≤ C3 .C2.

3 n n−1 n n−1 n n 3. b = 4 and K4 .C3 ≤ L ≤ K ≤ K4 .C3 .C2, K4 .C3 ≤ L ≤ K ≤ K4 .C3.C2,

n−1 n−1 or K4 .C3 ≤ L ≤ K ≤ K4 .C3.C2. Here K4 is the Klein 4−group. ∼ 4. Ab × An ≤ H < G ≤ Sb × Sn and K∆ = K.

5. Ab o An ≤ G ≤ Sb o Sn and if b∈ / {5, 6, 9} then Ab o An ≤ H.

As two interesting special cases, we determine the groups that are orbit equivalent to Sm × Sn and Sm o Sn with the standard imprimitive action on [m] × [n].

∆ In the ﬁrst half of Chapter 6, , we study the case when the restriction of G∆ is of type (α). If (GΣ, n) is not of type (γ), we have the following.

Theorem 1.3. Let G ≤ Sym(∆ × [n]) be a two-step imprimitive group and suppose

∆ Σ that (A = G∆, |∆|) ∈ (α) but (A, |∆|) 6= (C2, 2) and (B, n) = (G , n) 6∈ (γ). If H < G and H ≡ G then (G, H) must be one of the following:

(i) G = A o B with (A, |∆|) = (D14, 7) and (B, n) = (D10, 5), (AΓL(1, 9), 9), (AGL(2, 3), 9), or (PΓL(2, 9), 10). Moreover, H is an index 2 subgroup of G.

(ii) G = A o B with (A, |∆|) = (C3, 3) and

either (B, n) = (AΓL(1, 8), 8) or (PΓL(2, 8), 9). Moreover, H is an index 3

subgroup of G.

If G belongs to case (i) then H is unique while if G belongs to case (ii) then there are two possibilities for H.

If (GΣ, n) is of type (γ), we have the following result.

Σ ∆ Theorem 1.4. Suppose that (G , n) is of type (γ), (G∆, |∆|) is of type (α), n ≥ 2|∆| − 5 and |∆| > 9. Let H < G and H ≡ G and let K and L be the kernels of

4 G and H acting on Σ. Then K = L and |G : H| = 2. Moreover, G contains a complement to K isomorphic to Sn and H contains a complement to L isomorphic to An.

In the second half of Chapter 6, we give the following description of groups when

∆ (A, |∆|) = (G∆, |∆|) is of type (β).

Theorem 1.5. Let (GΣ, n) with n ≥ 3 be of type (α) and (A, |∆|) is of type (β). If there is a subgroup of H < G with H ≡ G, then A is either solvable or (A, |∆|) is one of (ASL(3, 2), 8), (PΓL(2, 8), 9), (PΓL(2, 9), 10), (PGL(2, 5), 6), (PGL(2, 7), 8), or (S6, 10).

For each of the six possibilities for (A, |∆|) listed in Theorem 1.5, there exists H <

G such that H ≡ G. If (A, |∆|) is one of (ASL(3, 2), 8), (PΓL(2, 8), 9), (PΓL(2, 9), 10), there exists a group A0 < A, A0 ≡ A such that H = A0 o B ≡ A o B.

If (A, |∆|) is one of (PGL(2, 5), 6), (PGL(2, 7), 8) and (S6, 10), then we can take G = A o B and H = (1/2)A o B. Then H ≡ G when B has a transitive index 2 subgroup.

Theorem 1.6. Let (GΣ, n) be of type (γ) and (A, |∆|) of type (β) with n ≥ 60. If there is a subgroup of H < G with H ≡ G, then one of the following holds.

1. A is solvable.

2. (A, |∆|) is one of (ASL(3, 2), 8), (PΓL(2, 8), 9), or (PΓL(2, 9), 10).

3. |G : H| = 2.

In the remaining chapters, we list all orbit equivalent permutation groups up to degree 15 and include the GAP code related to the computer computations.

5 Throughout this thesis, when we refer to concrete permutation representation, we assume the primitive groups are given as in the GAP library [11].

6 CHAPTER 2

PRELIMINARIES

2.1 Permutation groups

Let Ω be a nonempty set. We call a bijection of Ω onto itself a permutation of Ω.

The set of all permutations of Ω is a group, called the symmetric group on Ω. A permutation group on Ω is a subgroup of the symmetric group Sym(Ω) on Ω. The orbit of some α ∈ Ω under G is the set {αg|g ∈ G} ⊂ Ω. A permutation group G on

Ω is called transitive on Ω if G has only one orbit on Ω. An orbit O of G is called regular if |O| = |G|. A transitive group G is called regular on Ω if |G| = |Ω|.

Let G be a group acting on Ω and ∆ ⊂ Ω be a subset of Ω . The set of all elements

g g ∈ G with ∆ = ∆ is called the setwise stabilizer of ∆, denoted by G∆. The set of all elements g ∈ G which ﬁx each point of ∆ is call the pointwise stabilizer of ∆, denoted by G(∆).

2.2 Primitive groups

Let G be a group acting transitively on a set Ω. A nonempty subset ∆ of Ω is called a block for G if for each g ∈ G either ∆g = ∆ or ∆ ∩ ∆g = ∅. Obviously, Ω and a single element set α are blocks of G. These are called trivial blocks. Any other block

7 is called nontrivial. We say that G is primitive if G has no nontrivial blocks on Ω; otherwise G is called imprimitive. Let ∆ be a block of G on Ω and Σ = {∆g : g ∈ G}.

We call Σ = {∆g : g ∈ G} the system of blocks containing ∆.

If there exists no block system Σ0 such that |Σ0| < |Σ|, then Σ is called a maximal block system of G. Clearly Σ is a maximal block system if and only of GΣ is primitive on Σ.

2.3 Wreath products

Let K be a group acting on a set ∆ with |∆| > 1, and let H be a subgroup of Sn.

n −1 The wreath product K o H is the semidirect product K n H where h(k1, ..., kn)h =

n n (k1h , k2h , ..., knh ) for all (k1, k2, ..., kn) ∈ K and h ∈ H. The subgroup K is called the base group and H is called the top group.

There are two natural actions of a wreath product. The ﬁrst one is the imprimitive action.

Let K ≤ Sym(∆) and H ≤ Sym([n]) and let Ω = ∆ × [n]. The imprimitive action K o H on Ω is deﬁned by (i, δ)(k1,k2,...,kn)h = (ih, δki ) for all (i, δ) ∈ Ω and all

(k1, k2, ..., kn)h ∈ K o H. If K is transitive on ∆ and H is transitive on [n], then K o H is transitive on Ω under this action. Moreover, the partition Σ = {∆ × {1}, ∆ ×

{2}, ..., ∆ × {n}} is a block system of imprimitivity for K o H.

An imprimitive permutation group can be treated as a subgroup of a full wreath product.

Lemma 2.1. [3, Theorem 1.8] Let G be a group acting on ∆×[n] with a block system

8 Σ = {∆ × {1}, ∆ × {2}, ..., ∆ × {n}}. Let A = G∆1 and B = GΣ. Then G can be ∆1 identiﬁed with a subgroup of W = A o B acting on ∆ × Σ.

The second action of a wreath product is known as the product action. Let

Ω = ∆n, and we deﬁne the action of K o H on Ω by

k k −1 1h nh (k1,k2,...,kn)h =(δ h ,...,δ h ) (δ1, ..., δn) 1 n

−1 for all (δ1, ..., δn) ∈ Ω and all (k1, ..., kn)h ∈ K o H. The following well-known result gives a criterion for the primitivity of the product action.

Lemma 2.2. [9] Let K and H be two nontrivial groups acting on the sets ∆ and [n], respectively. The wreath product W = K o H is primitive in the product action on

Ω = ∆n if and only if: (i) K acts primitively but not regularly on ∆; and (ii) H acts transitively on [n].

2.4 The O’Nan-Scott Theorem

The socle of a group G is the subgroup generated by the set of all minimal normal subgroups of G, and it is denoted by soc(G). Obviously soc(G) is a characteristic subgroup of G.

The ﬁnite primitive groups can be divided into ﬁve classes by the O’Nan-Scott theorem.

Theorem 2.3. (O’Nan-Scott Theorem)[9] [13]

9 Let X be a ﬁnite primitive group of degree n, and let B be the socle of X. Then

B = T k with k ≥ 1, where T is a simple group and X is one of the following types.

I Aﬃne groups. T = Cp for some prime p, B is the unique minimal normal subgroup of X, and B is regular on Ω of degree n = pk. The set Ω can be identiﬁed

k with B = Cp so that X is a subgroup of the aﬃne group AGL(k, p). II Almost simple groups. Here k = 1, T is a nonabelian simple group and T ≤

X ≤ Aut(T ). Also Tα 6= 1. III(a) Simple diagonal action. Deﬁne

W = {(a1, a2, ..., ak)π : ai ∈ Aut(T ), π ∈ Sk, ai ≡ aj mod Inn(T ) for all i, j}, where π ∈ Sk permutes the components ai naturally. With the obvious multiplication

∼ k as a subgroup of Aut(T )oSn, W is a group with socle B = T , and W = B.(Out(T )×

Sk). The action of W is deﬁned by setting Wα = {(a, a, ..., a).π : a ∈ Aut(T ), π ∈

∼ ∼ k−1 Sk} = Aut(T ) × Sk, Bα = T and n = |T | . A subgroup X of W is of type III(a) if B ≤ X.

III(b) Product action. Let H be a primitive permutation group on the set Γ, of

l type II or III(a). For l > 1, let W = H o Sl with the natural product action on Γ . If K is the socle of H then the socle B of W is Kl. For γ ∈ Γ and α = (γ, γ, ..., γ)

l we have Bα = (Kγ) 6= 1. The subgroup X of W is of type III(b) if B ≤ X. III(c) Twisted wreath action. Let P be a transitive permutation group on {1, 2, ..., k} and let Q be the stabilizer P1. Suppose that there is a homomorphism ϕ : Q → Aut(T ) such that Im(ϕ) contains Inn(T ). Deﬁne B = {f : P → T : f(pq) = f(p)ϕ(q) for all p ∈ P, q ∈ Q}. Then B is a group under pointwise multiplication,

∼ k p and B = T . Let P act on B by f (x) = f(px) for p, x ∈ P . Deﬁne X = T twrϕP

10 be the semidirect product of B by P with this action and deﬁne an action of X on Ω by setting Xα = P . In this case B is the unique minimal normal subgroup of X and is regular on Ω.

2.5 Previous results

Two permutation groups H ≤ G acting on Ω are called orbit equivalent in notation

H ≡ G, if they have the same orbits on the power set of Ω.

Lemma 2.4. [19] Let H,G ≤ Sym(Ω) be orbit equivalent. Then H and G have the same block systems of imprimitivity. In particular, H is primitive if and only if G is primitive.

The primitive groups with regular orbits on the power set of the permutation domain were completely determined by Seress. The result is stated as follows.

Theorem 2.5. [17] The primitive groups G of degree n with no regular orbit on the power set and not containing An are (G, n) = (D10, 5), (AGL(1, 5), 5), (PSL(2, 5), 6), (PGL(2, 5), 6), (AGL(1, 7), 7), (PSL(3, 2), 7), (AΓL(1, 8), 8), (PSL(3, 2), 8), (PSL(3, 2).2, 8),

(ASL(3, 2), 8), (AGL(1, 9), 9), (AΓL(1, 9), 9), (ASL(2, 3), 9), (AGL(2, 3), 9), (PSL(2, 8), 9),

(PΓL(2, 8), 9), (S5, 10), (A6, 10), (S6, 10), (PGL(2, 9), 10), (M10, 10), (PΓL(2, 9), 10),

(PSL(2, 11), 11), (M11, 11), (PGL(2, 11), 12), (M11, 12), (M12, 12), (PSL(3, 3), 13),

4 4 4 4 (PGL(2, 13), 14), (A8, 15), (2 .(A5 × 3).2, 16), (2 .A6, 16), (2 .S6, 16), (2 .A7, 16),

(ASL(4, 2), 16), (PSL(2, 16).2, 17), (PSL(2, 16).4, 17), (PΓL(3, 4), 21), (M22, 22), (M22.2, 22),

(M23, 23), (M24, 24), (ASL(5, 2), 32).

11 By using the previous result and a computer search, the orbit equivalent pairs of primitive groups pairs are listed as follows.

Theorem 2.6. [17] For n ≥ 11, the only orbit equivalent pairs of primitive permutation groups of degree n are An and Sn. For n ≤ 10, the families of orbit equivalent groups are as follows. For n = 3 : {A3,S3}; for n = 4 : {A4,S4}; for n = 5 : {C5,D10}, {AGL(1, 5),A5,S5}; for n = 6 : {PGL(2, 5),A6,S6}; for n = 7 : {A7,S7}; for n = 8 : {AGL(1, 8), AΓL(1, 8), ASL(3, 2)}, {A8,S8}; for n =

9 : {AGL(1, 9), AΓL(1, 9)}, {ASL(2, 3), AGL(2, 3)}, {PSL(2, 8), PΓL(2, 8),A9,S9}; for n = 10 : {PGL(2, 9), PΓL(2, 9)}, {A10,S10}.

The distinguishing number D(G, Ω) of a permutation group G ≤ Sym(Ω) is the minimal number r with the property that there exists an ordered partition P of Ω into r parts such that only the identity element of G ﬁxes P . Equivalently, there exists a function f :Ω → {1, 2, . . . , r} such that if for some g ∈ G we have f(ωg) = f(ω) for all ω ∈ Ω then g = 1. The term distinguishing number was introduced in [1], but the notion was investigated earlier in [4, 6, 12, 16].

Clearly, D(G, Ω) = 2 if and only if the action of G on P(Ω) has a regular orbit,

D(An, [n]) = n − 1, and D(Sn, [n]) = n. For the primitive groups occurring in Theorem 2.5, the distinguishing number was determined by Dolﬁ [10].

Lemma 2.7. [10] The distinguishing number of the groups (M12, 12), (M11, 11), (ASL(3, 2), 8), (PSL(3, 2), 7), and (PGL(2, 5), 6) is four. For all other groups listed in Theorem 2.5, the distinguishing number is three.

12 CHAPTER 3

REGULAR ORBITS OF PRIMITIVE GROUPS

In this chapter,we focus on proving Theorem 1.1. We follow the approach in [17], which is based on the O’Nan-Scott Theorem and the classiﬁcation of ﬁnite simple groups. We examine the cases in the O’Nan-Scott theorem one by one.

3.1 Product action

Lemma 3.1. Let Ω = Ω1 × · · · × Ωr be the set of r-tuples (α1, α2, ..., αr) where

Ωi = {1, 2, . . . , n}, αi ∈ Ωi for some n ≥ 5 and let r ≥ 2. Then G = Sn o Sr in its product action on Ω has at least four regular orbits on P(Ω).

Proof. For r = 2, let

∆1 = {(1, 1), (1, 2), ...(1, n), (2, 2), (3, 3), ..., (n, n), (2, 3), (3, 4), ..., (n − 1, n)} and

∆2 = {(1, 2), ...(1, n), (2, 2), (3, 3), ..., (n, n), (2, 3), (3, 4), ..., (n − 1, n)}.

Furthermore, let ∆3 = Ω \ ∆1 and ∆4 = Ω \ ∆2. Then the four sets ∆i are in

diﬀerent G-orbits in P(Ω) because their cardinalities are diﬀerent. Moreover, G∆i = 1 for i = 1, 2, 3, 4. We prove this for ∆1; for ∆2, the proof is similar, and obviously

G∆i = G∆i−2 for i = 3, 4.

The elements of ∆1 correspond to the edges of a subgraph Γ of the complete

13 bipartite graph Kn,n and we have to show that Aut(Γ) = 1. Any automorphism of Γ must fix 1 ∈ Ω1 as the unique vertex of valency n and so the set E of edges incident to this vertex is also fixed (setwise). Deleting E, the remaining edges of Γ form a path. This path has to be fixed pointwise, because its endpoints 2 ∈ Ω2 and n ∈ Ω1 cannot be exchanged as only one of them is a neighbor of the fixed 1 ∈ Ω1. Hence all vertices of Γ must be fixed.

(i) For r ≥ 3 we proceed inductively. Suppose we have four subsets Γr ⊂ Ω1×· · ·×Ωr

(i) of diﬀerent cardinality such that (Sn o Sr) (i) = 1 for i = 1, 2, 3, 4. Let Γ := Γr r+1 (i) {(α1, α2, . . . , αr+1) | α1 = 1, (α2, α3, . . . , αr+1) ∈ Γr } ∪ {(α, α, . . . , α) | 2 ≤ α ≤ n}.

Let g ∈ (Sn o Sr+1) (i) be arbitrary and consider the natural imprimitive action of Γr+1 S Sn o Sr+1 on 1≤i≤r+1 Ωi. In this action, g must fix 1 ∈ Ω1 as the unique element (i) S covered by the most sequences in Γr+1. So g must fix setwise 2≤i≤r+1 Ωi as the set of (i) elements occurring in sequences together with 1 ∈ Ω1 in Γr+1 and so, by the inductive S (i) hypothesis, g fixes 2≤i≤r+1 Ωi pointwise. Finally, the n − 1 sequences in Γr+1 not containing 1 ∈ Ω1 uniquely determine the elements of Ω1 and g must fix all of them.

Hence g = 1 and so (Sn o Sr+1) (i) = 1. Thus Sn o Sr+1 has at least four regular orbits Γr+1 on Ω1 × · · · × Ωr+1.

Corollary 3.2. Primitive groups of type III(b) or III(c) have at least four regular orbits on the power set.

Proof. Primitive groups of type III(b) are subgroups of some Sm o Sr, in the product action. Primitive groups of type III(c) are subgroups of groups of type III(b) [13,

Remark 2(ii)].

14 3.2 An estimate

Let µ(G) be the minimal degree of G ≤ Sym(Ω) (that is, the minimal number of points in Ω moved by nonidentity elements of G).

Lemma 3.3. Let n > 8 and let G ≤ Sym(Ω), |Ω| = n, be transitive. If G has at most three regular orbits on P(Ω) then |G| > 2µ(G)/2.

Proof. We estimate the size of the set A := {(∆, g) : ∆ ⊂ Ω, g ∈ G, ∆g = ∆} two diﬀerent ways. For any nontrivial element g ∈ G, the number of cycles of g is at most

(n − µ(G)) + µ(G)/2 = n − µ(G)/2 and so the number of subsets ﬁxed by g is at most 2n−µ(G)/2. So |A| ≤ 2n−µ(G)/2(|G| − 1) + 2n.

On the other hand, G has at most three regular orbits on P(Ω). For any set

Γ ⊂ Ω not in the regular orbits, |GΓ| ≥ 2. Moreover, if |Γ| = 1 then |GΓ| = |G|/n and for Γ ∈ {∅, Ω} we have |GΓ| = |G|. Hence |G| |A| ≥ (3|G|) · 1 + (2n − 2 − 3|G| − n) · 2 + n + 2|G| = 2(2n − 2 − n). (3.1) n Comparing the estimates, we obtain

2(2n − 2 − n) ≤ 2n−µ(G)/2(|G| − 1) + 2n.

µ(G)/2 2n+4 µ(G)/2 Solving for |G| yields |G| ≥ 2 + 1 − ( 2n )2 . Since µ(G) ≤ n and n ≥ 9, 2n+4 µ(G)/2 µ(G)/2 we have 1 > ( 2n )2 and |G| > 2 .

15 3.3 Proof of Theorem 1.1

Lemma 3.4. Primitive groups of type III(a) have at least four orbits on the power set.

