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# Characteristically Simple Beauville Groups, I: Cartesian Powers of Alternating Groups

Gareth A. Jones

ABSTRACT. A Beauville surface (of unmixed type) is a complex algebraic sur­face which is the quotient of the product of two curves of genus at least 2 by a finite group G acting freely on the product, where G preserves the two curves and their quotients by G are isomorphic to the projective line, ramified over three points. Such a group G is called a Beauville group. We show that if a characterstically simple group G is a cartesian power of a finite alternating group, then G is a Beauville group if and only if it has two generators and is not isomorphic to A5.

1. Introduction

A Beauville surface 5 of unmixed type is a complex algebraic surface which is isogenous to a higher product (i. e. 5 = (C1 × C2)/G, where C1 and C2 are complex algebraic curves (compact Riemann surfaces) of genera GI 2 and G is a finite group acting freely on their product), and is rigid in the sense that G preserves the factors CI, with CI/G = P1 (C) (i = 1, 2) and the induced covering βI : CI P1(C) ramified over three points. (We will not consider Beauville surfaces of mixed type, where elements of G transpose the factors CI.) Since the first examples, with C1 = C2 Fermat curves, were introduced by Beauville [4, p. 159], these surfaces have been intensively studied by geometers such as Bauer, Catanese and Grunewald [2,3,6].

Group theorists have recently considered which groups G, called Beauville groups, and in particular which non-abelian finite simple groups, arise in this way [1,11,12,1419] ; see [23] for a survey. One easily shows that the smallest such group, the alternating group A5, is not a Beauville group. Bauer, Catanese and Grunewald [2] conjectured that all other non-abelian finite simple groups are Beauville groups. This was proved with finitely many possible exceptions by Gar – ion, Larsen and Lubotzky [18], proved completely by Guralnick and Malle [19], and extended to quasisimple groups (with A5 and A5 = SL2 (5) as the only ex­ceptions) by Fairbairn, Magaard and Parker [12]. Our aim here is to add to the catalogue of known Beauville groups by extending these results to some families of characteristically simple groups.

A finite group G is characteristically simple (that is, it has no characteristic subgroups other than G and 1) if and only if it is isomorphic to a cartesian power HK

2010 Mathematics Subject Classification. Primary 20D06; Secondary 14J29, 20B35, 30F10. Key words and phrases. Beauville group, Beauville surface, alternating group.

(£)2015 American Mathematical Society

Of a simple group [21, Satz I.9.12]. (Finiteness is essential here: for instance, the ad­ditive group of any division ring is characteristically simple, since the multiplicative group acts transitively on the non-zero elements.) The abelian characteristically simple finite groups are the elementary abelian P-groups G = Cp;, where P is prime. Catanese [6] has shown that an abelian group G is a Beauville group if and only if G = CN for some N coprime to 6, so the abelian characteristically simple Beauville groups are the groups CP for primes P ≥ 5. We therefore restrict our attention to cartesian powers G = HK of non-abelian finite simple groups H. Since simple groups have been dealt with, we may assume that K ≥ 2.

The ramification condition on the coverings βI implies that a Beauville group is a quotient of two triangle groups ∆I, so in particular it is a 2-generator group. For any finite group H and any integer N ≥ 1, there are only finitely many normal subgroups N in the free group FN of rank N with FN/N = H, so there is an integer CN(H) 0 such that HK is an N-generator group if and only if K ≤ cN(H). Then C2(H) is an upper bound on the values of K such that HK is a Beauville group.

Computing CN(H) can be difficult, even for N = 2, but if H is a non-abelian finite simple group then CN(H) is equal to the number DN(H) of orbits of Aut H On ordered N-tuples which generate H . In some small cases, methods introduced by P. Hall [20] allow DN(H) to be computed by hand: thus C2(A5) = D2(A5) = 19, for example. More recently, Dixon [9] and Maroti and Tamburini [31] have given asymptotic estimates and inequalities for D2(AN).

THEOREM 1.1. Let G = HK, where H = AN for n ≥ 5, and k ≥ 1. Then the following are equivalent:

G is a Beauville group;

G is a 2-generator group not isomorphic to A5;

N = 5 And 2 ≤ k ≤ d2(H) (= 19), or n > 5 And k ≤ d2(H).

It follows from Theorem 1.1 That a cartesian product of alternating groups of Arbitrary degrees N ≥ 5 is a Beauville group if and only if it is a 2-generator group (i. e. no AN appears as a direct factor more than D2 (AN) times) and is not isomorphic to A5; the only troublesome case is when A5 appears just once, and this can be dealt with by the methods developed here. Beauville structures in alternating groups have also been studied by Garion and Penegini in [17]; by a simple argument using [1, Lemma 3], their Theorem 1.2 yields an asymptotic version of this extension, namely that for each K there exists an integer N0(K) such that ANI × ∙ ∙ ∙ × ANK is a Beauville group provided each NI > n0(k). I am grateful to the referee for pointing this out.

The analogue of Theorem 1.1 Is also true when H is a simple group L2 (q), a Suzuki group Sz(2E), a ‘small’ Ree group R(3E), or one of the 26 sporadic simple groups [24]. This suggests the following conjecture:

CONJECTURE 1.2. Let G be a non-abelian finite characteristically simple group. Then G is a Beauville group if and only if G is a 2-generator group and G = A5.

In order to prove that any group G is a Beauville group, one needs to realise it as a quotient of two triangle groups ∆I, or equivalently to produce two generating triples (aI, bI, cI) (i = 1, 2) for G, images of the canonical generators of ∆I, which satisfy AIbIcI = 1. If G = HK then for I = 1, 2 the members of such a triple are K-tuples AI = (aIJ), BI = (bIJ) and CI = (cIJ) such that, for J = 1,…, k, the elements AIJ, BIJ and CIJ of H form K generating triples for H which satisfy AIJ bIJ cIJ = 1 and are mutually inequivalent under the action of AutH. Provided K ≤ d2(H) there are K-tuples of such triples in H. However, in order for G to be a Beauville group we need two such K-tuples, one for each I, satisfying the following extra conditions.

The curve CI has genus GI 2 provided the orders LI, mI and NI of AI, bI and CI satisfy L1 + M-1 + N1 < 1, a condition which is automatic unless H = A5 and K = 1. However, it is generally harder to satisfy the other condition, that G acts freely on C1 × C2. The elements of G with fixed points on CI are the conjugates of the powers of AI, bI and CI, so for I = 1, 2 these two subsets ΣI of G must intersect in only the identity element. This is equivalent to Σ1 Σ2 containing no elements of order P for each prime P dividing ∣G, a condition which can be ensured by a careful choice of the generating triples (aI, bI, cI) for H, and of how they are arranged to form K pairs for J = 1,… ,k. Also, as is typical when considering infinite families of finite groups, small groups sometimes require special treatment.

