# A Character Condition for Quadruple Transitivity

Authors: S. Aldhafeeri; R. T. Curtis
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

LetG be a permutation group of degree n viewed as a subgroup of the symmetric
group S≅Sn. We show that if the irreducible character of S corresponding to the partition of n into subsets of sizes n−2 and 2, that is, to say the character often
denoted by χ(n−2,2), remains irreducible when restricted to G, then n = 4, 5 or 9 and G≅S3, A5, or PΣL2(8), respectively, or G is 4-transitive.


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## Body

## 1. Introduction

LetG be a permutation group of degree n. The connection between the multiple transitivity of G and the irreducibility of certain irreducible characters of of the symmetric group Sn when restricted to G goes back to Frobenius, and a proof of his classical result can be found in Tsuzuku [1]. In [2], Saxl strengthens this result, limiting the irreducibility hypothesis to the characters χG(n-k,k), a special case of his theorem having been proved by Neumann [3]. In [4] Saxl uses the Classification of Finite Simple Groups to go much further. He obtains a complete list of pairs (χ,G) where χ is an irreducible character of the symmetric group Sn, and G is a subgroup of Sn such that the restriction of χ to G remains irreducible.The present paper is concerned with the case whenχG(n-2,2) is irreducible, proving that G is 4-transitive except in three small cases which are worked out explicitly. As such, the result, which appears as Theorem 3.3, can be deduced from [2, 3, 5]. However, the approach used here is self-contained and elementary.


## 2. Notation and Preliminary Results

### 2.1. Permutation Characters

We start by recalling two fundamental results from the elementary permutation character theory. If a groupG permutes a set X={1,2,…,n} then, we let B={e1,e2,…,en} be a basis of an n-dimensional complex vector space, which becomes a ℂG-module when we define eiν=eν(i), for ν∈G, and extend linearly. If πX denotes the character afforded by this representation and the permutation ν has cycle shape 1r.2s… when written as a product of disjoint cycles, then (2.1)πX(ν)=r=|FixX(ν)|,
where FixX(ν) denotes the set of points of X fixed by the permutation ν. The subscript X will be omitted when there is no fear of confusion. The standard inner product defined on the space of class functions when restricted to (generalized) characters becomes(2.2)〈χ,ψ〉=1|G|∑g∈Gχ(g)ψ(g)¯,
where χ and ψ are characters of the group G, and α¯ denotes the complex conjugate of the complex number α. The trivial character of G which takes the value 1 on every element of G will be denoted by 1G. Then we have the following result; it is often referred to as Burnside's Lemma but is actually due to Cauchy and Frobenius.Theorem 2.1.
IfG is a permutation group acting on the set X with permutation character πX, then we have the following result:
(2.3)〈πX,1G〉=orb(G,X),
where orb(G,X) denotes the number of orbits when G acts on X.Proof.
Firstly note that(2.4)|{(a,ν)∣a∈Xν∈G,aν=a}|=∑a∈X|StabG(a)|=∑ν∈G|Fix⁡(ν)|,
where StabG(a) denotes the stabilizer in G of a. So, if G has orbits O1,O2,…,Om when it acts on X, the Orbit-Stabilizer theorem implies that
(2.5)∑a∈X|StabG(a)|=∑i=1m∑ai∈Oi|StabG(ai)|=∑i=1m|Oi|⋅|StabG(bi)|=m|G|,
where bi is a representative of the orbit Oi. Thus,
(2.6)〈πX,1G〉=1|G|∑ν∈G|Fix⁡(ν)|⋅1=1|G||G|m=m.If the groupG acts on the set X and on the set Y with permutations characters πX and πY, respectively, then G also acts on the Cartesian product X×Y by defining (2.7)(x,y)ν=(xν,yν).An ordered pair is only fixed byν if both, its components are fixed, and so we see that |FixX×Y(ν)|=|FixX(ν)|·|FixY(ν)|. Then we have,Theorem 2.2.
If the groupG acts on the sets X and Y with permutation characters πX and πY respectively, then
(2.8)〈πX,πY〉=orb(G,X×Y).Proof.
(2.9)〈πX,πY〉=1|G|∑ν∈G|FixX(ν)|⋅|FixY(ν)|¯=1|G|∑ν∈G|FixX(ν)|.|FixY(ν)|=1|G|∑ν∈G|FixX×Y(ν)|=orb(G,X×Y).Corollary 2.3.
IfG acts transitively on X and a∈X, then 〈πX,πY〉=orb(Ga,Y), the number of orbits of the stabilizer of a point in X on Y.Corollary 2.4.
IfG acts transitively on X and a∈X, then 〈πX,πX〉=orb(Ga,X), the  rank  of the permutation action of G on X.In the case whenS≅Sn, the symmetric group acting on an n-element set Ω={1,2,…,n}, we denote the associated permutation character by π1. More generally πk will denote the permutation action of S≅Sn on unordered k-element subsets. In particular, π2 will denote the permutation character of the action of S on Δ={{a,b}∣a,b∈Ω}, the set of unordered pairs of elements of Ω which we shall call duads. Thus, if ν∈S has cycle shape 1r.2s…, then (2.10)π1(ν)=r,π2(ν)=(r2)+s.By Theorem2.1 and its corollaries we see that (2.11)〈π1,1S〉=〈π2,1S〉=1,〈π1,π1〉=2,〈π1,π2〉=2,〈π2,π2〉=3(forn>3),since the stabilizer of a point clearly has two orbits on duads, and the stabilizer of a duad has three orbits on duads, namely, the orbit containing just the fixed duad,δ say, the orbit containing duads which intersect δ in one point, and the orbit of duads disjoint from δ. We thus have that(2.12)π1=1G+χ,π2=1G+χ+ψ,
where χ and ψ are distinct irreducible characters of S of degrees n-1 and n(n-3)/2, respectively.