Proof. It is shown in [17, Lemma 6] that if G ≤ Sym(Ω), |Ω| = n, is of type III(a) √ √ then either |G| < 2 n/2 or |G| < 2µ(G)/2. It is observed in [4] that |G| < 2 n/2 implies |G| < 2µ(G)/2. Hence, in any case, Lemma 3.3 applies.

Lemma 3.5. If G ≤ Sym(Ω) is of type I and |Ω| ≥ 180 then G has at least four regular orbits on P(Ω).

Proof. If G ≤ AGL(d, p) ≤ Sym(Ω), |Ω| = pd, then µ(G) ≥ pd − pd−1 because the

ﬁxed points of any g ∈ GL(d, p) constitute a subspace of the vector space GF(p)d.

Moreover, |G| < pd2+d. Therefore, by Lemma 3.3, G has at least four regular orbits on P(Ω) if

2(pd−pd−1)/2 > pd2+d.

The only pairs (p, d) not satisfying this inequality are (p, 1) for p ≤ 17, (2, 2), (3, 2),

(5, 2), (2, 3), (3, 3), (2, 4), (3, 4), and (2, d) for 5 ≤ d ≤ 8. Among these pairs, only

(p, d) = (2, 8) yields |Ω| > 180.

In the case p = 2, d = 8, if G = AGL(8, 2) has no more than three regular orbits on P(Ω) then by equation (3.1) in the proof of Lemma 3.3, the set A := {(∆, g):

∆g = ∆, ∆ ⊂ Ω, g ∈ G} satisﬁes |A| ≥ 2(2256 − 2 − 28). However, we used a GAP program to compute the exact size of A (by ﬁrst computing the conjugacy classes of

G) and it turns out that |A| is less than this bound.

16 Lemma 3.6. Let m ≥ 9, 2 ≤ k ≤ m/2, and let G = Sm act on the k-subsets of {1, 2, ..., m}. Then G has at least four regular orbits on the power set.

Proof. For n ≥ 7, we deﬁne a graph Xn as the union of three paths of length 1, 2, and n−4, respectively (the length of a path is the number of edges). The three paths share one common endpoint and have no other common vertices. We also deﬁne a graph

Yn on {1, 2, ..., n} with edge set {{1, 2}, {2, 3},..., {n − 2, n − 1}} ∪ {{2, n}, {3, n}}.

For k = 2, take ∆1 as the edge set of the graph Xm on {1, 2, ..., m}. Clearly, the

automorphism group of Xm is trivial and so G∆1 = 1.

For k = 3, take Xm−1 on {1, 2, ..., m − 1} and denote the edges of Xm−1 by e1, e2, . . . , ep, where p = m − 2. Let ∆1 = {e1 ∪ {m}, e2 ∪ {m}, . . . , ep ∪ {m}}. Then

G∆1 = 1 because on its natural action on m points, any g ∈ G∆1 must ﬁx m and the graph Xm−1.

For k = 4, take Xm−2 on {1, 2, ..., m − 2} and denote the edges of Xm−1 by e1, e2, . . . , ep, where p = m − 3. Let ∆1 = {e1 ∪ {m − 1, m}, e2 ∪ {m − 1, m}, . . . , ep ∪ {m−1, m}, {m−2, m, a, b}} where {a, b} ⊂ {1, 2, . . . , m−2} but {a, b} is not an edge

of Xm−2. Then G∆1 = 1 because on its natural action on m points, any g ∈ G∆1 must

ﬁx {m − 1, m} and the graph Xm−2; moreover, the last element in ∆1 distinguishes m − 1 and m.

For m/2 ≥ k ≥ 5, take Xm−k+2 on the vertex set {k − 1, k, ..., m} and denote the edges of Xm−k+2 by e1, e2, . . . , ep, where p = m − k + 1. Let Γ1 = {e1 ∪ {1, 2, . . . , k −

2}, e2 ∪ {1, 2, . . . , k − 2}, . . . , ep ∪ {1, 2, . . . , k − 2}} and let Γ2 = {{k − 1, k − 2} ∪

{m, m − 1, . . . , m − k + 3},..., {2, 1} ∪ {m, m − 1, . . . , m − k + 3}}. Let ∆1 = Γ1 ∪ Γ2.

17 For any i ∈ {1, 2, . . . , k − 2}, i occurs p + 1 or p + 2 times in ∆1. On the other hand, any i > k − 2 occurs at most k − 2 + 3 = k + 1 times in ∆1. Notice that

p + 1 = m − k + 2 > k + 1. So {1, 2, . . . , k − 2} is ﬁxed setwise by G∆1 . This implies

Xm−k+1 is ﬁxed, so in the natural action of G∆1 on m points, the set {k − 1, . . . , m} is ﬁxed pointwise. Then the elements of Γ2 distinguish the points in {1, 2, . . . , k − 2}

and so G∆1 = 1.

In the constructions above, we can replace Xn by Yn to get another set ∆2 whose stabilizer is trivial. Let ∆3 = Ω \ ∆1 and ∆4 = Ω \ ∆2. It is easy to see ∆1, ∆2, ∆3 and ∆4 are of diﬀerent cardinality and so they are from four diﬀerent regular orbits on P(Ω).

For primitive groups of type II, we use the following estimates of Liebeck and Saxl

[15].

Theorem 3.7. [15] Let G ≤ Sn be almost simple and suppose that G 6= Am,Sm acting on the k−subsets of {1, 2, . . . , m}. Then

(a) µ(G) ≥ n/3.

(b) If soc(G) is alternating or sporadic, then µ(G) ≥ n/2.

− 4 PSp(4, 3), PΩ (4, 3), PSL(2, q) then µ(G) ≥ n(1 − 3q ).

(d) If soc(G) = PSL(2, q) then, with the exception of an explicit list, µ(G) ≥ n(1 −

4 3q ).

18 Lemma 3.8. If G ≤ Sym(Ω) is of type II, |Ω| ≥ 180, and G does not contain Alt(Ω) then G has at least four regular orbits on P(Ω).

Proof. Using Theorem 3.7, in [17, Lemma 12] it was established that with ﬁnitely many exceptions, almost simple primitive groups G ≤ Sym(Ω) not considered in

Lemma 3.6 and not containing Alt(Ω) satisfy |G| ≤ 2µ(G)/2 (and so, by Lemma 3.3, they have at least four regular orbits on P(Ω)). In the exceptional cases, soc(G) occurs on the following list.

(1) Higman-Sims group and Mathieu groups.

(2) Am for some m ≤ 8.

(3) Sz(8).

(4) PSL(2, q) for some q ≤ 32.

(5) One of the classical groups PSL(5, 3), PSL(4, 4), PSL(3, 3), PSL(3, 5), PSL(4, 3),

PSL(8, 2), PSL(7, 2), PSL(6, 2), PSL(5, 2) , PSL(4, 2), PSL(3, 2), PSp(6, 2),

PSp(4, 2), PSU(3, 3), PSU(3, 4), PSU(4, 2).

For the above exceptions, we use Theorem 3.7(a) and Lemma 3.3 to establish a lower bound on n := |Ω| which guarantees at least four regular orbits on P(Ω).

n/6 µ(G)/2 n/6 If |G| ≤ 2 , that is, if n ≥ 6 log2 |G|, then 2 ≥ 2 ≥ |G| and we are done. This estimate gives a lower bound less than 180 for n in all cases except

Soc(G) = PSL(6, 2), PSL(7, 2), PSL(8, 2). Moreover, for these three groups we only need to consider the natural action on the projective space due to the bounds for n.

19 Hence in the ﬁrst two cases |Ω| < 180 and in the case Soc(G) = PSL(8, 2) we are done by Lemma 3.5.

Proof of Theorem 1.1. By Corollary 3.2 and Lemmas 3.4, 3.5, 3.6, 3.8, it is enough to consider primitive groups of degree less than 180.

Primitive groups G of degree n < 180 are listed in the GAP library. For 32 < n <

180, we compute the conjugacy classes of G and then the exact size of the set A =

{(∆, g) : ∆g = ∆, ∆ ⊂ Ω, g ∈ G}. It turns out that in each case |A| < 2(2n − 2 − n) and so equation (3.1) in the proof of Lemma 3.3 implies that G has at least four regular orbits on P(Ω).

For 17 < n < 33, we also compute |A| and eliminate those groups where |A| <

2(2n − 2 − n). Only the following groups survive this test: (G, n) = (PGL(2, 17), 18),

(S7, 21), (PSL(3, 4), 21), (PΣL(3, 4), 21), (PGL(3, 4), 21), (PΓL(3, 4), 21), (M22, 22),

(M22 : 2, 22), (M23, 23), (M24, 24), ((S5 × S5).2, 25), (AGL(3, 3), 27), (PSp(4, 3) : 2, 27), (PSp(6, 2), 28), (PSL(5, 2), 31), (ASL(5, 2), 32),

The groups (PΓL(3, 4), 21), (M22, 22), (M22 : 2, 22), (M23, 23), (M24, 24), (ASL(5, 2), 32) are listed in Theorem 2.5 and hence have no regular orbits on P(Ω). The pair

((S5 × S5).2, 25) is covered by Lemma 3.1. For the other groups, we deﬁne certain subsets of Ω. For (G, n) = (PGL(2, 17), 18), take ∆1 = {1, 2, 3, 4, 5, 6, 7, 9, 10}, ∆2 = {1, 2, 3, 4, 5, 6, 7, 9}.

For (G, n) = (PGL(3, 4), 21), (PΣL(3, 4), 21) and (PSL(3, 4), 21), we take ∆1 =

{1, 2, 3, 4, 5, 6, 7, 9}, ∆2 = {1, 2, 3, 4, 5, 6, 7, 9, 11}.

20 For (G, n) = (S(7), 21), we take ∆1 = {1, 2, 3, 5, 6, 8, 10, 11, 12, 15, 16}. ∆2 = {1, 2, 3, 5, 6, 8, 10, 11, 12, 15, 16, 18}.

For (G, n) = (AGL(3, 3), 27), we take ∆1 = {1, 2, 3, 5, 6, 8, 10, 11, 12, 15, 16, 20},

∆2 = {1, 2, 3, 5, 6, 8, 10, 11, 12, 15, 16, 20, 22}.

For (G, n) = (PSL(5, 2), 31), we take ∆1 = {1, 2, 3, 4, 5, 9, 10, 11, 13, 16, 18, 21, 22},

∆2 = {1, 2, 3, 4, 5, 7, 9, 10, 11, 13, 16, 18, 21, 22}.

For (G, n) = (PSp(6, 2), 28), we take ∆1 = {1, 2, 3, 4, 5, 10, 12, 16, 21, 22, 23, 26},

∆2 = {1, 2, 3, 4, 5, 10, 12, 16, 18, 21, 22, 23, 26}.

For (G, n) = (PSp(4, 3) : 2, 27), we take ∆1 = {1, 2, 3, 4, 5, 10, 12, 16, 21, 22, 23, 26},

∆2 = {1, 2, 3, 4, 5, 8, 10, 12, 16, 21, 22, 23, 26}.

We also take ∆3 = Ω \ ∆1 and ∆4 = Ω \ ∆2. Then ∆i, for 1 ≤ i ≤ 4, are from four diﬀerent regular orbits of G on P(Ω).

For n ≤ 17, we compute the full orbit structure of G on P(Ω) by brute force, yielding the examples listed in Theorem 1.1.

21 CHAPTER 4

GENERAL RESULTS ON ORBIT EQUIVALENT GROUPS

In this chapter, we focus on the the general properties of orbit equivalent permutation groups.

4.1 Basic results

In this section we prove some general results about orbit equivalent permutation groups. Throughout this section, we suppose that G, H ≤ Sym(Ω), H < G, and

H ≡ G. Moreover, Σ = {∆1, ∆2,..., ∆n} is a block system of imprimitivity for G (and so, by Lemma 2.4, a block system for H).

Lemma 4.1. HΣ ≡ GΣ.

Proof. Let O be an orbit of GΣ in P(Σ) and let Γ ∈ O and Γ0 ∈ O be two subsets of

Σ. Then there exists g ∈ G such that Γg = Γ0. Let Γ be the union of all blocks in Γ g and Γ0 be the union of all blocks in Γ0. Hence we have Γ = Γ. Since H and G are h orbit equivalent, there exists h ∈ H such that Γ = Γ, which implies Γh = Γ0. So O is also an orbit of HΣ in P(Σ).

∆ ∆ Lemma 4.2. Let ∆ ∈ Σ(that is, ∆ is a block for G and H). Then G∆ ≡ H∆ .

22 G∆ H∆ Proof. Let Γ be a subset of ∆. We want to show that Γ ∆ = Γ ∆ .

Since ∆ is a block, for any g ∈ G we have Γg ⊂ ∆ or Γg ∩ ∆ = ∅, with the ﬁrst

∆ G G∆ 0 G of these possibilities occurring if and only if g ∈ G∆. Hence Γ ∆ = Γ = {Γ ∈ Γ |

0 H∆ 0 H 0 G H Γ ⊂ ∆}. Similarly, Γ ∆ = {Γ ∈ Γ | Γ ⊂ ∆}. However, H ≡ G, so Γ = Γ and

G∆ H∆ so Γ ∆ = Γ ∆ .

Lemma 4.3. [19] For any Θ ⊂ Ω, |GΘ : HΘ| = |G : H| and |G : H| is a divisor of

|GΘ|.

G H Proof. Since H ≡ G, we have |G : GΘ| = |Θ | = |Θ | = |H : HΘ|. Hence |G : H| =

|GΘ : HΘ| and so |G : H| divides |GΘ|.

Corollary 4.4. |G : H| divides gcd{|GΘ| | Θ ⊂ Ω}.

Lemma 4.5. [3, Ex. 1.1] Let K,G ≤ Sym(Ω) with K < G. Then K and G are orbit equivalent on Ω if and only if for all ∆ ⊂ Ω, KG∆ = G.

Proof. Clearly, KG∆ ⊆ G. Suppose K ≡ G and let ∆ ⊂ Ω. Then, since K is

G g k −1 transitive on ∆ , for any g ∈ G there exists k ∈ K such that ∆ = ∆ . So k g ∈ G∆ and g ∈ KG∆, implying KG∆ = G. Conversely, if KG∆ = G for ∆ ⊂ Ω, then K is transitive on ∆G, by reversing the steps of the previous argument.

4.2 Groups that are orbit equivalent to a regular group

As an application of the basic properties, we consider the groups that are orbit equivalent to a regular permutation group.

23 Lemma 4.6. Let H = (Cn, n) and H ≡ G but H 6= G, Then 3 ≤ n ≤ 5 and

(G, n) = (D2n, n).

Proof. Consider H acting a circle C with n vertices. Since H acts transitively on the edge set of C, we obtained that G is a subgroup of D2n = Aut(C) acting transitively on the edges. Hence, if H 6= G then G = D2n. Let 1, 2, 3, ..., n be the vertices of C, labelled so that i is adjacent to i + 1. If n > 5, then let S = {1, 2, 4}. Since

(D2n)S = 1, the proper subgroup H is not orbit equivalent to D2n. For 3 ≤ n ≤ 5, it is easy to check Cn is orbit equivalent to D2n.

Lemma 4.7. Let H be an elementary abelian 2-group acting regularly on Ω. If

H ≡ G, then H = G.

n Proof. Let H = C2 . Since hG, Hi is orbit equivalent to G and H, we can assume H ≤ G. We proceed by induction on n. The cases n = 1, 2 are easy to check, so assume n > 2. Note that H has a maximal block system {∆1, ∆2}. By Lemma

4.2 H∆1 is orbit equivalent to G∆1 and H∆1 is also an elementary abelian 2-group, ∆1 ∆1 ∆1 by induction H∆1 = G∆1 . Take a set S = {a, b, c} ⊂ ∆ and d ∈ ∆ and let ∆1 ∆1 1 2

T = S ∪ {d}. Clearly T is not a block of H and HT = 1, because HT must ﬁx d. we have |GT : HT | = |G : H|, since H and G are orbit equivalent. Also, we have

|GS : HS| = |G : H|. Since HS = 1 and GS ≤ GT , we obtain that GS = GT and GS

∆2 ﬁxes d. Similarly GS ﬁxes all points in ∆2, so GS = 1. We have HS = 1, because in a regular elementary abelian 2-group the invariant sets of nontrivial elements are of even size. Now H∆1 = G∆1 , implies G∆1 = H∆1 = 1. Combine with G∆2 = 1, we ∆1 ∆1 S S S obtain GS = 1 and so G = H.

24 Lemma 4.8. If H is not a elementary abelian 2-group and H acts regularly on Ω, then there exists a subset {a, b} ⊂ Ω which is not a block of H.

Proof. We can identify Ω by H, via H acting on itself by right multiplication. Since

H is not an elementary abelian 2-group, there exists an element ω whose order is not

2. Then the set {1, ω} is not a block.

For later use, we note that if T = {a, b} is not a block for H then HT = 1. Indeed,

h h if 1 6= h ∈ HT then a = b, b = a, and 1, h are the only elements of H that map either of a, b to a or b. Hence for all g ∈ H we have T g = T or T g ∩ T = ∅, a contradiction.

Theorem 4.9. Let H be a regular group acting on Ω and H ≡ G but H 6= G, Then

(H,G) = (A3,S3), (C4,D8) or (C5,D10).

Proof. Assume H < G. Let n = |Ω|. By Lemma 4.7, H is not an elementary abelian

2 group. By Lemma 4.8,there exists a subset T = {a, b} ⊆ Ω and |T H | = |H|.

Consider the undirected orbital graph K of H which has ab as an edge. In K each vertex has degree 2. If K is connected, then it must be a circle. Since G is orbit equivalent to H and H 6= G, we obtain G = D2n as in the proof of Lemma 4.6. By the same Lemma, (H,G) = (A3,S3), (C4,D8) or (C5,D10). If K is not connected, then each connected component is a circle. Let T ⊆

∆1, ..., ∆k be the all connected components. Clearly {∆1, ..., ∆k} is a block system of H and also for G by Lemma 2.4. Since T is not a block, we have |∆1| > 2. Take

H G c ∈ ∆1 and d ∈ ∆2 and let S = c ∪ (∆2 − {d}). Since H ≡ G we have |S | = |S | and |G : GS| = |H : HS|.

25 G It is easy to see GS ﬁxes c and d, so GS ≤ Gc and |S | = |G : GS| ≥ |G : Gc| = n.

G H G On the other hand, |S | = |S | ≤ n. Therefore, |S | = |G : Gc| = n and GS = Gc, implying Gc also ﬁxes d.

Similarly, Gc must fix all points in ∆2, ∆3, ..., ∆k and by the same argument Gd must fix all points in ∆1. Hence Gc = Gd and Gc fixes all points in Ω. This means

Gc = 1 and G is also regular on Ω. Noticing an abelian and transitive permutation group must be regular, we have the following corollary.

Corollary 4.10. If H is abelian and transitive on Ω and H ≡ G, then either H = G, or (H,G) = (C3,S3), (C4,D8), or (C5,D10).

4.3 Maximal blocks

We say that the block system Σ is maximal for G if and only if GΣ is primitive (and then, by Lemma 2.4,if H ≡ G then HΣ is primitive).

Lemma 4.11. Let n ≥ 2 and n 6= 4. Let G = An or Sn. If H < G and H 6= An, then |G : H| ≥ n.