The author is grateful to the referee for some very perceptive and helpful com­ments, and to the organisers of the ICTS programme Groups, Geometry and Dy­namics for the opportunity to publish these results here.

2. Beauville surfaces and structures

A finite group G is a Beauville group (of unmixed type) if and only if it has generating triples (aI, bI, cI) for i = 1, 2, of orders LI, MI and NI, such that

(1) AIbIcI = 1 for each I = 1, 2,

(2) L1 + M-1 + N-1 < 1 for each I = 1, 2, and

(3) no non-identity power of A1, B1 or C1 is conjugate in G to a power of A2, B2 or C2.

(In fact (2) is implied by (1) and (3), but it is useful to discuss it here.) We will call such a pair of triples (aI, bI, cI) a Beauville structure for G. Property (1) is equiva­lent to G being a smooth quotient ∆I/KI of a triangle group ∆I = ∆(LI, mI, nI) by a normal surface subgroup KI uniformising CI, with AI, bI and CI the local monodromy permutations for the covering CI → CI/G at the three ramification points; we call LI, MI and NI the Periods of ∆I. We call (aI, bI, cI) hyperbolic if it satisfies (2); this is equivalent to CI having genus at least 2, so that ∆I acts on the hyperbolic plane H, with CI = H∕KI and CI/G = HI = P1(C); note that if L-1 + M-1 + N1 1 then any quotient of ∆I is either solvable or isomorphic to A5, so most of the groups we shall consider automatically satisfy (2). Property (3), which is generally the most difficult to verify, is equivalent to G acting freely on C1 × C2, since the elements with fixed points in CI are the conjugates of the powers of AI, bI and CI.

By a Triple in a group G we will mean an ordered triple (A, b, c) of elements of G Such that Abc = 1; it is a Generating triple if A, b and C (and hence any two of them) generate G, and it has Type (l, m,n) if A, b and C have orders L, m and N (known as its Periods). Two triples are Equivalent if an automorphism of G takes one to the other; thus equivalent triples have the same type. A triple of type (l, m,n) Can be converted into one whose type is any permutation of (L, m, n) by permuting (and if necessary inverting) its elements. We say that a Beauville structure, in this notation, has Type (l1, m1, n1; L2, m2, n2); by the preceding remark, we can permute or transpose the two types (lI, mI, nI) while preserving properties (1), (2) and (3).

To show that a pair such as A1 and A2 satisfy (3) it is sufficient to verify that, for each prime P dividing both L1 and L2, the element Alf/p of order P is not conjugate to A^2/ for K = 1, 2,… ,p — 1; in particular, if L1 is prime it is sufficient to verify this for A1. Similar results apply to every other pair chosen from the two triples.

For I = 1, 2 let ΣI be the set of elements of G conjugate to powers of AI, bI or CI; This is a union of conjugacy classes of G, closed under taking powers. Condition (3) is satisfied if and only if Σι Σ2 = {1}. For any prime P, let Σ(P) denote the set of elements of order P in ΣI. The preceding remark then gives the following:

LEMMA 2.1. If generating triples (aI, bI, cI) (i = 1, 2) For a group G have types (lI, mI, nI), then they satisfy (3) If and only if ΣljP) ∩ Σ(p’) = 0 for each prime p dividing lPm-PnI and l2m2n2. In particular, they satisfy (3) If lPmPnI and l2m2n2 are mutually coprime.

3. Generating cartesian powers

Here we consider whether the K-th cartesian power of a finite group H can be a Beauville group for various K > 1. The observation that any Beauville group must be a 2-generator group immediately imposes restrictions on K, as follows.

If H is a finite group, let

B(H) = max{k ∣ HK is a Beauville group}

And let

C2(H) = max{k ∣ HK is a 2-generator group}, With B(H) = 0 or C2(H) = 0 if HK is not a Beauville group or a 2-generator group for any K ≥ 1. Then clearly

B(H) ≤ c2(H),

So any upper bound on C2(H) is also an upper bound on B(H).

LEMMA 3.1. Let H be a finite group. If H is not perfect then c2(H) 2.

Proof. Since H is not perfect it has CP as an epimorphic image for some prime P, so HK has C^ as an epimorphic image. If HK is a 2-generator group then so is CP, giving K ≤ 2. □

Thus in looking for Beauville groups among cartesian powers HK (K ≥ 2), it is sufficient to consider arbitrary powers of perfect groups H , and cartesian squares of imperfect groups H. (In the context of cartesian powers of finite groups, Wiegold [34] has shown that there is a more general dichotomy between perfect and imperfect groups H, with the rank D(HK) of HK respectively having essen­tially logarithmic or arithmetic growth as K → ∞. However Tyrer Jones [33] has constructed finitely-generated infinite groups satisfying H = H2, so that D(HK) is constant.)

When H is finite and perfect, arguments due to P. Hall [20] bound the values of K one needs to consider. For any finite group H, let D2(H) denote the number of normal subgroups N of the free group F2 = (X, Y ∣ —) of rank 2 with F2/N = H, and let φ2(H) be the number of 2-bases (ordered generating pairs) for H. Any 2­base (x, y) for H determines an epimorphism θ : F2 → H, sending X and Y to X and Y, so it determines a normal subgroup N = ker θ of F2 with F2/N = H; conversely, every such normal subgroup arises in this way, with two 2-bases yielding the same
Normal subgroup N if and only if they are equivalent under an automorphism of H. Thus D2(H) is equal to the number of orbits of Aut H on 2-bases for H. Since only the identity automorphism can fix a 2-base, this action is semiregular, so we have

LEMMA 3.2. If H is a finite group then

 Φ2(H) = D2(H )IAut H I. If H is a non-identity finite group then φ2 (H) H) ≤ d2(H)= .

Proof. If HK is a 2-generator group, then it is a quotient of F2, so there are at least K normal subgroups of F2 with quotient group H. This gives C2(H) ≤ d2(H), and Lemma 3.2 Completes the proof. □

For some groups we have C2(H) < d2(H): for instance, if P is prime then C2(CP) = 2 whereas D2(CP) = P + 1. However, if H is a non-abelian finite simple group we have equality in Corollary 3.3. This uses the following well-known result:

LEMMA 3.4. Let H be a non-abelian finite simple group, and let a group Γ Have distinct normal subgroups NI,…, NK with Γ/NI = H. Then Γ/ ∩K=1 NI = HK.

