Orbits of Sylow subgroups of finite permutation groups
John Bamberg, Alexander Bors, Alice Devillers, Michael Giudici, Cheryl E. Praeger, Gordon F. Royle
aa r X i v : . [ m a t h . G R ] F e b ORBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS
JOHN BAMBERG, ALEXANDER BORS, ALICE DEVILLERS, MICHAEL GIUDICI,CHERYL E. PRAEGER, AND GORDON F. ROYLE
Abstract.
We say that a finite group G acting on a set Ω has Property ( ∗ ) p for a prime p if P ω is a Sylow p -subgroup of G ω for all ω ∈ Ω and Sylow p -subgroups P of G . Property ( ∗ ) p arose in the recent work of Tornier (2018) on local Sylow p -subgroups of Burger-Mozes groups,and he determined the values of p for which the alternating group A n and symmetric group S n acting on n points has Property ( ∗ ) p . In this paper, we extend this result to finite 2-transitivegroups and we give a structural characterisation result for the finite primitive groups that satisfyProperty ( ∗ ) p for an allowable prime p . Introduction
Given G Sym(Ω) with | Ω | = n , the Burger-Mozes group U ( G ) [3] is the group of allautomorphisms of the n -regular tree T n such that, for all vertices v , the stabiliser of v inducesthe group G on the set of all neighbours of v . Tornier [17] introduced the following propertywhen studying local Sylow p -subgroups of Burger-Mozes groups. We use Syl p ( G ) to denote theset of all Sylow p -subgroups of a group G and for a group G acting on a set Ω we use G ω todenote the stabiliser in G of the point ω ∈ Ω. Definition 1.1.
Let G be a finite group acting on a set Ω with | Ω | = n , let p be a prime andlet P ∈ Syl p ( G ). We say that G has Property ( ∗ ) p if P ω ∈ Syl p ( G ω ) for all ω ∈ Ω.Note that since G acts transitively by conjugation on the set of Sylow p -subgroups of G ,Property ( ∗ ) p does not depend on the choice of P . From the definition it is immediate that any semiregula r permutation group, (i.e., one where G ω = 1 for all ω ∈ Ω) has Property ( ∗ ) p .Tornier’s motivation for studying Property ( ∗ ) p was as follows: Let T be a finite subtree of T n and for any H Aut( T n ) let H ( T ) denote the pointwise stabiliser of T in H . Then Tornier[17, Proposition 10] showed that for G Sym(Ω) and P ∈ Syl p ( G ), the group U ( P ) ( T ) is a localSylow p -subgroup of U ( G ) ( T ) if and only if G has Property ( ∗ ) p . Tornier [17, Proposition 11]showed that if G is the alternating group A n or symmetric group S n acting on n points then G has Property ( ∗ ) p , with p dividing | G | , if and only if n p p > n , where n p is the highest power of p that divides n . Moreover, if either P has the same orbits as G or | Ω | = p n , then ( ∗ ) p holds[17, Propositions 12 and 13].The purpose of this paper is to undertake a thorough investigation of permutation groups withProperty ( ∗ ) p . In Section 2 we start by showing that a permutation group has Property ( ∗ ) p ifand only if its induced action on each orbit has Property ( ∗ ) p , thereby reducing the problem totransitive groups. In the remainder of Section 2 we derive some basic structural results leadingup to the fact that a transitive permutation group of degree n has Property ( ∗ ) p if and only ithas a Sylow p -subgroup all of whose orbits are the same size. In this case, the common size ofthe orbits is the highest power of p dividing n , that is, n p . This also allows us to deduce that if pn p > n then G has Property ( ∗ ) p (Lemma 2.4(5)).A transitive group can either be imprimitive or primitive, and so we consider each case inturn. In Section 3 we consider imprimitive groups, where we prove the following: Theorem 1.2.
Let G act transitively on a set Ω with system of imprimitivity B . Let B ∈ B ,let G BB be the permutation group induced on B by the setwise stabiliser G B and let G B be thepermutation group induced by G on B . Date : February 10, 2021. (1) If G BB and G B have Property ( ∗ ) p then G has Property ( ∗ ) p .(2) If G has Property ( ∗ ) p then G BB has Property ( ∗ ) p .(3) If G has Property ( ∗ ) p then G B does not necessarily have Property ( ∗ ) p . Then we turn to primitive groups, and in Section 4 and Section 5 we develop, and then prove,the following structural characterisation of primitive groups with Property ( ∗ ) p . Theorem 1.3.
Let G be a primitive permutation group on Ω and suppose that G has Prop-erty ( ∗ ) p for some prime p dividing | Ω | . Then one of the following holds:(1) G is an almost simple group;(2) G is of Affine type and | Ω | = p k ;(3) Ω = ∆ k for some k > and G H wr K where H is an almost simple group actingprimitively on ∆ with Property ( ∗ ) p and K S k . Moreover, either p is coprime to | K | ,or p divides | K | and | ∆ | is a power of p .Moreover, any primitive group in cases (2) and (3) has Property ( ∗ ) p . Not all almost simple groups satisfy Property ( ∗ ) p and indeed examples for which pn p < n and p divides | G ω | seem rare. Indeed, apart from examples from one infinite family satisfyingProperty ( ∗ ) , there are only 6 such examples of degree at most 4095. See Subsection 4.5. Thissuggests the following problem: Problem 1.4.
Determine all the almost simple primitive permutation groups G of degree n that have Property ( ∗ ) p for some prime p dividing n and for which pn p < n and p divides | G ω | .As a contribution towards Problem 1.4, we determine all the 2-transitive permutation groupswith Property ( ∗ ) p . This generalises Tornier’s result for A n and S n , while all 2-transitive groupswith Property ( ∗ ) p and n p = p have already been determined by the fourth author [15, Theorem]. Theorem 1.5.
Let G be a -transitive permutation group of degree n on a set Ω with ω ∈ Ω ,and let p a prime dividing n . Then G has Property ( ∗ ) p if and only if one of the following holds: (a) pn p > n ; (b) p does not divide | G ω | ; (c) G = A with n = 6 and p = 2 ; (d) G = M with n = 12 and p = 3 ; (e) G = PΓL (8) with n = 28 and p = 2 ; We note that in case (e) the subgroup PSL (8) of G also has Property ( ∗ ) p . However, PSL (8)is not 2-transitive of degree 28, and so does not appear as a possibility.We finish the paper in Section 6 with a survey of the permutation groups of degree at most39 that have Property ( ∗ ) or ( ∗ ) . 2. Preliminaries
Our first observation (Lemma 2.1) shows that, in order to study groups with Property ( ∗ ) p , itis sufficient to restrict our attention to transitive group actions. The proof of Lemma 2.1 followsimmediately from Definition 1.1. Lemma 2.1.
Let G act on a set Ω . Then G has Property ( ∗ ) p if and only if G acts on each ofits orbits with Property ( ∗ ) p . The following is a generalisation of a result of Wielandt [18, Theorem 3.4 ′ ] to arbitrary groupactions. Lemma 2.2.
Let G be a group acting transitively on a set Ω of size n and let P ∈ Syl p ( G ) .Then the minimum length of a P -orbit is n p .Proof. Let ω ∈ Ω and let Q ∈ Syl p ( G ω ). Then by Sylow’s Theorems, there exists a Sylow p -subgroup P ′ of G such that Q P ′ . Then Q = P ′ ∩ G ω = P ′ ω and so P ′ has an orbit oflength | P ′ : Q | = | G | p / | G ω | p = n p | G ω | p / | G ω | p = n p . Since all Sylow p -subgroups are conjugate RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 3 in G it follows that P has an orbit of length n p . Moreover, if P has a shorter orbit ω P then | P ω | = | P | / | ω P | > | P | /n p = | Q | , contradicting | Q | being the order of a Sylow p -subgroup of apoint stabiliser. (cid:3) This leads to the following easy test for Property ( ∗ ) p . Lemma 2.3.
Let G act transitively on Ω , let p be a prime, and let P ∈ Syl p ( G ) . Then Prop-erty ( ∗ ) p holds if and only if all orbits of P on Ω have the same length, and that length is n p .Proof. By Lemma 2.2, P has an orbit of length n p . Also if | ω P | = n p then | P ω | = | P | /n p = | G | p /n p = | G ω | p . Hence all orbits of P on Ω have the same length if and only if P ω has order | G ω | p for all ω ∈ Ω, that is, if and only if P ω ∈ Syl p ( G ω ) for all ω ∈ Ω. (cid:3) It is easy to see from Lemma 2.3 combined with Lemma 2.1 that, if G acts on Ω with Property( ∗ ) p , then the permutation group G Ω Sym(Ω) induced by G also has Property ( ∗ ) p . We collectsome other easy observations in the following lemma. Lemma 2.4.