Proof. For G = Sn and n ≥ 5, this statement is proved in [9](Lemma 2.9A). It is easy to see the lemma holds when G = S3 or S2. Next, we consider the case G = An. The case n = 3, it is trivial. If n ≥ 5, then An is a simple group. Let d = |An : H|, then

An has a transitive representation on a set of d elements. The kernel of the action

26 is a normal subgroup of An, so it must be 1. Then An is a subgroup of Sd, implying n!/2 ≤ d!. So d ≥ n.

Theorem 4.12. If H ≤ Sym(Ω) acting transitively and faithfully on a maximal block system Σ of H, then H has a regular orbit on the power set of Ω.

Proof. Let Σ = {∆1, ∆1, ..., ∆n} be a maximal block system and assume H acts on Σ primitively. and faithfully. If H has a regular orbit on the power set of Σ, then H has a regular orbit on the power set of Ω. We only need to consider the groups H with no regular orbits.

Case 1: (H, n) is (An, n) or (Sn, n). In this case H∆1 = An−1 or Sn−1. Let a ∈ ∆1;

then Ha ≤ H∆1 . If n 6= 5, then by Lemma 4.11, |∆1| = |H∆1 : Ha| ≥ n − 1. Take

S ⊂ Ω such that |S ∩ ∆i| = i − 1 for i = 1, 2, ..., n. Clearly HS ﬁxes Σ pointwise.

Therefore HS = 1 and H has a regular orbit on the power set of Ω.

If n = 5, (H, n) = (A5, n) or (S5, n). So H∆1 = A4 or S4. It is enough to look at the imprimitive transitive representations of degree 10 or 15. There are only two possible candidates (A5, 15) and (S5, 15). By GAP computation, their setwise stabilizer of {13, 14, 15} are trivial.

Case 2: The distinguishing number of (H, n) on Σ is 3. In this case there exists a

3-partition P = (S1,S2,S3) of Σ such that only the identity preserves P . We take

S ⊆ Ω such that |S ∩ ∆i| = k − 1 if ∆i ∈ Sk k = 1, 2, 3. Then HS ﬁxes the partition

P and HS = 1. Case 3: The distinguishing number of (H, n) is 4. In this case there exists a

4-partition P = (S1,S2,S3,S4) of Σ such that only the identity preserves P . If

27 |∆1| ≥ 3, then we take S ⊆ Ω such that |S ∩ ∆i| = k − 1 if ∆i ∈ Sk k = 1, 2, 3, 4.

Then HS ﬁxes partition P and HS = 1. Hence we only need to consider the subcase

|∆1| = 2. By Theorem 2.7, (H, n) must be one of the following: (S5, 6), (PSL(3, 2), 7),

3 (2 PSL(3, 2), 8), (M11, 11) or (M12, 12). Notice that |H∆1 : H1| = 2. For (H, n) =

3 (2 PSL(3, 2), 8) and (M12, 12), H∆1 has no subgroup of index 2. In the remaining cases, we use GAP to ﬁnd regular orbit on P(Ω). If H = S5 acting on 12 elements, then the setwise stabilizer of {1, 2, 4, 5} is trivial. If H = PSL(3, 2) acting on 14 elements, then the setwise stabilizer of {1, 2, 3} is trivial. If H = M11 acting on 22 elements, then the setwise stabilizer of {1, 2, 3, 5, 7, 11, 12} is trivial. In these computations, we used permutation representations with ∆i = {2i − 1, 2i} for i = 1, 2, ..., k and primitive action on Σ as listed in the GAP library.

Regarding the action on the maximal block system of two orbit equivalent permutation groups, we have the following theorem.

Theorem 4.13. If Σ is a maximal block system for H ≡ G then either (HΣ,GΣ) =

Σ Σ (An,Sn) or H = G .

Proof. Since Σ is a maximal block system, HΣ and GΣ are primitive on Σ. Moreover, by Lemma 4.1, HΣ ≡ GΣ. Hence, if the conclusion of the theorem does not hold then (HΣ,GΣ) must be one of the orbit equivalent pairs of primitive groups listed in Theorem 2.6. We process the pairs of groups listed in Theorem 2.6 one-by-one, and in each case we reach a contradiction. We suppose that the primitive groups are given as in the GAP library.

Σ Σ Case 1: n = 5 and (H ,G ) = (C5,D10). Choose S ⊂ Ω such that S = S1 ∪ S2 with

28 S1 ⊂ ∆1 , S2 ⊂ ∆2, |S1| = 1, |S2| = 2. Then GS must ﬁx ∆1 and ∆2 setwise. Notice

Σ Σ Σ Σ Σ Σ that G = D10, and so (GS) = 1. Therefore, (HGS) = H (GS) < G , implying

Σ Σ HGS 6= G. However, this contradicts Lemma 4.5, and so (H ,G ) 6= (C5,D10).

Σ Σ Case 2: n = 5 and (H ,G ) = (AGL(1, 5),A5) or (AGL(1, 5),S5). Choose S ⊂ Ω such that S = S1 ∪ S2 ∪ S3 ∪ S4 with Si ⊂ ∆i (i = 1, 2, 3, 4), |S1| = |S2| = 2,

Σ Σ 2 |S3| = |S4| = 1. Then GS ﬁxes {∆1, ∆2} and {∆3, ∆4}, and so (GS) ≤ C2 .

Σ Σ Σ Σ Σ Σ Σ By assumption G = HGS and so |G | = |H (GS) | = |H ||(GS) |/|(GS) ∩H |. However, the right side of this equation is a number not divisible by 3, contradicting that |GΣ| is divisible by 3.

Σ Σ Case 3: n = 6 and (H ,G ) = (PGL(2, 5),A6) or (PGL(2, 5),S6). Note that

Σ Σ PGL(2, 5) is not a subgroup of A6, so the case (H ,G ) = (PGL(2, 5),A6) does

Σ Σ not occur at all. For the case (H ,G ) = (PGL(2, 5),S6), choose S ⊂ Ω such that

S = S1∪S2∪S3∪S4 with Si ⊂ ∆i (i = 1, 2, 3, 4), |S1| = |S2| = 2, |S3| = |S4| = 1. Then

Σ Σ 3 GS ﬁxes {∆1, ∆2}, {∆3, ∆4} and {∆5 ∆6}. So (GS) ≤ C2 and 3 does not divide

Σ Σ Σ Σ Σ Σ Σ Σ |(GS) |. Thus 9 does not divide |(HGs) | = |H (GS) | = |H ||(GS) |/|(GS) ∩H |, which contradicts the fact that |GΣ| is divisible by 9.

Case 4: n = 8 and (HΣ,GΣ) = (AGL(1, 8), AΓL(1, 8). Choose S ⊂ Ω such that

S = S1 ∪ S2 ∪ S3 ∪ S4 with Si ⊂ ∆i (i = 1, 2, 3, 4), |S1| = |S2| = 2, |S3| = |S4| = 1.

Σ Σ Σ Then GS ﬁxes A = {∆1, ∆2} and B = {∆3, ∆4}. Let (G )AB be the subgroup of G

Σ Σ which ﬁxes A and B setwise. Clearly (GS) ≤ (G )AB.

Σ Σ Σ Σ By a GAP computation, we have |(G )AB| = 4. So |(HGS) | = |H (GS) | =

Σ Σ Σ Σ Σ Σ |H ||(GS) |/|(GS) ∩ H | is not divisible by 3 which implies that |G |= 6 |(HGS) |. Case 5: n = 8 and (HΣ,GΣ) = (AGL(1, 8), ASL(3, 2) or (AΓL(1, 8), ASL(3, 2)).

29 Choose S ⊂ Ω such that S = S1 ∪S2 ∪S3 ∪S4 ∪S5 ∪S6 with Si ⊂ ∆i (i = 1, 2, 3, 4, 5, 6),

Σ |S1| = |S2| = |S3| = 2, |S4| = |S5| = |S6| = 1. Then GS ﬁxes A = {∆1, ∆2, ∆3} and

B = {∆4, ∆5 ∆6}.

Σ Σ Σ Σ By a GAP computation, we have |(G )AB| = 4. Hence |(HGS) | = |H (GS) | ≤

Σ Σ Σ |H ||(GS) | < |G |, a contradiction. Case 6: n = 9 and (HΣ,GΣ) = (AGL(1, 9), AΓL(1, 9)). Choose S ⊂ Ω such that

S = S1 ∪ S2 ∪ S4 ∪ S5 with Si ⊂ ∆i (i = 1, 2, 4, 5, ), |S1| = |S2| = 2, |S4| = |S5| = 1.

Σ Then GS ﬁxes A = {∆1, ∆2} and B = {∆4, ∆5}.

Σ Σ Σ Σ By a GAP computation, we have |(G )AB| = 1. Hence |(HGS) | = |H | < |G |, a contradiction.

Case 7: n = 9 and (HΣ,GΣ) = (ASL(2, 3), AGL(2, 3)) or (PSL(2, 8), PΓL(2, 8)).

Choose S ⊂ Ω such that S = S1 ∪S2 ∪S4 ∪S6 ∪S7 ∪S9 with Si ⊂ ∆i (i = 1, 2, 4, 6, 7, 9),

Σ |S1| = |S2| = |S7| = 2, |S4| = |S6| = |S9| = 1. Then GS ﬁxes A = {∆1, ∆2, ∆7} and

B = {∆4, ∆6, ∆9}.

Σ Σ Σ Σ By a GAP computation, we have |(G )AB| = 1. Hence |(HGS) | = |H | < |G |, a contradiction.

Σ Σ Case 8: n = 9 and (H ,G ) = (PSL(2, 8),A9), (PSL(2, 8),S9), (PΓL(2, 8),A9), or

(PΓL(2, 8),S9). Choose S ⊂ Ω such that S = S1 ∪ S2 ∪ S3 ∪ S4 ∪ S5 ∪ S6 with Si ⊂ ∆i

Σ (i = 1, 2, 3, 4, 5, 6), |S1| = |S2| = |S3| = 2, |S4| = |S5| = |S6| = 1. Then GS ﬁxes

A = {∆1, ∆2, ∆3} and B = {∆4, ∆5, ∆6}.

Σ Σ Σ Σ Σ Σ Σ We have |(G )AB| ∈ {108, 216}. Hence |(HGS) | = |H (GS) | = |H ||(GS) |/|(GS) ∩ HΣ| is not divisible by 5, contradicting the fact that 5 divides |GΣ|..

Case 9: n = 10 and (HΣ,GΣ) = (PGL(2, 9), PΓL(2, 9)). Choose S ⊂ Ω such that

30 S = S1 ∪S2 ∪S3 ∪S4 ∪S5 ∪S7 with Si ⊂ ∆i (i = 1, 2, 3, 4, 5, 7), |S1| = |S2| = |S3| = 2,

Σ |S4| = |S5| = |S7| = 1. Then GS ﬁxes A = {∆1, ∆2, ∆3} and B = {∆4, ∆5, ∆7}.

Σ Σ Σ Σ By a GAP computation, we have |(G )AB| = 1. Hence |(HGS) | = |H | < |G |, a contradiction.

Σ Σ Σ Σ Therefore, the only possibility is that H = G or (H ,G ) = (An,Sn) for some n.

4.4 Orbit equivalent pairs of wreath products

In this section, we give some inﬁnite families of orbit equivalent permutation groups.

In the case when the top group GΣ is primitive, we have GΣ = HΣ or (GΣ,HΣ) =

(An,Sn) by Theorem 4.13. Clearly, if P ≡ Q, then P o B ≡ Q o B. When the top groups HΣ and GΣ are not equal, we have the following.

Lemma 4.14. Let P ≤ Q ≤ Sym(Γ) and P ≡ Q. Then H = P o An ≡ G = Q o Sn on Γ × ∆ if and only if n ≥ u(P ) + 1, where u(P ) is the number of orbits of P acting on P(Γ).

Proof. Suppose n ≥ u(P ) + 1. For any S ⊆ Γ × ∆, let Si = {x;(x, i) ∈ S} ⊆ Γ. Then Sn S = i=1(Si, i). Let Σ = {∆1, ∆2, ..., ∆k} be a partition of {1, 2, ..., n} such that i and j are in the same part if and only if Si and Sj are in the same orbit of P on P(Γ).

Then HS = (PS1 × PS2 × ... × PSn ):(An)Σ, GS = (QS1 × QS2 × ... × QSn ):(Sn)Σ.

There exists ∆i ∈ Σ with |∆i| > 1 because u(P ) < n and | (Sn)Σ :(An)Σ |= 2. Thus,

n H G |GS : HS| = 2.|Q : P | = |G : H|, which implies that S = S for any S ⊂ Γ × ∆. Hence, G and H are orbit equivalent on Γ × ∆.

31 Conversely, if n ≤ u(P ), then let S1,S2, ..., Sn be n subsets of Γ which belong to

diﬀerent orbits of P on P(Γ). Let S = {(x, i); x ∈ Si}. Then HS = PS1 ×PS2 ×...×PSn n and GS = QS1 × QS2 × ... × QSn . Hence |G : H| = 2|Q : P | 6= |GS : HS| and so SH 6= SG. Hence, G and H are not orbit equivalent.

Lemma 4.15. Suppose P ≤ Q are two primitive groups acting on ∆. Let A ≤ B be two transitive permutation groups on Γ. Then, A o P ≡ B o Q on Γ × ∆ if and only if one of the following holds.

1. P = Q and A ≡ B on Γ.

2. (P,Q) = (An,Sn), and A ≡ B on Γ and n ≥ u(A) + 1 where u(A) is the number of orbits of A on P(Γ)

Proof. If A, B, P, Q satisfy either of the conditions 1, 2 then, by Lemma 4.14 and the remark before the lemma, we have AoP ≡ B oQ. Conversely, let H = AoP , G = B oQ and suppose that H ≡ G. Since P,Q are primitive and P ≡ Q on ∆, by Theorem

4.13 we obtain P = Q or (P,Q) = (An,Sn). Also by Lemma 4.2 A ≡ B. If P = Q then we are in case 1. If (P,Q) = (An,Sn) then we are in case 2 by Lemma 4.14.

32 CHAPTER 5

A ∈ (γ) AND B ∈ (γ)

Throughout this chapter and the next chapter, we suppose that G ≤ Sym(Ω), and

Σ = {∆ := ∆1, ∆2,..., ∆n} is an nontrivial block system of G. We identify Ω by ∆ × [n]. Let K be the kernel of G acting on Σ. The group G is called two-step

∆ Σ imprimitive if A := G∆ and B := G are primitive on ∆ and Σ, respectively. In this chapter we focus on a special class of 2-step imprimitive permutation groups. Namely,we assume that (A, |∆|) and (B, |Σ|) are of type (γ). Throughout this chapter, we consider the standard imprimitive action of wreath products, unless explicitly stated otherwise.

5.1 Clean subgroups isomorphic to An

The deﬁnition of clean subgroup is introduced in [2]. For G ≤ Sym(Ω) and a

G−invariant partition Σ = {∆ := ∆1, ∆2,..., ∆n} of Ω, ω ∈ G is called clean if

ω for every i ≤ n, if ∆i = ∆i then ω ﬁxes ∆i pointwise. A clean subgroup is a group whose elements are all clean.

Lemma 5.1. Let GΣ be primitive on Σ and T ≤ A = G∆1 ≤ Aut(T ) where T is a ∆1 nonabelian simple group acting on ∆1. Then one of the following holds.

33 1. K = 1.

2. K acts faithfully on ∆1 and T ≤ K ≤ Aut(T ).

n n 3. K does not act faithfully on ∆1 and T ≤ K ≤ Aut(T ) .

Proof. Let S ∼= GΣ be the primitive permutation group acting on {1, 2, ..., n}. Then ∼ G ≤ W = {(t1, t2, ..., tn)s : ti ∈ A, s ∈ S} = AoS. Let W0 = {(t1, ..., tn): ti ∈ A} and

K = G ∩ W0. The projection maps πi : W0 → A are deﬁned as πi((t1, ..., tn)) = ti.

The map π is surjective on A for each i. Let A∗ = K∆1 . Obviously A∗ ¡ A, so i ∆1 A∗ = 1, or T ≤ A∗ ≤ Aut(T ). If A∗ = 1, then K = 1.

In the case A∗ 6= 1, for two coordinates i and j, we say that i is not linked to j, if there exists g ∈ K such that πi(g) 6= 1 and πj(g) = 1. Otherwise, we say that i is linked to j, and denote this fact by i ∼ j. The basic properties of the relation ∼ are as follows.

For all i, i ∼ i.

If i ∼ j, then j ∼ i.

If i ∼ j and j ∼ k, then i ∼ k.

Thus, being linked is an equivalence relation on Σ.

If any two coordinates of i1, i2, ..., ik are not linked, then for any t ∈ T there exists

g ∈ K such that πi1 (g) = t and πil (g) = 1 for l = 2, 3, ..., k. It is easy to see i ∼ j if and only if is ∼ js for all s ∈ S. Such relation ∼ is a called a G-congruence [9]. The equivalence classes of ∼ form a system of blocks for

S. Since S is primitive, each equivalence class contains either one single coordinate or the entire set Σ.

34 If Σ consists of one equivalence class, then any two coordinates of 1, 2, ..., n are

∗ linked to each other. Hence for each i, πi : K → A is surjective and injective. In

∼ ∗ this case K = A and K acts faithfully on ∆1. If each equivalence class if Σ contains one single coordinate, then any two coordinates of 1, 2, ..., n are not linked to each other. For any t ∈ T , there exists an element g ∈ K such that π1(g) = t and πi(g) = 1 for i = 2, 3, ..., n. Then

T1 = {(t, 1, 1, ..., 1) : t ∈ T } ≤ K. Similarly, Ti = {(1, ...1, t, 1, ..., 1) : t ∈ T } ≤ K. So

n T ≤ K and K is not faithful on ∆1. Clean subgroups play an important role in our investigation of orbit equivalent permutation groups. The following Giant Action Theorems are central in our investigation.

Theorem 5.2. [2] Let G ≤ Sym(Ω) and let Σ = {∆1, ..., ∆n} be a G-invariant

Σ partition of Ω. Suppose G = An and let K be the kernel of this action. Assume ∼ that n > 4b where b = |∆1|. Then K has a clean complement R = An.

Theorem 5.3. [2] Let G ≤ Sym(Ω) and let Σ = {∆1, ..., ∆n} be a G-invariant

Σ partition of Ω. Suppose G = An and let K be the kernel of this action. Assume G

∼ ∗ ∆ ∗ has a clean subgroup R = An. Let A = K∆ and let N = {ω ∈ A : ∃ω3, ..., ωn ∈

∗ ∗ ∗ A : (1, ω, ω3, ..., ωn) ∈ K} and N = {ω ∈ A :(ω, 1, ..., 1) ∈ K}.

∗ ∗ ∆ ∗ Then, N < N < A = K∆ are normal subgroups of A . Moreover, G has normal subgroups M and M ∗ ≤ M ≤ K, such that

1. M = N n ∩ K and M ∗ = (N ∗)n.

35 n−1 2. The restriction of M to Ω \ ∆i is N = N1 × ... × Ni−1 × Ni+1 × ... × Nt. 3. A∗/N ∼= K/M. 4. N/N ∗ is abelian.

∗ ∗ Qn 5. M/M = {(ω1, ..., ωt): ωi ∈ N/N , i=1 ωi = 1}.

The following lemmas guarantee in some cases that G has a clean subgroup R ∼=

An.

∆ Σ Lemma 5.4. Let K∆ = T for some nonabelian simple group T and let (G , n) =

(An, n) with n ≥ 9. Then G = T o An or T × An, and G contains a clean subgroup ∼ R = An.

Proof. By Lemma 5.1, we have K = T or K = T n. Let Σ0 = Σ \{1} and let M be

0 the induced group of G∆ acting on ∆ ∪ Σ . So M = T.An−1. Let M1 = M(∆) and

M2 = M(Σ0). Clearly, M1 ¡ M, M2 ¡ M, and M1 ∩ M2 = 1.

∼ ∆ We claim that M1 6= 1. Otherwise, M is faithful on ∆ and M = M ≤ Aut(T ), a

∼ ∆ ∼ Σ0 ∼ ∆ contradiction. Also, M2 = K∆ 6= 1. We have M1 = M1 = An−1 and M2 = M2 = T ,

∆ hence M = M1 × M2 = T × An−1. So G∆ = T . By Lemma 2.1, G ≤ T o An.

n If K = T , then G = T o An. If K = T , then G = T × An. In both cases, G ∼ contains a clean subgroup R = An. By using the idea of the proof of Giant Action Theorem, we prove the following lemmas.

Lemma 5.5. If (GΣ, n) is of type (γ) and there exists a prime p ≥ 3 such that p - |A| ∼ and 2p − 1 ≤ n, then G contains a clean subgroup R = An.

36 Σ Proof. We can assume G = An. There exists an element τ ∈ G such that τ =

(1, 2, ..., p). Since p - |A|, there exists an m such that τ m is clean. The rest of the proof is the same as the proof of the Giant Action Theorem [2].

Lemma 5.6. Let (GΣ, n) and (A, |∆|) be of type (γ), n ≥ 9, and K 6= 1. Then G contains a clean subgroup An.

Σ a b Proof. We assume G = An. Firstly, if |∆| ≤ 4, then |A| = 2 · 3 and we can apply Lemma 5.5 with p = 5.

n In the case |∆| ≥ 5, by Lemma 5.1 we have K ≥ Am or Am ≤ K ≤ Sm. For

K0 = soc(K), G contains a subgroup K0.An. Hence by Lemma 5.4, G contains a clean subgroup isomorphic to An.

5.2 Proof of Theorem 1.2

We prove the following lemmas ﬁrst.

Σ ∆ Lemma 5.7. Let H < G. Suppose that (G , n) and (G∆, |∆|) are of type (γ). Suppose b := |∆| ≥ 5 and n := |Σ| ≥ 9 and let K and L be the kernels of the

n n action of G and H on Σ, respectively. If (soc(K), soc(L)) = (Ab ,Ab), (Ab , 1) or

(soc(K), soc(L)) = (Ab, 1), then G and H are not orbit equivalent.

Σ Σ Σ Proof. We assume G ≡ H. So G ≡ H . By Theorem 4.13, H = An or Sn. By ∼ Lemma 5.6, H contains a clean subgroup R = An. Let O be an orbit of R on Ω.

n Clearly, we have |O ∩ ∆i| = 1 for each i and HO ≥ R. By Lemma 5.1, soc(K) = Ab or Ab.

37 If soc(K) = Ab and soc(L) = 1, then L = 1 and |G : H| = |K : L| = |K| or 2|K :

L| = 2|K|. On the other hand, we have |G : H| = |GO : HO| = |KO : LO| = |Ab−1| or

2|KO : LO| = 2|Ab−1|. This is a contradiction.

n If soc(K) = Ab and soc(L) = Ab or 1. By the same argument as in the previous case, soc(L) 6= 1. So soc(L) = Ab. In this case L ≤ Sb and Ab o An ¡ G. Let x ∈ ∆1.

Obviously, Lx ﬁxes a point y in ∆2. Let Γ1 = {x, y} and let z ∈ ∆2 and z 6= y. Let

Γ2 = {x, z}. Clearly, GΓ2 = GΓ1 but HΓ2 < HΓ1 . This is a contradiction.

Σ ∆ Lemma 5.8. Let H < G and H ≡ G. Suppose that (G , n) and (G∆, |∆|) are of type (γ). Suppose b := |∆| ∈ {5, 6, 9} and n := |Σ| ≥ 9 and let K and L be the kernels of the action of G and H on Σ, respectively. If soc(K) = Ab, then soc(L) = Ab.

Proof. Since soc(K) = Ab, then K ≤ Sb. Applying Lemma 5.5 with p = 3, we obtain ∼ that G/soc(K) contains a subgroup S = An. Then the pre-image of S in G is Ab ×An.

Let Γ = S1 ∪S2 ∪...∪Si such that Si ⊂ ∆i and |Si| = i. Then |G : H| = |GΓ : HΓ| = 2.

∆ For b := |∆| ∈ {5, 6, 9}, H∆ must be Ab or Sb. Otherwise, |K : L| is not a power of

2. Hence L = Ab or Sb and soc(L) = Ab.

Σ ∆ Theorem 5.9. Let H < G and H ≡ G. Suppose that (G , n) and (G∆, |∆|) are of type (γ). Suppose b := |∆| ≥ 2 and n := |Σ| ≥ 9 and let K and L be the kernels of the action of G and H on Σ, respectively. Then one of the following holds.

n−1 n 1. b = 2 and C2 ≤ L ≤ K ≤ C2 or 1 ≤ L ≤ K ≤ C2.

n−1 n−1 2. b = 3 and C3 ≤ L ≤ K ≤ C3 .C2.

n n−1 n n−1 n n 3. b = 4 and K4 .C3 ≤ L ≤ K ≤ K4 .C3 .C2, K4 .C3 ≤ L ≤ K ≤ K4 .C3.C2,

n−1 n−1 or K4 .C3 ≤ L ≤ K ≤ K4 .C3.C2. Here K4 is the Klein 4−group.

38 ∼ 4. Ab × An ≤ H < G ≤ Sb × Sn and K∆ = K.

5. Ab o An ≤ G ≤ Sb o Sn and if b∈ / {5, 6, 9} then Ab o An ≤ H.

Proof. Since G has no regular orbits, by Theorem 4.12 we obtain K 6= 1. By Lemma

∼ Σ Σ 5.6, H contains a clean subgroup R = An. We have G ≡ H since G ≡ H. By

Σ Theorem 4.13, H = An or Sn. Let O be an orbit of R on Ω. Clearly, we have

|O ∩ ∆i| = 1 for each i and HO ≥ R.

∗ First, we look at the case 2 ≤ b ≤ 4. Let N = {ω ∈ A : ∃ω3, ..., ωn ∈ A :

∗ ∗ n (1, ω, ω3, ..., ωn) ∈ K}, N = {ω ∈ A :(ω, 1, ..., 1) ∈ K}, M = N ∩ K and M ∗ = (N ∗)n. By the Giant Action Theorem 5.3, N and N ∗ are normal subgroups of

A∗.

Case 1: b = 2. Since G ≡ H, we have |G : H| = |GO : HO|. Noticing An ≤ GO ≤ Sn, we obtain GO = Sn and HO = An. Hence |G : H| = 2. Considering that K and

n n L are two G−invariant subspaces of C2 and the An−invariant subspaces of C2 are

n−1 n 0, 1, n − 1, or n−dimensional, we have C2 ≤ L ≤ K ≤ C2 or 1 ≤ L ≤ K ≤ C2.

Case 2: b = 3. In this case An ≤ HO ≤ GO ≤ C2 o Sn, because the action of HO on

Ω \ O is isomorphic to a subgroup of C2 o Sn and HO and GO act faithfully on Ω \ O.

So |G : H| = |G : : H | is a power of 2. Recall that A∗ = K∆1 ¡ A = G∆1 , and N O O ∆1 ∆1 and N ∗ are normal subgroups of A∗. By the Giant Action theorem 5.3, we have the following subcases.

∗ n (a). N ≥ C3. In this case, K ≥ A3 and G ≥ A3 o An. Since |G : H| is a power of

n 2, we obtain L ≥ C3 and H ≥ C3 o An. ∗ ∼ ∗ (b). N = 1 and N = 1. In this case, K = A ≤ S3. Noticing that CR(K) ¡

39 ∼ NR(K) = R, we have CR(K) = 1 or CR(K) = An. If CR(K) = 1, then NR(K)/CR(K) =

R ≤ Aut(K) ≤ S3, which is not possible. Hence CR(K) = R and G ≥ A3 × An. In this case, since K acts faithfully on ∆1, we have G ≤ S3 × Sn. Noticing that |G : H| is a power of 2, we have L ≥ A3 and H ≥ A3 × An.

∗ ∗ (c). N = 1 and N = C3. In this case, by the Giant Action Theorem 5.3, M = 1,

∗ ∗ Qn n−1 ∼ ∗ M/M = {(ω1, ..., ωt): ωi ∈ N/N , i=1 ωi = 1} = C3 and K/M = A /N ≤ C2. n−1 n−1 n−1 Then, we have K = C3 .C2 or C3 and L ≥ C3 .

∗ ∗ (d).N = 1 and N = S3. This case is impossible because N/N must be abelian.

Case 3: b = 4. Let O1 and O2 be two orbits of R on Ω. Then R = An ≤ HO1cupO2 ≤

∗ GO1∪O2 ≤ (K4) o Sn and |G : H| is a power of 2. Noticing that N and N are normal subgroups of A∗, N/N ∗ is abelian, we have the following subcases.

∗ (a). N ≥ A4. Then G ≥ A4 o An and H ≥ A4 o An.

∗ ∗ ∗ n n ∗ (b). N = K4 and N = A4. In this case M = (N ) = K4 , M/M = ∗ Qn n−1 ∼ ∗ {(ω1, ..., ωt): ωi ∈ N/N , i=1 ωi = 1} = C3 and K/M = A /N ≤ C2, then

n n−1 n n−1 K = K4 .C3 or K4 .C3 .C2.

∗ ∗ ∗ n n ∗ (c). N = K4 and N = K4. In this case M = (N ) = K4 , M/M = ∗ Qn ∼ ∗ {(ω1, ..., ωt): ωi ∈ N/N , i=1 ωi = 1} = 1 and K/M = A /N = S3 or C3,

n n then K = K4 .C3 or K4 .C3.C2.

∗ ∗ ∗ n ∗ (d). N = 1 and N = K4. In this case M = (N ) = 1, M/M = {(ω1, ..., ωt):

∗ Qn n−1 ∼ ∗ n−1 ωi ∈ N/N , i=1 ωi = 1} = K4 and K/M = A /N = S3 or C3, then K = K4 .C3 n−1 or K4 .C3.C2.

∗ ∗ ∗ n ∗ (e). N = 1 and N = 1. In this case M = (N ) = 1, M/M = {(ω1, ..., ωt):

∗ Qn ∼ ∗ ωi ∈ N/N , i=1 ωi = 1} = 1, then K = A . Noticing CR(K) ¡ NR(K) = R, we have

40 CR(K) = 1 or CR(K) = R. If CR(K) = 1, R = NR(K)/CR(K) ≤ Aut(K) ≤ S4.

∗ This is a contradiction. Hence CR(K) = R. Thus, G ≥ R × A . In this case, K acts faithfully on ∆1, A4 o An ≤ H ≤ G ≤ S4 × Sn.

n Case 4: b ≥ 5. By Lemma 5.1, soc(K) = Ab or Ab . We consider G/soc(K) ≤ S2 o Sn. Applying Lemma 5.5 with p = 3, we obtain that G/soc(K) contains a subgroup

∼ n S = An. Then the pre-image of S in G is either Ab × An or Ab o An. If soc(K) = Ab

∆ and b∈ / {5, 6, 9}, then by Lemma 2.6 and 4.2 we have L∆ ≥ Ab. Hence by Lemma

5.1 and 5.7 Ab o An ¡ H < G ≤ Sb o Sn. If soc(K) = Ab, by Lemma 5.7 and 5.8, soc(L) = Ab and Ab × An ¡ H ≤ G ≤ Sb × Sn.

5.3 Sm o Sn

In this section we consider G = Sn o Sm with the imprimitive action acting on Ω = [n] × [m].

Lemma 5.10. Let H < G = A o Sn and H ≡ G, n ≥ 5 and (A, |∆|) = (F20, 5) or

(PGL(2, 5), 6). Then H = (1/2)A o Sn.

Σ Proof. By Theorem 4.13, we have H = An or Sn.

In the case (A, |∆|) = (F20, 5), applying Lemma 5.5 with p = 3, we have H ∼ contains a clean subgroup R = An. Take a set Γ ⊂ Ω such that HΓ ≥ R and

|Γ ∩ ∆i| = 2.

Then as KΓ and LΓ are elementary abelian 2−groups which can be considered as

n−1 n R−invariant GF (2)−subspaces, we obtain LΓ = 1, C2, C2 or C2 . Let us consider the G-orbit containing Γ on P(Ω). Since KSn = G ≥ K.R and Γ is R and Sn

41 invariant, ΓG = ΓK . Since H ≡ G, H is transitive on ΓG. We consider the group L acting on this orbit ΓG.

K If L is transitive on Γ , then we claim that L = K, and H = F20 o An. If not, let

n n M be a maximal subgroup of K such that L ≤ M < K = F20. Clearly M ≥ D10 and

MΓ = KΓ because the stabilizer of two-element sets are isomorphic to C2 in both M and K. However, this implies |ΓL| ≤ |ΓM | < |ΓK |, which is a contradiction.

If L is not transitive on ΓG, then the orbits of L on ΓG form a block system of H

G G acting on Γ . Let O be the orbit of L on Γ containing Γ. Clearly R ≤ HΓ ≤ HO and L ≤ HO. Hence L.R ≤ HO ≤ H ≤ L.Sn. It follows that |H : HO| ≤ 2. We

H G HO G have Γ = Γ because H ≡ G, and Γ = O 6= Γ . Now |H : HO| = 2 implies that |ΓG| = 2|O| and L has only two orbits on ΓG.

Let M be a maximal subgroup of K such that |K : M| = 2 and L ≤ M <

n L M K L M K = F20. In this case, MΓ = KΓ and |Γ | ≤ |Γ | < |Γ | implies Γ = Γ . We

n n n also have MΓ = C2 . Noticing M = (1/2)A and |M : L| = |MΓ : LΓ| ≤ 2 , we

n n−1 n−1 have L ≥ C5 .C2 and LΓ ≥ C2 . Therefore |M : L| = |MΓ : LΓ| ≤ 2. By the

n n−1 Giant Action Theorem 5.3, in all possibilities of L, except L = D10.C2 we always

n have |M : L| > 2. Hence, we obtain L = M = (1/2)A and LΓ = KΓ. Now

Σ |G : H| = |GΓ : HΓ| ≤ |KΓ.Sn : LΓ.An| ≤ 2 implies H = Sn and H = (1/2)F20 o Sn. In the case (A, |∆|) = (PGL(2, 5), 6), by Lemma 5.1, H/soc(L) contains a sub- ∼ group S = An. The pre-image of S is either PSL(2, 5) × An or PSL(2, 5) o An. In ∼ either case, H contains a clean subgroup R = An. Next, we use the same argument in the case (A, |∆|) = (F20, 20). Then H = (1/2)(PGL(2, 5) o Sn).

42 Theorem 5.11. Let H < G = Sm o Sn with m ≥ 5 ,n ≥ 5 and H ≡ G. Then one of the following holds.

1. m = 5, either A5 o An ≤ H < G ≤ S5 o Sn or (1/2)F20 o Sn ≤ H < G ≤ S5 o Sn.

2. m = 6, either A6 o An ≤ H < G ≤ S6 o Sn or (1/2)PGL(2, 5) o Sn ≤ H < G ≤

S6 o Sn.

3. m = 9, and PSL(2, 8) o An ≤ H < G ≤ S9 o Sn.

4. m 6= 5, 6, 9 and Am o An ≤ H < G ≤ Sm o Sn.

∆ Proof. Let A0 = H∆ . If m∈ / {5, 6, 9}, then since Am is the only group orbit equivalent to Sm, then we have A0 ≥ Am. Hence by Lemma 5.7, Am o An ≤ H < G ≤ Sm o Sn.

If m = 5 and A0 ≥ A5, then by Lemma 5.7, H ≥ A5 o An. Otherwise we have

A0 = F20 and H ≤ F20 o Sn. By Lemma 5.10, H ≥ (1/2)F20 o Sn.

If m = 6 and A0 ≥ A6, then by Lemma 5.7, H ≥ A6 o An. For A0 = PGL(2, 5), we have H ≤ PGL(2, 5) o Sn. By Lemma 5.10, H ≥ (1/2)PGL(2, 5) o Sn.

If m = 9, then A0 = A9 or A0 ≥ PSL(2, 8). In either case we have H ≥ PSL(2, 8)o

An.

5.4 Sn × Sm

In this section we consider G = Sn × Sm with the imprimitive action acting on Ω = [n] × [m].

Suppose Sn and Sm act on Γ and ∆, respectively. Then Sn ×Sm acts on Ω = Γ×∆ with the standard action. It is often useful to treat Γ × ∆ as a matrix with n rows and m columns with Sn and Sm permuting the rows and columns, respectively. We

43 give a necessary and suﬃcient condition when the group G has a regular orbit and

ﬁnd all groups that are orbit equivalent to G.

5.4.1 Regular orbits of Sn × Sm

In this section, we give all the pairs (m, n) such that Sn × Sm has a regular orbit on P(Ω).

Lemma 5.12. Let m1 > m2 ≥ n if Sn × Sm1 has a regular orbit on the power set of

[n] × [m1] then Sn × Sm2 has a regular orbit on the power set of [n] × [m2].

Proof. Take a subset X ⊆ [m1] such that |X| = m2. Let S be a set in a regular

orbit of Sn × Sm1 . Then the setwise stabilizer of S ∩ ([n] × X) is trivial in the group

Sn × Sym(X). So Sn × Sm2 has a regular orbit.

Lemma 5.13. Acting on the power set of Ω = [n] × [n], Sn × Sn has a regular orbit.

Proof. Take S = {(i, j): i ≤ j} ⊂ [n] × [n]. Only identity of Sn × Sn ﬁxes S setwise.

n Lemma 5.14. Acting on the power set of Ω = [n] × [2 ], Sn × S2n has no regular orbit.

n Proof. Let us consider Sn × S2n acting on an n × 2 matrix Γ × ∆ by permuting the rows and columns. Let S be a subset. If there are two columns which are identical in

S, then we can switch those two columns and keep S unchanged. This implies that there exists a nontrivial permutation ﬁxes S setwise. If there are no two identical columns, and we can label the columns with elements of P(Γ), then each subset of

44 the n rows occurs exactly once in the columns in S. Take a nontrivial permutation σ from Sn which permutes Γ. Then σ induces a permutation ω of P(Γ) which can be considered as a permutation of the columns. Then, σω−1 ﬁxes S setwise.

To determine all the pairs (n, m) such that Sn × Sm has a regular orbit on P(Ω), it suﬃce to ﬁnd the maximal value of m such that Sn × Sm has a regular orbit.

Let f(n) = max{m : m ≥ n and Sn × Sm has a regular orbit }. The above

n lemmas tell us n ≤ f(n) < 2 and Sn × Sm (n ≤ m) has a regular orbit if and only if m ≤ f(n).

n Lemma 5.15. [7] Let m < 2 . Then Sn × Sm has a regular orbit if and only if

Sn × S(2n−m) has a regular orbit.

n Proof. Suppose Sn ×Sm has a regular orbit including S. By adding (2 −m) columns to the matrix, we will have an n × 2n matrix. Clearly there are no two identical columns in S. Order the rows from 1 to n. We can take a set T from the just added n × (2n − m) matrix such that the columns in T are diﬀerent from each other and from the columns in S. Each element of P([n]) occurs exactly once as a column of

S or T . If (σ, ω) ∈ (Sn × S2n−m) fixes T setwise, then ω must be the restriction of the permutation ϕ induced by σ on P([n]), so (σ, ϕ) fixes S ∪ T as well. This means that for the restriction π of ϕ to the first m columns, (σ, π) fixes S. This is a contradiction.