Applying this with Γ = F2 we have the following:

COROLLARY 3.5. If H is a non-abelian finite simple group then

Since B2(H) ≤ c2(H), and since finite simple groups have been dealt with, this means that in order to prove Theorem 1.1 It is sufficient to show that if H = AN Then HK is a Beauville group for each K = 2,…, d2(H). We will do this by repeated use of the following immediate consequence of Lemma 3.4:

COROLLARY 3.6. Let H be a non-abelian finite simple group. Then k-tuples a = (aJ), b = (bJ) And c = (cJ) Form a generating triple for G = HK if and only if their components (aJ, bJ, CJ) For j = 1, .. . ,k form k generating triples for H which are mutually inequivalent under Aut H.

Although we have concentrated on 2-generator groups, most of the results dis­cussed here have obvious extensions to N-generator groups for all N N [20, §1.6].

4. Evaluating φ2(H), D2(H) and C2(H)

In [20], Hall gave a method which, among many other applications, gives a formula for φ2(H) for any finite group H. From this one can deduce the value of D2 (H), though in practice one often has to be content with approximations.

Since each pair of elements of H generate a unique subgroup K we have

IHI2 = ∑ φ2(K).

Applying Mobius inversion in the lattice Λ of subgroups of H, we therefore have Φ2(H) = ∑ μ(K)∣KI2, K<H

Where μ is the Mobius function for Λ. This is defined recursively by the formula L μ(^) = δκ,H L>K

For each K ≤ H, where δ denotes the Kronecker delta-function. For instance, Hall used this to show that D2(A5) = 19 and D2(A6) = 53.

For any finite group H with trivial centre, we have

(H) = (H) = ⅛SL < H = JHL

2 ) 2( ) |AutH| < |InnH|.|Out H| |Out H

Results of Dixon [8], of Kantor and Lubotzky [27], and of Liebeck and Shalev [28] show that a randomly-chosen pair of elements generate a non-abelian finite simple group H with probability approaching 1 as H| → ∞, so for such groups this upper bound is asymptotically sharp, that is,

D2(H) ~ as H| → ∞.

21 |Out H | 1 1

The values of |H | and |Out H | for all non-abelian finite simple groups H are given in [7]. They show that for each of the infinite families of such groups, |Out H| grows much more slowly than H|, so that in fact D2(H) grows almost as quickly as H|. For instance, D2(AN) ~ n!/4 as N → ∞ (see [9,31] for more detailed results).

5. Beauville structures in cartesian powers

We saw in Corollary 3.6 How to form generating triples in a cartesian power G = HK of a non-abelian finite simple group H. In this section we consider sufficient conditions for two such triples to satisfy the hypothesis Σ (∣P) ∩ ΣP = 0 of Lemma 2.1 for all primes P, so that condition (3) of a Beauville structure is satisfied.

For any prime P, and any G = (gJ) ∈ G = HK, define the P-proßle of G to be the K-tuple PP(g) = (EJ∙), where PEJ is the highest power of P dividing the order O(gJ) of GJ. Define the P-summit of G to be the set SP(g) of J NK := {1,…, k} for which βJ attains its maximum value, provided this is not 0, and define SP(g) = 0 If O(g) is not divisible by P. If an element G’ = (gJ) ∈ G of order P is conjugate to a power of G, then it has coordinates GJ of order P at all J ∈ SP(g), with GJ = 1 elsewhere; in other words, the support supp(G‘) := {j | gJ = 1} of G’ is equal to SP(g). If T = (a, b,c) is a triple in G, define the P-summit SP(T) to be the set {SP(a), SP(b), SP(c)} of subsets of NK. The first part of the following result is now obvious, and the second follows from Lemma 2.1:

LEMMA 5.1. Let H be a finite group. If two triples (aI, bI, cI) (i = 1, 2) In G = HK have disjoint p-summits SP(TI) for some prime p, then Σ∣P) ∩ Σ^p’) = 0. If this happens for each prime p dividing any of their periods, then Σι Σ2 = {1}.

One way of ensuring that certain elements of NK are or are not in SP (G), without needing to know all the coordinates of G, is to make at least one coordinate of G p-full, meaning that its order is divisible by the highest power of P dividing the exponent of H. In this case, SP(g) is the set of all J such that GJ is P-full.

For any triple T = (a, b,c) in H, let νP(T) be the number of P-full elements among A, B and C. Define two triples TI (i = 1, 2) to be P-distinguishing if νP(T1) = νP(T2), and to be Strongly p-distinguishing if, in addition, whenever νP(TI) = 0 then either P2 does not divide exp(H) or P does not divide any of the periods of TI.

LEMMA 5.2. Suppose that a non-abelian finite simple group H has a set T = {(T1,s, T2,s) I s = 1, .. ., t} of pairs (T1,s, T2,s) of generating triples for H such that

(1) For each prime p dividing ∣H∣ there is some s = s(p) ∈ {1, .. . ,t} such that T1,s and T2,s are a strongly p-distinguishing pair;

(2) For each i = 1, 2 The 3T triples consisting of TI,ι,. .., TI,t and their cyclic permutations are mutually inequivalent under Aut H.

Then G := HK is a Beauville group for each k = 3t,. .., d2(H).

Proof. Let TI,s = (xI,s, yI,s, zI,s) for each I = 1, 2 and S = 1,…,t. Define elements αI, BI and CI of G by using XpS, hI, s,zI, s or Y,…, p…, x,… or ZI, s, xI, s, yI, s Respectively in their J-th coordinate positions where J = 3S — 2, 3S — 1 or 3S for S = 1,… ,t, so that the J-th coordinates of these three elements form a cyclic permutation of TI,s. Thus

A = (xI,ι, yI,ι, zI,ι,… xI, t, yI, t, zI, t,…),

BI = (yI,ι,zI,ι,xI,ι,.. .yI, t,zI, t,xI, t,…),

CI = (zI,1,xp1,yI,1,.. .zI, t,xI, t,yI, t,…).

If K > 3t then in coordinate positions J for J = 3T + 1,… k we use K — 3t further generating triples for H which are inequivalent to each other and to those already used for J = 1,…, 3T. (These exist since H has D2 (H) mutually inequivalent generating triples, and K ≤ d2(H).) Thus AI, bI and CI have K mutually inequivalent generating triples for H in their coordinate positions, so they form a generating triple for G. Since T ≥ 1 we have K ≥ 3, so G = A5 and this triple is hyperbolic.

Now let P be a prime dividing ∣H, so there is some S = S(P) ∈ {1,… ,t} such that T1,s and T2,s are a strongly P-distinguishing pair. If G Σ(P) for some I then supp(G) = SP(dI) where DI = AI, bI or CI, so

Supp(G) ∩ {3S — 2, 3S — 1, 3S} = νP(TI, s).