Let G act transitively on a set Ω of size n , let ω ∈ Ω and let p be a prime.(1) If p does not divide | G ω | then G has Property ( ∗ ) p .(2) If G acts faithfully on Ω with Property ( ∗ ) p such that p divides | G | , then p divides n .(3) If | Ω | = p k then G has Property ( ∗ ) p but not Property ( ∗ ) r for any prime r = p whichdivides | G Ω | , where G Ω is the permutation group on Ω induced by G .(4) If pn p > n then G has Property ( ∗ ) p .(5) If H is a transitive subgroup of G and G has Property ( ∗ ) p then H has Property ( ∗ ) p .Proof. (1) If p does not divide | G ω | for some ω then p does not divide | G ω | for all ω . Since P ω is a p -subgroup of G ω it follows that P ω = 1 for all ω ∈ Ω and is a Sylow p -subgroup of G ω .(2) Suppose that G has Property ( ∗ ) p with p coprime to n . Then, by Lemma 2.3, all orbitsof a Sylow p -subgroup of G have length n p = 1. Since G acts faithfully on Ω it followsthat a Sylow p -subgroup of G is trivial and so p does not divide | G | .(3) By Lemma 2.2 a Sylow p -subgroup is transitive and so Property ( ∗ ) p holds while therest of the assertion follows from part (2).(4) By Lemma 2.2, P has an orbit of length n p and all orbits of P have length at least n p .Since orbits of P have length a power of p and pn p > n it follows that all orbits of P have length n p and so by Lemma 2.3 we have that G has Property ( ∗ ) p .(5) Let P ∈ Syl p ( H ). Then by Lemma 2.2, all orbits of P have size at least n p . However,since G has Property ( ∗ ) p and P is contained in a Sylow p -subgroup of G , Lemma 2.3implies that every orbit of P has size at most n p . Thus all orbits of P have length n p and the result follows from applying Lemma 2.3 again. (cid:3) Imprimitive groups
In this section, we investigate imprimitive permutation groups with Property ( ∗ ) p . Supposethat G is an imprimitive group with a non-trivial system of imprimitivity B and that B ∈ B . Asin Theorem 1.2, we let G BB denote the permutation group induced on a block B by the setwisestabiliser G B , and G B denote the permutation group induced by G on B . Then G G BB wr G B and the assumption that B is non-trivial ensures that both G BB and G B have strictly lower degreethan G . Next we consider how Property ( ∗ ) p interacts with these three groups. Lemma 3.1.
Let H act transitively on Ω and K act transitively on ∆ . Then H wr K actingimprimitively on Ω × ∆ has Property ( ∗ ) p if and only if both H and K have Property ( ∗ ) p .Proof. Let G = H wr K = H k ⋊ K where k = | ∆ | and let Σ = Ω × ∆. Then R = P k ⋊ Q isa Sylow p -subgroup of G , where P is a Sylow p -subgroup of H and Q is a Sylow p -subgroup of K . Let ω ∈ Ω and δ ∈ ∆. Then G ( ω,δ ) = ( H ω × H k − ) ⋊ K δ and R ( ω,δ ) = ( P ω × P k − ) ⋊ Q δ . BAMBERG, BORS, DEVILLERS, GIUDICI, PRAEGER, AND ROYLE If H and K both have Property ( ∗ ) p , then | P ω | does not depend on the choice of ω and | Q δ | does not depend on the choice of δ . Thus | R ( ω,δ ) | is constant and so Lemma 2.3 implies that G has Property ( ∗ ) p .Conversely, suppose that G has Property ( ∗ ) p . Then by Lemma 2.3, | R ( ω,δ ) | does not dependon ( ω, δ ) and | ( ω, δ ) R | = | Σ | p = | Ω | p | ∆ | p . Now | ( ω, δ ) R | = | ω P || δ Q | . By Lemma 2.2 we havethat | ω P | > | Ω | p and | δ Q | > | ∆ | p . Thus, as | ( ω, δ ) R | = | Ω | p | ∆ | p for all ω ∈ Ω and δ ∈ ∆ itfollows that | ω P | = | Ω | p for all ω ∈ Ω and | δ Q | = | ∆ | p for all δ ∈ ∆. Hence both H and K haveProperty ( ∗ ) p . (cid:3) Corollary 3.2.
Let G be an imprimitive group with system of imprimitivity B . Let B be a blockof B . Let H = G BB be the permutation group induced on B by the setwise stabiliser G B , andlet K = G B be the group induced by G on B . If H and K have Property ( ∗ ) p , then G also hasProperty ( ∗ ) p .Proof. The wreath product H wr K has Property ( ∗ ) p by Lemma 3.1. Since G is a subgroup of H wr K , the statement follows immediately from Lemma 2.4(5). (cid:3) We have the following partial converse to Corollary 3.2.
Lemma 3.3.
Suppose that G acts transitively on Ω with Property ( ∗ ) p , that it is imprimitivewith system of imprimitivity B , and that B ∈ B . Then G BB has Property ( ∗ ) p .Proof. Suppose that G has Property ( ∗ ) p and let P ∈ Syl p ( G B ). By Sylow’s Theorems thereexists a Sylow p -subgroup Q of G that contains P . Thus Q B = P . Suppose first that Q = P .Since Q ω ∈ Syl p ( G ω ) for all ω ∈ Ω and ( G B ) ω = G ω , then Q ω ∈ Syl p (( G B ) ω ) for all ω ∈ B .Hence G BB has Property ( ∗ ) p . Thus from now on suppose that P < Q . Let α ∈ B . Then Q acts imprimitively on α Q with block α Q ∩ B = α P . Suppose that B ∈ B and there exists q ∈ Q such that B q = B . Then α q ∈ B and so B ∩ α Q = ∅ . Hence | α Q | = | α P || B Q | . Since G hasProperty ( ∗ ) p on Ω, Lemma 2.3 implies that | α Q | is independent of the choice of α ∈ B andhence so is | α P | . Thus by Lemma 2.3, G B has Property ( ∗ ) p on B . (cid:3) We note that the full converse of Corollary 3.2 does not hold. For example, let G = A actingon 20 points. Then G has Property ( ∗ ) , by Lemma 2.4(1), and G acts imprimitively with asystem of imprimitivity B consisting of 10 blocks of size 2. However, G B does not have Property( ∗ ) , by Lemma 2.3, since a Sylow 2-subgroup of G has orbit lengths 2 , , , B . We generalisethis example below. Corollary 3.2 and Lemma 3.3 then imply Theorem 1.2. Lemma 3.4.
Let p be a prime, and let G Sym( X ) be a transitive permutation group of degree d = kp + f with < f < p . Let P ∈ Syl p ( G ) and let fix X ( P ) be the set of fixed points of P .Assume that | fix X ( P ) | = f and P acts semiregularly on X \ fix X ( P ) . Let Y be an orbit of G on the set of elements of X p with pairwise distinct entries. Then the natural action G induceson Y is faithful and imprimitive with Property ( ∗ ) p . Take B to be the partition of Y where two p -tuples lie in the same block if they correspond to the same unordered p -set. If the blocks in thispartition have size a multiple of p , then G B does not have Property ( ∗ ) p .Proof. Since P is a p -group, all its orbits on X have size a power of p , so P fixes at least f points.We assume P fixes exactly f points. By definition G acts transitively on Y , and it is clear that B is a G -invariant partition of Y . Since G is transitive on X , every element of X occurs as thefirst element of a p -tuple in Y , so the subgroup of G fixing each p -tuple in Y fixes X pointwise.Thus G acts faithfully on Y . To show that G acts on Y with Property ( ∗ ) p , we need to showthat all P -orbits on Y have the same size. Consider the p -tuple ( x , x , . . . , x p ) ∈ Y . Since | fix X ( P ) | = f < p , there is an index i such that x i fix X ( P ). Since P acts semiregularly on X \ fix X ( P ), x Pi has size | P | , and hence ( x , x , . . . , x p ) P also has size | P | . Therefore G on Y has Property ( ∗ ) p and | P | = | Y | p by Lemma 2.3.Let K be the kernel of the action of G on B , so that the induced group G B ∼ = G/K . Now G B is the natural action of G on an orbit of unordered p -sets of X (the subsets that correspond RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 5 to p -tuples in Y ). Let Q = P B = P K/K , so Q ∈ Syl p ( G B ). Then Q ∼ = P/ ( P ∩ K ), and inparticular | Q | | P | .Now assume each block of B has size ℓp for some integer ℓ >
1. Note that |B| = | Y | / ( ℓp ), so |B| p = | Y | p / ( p ℓ p ) | Y | p /p = | P | /p . If G B had Property ( ∗ ) p , then by Lemma 2.3 all Q -orbitson B would have size |B| p | P | /p . We will display an orbit of size | P | , yielding a contradiction(and also implying P ∩ K = 1). Since G is transitive on X and since f > Y containsa p -tuple x = ( x , x , . . . , x p ) with x p ∈ fix X ( P ). Further, since f < p , not all the x i are infix X ( p ). Consider the corresponding p -set b x := { x , x , . . . , x p } where (without loss of generality) x , x , . . . , x p − a are not fixed by P and x p − a +1 , . . . , x p lie in fix X ( P ), so 0 < a f < p . Let g ∈ P fix b x setwise. Then g fixes each of the points x p − a +1 , . . . , x p of fix X ( P ), and also fixes { x , x , . . . , x p − a } setwise. Since g is a p -element and 0 < p − a < p it follows that g fixes each x i for 1 i p . Thus g fixes the p -tuple x = ( x , . . . , x p ) ∈ Y . We have already shown thateach P -orbit in Y has size | P | , and so we conclude that g = 1. Thus the only element of P fixing b x setwise is the identity, so the P -orbit in B containing b x has size | P | . This completes theproof. (cid:3) If we take G = A acting on a set X of size 5, then a Sylow 2-subgroup P is isomorphic tothe Klein four-group, fixes one point x , and is semiregular on X \ { x } . Since G is 2-transitiveon X , G has a single orbit Y on the set of ordered pairs with distinct entries from X . Thus theblocks in B have size 2. All the conditions in Lemma 3.4 hold, and this yields the example with A of degree 20 mentioned above. Corollary 3.5.