### 2.2. The Irreducible Characters of the Symmetric GroupSn

Definition 2.5.
Apartitionλ of the integer n is a sequence λ=(λ1,λ2,…,λk) where λi≥λi+1 for i=1,…,k-1 and ∑i=1kλi=n.The irreducible characters ofSn are in one-to-one correspondence with the partitions of n and we denote the character corresponding to λ by χλ. In this notation, the trivial character of Sn corresponds to the partition of n into a single subset of size n and is thus denoted by 1S=χ(n). The permutation character of degree n is given by π1=χ(n)+χ(n-1,1). Indeed we have the general result.Theorem 2.6.
Ifk≤n/2 and πk denotes the permutation character of the action of Sn on unordered k-subsets, then
(2.13)πk=χ(n)+χ(n-1,1)+⋯+χ(n-k,k).That this permutation character is a sum of irreducibles of the right degree is readily shown and a proof may be seen in James and Liebeck [6, page 344]. That these irreducibles are indeed those corresponding to the above partitions may be verified using the celebrated Murnaghan-Nakayama rule, see [7, page 79], which enables one to evaluate the character value of the irreducible χλ on any conjugacy class of the symmetric group. We see that for k≤n/2 we have (2.14)χ(n-k,k)=πk-πk-1,and so, ifσ∈Sn, then(2.15)χ(n-k,k)(σ)=|{k-subsets fixed byσ}|-|{(k-1)-subsets fixed byσ}|.
In particular, the degree of χ(n-k,k) is given by (2.16)χ(n-k,k)(1)=(nk)-(nk-1).Applying (2.15) to a permutation σ with cycle shape 1r.2s…, we have(2.17)χ(n-2,2)(σ)=(r2)+s-r=r(r-3)2+s.


### 2.3. Multiple Transitivity

Definition 2.7.
The groupG is said to act k-transitively on the set X if, and only if, given (x1,x2,…,xk) and (y1,y2,…,xk), two ordered k-tuples of distinct points of X, there exists a permutation σ∈G such that xiσ=yi for i=1⋯k.Definition 2.8.
The groupG is said to act k-homogeneously on the set X if, and only if, given A,B⊂X with |A|=|B|=k, there exists a permutation σ∈G such that Aσ=B.In other words,G is k-homogeneous if it acts transitively on the k-subsets. Those fascinating groups which are k-homogeneous without being k-transitive have been investigated by Kantor, see [8, 9], and the Livingstone-Wagner theorem of [10] is a highly significant result in this area. In this paper, we shall be concerned with the cases k=2,3,and4 and thus with the action of subgroups of Sn on unordered pairs of points, which we shall refer to as duads, and on triples and tetrads. We can immediately see that the following applies.Lemma 2.9.
A group which is 2-homogeneous without being 2-transitive must have odd order.Proof.
Suppose thatG has even order and acts 2-homogeneously on X. Let α={a1,a2} and β={b1,b2} be two duads; we must produce a permutation which maps ai to bi for i=1 and 2. By 2-homogeneity, there exists a σ∈G such that ασ=β, and since G is even and again using the 2-homogeneity, there exists an element τ∈G of order 2 which acts as τ=(b1b2)⋯. If a1σ=b1, then we are done, otherwise στ will suffice.


### 2.4. Restriction to a Subgroup ofSn

In what follows, the subgroupG will be a fixed subgroup of S≅Sn, and we shall denote the restriction of the character χ of S to G by (2.18)χ|G.
In order to simplify the notation, we shall denote (2.19)π1|Gbyψ1,π2|Gbyψ2,χ(n-1,1)|Gbyψ(n-1,1),andχ(n-2,2)|Gbyψ(n-2,2).
So we shall have (2.20)ψ1=1G+ψ(n-1,1),ψ2=1G+ψ(n-1,1)+ψ(n-2,2).
The restriction to G of an irreducible character of S may or may not remain irreducible. However, if G≤S≅Sn, then using Theorems 2.1 and 2.2 we have  (1)
〈ψ1,1G〉 is the number of orbits of G on points,(2)
〈ψ2,1G〉 is the number of orbits of G on duads,(3)
〈ψ1,ψ1〉 is the number of orbits of G on Ω×Ω,(4)
〈ψ1,ψ2〉 is the number of orbits of G on Ω×Δ, and(5)
〈ψ2,ψ2〉 is the number of orbits of G on Δ×Δ.Note that ifχ and ψ are as in (2.12), then χ|G=ψ(n-1,1) and ψ|G=ψ(n-2,2). We shall use their correct labels in what follows but shall not make use of the correspondence with partitions in our argument.