Since νP(T1,s) = νP(T2,s) it follows that G cannot be a member of Σ(P) for both I = 1 and I = 2. Thus Σ’∣P Σ.2∕’) = 0, as required. □

If we assume only that T1,s and T2,s are P-distinguishing, we have the following:

LEMMA 5.3. Suppose that a non-abelian finite simple group H has a set T = {(Tι,s, T2,s) ∣ s = 1, .. ., t} of pairs (T2,s, T2,s) of generating triples for H such that

(1) For each prime p dividing ∣H∣ there is some s = S(P) ∈ {1, .. . ,t} such that Tι,s and T2,s are a p-distinguishing pair;

(2) The 6T triples consisting of the 2t triples TI,s and their cyclic permutations are mutually inequivalent under Aut H.

Then G := HK is a Beauville group for each k = 6t,. .., d2(H).

Proof. The proof is similar, with the first 3T coordinates of AI, bI and CI defined as before. For J = 3T + 1,…, 6t we use the coordinates of A3I, b3I or C3I in position J — 3T, and if K > 6t we use further mutually inequivalent generating triples for H in the remaining coordinate positions, so each triple AI , BI or CI generates G. For each prime P dividing ∣H, each of the six generators AI, BI, CI has at least one P-full coordinate, so again if G ∈ Σ(p’) for some I then supp(G) = SP(dI) where DI = AI, bI or CI. The proof now continues as before. □

In many applications of these lemmas one can take T = 1, so that a single pair of generating triples suffices. This often happens when ∖H| is divisible by a small number of primes, as with the smaller alternating groups. The small values of K not covered by these lemmas can often be dealt with by applying the following result:

LEMMA 5.4. Let H be a non-abelian finite simple group with r generating triples of type (l, m,n), mutually inequivalent under Aut H, where r ≥ 2. If l, m and n are mutually coprime then HK is a Beauville group for each k = 2, .. ., 6r.

Proof. Let the specified generating triples for H be (xJ, yJ, ZJ) for J = 1,…, r. The 6R generating triples formed by cyclically permuting the entries of each triple (xI, yI, zI) and (z-1 ,y-1 ,x-1) are then mutually inequivalent. Since H is non – abelian and simple and G = AG, it follows that for each K = 2,…, 6r one can form a hyperbolic generating triple for HK by using any set of K of these triples in different coordinate positions. In particular, one can choose two such triples of the forms

A1 = (X1,x2,…), bI = (Y1,y2,.. .),cι = (Z1,z2,…) and

«2 = (XI,y2,…), b2 = (YI,z2,…), C2 = (Z1,x2,…),

Where in both cases the dots represent arbitrary choices of K — 2 generating triples for H from the remaining 6R — 2. If L, M and N are mutually coprime then any prime P dividing Lmn must divide exactly one of L, m and N. Thus if G 5∑1P) then supp(G) contains both 1 and 2, whereas if G ∈ Σ2p) then supp(G) contains only one of them. Hence Σc1P) xT = 0. □

Note that if a power G = HK of some group H is a Beauville group for some K ≥ 2, then although each direct factor HJ = H of G acts freely on C1 × C2, Preserving the curves CI, this does not imply that (C1 × C2)/HJ is a Beauville surface: the case where G = AG is an obvious counterexample. The point is that each direct factor HJ acts on each CI as a quotient of some Fuchsian group which is a subgroup of a triangle group ∆I, but which need not itself be a triangle group; thus the curve CI∕HJ need not be the projective line, and the covering CI → CI∕HJ Need not be ramified over three points. Studying the interesting geometry of this situation is a pro ject for another time and place.

6. Primitive permutation groups

In preparation for dealing with the groups H = AN, here we present some results which imply that certain triples generate the alternating group.

PROPOSITION 6.1. Let H be a primitive permutation group of degree n. Then H ≥ AN if any of the following conditions is satisfied:

(1) H has a subgroup with an orbit of length m, where 1 < m < n∕2, fixing the remaining n — m points;

(2) H contains an m-cycle, where m = 2, .. . ,n — 3;

(3) H contains a double transposition, with n ≥ 9.

Proof. (1) This is Margraff’e extension [29,30] of a theorem of Jordan, that if 1 ≤ m ≤ n/2 then H is 3-transitive (see also [35, Theorems 13.4, 13.5]).

(2) This is a recent extension, by the author [22], of a theorem of Jordan [26] which deals with the case where M is prime. (Although Wielandt, in [35, Theo­rem 13.9], refers to Jordan’s paper for this, it is not explicitly stated there. How­ever, it follows easily from Theoreme I of his paper [25], together with Theoreme I of [26].)

(3) A proof of this can be found on Peter Cameron’s blog [5]. The natural representation of AGL3(2) shows that the lower bound on N cannot be relaxed. □

Proposition 6.1(2) follows from a more general result [22], classifying the prim­itive groups containing an M-cycle for any M; this uses the classification of finite simple groups, and results of Müller [32] and Feit [13] for M = N — 1 and M = N.

In some situations, the assumption of primitivity can be replaced with transi­tivity, which is easier to verify, by using the following:

LEMMA 6.2. Let H be a transitive permutation group of degree n, containing an m-cycle. If m is coprime to n and m > n/2, then H is primitive. In particular, if m is prime and m > n/2 then H is primitive.

Proof. For the first assertion, if H is imprimitive, then being transitive it has blocks of the same size B for some proper divisor B of N. If the cycle acts trivially on the blocks, its support is contained in a single block, so M ≤ b ≤ n/2, against our hypothesis. If the cycle acts non-trivially on the blocks, its support is a union of blocks, again contradicting our hypothesis.

COROLLARY 6.3. Let H be a transitive permutation group of degree n, contain­ing an m-cycle. If m is coprime to n and n/2 < m < n — 2, then H ≥ AN.

Proof. This follows immediately from Lemma 6.2 And Proposition 6.1(2). □

The following result, concerning elements with two or three cycles, is also useful:

LEMMA 6.4. Let H be a transitive permutation group of degree n. If H has an element h with either of the following cycle structures, then H ≥ AN:

(1) Cycle structure c, d for coprime integers c, d> 1;

(2) Cycle structure 1, c, d for coprime integers c, d> 1 Such that neither 1 + C nor 1 + D divides n.

Proof. (1) We first show that H is primitive, so suppose that it is imprimitive, with blocks of size B, a proper divisor of N. Let the cycles of H be C and D, of lengths C and D. If C contains a block then it is a disjoint union of blocks, so B divides C and hence divides N — c = D, contradicting the fact that C and D are coprime. The same argument applies to D, so every block B meets both C and D. It follows that (h) permutes the blocks transitively, so they meet C in the same number R = ∖B ∩ C| of points, giving C = Rn/b. Similarly, the blocks all satisfy ∖B ∩ D = S for some S, so D = Sn/b. Thus N/b divides both C and D, again a contradiction. Hence H is primitive.