Let p be a prime. There are infinitely many imprimitive groups G that haveProperty ( ∗ ) p for which the induced action on some G -invariant block system does not haveProperty ( ∗ ) p .Proof. There are infinitely many p -powers q , and for each q we will construct an example. Take G = PGL ( q ), in its natural action of degree d = q + 1 on F q ∪ {∞} . The subgroup P oftranslations (with unique fixed point ∞ ) is a Sylow p -subgroup of G and is semiregular on F q . Let Y be the orbit under G of the p -tuple x = (0 , , , . . . , p −
1) (corresponding to thesubfield F p of F q ). By Lemma 3.4, G has a natural faithful imprimitive action on Y which hasProperty ( ∗ ) p . Take B to be the partition of Y where two p -tuples lie in the same block if theycorrespond to the same unordered p -set. The size of a block of B is | G { , , ,...,p − } | / | G (0 , , ,...,p − | .Now G { , , ,...,p − } contains the group of translations by elements of F p , and hence is transitiveon { , , , . . . , p − } . Thus the size of a block is divisible by p , and hence G B does not haveProperty ( ∗ ) p by Lemma 3.4. (cid:3) Note these are not the only examples. We display here a second family satisfying the hypothe-ses of Lemma 3.4 (this family has only p − p ). Take n = p + f where p is a prime and 1 < f < p . Let G = A n acting naturally on X of size n and let P ∈ Syl p ( G ).Then | fix X ( P ) | = f , and P is cyclic of order p and hence semiregular on X \ fix X ( P ). Now G is transitive on the set Y of p -tuples of distinct points; for any such p -tuple there are p ! tuplesin Y corresponding to the same underlying subset β , and G β is transitive on these p ! tuplessince n > p + 2. Thus the action of G on Y is imprimitive with a system of imprimitivity B consisting of (cid:0) np (cid:1) blocks of size p !, so all the conditions of Lemma 3.4 hold and it follows that G Y has Property ( ∗ ) p , but G B does not have Property ( ∗ ) p .We also note that the examples produced by Lemma 3.4 have G ∼ = G B . There are also exam-ples of groups G having Property ( ∗ ) p such that G = G B and G B does not have Property ( ∗ ) p .For example, let G = D acting regularly on itself. Then G has Property ( ∗ ) . Since G hasa subgroup of order 4, this action has a system of imprimitivity B with blocks of size 4. Then G B = S , which does not have Property ( ∗ ) by Lemma 2.4(2).4. Primitive groups
By the O’Nan-Scott Theorem, see for example [13], the finite primitive permutation groupscan be categorised into five types: (i) Affine type, (ii) Almost simple type, (iii) Diagonal type,
BAMBERG, BORS, DEVILLERS, GIUDICI, PRAEGER, AND ROYLE (iv) Product action type, and (v) Twisted wreath type. For more information about these typessee [5] or [6]. We discuss the last three types in the next three sections.4.1.
Diagonal type groups.
We show that primitive groups of Diagonal type do not haveProperty ( ∗ ) p . Lemma 4.1.
Let G be a primitive group of Diagonal type acting primitively on a set Ω and let p be a prime dividing | Ω | . Then G does not have Property ( ∗ ) p .Proof. Suppose that G has Property ( ∗ ) p and let N = soc( G ). Then N = T k for some finitenonabelian simple group T and integer k >
2. Let D = { ( t, t, . . . , t ) | t ∈ T } N . Then wemay assume that Ω is the set of right cosets of D in N . Letting [ t , t , . . . , t k − ] denote the coset D ( t , t , . . . , t k − ,
1) we have that Ω can be identified with { [ t , t , . . . , t k − ] | t , . . . , t k − ∈ T } .Let p be a prime dividing | Ω | = | T | k − and so p divides | T | . Let Q ∈ Syl p ( T ). Then b Q = Q k is a Sylow p -subgroup of N and by Sylow’s Theorems, is contained in a Sylow p -subgroup P of G . Now b Q = Q × Q where Q = Q k − × Q = 1 k − × Q . For each( g , . . . , g k − , ∈ Q and t ∈ T we have that [ t, , . . . , ( g ,...,g k − , = [ tg , g , . . . , g k − ]. Thus[ t, , . . . , Q = { [ x , x , . . . , x k − ] | x ∈ tQ, x , . . . , x k − ∈ Q } , which has size | Q | = n p . ByLemma 2.4(5), N has Property ( ∗ ) p and so it follows from Lemma 2.3 that all orbits of b Q have size n p , and hence that [ t, , . . . , b Q = [ t, , . . . , Q . In particular, [ t, , . . . , Q ⊆ [ t, , . . . , Q .Now [ t, , . . . , (1 , ,..., ,g ) = [ g − t, g − , . . . , g − ] and so [ t, , . . . , Q = { [ x , x , x , . . . , x ] | x = x t, x ∈ Q } . Thus Qt = tQ for all t ∈ T and so Q P T , a contradiction. (cid:3) Product action.
Let k > H Sym(∆) for a finite set ∆. ThenSym(∆) wr S k = Sym(∆) k ⋊ S k acts on Ω = ∆ k such that for each ( δ , δ , . . . , δ k ) ∈ Ω we have( δ , δ , . . . , δ k ) ( h ,h ,...,h k ) = ( δ h , δ h , . . . , δ h k k ) for all h , h , . . . , h k ∈ Sym(∆)and ( δ , δ , . . . , δ k ) σ = ( δ σ − , δ σ − . . . , δ kσ − ) for all σ ∈ S k . Note that any G Sym(∆) wr S k acts on { , , . . . , k } and the permutation group G { , ,...,k } induced is the projection of G onto S k . We denote the stabiliser in G of 1 in this action by G .Note that we have a homomorphism from G onto a subgroup of Sym(∆) given by( h , h , . . . , h k ) σ h and so the group G acts naturally on ∆.We begin with a result about the full wreath product. Lemma 4.2.
Suppose that p is a prime, k is a positive integer such that k > , and K S k suchthat | K | is coprime to p . Suppose also that H acts transitively on a set ∆ , and let G = H wr K in product action on Ω = ∆ k . Then G has Property ( ∗ ) p on Ω if and only if H has Property ( ∗ ) p on ∆ .Proof. Since p does not divide | K | , the group R = P k is a Sylow p -subgroup of G , where P is aSylow p -subgroup of H . Let ω = ( δ , . . . , δ k ) ∈ Ω. Then R ω = P δ × · · · × P δ k . Suppose first that H has Property ( ∗ ) p . Then Lemma 2.3 implies that | R δ i | does not depend on the choice of δ i and so | R ω | does not depend on ω . Lemma 2.3 then implies that G has Property ( ∗ ) p . Supposeconversely that G has Property ( ∗ ) p . By Lemma 2.3, | R : R ω | = | ∆ | kp for all ω ∈ Ω. Moreover,by Lemma 2.2 we have that | P : P δ i | > | ∆ | p for all δ i ∈ ∆. Thus | P : P δ i | = | ∆ | p for all δ i ∈ ∆and so again by Lemma 2.3, H has Property ( ∗ ) p . (cid:3) Throughout the remainder of this section we make the following assumption on the group G . Hypothesis 4.3.
Let p be a prime, k be a positive integer such that k > G Sym(∆) wr S k acts transitively on Ω = ∆ k in product action. Let K = G { , ,...,k } and H = G ∆1 . Suppose that H has a transitive subgroup N such that N k G H wr K .We first give a necessary condition for a group G in product action to have Property ( ∗ ) p . RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 7
Lemma 4.4.
Suppose that Hypothesis 4.3 holds. If G has Property ( ∗ ) p on Ω then H hasProperty ( ∗ ) p on ∆ .Proof. Suppose that G has Property ( ∗ ) p on Ω and that H does not have Property ( ∗ ) p on ∆.Then by Lemma 2.4(5), N k has Property ( ∗ ) p on Ω and by Lemma 4.2, N has Property ( ∗ ) p on∆ (take K = 1). Let Q be a Sylow p -subgroup of N and let R = Q k . Then by Lemma 2.3, allorbits of Q on ∆ have size | ∆ | p and all orbits of R on Ω have size | Ω | p . Note that R G .Let P be a Sylow p -subgroup of H . Since G ∆1 = H , there exists a Sylow p -subgroup b P of G that contains R and such that b P ∆ = P . Since G has Property ( ∗ ) p and G is transitive on Ω,Lemma 2.4(5) implies that G has Property ( ∗ ) p and so by Lemma 2.3 all orbits of b P have size | Ω | p .As H does not have Property ( ∗ ) p on ∆, there exists δ ∈ ∆ such that | δ P | > | ∆ | p . Let ω = ( δ, . . . , δ ). Then ω R = δ Q × δ Q × · · · × δ Q has size | Ω | p . However, δ Q ⊂ δ P and so ω R ⊂ ω b P as b P G and induce P on the first factor. This contradicts the fact that G has Property ( ∗ ) p and so H must have Property ( ∗ ) p . (cid:3) For sufficient conditions we need to split our analysis according to whether or not p divides | K | . Lemma 4.5.
