A Character Condition for Quadruple Transitivity
Abstract
Let G 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.
1. Introduction
Let G 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 , 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 when 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
Theorem 2.1. If G is a permutation group acting on the set X with permutation character πX, then we have the following result:
Proof. Firstly note that
An ordered pair is only fixed by ν 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 group G acts on the sets X and Y with permutation characters πX and πY respectively, then
Proof.
Corollary 2.3. If G 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. If G acts transitively on X and a ∈ X, then 〈πX, πX〉 = orb(Ga, X), the rank of the permutation action of G on X.
2.2. The Irreducible Characters of the Symmetric Group Sn
Definition 2.5. A partition λ of the integer n is a sequence λ = (λ1, λ2, …, λk) where λi ≥ λi+1 for i = 1, …, k − 1 and .
The irreducible characters of Sn 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. If k ≤ n/2 and πk denotes the permutation character of the action of Sn on unordered k-subsets, then
2.3. Multiple Transitivity
Definition 2.7. The group G 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 for i = 1 ⋯ k.
Definition 2.8. The group G 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, and 4 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 that G 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 = (b1 b2)⋯. If , then we are done, otherwise στ will suffice.
2.4. Restriction to a Subgroup of Sn
- (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 if χ 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. Let K 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.
If K 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 group GLm(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
Let G ≤ 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 Corollary 2.4.
Lemma 3.2. G is 4-transitive, n ≥ 4⇒ψ(n−1,1), ψ(n−2,2) ∈ Irr(G).
Proof. If G 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,
It is thus natural to ask in what circumstances we can have ψ(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. Let G 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
- (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 character χ(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.
A4 | 12 | 4 | 3 | 3 |
---|---|---|---|---|
14 | 22 | 1.3 | 1.3 | |
ψ(2,2) | 2 | 2 | −1 | −1 |
1 | 1 | ω | ||
1 | 1 | ω |
D8 | 8 | 8 | 4 | 4 | 4 |
---|---|---|---|---|---|
14 | 22 | 4 | 22 | 12.2 | |
ψ(2,2) | 2 | 2 | 0 | 0 | 2 |
1 | 1 | 1 | 1 | 1 | |
1 | 1 | −1 | −1 | 1 |
S3 | 6 | 3 | 2 |
---|---|---|---|
14 | 1.3 | 12.2 | |
ψ(2,2) | 2 | −1 | 0 |
ψ1 | 4 | 1 | 2 |
We now wish to show that G 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 . So 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 and . Thus, and as required.
That G 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.
Let T = {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.
Note further that if G 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 stabilizer Gab 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.
Lemma 3.4. No two involutions of G 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 involutions σ1 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 tetrad T = {a, b, c, d} there is a unique involution acting as (a b)(c d) on the points of T and a unique Klein four-group acting on T.
Proof. We have seen that the factor group F acting on the four points of T is isomorphic to A4; by the Lemma the element π1 acting as (a b)(c d) is unique. But if we conjugate this element by the element π2 which acts as (a c)(b d), we wil obtain an involution acting as (a b)(c d) on T which can only be π1. Thus, and π1π2 = π2π1 = π3, the unique involution which acts as (a d)(b c) on T.
Lemma 3.6. If 2m = pd + 1 where p is prime, then d = 1 and m is prime.
Proof. If p ≡ 1 mod 4, 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 ≡ 3 mod 4, and d must be odd, since even d would result in the left-hand side being again congruent to 2 modulo 4. But now
The point stabilizing subgroup Ga thus has the shape
But G is transitive on tetrads, and the stabilizer of a tetrad acts as A4 on the 4 points of the tetrad. Thus, divides |G|, and we have
We shall take the set Ω 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 = (0 1 2 3 4 5 6) normalized within Ga = G⋆ by y = (1 2 4)(3 6 5). 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
We now consider the number of Sylow 7-subgroups which G 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 which has the index 9.8.7.3/7.3 = 72 in G. Now 72 ≡ 2 modulo7, 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 = (⋆ ∞)(1 6)(2 5)(3 4). 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
ηi | ∞ | 0 | 1 | η | η2 | η3 | η4 | η5 | η6 |
---|---|---|---|---|---|---|---|---|---|
f(η) | ∞ | 0 | 1 | η | η2 | η + 1 | η2 + η | η2 + η + 1 | η2 + 1 |
Ω | ⋆ | ∞ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |