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Publicly Available Published by De Gruyter July 27, 2016

Beauville structures in p-central quotients

Şükran Gül EMAIL logo
From the journal Journal of Group Theory

Abstract

We prove a conjecture of Boston that if p5, all p-central quotients of the free group on two generators and of the free product of two cyclic groups of order p are Beauville groups. In the case of the free product, we also determine Beauville structures in p-central quotients when p=3. As a consequence, we give an infinite family of Beauville 3-groups, which is different from the ones that were known up to date.

1 Introduction

A Beauville surface of unmixed type is a compact complex surface isomorphic to (C1×C2)/G, where C1 and C2 are algebraic curves of genus at least 2 and G is a finite group acting freely on C1×C2 and faithfully on the factors Ci such that Ci/G1() and the covering map CiCi/G is ramified over three points for i=1,2. Then the group G is said to be a Beauville group.

It is easy to formulate the condition for a finite group G to be a Beauville group in purely group-theoretical terms. For a couple of elements x,yG, we define

Σ(x,y)=gG(xgygxyg),

that is, the union of all subgroups of G which are conjugate to x, to y or to xy. Then G is a Beauville group if and only if the following conditions hold:

  1. G is a 2-generator group.

  2. There exists a pair of generating sets {x1,y1} and {x2,y2} of G such that Σ(x1,y1)Σ(x2,y2)=1.

Then {x1,y1} and {x2,y2} are said to form a Beauville structure for G. We call {xi,yi,xiyi} the triple associated to {xi,yi} for i=1,2. The signature of a triple is the tuple of orders of the elements in the triple.

In 2000, Catanese [6] proved that a finite abelian group is a Beauville group if and only if it is isomorphic to Cn×Cn, where n>1 and gcd(n,6)=1. On the other hand, all finite quasisimple groups other than A5 and SL2(5) are Beauville groups [8, 9] (see also [11] and [13]).

If p is a prime, Barker, Boston and Fairbairn [1] have shown that the smallest non-abelian Beauville p-groups for p=2 and p=3 are of order 27 and 35, respectively. They have also proved that there are non-abelian Beauville p-groups of order pn for every p5 and every n3. The existence of infinitely many Beauville 2-groups and 3-groups has been settled in the affirmative in [2], and in [17] and [12], respectively. In particular, by [17, Theorem 2], there are quotients of the ordinary triangle group T=x,yx3=y3=(xy)9=1 which are Beauville 3-groups of every order greater than or equal to 35. Among them, one can find the 3-central quotients T/λn(T) for all n4. In all these groups, the signature of one of the triples of the Beauville structure takes the constant value (3,3,9). On the other hand, by considering quotients of the Nottingham group over 𝔽3, Fernández-Alcober and Gül [10] have recently given an infinite family of Beauville 3-groups, for all orders at least 35, and in this case the signatures of the triples are not bounded.

In [4], Boston conjectured that if p5 and F is either the free group on two generators or the free product of two cyclic groups of order p, then its p-central quotients F/λn(F) are Beauville groups. The goal of this paper is to prove Boston’s conjecture. In fact, in the case of the free product, we extend the result to p=3.

The main results of this paper are as follows.

Theorem A

Let F=x,y be the free group on two generators. Then a p-central quotient F/λn(F) is a Beauville group if and only if p5 and n2.

Theorem B

Let F=x,yxp,yp be the free product of two cyclic groups of order p. Then a p-central quotient F/λn(F) is a Beauville group if and only if p5 and n2 or p=3 and n4.

We will see that the signatures of the triples in the Beauville structures arising from Theorem B are unbounded as n goes to infinity. As a corollary, we get examples of Beauville 3-groups of every order greater than or equal to 35 which are completely different from the ones given by Stix and Vdovina. We also compare these examples with the Beauville quotients of the Nottingham group over 𝔽3 given in [10], and we show that the two infinite families only coincide at the group of order 35, see Theorem 3.7.

Notation.

We use standard notation in group theory. If G is a group, then we denote by ClG(x) the conjugacy class of the element xG. Also, if p is a prime, then we write Gpi for the subgroup generated by all powers gpi as g runs over G and Ωi(G) for the subgroup generated by the elements of G of order at most pi. The exponent of G, denoted by expG, is the maximum of the orders of all elements of G.