Suppose that Hypothesis 4.3 holds and that | K | is coprime to p . Then G hasProperty ( ∗ ) p on Ω if and only if H has Property ( ∗ ) p on ∆ .Proof. Suppose first that H has Property ( ∗ ) p . Then by Lemma 4.2, H wr K has Property ( ∗ ) p and so by Lemma 2.4(5) so does G .Conversely, suppose that G has Property ( ∗ ) p . We assume that H does not have Property ( ∗ ) p on ∆ and seek a contradiction. By Lemma 2.4(5), N k has Property ( ∗ ) p on Ω and by Lemma4.2, N has Property ( ∗ ) p on ∆ (take K = 1). Let Q be a Sylow p -subgroup of N and let R = Q k .Then by Lemma 2.3, all orbits of Q on ∆ have size | ∆ | p and all orbits of R on Ω have size | Ω | p .Note that R G .Let P be a Sylow p -subgroup of H . Since G ∆1 = H , there exists a Sylow p -subgroup b P of G that contains R and such that b P ∆ = P . Since G has Property ( ∗ ) p and G is transitive on Ω,Lemma 2.4(5) implies that G has Property ( ∗ ) p and so by Lemma 2.3 all orbits of b P have size | Ω | p .As H does not have Property ( ∗ ) p on ∆, there exists δ ∈ ∆ such that | δ P | > | ∆ | p . Let ω = ( δ, . . . , δ ). Then ω R = δ Q × δ Q × · · · × δ Q has size | Ω | p . However, δ Q ⊂ δ P and so ω R ⊂ ω b P as b P G and induce P on the first factor. This contradicts the fact that G has Property ( ∗ ) p and so H must have Property ( ∗ ) p . (cid:3) Finally, we deal with the case where p divides the order of K . Lemma 4.6.
Suppose that Hypothesis 4.3 holds and that p divides | K | . Then G has Property ( ∗ ) p on Ω if and only if | ∆ | is a power of p . We recall from Lemma 2.4(3) that, if | ∆ | is a power of p , then H has Property ( ∗ ) p on ∆. Proof. If | ∆ | is a power of p then so is | Ω | . Hence as G acts transitively on Ω it follows fromLemma 2.4(3) that G has Property ( ∗ ) p .Suppose conversely that G has Property ( ∗ ) p on Ω. Then by Lemma 2.4(5) the transitivesubgroup N k has Property ( ∗ ) p on Ω and hence by Lemma 4.2 the group N has Property ( ∗ ) p on ∆. Let Q be a Sylow p -subgroup of N . Then by Lemma 2.3 all orbits of Q on ∆ have size | ∆ | p and all orbits of Q k ∈ Syl( N k ) on Ω have size | Ω | p = | ∆ p | k . Let R be a Sylow p -subgroupof G containing Q k . We may assume that R P wr S where P is a Sylow p -subgroup of H containing Q and S is a Sylow p -subgroup of K . As G has Property ( ∗ ) p all orbits of R havesize | Ω | p = | ∆ | kp . Choose δ , δ ∈ ∆ such that δ = δ , and consider ω ′ := ( δ , δ , δ , . . . , δ ) ∈ Ω.Then | ( ω ′ ) R | = | ∆ | kp = | ( ω ′ ) Q k | , and hence Q k acts transitively on ( ω ′ ) R . Hence R = Q k R ω ′ andso R { , ,...,k } ω ′ = R { , ,...,k } S . Since p divides | K | we have that R { , ,...,k } = 1 and so we may BAMBERG, BORS, DEVILLERS, GIUDICI, PRAEGER, AND ROYLE assume that there exists g = ( h , h , . . . , h k ) σ ∈ R ω ′ such that 1 σ = 2 and each h i ∈ P . By thedefinition of the action, the image ( ω ′ ) g has second entry δ h , and since g fixes ω ′ it follows that δ h = δ . Since δ , δ were arbitrary distinct points of ∆, we conclude that P acts transitivelyon ∆ and so | ∆ | is a power of p . (cid:3) Lemmas 4.5 and 4.6 can be combined to get the following result.
Theorem 4.7.
Let G be a group satisfying Hypothesis 4.3.(a) If p does not divide | K | , then G has Property ( ∗ ) p on Ω if and only if H has Property ( ∗ ) p on ∆ .(b) If p divides | K | , then G has Property ( ∗ ) p on Ω if and only if | ∆ | is a power of p . Twisted wreath type groups.
We show that primitive groups of Twisted wreath typedo not have Property ( ∗ ) p . Lemma 4.8.
Let G be a primitive group of Twisted wreath type acting on Ω and let p be a primedividing | Ω | . Then G does not have Property ( ∗ ) p .Proof. Suppose that G is primitive of twisted wreath type with socle N = T k . Then N actsregularly on Ω. Thus we can identify Ω with N and we can identify G as the twisted wreathproduct T twr ϕ P for some homomorphism ϕ : Q → Aut( T ) where Q is a subgroup of P of index k . We follow the treatment in [16, Section 6] and let ω ∈ Ω correspond to 1 N . Then G ω = P .By [16, Lemma 6.4] we have that that T k G H wr S k where H = G ∆1 Sym(∆) where∆ = T . Now N P G and so H contains T as a regular normal subgroup. Now ( G ) ω = Q and(( G ) ω ) ∆ = Qϕ . Thus H T ⋊ Aut( T ). By [16, Corollary 6.3] we have that Inn( T ) ϕ ( Q )and so H = G ∆1 is a diagonal group with socle T acting on ∆. By Lemma 4.1, T × T does nothave Property ( ∗ ) p on ∆ and so by Lemma 2.4(5), H does not have Property ( ∗ ) p on ∆. HenceLemma 4.4 implies that G does not have Property ( ∗ ) p on Ω. (cid:3) The following example shows that we cannot replace the primitive assumption in Lemma 4.8with quasiprimitive.
Example 4.9.
Let T be a finite nonabelian simple group with order divisible by a prime p andlet ∆ = T . Then T acts regularly on ∆ with Property ( ∗ ) p . Let k be a positive integer coprimeto p and let G = T wr C k acting in product action on Ω = T k . Then by Lemma 4.2, G hasProperty ( ∗ ) p on Ω. Note that N = T k is the unique minimal normal subgroup of G and actsregularly on Ω. Thus G is quasiprimitive of Twisted wreath type [16, p156].4.4. Proof of Theorem 1.3.
Let G be a primitive permutation group on Ω and suppose that G has Property ( ∗ ) p for some prime p dividing | Ω | . By the O’Nan-Scott Theorem G has exactlyone of the following five types: (i) Affine type, (ii) Almost simple type, (iii) Diagonal type, (iv)Product action type, and (v) Twisted wreath type.By Lemma 4.1, G is not of Diagonal type and by Lemma 4.8, G is not of Twisted wreathtype. Moreover, if G is of Almost simple type then case (1) of the theorem holds.If G is of Affine type then G AGL( d, p ) and | Ω | = p d . Thus by Lemma 2.4(3), G hasProperty ( ∗ ) p but does not have Property ( ∗ ) r for any r = p and r dividing | G | . Hence case (2)of the theorem holds.It remains to consider the case where G is of Product action type. Then following thedescription in Case III(b) of [13] we have that Ω = ∆ k and there is a primitive group H Sym(∆) of Almost simple or Diagonal type and with socle N , such that N k G H wr K for some K S k . Since N is transitive on ∆ it follows that G satifies Hypothesis 4.3. Henceby Lemma 4.4, H has Property ( ∗ ) p on ∆. Thus by Lemma 4.1 H is of Almost simple type.Moreover, by Lemma 4.6, if p divides | K | then | ∆ | is a power of p . Thus case (3) of thetheorem holds. Moreover, by Lemmas 4.5 and 4.6 any group of satisfying these conditions hasProperty ( ∗ ) p . RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 9
Degree
G p
Rank6 PSL (5) 2 212 M (3) 3 436 PΓU (3) 3 3112 PSU (3) 2 3135 PSp (2) 3 4 Table 1.
Some primitive groups G of degree less than 4095 and with property ( ∗ ) p .4.5. Almost-simple primitive groups of low degree.
Note that PSL ( q ) acts on the non-singular conic of the Desarguesian projective plane of order q . This is the natural 2-transitiveaction of PSL ( q ), but for q = 7 ,
9, it also has a (unique) primitive action of degree (cid:0) q (cid:1) , whichis the action on the external lines to the nonsingular conic. In this action, for q even, the pointstabiliser is isomorphic to the dihedral group D q +1) of order 2( q + 1). (Note: there are q + 1tangents to the conic, (cid:0) q +12 (cid:1) secants, and q + q + 1 lines in total.) Lemma 4.10.
The action of
PSL ( q ) , q even, on n = (cid:0) q (cid:1) points has Property ( ∗ ) . Moreover, n < n for q > and the stabiliser of a point has even order.Proof. The stabiliser in G := PSL ( q ) of an external line is isomorphic to the dihedral group D q +1) . Let P be a Sylow 2-subgroup of G . Then P has order q . Indeed, P fixes a uniquepoint X of the conic, and the stabiliser of X in G is of the form P ⋊ C q − . Let ℓ be an externalline. We claim that | G X ∩ G ℓ | = 2, and hence | P ∩ G ℓ | = 2. Let t X be the tangent line at X .Since q is even, t X passes through the nucleus N of the conic. So G X ∩ G ℓ fixes X , N , and thepoint Y := t X ∩ ℓ . Moreover, G X ∩ G ℓ fixes every point of t X (because it fixes three points of t X and preserves cross-ratio). Hence a nontrivial element of G X ∩ G ℓ is the unique elation withaxis t X and centre Y . Therefore, the intersection of P with the stabiliser of any external linehas order 2, and so every P -orbit on external lines has size q/ n . Since n = n ( q −
1) andthe stabiliser of a point is a dihedral group, the result follows. (cid:3)
We remark that PΓL ( q ), where q = 2 f , acting on (cid:0) q (cid:1) points, satisfies ( ∗ ) when f is odd asa Sylow 2-subgroup of PΓL ( q ) is a Sylow 2-subgroup of PSL ( q ).By a straight-forward computation in GAP [8], we have enumerated the primitive groups ofAlmost Simple type with Property ( ∗ ) p of degree n at most 4095 such that pn p < n and p dividesthe order of the stabiliser of a point. All such groups apart from when the socle is PSL ( q ), q even, acting in degree (cid:0) q (cid:1) , are listed in Table 1. It is curious that the largest degree exampleknown that does not have a socle isomorphic to PSL ( q ) has degree 135. Question 4.11.
Are there any Almost simple type primitive permutation groups of degree n >
135 with Property ( ∗ ) p such that pn p < n and p divides the order of the stabiliser of apoint, other than those of degree (cid:0) q (cid:1) and with socle PSL ( q )?5. Proof of Theorem 1.5
We note the following classic result which, together with Tornier’s result [17] for the alternat-ing and symmetric groups, deals completely with the case where n p = p . Theorem 5.1. [15, Theorem]
Let p be a prime and n a positive integer with n p = p . Supposethat G is a -transitive permutation group on n points that does not contain A n and let P be aSylow p -subgroup of G . If G has Property ( ∗ ) p then one of the following holds: (a) p does not divide | G ω | ; (b) | P | = 4 and G = A with n = 6 ; (c) | P | = 9 and G = M with n = 12 ; We need the following three preliminary results. The first concerns primitive prime divisorsand can be found in [10, Lemma 2.5] (for p = 2) and [14, Lemma 4.1(iii)]. Lemma 5.2.
Let q be a prime power and let p be a prime. Suppose that p divides q m − , andlet e = o ( q mod p ) , the least positive integer such that p divides q e − . Then p ≡ e ) and m = eb , for some b , Moreover,(a) if p is odd then ( q m − p = ( q e − p · b p ; and(b) if p = 2 then e = 1 and ( q m − = ( q − · ( m/ if m is even, and ( q − if m is odd. The second is a technical result about semilinear groups.
Lemma 5.3.
Let p and r be distinct primes, f a positive integer divisible by p and d > apositive integer coprime to p such that p divides r df − . Let Σ = GF( r df ) \{ } and let X = h b ξ, φ i be a Sylow p -subgroup of ΓL ( r df ) where b ξ : x ξx for some ξ ∈ GF( r df ) of order ( r df − p and φ : x x r df/fp . Let Y = h b ξ i and σ = φ f p /p . Then(a) p divides r df/p − .(b) The group X has exactly p subgroups of order p that intersect trivially with Y , and each isconjugate to h σ i .(c) | C Y ( σ ) | = ( r df − p /p .(d) σ fixes at most r df/p − one-dimensional GF( r f ) -subspaces of GF( r df ) .(e) Suppose that P X with Y < P and P acts on a set Ω with all orbits having size ( r df − p and Y acting semiregularly. Then h Y, σ i P and σ has | Ω | /p fixed points.Proof. Let e = o ( r mod p ). Then by Lemma 5.2, p ≡ e ) and e divides df . Hence e divides df /p and so p divides r df/p − X of order p that intersect trivially with Y have the form h b ξ j σ i for someinteger j . Since σ − b ξσ = b ξ r df/p we have that( b ξ j σ ) p = b ξ j (1+ x + x + ··· + x p − ) = b ξ j ( x p − / ( x − where x = r df ( p − /p . By Fermat’s Little Theorem, x p − ≡ p ) and so p divides x p − p divides x −
1. Moreover, in this case Lemma 5.2 implies that (( x p − / ( x − p = p . Thus if b ξ j σ has order p then ( r df − p /p divides j and so there are at most p subgroups oforder p of the form h b ξ j σ i in X . Moreover, by Lemma 5.2 and part (a), ( r df − p = ( r df/p − p p and so ξ p ∈ GF( r df/p ), which is centralised by σ . Since | ξ | does not divide r df/p − ξ / ∈ GF( r df/p ). Thus C X ( σ ) = h b ξ p , φ i and so h σ i has p conjugates in X . Hence all subgroups of X of order p and that intersect trivially with Y are conjugate in X and parts (b) and (c) hold.If σ fixes the GF( r f )-subspace spanned by η then η r df/p = λη for some λ ∈ GF( r f ). Hence η ( r df/p − r f − = 1 and so there are at most ( r df/p − r f −
1) such elements η . As each suchone-dimensional subspace contains r f − r df/p − r f )-subspaces fixed by σ . Thus (d) holds.Suppose now that P X with Y < P and P acts on a set Ω with all orbits having size ( r df − p and Y acting semiregularly. Since | Y | = ( r df − p , this means that Y is transitive (regular) oneach of the P -orbits in Ω. Then for all ω ∈ Ω we have that | P ω | = | P | / ( r df − p = | P : Y | > p .Hence P ω contains a subgroup of order p that intersects trivially with Y and so by part (b) wemay assume that h Y, σ i P and for each orbit O of Y there exists ω ∈ O such that h σ i P ω .Moreover, the elements of O can be identified with the elements of Y such that ω correspondsto 1 and σ acts on O as an automorphism of Y . Hence by (b), σ fixes ( r df − p /p elements of O and so fixes | Ω | /p elements of Ω as in part (e). (cid:3) The third preliminary result is the following number theoretic lemma.
Lemma 5.4.
Suppose p and r are prime numbers, with p odd, and let f be an integer at least , such that p divides f and r f + 1 . Then r f + 1 p > r f/p − , RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 11 with equality if and only if ( r, p, f ) = (2 , , .Proof. The proof is by induction on f . If f = 3, then p = 3 and, for all p[rimes r > r f + 1 p = r + 13 > r − r f/p − . Suppose now that the result holds for some f >
3. Then r f +1 + 1 p = r · r f + 1 p + 1 p (1 − r ) > r · ( r f/p −
1) + 1 p (1 − r )= r f +1) /p − r f/p ( r − r /p ) + (1 − r )(1 + 1 /p ) . We will show that r f/p ( r − r /p ) + (1 − r )(1 + 1 /p ) > . By elementary calculus, the function x ( x − x / ) − ( x −
1) is positive except for values of x inthe interval [1 , ε ] where ε ≈ . r >
2, we have r ( r − r / ) > ( r −
1) and hence r f/p ( r − r /p ) >
43 ( r − > ( r − /p ) . Therefore, r f/p ( r − r /p ) + (1 − r )(1 + 1 /p ) > f + 1, and hence,by induction, for all f > r f +1 p = r f/p −
1. Let q := r f/p . Then q − q p + 1 and hence q = 2. So r = 2 and f = p . Therefore, 2 p + 1 = 3 p . By Fermat’s littletheorem 2 p ≡ p ) and so 0 ≡ p (mod p ) ≡ p + 1 (mod p ) ≡ p ). Hence p = 3. (cid:3) We now prove Theorem 1.5.
Proof of Theorem 1.5.