### 2.5. General Preliminary Results

A result which will be appealed to several times in what follows is the following.Lemma 2.10.
The only permutation which commutes with a transitive permutation group and which fixes a point is the identity. In particular, a transitive, abelian permutation group acts regularly.Proof.
LetK be a transitive subgroup of the symmetric group Sn acting on the set Ω, and let the permutation σ∈Sn commute with K and fix a point a∈Ω. Then F= Fix(σ)⊂Ω is invariant under the action of K and is thus the empty set ϕ or Ω. But a∈F≠ϕ, and so F=Ω and σ=1.
IfK is abelian and σ∈Ka, the stabilizer in K of a, then σ commutes with K and fixes the point a. Thus, σ=1, from the above, and so Ka=〈1〉, and the transitive group K acts regularly.A well-known result about linear groups has been taken from Theorem 7.3 of Huppert [11, page 187].Theorem 2.11.
The general linear groupGLm(pf) contains a cyclic subgroup K of order pmf-1. We have CG(K)=K and that NG(K)/K is cyclic of order m.  (Such cycles of maximal possible length are referred to as Singer cycles.)


## 2.1. Permutation Characters

We start by recalling two fundamental results from the elementary permutation character theory. If a groupG permutes a set X={1,2,…,n} then, we let B={e1,e2,…,en} be a basis of an n-dimensional complex vector space, which becomes a ℂG-module when we define eiν=eν(i), for ν∈G, and extend linearly. If πX denotes the character afforded by this representation and the permutation ν has cycle shape 1r.2s… when written as a product of disjoint cycles, then (2.1)πX(ν)=r=|FixX(ν)|,
where FixX(ν) denotes the set of points of X fixed by the permutation ν. The subscript X will be omitted when there is no fear of confusion. The standard inner product defined on the space of class functions when restricted to (generalized) characters becomes(2.2)〈χ,ψ〉=1|G|∑g∈Gχ(g)ψ(g)¯,
where χ and ψ are characters of the group G, and α¯ denotes the complex conjugate of the complex number α. The trivial character of G which takes the value 1 on every element of G will be denoted by 1G. Then we have the following result; it is often referred to as Burnside's Lemma but is actually due to Cauchy and Frobenius.Theorem 2.1.
IfG is a permutation group acting on the set X with permutation character πX, then we have the following result:
(2.3)〈πX,1G〉=orb(G,X),
where orb(G,X) denotes the number of orbits when G acts on X.Proof.
Firstly note that(2.4)|{(a,ν)∣a∈Xν∈G,aν=a}|=∑a∈X|StabG(a)|=∑ν∈G|Fix⁡(ν)|,
where StabG(a) denotes the stabilizer in G of a. So, if G has orbits O1,O2,…,Om when it acts on X, the Orbit-Stabilizer theorem implies that
(2.5)∑a∈X|StabG(a)|=∑i=1m∑ai∈Oi|StabG(ai)|=∑i=1m|Oi|⋅|StabG(bi)|=m|G|,
where bi is a representative of the orbit Oi. Thus,
(2.6)〈πX,1G〉=1|G|∑ν∈G|Fix⁡(ν)|⋅1=1|G||G|m=m.If the groupG acts on the set X and on the set Y with permutations characters πX and πY, respectively, then G also acts on the Cartesian product X×Y by defining (2.7)(x,y)ν=(xν,yν).An ordered pair is only fixed byν if both, its components are fixed, and so we see that |FixX×Y(ν)|=|FixX(ν)|·|FixY(ν)|. Then we have,Theorem 2.2.
If the groupG acts on the sets X and Y with permutation characters πX and πY respectively, then
(2.8)〈πX,πY〉=orb(G,X×Y).Proof.
(2.9)〈πX,πY〉=1|G|∑ν∈G|FixX(ν)|⋅|FixY(ν)|¯=1|G|∑ν∈G|FixX(ν)|.|FixY(ν)|=1|G|∑ν∈G|FixX×Y(ν)|=orb(G,X×Y).Corollary 2.3.
IfG acts transitively on X and a∈X, then 〈πX,πY〉=orb(Ga,Y), the number of orbits of the stabilizer of a point in X on Y.Corollary 2.4.
IfG acts transitively on X and a∈X, then 〈πX,πX〉=orb(Ga,X), the  rank  of the permutation action of G on X.In the case whenS≅Sn, the symmetric group acting on an n-element set Ω={1,2,…,n}, we denote the associated permutation character by π1. More generally πk will denote the permutation action of S≅Sn on unordered k-element subsets. In particular, π2 will denote the permutation character of the action of S on Δ={{a,b}∣a,b∈Ω}, the set of unordered pairs of elements of Ω which we shall call duads. Thus, if ν∈S has cycle shape 1r.2s…, then (2.10)π1(ν)=r,π2(ν)=(r2)+s.By Theorem2.1 and its corollaries we see that (2.11)〈π1,1S〉=〈π2,1S〉=1,〈π1,π1〉=2,〈π1,π2〉=2,〈π2,π2〉=3(forn>3),since the stabilizer of a point clearly has two orbits on duads, and the stabilizer of a duad has three orbits on duads, namely, the orbit containing just the fixed duad,δ say, the orbit containing duads which intersect δ in one point, and the orbit of duads disjoint from δ. We thus have that(2.12)π1=1G+χ,π2=1G+χ+ψ,
where χ and ψ are distinct irreducible characters of S of degrees n-1 and n(n-3)/2, respectively.