2 The free group on two generators

In this section, we give the proof of Theorem A. We begin by recalling the definition of p-central series for the convenience of the reader.

Definition 2.1

For any group G, the normal series

G=λ1(G)λ2(G)λn(G)

given by λn(G)=[λn-1(G),G]λn-1(G)p for n>1 is called the p-central series of G.

Then a quotient group G/λn(G) is said to be a p-central quotient of G. To prove the main theorems, we need the following properties of the subgroups λn(G) (see [15, Definition 1.4 and Theorem 1.8, respectively]): we have

(2.1)λn(G)=γ1(G)pn-1γ2(G)pn-2γn(G),

and any element of λn(G) can be written in the form

(2.2)a1pn-1a2pn-2anfor some aiγi(G).

Also observe that if expG/G=p, then λn(G)=γn(G), since for any i,j1 we have γi(G)pjγi+j(G).

Lemma 2.2

Let G be a group and x,yG. For n3, we have

(xy)pn-2xpn-2ypn-2(modλn(G)).

Proof.

By the Hall–Petrescu formula (see [15, Lemma 1.1]), we have

(xy)pn-2xpn-2ypn-2(modγ2(G)pn-2r=1n-2γpr(G)pn-2-r).

Now the result follows, since by (2.1), γ2(G)pn-2λn(G) and for 1rn-2 we have

γpr(G)pn-2-rλpr+n-2-r(G)λn(G).

Note that if yλ2(G) in Lemma 2.2, then

(2.3)(xy)pn-2xpn-2(modλn(G)).

Before we proceed to prove Theorem A, we will need to introduce a lemma.

Let F=x,y be the free group on two generators. Notice that for n2, Φ(F/λn(F)) coincides with λ2(F)/λn(F), and thus elements outside λ2(F) are potential generators in F/λn(F). In order to determine Beauville structures in the quotients F/λn(F), it is fundamental to control pn-2nd powers of elements outside λ2(F) in these quotients groups.

Lemma 2.3

Let F=x,y be the free group on two generators. Then xpn-2 and ypn-2 are linearly independent modulo λn(F) for n2.

Proof.

We argue by way of contradiction. Suppose that

yipn-2xpn-2(modλn(F)).

It follows from (2.2) that

x-pn-2yipn-2=a1pn-1a2pn-2anfor some ajγj(F),

and then we have

y-ipn-2xpn-2a1pn-1γ2(F).

Write a1=xkylz for some zγ2(F) and some k,l. Then

xpn-2(1+kp)ypn-2(lp-i)γ2(F).

On the other hand, an element of the free group F belongs to γ2(F) if and only if the exponent sum of both generators is zero. Hence we get pn-2(1+kp)=0, which is a contradiction. ∎

As a consequence of Lemma 2.3, x and y have order pn-1 modulo λn(F).

By equation (2.3), if we want to know pn-2nd powers of all elements outside λ2(F) in F/λn(F), it is enough to know the power of each element in the set {y,xyi0ip-1}. Also, by Lemma 2.2, we have

(xyi)pn-2xpn-2yipn-2(modλn(F))for 1ip-1,

and since xpn-2 and ypn-2 are linearly independent modulo λn(F) by Lemma 2.3, the following lemma is straightforward.

Lemma 2.4

If G=F/λn(F), the power subgroups Mpn-2 are all different and of order p in λn-1(F)/λn(F), as M runs over the p+1 maximal subgroups of G. In particular, all elements in MΦ(G) are of order pn-1.

After these preliminaries, we can now prove Theorem A.

Theorem 2.5

A p-central quotient F/λn(F) is a Beauville group if and only if p5 and n2.

Proof.

For simplicity let us call G the quotient group F/λn(F). We first show that if p=2 or 3, then G is not a Beauville group. By way of contradiction, suppose that {u1,v1} and {u2,v2} form a Beauville structure for G. Since G has p+14 maximal subgroups, we may assume that u1 and u2 are in the same maximal subgroup. Then by (2.3), we have

u1pn-2=u2pn-2,

which is a contradiction.