Let p be a prime such that G is a 2-transitive group with Property ( ∗ ) p of degree n . By Burnside’s Theorem [6, Theorem 4.1B], G is either affine or almost simple. If G is affine then n = r k for some prime r , and by Lemma 2.4(3) G has Property ( ∗ ) p if and onlyif p = r . Thus pn p > n and so (a) holds. Hence it remains to consider the case where G isalmost simple. The almost simple 2-transitive permutation groups are listed in [5, Table 7.4].If soc( G ) = A n then by Tornier [17] we have that pn p > n and again case (a) holds. Also, if n p = p then one of (b), (c) or (d) holds by Theorem 5.1, so we also assume that n p > p . Wenow work through the remaining possibilities for soc( G ).(1) soc( G ) = PSL d ( q ) with n = q d − q − and d > : Suppose first that soc( G ) = PSL d ( q ) and let T = soc( G ). By Lemma 2.4(5), T alsohas Property ( ∗ ) p , so we will assume to start with that G = T . If p n p > n then case (a)holds (for both G and T ), so we may assume that n = n p c with c > p (since c = p bythe definition of n p ). (We will try not to use case (b) since it may hold for T but not for G .) In particular n p > p and p divides ( q d − / ( q − e = o ( q mod p ), the leastpositive integer such that p divides q e −
1. Let ( q e − p = p a . By Lemma 5.2, d = eb ,for some b , and if p is odd then ( q d − p = p a · b p , while if p = 2 then ( q d − p is givenby Lemma 5.2(b). Note that, if e > p is odd, for if p = 2 divides q d − q isodd and so e = 1. • Case both b, e > . As we just noted, p is odd, and since in this case p doesnot divide q −
1, we will argue in GL d ( q ) acting on the set of 1-spaces of theunderlying vector space V = F dq . Then P preserves a decomposition V = ⊕ bi =1 U i with dim( U i ) = e for each i , and P contains Q bi =1 P i with P i = C p a a Sylow p -subgroup of GL( U i ). Now P i acts semiregularly on the nonzero vectors of U i .Choose a nonzero u i ∈ U i for each i , and set u = P bi =1 u i . Then the P -orbit on h u i has size at least p ab . On the other hand n p = ( q d − p (since e > p a .b p which is strictly less than p ab since p is odd, acontradiction to Lemma 2.3. • Case b > but e = 1 . Here d = b and the pre-image e P of P in SL d ( q ) has order p a ( d − and acts on the set of 1-spaces of V with kernel of order gcd( d, ( q − p ) =gcd( d, p a ). There is a e P -invariant decomposition V = ⊕ di =1 U i with this time eachdim( U i ) = 1. Choosing u i and u as in the previous case we obtain an orbit of e P onnonzero vectors of size at least p a ( d − , and hence a P -orbit on 1-spaces of length atleast p a ( d − / gcd( d, p a ) > p a ( d − . This time n p = ( q d − p / ( q − p = ( q d − p · p − a (since e = 1). Since n p > n p = d p if p is odd,while if p = 2 then d is even and n = ( q + 1) · ( d/ . Suppose first that either p is odd, or p = 2 and q ≡ n p = d p . If d > d = 3 and p > p a ( d − > d > d p = n p , which is a contradiction to Lemma 2.3. If d = 3 and p = 2 then p a ( d − / gcd( d, p a ) = 2 a > d = 1, another contradiction. Similarly,if d = 2 and p is odd then p a ( d − / gcd( d, p a ) = p a > d p = 1 while if d = 2 and q ≡ p a ( d − / gcd( d, p a ) = 2 a − > > d = 2 and we again get acontradiction. When d = p = 2 and q ≡ n = 2, whichcontradicts the fact that n p > p .It remains to consider the case where d is even, p = 2 and q ≡ n = ( q + 1) · ( d/ = x · ( d/ , where x = ( q − . For this case we observethat e P contains a subgroup Q of index 2 in Q d/ i =1 Q i with each Q i ∼ = C x , and Q leaves invariant a decomposition V = ⊕ d/ i =1 W i with each W i of dimension 2. Theusual argument produces a Q -orbit on 1-spaces of size at least x d/ . This is strictlygreater than n if d >
2, contradicting Lemma 2.3, so we are left with d = 2 and q ≡ q = r f for some odd prime r , and f must alsobe odd. Hence | Out( T ) | = 2 f . If | G/T | is odd, then | G | = n so | G ω | is oddand case (b) holds. So we may assume that G contains PGL ( q ) and hence that | P | > ( q − = 2 n . By Lemma 2.4(5), H = PGL ( q ) has Property ( ∗ ) also.As we are assuming that n > n we have that n = q + 1 is not a power of 2. Nowa Sylow 2-subgroup of PGL ( q ) is a dihedral subgroup of D q +1) , and has at leastone orbit of each length n and 2 n , contradicting Lemma 2.3. • Case b = 1 and so e = d . The remaining case is b = 1 , e = d , so p is a primitiveprime divisor of q d −
1. Since p does not divide q − d ( q ).Let q = r f with r prime and f >
1. Then the p -part of | Out( T ) | is f p . If p does not divide | G/T | then | G ω | is not divisible by p and we have case (b). Sowe may assume that f p > p divides | G/T | . Moreover, we may assumethat ( r, f, d ) = (2 , ,
2) as in this case n = 9 and hence p = 3 and n = p , so weare in case (a). Furthermore, since p ≡ e ) we have that p is coprime to d . Thus p, r, f and d satisfy the hypotheses of Lemma 5.3 and we consider V asGF( q d ). Moreover, since G has Property ( ∗ ) p we may assume that P ΓL ( r df )and satisfies the hypotheses on the group P in Lemma 5.3(e) with Y = P ∩ GL d ( q ).In particular, σ , as defined in the statement of Lemma 5.3, lies in P and has n/p fixed points on Ω. However, by Lemma 5.3(d), σ has at most r df/p − d = 2 then n = r f + 1 and so this contradicts Lemma 5.4. Thus d > p > d it follows that r df/p − < q and since p divides r df/p − p < q . Hence n = q d − + q d − + · · · + q + 1 > pq , andso n/p > q > r df/p −
1. Thus we again get a contradiction to the fact that σ has n/p fixed points.(2) soc( G ) = PSU ( q ) with n = q + 1 : RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 13
Suppose that soc( G ) = PSU ( q ) and let T = soc( G ) = PSU ( q ). Then G PΓU ( q )with q = r f . Let V be a 3-dimensional vector space over GF( q ) equipped with a G -invariant nondegenerate hermitian form B . Moreover, Ω is the set of totally isotropic1-subspaces of V with respect to the form B . Now | PSU ( q ) | = ,q +1) q ( q − q + 1)and PΓU ( q ) = q ( q − q + 1)2 f . Suppose that we are not in case (b) and so p divides | G ω | . Then p divides ( q − f . Note that if p = 2 then q is odd and p divides q − p divides q − p divides f . • Case p divides q − . Nowgcd( q + 1 , q −
1) = gcd(( q + 1)( q − q + 1) , q − q + 1)( q −
1) + ( q + 1)( − q + 2) , q − q + 1) gcd( − q + 2 , q − q + 1and so p divides q + 1. Thus by [4, Lemma A.4] we have n p = (cid:26) ( q + 1) p if p = 33( q + 1) if p = 3.Let v , v , v be an orthonormal basis for V and consider the subgroup X of PGU ( q )whose preimage in GU ( q ) is e X = λ λ
00 0 λ | λ q +1 i = 1 . Then | X | = ( q + 1) . Consider the set Σ of 1-dimensional subspaces of V of theform h a v + a v + v i for a , a ∈ GF( q ). If g = λ λ
00 0 λ ∈ e X then h a v + a v + v i g = h λ λ − a v + λ λ − a v + v i . Thus g fixes an element ofΣ precisely if λ = λ = λ , in which case g corresponds to the identity element of X . Hence X acts semiregularly on Σ. Now B ( a v + a v + v , a v + a v + v ) =1 + a q +11 + a q +12 . Since − − a q +12 ∈ GF( q ) for all a ∈ GF( q ), it follows that for any a there exists a ∈ GF( q ) such that a q +11 = − − a q +12 and so Σ contains totallyisotropic 1-subspaces. In particular, X has a regular orbit on Ω.Let Q be the unique Sylow p -subgroup of X ∩ T and let P be a Sylow p -subgroupof G containing Q . Note that the preimage of X ∩ T in e X consists of all elementsof e X such that λ λ λ = 1. Suppose first that p = 3. Then | Q | = ( q + 1) p .Since X has a regular orbit on Ω it follows that P has an orbit of size at least(( q + 1) p ) > n p = ( q + 1) p , contradicting Property ( ∗ ) p . Thus p = 3. Then | Q | = ( q + 1) /
3. Let σ ∈ PSU ( q ) be the element of order 3 whose preimagein SU ( q ) cyclically permutes v , v and v . Then σ normalises X and Q . Inparticular, b Q = h Q, σ i is a 3-subgroup of T of order ( q + 1) , and we may assumethat b Q P . As discussed in the previous paragraph, there exists a ∈ GF( q ) suchthat a q +11 = − ω = h a v + v + v i is a totally singular subspace in Σ.Now if h b v + b v + v i ∈ ω Q then b q +12 = 1. However, ω σ = h v + a v + v i and a q +11 = − = 1. Hence ω Q ⊂ ω b Q and so | ω b Q | = ( q + 1) . If ( q + 1) > P has an orbit of length greater than n = 3( q + 1) , contradicting G having Property ( ∗ ) . Thus ( q + 1) = 3. Since 3 divides | G ω | it follows that eitherPGU ( q ) G or 3 divides f . In the first case, taking Q to be the unique Sylow3-subgroup of X and b Q = h Q, σ i , we again see that | ω b Q | = | b Q | = ( q + 1) > n ,contradicting Property ( ∗ ) . Finally, if 3 divides f and q + 1 = r f + 1, note that 3 does not divide r f − r −