Without loss of generality, suppose that C < d, so C < n/2 since C+D = N. Since C and D are coprime, HD is a cycle of length C. Since H is primitive and 1 < c < n/2, Proposition 6.1(1) implies that H ≥ AN.

(2) If H were imprimitive, the block containing the unique fixed point of H Would be a union of cycles of H. This is impossible since neither 1 + C nor 1 + D Divides N, so H is primitive. As in (1), Proposition 6.1(1) completes the proof. □

In this result, H ∈ AN if and only if C ≡ d mod (2). Provided N ≥ 8, if N = 2m is even, or if N = 2m +1 is odd, we obtain an element H ∈ AN satisfying hypothesis (1) or (2) respectively by taking C, d = M ± 1 or M ± 2 as M is even or odd.

The following result [10, Lemma 2.3] is also useful in proving primitivity:

LEMMA 6.5. Let H = (hN .. ., hr/ be a transitive permutation group containing a cycle h of prime length. Suppose that for each i = 1,. .., r there is an element of Supp(H) Whose image under hI is also in supp(H). Then H is primitive.

7. Small alternating groups

We will now prove Theorem 1.1. The result is already known for K = 1 (see §1), so it is sufficient to show that if H = AN with N ≥ 5 then HK is a Beauville group for each K = 2, . . . , d2(H). In this section we will consider the groups H = AN For N = 5,…, 11. The remaining alternating groups will be considered in the next section, using general methods which do not always apply when N is small.

In this section and the next we will consider various triples (x, y, z) in AN; the reader may find it helpful to represent these as directed graphs on N vertices, with arcs corresponding to the actions of two of the generators (usually X and Y).

PROPOSITION 7.1. If n = 5,…, 11 Then AN is a Beauville group for each k = 2,…,d2(AN).

Proof. We will deal with these seven groups individually.

7.1. H = A5. It is well known that A5 is not a Beauville group: for instance, any generating triple for this group must contain an element of order 5 (otherwise it generates a solvable and hence proper subgroup); Sylow’s Theorems imply that all elements of order 5 are conjugate to powers of each other, so no pair of gen­erating triples can satisfy condition (3). Nevertheless we will prove that all other 2-generator cartesian powers of A5 are Beauville groups.

Hall [20] showed that D2 (A5) = 19, giving 19 equivalence classes of generating triples for A5. Note that A5 has two conjugacy classes of elements of order 5, transposed under conjugation by odd permutations, and also under squaring.

PROPOSITION 7.2. Thegroup AK is a Beauville group for each k = 2,. .., d2(A5).

Proof. The existence of Beauville structures for K = 3,…, 19 may be deduced immediately from Lemma 5.2 With T = 1, by using generating triples TI = (xI, yI, zI) (i = 1, 2) of types (2, 5, 5) and (3, 3, 5), such as

X1 = (1, 2)(3, 4), y1 = (1,4, 2, 3, 5), z1 = (1, 5, 4, 2, 3)

And

X2 = (1, 2, 3), y2 = (3,4, 5), Z2 = (1, 3, 5,4, 2).

The only maximal subgroup of A5 containing a subgroup {zi’) = C5 is its normaliser, isomorphic to D5, and in neither case does this contain YI, so each TI generates A5. Since A5 has exponent 30, this pair of triples are strongly Р-distinguishing for each of the relevant primes Р = 2, 3 and 5, so AK is a Beauville group for each K = 3,…, 19.

If K = 2 we cannot use Lemma 5.4, since A5 has no pairs of generating triples satisfying its hypotheses. Instead we use the following more specific approach.

Let (a1J, b1J, c1J) be generating triples for A5 of types (5, 5, 5) and (3, 5, 5) respectively for J = 1, 2, with all five generators of order 5 conjugate in A5. For instance we could take

αιι = (1, 2, 3,4, 5), b11 = (1, 4, 5, 2, 3), c11 = (1, 2,4, 5, 3), And

A12 = (1, 2,4), b12 = (1, 2, 3,4, 5), c12 = (1, 5, 2, 4, 3).

These triples are inequivalent, so the elements α1 = (A11,a12), B1 = (B11,b12) and C1 = (C11, c12) form a generating triple of type (15, 5, 5) for the group G = A2.

Now let A2 = (A21,a22), B2 = (B21,b22) and C2 = (C21,c22) in G, where (a2J, b2J, c2J) is a generating triple for A5 of type (3, 5, 5) or (5, 5, 5) respectively for J = 1, 2, with B21 and C21 conjugate to each other in A5, but not conjugate to A22, B22 or C22. For instance, we could take (A21, b21, c21) = (A12, b12, c12) and (A22, b22, c22) = (A11, b11, c11)S for some odd permutation G ∈ S5. Then (a2,b2,c2) Is a generating triple for G, also of type (15, 5, 5).

In order to use Lemma 5.1 To show that these triples (aI, bI, cI) satisfy condi­tion (3), it is sufficient to consider the primes P = 3 and 5. Elements of Σ13) or ς have their first or second coordinates respectively equal to the identity, so they cannot be equal. Elements of Σ^) have the form (G1,g2) with G1 conjugate to G2 vA5)

Or with G2 = 1, whereas elements of Σ2 have G1 = 1 or G1 not conjugate to G2, so again they cannot be equal. Thus Σ1 Σ2 = {1}, so these two triples form a Beauville structure for G. □

7.2. H = A6. When K = 2 we can use Lemma 3.4. Up to conjugacy in S6, the group H = A6 has four equivalence classes of generating triples (x, y,z) Of type (3,4, 5), given by taking Z to be a 5-cycle (a, b, c, d, e) and X = (a, f,c), (a, f, e), (A, b, f )(c, d, e) or (A, f, d)(b, c, e). Each of these triples generates H since no maximal subgroup contains elements of orders 3, 4 and 5 (see [7]). Now Aut A6 contains S6 with index 2, and acts semi-regularly on generating triples, so it has two orbits on triples of this type. It follows from Lemma 5.4 That AK is a Beauville group for K = 2,…, 12.

For K = 3,… ,d2(A6) = 53 (see [20]) we can apply Lemma 5.2 With T = 1, using triples of types (3, 5, 5) and (4, 4, 5) such as

X1 = (1, 2, 3), y1 = (1, 3, 4, 5, 6), z1 = (1, 6, 5,4, 2)

And

X2 = (1, 2, 3,4)(5, 6), y2 = (1, 3)(2, 5, 4, 6), z2 = (1, 2, 3, 4, 5).