Thus we assume that p5. First of all, notice that if n=2, GCp×Cp is a Beauville group, by Catanese’s criterion. So we will deal with the case n3. Let u and v be the images in G of x and y, respectively. We claim that {u,v} and {uv2,uv4} form a Beauville structure for G. If we have A={u,v,uv} and B={uv2,uv4,uv2uv4}, we need to show that

(2.4)agbh=1

for all aA, bB, and g,hG. Observe that ag and bh lie in different maximal subgroups of G in every case, since u and v are linearly independent modulo Φ(G) and p5.

Now, all elements aA and bB are of order pn-1, by Lemma 2.4. If (2.4) does not hold, then

(ag)pn-2=(bh)pn-2,

and again by Lemma 2.4, ag and bh lie in the same maximal subgroup of G, which is a contradiction. We thus complete the proof that G is a Beauville group. ∎

3 The free product of two cyclic groups of order p

Now we focus on the free product F=x,yxp,yp of two cyclic groups of order p. Notice that since F/F has exponent p, we have λn(F)=γn(F) for all n1.

We start with an easy lemma whose proof is left to the reader.

Lemma 3.1

Let ψ:G1G2 be a group homomorphism, let x1,y1G1 and x2=ψ(x1), y2=ψ(y1). If o(x1)=o(x2), the condition x2ψ(g)y2ψ(h)=1 implies that x1gy1h=1 for g,hG1.

To prove the main theorem we also need a result of Easterfield [7] regarding the exponent of Ωi(G). More precisely, if G is a p-group, then for every i,k1, the condition γk(p-1)+1(G)=1 implies that

expΩi(G)pi+k-1.

A key ingredient of the proof of Theorem B will be based on p-groups of maximal class with some specific properties. Let G=sA, where s is of order p and App-1. The action of s on A is via θ, where θ is defined by the companion matrix of the pth cyclotomic polynomial xp-1++x+1. Then G is the only infinite pro-p group of maximal class. Since sp=1 and θp-1++θ+1 annihilates A, this implies that for every aA,

(sa)p=spasp-1++s+1=1.

Thus all elements in GA are of order p. An alternative construction of G can be given by using the ring of cyclotomic integers (see [16, Example 7.4.14]).

Let P be a finite quotient of G of order pn for n3. Let us call P1 the abelian maximal subgroup of P and Pi=[P1,P,i-1,P]=γi(P) for i2. Then one can easily check that

expPi=pn-ip-1

and every element in PiPi+1 is of order pn-ip-1.

Now we can begin to determine which p-central quotients of F are Beauville groups. We first assume that p=2. The free product F of two cyclic groups of order 2 is the infinite dihedral group D. Then by [3, Lemma 3.7], no finite quotient of F is a Beauville group. In the remainder, we consider the case where p is an odd prime.

Lemma 3.2

Let G=F/λn(F) for n2. If u and v are the images of x and y in G, then for any i,j0(modp) all elements in the coset uivjΦ(G) have order pn-1p-1.

Proof.

Let P be the p-group of maximal class of order pn which is mentioned above and let sPP1 and s1P1P. Since all elements in PP1 are of order p and λn(P)=1, the map

ψ:GP,uis-1,vjss1,

is well-defined and an epimorphism. Set k=n-1p-1. Since ψ is an epimorphism, we have ψ(uivjΦ(G))=ψ(uivj)Φ(P)=s1Φ(P), where every element in the coset s1Φ(P) has the same order as s1, namely pk. Then for every guivjΦ(G), we have o(g)o(s1)=pk. On the other hand, γk(p-1)+1(G)γn(G)=1. Then by (6), we get expGpk, and consequently o(g)=pk. ∎

We deal separately with the cases p5 and p=3.

Theorem 3.3

If p5, then the p-central quotient F/λn(F) is a Beauville group for every n2.

Proof.

For simplicity let us call G the quotient group F/λn(F). Observe that Ω1(G)=G.