1. Hence 3 divides r + 1 and soby [4, Lemma A.4], ( q + 1) = ( r + 1) f >
3, contradicting our assumption that( q + 1) = 3. • Case p divides f but not q − . Recall that p divides n = q + 1. Also, inthis case p = 3 as (3 , q ) = 1 and so 3 divides either q − q + 1. In particular, aSylow p -subgroup of PGU ( q ) is contained in PSU ( q ). Moreover, as p divides | G ω | we have that p divides | G : PSU ( q ) | . Let V = GF( q ), treated as a vector spaceover GF( q ). Then the map B ( v, w ) = Tr q → q ( vw q ) is a nondegenerate hermitianform on V and the totally singular vectors are all the elements v ∈ GF( q ) suchthat Tr q → q ( v q +1 ) = 0. Letting d = 6 we have that p, r, f and d satisfy thehypotheses of Lemma 5.3. Moreover, as G satisfies Property ( ∗ ) p we have that P isa subgroup of ΓL ( r f ) and satisfies the hypotheses of part Lemma 5.3(e) with Y being the Sylow p -subgroup of PSU ( q ). In particular, σ , as defined in the statementof Lemma 5.3, lies in P and has at most n/p fixed points on Ω. However, p < q and so n/p = ( q + 1) /p > q while by Lemma 5.3(d) we have that σ fixes at most r f/p − < q elements of Ω, a contradiction.(3) soc( G ) = Ree( q ) with n = q + 1 : Suppose that soc( G ) = Ree( q ) and let T = Ree( q ).In this case we have q = 3 f for some odd integer f >
3. Now | T | = q ( q + 1)( q −
1) and | Out( T ) | = f . Now gcd( q + 1 , q −
1) = 2. Thus if (b) does not hold then there are twocases to consider: p = 2, or p divides f and | G : T | is divisible by p . Note that p = 3.Let P be a Sylow p -subgroup of G . • Case p = 2 : Note that since q = 3 f for f odd, [4, Lemma A.4] implies that( q + 1) = ( q + 1) = 4 and ( q − = 2. Thus P T and | P | = 8. Now P is elementary abelian, all involutions in G are conjugate and the centraliser of aninvolution in T is C × PSL ( q ), see [12]. Thus T and hence G contains q ( q − q + 1)involutions. Moreover, given a point ω ∈ Ω, we have that T ω = [ q ] : C q − where[ q ] denotes a subgroup of order q . Since q does not divide the centraliser of aninvolution, it follows that T ω and hence G ω contains q involutions. Hence doublecounting shows that each involution in G fixes q + 1 points. Let O be an orbit of P of length 4. Then there is a unique involution in P that fixes O pointwise, andthe remaining involutions are derangements on O . Since P contains 7 involutionsit follows that P has at most 7( q + 1) / q > n/ > q + 1) / G does not have Property ( ∗ ) ,a contradiction. It is interesting to note that in the excluded case when q = 3 wehave that G = Ree(3) = PΓL (8) in its action on 28 points. Here n/ P indeed has 7 orbits of length 4 and G has Property ( ∗ ) , as in part (e). • Case p divides f : In this case we have p > p is a primitive prime divisorof q − q −
1. In both cases, | Aut( T ) | p = n p f p and a Sylow p -subgroup of T acts semiregularly on Ω. Now Ree( q ) GL ( q ) and Ω can be identified with a setof 1-dimensional subspaces of V = GF( q ) of size q + 1.Suppose first that p is a primitive prime divisor of q −
1. Then p ≡ p >
7. Letting d = 6 and r = 3 we have that p, r, f and d satisfy thehypotheses of Lemma 5.3. As G satisfies Property ( ∗ ) p we have that P is a subgroupof ΓL ( r f ) that satisfies the hypotheses of part Lemma 5.3(e) with Y being theSylow p -subgroup of Ree( q ). In particular, σ , as defined in the statement of Lemma5.3 lies in P and has at most n/p fixed points on Ω. However, p < q and so n/p =( q + 1) /p > q while by Lemma 5.3(d) we have that σ fixes at most r f/p − < q elements of Ω, a contradiction. Thus p is a primitive prime divisor of q − T has a maximal subgroup H = C × PSL ( q ) which contains a Sylow p -subgroup of T . Moreover, as the centraliser of a field automorphism of order p is Ree(3 f/p ), field automorphisms of Ree( q ) of order a power of p normalising H induce field automorphisms on PSL ( q ). Hence letting d = 2 and r = 3, we can RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 15 view P as a subgroup of ΓL ( r f ) that satisfies the hypotheses of part Lemma 5.3(e)with Y being a Sylow p -subgroup of Ree( q ) contained in H . Hence σ , as definedin the statement of Lemma 5.3, lies in P and has at most n/p fixed points on Ω.Note that n/p = ( q + 1) /p > q as p < q . By [9, Proposition 4.91(d)], subgroupsof Aut( T ) of order p and which intersect T trivially are all conjugate. Hence thenumber of fixed points of σ on Ω is the same as the number of fixed points as theelement σ in the previous paragraph. However, this number is at most r f/p − < q ,yielding a final contradiction.(4) soc( G ) = Sz( q ) with n = q + 1 : Suppose that soc( G ) = Sz( q ) and let T = Sz( q ). Inthis case we have q = 2 f for some odd integer f >
3. If p divides n then p divides q − p is a primitive prime divisor of q − q − q −
1. Since q is even, it is notpossible for p to divide q + 1 and either q − q −
1. Hence p is a primitive primedivisor of q −
1. Now | G ω | divides q ( q − f . Thus either (b) holds, or p divides f and | G : T | is divisible by p . We consider the latter case. Letting d = 4 and r = 2then d , r , f and p satisfy the hypotheses of Lemma 5.3. Now Sz( q ) Sp ( q ) and Ωcan be identified with a set of 1-dimensional subspaces of V = GF( q ) of size q + 1.As G satisfies Property ( ∗ ) p we have that P is a subgroup of ΓL ( r f ) that satisfiesthe hypotheses of part Lemma 5.3(e) with Y being the Sylow p -subgroup of Sz( q ). Inparticular, σ , as defined in the statement of Lemma 5.3, lies in P and has at most n/p fixed points on Ω. However, p < q and so n/p = ( q + 1) /p > q while by Lemma 5.3(d)we have that σ fixes at most r f/p − < q elements of Ω, a contradiction.(5) soc( G ) = Sp d (2) with n = 2 d − ± d − and d > : Suppose that soc( G ) = Sp d (2).Then G = soc( G ). In this case, for ǫ ∈ {±} there are sets Ω ǫ with | Ω ǫ | = 2 d − + ǫ d − such that G acts 2-transitively on each Ω ǫ .Let V be a 2 d -dimensional vector space over GF(2) and let B be a nondegeneratealternating form on V preserved by G . We say that a quadratic form Q : V → GF(2) polarises to B if B ( v, w ) = Q ( v ) + Q ( w ) − Q ( v + w ) for all v, w ∈ V . Then G acts onthe set of all such quadratic forms via Q g ( v ) = Q ( v g − ) for all v ∈ V, g ∈ G . Moreover, G has two orbits, namely Ω + and Ω − where Ω + is the set of all hyperbolic quadraticforms polarising to B and Ω − is the set of all elliptic quadratic forms polarising to B .In particular, if Q ∈ Ω + then G Q ∼ = GO +2 d (2) while if Q ∈ Ω + then G Q ∼ = GO − d (2).Let g ∈ G and let Q ∈ Ω + ∪ Ω − such that Q g = Q . Then following [7] and notingthat since x = x for all x ∈ GF(2), the elements of Ω + ∪ Ω − are precisely the quadraticforms Q a , for a ∈ V given by Q a ( v ) = Q ( v ) + B ( a, v ) . Moreover, ( Q a ) g ( v ) = Q a ( v g − ) = Q ( v g − ) + B ( a, v g − ) = Q ( v ) + B ( a g , v ) as g preserves Q and B . Now GO ǫ d (2) has two orbits on the nonzero vectors of V and these havesize 2 d − (2 d − ǫ ) (the nonsingular vectors) and (2 d − ǫ )(2 d − + ǫ ) (the singular vectors).Suppose that GO ǫ d (2) fixes Q . Then the action of GO ǫ d (2) on V is equivalent to theaction of GO ǫ d (2) on Ω + ∪ Ω − . Thus the quadratic forms Q a lying in the same G -orbitas Q are those of the form Q a such that Q ( a ) = 0. In particular, if H GO ǫ d (2) G ,then the orbits of H on the set of singular vectors of Q correspond to the orbits of H on Ω ǫ \{ Q } . Let e , f ∈ V such that Q ( e ) = Q ( f ) = 0 and B ( e , f ) = 1. By [11,4.1.20], GO ǫ (2) e = b P ⋊ GO ǫ d − (2) where | b P | = 2 d − and acts trivially on h e , f i ⊥ .Thus b P f acts trivially on V and so | f b P | = | b P | = 2 d − . Since Q ( f ) = 0 we have that Q f ∈ Ω ǫ and so b P has an orbit of length 2 d − > n = 2 d on Ω ǫ . (Recall that d > p = 2.To deal with the p odd case we need to set up some more notation. Let V = V ⊥ V ⊥ . . . ⊥ V k be an orthogonal decomposition of V such that B i = B | V i × V i is a nondegeneratealternating form. Let Q i : V i → GF(2) be a nondegenerate quadratic form on V i thatpolarises to B i such that sign( Q i ) = ǫ i . For each v ∈ V write v = v + v + · · · + v k where each v i ∈ V i and define ( Q ⊕ Q ⊕ · · · ⊕ Q k )( v ) = Q ( v ) + Q ( v ) + · · · + Q ( v k ).Then Q = Q ⊕ Q ⊕ · · · ⊕ Q k is a nondegenerate quadratic form that polarises to B and Q ∈ Ω ǫ for ǫ = Π ki =1 ǫ i . Moreover, for g i ∈ Sp( V i ) we have that g = ( g , g , . . . , g k ) ∈ G and Q g = Q g ⊕ Q g ⊕ · · · ⊕ Q g k k ∈ Ω ǫ . See [1] for more details.Suppose first that G has Property ( ∗ ) p on Ω + with p odd. Since n = 2 d − (2 d + 1)we have that p divides 2 d + 1 and hence divides 2 d −