From the following result f(n) can be computed recursively.

Lemma 5.16. f(1) = 2 and f(n) = 2n − min{k : f(k) ≥ n} for n ≥ 2.

45 Proof. By Lemma 5.14, 5.16, Sn × Sn and Sn × S2n−n have regular orbits on P(Ω).

n This implies f(n) ≥ 2 − n. By Lemma 5.15, f(n) = max{m : Sn × S(2n−m) has a regular orbit } =max{m : f(2n − m) ≥ n} =2n − min{k : f(k) ≥ n}.

n Corollary 5.17. f(n) = 2 − dlog2 ne − (n) and (n) ∈ {0, 1}. Moreover if n ≥

dlog2 ne dlog2 ne 2 − dlog2dlog2 nee + 1, then (n) = 1. If n ≤ 2 − dlog2dlog2 nee − 1, then

dlog2 ne (n) = 0. If n = 2 − dlog2dlog2 nee, then (n) = 1 if and only if (dlog2 ne) = 1.

n n Proof. First, we use induction on n to prove: 2 −dlog2 ne−1 ≤ f(n) ≤ 2 −dlog2 ne. For n = 2 and 3, the statement is trivial.

k k Assume 2 − dlog2 ke − 1 ≤ f(k) ≤ 2 − dlog2 ke for 3 ≤ k ≤ n − 1. Let m = min{k : f(k) ≥ n}. Then f(m) ≥ n and f(m − 1) ≤ n − 1. So

m 2 − dlog2 me ≥ f(m) ≥ n,

implying m ≥ dlog2 ne. On the other hand,

m−1 2 − dlog2 m − 1e − 1 ≤ f(m − 1) ≤ n − 1

m−1 n implying 2 − dlog2 (m − 1)e ≤ n and m − 1 ≤ dlog2 ne. Then 2 − dlog2 ne − 1 ≤

n f(n) ≤ 2 − dlog2 ne.

r−1 r Let r = dlog2 ne. Clearly, 2 < n ≤ 2 .

r r r Case 1: 2 − dlog2 re + 1 ≤ n ≤ 2 . Then we have f(r) ≤ 2 − dlog2 re < n implying that min{k : f(k) ≥ n} ≥ r + 1. So f(n) = 2n − min{k : f(k) ≥ n} ≤ 2n − r − 1.

On the other hand, we have f(n) ≥ 2n − r − 1. So f(n) = 2n − r − 1 and (n) = 1.

46 r−1 r r Case 2: 2 ≤ n ≤ 2 −dlog2 re−1. Then we have f(r) ≥ 2 −dlog2 re ≥ n implying that min{k : f(k) ≥ n} ≤ r. So f(n) = 2n − min{k : f(k) ≥ n} ≥ 2n − r. On the other hand, we have f(n) ≤ 2n − r. So f(n) = 2n − r and (n) = 0.

r r Case 3: n = 2 − dlog2 re. If (r) = 0, then f(r) = 2 − dlog2 re = n implying that min{k : f(k) ≥ n} ≤ r. So f(n) = 2n − min{k : f(k) ≥ n} ≥ 2n − r. Noticing f(n) ≤ 2n − r, we have f(n) = 2n − r and (n) = 0.

r If (r) = 1, then f(r) = 2 − dlog2 re − 1 < n implying that min{k : f(k) ≥ n} ≥ r + 1. So f(n) = 2n − min{k : f(k) ≥ n} ≤ 2n − r − 1. Noticing f(n) ≥ 2n − r − 1, we have f(n) = 2n − r − 1 and (n) = 1.

5.4.2 Groups that are orbit equivalent to Sn × Sm

In this section, we start to investigate the groups that are orbit equivalent to Sn ×Sm.

Lemma 5.18. Suppose H ≡ G = Sn × Sm ≤ Smn.Then H ≤ G = Sn × Sm.

Proof. Note H and G have the same block systems and G has two block systems

A = {∆1, ..., ∆m} and B = {Γ1, ..., Γn} with |∆i| = n, |Γj| = m, A and B are also block systems of H.

Let K be the kernel of the action of H on {∆1, ..., ∆m, Γ1, ..., Γn}. Since K ﬁxes

{∆1, ..., ∆m, Γ1, ..., Γn} pointwise, K also ﬁxes {∆i ∩Γj : i = 1, 2, ..., n, j = 1, 2, ..., m}, which is the original permutation domain. Hence K = 1. Since the action of H is faithful and intransitive on {∆1, ..., ∆m, Γ1, ..., Γn} with two orbits {∆1, ..., ∆m} and

{Γ1, ..., Γn}. We obtain H ≤ Sn × Sm = G.

47 Lemma 5.19. Let H ≡ G = Sn × Sm ≤ Smn with m ≥ 5 and m ≥ n. Then

An × Am ≤ H.

Proof. By Lemma 5.18, H ≤ G.

Let A = {∆1, ..., ∆m} and B = {Γ1, ..., Γn} with |∆i| = n and |Γj| = m. Let K and L be the kernel of the action of H on A and B respectively. By Theorem 4.13,

H/K ≥ Am and H/L ≥ An. Note that H is faithful on {∆1, ..., ∆m, Γ1, ..., Γn}, and so K ∩ L = 1.

Case 1: m ≥ 5 and n 6= 2, 4. In this case, both Am and An are simple groups.

Obviously K 6= 1. Because otherwise H = Am or Sm which contradicts the fact ∼ that H has a quotient group An or Sn. Hence KL/L = K/K ∩ L 6= 1. Notice ∼ ∼ that KL/L ¡ H/L, so K = KL/L = Am or Sm. Similarly, L = An or Sn. So

An × Am ≤ KL ≤ H.

Case 2: m ≥ 5 and n = 4. In this case, Am ≤ H/K ≤ Sm and H/L = S4 or A4 by ∼ Theorem 4.13. Since KL/K ¡ H/K, L = KL/K = Am or Sm. On the other hand, ∼ K = KL/L ¡ H/L = S4. As before K 6= 1, otherwise H = Am or Sm and H has a factor A4 or S4, which is not possible. So K = S4,A4 or K4. If K = K4, then

|H/L| = |K||H/K|/|L| is not divisible by 3, which is not possible. Hence A4 ≤ K and A4 × Am ≤ KL ≤ H.

Case 3: m ≥ 5 and n = 2. As in the previous cases H/L = S2 and Am ≤ H/K ≤ Sm.

We have L 6= 1, otherwise H = S2 which is not possible. As L = KL/K ¡ H/K,

L = Am or Sm. So Am ≤ KL ≤ H.

Suppose H ≡ G = Sn ×Sm ≤ Smn and m > n ≥ 5. By the previous two lemmas H

48 can be An×Am, An×Sm, Sn×Am, Sn×Sm or Sm,n where Sm,n = {(σ, ω) ∈ Sn×Sm : σ and ω have the same parity }.

Lemma 5.20. Let H = An × Am and G = Sn × Sm acting on Ω = [m] × [n]. Then H is not orbit equivalent to G.

Proof. Suppose m ≥ n and that Sn × Sm acts on an n × m matrix by permuting the rows and columns.

Let the set S contains exactly the ﬁrst i entries in row i, for i = 1, 2, ..., n. Then the setwise stabilizers of G = Sn × Sm and H = An × Am are GS = S(m−n) and

G H HS = A(m−n) respectively. So |S | = |G : GS|= 6 |H : HS| = |S |, implying that H and G are not orbit equivalent.

Lemma 5.21. Suppose n ≥ 3 and that σ is a permutation of Sn on Ω. Then the induced permutation of σ on the power set of Ω is an even permutation.

Proof. It is suﬃcient to prove the statement when σ is a transposition. In the case n − 2 n − 2 n − 2 the induced permutation is product of 1+ + +...+ = 2n−2 1 2 n − 2 transpositions which is an even permutation.

n Lemma 5.22. Suppose 3 ≤ n ≤ m ≤ 2 . Then Sn × Am is not orbit equivalent to

Sn × Sm.

Proof. If m ≤ f(n), then Sn × Sm has a regular orbit and so Sn × Am is not orbit equivalent to Sn × Sm. Suppose m ≥ f(n) + 1 ≥ 2n − n and let k = 2n − m ≤ n. As usual, we consider

Sn × Sm acting on the entries of an n × m matrix.

49 For each column, the set of entries can be regarded as a subset of the n rows. We construct a set S of entries in the following way. Order the n rows from 1 to n. Each column in S corresponds to a subset of {1, 2, ..., n}.

We define S so that with the exception of the k sets {1}, {1, 2}, ....,{1, 2, ..., k}, each subset of {1, 2, ..., n} occurs exactly once as a column in S. Consider the setwise stabilizer GS of G = Sn × Sm. The group GS must fix the first k rows and the last n − k rows can be permuted freely by GS.

We have GS = {(σ, ω) ∈ Sn × Sm : σ ﬁxes the ﬁrst k rows and ω is induced by σ ∼ on the columns } = S(n−k).

Similarly, the setwise stabilizer HS in H = Sn × Am fixes the first k rows. We have HS = {(σ, ω) ∈ Sn × Am : σ fixes the first k rows and ω is induced by σ on the columns }.

For each (σ, ω) ∈ GS, ω is induced by σ on the columns. By Lemma 5.21, ω is

G an even permutation of Sm. So (σ, ω) ∈ HS. Hence GS = HS and |S | = [G : GS] =

H 2|H : HS| = 2|S |, implying that H are not orbit equivalent to G.

Lemma 5.23. Suppose 3 ≤ n ≤ m. Then An ×Sm is not orbit equivalent to Sn ×Sm.

Proof. Let ∆ = {(i, j) : 1 ≤ j ≤ i ≤ n}. Then (Sn × Sm)∆ ≤ An × Sm. By Lemma

4.5, An × Sm is not orbit equivalent to Sn × Sm. The following lemma is similar to Lemma 5.15.

n Lemma 5.24. Let n > 2 and 1 ≤ m < 2 . Then Sn,m is not orbit equivalent to

Sn × Sm if and only if Sn,(2n−m) is not orbit equivalent to Sn × S(2n−m).

50 Proof. By Lemma 4.5 Sn,m ≡ Sn × Sm, if and only if for all subsets ∆ ⊆ Ω, (Sn ×

Sm)∆Sn,m = Sn × Sm. So Sn,m is not orbit equivalent to Sn × Sm if and only if there exists a subset ∆ ⊆ Ω such that (Sn × Sm)∆ ≤ Sn,m.

Assume there exists ∆ ⊂ Ω = [m]×[n] such that (Sn×Sm)∆ ≤ Sn,m. Clearly, there are no two identical columns in ∆. because otherwise we can take a transposition

ω ∈ Sm permuting two identical columns. Then (1, ω) ∈ (Sn × Sm)∆,which is not possible because (Sn × Sm)∆ ≤ Sn,m.

Let ∆j = {i :(i, j) ∈ ∆}. We have shown that ∆1, ∆2, ..., ∆m are diﬀerent subsets of [n]. Let {∆m+1, ∆m+2, ..., ∆2n } be the complement of {∆1, ..., ∆m} in the

S2n n power set of [n]. Let Γ = i=m+1(∆i, i) ⊂ [n] × [2 − m].

Claim: (Sn × S(2n−m))Γ ≤ Sn,2n−m.

0 Proof of claim: For any (σ, ω) ∈ (Sn × S(2n−m))Γ, there exists a unique ω ∈ Sm

0 such that (σ, ω ) ∈ (Sn × Sm)∆. Let σ induce permutation τ on P([n]). By Lemma 5.21, τ is an even permutation. By the construction of τ, we have τ = ω−1(ω0)−1. So

(σ, ω) ∈ Sn,m if and only if (σ, ω) ∈ Sn,2n−m.

n Lemma 5.25. Suppose 3 ≤ n ≤ m ≤ 2 . If Sn × Sm is not orbit equivalent to Sn,m, then Sn × Sm−1 is not orbit equivalent to Sn,m−1.

Proof. If Sn × Sm−1 has no regular orbit on the power set of [n] × [m − 1], then the statement is true. It suﬃces to consider the case when Sn × Sm−1 has no regular

n n orbit.Let k = 2 − m. By Lemma 5.17, m − 1 ≥ f(n) + 1 ≥ 2 − dlog2 ne implying

n k = 2 − m ≤ dlog2 ne − 1 ≤ n − 1.

51 Since Sn × Sm is not orbit equivalent to Sn,m, by Lemma 5.24 Sn,k is not orbit equivalent to Sn × Sk. There exists an ∆ ⊂ [n] × [k] such that (Sn × Sk)∆ ≤ Sn,k.

Let Si = {j :(j, i) ∈ ∆} for i = 1, 2, ..., k.

There exists a subset S ⊂ P([n]) such that |S|= 6 |Si| for all i = 1, 2, ..., k.

0 Let ∆ = ∆ ∪ (S, k + 1). We have (Sn × S(k+1))∆0 ≤ Sn,(k+1). Since |S|= 6

|Si| for all i = 1, 2, ..., k,(Sn × S(k+1))∆0 must ﬁx the (k + 1)-th row. So we have

(Sn × S(k+1))∆0 ≤ (Sn × Sk)∆ ≤ Sn,k ≤ Sn,(k+1) implying that Sn × S(k+1) is not orbit equivalent to Sn,(k+1). By Lemma 5.24, Sn × Sm−1 is not orbit equivalent to

Sn,m−1.

n Lemma 5.26. If n > 2 and m ≥ 2 , then Sn × Sm is not orbit equivalent to Sn,m.

Proof. For n > 2, we only need to show Sn × S2n is orbit equivalent to Sn,2n . For any subset S from a n × 2n matrix, if there are two identical columns in S, then there exists a transposition ω switching two identical columns. so (1, ω) ∈ (Sn × S2n )S.

Then (Sn × S2n )SSn,m = Sn × Sm. So every two columns in S are diﬀerent. So we can take an odd permutation

σ from Sn which induces an even permutation ω on the power set of [n]. Then

−1 (σ, ω ) ∈ (Sn × S2n )S and (Sn × S2n )SSn,m = Sn × Sm. Hence S2 × S2n is orbit equivalent to Sn,2n .

By the previous lemmas, to determine the pairs (m, n) such that Sn × Sm ≡ Sn,m with m > n ≥ 2, we only need to determine the following

g(n) = max{m : m ≥ n and Sn,m is not orbit equivalent to Sn × Sm}.

52 Since S2 ×S4 is not orbit equivalent to S2,4 but S2 ×S5 ≡ S2,5, we obtain g(2) = 4. The following lemma is directly from Lemma 5.24.

Lemma 5.27. g(n) = 2n − min{k : g(k) ≥ n} for n > 2.

The next example tell us that there exists (m, n) such that Sn ×Sm has no regular orbit but no group other than itself is orbit equivalent to it.

Example. Let G = S4 × S14 acting on [4] × [14] . Since f(4) = 13 and g(4) = 14, G has no regular orbit but G is not orbit equivalent to any other group.

From the corollary, one can see f(n) and g(n) is very close. There are not too many (m, n) satisfying the following two conditions: 1. Sn × Sm has no regular orbit

2. Beside itself Sn × Sm, no group is orbit equivalent to it.

Theorem 5.28. Let H 6= G and H ≡ G = Sn ×Sm ≤ Sym([m]×[n]) and m ≥ n ≥ 2, Then, one of the following holds:

1) (n, m) = (2, 4) and H = S2 × A4.

2) n = 2 and m ≥ 5, H = S2 × Am or S2,m.

n 3) n ≥ 3 and m ≥ 2 + 1, H = S2 × Am or Sn,m

n 4) n ≥ 3 and m = 2 , H = Sn,m.

n 5) n > 2 and f(n) < m ≤ 2 − 1, H = Sn,m if and only if Sn × S2n−m is orbit equivalent to Sn,2n−m.

Proof. By Lemma 5.18, we have H ≤ Sn × Sm. Case 1: 2 ≤ m ≤ 4. If (n, m) = (2, 2) , (n, m) = (2, 3) , (n, m) = (3, 4), or (n, m) = (4, 4), then Sn × Sm has a regular orbit on P(Ω). So, in this cases

53 Sn × Sm is not orbit equivalent to any group other than itself. If (n, m) = (2, 4), then

S2 × S4 ≡ S2 × A4 and is not orbit equivalent to any other group.

Case 2: m ≥ 5. By Lemma 5.19,we have An × Am ≤ H. Then H = An × Am

,An × Sm , Sn × Am or Sn,m. Notice An × Am and An × Sm are not orbit equivalent

n to Sn × Sm. If m ≥ 2 + 1, by Lemma 5.26, Sn × Sm is orbit equivalent to Sn × Am

n and Sn,m. If m ≤ 2 , then Sn × Sm is not orbit equivalent to Sn × Am and Sn × Sm is orbit equivalent to Sn,m is and only if Sn × S2n−m is orbit equivalent to Sn,2n−m.

54 CHAPTER 6

MISCELLANEOUS ORBIT EQUIVALENT 2-STEP

IMPRIMITIVE PERMUTATION GROUPS

In this chapter we use the following notation. Let Ω = ∆ × {1, 2, . . . , n}, and

G ≤ Sym(Ω) with block system Σ = {∆1,..., ∆n}, ∆i := {(δ, i) | δ ∈ ∆} for i = 1, 2, . . . , n. For Γ ⊂ ∆, we shall abbreviate {(γ, i) | γ ∈ Γ} by (Γ, i). Let

A ≤ Sym(∆) and B ≤ Sym({1, 2, . . . , n}) such that G∆1 ∼ A and GΣ ∼ B. ∆1 = = Let H < G and H ≡ G. Moreover, let K and L be the kernels of the action of G and H on Σ, respectively.

6.1 A ∈ (α) and B ∈ (β)

We write the elements of W := A o B as (n + 1)-tuples (a1, . . . , an; b) where ai ∈ A

n for 1 ≤ i ≤ n and b ∈ B. For a := (a1, . . . , an; 1) ∈ A , we deﬁne the support supp(a) of a as the set of indices I ⊂ {1, . . . , n} with ai 6= 1 for i ∈ I and ai = 1 for i 6∈ I. The following lemmas are basic.

Lemma 6.1. Suppose that A ∈ (α), and let S ⊂ ∆ so that AS = 1 and ∆ \ S is not

Sn Σ Σ in the same orbit as S. Then T = i=1(S, i), we have GT ≡ HT .

55 0 S S Proof. For any subset Γ ⊂ Σ, let Γ = i∈Γ(S, i)∪ i∈∆\Γ(∆\S, i). We have (GT )Γ =

GΓ0 and (HT )Γ = HΓ0 .

Since H ≡ G, we have |G : H| = |GT : HT | and |G : H| = |GΓ0 : HΓ0 |. Therefore,

Σ Σ |(GT )Γ :(HT )Γ| = |GT : HT | and GT ≡ HT .

Corollary 6.2. Let H ≤ G = A o Sn for n ≥ 6 and A ∈ (α). Suppose that there exists S ⊂ ∆ such that ∆ \ S are in two diﬀerent regular orbits of A on P(∆) and

suppose H ≡ G. Then either H = G or |G : H| = 2.

Sn n Σ Σ Proof. Let T = i=1(S, i). Clearly GT ∩ A = 1 and GT = Sn, and HT ≡ GT by

Σ Lemma 6.1. So HT is either Sn or An, and G = H or |G : H| = 2.