If n=2, then GCp×Cp is a Beauville group, by Catanese’s criterion. Thus we assume that n3. Let u and v be the images of x and y in G, respectively. We claim that {u,v} and {uv2,uv4} form a Beauville structure for G. Let A={u,v,uv} and B={uv2,uv4,uv2uv4}. Assume first that a=u or v, which are elements of order p, and bB. If agbh1 for some g,hG, then agbh, and hence aΦ(G)=bΦ(G), which is a contradiction since p5. Next we assume that a=uv. Since p5, for every bB we have ψ(b)PP1, which is of order p. Thus for all g,hG we have

s1ψ(g)ψ(b)ψ(h)=1.

Since o(uv)=o(s1), it then follows from Lemma 3.1 that agbh=1. This completes the proof. ∎

In order to deal with the prime 3, we need the following lemmas.

Lemma 3.4

Let G be a p-group which is not of maximal class such that d(G)=2. Then for every xG there exists tΦ(G){[x,g]gG}.

Proof.

Note that a p-group has maximal class if and only if it has an element with centralizer of order p2 (see [14, III.14.23]). Thus for every xG we have |CG(x)|p3, and hence

|{[x,g]gG}|=|ClG(x)|=|G:CG(x)|pn-3.

Since |Φ(G)|=pn-2, there exists tΦ(G) such that t{[x,g]gG}. ∎

Lemma 3.5

Lemma 3.5 ([10, Lemma 3.8])

Let G be a finite p-group and let xGΦ(G) be an element of order p. If tΦ(G){[x,g]gG}, then

(gGxg)(gGxtg)=1.
Theorem 3.6

Let p=3. Then the following hold:

  1. The p-central quotient F/λn(F) is a Beauville group if and only if n4.

  2. The series {λn(F)}n4 can be refined to a normal series of F such that two consecutive terms of the series have index p and for every term N of the series F/N is a Beauville group.

Proof.

Since the smallest Beauville 3-group is of order 35, the quotient F/λn(F) can only be a Beauville group if n4. We first assume that n=4. Now consider the group

H=a,b,c,d,ea3=b3=c3=d3=e3=1,[b,a]=c,[c,a]=d,[c,b]=e,

where we have omitted all commutators between generators which are trivial. This is the smallest Beauville 3-group. Since λ4(H)=1, F/λ4(F) maps onto H. On the other hand, it is clear that |F/λ4(F)|35 and so F/λ4(F)H. Consequently, F/λ4(F) is a Beauville group. Thus we assume that n5.

Now let us call G the quotient group F/λn(F). Consider the map ψ:GP defined in the proof of Theorem 3.3. Since ψ is an epimorphism, we have

ψ(λn-1(G))=λn-1(P).

Observe that the subgroup Kerψλn-1(G) has index 3 in λn-1(G), since the subgroup λn-1(P) is of order 3. Choose a normal subgroup N of F such that λn(F)N<λn-1(F) and N/λn(F)Kerψ. Then ψ induces an epimorphism ψ¯ from F/N to P.

We will see that L=F/N is a Beauville group, which simultaneously proves (i) and (ii). Let u and v be the images of x and y in L, respectively. Set k=n-12. Then o(uv)o(xyλn(F))=3k. On the other hand, since ψ¯(uv)=s1, we have o(uv)o(s1)=3k, and consequently we get o(uv)=3k in L. Since the subgroup F/λ4(F)H is not of maximal class, L is not of maximal class. Thus, by Lemma 3.4, there exist elements z,tΦ(L) such that z{[u,l]lL} and t{[v,l]lL}. We claim that {u,v} and {(uz)-1,vt} form a Beauville structure for L. Let A={u,v,uv} and B={(uz)-1,vt,(uz)-1vt}.

If a=u, which is of order 3, and b=vt or (uz)-1vt, we get agbh=1 for every g,hL, as in the proof of Theorem 3.3. When a=v and b=(uz)-1 or (uz)-1vt, the same argument applies. If we are in one of the following cases: a=u and b=(uz)-1, or a=v and b=vt, then the condition agbh=1 follows from Lemma 3.5.

It remains to check the case when a=uv and bB. For every element bB, we have ψ¯(b)PP1, which has order 3. Since o(uv)=o(s1), the condition agbh=1 follows from Lemma 3.1, as in the proof of Theorem 3.3. This completes the proof. ∎

Thus the quotients in Theorem 3.6 constitute an infinite family of Beauville 3-groups of order 3n for all n5.