## 2.2. The Irreducible Characters of the Symmetric GroupSn

Definition 2.5.
Apartitionλ of the integer n is a sequence λ=(λ1,λ2,…,λk) where λi≥λi+1 for i=1,…,k-1 and ∑i=1kλi=n.The irreducible characters ofSn are in one-to-one correspondence with the partitions of n and we denote the character corresponding to λ by χλ. In this notation, the trivial character of Sn corresponds to the partition of n into a single subset of size n and is thus denoted by 1S=χ(n). The permutation character of degree n is given by π1=χ(n)+χ(n-1,1). Indeed we have the general result.Theorem 2.6.
Ifk≤n/2 and πk denotes the permutation character of the action of Sn on unordered k-subsets, then
(2.13)πk=χ(n)+χ(n-1,1)+⋯+χ(n-k,k).That this permutation character is a sum of irreducibles of the right degree is readily shown and a proof may be seen in James and Liebeck [6, page 344]. That these irreducibles are indeed those corresponding to the above partitions may be verified using the celebrated Murnaghan-Nakayama rule, see [7, page 79], which enables one to evaluate the character value of the irreducible χλ on any conjugacy class of the symmetric group. We see that for k≤n/2 we have (2.14)χ(n-k,k)=πk-πk-1,and so, ifσ∈Sn, then(2.15)χ(n-k,k)(σ)=|{k-subsets fixed byσ}|-|{(k-1)-subsets fixed byσ}|.
In particular, the degree of χ(n-k,k) is given by (2.16)χ(n-k,k)(1)=(nk)-(nk-1).Applying (2.15) to a permutation σ with cycle shape 1r.2s…, we have(2.17)χ(n-2,2)(σ)=(r2)+s-r=r(r-3)2+s.


## 2.3. Multiple Transitivity

Definition 2.7.
The groupG is said to act k-transitively on the set X if, and only if, given (x1,x2,…,xk) and (y1,y2,…,xk), two ordered k-tuples of distinct points of X, there exists a permutation σ∈G such that xiσ=yi for i=1⋯k.Definition 2.8.
The groupG is said to act k-homogeneously on the set X if, and only if, given A,B⊂X with |A|=|B|=k, there exists a permutation σ∈G such that Aσ=B.In other words,G is k-homogeneous if it acts transitively on the k-subsets. Those fascinating groups which are k-homogeneous without being k-transitive have been investigated by Kantor, see [8, 9], and the Livingstone-Wagner theorem of [10] is a highly significant result in this area. In this paper, we shall be concerned with the cases k=2,3,and4 and thus with the action of subgroups of Sn on unordered pairs of points, which we shall refer to as duads, and on triples and tetrads. We can immediately see that the following applies.Lemma 2.9.
A group which is 2-homogeneous without being 2-transitive must have odd order.Proof.
Suppose thatG has even order and acts 2-homogeneously on X. Let α={a1,a2} and β={b1,b2} be two duads; we must produce a permutation which maps ai to bi for i=1 and 2. By 2-homogeneity, there exists a σ∈G such that ασ=β, and since G is even and again using the 2-homogeneity, there exists an element τ∈G of order 2 which acts as τ=(b1b2)⋯. If a1σ=b1, then we are done, otherwise στ will suffice.


## 2.4. Restriction to a Subgroup ofSn

In what follows, the subgroupG will be a fixed subgroup of S≅Sn, and we shall denote the restriction of the character χ of S to G by (2.18)χ|G.
In order to simplify the notation, we shall denote (2.19)π1|Gbyψ1,π2|Gbyψ2,χ(n-1,1)|Gbyψ(n-1,1),andχ(n-2,2)|Gbyψ(n-2,2).
So we shall have (2.20)ψ1=1G+ψ(n-1,1),ψ2=1G+ψ(n-1,1)+ψ(n-2,2).
The restriction to G of an irreducible character of S may or may not remain irreducible. However, if G≤S≅Sn, then using Theorems 2.1 and 2.2 we have  (1)
〈ψ1,1G〉 is the number of orbits of G on points,(2)
〈ψ2,1G〉 is the number of orbits of G on duads,(3)
〈ψ1,ψ1〉 is the number of orbits of G on Ω×Ω,(4)
〈ψ1,ψ2〉 is the number of orbits of G on Ω×Δ, and(5)
〈ψ2,ψ2〉 is the number of orbits of G on Δ×Δ.Note that ifχ and ψ are as in (2.12), then χ|G=ψ(n-1,1) and ψ|G=ψ(n-2,2). We shall use their correct labels in what follows but shall not make use of the correspondence with partitions in our argument.