1. Let e = o (2 (mod p )) and let( q e − p = p a . Then by Lemma 5.2, 2 d = eb and ( q d − p = p a b p . If e = 2 d then n p = | G p | and so case (b) holds. Thus suppose that e < d . If e divides d then p divides2 d −
1, which is coprime to n . Thus e is even. Let V = V ⊥ V ⊥ . . . ⊥ V b where each V i is a nondegenerate subspace of dimension e and let B i be the restriction of B to V i × V i .Then Sp( V i ) acts transitively on the set Ω + i of 2 e/ − (2 e/ +1) hyperbolic quadratic formson V i that polarise to B i . Let P i ∈ Syl p (Sp( V i )). By Lemma 2.2 there exists a hyperbolicquadratic form Q i ∈ Ω + i such that | Q P i i | = p a . Then Q := Q ⊕ Q ⊕ · · · ⊕ Q b ∈ Ω + .Moreover, letting b P = Π bi =1 P i we see that | Q b P | = p ab > n p = p a b p . Thus G has a Sylow p -subgroup with an orbit of length greater than n p , contradicting Lemma 2.3. Thus G does not have Property ( ∗ ) p on Ω + .Next suppose that G has Property ( ∗ ) p on Ω − with p odd. Then n = 2 d − (2 d − p divides 2 d −
1. Let e = o (2 (mod p )) and let ( q e − p = p a . Then by Lemma5.2, d = eb and ( q d − p = p a b p . Recall that dim( V ) = 2 d = 2 eb in this case. Supposefirst that e = d and d is odd. Then n p = | G | p and so case (b) holds. Next suppose that e is even. Let V = V ⊥ V ⊥ . . . ⊥ V b where each V i is a nondegenerate subspace ofdimension e and let B i be the restriction of B to V i × V i . For i = 1 , . . . , b −
1, Sp( V i )acts transitively on the set Ω + i of 2 e/ − (2 e/ + 1) hyperbolic quadratic forms on V i thatpolarise to B i . Note that (2 e/ + 1) p = p a . Let P i ∈ Syl p (Sp( V i )). Then by Lemma 2.2there exists a hyperbolic quadratic form Q i ∈ Ω + i such that | Q P i i | = p a . Finally, notethat Sp( V b ) acts transitively on the set Ω − b of 2 e/ − (2 e/ −
1) hyperbolic quadraticforms on V b that polarise to B b . Let P b ∈ Syl(Sp( V b )) and choose Q b ∈ Ω − b suchthat | Q P b b | > p . Then Q := Q ⊕ Q ⊕ · · · ⊕ Q b ∈ Ω − . Moreover, letting b P = Π bi =1 P i we see that | Q b P | > p a (2 b − p > n p = p a b p . Thus G has a Sylow p -subgroup with anorbit of length greater than n p , contradicting Lemma 2.3. Next suppose that e < d and e is odd. Note that b >
2. In this case, let V = V ⊥ V ⊥ . . . ⊥ V b where each V i is a nondegenerate subspace of dimension 2 e and let B i be the restriction of B to V i × V i . For i = 1 , . . . , b −
1, Sp( V i ) acts transitively on the set Ω + i of 2 e − (2 e −
1) ellipticquadratic forms on V i that polarise to B i . Let P i ∈ Syl p (Sp( V i )). Then by Lemma2.2 there exists a hyperbolic quadratic form Q i ∈ Ω − i such that | Q P i i | = p a . Finally,choose ǫ b so that ( − b − ǫ b = − . Let P b ∈ Syl(Sp( V b )) and choose Q b ∈ Ω ǫ b b such that | Q P b b | > p . Then Q := Q ⊕ Q ⊕ · · · ⊕ Q b ∈ Ω − . Moreover, letting b P = Π bi =1 P i we seethat | Q b P | > p a ( b − p > n p = p a b p , since p >
3. Thus G has a Sylow p -subgroup withan orbit of length greater than n p , contradicting Lemma 2.3. Thus G does not haveProperty ( ∗ ) p on Ω − .(6) Others
The isolated examples of 2-transitive groups were dealt with by computer. First,PSL(2 ,
11) on 11 points, M on 11 points, and M on 23 points have prime degree andthe result follows from Lemma 2.4(3). For the remainder, see Table 2 for details.6. Low degree transitive groups
This section contains computational results on which transitive groups of low degree haveProperty ( ∗ ) or Property ( ∗ ) . This gives an overview of how the theoretical results above arerealized in practice. Table 3 gives summary data for the transitive groups of degrees at most 47.In this table, t ( n ) denotes the number of transitive groups of degree n . Note that by Lemma 2.4, RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 17 n G p
Orbit lengths of a Sylow p -subgroup12 M , M , , A , M , , M : 2 2 2 , , M , ,
28 PΓL(2 ,
8) 2 4
176 HS 2 16 , ,
276 Co , , , , , , , Table 2.
Sporadic almost simple 2-transitive actions. We only consider theprimes p such that pn p < n in each case.a transitive group G of odd degree has Property ( ∗ ) precisely when G has odd order. Similarly,a transitive group of degree n with n coprime to 3 has Property ( ∗ ) precisely when the orderof G is coprime to 3. Moreover, by Lemma 2.4(4), if pn p > n then all transitive permutationgroups of degree n have Property ( ∗ ) p . n t ( n ) ( ∗ ) ( ∗ ) n t ( n ) ( ∗ ) ( ∗ ) Table 3.
Counts of low-degree transitive groups with Property ( ∗ )
28 BAMBERG, BORS, DEVILLERS, GIUDICI, PRAEGER, AND ROYLE
Maximal ( ∗ ) groups. As Property ( ∗ ) p is closed under taking transitive subgroups oftransitive groups, another characterization of the transitive groups with Property ( ∗ ) p is to listthe maximal transitive groups with Property ( ∗ ) p . Then a transitive group has Property ( ∗ ) p ifand only if it is a subgroup of one of the listed maximal transitive groups with Property ( ∗ ) p .To illustrate this, Table 4 gives the maximal transitive groups with Property ( ∗ ) of de-grees 2 n
39. Most of these groups are wreath products of primitive groups and theirstructure can be given explicitly. The few exceptions are given by their indices in the lists oftransitive groups in
Magma [2] and
GAP [8]. So 20 is the group returned by the expression TransitiveGroup(20,89) in either
Magma or GAP .We note that for n = 2 k , every transitive group has Property ( ∗ ) and hence the uniquemaximal transitive ( ∗ ) is the symmetric group S k . Every transitive group of odd order hasodd degree and has Property ( ∗ ) . Hence for each odd degree, the table simply lists the maximaltransitive groups of odd order.Degree Structures of the maximal ( ∗ ) groups2 23 34 S , S , , S , PSL(2 ,
5) wr 2, S wr 313 13 : 314 (7 : 3) wr 2, 2 wr (7 : 3)15 5 wr 3, 3 wr 516 S
17 1718 3 wr 3 wr 2, 3 wr 2 wr 3, 2 wr 3 wr 3, 3 wr PSL(2 , ,
5) wr 319 19 : 920 20 , , S , 2 wr 5 wr 2, S wr 521 (7 : 3) wr 3, 3 wr (7 : 3)22 (11 : 5) wr 2, 2 wr (11 : 5)23 23 : 1124 PSL(2 , ,
11) wr 2, 2 wr PSL(2 , S , 2 wr PSL(2 ,
5) wr 2,3 wr S , PSL(2 ,
5) wr S , S wr 3 wr 2, S wr PSL(2 , S wr 325 5 : 3, 5 wr 526 (13 : 3) wr 2, 2 wr (13 : 3)27 3 : 13 .
3, 3 wr 3 wr 328 28 , , ,
27) : 3, (7 : 3) wr S S wr (7 : 3)29 29 : 730 5 wr 3 wr 2, 5 wr 2 wr 3, 5 wr PSL(2 , ,
5) wr 531 31 : 1532 S
33 (11 : 5) wr 3, 3 wr (11 : 5)34 17 wr 2, 2 wr 1735 5 wr (7 : 3), (7 : 3) wr 5
RBITS OF SYLOW SUBGROUPS OF FINITE PERMUTATION GROUPS 19
Degree Structures of the maximal ( ∗ ) groups36 3 wr PSL(2 , ,
11) wr 3, 3 wr 3 wr S , 3 wr 2 wr PSL(2 , ,
5) wr 2, 2 wr 3 wr 2 wr 3, 2 wr 3 wr PSL(2 , S wr 3, 2 wr PSL(2 ,
5) wr 3, PSL(2 ,
5) wr 3 wr 2,PSL(2 ,
5) wr 2 wr 3, PSL(2 ,
5) wr PSL(2 , S wr 3 wr 337 37 : 938 (19 : 9) wr 239 (13 : 3) wr 3, 3 wr (13 : 3)Table 4: Structure of maximal ( ∗ ) groupsMany of these maximal groups are wreath products involving maximal ( ∗ ) -groups of smallerdegrees. By Lemma 3.1, if a n and b m have Property ( ∗ ) then a n wr b m has Property ( ∗ ) in itsimprimitive action of degree ab . However, even if a n and b m are maximal ( ∗ ) -groups of degree a and b respectively, it does not necessarily follow that a n wr b m is a maximal ( ∗ ) group in S ab . The smallest example where this occurs is (3 wr 2) wr 2 which has Property ( ∗ ) , but isnot maximal. By associativity of the wreath product (3 wr 2) wr 2 = 3 wr (2 wr 2), but 2 wr 2is not maximal, as it is contained in S .In some sense the most interesting maximal transitive ( ∗ ) groups are the imprimitive onesthat are not wreath products as these are examples where the permutation group induced onblocks does not have Property ( ∗ ) . This occurs just twice in the table — the group 20 is a subgroup of the group 2 wr PSL(2 , is a subgroup of the group2 wr PSL(2 , ,
9) or PSL(2 ,
13) have Property ( ∗ ) . Acknowledgements.
The work on this paper began at the 2020 annual research retreat of theCentre for the Mathematics of Symmetry and Computation. The authors thank the participantsfor the enjoyable and stimulating environment.
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All but second author: Centre for the Mathematics of Symmetry and Computation, Depart-ment of Mathematics and Statistics, The University of Western Australia, Crawley, WA 6009,Australia.Second author: School of Mathematics and Statistics, Carleton University, 1125 Colonel ByDrive, Ottawa ON K1S 5B6, Canada.
Email address : All but second author: [email protected]
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