Each triple generates a subgroup of A6 which is doubly transitive and therefore primitive, and which contains a 3-cycle (X1 or X2z-1 respectively), so by Proposi­tion 6.1(2) it generates A6. Since A6 has exponent 60, these two triples are strongly P-distinguishing for each of the relevant primes P = 2, 3 and 5.

7.3. H = A7. For small K we can use Lemma 5.4, with triples

X =(1, 2, 3), y = (3,4, 5, 6, 7), z = (1, 3, 7, 6, 5,4, 2) and

X =(1, 2, 3)(4, 5, 6), y =(1, 6, 7, 3,4), z = (1, 6, 3, 7, 5,4, 2) of type (3, 5, 7). No maximal subgroup of A7 contains elements of orders 3, 5 and 7, so they are both generating triples. They are inequivalent since Aut A7 = S7 and the elements of order 3 have different cycle structures, so ALf is a Beauville group for each K = 2,…, 12.

For larger K we can use Lemma 5.2 With T = 1. We choose one of the above triples of type (3, 5, 7), together with a triple of type (4, 7, 7), such as

X =(1, 2, 3, 4)(5, 6), y =(1, 5, 2,4, 6, 7, 3), г =(1, 2, 6, 3, 7, 5, 4).

This generates a transitive group, which is primitive since its degree is prime, and which must therefore be A7 since it contains the 3-cycle Xy3 = (1, 7, 5). Since A7 has exponent 22.3.5.7, these two triples deal with the cases K = 3,…, d2(A7).

7.4. H = A8,…, A11. The arguments when N = 8, 9, 10 and 11 are similar to those used for A6 and A7: when K = 3,…, d2(AN) we apply Lemma 5.2 With T = 1, and when K = 2 we apply Lemma 5.4. In each case, we will simply state some triples which can be used. Verifying that these are generating triples is straightforward, using results from the preceding section. Pairs of triples of the same type can easily be shown to be inequivalent by representing them as edge-labelled directed graphs on N vertices, with arcs representing the actions of X and Y: inequivalence of triples corresponds to non-isomorphism of their corresponding graphs.

When H = A8, of exponent 22.3.5.7, a pair of generating triples

X =(1, 2)(3,4, 5, 6), y = (1, 4, 3, 7, 8), г = (1, 8, 7, 6, 5,4, 2),

X =(1, 2, 3), y =(3, 2,4, 5, 6, 7, 8), г =(1, 3, 8, 7, 6, 5, 4) of types (4, 5, 7) and (3, 7, 7) satisfy the hypotheses of Lemma 5.2, so this deals with K = 3,…, d2(A8). For K = 2 we can use Lemma 5.4, with an inequivalent generating triple of type (4, 5, 7), such as

X =(1, 2)(3,4, 5, 6), y = (1, 7,4, 3, 8), г = (1, 8, 6, 5,4, 7, 2).

When H = AG, of exponent 22.32.5.7, we can use generating triples

X =(1, 2, 3,4)(5, 6, 7, 8), y = (1,4, 5, 8, 9), г =(1, 9, 7, 6, 5, 3, 2),

X =(1, 2, 3, 4, 5), y =(1, 2, 5)(3, 6, 7, 8, 9), г = (1,4, 3, 9, 8, 7, 6, 2, 5) of types (4, 5, 7) and (5,15, 9) for K = 3,…, d2(AG). For K = 2 we can use an inequivalent generating triple of type (4, 5, 7), such as

X =(1, 2, 3,4)(5, 6, 7, 8), y = (1,4, 3, 5, 9), г =(1, 9, 8, 7, 6, 5, 2).

When H = A10, of exponent 23.32.5.7, we can use generating triples

X =(1,…, 8)(9,10), y =(1, 8, 7, 9,10), г =(1, 9, 6, 5,4, 3, 2),

X =(1,…, 9), y =(1, 3, 5, 7,10)), г = (1,10, 6, 5, 2)(3, 9, 8, 7, 4) of types (8, 5, 7) and (9, 5, 5) for K = 3,…, d2(A10). For K = 2 we can use an inequivalent generating triple of type (8, 5, 7), such as

X =(1,…, 8)(9,10), y =(1, 8, 7, 6, 9), г =(1,10, 9, 5,4, 3, 2).

When H = A11, of exponent 23.32.5.7.11, we can use generating triples

X = (1, 2, 3, 4, 5), y = (2,11)(3, 6, 7, 8, 9,10, 5, 4), г = (1, 5,10, 9, 8, 7, 6, 2,4),

X = (1, 2, 3, 4, 5, 6, 7), y = (1, 6, 7, 2,4, 5, 8, 9,10,11, 3)), г = (1, 2, 6, 7, 5, 3,11,10, 9, 8,4) of types (5, 8, 9) and (7,11,11) for K = 3,…, d2 (A11). For K = 2 we can use two inequivalent generating triples of type (11, 3, 8) given by taking

X =(1,…, 11) and Y = (1,4, 2) or (1,10, 2).

Thus, for each N = 5,…, 11 we have shown that A1fl is a Beauville group for K = 2,…, d2(AN). This completes the proof of Proposition 7.1. □

8. Larger alternating groups

If we try to apply the preceding method to AN for larger N, then as N increases we need to consider more primes dividing the group order, namely the π(n) ~ n/ log N primes P ≤ n. The irregular distribution of primes means that the Ad hoc approach used for N = 5,…, 11 will not work in general, so we need a more systematic method of proof to deal with AN for N ≥ 12 (an assumption which will be maintained throughout this section).

PROPOSITION 8.1. If n ≥ 12 Then AN is a Beauville group for each k = 2,…, (n — 5)(N — 6)(N — 7)/4.

Proof. We will use Lemma 5.4, which requires a number R ≥ 2 of inequivalent generating triples of a type with mutually coprime periods. For odd N we take

X =(1, 2,…, n — 4) and Y = (s, n — 3)(t, n — 2)(u, n — 1)(V, n)

Where S, t,u and V are distinct elements of {1,…,n — 4}, so that Z := (xy)-1 Is an N-cycle. Since X and Y are even permutations they generate a subgroup H = (x, y) of AN. Since it contains Z, H is transitive, and since N — 4 is coprime to N and greater than N/2, Corollary 6.3 Implies that H = AN. Thus (x, y,z) Is a generating triple of type (N — 4, 2, n) for AN, with mutually coprime periods. By representing these triples as directed graphs on N vertices, with arcs labelled X or Y, we see that their equivalence classes (under Aut H = SN) correspond to the orbits of the additive group Z,, 4 on its 4-element subsets {s, t,u, v}. Since N — 4 is odd, this action is semi-regular, so the number of equivalence classes is (V)/(N — 4) = (N — 5)(N — 6)(N — 7)/24.