## 2.5. General Preliminary Results

A result which will be appealed to several times in what follows is the following.Lemma 2.10.
The only permutation which commutes with a transitive permutation group and which fixes a point is the identity. In particular, a transitive, abelian permutation group acts regularly.Proof.
LetK be a transitive subgroup of the symmetric group Sn acting on the set Ω, and let the permutation σ∈Sn commute with K and fix a point a∈Ω. Then F= Fix(σ)⊂Ω is invariant under the action of K and is thus the empty set ϕ or Ω. But a∈F≠ϕ, and so F=Ω and σ=1.
IfK is abelian and σ∈Ka, the stabilizer in K of a, then σ commutes with K and fixes the point a. Thus, σ=1, from the above, and so Ka=〈1〉, and the transitive group K acts regularly.A well-known result about linear groups has been taken from Theorem 7.3 of Huppert [11, page 187].Theorem 2.11.
The general linear groupGLm(pf) contains a cyclic subgroup K of order pmf-1. We have CG(K)=K and that NG(K)/K is cyclic of order m.  (Such cycles of maximal possible length are referred to as Singer cycles.)


## 3. Statement and Proof of Main Theorem

LetG≤Sn. The transitivity of G is related to the reducibility of certain irreducible characters of S upon restriction to G as follows.Lemma 3.1.
G is doubly transitive⇔ψ(n-1,1)∈Irr(G).Proof.
Follows immediately from Corollary2.4.Lemma 3.2.
G is 4-transitive,n≥4⇒ψ(n-1,1),ψ(n-2,2)∈Irr(G).Proof.
IfG is 4-transitive, then it is certainly doubly transitive, and so ψ(n-1,1)∈Irr(G). Moreover, if δ is a duad, then Gδ certainly has just three orbits on duads, as described towards the end of Section 2.1. Thus,
(3.1)3=〈ψ2,ψ2〉=〈1G+ψ(n-1,1)+ψ(n-2,2),1G+ψ(n-1,1)+ψ(n-2,2)〉≥2+〈ψ(n-2,2),ψ(n-2,2)〉,
and so 〈ψ(n-2,2),ψ(n-2,2)〉=1 and ψ(n-2,2)∈Irr(G).It is thus natural to ask in what circumstances we can haveψ(n-2,2)∈Irr(G) without G being 4-transitive. Our main result is the following theorem which appeals to the Odd-Order theorem of Feit and Thompson and uses part of an argument due to Kantor [8] but is otherwise self-contained and based on Aldhafeeri [12].Theorem 3.3.
LetG be a subgroup of the symmetric group of degree n with n≥4, thus G≤S≅Sn, and let χ(n-2,2) represent the irreducible character of S corresponding to the partition of n into a pair and a subset of cardinality n-2. Then
(3.2)χ(n-2,2)|G∈Irr(G)
(i.e., the restriction of χ(n-2,2) to G is an irreducible representation of G) if, and only if,  (i)
n=4 and G≅S3,(ii)
n=5 and G≅A5,(iii)
n=9 and G≅PΣL2(8), or(iv)
G is 4-transitive on n letters.Proof.
Note that the characterχ(n-2,2) of Sn only exists for n≥4, and so the restriction on n is only required to make the theorem meaningful. In case (i), the subgroup G is not transitive, but no restriction was placed on the transitivity of G.We shall assume thatG is not 4-transitive and that ψ(n-2,2)∈Irr(G). We have ψ(n-2,2)(1)=n(n-3)/2≥2 for n≥4. Thus,
(3.3)〈ψ1,1G〉=〈ψ2,1G〉=1+〈ψ(n-1,1),1G〉,
and so orb(G,Ω)=orb(G,Δ)=k, say. But if orb(G,Ω)=k then orb(G,Δ)≥(k2), as we may choose a duad from each of two orbits in (k2) ways. Now (k2)>k for k>3, and if k=3, at least one of the three orbits must have more than one point in it (since n≥4), and so orb(G,Δ)≥(32)+1=4. Thus, k≤2. If k=2 and both orbits are of cardinality greater than 1 then orb(G,Δ)≥3, a contradiction, so we must have one orbit of length 1 and the other of length n-1.Nowψ(n-2,2)(1)=n(n-3)/2>n-1=ψ(n-1,1)(1) provided n2-5n+2>0, which holds for n≥5. So provided n>4, we have 〈ψ(n-2,2),ψ(n-1,1)〉=0. Thus,
(3.4)orb(G,Ω×Ω)=〈ψ1,ψ1〉=〈ψ1,ψ2〉=orb(G,Ω×Δ)=m,say.
For this reason, we treat n=4 separately and consider the circumstances in which this 2-dimensional representation of S4 can remain irreducible when restricted to a subgroup G. Since all irreducible representations of abelian groups are linear, we must have that G is nonabelian, and the only proper non-abelian subgroups of S4 are A4,D8, and S3, fixing a point. In the following diagram we show the restrictions of χ(2,2) to subgroups of S4 isomorphic to A4,D8, and S3, respectively. The top line in Tables 1, 2, and 3 gives the order of the centralizer of elements in the conjugacy class of that column, and the second row gives the cycle shape of elements in that class (viewed as elements in S4). It is clear that in the A4 and D8 cases the restricted character is the sum of the two linear characters shown, but in the S3 case the restricted character is indeed irreducible.Table 1A41243314221.31.3ψ(2,2)22−1−111ωω¯11ω¯ωTable 2D888444142242212.2ψ(2,2)220021111111−1−11Table 3S3632141.312.2ψ(2,2)2−10ψ1412We may now assume thatn≥5 and that (3.4) holds. Let us return to the case reached above where G has two orbits on Ω, of lengths 1 and n-1, respectively. In order to dismiss this case (for n>4), we may assume that G≅Sn-1, as a character which remains irreducible when restricted to a subgroup and certainly remains irreducible when restricted to a supergroup of that subgroup. Suppose the two orbits of G are {a} and {b1,b2,…,bn-1}. But now orb(G,Ω×Ω)=5 with orbit representatives {(a,a),(a,b1),(b1,a),(b1,b1),(b1,b2)}. However, orb(G,Ω×Δ)=6 with orbit representatives
(3.5){(a,{a,b1}),(a,{b1,b2}),(b1,{a,b1}),(b1,{a,b2}),(b1,{b1,b2}),(b1,{b2,b3})}.
(Note that when n=4 and G≅S3, we do indeed have 〈ψ1,ψ1〉=5 and that 〈ψ1,ψ2〉=〈ψ1,ψ1+ψ(2,2)〉=6). However, if n>4, this contradicts (3.4), and so we must have that G acts transitively on Ω and on Δ.Now letΩ={a,b,c,…}. The transitivity of G together with (3.4) imply that the stabilizer of a point, Ga, has the same number of orbits on points as it has on duads. As above, if Ga has k orbits on Ω, then it has at least (k2) orbits on duads, so we must have that 2≤k≤3. If k=3, then at least one of the orbits of Ga must have length greater than 1 (since n≥5), and so the number of orbits of Ga on duads is at least 4, a contradiction. Thus, k=2 and G is doubly transitive. In particular, ψ(n-1,1),ψ(n-2,2)∈Irr(G), and we have
(3.6)orb(G,Ω×Δ)=2,orb(G,Δ×Δ)=3.
Since Gδ, the stabilizer in G of the duad δ={a,b}, has just two orbits on Ω, these must be {a,b} and {c,d,…}, and so Gδ acts transitively on Ω∖{a,b}. The transitivity of G on duads thus implies that G is transitive on triples. Moreover, by (3.6) Gδ has just three orbits on duads which must consist of {a,b} itself, of length 1, the 2(n-2) duads such as {a,c} which intersect δ in one point, and the (n-22) duads disjoint from δ.We now wish to show thatG acts triply transitively on Ω. So let {a,b,c} and {x,y,z} be two triples of points in Ω; we must produce a permutation π∈G such that xπ=a,yπ=b,zπ=c. Now by the transitivity on duads, there exists a σ1∈G with {x,y}σ1={a,b}. So {x,z}σ1 is a duad intersecting δ in 1 point. But Gδ acts transitively on the set of 2(n-2) such duads, and so there exists a σ2∈Gδ such that {xσ1,zσ1}σ2={a,c}={x,z}σ1σ2 and {x,y}σ1σ2={a,b}. Thus, xσ1σ2=a,yσ1σ2=b and zσ1σ2=c as required.ThatG acts transitively on tetrads, that is, that G is 4-homogeneous, follows immediately from the fact that G is transitive on duads, and Gδ is transitive on duads disjoint from δ. We now investigate from the first principles the circumstances in which G is not 4-transitive.LetT={a,b,c,d} denote a tetrad, then the GT denotes the (set) stabilizer of T in G, and G[T] denotes the point-wise stabilizer of T. The factor group F=GT/G[T] gives the action of GT on the four points of T and is thus isomorphic to a subgroup of S4. If F≅S4, then G is 4-transitive, since any tetrad can be mapped into T, and then the four points {a,b,c,d} may be rearranged as we wish by elements of GT. We shall now show that if F≇S4, then F≅A4.Ifδ1 and δ2 are two disjoint duads, then the transitivity of G on duads together with the transitivity of Gδ1 on duads disjoint from δ1 show that
(3.7)|G|=(n2)(n-22)|L|=n(n-1)(n-2)(n-3)4|L|,
where L denotes the stabilizer of δ1 and δ2. Then, Ga, the stabilizer of a point, has, order
(3.8)|Ga|=(n-1)(n-2)(n-3)4|L|,
which is visibly divisible by 3, and so G contains elements of order 3 which fix a point. Thus, the factor group F contains elements of cycle shape 1.3. Now by the triple transitivity, we see that the point stabilizer Ga acts double transitively on Ω∖{a} and thus has even order. So G contains involutions of cycle shape 1r.2s where r≥1 and s≥2, for if s=1, then G contains transpositions, and the double transitivity of G will then force G≅Sn. Thus, F contains elements which act as 22 on the four points of T. The only subgroups of S4 which contain elements of order 3 and such involutions are A4 and S4.Note further that ifG contains involutions which fix more than 1 point, then F will contain transpositions; we shall then have F≅S4, and G will be 4-transitive. So we may assume that every involution of G has just one fixed point. Moreover, no element of G may contain a 4-cycle as we could conjugate this 4-cycle into the positions of T and would again have F≅S4. Thus every nontrivial element in a Sylow 2-subgroup of G has order 2 and so Sylow 2-subgroups of G are elementary abelian.We have seen that the 2-point stabilizerGab in G is of odd order,and, therefore, by the Feit-Thompson theorem, it is soluble. A minimal normal subgroup of a soluble group is an elementary abelian p-group, see, for example, Scott [13, page 74], where in this case p is odd. If K≤Gab is such a subgroup, then, since Gab acts transitively on the duads disjoint from {a,b}, K must act transitively on Ω∖{a,b}, otherwise duads in different orbits of K could not be mapped to duads in the same orbit. Moreover, K is abelian and so by Lemma 2.10  K acts regularly. Any other minimal normal subgroup L would have to intersect K trivially as the intersection will also be normal. But if l∈L,k∈K then [l,k]=l-1k-1lk∈L∩K=〈1〉, and so l commutes with the transitive subgroup K. Thus 〈L,K〉 is abelian and transitive, and so regular. Thus, 〈L,K〉=K=L, and K is unique. |K|=pd=n-2.We now investigate the structure of the point stabilizerGa and will in fact show that it has a regular normal subgroup which is an elementary abelian 2-group. At this point, our work is close to that of Kantor [9], and we include, part of his argument: let τ∈G{a,b} be a permutation which interchanges a and b; since τ2∈Gab, a group of odd order, we may assume that τ is an involution. By the uniqueness of K we see that Kτ=K and so |〈K,τ〉|=2|K|=2(n-2). Moreover, τ has a unique fixed point, and K acts transitively on Ω∖{a,b}, and so |τK|=n-2. Thus, we have
(3.9)τK={τ1=τ,τ2,…,τn-2}=Kτ,
where each τi is an involution interchanging a and b. So K={τ1τ,τ2τ,…,τn-2τ}, and it is clear that conjugation by τ inverts every element of K. But now we can prove the following.Lemma 3.4.
No two involutions ofG which contain the same transposition can fix the same point, nor can they have a further transposition in common.Proof.
We may assume that the involutionsσ1 and σ2 both contain the transposition (a,b). Then both σ1 and σ2 invert every element of K, and so σ1σ2 commutes with every element of K, and fixes a and b. If σ1 and σ2 either have a fixed point in common or have a further transposition in common, then σ1σ2 fixes a point and commutes with a transitive group (on Ω∖{a,b}). Thus, σ1σ2=1, and so σ1=σ2.Corollary 3.5.
For any tetradT={a,b,c,d} there is a unique involution acting as (ab)(cd) on the points of T and a unique Klein four-group acting on T.Proof.
We have seen that the factor groupF acting on the four points of T is isomorphic to A4; by the Lemma the element π1 acting as (ab)(cd) is unique. But if we conjugate this element by the element π2 which acts as (ac)(bd), we wil obtain an involution acting as (ab)(cd) on T which can only be π1. Thus, π1π2=π1 and π1π2=π2π1=π3, the unique involution which acts as (ad)(bc) on T.We are now in a position to count the number of involutions inG and, in particular, to count the number of involutions which commute with a given involution. From Lemma 3.4, we see that there is a unique involution fixing a given point and containing a given transposition; thus, the number of involutions fixing a given point is
(3.10)(n-12)/(n-1)2=n-2.
It is clear that commuting involutions must fix one another's set of fixed points, and so, since our involutions have just one fixed point, commuting involutions have the same fixed points. Now a given involution is in
(3.11)(n-122)=(n-1)(n-3)8
Klein four-group, and each pair of commuting involutions fixes (n-1)/4 tetrads. Thus, the number of involutions commuting with a given involution is
(3.12)(n-1)(n-3)8×2÷(n-1)4=n-3.
Thus the (n-2) involutions which fix a given point all commute with one another and so must form the nontrivial elements of an elementary abelian 2-group which must be normal in the point stabilizer. Thus, n=2m+1 for some m≥2, and 2m-1=pd for some d where p is an odd prime. We have the following.Lemma 3.6.
