Research Article  Open Access
A Character Condition for Quadruple Transitivity
Abstract
Let be a permutation group of degree viewed as a subgroup of the symmetric group . We show that if the irreducible character of corresponding to the partition of into subsets of sizes and 2, that is, to say the character often denoted by , remains irreducible when restricted to , then = 4, 5 or 9 and , A_{5}, or PΣL_{2}(8), respectively, or is 4transitive.
1. Introduction
Let be a permutation group of degree . The connection between the multiple transitivity of and the irreducibility of certain irreducible characters of of the symmetric group when restricted to 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 where is an irreducible character of the symmetric group , and is a subgroup of such that the restriction of to remains irreducible.
The present paper is concerned with the case when is irreducible, proving that is 4transitive 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 selfcontained 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 group permutes a set then, we let be a basis of an dimensional complex vector space, which becomes a module when we define , for , and extend linearly. If denotes the character afforded by this representation and the permutation has cycle shape when written as a product of disjoint cycles, then where denotes the set of points of fixed by the permutation . The subscript 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 where and are characters of the group , and denotes the complex conjugate of the complex number . The trivial character of which takes the value 1 on every element of will be denoted by . 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. If is a permutation group acting on the set with permutation character , then we have the following result: where orb denotes the number of orbits when acts on .
Proof. Firstly note that where denotes the stabilizer in of . So, if has orbits when it acts on , the OrbitStabilizer theorem implies that where is a representative of the orbit . Thus,
If the group acts on the set and on the set with permutations characters and , respectively, then also acts on the Cartesian product by defining
An ordered pair is only fixed by if both, its components are fixed, and so we see that . Then we have,
Theorem 2.2. If the group acts on the sets and with permutation characters and respectively, then
Proof.
Corollary 2.3. If acts transitively on and , then , the number of orbits of the stabilizer of a point in on .
Corollary 2.4. If acts transitively on and , then , the rank of the permutation action of on .
In the case when , the symmetric group acting on an element set , we denote the associated permutation character by . More generally will denote the permutation action of on unordered element subsets. In particular, will denote the permutation character of the action of on , the set of unordered pairs of elements of which we shall call duads. Thus, if has cycle shape , then
By Theorem 2.1 and its corollaries we see that
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, say, the orbit containing duads which intersect in one point, and the orbit of duads disjoint from . We thus have that where and are distinct irreducible characters of of degrees and , respectively.
2.2. The Irreducible Characters of the Symmetric Group
Definition 2.5. A partition of the integer is a sequence where for and .
The irreducible characters of are in onetoone correspondence with the partitions of and we denote the character corresponding to by . In this notation, the trivial character of corresponds to the partition of into a single subset of size and is thus denoted by . The permutation character of degree is given by . Indeed we have the general result.
Theorem 2.6. If and denotes the permutation character of the action of on unordered subsets, then
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 MurnaghanNakayama rule, see [7, page 79], which enables one to evaluate the character value of the irreducible on any conjugacy class of the symmetric group. We see that for we have
and so, if , then In particular, the degree of is given by
Applying (2.15) to a permutation with cycle shape , we have
2.3. Multiple Transitivity
Definition 2.7. The group is said to act transitively on the set if, and only if, given and , two ordered tuples of distinct points of , there exists a permutation such that for .
Definition 2.8. The group is said to act homogeneously on the set if, and only if, given with , there exists a permutation such that .
In other words, is homogeneous if it acts transitively on the subsets. Those fascinating groups which are homogeneous without being transitive have been investigated by Kantor, see [8, 9], and the LivingstoneWagner theorem of [10] is a highly significant result in this area. In this paper, we shall be concerned with the cases and thus with the action of subgroups of 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 2homogeneous without being 2transitive must have odd order.
Proof. Suppose that has even order and acts 2homogeneously on . Let and be two duads; we must produce a permutation which maps to for and 2. By 2homogeneity, there exists a such that , and since is even and again using the 2homogeneity, there exists an element of order 2 which acts as . If , then we are done, otherwise will suffice.
2.4. Restriction to a Subgroup of
In what follows, the subgroup will be a fixed subgroup of , and we shall denote the restriction of the character of to by In order to simplify the notation, we shall denote So we shall have The restriction to of an irreducible character of may or may not remain irreducible. However, if , then using Theorems 2.1 and 2.2 we have (1) is the number of orbits of on points, (2) is the number of orbits of on duads, (3) is the number of orbits of on , (4) is the number of orbits of on , and (5) is the number of orbits of on .
Note that if and are as in (2.12), then and . 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 be a transitive subgroup of the symmetric group acting on the set , and let the permutation commute with and fix a point . Then Fix is invariant under the action of and is thus the empty set or . But , and so and .
If is abelian and , the stabilizer in of , then commutes with and fixes the point . Thus, , from the above, and so , and the transitive group acts regularly.
A wellknown result about linear groups has been taken from Theorem 7.3 of Huppert [11, page 187].