Lemma 6.3. If A has at least k regular orbits on P(∆) and the distinguishing number

of B is k then G has a regular orbit on P(Ω).

Proof. Let S = (S1,S2,...,Sk) be an ordered partition of {1, 2, . . . , n} such that only

the identity element in B fixes S. Let R1,R2,...,Rk be k subsets of ∆, in k different regular orbits of A on P(∆). Let Γ = Sk (S (R , i)). Then only the identity of j=1 i∈Sj i A o B fixes Γ setwise. So, by Lemma 2.1, G has a regular orbit on P(Ω).

We need the following variant of Lemma 2.7.

Lemma 6.4. We divide the groups (H, k) listed in Theorem 2.5 into the following

three categories.

Type I: (D10, 5), (AGL(1, 5), 5), (AGL(1, 7), 7), (AΓL(1, 8), 8), (PSL(3, 2), 8),

(AGL(1, 9), 9), (AΓL(1, 9), 9), (ASL(2, 3), 9), (PSL(2, 8), 9); (S5, 10),

56 (PGL(2, 9), 10), (M10, 10), (A6, 10), (PSL(2, 11), 11), (PGL(2, 11), 12),

4 4 4 4 (PGL(2, 13), 14), (2 .(A5 × 3).2, 16), (2 .A6, 16), (2 .S6, 16), (2 .A7, 16),

(PSL(2, 16).2, 17), (PSL(2, 16).4, 17), (PΓL(3, 4), 21), (M22, 22), (M22.2, 22),

(ASL(5, 2), 32).

Type II: (PSL(2, 5), 6), (PGL(2, 5), 6), (PSL(3, 2), 7), (PSL(3, 2).2, 8),

(AGL(2, 3), 9), (P ΓL(2, 8), 9), (PΓL(2, 9), 10), (S6, 10); (M11, 11),

(M11, 12), (PSL(3, 3), 13), (A8, 15), (M23, 23).

Type III: (ASL(3, 2), 8), (M12, 12), (ASL(4, 2), 16), (M24, 24).

For each group H of type I, there is an ordered partition Q = (Q1,Q2,Q3) of the

permutation domain with |Q1| = 1 such that only the identity of H ﬁxes Q. For

each group H of type II, there is an ordered partition Q = (Q1,Q2,Q3,Q4) of the

permutation domain with |Q1| = |Q2| = 1 such that only the identity of H ﬁxes Q.

For each group H of type III, there is an ordered partition Q = (Q1,Q2,Q3,Q4,Q5)

of the permutation domain with |Q1| = |Q2| = |Q3| = 1 such that only the identity of H ﬁxes Q.

Proof. It is done by a GAP computation. In the cases k ≥ 18, we list the partitions

below; for k ≤ 17, a brute force search yields appropriate partitions. As usual, the

set stabilizers are taken in the permutation representation given in the GAP library.

For (H, k) = (P ΓL(3, 4), 21), take Q1 = {1},Q2 = {2, 4, 5, 8, 9, 10, 11, 15}. For

(H, k) = (M22.2, 22) or (M22, 22),take Q1 = {1},Q2 = {2, 4, 5, 8, 9, 10, 11, 15, 16, 19}.

57 For (H, k) = (ASL(5, 2), 32), the one point stabilizer is primitive but it is not in the list of Theorem 2.5, so the required partition (Q1,Q2,Q3) exists. For (H, k) =

(M23, 23), take Q1 = {23},Q2 = {1},Q3 = {2, 4, 5, 8, 9, 10, 11, 15, 16, 19}. For (H, k) =

(M24, 24), take Q1 = {24},Q2 = {23},Q3 = {1},Q4 = {2, 4, 5, 8, 9, 10, 11, 15, 16, 19}.

Lemma 6.5. For the groups (H, k) listed in Theorem 2.5, let MH (q) denote the k-dimensional permutation module of H over GF(q), for q ∈ {2, 3}. Then MH (q) has a unique (k − 1)-dimensional H-invariant submodule. Moreover, if (H, k) 6=

(PSL(3, 2), 7) then MH (2) has no (k − 3) or (k − 2)-dimensional H-invariant submodule.

Proof. For each group of type (β), we construct the permutation matrices of gen- erators and compute the submodule structure by a GAP program. Checking the dimensions of submodules, we obtain the statement of the lemma.

4 Lemma 6.6. Let (A, |∆|) = (C2 .S5, 16) or (D14, 7) and B ∈ (β). Then, if G has no regular orbit on Ω then G = W or |W : G| = 2.

Proof. We identify ∆ with {1, 2,..., |∆|}. Let R1 and R2 be two subsets of ∆, from two diﬀerent regular orbits of A on P(∆). We also deﬁne three subsets T1, T2, and T3

4 of ∆ the following way. In the case (A, |∆|) = (C2 .S5, 16), let T1 = {1, 2, 3, 4, 5, 6, 9},

T2 = ∆ \ T1, and T3 = {1, 2, 3, 5, 7, 8, 9}. In the case (A, |∆|) = (D14, 7), let T1 = {1},

T2 = ∆ \ T1, and T3 = {2, 5}.

A small GAP computation veriﬁes the following facts. First of all, the Ti are in

three diﬀerent orbits of A on P(∆) and |ATi | = 2 for each i. Moreover, in the case

58 A = D14, for each i we have ATi = ha1i for the permutation a1 = (2, 5)(3, 6)(4, 7),

0 and a1 ∈ A \ A . The permutation a2 = (1, 3)(2, 4)(5, 6) is an A-conjugate of a1, the aj are noncommuting, and there is no nontrivial element of A centralizing both a1 and a2. Finally, the A-conjugates of a1 generate A, and any two A-conjugates of a1 either commute, or the normal closure of their commutator is A0.

4 In the case A = C2 ,S5, for each i we have ATi = ha1i for the permutation a1 =

0 (3, 5)(4, 6)(11, 13)(12, 14), and a1 ∈ A\A . The permutations a2 = (5, 9)(6, 10)(7, 11)(8, 12), a3 = (5, 16)(6, 15)(7, 14)(8, 13), a4 = (2, 5)(4, 7)(10, 13)(12, 15) are A-conjugates of a1, the aj are pairwise noncommuting for 1 ≤ j ≤ 4, and there is no nontrivial element of A centralizing all aj. Finally, the A-conjugates of a1 generate A and any two

0 A-conjugates of a1 either commute, or the normal closure of their commutator is A . Suppose ﬁrst that B is of type I in Lemma 6.4. Then there exists an ordered partition Q = (Q1,Q2,Q3) of {1, 2, . . . , n}, with Q1 = {j} for some j ≤ n, such that only the identity of B ﬁxes Q. Let Γ = S (T , i) ∪ S (R , i) ∪ S (R , i). i∈Q1 1 i∈Q2 1 i∈Q3 2 n Then |WΓ| = 2 and WΓ = hx1 := (1, 1,..., 1, a1, 1,..., 1; 1)i for the element x1 ∈ A with support {j}. Since G has no regular orbit, GΓ = WΓ and x1 ∈ G. The elements G of x ∆j ⊂ G also have support {j}. Since G∆j ∼ A, these conjugates generate a 1 ∆j = subgroup Aj ≤ G isomorphic to A by the last observations in the previous paragraphs and the nontrivial elements of Aj all have support {j}. Also, since G acts transitively

n n on Σ, the G-conjugates of Aj generate A . Hence K = A and G = W . Suppose next that B is of type III. In this case, B is 3−transitive on {1, 2, . . . , n} and there exists an ordered partition Q = (Q1,Q2,Q3,Q4,Q5) of {1, 2, . . . , n}, with

59 |Q1| = |Q2| = |Q3| = 1 such that only the identity of B ﬁxes Q. First, we show that

n 4 n 4 K ≥ C7 if A = D14 and K ≥ (C2 .A5) if A = C2 .S5. Let Γ = S (T , i) ∪ S (T , i) ∪ S (T , i) ∪ S (R , i) ∪ S (R , i). i∈Q1 1 i∈Q2 2 i∈Q3 3 i∈Q4 1 i∈Q5 2 ∼ Then WΓ = C2 × C2 × C2, and the supports of the elements of WΓ are subsets of

Q1 ∪ Q2 ∪ Q3. In each w ∈ WΓ, the nontrivial coordinates are equal to a1. Let

x1 ∈ GΓ be a nontrivial element of GΓ; since ATi is the same for all i, without loss of

th generality we can suppose that the j coordinate of x1 is a1, for Q1 = {j}.

Conjugating x1 by appropriate elements of G∆j , we obtain x2, x3, x4 ∈ G with

th support size at most three such that the j coordinate of xi is ai, for 2 ≤ i ≤ 4. If there exists i ∈ {2, 3, 4} such that supp(x1) ∩ supp(xi) = {j} then the commutator

0 [x1, xi] 6= 1 and has support {j}. Its conjugates by G∆j generate a subgroup Aj ≤ G, 0 ∼ 0 0 Aj = A , and the nontrivial elements of Aj all have support {j}. Also, since G acts

0 0 n 0 n transitively on Σ, the G-conjugates of Aj generate (A ) . Hence K ≥ (A ) . On the other hand, if |supp(x1) ∩ supp(xi)| ≥ 2 for all i ≤ 4 then the union of the four supports have size at most six. Since we are in case III, n ≥ 8 and B is 3-transitive.

Hence there exists a G∆j -conjugate y of x1 with supp(y) ∩ supp(xi) = {j} for all

th i ≤ 4. Since the j coordinate of y cannot commute with all four ai, there exists

0 0 n i ≤ 4 with [xi, y] 6= 1 and supp([xi, y]) = {j}. So Aj ≤ G and K ≥ (A ) , as above. The factor group K/(A0)n can be identiﬁed with an n-dimensional vector space over GF(2). The element x1 ∈ K corresponds to a vector with at most three nontrivial coordinates; if it has three nontrivial coordinates then the product of x1 with an appropriate G-conjugate has only two (we use the 3-transitivity of B in this step).

Hence, using 2-transitivity of B, K/(A0)n contains all vectors with two nontrivial

60 coordinates and these vectors generate an (n − 1)-dimensional subspace. Therefore

0 n n−1 K ≥ (A ) .C2 and |W : G| ≤ 2. The proof in the case when B is of type II is similar. We start with an ordered partition Q = (Q1,Q2,Q3,Q4) of {1, 2, . . . , n}, with |Q1| = |Q2| = 1 such that only the identity of B ﬁxes Q and with Γ = S (T , i) ∪ S (T , i) ∪ S (R , i) ∪ i∈Q1 1 i∈Q2 2 i∈Q3 1 S (R , i). Then W ∼ C ×C , and the supports of the elements of W are subsets i∈Q4 2 Γ = 2 2 Γ of Q1 ∪ Q2. We deﬁne x1, x2, x3, x4 as in the type III case, and if all four xi have the same 2-element support then, noting that all type II groups B are 2-transitive, we can construct y ∈ G with supp(xi) ∩ supp(y) = Q1 for all i. The rest of the proof is identical to the type III case.

Lemma 6.7. (i) Let (A, |∆|) = (D14, 7) and B ∈ (β). Then W has an index 2 subgroup H orbit equivalent with W if and only if B has an index 2 subgroup C orbit equivalent with B. Moreover, if such orbit equivalent H exists then it is unique.

(ii) Let (A, |∆|) = (C3, 3). Then W has an index 3 subgroup H orbit equivalent with W if and only if B has an index 3 subgroup C orbit equivalent with B. Moreover, if such orbit equivalent H exists then there are exactly two possibilities for H.

Proof. We prove parts (i) and (ii) simultaneously. Let m = 2 if A = D14 and m = 3 if A = C3. Let R1 and R2 be two subsets of ∆ in diﬀerent regular orbits of A on P(∆). For D ≤ B, we denote by D∗ the subgroup {(1, 1,..., 1; b) | b ∈ D} of W .

Suppose that there exists an index m subgroup H of W that is orbit equivalent

S ∗ ∗ ∗ to W . Let Γ = 1≤i≤n(R1, i). Then WΓ = B and HΓ = B ∩ H = C for some

61 subgroup C ≤ B. By Lemma 4.3, |WΓ : HΓ| = |W : H| = m and so |B : C| = m. We shall prove that B ≡ C. S S For any subset S ⊂ {1, 2, . . . , n}, take Γ(S) = i∈S(R1, i) ∪ i/∈S(R2, i). Then

∗ ∗ ∗ WΓ(S) = (BS) and HΓ(S) = H ∩ (BS) = (CS) . Using Lemma 4.3 again, |WΓ(S) :

Conversely, suppose that an index m subgroup C of B is orbit equivalent to B.

We construct an index m subgroup of H of W the following way. The factor group

An/(A0)n can be identiﬁed with an n-dimensional vector space over GF(m). Let V be the (n − 1)-dimensional subspace consisting of vectors whose sum of coordinates is 0 (in GF(m)) and let L be the index m subgroup of An containing (A0)n and

L/(A0)n corresponding to V . Let a ∈ A \ A0 and b ∈ B \ C. We deﬁne H as

H = hL, C∗, (a, 1, 1,..., 1; b)i.

We claim that H ≡ W . Let Γ ⊂ Ω be arbitrary. By Lemma 4.5, it is enough to prove that HWΓ = W ; since |W : H| is a prime number, this is equivalent to

WΓ \ H 6= ∅. We distinguish two cases.

Case 1: There exists some i ≤ n such that Γi := Γ∩∆i has nontrivial stabilizer in

A. In this case, there exists w := (1,..., 1, t, 1,..., 1; 1) ∈ WΓ with support {i}. The crucial observation is (and this is the feature that distinguishes these two groups A from the other groups listed in Theorem 1.1) that t can be chosen in A\A0. This fact is obvious for A = C3. In the case A = D14, if Γi ∈ {∅, ∆i} then its stabilizer is D14 so t can be chosen as desired. If Γi is a proper subset of ∆i with nontrivial stabilizer

62 then |(D14)Γi | = 2 and the nontrivial element of the stabilizer is a good choice for t.

n In any case, since w 6∈ L and H ∩ A = L, we have w ∈ WΓ \ H.

Case 2: For all i ≤ n,Γ ∩ ∆i has trivial stabilizer in A. Since A has only two regular orbits on P(∆), this means that for any i,Γ ∩ ∆i is in the same orbit as

ai R1 or R2. Deﬁne the set S ⊂ {1, 2, . . . , n} so that for i ∈ S,Γ ∩ ∆i = (R1 , i) and

ai ∗ for i ∈ {1, 2, . . . , n}\ S,Γ ∩ ∆i = (R2 , i), for some ai ∈ A. Then WΓ = (BS) and

∗ HΓ = (CS) . Since B ≡ C, there exists b ∈ BS \CS. This implies that (1, 1,..., 1; b) ∈

WΓ \ H.

Finally, we prove the claims about the possibilities for H. In the case A = C3, hL, C∗, (a, 1, 1,..., 1; b)i and hL, C∗, (a2, 1, 1,..., 1; b)i are two diﬀerent H < W that are orbit equivalent with W . Conversely, suppose that H ≡ W and |W : H| = m.

Examining the list of Theorem 2.6, we see that for all groups B ∈ (β) which have an orbit equivalent subgroup C of order m, C = B0 so B determines C uniquely.

∗ Moreover, we have shown above that H must contain C (by examining HΓ for Γ =

S 0 n 1≤i≤n(R1, i)). Also, since H acts as B on the vector space (A/A ) by Theorem 4.13 and (A/A0)n has a unique (n − 1)-dimensional B-invariant subspace by Lemma 6.5,

∗ ∼ H must contain L. Since W/hC ,Li = Cm × Cm, we have to determine which subgroups of order m of this factor group can be included in H. The group Cm × Cm has m + 1 subgroups of order m. In our setting, we cannot add (a, 1, 1,..., 1; 1) to hC∗,Li because H does not contain An. We also cannot add (1, 1,..., 1; b) to hC∗,Li because H does not contain B∗. This leaves m − 1 subgroups, and we have seen that these m − 1 possibilities lead to orbit equivalent H < W .

Proof of Theorem 1.3. Suppose that A ∈ (α) but (A, |∆|) 6= (C2, 2) and B 6∈ (γ).

63 If A has at least four regular orbits on P(∆) then, by Lemmas 2.7 and 6.3, W has a regular orbit on P(Ω). Similarly, if B ∈ (α) then by Lemma 6.3 W has a regular orbit on P(Ω). Hence in these cases there is no H < G ≤ W with H ≡ G and it is enough to consider the case when A is listed in Theorem 1.1 and B ∈ (β).

From now on, suppose that (A, |∆|) is one of (C3, 3) (D14, 7), (AGL(1, 8), 8),

4 (PSL(2, 11), 12), (PSL(2, 13), 14), (A7, 15), (C2 .S5, 16) and B ∈ (β). Moreover, let H < G and H ≡ G (note that this implies that G has no regular orbit on P(Ω)). Let

M be the kernel of action of H on Σ. By Theorem 4.13, HΣ = GΣ, so |K : M| =

|G : H|. Let R1 and R2 be two subsets of ∆ in two diﬀerent regular orbits of A on

P(∆) and let S = (S1,S2,S3,S4) be an ordered partition of {1, 2, . . . , n} such that only the identity of B ﬁxes S.

Case 1: (A, |∆|) is one of (PSL(2, 11), 12), (PSL(2, 13), 14), (A7, 15). A small

GAP computation shows that there exist T1 ⊂ ∆ and T2 ⊂ ∆ such that |Ti| < |∆|/2 for i = 1, 2, |A | = 2, and |A | = 3. Let Γ = S (T , i) ∪ S (∆ \ T , i) ∪ T1 T2 1 i∈S1 1 i∈S2 1 S (R , i) ∪ S (R , i). Then |G | divides 2|S1∪S2|. If we replace T by T , we i∈S3 1 i∈S4 2 Γ1 1 2

|S1∪S2| obtain a subset Γ2 ⊂ Ω such that |GΓ2 | divides 3 . So Corollary 4.4 implies |G : H| = 1, a contradiction.

Case 2: (A, |∆|) = (AGL(1, 8), 8). If T1,T2 ⊂ ∆, |Tj| = j for j = 1, 2 then |A | = 7 and |A | = 2. Hence for the sets Γ = S (T , i) ∪ S (∆ \ T , i) ∪ T1 T2 j i∈S1 j i∈S2 j S (R , i)∪S (R , i), we have that |G | divides 7|S1∪S2| and |G | divides 2|S1∪S2|. i∈S3 1 i∈S4 2 Γ1 Γ2 So Corollary 4.4 leads to a contradiction.

4 Case 3: (A, |∆|) is (D14, 7) or (C2 .S5, 16). By Lemma 6.6, G = W or |W : G| = 2.

Let T1 , T2, and T3 be the three subsets of ∆ deﬁned in the ﬁrst paragraph of

64 the proof of Lemma 6.6. By Lemma 6.4, there exists an ordered partition Q =

(Q1,Q2,Q3,Q4,Q5) of {1, 2, . . . , n} with |Q1| = |Q2| = |Q3| = 1 such that only the identity of B ﬁxes Q. Let Γ = S (T , i)∪S (T , i)∪S (T , i)∪S (R , i)∪ i∈Q1 1 i∈Q2 2 i∈Q3 3 i∈Q4 1 S (R , i). Then |W | = 8. Therefore, by Lemma 4.3, if G = W then |W : H| i∈Q5 2 Γ ∼ 0 n n−1 divides 8. If |W : G| = 2 then |GΓ| = 4 (because K = (A ) .C2 is uniquely determined by Lemma 6.5 and this unique K does not contain elements from An\(A0)n with one nontrivial coordinate) and so |G : H| divides 4 by Lemma 4.3. Thus |W : H| is a divisor of 8 in this case as well.

n ∼ 0 n k Since |W : H| = |A : M|, we obtain that M = (A ) .C2 , for some k ≥ n − 3. However, if (B, n) 6= (PSL(3, 2), 7)) then, by Lemma 6.5, there is no (n − 2) or

(n − 3)-dimensional B-invariant subspace of GF(2)n. Hence the only possibility is that k = n − 1, which means G = W and |G : H| = 2. If (B, n) = (PSL(3, 2), 7)) then let [ [ Γ(S) = (R1, i) ∪ (R2, i) and d(B) = gcd{|WΓ|, |WΓ(S)| | S ⊂ [n]}. i∈S i∈[n]\S We obtain d(B) = 2 by a GAP computation. Now |G : H| divides d(B), so |G : H| divides 2 and |W : H| is a divisor of 4. There is no 5-dimensional B-invariant subspace of GF (2)7, which means the only possibility is W = G and |G : H| = 2.

In the case A = D14, we are in the situation described in Lemma 6.7. Examining the list of Theorem 2.6, we obtain the four B ∈ (β) with an orbit equivalent index 2 subgroup as listed in Theorem 1.3(i).

4 In the case A = C2 .S5 there are no examples. Identify ∆ with {1, 2,..., 16} and let U1 = {1, 2, 3, 6, 11}, U2 = ∆ \ U1, U3 = {1, 2, 3, 6, 11, 13}, and U4 = ∆ \ U3. The

65 Ui are in diﬀerent orbits of A on P(∆), and a small GAP computation shows that A ≤ A0 = C4.A for 1 ≤ i ≤ 4. Hence for the set Γ = S (U , i) ∪ S (U , i) ∪ Ui 2 5 i∈S1 1 i∈S2 2 S (U , i) ∪ S (U , i), we have G ≤ (A0)n ≤ M ≤ H. This implies G = H , a i∈S3 3 i∈S4 4 Γ Γ Γ contradiction.

Case 4: (A, |∆|) = (C , 3). Let Γ = ∪ (∆, i) ∪ S (R , i) ∪ S (R , i). 3 i∈S1 i∈S3 1 i∈S4 2

|S1∪S2| S Then |WΓ| divides 3 . Moreover, for S ⊂ {1, 2, . . . , n}, let Γ(S) = i∈S(R1, i) ∪ S i∈{1,...,n}\S(R2, i). We deﬁne d(B) as d(B) = gcd{|WΓ|, |WΓ(S)| | S ⊂ {1, 2, . . . , n}}. We claim that d(B) = 3 if (B, n) = (ASL(3, 2), 8), (AΓL(1, 8), 8), (P ΓL(2, 8), 9),

(M12,, 12), d(B) divides 3 if (B, n) = (M24, 24), and d(B) = 1 for all other groups B ∈ (β). We prove this claim by a GAP computation. For n ≤ 17, we compute

|WΓ(S)| for a representative S from each B-orbit on the power set of {1, 2, . . . , n}.

For n ≥ 18, it is enough to observe that for S = {1, 2, 3, 4, 5, 8}, |WΓ(S)| = 8, 8, 16 for (B, n) = (PΓL(3, 4), 21), (M22, 22), and (M22.2, 22), respectively and for S =

{1, 4, 5, 6, 8, 10, 11, 13, 15, 17, 18, 20}, |WΓ(S)| = 20, 660, 8 for (B, n) = (M23, 23), (M24, 24), and (ASL(5, 2), 32), respectively.

Since gcd{|GΘ| | Θ ⊂ Ω} divides d(B), we must have d(B) = 3 by Corollary 4.4. Hence it is enough to consider the cases (B, n) = (ASL(3, 2), 8), (AΓL(1, 8), 8),

(P ΓL(2, 8), 9), (M12,, 12), and (M24, 24). Moreover, in these five cases, |G : H| = 3 and so |K : M| = 3. The subgroups K and M correspond to B-invariant subspaces of GF(3)n, and the dimensions of these subspaces differ by 1. We check by a GAP computation that in all five cases there are exactly four B-invariant subspaces, of dimensions 0, 1, n − 1, and n, respectively. So either |K| = 3 or K = An, the latter implying G = W .

66 In the case |K| = 3, the nontrivial elements of K have support {1, 2, . . . , n}.

Hence KΓ = GΓ = 1, contradicting the fact that G has no regular orbit. So the only possibility is G = W and |G : H| = 3. This is the the situation described in

Lemma 6.7. Examining the list of Theorem 2.6, we obtain the two B ∈ (β) with orbit equivalent index 3 subgroups as listed in Theorem 1.3(i).

6.2 A ∈ (α) and B ∈ (γ)

In this section, we investigate the case when the top group B ∈ (γ) and A ∈ (α).

The following important theorem gives us an easy way to identify a giant group

An or Sn.

Theorem 6.8. (Jordan) [9] Let X ≤ Sym(Ω) be a primitive group which contains a cycle x of prime length p. Then either X ≥ Alt(Ω) or |Ω| ≤ p + 2.

Based the previous theorem, we prove the following result for a transitive group.

Lemma 6.9. Let P be a transitive permutation group on Ω, and suppose that there exists an element g of prime order p where n/2 < p < n − 2. Then P ≥ An.

Proof. We claim P is primitive. If not, then there exists a block ∆ such that P ≤

Sym(∆) o Sm where m = |Ω|/|∆|. However Sym(∆) o Sm has no element of prime order p, a contradiction. By Jordan’s Theorem, we have P ≥ An. We shall use the following slight generalization of the Giant Action Theorem 5.2.

67 Theorem 6.10. Let G ≤ Sym(Ω) and let Σ = {∆1, ..., ∆n} be a G-invariant partition of Ω. Suppose that the image of the G-action G → Sym(Σ) is Alt(Σ) and let K be the kernel of this action. Let b = |∆i| and assume further that n > 2b − 5 and b > 9. ∼ Then K has a clean complement R = An.

Proof. By Bertrand’s postulate, for b > 9 there exists a prime p such that b/2 < p < b − 2. Take π ∈ G such that πΣ = (1, 2, ..., p). There exists an m such that the order of πm is a power pk for some k. We can assume m = 1.

If π is not a clean element, then there exists i such that the restriction of π on ∆i is not identity is a p-cycle. By Lemma 6.9, G∆i contains Alt(∆ ). Then by Lemma ∆i i n 5.1 K ≥ Ab or Ab ≤ K ≤ Sb. By Lemma 5.5 with p = 3, G/soc(K) has a subgroup 0 ∼ 0 0 0 R = An. The pre-image of R is either Ab o R or Ab × R . In either case, K has a clean complement.

The rest of the proof is similar to the proof of Giant Action Theorem. If π is a clean element, then there exists another clean (π0)Σ = (p, p + 1, ..., 2p − 1). Thus the commutator τ = [π, π0] is a clean permutation such that τ Σ is a 3-cycle. Let

Σ τi be a conjugate of τ such that (τi) = (i, i + 1, n). If n is odd, then the group

hτ1, τ3, ..., τn−2i is a clean complement to K. If n is even then let C = hτ1, τ3i∆n . ∼ Clearly C = A4. Then hC, τ4, τ6, ..., τn−2i is a clean complement to K.

Lemma 6.11. Let G be the group in the last theorem. n > 2b − 5 and b > 9. For ∼ any subset S ∈ ∆1, there exists a set Γ ∈ Ω such that Γ ∩ ∆1 = S and GΓ ≥ R = An.

Proof. Let R be the clean complement to K. For a ∈ ∆1, let O(a) be the orbit of R containing a. We claim that |O(a)| = n. Indeed if |O(a)| > n, then |O(a) ∩ ∆1| > 1

68 and RO(a)∩∆1 > Ra. This is a contradiction with R is clean. So, for Γ = ∪x∈SO(x), we have GΓ ≥ R and Γ ∩ ∆1 = S.

Theorem 6.12. Let (B, n) be of type (γ), (A, b) be of type (α), n ≥ 2b − 5, and b > 9. If H ≡ G, then K = L, |G : H| = 2 and G contains a clean complement to

Σ Σ ∆ Proof. By Theorem 4.13, H = Sn or H = An. Let A0 = H∆ . Since A ∈ α, there exists an S ⊂ ∆ such that AS = 1. By Lemma 6.11 there exists Γ ⊂ Ω such that

HΓ ≥ An and AS = 1 implies HΓ = An or Sn. Since GΓ 6= HΓ, we obtain H = An and GΓ = Sn, and the latter is a clean complement of K. The previous theorem is not a suﬃcient condition for the orbit equivalence of

G and H. The following theorem shows when n is large enough, the condition is suﬃcient.

Theorem 6.13. Let G and H be two imprimitive groups acting on Ω = ∆ × [n], with block system Σ = {∆1, ∆2, ..., ∆n}. If G = K o Sn and H = K o An and b = |∆1|, then there exists an integer h(b) such that H ≡ G when n > h(b).

b Proof. Note H contains a clean subgroup An but not Sn. Let h(b) = 2 . For any

Γ ⊂ Ω, there must exist i and j such that Γ ∩ ∆i and Γ ∩ ∆j are identical. Then

GΓ contains an odd permutation in Sn. Therefore GΓH = G and, by Lemma 4.5, H ≡ G.

69 6.3 A ∈ (β) and B ∈ (α)

In this section,we focus on the case when the top group B is in α and A is in β.

If H < G and H ≡ G, then for any Γ ⊂ Ω the index |G : H| = |GΓ : HΓ| divides the order of the setwise stabilizer GΓ. Therefore, it is useful to compute the greatest common divisors of the setwise stabilizers of groups of type (β). Based on

GAP computations, we obtain the following table.

Lemma 6.14. The greatest common divisors of the setwise stabilizers of a group with degree up to 17 of type (β) are listed in Table 6.1. The entries in the notation

(n, G, i) denote the degree of the group, the group and the position of the group in the primitive group of GAP respectively.

For the groups with degree greater than 17, we just compute some of the set stabilizers in GAP and have the following.

Lemma 6.15. If G is one of (PΓL(3, 4), 21), (M22.2, 22), (M22, 22), (M23, 5), (M24, 24), (ASL(5, 2), 32), the greatest common divisors of the size of the setwise stabilizers are divisors of 2, 4, 2, 4, 8, 2 respectively.

Lemma 6.16. The almost simple groups A in Theorem 2.5 can be divided into the following four categories.

Type I: Simple groups.

70 (n, G, i) GCD of |G∆| (n, G, i) GCD of |G∆|

(5,D10, 2) 2 (5, AGL(1, 5), 3) 2 (6, PSL(2, 5), 1) 2 (6, PGL(2, 5), 2) 2

(7, AGL(1, 7), 4) 1 (7, PSL(3, 2), 5) 2

(8, PGL(2, 7), 5) 2 (8, PSL(3, 2), 4) 1

(8, AΓL(1, 8), 2) 3 (8, ASL(3, 2), 3) 24

2 (9, 3 : D8, 2) 2 (9, AΓL(1, 9), 5) 2 (9, PSL(2, 8), 8) 2 (9, PΓL(2, 8), 9) 6

(9, AGL(2, 3), 7) 2 (9, ASL(2, 3), 6) 1

(10, PΓL(2, 9), 7) 4 (10, PGL(2, 9), 4) 2

(10,M10, 6) 2 (10,S6, 5) 2

(10,A6, 3) 2 (10,S5, 2) 2

(11, PSL(2, 11), 5) 2 (11,M11, 6) 4

(12,M11, 1) 2 (12,M12, 2) 24 (12, PGL(2, 11), 4) 1 (13, PSL(3, 3), 7) 2

(14, PGL(2, 13), 2) 1 (15, PSL(4, 2) = A8, 4) 4 (16, 24.PSL(4, 2), 11) 8 (16, AΓL(2, 4), 12) 1

4 4 (16, 2 .A7, 20) 1 (16, 2 .A6, 17) 1

4 4 (16, 2 .S6, 16) 2 (17, PΓL(2, 2 ), 8) 2 (17, PSL(2, 24) : 2, 7) 2

Table 6.1: Groups with no regular orbit

71 Type II: (PGL(2, 5), 6), (PGL(2, 7), 8), (S6, 10).

Type III: (PGL(2, 11), 11), (PΓL(3, 4), 21).

4 4 Type IV: (PGL(2, 9), 10), (S5, 10), (PGL(2, 13), 14), (PΓL(2, 2 ), 17), (PSL(2, 2 ), 17) :

2, 17), (M22.2, 22).

Type V: (PΓL(2, 8), 9), (PΓL(2, 9), 10).

Let T = soc(A). For each group A of type II, there exists a subset ∆1 of Ω

such A∆1 ≤ T . For each group of type III, there exist two subsets ∆1 and ∆2 from

two diﬀerent orbits of A such that A∆i ≤ T (i = 1, 2). For each group of type IV, there exists four subsets ∆i (i = 1, 2, 3, 4) from four diﬀerent orbits of A such that

A∆i ≤ T (i = 1, 2, 3, 4).

Proof. For n ≤ 17, a brute force search yields the required subsets. For A =

(PΓL(3, 4), 21), take ∆1 = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11} and ∆2 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 18}.

For A = (M22.2, 22), take ∆1 = {1, 2, 3, 5, 8}, ∆2 = {1, 2, 3, 5, 8, 13}, ∆3 = Ω \ ∆1, and ∆4 = Ω \ ∆2.

Lemma 6.17. Let B = (GΣ, n) with n ≥ 3 be of type (α) and A = (G∆1 , |∆|) be of ∆1 type (β) and almost simple. Let T = soc(A). If G has no regular orbit on P(Ω), then

K ≥ T n.

Proof. By Theorem 4.12, K 6= 1. By Lemma 5.1, K ≥ T n or T ≤ K ≤ A.

If T ≤ K ≤ A, then we shall prove that G must have a regular orbit. By Theorem

2.7, the distinguishing number of the groups in Theorem 2.5 is no more than four.

There exists a partition (U1,U2,U3,U4) of ∆1 such that only identity of A ﬁxes it.

72 Let S ⊂ Σ such that BS = 1. Let Q1 = U1 ∪ U2, Q2 = U2 ∪ U3 and Q3 = U3 ∪ U4.

g1 g2 Let g1, g2 ∈ G such that ∆1 = ∆2 and ∆1 = ∆3 and let

g1 g2 [ Γ = Q1 ∪ Q2 ∪ Q3 ∪ (∆, i). i∈S\{1,2,3}

Σ We claim that GΓ = 1. First, we have GΓ = 1 since BS = 1. Since KΓ ﬁxes

g1 g2 ∆1 ∆1 ∆1 ∆1 (Q1,Q ,Q ) , so (KΓ) ≤ (KQ ) ∩ (K g1 ) ∩ (K g2 ) = AQ ∩ AQ ∩ AQ = 1 2 3 1 Q2 Q3 1 2 3

Hence KΓ ﬁxes (Q1,Q2,Q3). Noticing that K is faithful on ∆1, we obtain that KΓ = 1 and GΓ = 1.

Lemma 6.18. Let B = (GΣ, n) with n ≥ 3 be of type (α) and A = (G∆1 , |∆|) be of ∆1 type (β) and almost simple. Let T = soc(A) and suppose no subgroup of A is orbit equivalent to A. If H < G and H ≡ G, then L ≥ T n.

Proof. By Lemma 6.17, we have K ≥ T n. Since no subgroup of A is orbit equivalent

∆ n n to A, H∆ = A. By Lemma 5.1, we have L = 1, T ≤ L ≤ A, or T ≤ L ≤ A . Let Sn S Q ⊂ Σ such that BQ = 1. For S ⊆ ∆, let Γ(S) = i∈Q(S, i) ∪ i/∈Q(∆ \ S, i) and d(S) = gcd{|GΓ(S)| : S ⊂ ∆}. By Lemma 6.14, d(S) is a power of 2 except in the case

a b (A, |∆|) = (M12, 12). For the case (A, |∆|) = (M12, 12), d(S) is of the form 2 · 3 . By Lemma 4.4, we have the index |G : H| divides d(S).

On the other hand, by Theorem 4.13 we have |G : H| = |K : L|. If T ≤ L ≤ A or

L = 1, then |T | divides |K : L|. Noticing that T is nonsolvable, we have |T | is not of the form 2a · 3b. This is a contradiction.

Lemma 6.19. Let B = (GΣ, n) with n ≥ 3 be of type (α) and A = (G∆1 , |∆|) ∆1 be of type (β) and almost simple. If there exists a subgroup H < G and H ≡ G,

73 then (A, |∆|) must be one of the following: (PGL(2, 5), 6), (PGL(2, 7), 8), (S6, 10), (PΓL(2, 8), 9), or (PΓL(2, 9), 10).

Proof. Again by Theorem 4.12 and Theorem 4.13, we have K 6= 1 and HΣ = GΣ.

We shall prove (A, |∆|) must be of type II or V in Lemma 6.16.

If A is of type I in Lemma 6.16, then by Lemma 6.17 and 6.18, K = L = soc(A)n and G = H.

If A is of type III or IV, then there exist two subsets R1 and R2 such that AR1 ≤

soc(A) and AR2 ≤ soc(A). Let S1 be a subset of [n] such that BS = 1 and let S = [n]\S , and let Γ = S (R , i)∪S (R , i). Noticing that H ≥ L ≥ soc(A)n 2 1 i∈S1 1 i∈S2 2 n by Lemmas 6.17 and 6.18, we obtain GΓ ≤ soc(A) ≤ H which implies that G and H are not orbit equivalent.

4 4 Lemma 6.20. If (A, |∆|) = (2 .PSL(4, 2), 16), (2 .S6, 16), (ASL(5, 2), 32), there exist two subsets Γ1 and Γ2 such that

1. (AΓ1 soc(A))Γ2 = 1.

2. (AΓ2 soc(A))Γ1 = 1.

If (A, |∆|) = (ASL(3, 2), 8), then there exist three subsets Γ1, Γ2 and Γ3 such that

((AΓ1 soc(A))Γ2 soc(A))Γ3 = 1.

Proof. It is done by GAP computations.

For (A, |∆|) = (24.PSL(4, 2), 16), we take

Γ1 = {1, 3, 5, 7, 8, 9, 11, 12} and

Γ2 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13}.

74 4 For (A, |∆|) = (2 .S6, 16), we take

Γ1 = {1, 3, 5, 7, 8, 9, 11, 12} and

Γ2 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}.

For (A, |∆|) = (ASL(5, 2), 32), we take

Γ1 = {1, 3, 5, 7, 8, 9, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23} and

Γ2 = {1, 22, 14, 15, 16, 25, 26, 27, 30}.

For (A, |∆|) = (ASL(3, 2), 8), we take Γ1 = {1, 2, 3},Γ2 = {3, 5, 7}, and Γ3 = {1, 2, 3, 5, 6}.

4 4 Lemma 6.21. Let (A, |∆|) = (ASL(3, 2), 8), (2 .PSL(4, 2), 16), (2 .S6, 16), (ASL(5, 2), 32) and K 6= 1. If B is of type (α) and G has no regular orbit, then K ≥ (A0)n.

n Proof. Let E = A/soc(A) and let K0 = K ∩ soc(A) . Noticing that E is an almost

n simple group, by Lemma 5.1, we have K/K0 ≤ E or K/K0 ≥ soc(E) .

We claim that if K ≤ K0.E then G has a regular orbit. Let S be a subset of [n] such that BS = 1. We also assume 1 ∈ S and 2 ∈/ S.