If2m=pd+1 where p is prime, then d=1 and m is prime.Proof.
Ifp≡1mod4, the right-hand side is congruent to 2 modulo 4 which is a contradiction since m≥2, and so the left-hand side is congruent to 0 modulo 4. So p≡3mod4, and d must be odd, since even d would result in the left-hand side being again congruent to 2 modulo 4. But now
(3.13)2m=pd+1=(p+1)(pd-1-⋯-p+1)︸dterms.
The underbracketed term is a sum of an odd number of odd integers and so is odd, which is a contradiction unless d=1. But then 2m-1 cannot be prime unless m is prime.
The point stabilizing subgroupGa thus has the shape
(3.14)Ga≅2m:P:H,
where P is a cyclic subgroup of order p=2m-1 and H normalizes P. If P=〈x〉, then x acts on the elementary abelian group of order 2m, which may be regarded as an m-dimensional vector space over ℤ2, as a cycle of maximal length—known as a Singer cycle. The subgroup P:H is thus acting faithfully as a subgroup of the general linear group GLm(2). We now appeal to Theorem 7.3 in Huppert [11, page 187], which for convenience we have included here as Theorem 2.11. For our purposes, we have f=1 and p=2; the Theorem says that our group P:H is a subgroup of a Frobenius group of shape p:m, so either H≅Cm or H is trivial as m is prime. We thus have
(3.15)|G|=(2m+1)(2m)(2m-1)k,
where k=m or 1, m is prime, and n=2m+1.
ButG is transitive on tetrads, and the stabilizer of a tetrad acts as A4 on the 4 points of the tetrad. Thus, (n4)×12 divides |G|, and we have
(3.16)n(n-1)(n-2)k=n(n-1)(n-2)(n-3)4!.12.s,
for some integer s; so (n-3)/2 divides k. If k=1, we must have n=5, when G is a triply transitive subgroup of S5 of the order 5.4.3=60. Thus G≅A5. If k=m, we have 2m-1-1 divides m, and since m is prime we have 2m-1=m+1, and so m=3. In this case, |G|=9.8.7.3. We now construct a permutation group of this order and degree 9 and show that it is unique.
We shall take the setΩ to be Ω={a=⋆,b=∞,0,1,…,6}, that is, to say the projective line P1(7) supplemented by the symbol ⋆. Without loss of generality, we take x=(0123456) normalized within Ga=G⋆ by y=(124)(365). We now seek an elementary abelian group E of order 8 which is normalized by 〈x,y〉≅7:3. Now one of the 7 involutions of E must commute with y, and since it has cycle shape 1.24, it must be one of
(3.17)(i)σ1=(∞0)(13)(26)(45),(ii)σ2=(∞0)(15)(23)(46),(iii)σ3=(∞0)(16)(25)(43).
Both σ1〈x,y〉 and σ2〈x,y〉 are indeed elementary abelian groups of order 8; however, σ3x does not commute with σ3. Moreover, the the permutation (∞)(0)(16)(25)(34) of the symmetric group S8, which negates the elements of ℤ7, normalizes 〈x,y〉 and interchanges σ1 and σ2. So without loss of generality, we may assume that G⋆=〈x,y,σ1〉≅23:7:3.
We now consider the number of Sylow 7-subgroups whichG contains. The only permutation of S9 outside P which commutes with P is the transposition (⋆∞) which does not have the desired cycle shape (and, in any case, would generate S9). Thus, CG(P)=P. We know that NG⋆(P)=〈x,y〉≅7:3 which has the index 9.8.7.3/7.3=72 in G. Now 72≡2modulo7, and so NG(P) must have the order 42 and contain an involution commuting with y and inverting x. This element must have, cycle shape 1.24 as usual, and so without loss of generality it is z=(⋆∞)(16)(25)(34). Since generators for G were reached without loss of generality, and since the group PΣL2(8)—which consists of the projective special linear group PSL2(8) with the field automorphism adjoined—is well known to have the required properties, we must have
(3.18)〈x,y,z〉≅PΣL2(8).
However, for completeness, we exhibit the identification as permutations of the projective line P1(8). The field of order 8 is obtained by extending ℤ2 by a root of η3+η+1. In Table 4 we display elements of P1(8) both as powers of η and as quadratic expressions in η. We also show the correspondence with elements of Ω as used above. Our generators now become
(3.19)x:u⟼ηu,y:u⟼u2,σ1:u⟼u+1,z:u⟼1/u.
This completes the proof of the main theorem.Table 4
The projective planeP1(8).
ηi∞01ηη2η3η4η5η6f(η)∞01ηη2η+1η2+ηη2+η+1η2+1Ω⋆∞0123456



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*Source: 757134-2011-08-23.xml*