The argument is similar for even N. We take

X = (1, 2,…, n — 3) and Y = (s, s + 1)(T, n — 2)(u, n — 1)(V, n)

Where S, s + 1, t, u and V are distinct elements of {1,…, n — 3}, so that Z is an (N —1)-cycle fixing S. In this case H = Fx, y) is doubly transitve, since it is transitive and the stabiliser of S contains Z, so it is primitive; since X ∈ H, Proposition 6.1(2) implies that H = AN. We therefore have (N5) = (N — 5)(N — 6)(N — 7)/6 equivalence classes of generating triples of type (N — 3, 2, n — 1), with mutually coprime periods.

In either case the number R of equivalence classes is at least (N — 5)(N — 6) (N — 7)/24, so Lemma 5.4 Implies that AN is a Beauville group for K = 2,…, 6R. □

To deal with larger K we will use Lemma 5.3. If N is large there is no single pair of generating triples which are P-distinguishing for all primes P ≤ n, as there is for N = 5,…, 11: consideration of cycle structures shows that a single pair can distinguish at most six primes P ≥ n/2. Instead we will find one such pair (TP, T!P) for each P, with TP having at least one P-full element, while Tp has none, so that the pair is P-distinguishing.

It is easy to find such generating triples TP and Tp for each P. However, in order to apply Lemma 5.3 We also need the triples TP, TP and their cyclic permutations to be mutually inequivalent under the action of Aut AN = SN. We do this by constructing the triples TP and TP in such a way that the corresponding prime P Can be recognised from their three elements.

LEMMA 8.2. If n ≥ 12 There is a set TN of generating triples TP for AN, one for each prime p ≤ n, such that

(1) At least one of the three elements of TP is p-full, and

(2) The triples TP (p ≤ n) and their cyclic permutations are mutually inequiv­alent under SN.

Proof. First let N be even, say N = 2m. If P is odd take X to be a cycle of length PE, where PE ≤ n < pE+1, so that X is P-full. Take Y to consist of two cycles of coprime lengths M ± 1 or M ± 2 as M is even or odd, with both cycles meeting the support of X. Then H = Fx, y) is a transitive subgroup of AN, and applying Lemma 6.4(1) to Y shows that H = AN. Given X, we can choose such an element Y So that Z := (xy)1 has at least two non-trivial cycles, so X is the only cycle in the generating triple TP = (X, y, z). Thus P can be recognised from {x, y, z} as the only prime dividing the order of this element, so for the odd primes P ≤ n the triples TP And their cyclic permutations are mutually inequivalent. If P = 2 take X to have two cycles of lengths 2E and 2 where 2E + 2 ≤ n < 2E+1 + 2 (so E ≥ 2), together with N — 2E 2 fixed points, so X is 2-full. Take Y as before, ensuring that some cycle of X meets both cycles of Y, and that Z again has at least two non-trivial cycles. The resulting triple T2 is not equivalent to any cyclic permutation of a triple TP for P Odd, since none of its members is a cycle, nor to any cyclic permutation of itself, since X and Z have even and odd orders respectively.

If N = 2m + 1 is odd, let X and Y be as above, but with Y having an extra fixed point, and all three of its cycles meeting the (or a) cycle of X. It is easy to check that Y satisfies the conditions of Lemma 6.4(2), so Fx, y) = AN, and the proof proceeds as before. □

Although the cycle structures of X and Y are completely specified in the above construction, there is usually some freedom of choice for that of Z .

LEMMA 8.3. If n ≥ 12 There is a set TN of generating triples TP for AN, one for each prime p ≤ n, such that

(1) None of the three elements of TP is p-full, and

(2) The triples Tp (p ≤ n) and their cyclic permutations are mutually inequiv­alent under SN.

Proof. We shall construct the triples Tp so that the corresponding prime P can be recognised from each of them, as follows. Given a prime P such that 7 ≤ p ≤ n, let Q be the largest prime less than P (so Q ≥ 5), and let X be the Q-cycle (1, 2,… ,q) ∈ AN. We will choose Y ∈ AN so that (x, y) = AN, giving a generating triple Tp = (X, y, z) for AN where Z = (xy)1. We will arrange that Y is not a cycle, and if Z is a cycle then it is at least as long as X; thus P is the smallest prime number greater than the length of a shortest cycle among X, y and Z. The primes P = 2, 3 and 5 will be dealt with later, using separate constructions.

Let P ≥ 7, and first suppose that N is even, soP < n. We take Y = (1, 2)C3 … cR, Where C3,…, cR are disjoint cycles, and each CI has length MI, joining MI 1 points from {q + 1,. .., n} to the point I ∈ {3,. .. ,q}, so r ≤ q. We take M3 + ∙ ∙ ∙ + MR = N — q + R — 2, so all such points are joined, and Z = (xy)1 is a cycle of length N — 1, fixing 1; thus Z is even, and hence so is Y, so the triple TP = (x, y, z) lies in AN. It generates a transitive subgroup H, which is primitive by Lemma 6.5 With H = X, so Proposition 6.1(2) gives H = AN since Q ≤ p — 2 < n — 2. This triple satisfies (1) if M3,…, mR and N — 1 are coprime to P, so that X, y and Z have orders coprime to P. If N ≡ 1 mod (P) then provided N ≡ q — 1 mod (P) we can take R = 3, with a single cycle C3 of length N — q + 1 coprime to P; if N ≡ q — 1 mod (P) we can take R = 4, with cycles C3 and C4 of lengths M3 = 2 and M4 = N — q (≡ —1 mod (P)). (It is here that we need P > 5, so that Q > 3, otherwise we cannot take R = 4.) If N ≡ 1 mod (P) we instead define Y = (2, q)C3 … cR, fixing 1, with the cycles CI as above; then Z transposes 1 and Q, and acts as a cycle of length N — 2 on the remaining points, so again Tp is contained in AN, and generates this group. In this case we can take R = 3, with C3 a cycle of length M3 = N — q +1 2 — q mod (P), so M3 is coprime to P since 2 < q < p.

Now suppose that N is odd. We take Y = (1, q, 2)C3 … cR, with the cycles CI As above, so Z fixes 1 and Q and acts as a cycle of length N — 2 on the remaining points. Again TP generates AN (if Q = P — 2 = N — 2 we can take R = 4 with C3 and C4 transpositions, so Y2 is a 3-cycle and hence Proposition 6.1(2) applies). This triple has periods coprime to P, and hence satisfies (1), provided N ≡ 2 mod (P): we can choose C3, or C3 and C4, as when N is even, if N ≡ q — 1 mod (P) or N ≡ q — 1 mod (P) respectively. If N ≡ 2 mod (P) we instead define Y = (1, 2, q)C3 … cR, with the cycles CI as before, so that Z is now an N-cycle; the same argument then applies.