Observe that as a consequence of Lemma 3.2, the signatures of the triples in the Beauville structures arising from Theorems 3.3 and 3.6 are unbounded as n goes to infinity. Consequently, these examples are different from those of Stix and Vdovina, since in their examples the signatures of one of the triples of the Beauville structures take the constant value.

We next compare the infinite family of Beauville 3-groups in Theorem 3.6 with the ones given in [10], by considering quotients of the Nottingham group over 𝔽3. We will show that these two infinite families of Beauville 3-groups only coincide at the group of order 35.

Before proceeding we recall the definition of the Nottingham group and some of its properties. The Nottingham group𝒩 over the field 𝔽p, for odd p, is the (topological) group of normalised automorphisms of the ring 𝔽p[[t]] of formal power series. For any positive integer k, the automorphisms f𝒩 such that f(t)=t+ik+1aiti form an open normal subgroup 𝒩k of 𝒩 of index pk-1. The lower central series of 𝒩 is given by

(3.1)γi(𝒩)=𝒩r(i),where r(i)=i+1+i-2p-1,

and

(3.2)𝒩kp=𝒩kp+r,where 0rp-1 is the residue of k modulo p

(see [5, Remark 1 and Theorem 6, respectively]).

Also, each non-trivial normal subgroup of 𝒩 lies between some γi(𝒩) and γi+1(𝒩) (see [5, Remark 1 and Proposition 2]).

By [10, Theorem 3.10], if p=3, a quotient 𝒩/𝒩k is a Beauville group if and only if k6 and kzm for all m1, where zm=pm+pm-1++p+2. Furthermore, by [10, Theorem 3.11], for i1 there exists a normal subgroup 𝒲 between 𝒩ip+3 and 𝒩ip+1 such that 𝒩/𝒲 is a Beauville group. This gives quotients of 𝒩 which are Beauville groups of every order 3n with n5.

Theorem 3.7

Let Nγ4(F) be a normal subgroup of F such that F/N is a Beauville group. Then F/N is not isomorphic to any quotient of N which is a Beauville group. On the other hand, F/γ4(F) is isomorphic to N/γ4(N).

Proof.

Since there is only one Beauville group of order 35 [1], it follows that F/γ4(F) is isomorphic to 𝒩/γ4(𝒩). Now suppose that F/N𝒩/𝒲, where γn(F)N<γn-1(F) for n5 and F/N is a Beauville group. Since F/N is of class n-1 and 𝒲 lies between two consecutive terms of the lower central series, we have γn(𝒩)𝒲<γn-1(𝒩). Note that if n=5, then

𝒩7=γ5(𝒩)𝒲<γ4(𝒩)=𝒩6

and so 𝒲=γ5(𝒩). If n>5, then 𝒲γ5(𝒩). Consequently, the isomorphism F/N𝒩/𝒲 implies that F/γ5(F)N𝒩/γ5(𝒩). We next show that this is not possible.

Note that by (3.1), we have γ2(𝒩)=𝒩3 and by (3.2), 𝒩33=𝒩9. Thus the exponent of γ2(𝒩/γ5(𝒩)) is 3. On the other hand, as in the proof of Theorem 3.6, there is an epimorphism from F/γ5(F)N to a p-group of maximal class P of order 35 with expP=32. It follows that 𝒩/γ5(𝒩) cannot be isomorphic to F/γ5(F)N. ∎


Communicated by Nigel Boston


Funding statement: The author is supported by the Spanish Government, grant MTM2014-53810-C2-2-P, the Basque Government, grants IT753-13 and IT974-16, and TÜBİTAK-BİDEB-2214/A

Acknowledgements

I would like to thank G. Fernández-Alcober, N. Gavioli and C. M. Scoppola for helpful comments and suggestions. Also, I would like to thank the Department of Mathematics at the University of the Basque Country for its hospitality while this paper was being written.

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Received: 2016-4-20
Revised: 2016-5-25
Published Online: 2016-7-27
Published in Print: 2017-3-1

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