Theorem 2.11. The general linear group contains a cyclic subgroup of order . We have and that is cyclic of order . (Such cycles of maximal possible length are referred to as Singer cycles.)
3. Statement and Proof of Main Theorem
Let . The transitivity of is related to the reducibility of certain irreducible characters of upon restriction to as follows.
Lemma 3.1. G is doubly transitive .
Proof. Follows immediately from Corollary 2.4.
Lemma 3.2. G is 4transitive, .
Proof. If is 4transitive, then it is certainly doubly transitive, and so . Moreover, if is a duad, then certainly has just three orbits on duads, as described towards the end of Section 2.1. Thus, and so and .
It is thus natural to ask in what circumstances we can have without being 4transitive. Our main result is the following theorem which appeals to the OddOrder theorem of Feit and Thompson and uses part of an argument due to Kantor [8] but is otherwise selfcontained and based on Aldhafeeri [12].
Theorem 3.3. Let be a subgroup of the symmetric group of degree with , thus , and let represent the irreducible character of corresponding to the partition of into a pair and a subset of cardinality . Then (i.e., the restriction of to is an irreducible representation of ) if, and only if, (i) and , (ii) and , (iii) and , or (iv) is 4transitive on letters.
Proof. Note that the character of only exists for , and so the restriction on is only required to make the theorem meaningful. In case , the subgroup is not transitive, but no restriction was placed on the transitivity of .
We shall assume that is not 4transitive and that . We have for . Thus, and so orb, say. But if orb then orb, as we may choose a duad from each of two orbits in ways. Now for , and if , at least one of the three orbits must have more than one point in it (since ), and so orb. Thus, . If and both orbits are of cardinality greater than 1 then orb, a contradiction, so we must have one orbit of length 1 and the other of length .
Now provided , which holds for . So provided , we have . Thus, For this reason, we treat separately and consider the circumstances in which this 2dimensional representation of S_{4} can remain irreducible when restricted to a subgroup . Since all irreducible representations of abelian groups are linear, we must have that is nonabelian, and the only proper nonabelian subgroups of S_{4} are , and S_{3}, fixing a point. In the following diagram we show the restrictions of to subgroups of S_{4} isomorphic to , and S_{3}, 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 S_{4}). It is clear that in the A_{4} and D_{8} cases the restricted character is the sum of the two linear characters shown, but in the S_{3} case the restricted character is indeed irreducible.



We may now assume that and that (3.4) holds. Let us return to the case reached above where has two orbits on , of lengths 1 and , respectively. In order to dismiss this case (for ), we may assume that , 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 are and . But now orb with orbit representatives . However, orb with orbit representatives (Note that when and , we do indeed have and that ). However, if , this contradicts (3.4), and so we must have that acts transitively on and on .
Now let . The transitivity of together with (3.4) imply that the stabilizer of a point, , has the same number of orbits on points as it has on duads. As above, if has orbits on , then it has at least orbits on duads, so we must have that . If , then at least one of the orbits of must have length greater than 1 (since ), and so the number of orbits of on duads is at least 4, a contradiction. Thus, and is doubly transitive. In particular, , and we have Since , the stabilizer in of the duad , has just two orbits on , these must be and , and so acts transitively on . The transitivity of on duads thus implies that is transitive on triples. Moreover, by (3.6) has just three orbits on duads which must consist of itself, of length 1, the duads such as which intersect in one point, and the duads disjoint from .
We now wish to show that acts triply transitively on . So let and be two triples of points in ; we must produce a permutation such that . Now by the transitivity on duads, there exists a with . So is a duad intersecting in 1 point. But acts transitively on the set of such duads, and so there exists a such that and . Thus, and as required.
That acts transitively on tetrads, that is, that is 4homogeneous, follows immediately from the fact that is transitive on duads, and is transitive on duads disjoint from . We now investigate from the first principles the circumstances in which is not 4transitive.
Let denote a tetrad, then the denotes the (set) stabilizer of in , and denotes the pointwise stabilizer of . The factor group gives the action of on the four points of and is thus isomorphic to a subgroup of S_{4}. If , then is 4transitive, since any tetrad can be mapped into , and then the four points may be rearranged as we wish by elements of . We shall now show that if , then .
If and are two disjoint duads, then the transitivity of on duads together with the transitivity of on duads disjoint from show that where denotes the stabilizer of and . Then, , the stabilizer of a point, has, order which is visibly divisible by 3, and so contains elements of order 3 which fix a point. Thus, the factor group contains elements of cycle shape 1.3. Now by the triple transitivity, we see that the point stabilizer acts double transitively on and thus has even order. So contains involutions of cycle shape where and , for if , then contains transpositions, and the double transitivity of will then force . Thus, contains elements which act as 2^{2} on the four points of . The only subgroups of S_{4} which contain elements of order 3 and such involutions are A_{4} and S_{4}.