Note that in all of the above triples Tp (p ≥ 7), X is a Q-cycle for a prime Q Such that 5 ≤ q ≤ n — 2; Y has a cycle contained in supp(X), of length 2 or 3 as N Is even or odd, it has one or two more non-trivial cycles, each meeting supp(X) in one point, and it fixes the rest of supp(X); finally Z is a cycle of length N, N — 1 or N — 2, or has cycle structure N — 2, 2. In particular, X has a single non-trivial cycle, whereas Y has at least two. Given such a triple, or any cyclic permutation of it, one can use this general information to identify which elements of the triple are X, Y and Z , and hence to identify the prime P from the length Q of the cycle X.

When P = 5 we will take X = (1, 2,… ,q) again, but with Q = 7 rather than 3. (Using the preceding construction, and taking Q = 3, would force R = 3, whereas we need to allow R = 4 for some N.) We must therefore ensure that ‘Γ5ι is not equivalent to T(1 or its cyclic permutations, since this triple (but no other TJ)) uses the same element X. If N is even we define Y = (2, 6)C3 … cR, with R = 3 or 4 and CI defined as before, so that Z has cycle structure 3,n — 3. We can do this, with each MI and N — 3 coprime to 5, provided N ≡ 3 mod (5); if N ≡ 3 mod (5) we take Y = (2, 5)C3 … cR instead, so that Z has cycle structure 4, n — 4. Now suppose that N is odd. If N ≡ 2 mod (5) we define Y = (1, 2)(6, 7)C3 … cR, so that Z is an (N — 2)-cycle, and if N ≡ 2 mod (5) we define Y = (1, 2)(5, 7)C3 … cR, so that Z has cycle structure 1, 2, n — 3. In all cases, applying Lemma 6.5 And Proposition 6.1(2) to the 7-cycle X shows that the resulting triple T5′ generates AN. Moreover, if N is even then Z has a cycle of length 3 or 4, while if N is odd then Y has two non-trivial cycles contained in supp(X); in either case this shows that T5′ is not equivalent to any cyclic permutation of T1’1, and hence of any Tp for P > 5.

For P = 2 or 3, since N ≥ 12 we can satisfy (1) by imposing the weaker condition that none of X, y and Z has a cycle of length divisible by P2, rather than the more restrictive P.

Let P = 3. For odd N ≡ 0 mod (9) let

X = (1, 2,…,n), y = (1, 2, 3), z =(1, 2,n, n — 1,…, 3).

Then H is primitive since the only (^-invariant proper equivalence relations are given by congruence mod (m) for some proper divisor M of N, and Y preserves none of these. Thus H = AN by Proposition 6.1(2), giving a generating triple T3 of type (N, 3, n). For odd N ≡ 0 mod (9) let

X = (1, 2,…, n — 4)(N — 3, n — 2, n — 1), y =(1, 3, 2)(N — 4, n — 3, n), So that Z is an (N — 2)-cycle. Then H is AN by Corollary 6.3 Since it contains the (N — 4)-cycle X3 with N — 4 > n/2 and gcd(N, n — 4) = 1, so we have a generating triple of type (3(N — 4), 3, n — 2). Now let N be even. If N ≡ 1 mod (9) let

X =(1, 2,…,n — 1), y =(1,n, 2),

So that Z is an (N — 1)-cycle, giving a generating triple of type (N — 1, 3, n — 1). For even N ≡ 1 mod (9) let

X = (1, 2,…, n — 3), y = (1, n, 2)(3, 5, 4)(N — 3, n — 2, n — 1),

Z = (2,n, n — 3,n — 1, n — 2, n — 4, n — 5,…, 5).

Then H = AN by Corollary 6.3 Since X ∈ H ,so we have a generating triple of type (N — 3, 3, n — 3). In all these cases, Y has order 3, whereas no triple TP for P > 3 contains such an element.

Now let P = 2. If N is odd let

X = (1,…, n — 2), y = (N — 3, n — 1)(N — 2, n).

Then H is primitive by Lemma 6.2 Since X ∈ H ,so H = AN by Proposition 6.1(3) since Y ∈ H. This gives a generating triple T2′ of type (N — 2, 2, n). If N is even let

X =(1,…,n — 1), y =(1, 2)(N — 1,n),

So a similar argument shows that this is a generating triple of type (N — 1, 2, n — 1). In both cases, the presence of an element Y of order 2 distinguishes T2′ from all other triples TP and their cyclic permutations, so conclusion (2) is satisfied. □

LEMMA 8.4. If n ≥ 12 No triple in TN is cyclically equivalent to a triple in T^

Proof. Each triple (x, y, z) TN contains an element Y with cycle structure C, d or 1, c, d as N is even or odd, where C, d ≥ 5. However, if (x’, y’, z’) TN then no element X’,y’ or Z’ has such a cycle structure: by the construction of TP, each of X’ and Z’ is a single cycle (possibly with fixed points), or has a cycle of length 2 or 3, while Y’ always has a cycle of length 2 or 3. Thus the triples (X, y, z) and (x’, y’, z‘) cannot be cyclically equivalent. □

Proposition 8.5. If n ≥ 12 Then AN is a Beauville group for each k = 6π(n), .. ., d2(AN).

Proof. The preceding three lemmas show that the π(N) ordered pairs (TP, TP) of generating triples for AN, where P ≤ n, satisfy the hypotheses of Lemma 5.3. □

Proposition 8.1 Deals with AN where K = 2,…, (n — 5)(N — 6)(N — 7)/4, while Proposition 8.5 Does this for K = 6π(N),…, d2(AN). Easy estimates show that

6π(N) 3(N + 1) (N — 5)(N — 6)(N — 7)/4 for all N ≥ 12, so these two results cover each K = 2,…, d2(AN) for such N. Com­bining this with Proposition 7.1, which deals with N = 5,…, 11, we have proved:

THEOREM 8.6. If n ≥ 5 Then AN is a Beauville group for each k = 2, .. ., d2 (AN).

This completes the proof of Theorem 1.1.

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School of Mathematics, University of Southampton, Southampton SO17 1BJ, United Kingdom

E-mail address: G. A.Jones@maths. soton. ac. uk

Contemporary Mathematics

Volume 639, 2015

Http://dx. doi. org/10.1090/conm/639/12792