4 4 If (A, |∆|) = (2 .PSL(4, 2), 16), (2 .S6, 16) or (ASL(5, 2), 32), by Lemma 6.20 there

exist two subsets Γ1 and Γ2 such that (AΓ1 soc(A))Γ2 = 1 and (AΓ2 soc(A))Γ1 = 1.

0 S 0 S 0 Let Γ = {1} and let Γ = (Γ1, 1) ∪ (Γ2, 2) ∪ ( k∈S,k6=1(Γ , k) ∪ k∈ /S,k6=2(∆ \ Γ , k).

Σ We claim that GΓ = 1. First, since BS = 1, GS = 1 and GΓ ≤ KΓ.

75 ∆2 ∆1 We have KΓ ≤ (AΓ1 soc(A))Γ2 = 1 and KΓ ≤ (AΓ2 soc(A))Γ1 = 1. For i 6= 1, 2,

∆i (KΓ ) ≤ (soc(A))Γ0 = 1. Now, we have KΓ = 1 and GΓ = 1.

If (A, |∆|) = (ASL(3, 2), 8), then for the sets Γ1 = {1, 2, 3},Γ2 = {3, 5, 7} and

Γ3 = {1, 2, 3, 5, 6} we have ((AΓ1 soc(A))Γ2 soc(A))Γ3 = 1. Let Γ0 = {1} and let

[ 0 [ 0 Γ = (Γ1, 1) ∪ (Γ2, 2) ∪ (Γ3, 3) ∪ ( (Γ , k) ∪ ( (∆ \ Γ , k)). k∈S−{1,2,3} k∈[n]−S−{1,2,3}

As in the previous case, we have GΓ = 1. This is a contradiction.

n Hence, we have K/K0 ≥ soc(E) . There exists a = (a1, a2, ..., an) ∈ K such that a1 ∈/ soc(A) and ai ∈ soc(A) for i ≥ 2. Let b = (b1, b2, ..., bn) ∈ K0 such that b1 ∈/ soc(A), [a1, b1] ∈/ soc(A), and bi ∈ soc(A) for i ≥ 2. So c = [a, b] = (c, 1..., 1) where c∈ / soc(A). Then hcg : g ∈ Gi contains (A0, 1, ..., 1). It follows that hcg : g ∈

Gi ≥ (A0)n. Therefore K ≥ (A0)n.

Theorem 6.22. Let (GΣ, n) with n ≥ 3 be of type (α) and (A, |∆|) is of type (β). If there is a subgroup H < G, H ≡ G, then A is either solvable or one of (ASL(3, 2), 8),

(PΓL(2, 8), 9), (PΓL(2, 9), 10), (PGL(2, 5), 6), (PGL(2, 7), 8), or (S6, 10).

Proof. If A is almost simple, then by Lemma 6.19, (A, |∆|) must be one of (PΓL(2, 8), 9),

(PΓL(2, 9), 10), (PGL(2, 5), 6), (PGL(2, 7), 8), or (S6, 10). Next, we shall examine the case when A is neither almost simple nor solvable. Let

S ⊂ [n] such that BS = 1.

4 0 n 0 n In the case (A, |∆|) = (2 .S6, 16), we have K ≥ (A ) . First we show L ≥ (A ) . By Theorem 4.13, |G : H| = |K : L|.

76 n n Let L0 = L ∩ soc(A) and E = A/soc(A). By Lemma 5.1 L/L0 ≥ soc(E) or

L/L0 ≤ E. Sn Again, we shall use the gcd argument. For S ⊆ ∆, let Γ(S) = i∈Q(S, i) ∪ S i/∈Q(∆ \ S, i) and d(S) = gcd{|GΓ(S)| : S ⊂ ∆}. By Lemma 6.14, d(S) is a power of 2. By Lemma 4.4, we have the index |G : H| divides d(S). Noticing K ≥ (A0)n,

n and |K : L| is a power of 2, we have L/L0 ≥ soc(E) . By the argument in the last lemma, we have L ≥ (A0)n.

0 We take R1 = {1, 2, 3, 4, 5, 9, 13} and R2 = ∆ \ R1. We have AR1 ≤ A and

0 AR2 ≤ A . Let [ [ Γ = (R1, i) ∪ (R2, i). i∈S i∈[n]\S 0 n Then by Lemma 6.21, we have GΓ ≤ (A ) ≤ H, which implies G and H are not orbit equivalent.

In the case (A, |∆|) = (24.PSL(4, 2), 16) or (ASL(5, 2), 32), by Lemma 6.21, we have K ≥ (A0)n.

∆ There is no subgroup of A which is orbit equivalent to A. So H∆ = A. Let

n n L0 = L∩soc(A) . Again by lemma 5.1, we have L/L0 is 1, A/soc(A) or (A/soc(A)) . Let [ [ Γ(R) = (R, i) ∪ (∆ \ R, i) i∈S i∈[n]\S a b and d(A) = gcd{|WΓ(R)| : R ⊂ ∆}. By Lemma 6.14 d(A) is of the form 2 · 3 .

If L/L0 = A/soc(A) or 1, then |G : H| = |K : L| divides |A/soc(A)| which is not

77 a b n of the form 2 · 3 . This is a contradiction. Now, we have L/L0 = (A/soc(A)) and L ≥ An which implies K = L and G = H.

4 4 In the case (A, |∆|) = (AΓL(2, 4), 16), (2 .A7, 16) or (2 .A6, 16). Let W = A o B and let [ [ Γ(R) = (R, i) ∪ (∆ \ R, i) i∈S i∈[n]\S and d(A) = gcd{|WΓ(R)| : R ⊂ ∆}. By Lemma 6.14 d(A) = 1 which implies G = H by Lemma 4.4.

6.4 A ∈ (β) and B ∈ (γ)

In this section, we come back to the case when the top group (B, n) is of type (γ) but (A, |∆|) is of type (β).

Lemma 6.23. Let (B, n) be of type (γ) with n ≥ 60. If (A, |∆|) is one of (AΓL(2, 4), 16),

4 4 (2 .A7, 16), (2 .A6, 16) and (B, n) is of type (γ), then |G : H| = 2.

Σ ∼ Proof. We ﬁrst observe H is either An or Sn, so H has a clean subgroup R = An by

S g Theorem 6.10. For any subset S ⊆ ∆, we deﬁne Γ(S) = {S : g ∈ R}. So R ≤ HΓ(S) and R ≤ GΓ(S). Then |GΓ(S) : HΓ(S)| = |KΓ(S) : LΓ(S)| or 2|KΓ(S) : LΓ(S)|. By Lemma

6.14, we have gcd{|KΓ(S) : LΓ(S)| : S ⊂ ∆} = 1. This implies |G : H| = 2. The following lemma is similar.

Lemma 6.24. Let (B, n) be of type (γ) with n ≥ 60 and let (A, |∆|) be of type (β). If there exists no subgroup of A which is orbit equivalent to A, then |G : H| is a power of 2.

78 Proof. Noticing |∆| ≤ 32 and n ≥ 60, by Theorem 6.10 H has a clean subgroup ∼ R = An.

S g For any subset S ⊂ ∆, let Γ(S) = {S : g ∈ R}. Clearly HΓ(S) ≥ R. We have

|GΓ(S) : HΓ(S)| = |KΓ(S) : LΓ(S)| or 2|KΓ(S) : LΓ(S)|.

If (A, |∆|) = (M12, 12), then the gcd of the size of the setwise stabilizers of subsets

a b of A is 24. Hence gcd{|KΓ(S) : LΓ(S)| : S ⊂ ∆} = 2 · 3 for some integer a, b. On the other hand, |G : H| = |K : L| or 2|K : L|. Considering that A is simple in this case, by Lemma 5.1 we have K = An or A, and L is 1, A, or An. In either case we have

K = L. Hence |G : H| = 2.

a If (A, |∆|) 6= (M12, 12), then by Lemma 6.14, gcd{|KΓ(S) : LΓ(S)| : S ⊂ ∆} = 2 for some integer a.

Lemma 6.25. Let (GΣ, n) with n ≥ 60 be of type (γ) and (A, |∆|) be one of (24.PSL(2, 4), 16),

4 (ASL(5, 2), 32), or (2 .S6, 16). Suppose there is a subgroup H < G with H ≡ G. Let

n K0 = K ∩ soc(A) and let E = A/soc(A). If K/K0 ≤ E, then |G : H| = 2.

Proof. Noticing 16 ≤ |∆| ≤ 32 and n ≥ 60, by Theorem 6.10 H has a clean subgroup ∼ R = An.

By Lemma 6.20, we can take two subsets Γ1 and Γ2 of ∆ such that (AΓ1 soc(A))Γ2 =

1 and (AΓ2 soc(A))Γ1 = 1.

Let O be an orbit of R on Ω and let Qi = O ∩ ∆i. Sn Let Γ = (Γ1, 1) ∪ (Γ2, 2) ∪ ( i=3 Qi).

∆1 ∆2 If K/K0 ≤ E, then (KΓ) ≤ (AΓ2 soc(A))Γ1 = 1 and (KΓ) ≤ (AΓ1 soc(A))Γ2 =

1. Hence KΓ = 1.

79 On the other hand, GΓ ≥ HΓ ≥ R∆1∪∆2 ≥ An−2. Noticing |G : H| = |GΓ : HΓ|, we obtain GΓ = Sn−2 and HΓ = An−2 and |G : H| = 2.

Theorem 6.26. Let (GΣ, n) be of type (γ) and (A, |∆|) of type (β) with n ≥ 60. If there is a subgroup of H < G with H ≡ G, then one of the following holds.

1. A is solvable.

2. (A, |∆|) is one of (ASL(3, 2), 8), (PΓL(2, 8), 9), or (PΓL(2, 9), 10).

3. |G : H| = 2.

Σ ∼ Proof. Again, we have H = An or Sn and H has a clean subgroup R = An. In the case when A is simple, there is no subgroup of A which is orbit equivalent

∆ n to A. By Lemma 4.2, we have H∆ = A. Then by Lemma 5.1, K, L are A , A or 1. Noticing |G : H| = |K : L| or 2|K : L| is of the form 2a · 3b, we obtain K = L. It follows that |G : H| = 2.

In the case when (A, |∆|) is one of (24.PSL(2, 4), 16) ,(ASL(5, 2), 32), A/soc(A) is

n n simple. Let E = A/soc(A) and K0 = K ∩ soc(A) . By Lemma 5.1, K/K0 ≥ soc(E) or K/K0 ≤ E. If K/K0 ≤ E, then by Lemma 6.25 we have |G : H| = 2. If

n n K/K0 ≥ soc(E) = E , then since |K : L| is a power of 2, we obtain K = L and |G : H| = 2.

4 0 In the case (A, |∆|) is (2 .S6, 16), we take S ⊂ ∆ such that AS ≤ A . Let

S g n Γ(S) = {S : g ∈ R}. Then HΓ(S) ≥ An. Let E = A/soc(A), K0 = K ∩ soc(A)

n n and L0 = L∩soc(A) . By Lemma 5.1, K/K0 ≥ soc(E) or K/K0 ≤ E. If K/K0 ≤ E,

n by Lemma 6.25, we have |G : H| = 2. If K/K0 ≥ soc(E) then, since |K : L| is a

n power of 2 and L ≤ K, we obtain L/L0 ≥ soc(E) . Using the same argument as the

80 0 n proof of Lemma 6.21, we obtain K ≥ L ≥ (A ) . Hence KΓ(S) ≤ L and KΓ(S) = LΓ(S).

Since |G : H| = |KΓ(S) : LΓ(S)| or 2|KΓ(S) : LΓ(S)|, we obtain |G : H| = 2.

4 4 In the case when A is one of (AΓL(2, 4), 16), (2 .A7, 16), (2 .A6, 16), by Lemma 6.23 we have |G : H| = 2.

In the remaining cases, A is not one of (ASL(3, 2), 8), (PΓL(2, 8), 9), or (PΓL(2, 9), 10) and A is almost simple. Noticing there is no subgroup of A orbit equivalent to A, we

∆ obtain H∆ = A. Since |G : H| = |K : L| or 2|K : L| is a power of 2 by Lemma 6.24, then we have

0 n 0 0 either K ≥ L ≥ (A ) or A ≤ L ≤ K ≤ A. There exists S ⊂ ∆ such that AS ≤ A .

S g Let Γ(S) = {S : g ∈ R} and HΓ(S) ≥ R. So we have KΓ(S) = LΓ(S) which implies |G : H| = 2.

If (A, |∆|) is one of (ASL(3, 2), 8), (PΓL(2, 8), 9) and (PΓL(2, 9), 10), then there is a subgroup A0 ≤ A such that A0 ≡ A. Then by Lemma 4.15 we have H = A0 o Sn ≡

G = A o Sn and the index |G : H| is not 2.

81 CHAPTER 7

ORBIT EQUIVALENT PERMUTATION GROUPS UP TO

DEGREE 15

We list all orbit equivalent permutation groups up to degree 15. The numbers in the brackets are the numbers of the groups in the GAP library.

n=6: [ 5, 9, 10, 13 ], [ 8, 11 ], [ 14, 15, 16 ]

n=8: [ 13, 24 ], [ 16, 27 ], [ 19, 29 ], [ 21, 31 ], [ 25, 36, 48 ], [ 26, 28,

30, 35 ], [ 32, 39 ], [ 33, 41 ], [ 38, 40, 44 ], [ 42, 45, 46, 47 ], [ 49, 50 ]

n=9: [ 11, 18 ], [ 15, 19 ], [ 17, 22, 25, 28 ], [ 20, 21, 24, 29, 30, 31 ], [

23, 26 ], [ 27, 32, 33, 34 ]

n=10: [ 6, 9, 10, 21 ], [ 11, 12, 22 ], [ 16, 23 ], [ 17, 19, 20, 27 ], [ 18, 28

], [ 25, 29 ], [ 30, 35 ], [ 33, 40, 41, 42, 43 ], [ 34, 37 ], [ 36, 38, 39 ], [ 44,

45 ]

n=12: [ 93, 188 ], [ 94, 141, 189, 222 ], [ 105, 134 ], [ 107, 135 ], [ 120,

156 ], [ 130, 168, 171, 172, 174, 210, 214, 242, 246, 261 ], [ 131, 170, 173,

211, 215, 244, 245, 264 ], [ 145, 193 ], [ 149, 192, 224 ], [ 150, 151, 185, 221,

223, 225, 250 ], [ 152, 153 ], [ 154, 155 ], [ 167, 169, 209, 217, 243, 247,

248, 249, 262, 263, 266, 267, 274 ], [ 178, 213 ], [ 180, 183, 219 ], [ 190, 227

], [ 194, 232, 234, 271, 280 ], [ 196, 197, 240 ], [ 198, 235, 237, 238, 241 ], [

82 199, 202, 236 ], [ 200, 260 ], [ 212, 216 ], [ 228, 265, 273, 284, 292 ], [ 231,

233, 258, 259, 281, 282, 289 ], [ 254, 275, 276, 283, 290, 291, 294 ] , [ 256,

270 ], [ 277, 285 ], [ 278, 288, 296, 297, 298, 299 ], [ 286, 287, 293 ], [ 300,

301 ].

n=14: [ 27, 38 ], [ 40, 48 ], [ 46, 47, 49 ], [ 53, 55 ], [ 54, 56, 57 ], [ 58,

59, 60, 61 ], [ 62, 63 ]

n=15: [ 25, 30, 39, 50 ], [ 31, 32, 40, 48, 51, 60 ], [ 35, 43 ], [ 36, 44, 71,

81 ], [ 37, 58 ], [ 38, 59 ], [ 45, 46, 55, 79, 80, 86 ], [ 49, 68 ], [ 53, 63 ], [

54, 56, 64, 84, 85, 87 ], [ 57, 67 ], [ 61, 62, 70 ], [ 65, 66, 74 ], [ 69, 76, 77,

78, 83, 88, 89, 90, 91, 93 ], [ 73, 82, 94, 96, 97, 99, 100, 102 ], [ 75, 92, 95,

98, 101 ], [ 103, 104 ]

83 CHAPTER 8

GAP CODE

In this chapter,we list the GAP code we have used.

1. The following code was used to generate all orbit equivalent permutation groups of degree d.

result:=[];

order:=[];

n:=NrTransitiveGroups(d);

for i in [1..(n-1)] do

small:=[i];

s:=TransitiveGroup(d,i);

for j in [(i+1)..n] do

p:=1; for k in [1..Size(result)] do

for m in [1..Size(result[k])] do

if j=result[k][m] then p:=0; ﬁ;

od;

if p=1 then

t:=TransitiveGroup(d,j);

84 if compare(s,t,d)=1 then

Add(small,j);

ﬁ; ﬁ;

od;

if Size(small)>1 then

result:=Concatenation(result,[small]);

ﬁ;

od;

2. The following code was used to compute the greatest common divisor of the order of the stabilizers of a permutation group.

gcd:=function(g,d)

local Nr,O,s,result,i;

O:=OrbitsDomain(g,Combinations([1..d]),OnSets);

Nr:=Size(O);

result:=Size(Stabilizer(g,O[1][1],OnSets));

for i in [2..Nr] do

s:=Size(Stabilizer(g,O[i][1],OnSets));

result:=Gcd(result,s);

od;

return result;

end;

3. The following code was used to determine if two groups are orbit equivalent.

compare:=function(g,h,d)

85 local c,Og,Nr,a,i,f,r;

r:=Int(d/2);

c:=1;

Og:=OrbitsDomain(g,Combinations([1..d],r),OnSets);

Nr:=Size(Og);

f:=ClosureGroup(h,GeneratorsOfGroup(g));

for i in [1..Nr] do

a:=Og[i][1];

if OrbitLength(g,a,OnSets)

return c;

ﬁ;

if OrbitLength(h,a,OnSets)

return c;

ﬁ;

od;

return c;

end;

4. The following code was used to determine if a group has a regular orbit on the power set.

regular:=function(g,d)

local a;

for a in Combinations([1..d]) do

if Size(Stabilizer(g,a,OnSets))=1 then return 1;

86 ﬁ; od;

return 0;

end;

5. The following code was used to compute the primitive groups with no more than four regular orbits up to degree 17.

result:=[];

i:=1;

for n in [1..17] do

N:=NrPrimitiveGroups(n);

for r in [1..N] do

g:=PrimitiveGroup(n,r);

if (Size(g)-Factorial(n)/2)<0 then

a:=OrbitLengthsDomain(g,Combinations([1..n]),OnSets);

j:=0;

for i in [1..Size(a)] do

if a[i]=Size(g) then j:=j+1; ﬁ;

od;

if j<4 then

if j>0 then Add(result,[n,r,j]); ﬁ;

ﬁ;

od;

87 6. The following code was used to determine all primitive groups which satisfy

Formula 3.1 up to degree 180. In this code, for each conjugacy class of a group, we take an element from it. Then we count the number of subsets ﬁxed by this element.

result:=[];

for n in [18..180] do

N:=NrPrimitiveGroups(n);

for r in [1..N] do

g:=PrimitiveGroup(n,r);

if (Size(g)-Factorial(n)/2)<0 then

c:=ConjugacyClasses(g);

a:=0;

for i in [1..Size(c)] do

h:=Group(Representative(c[i]));

Cys:=Size(Orbits(h,[1..n]));

a:=a+(2 ˆCys)*Size(c[i]);

od;

b:=2*(2 ˆn-2-n);

c:=a-b;

if c >0 then Add(result,[n,r]); ﬁ;

ﬁ;

od;

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