Note further that if contains involutions which fix more than 1 point, then will contain transpositions; we shall then have , and will be 4transitive. So we may assume that every involution of has just one fixed point. Moreover, no element of may contain a 4cycle as we could conjugate this 4cycle into the positions of and would again have . Thus every nontrivial element in a Sylow 2subgroup of has order 2 and so Sylow 2subgroups of are elementary abelian.
We have seen that the 2point stabilizer in is of odd order,and, therefore, by the FeitThompson theorem, it is soluble. A minimal normal subgroup of a soluble group is an elementary abelian group, see, for example, Scott [13, page 74], where in this case is odd. If is such a subgroup, then, since acts transitively on the duads disjoint from , must act transitively on , otherwise duads in different orbits of could not be mapped to duads in the same orbit. Moreover, is abelian and so by Lemma 2.10 acts regularly. Any other minimal normal subgroup would have to intersect trivially as the intersection will also be normal. But if then , and so commutes with the transitive subgroup . Thus is abelian and transitive, and so regular. Thus, , and is unique. .
We now investigate the structure of the point stabilizer and will in fact show that it has a regular normal subgroup which is an elementary abelian 2group. At this point, our work is close to that of Kantor [9], and we include, part of his argument: let be a permutation which interchanges and ; since , a group of odd order, we may assume that is an involution. By the uniqueness of we see that and so . Moreover, has a unique fixed point, and acts transitively on , and so . Thus, we have where each is an involution interchanging and . So , and it is clear that conjugation by inverts every element of . But now we can prove the following.
Lemma 3.4. No two involutions of 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 and both contain the transposition . Then both and invert every element of , and so commutes with every element of , and fixes and . If and either have a fixed point in common or have a further transposition in common, then fixes a point and commutes with a transitive group (on ). Thus, , and so .
Corollary 3.5. For any tetrad there is a unique involution acting as on the points of and a unique Klein fourgroup acting on .
Proof. We have seen that the factor group acting on the four points of is isomorphic to A_{4}; by the Lemma the element acting as is unique. But if we conjugate this element by the element which acts as , we wil obtain an involution acting as on which can only be . Thus, and , the unique involution which acts as on .
We are now in a position to count the number of involutions in 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 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 Klein fourgroup, and each pair of commuting involutions fixes tetrads. Thus, the number of involutions commuting with a given involution is Thus the involutions which fix a given point all commute with one another and so must form the nontrivial elements of an elementary abelian 2group which must be normal in the point stabilizer. Thus, for some , and for some where is an odd prime. We have the following.
Lemma 3.6. If where is prime, then and is prime.
Proof. If , the righthand side is congruent to 2 modulo 4 which is a contradiction since , and so the lefthand side is congruent to 0 modulo 4. So , and must be odd, since even would result in the lefthand side being again congruent to 2 modulo 4. But now
The underbracketed term is a sum of an odd number of odd integers and so is odd, which is a contradiction unless . But then cannot be prime unless is prime.
The point stabilizing subgroup thus has the shape
where is a cyclic subgroup of order and normalizes . If , then acts on the elementary abelian group of order , which may be regarded as an dimensional vector space over , as a cycle of maximal length—known as a Singer cycle. The subgroup is thus acting faithfully as a subgroup of the general linear group . 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 and ; the Theorem says that our group is a subgroup of a Frobenius group of shape , so either or is trivial as is prime. We thus have
where or 1, is prime, and .
But is transitive on tetrads, and the stabilizer of a tetrad acts as A_{4} on the 4 points of the tetrad. Thus, divides , and we have
for some integer ; so divides . If , we must have , when is a triply transitive subgroup of S_{5} of the order . Thus . If , we have divides , and since is prime we have , and so . In this case, . We now construct a permutation group of this order and degree 9 and show that it is unique.
We shall take the set to be , that is, to say the projective line supplemented by the symbol . Without loss of generality, we take normalized within by . We now seek an elementary abelian group of order 8 which is normalized by . Now one of the 7 involutions of must commute with , and since it has cycle shape , it must be one of
Both and are indeed elementary abelian groups of order 8; however, does not commute with . Moreover, the the permutation of the symmetric group S_{8}, which negates the elements of , normalizes and interchanges and . So without loss of generality, we may assume that .
We now consider the number of Sylow 7subgroups which contains. The only permutation of S_{9} outside which commutes with is the transposition which does not have the desired cycle shape (and, in any case, would generate S_{9}). Thus, . We know that which has the index in . Now , and so must have the order 42 and contain an involution commuting with and inverting . This element must have, cycle shape as usual, and so without loss of generality it is . Since generators for were reached without loss of generality, and since the group —which consists of the projective special linear group with the field automorphism adjoined—is well known to have the required properties, we must have
However, for completeness, we exhibit the identification as permutations of the projective line . The field of order 8 is obtained by extending by a root of . In Table 4 we display elements of both as powers of and as quadratic expressions in . We also show the correspondence with elements of as used above. Our generators now become
This completes the proof of the main theorem.

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Copyright
Copyright © 2011 S. Aldhafeeri and R. T. Curtis. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.