Metabelian thin Beauville $p$-groups

0.15 MB·

by  Norberto Gavioli

.
METABELIAN THIN BEAUVILLE p-GROUPS
NORBERTOGAVIOLI,S¸U¨KRANGU¨L,ANDCARLOSCOPPOLA
Abstract. A non-cyclic finite p-group G is said to be thin if every normal
subgroup of G lies between two consecutive terms of the lower central series
and|γi(G):γi+1(G)|≤p2foralli≥1. Inthispaper,wedetermineBeauville
structuresinmetabelianthinp-groups.
7
1
0
2
1. Introduction
n
a Beauville surfaces [2,page159]arecomplexsurfacesofgeneraltypeconstructed
J
fromtwoorientableregularhypermapsofgenusatleast2,withthe sameautomor-
4
phism groups. A finite group which arises as an automorphism group linked to a
2
Beauville surface is called a Beauville group.

] A group-theoretical formulation of Beauville groups can be given as follows.
R
Given a finite group G and a couple of elements x,y ∈G, we define
G
Σ(x,y)= hxig ∪hyig ∪hxyig ,
.
h
g[∈G (cid:16) (cid:17)
t
a that is, the union of all subgroups of G which are conjugate to hxi, to hyi or to
m
hxyi. Then G is a Beauville group if and only if the following conditions hold:
[ (i) G is a 2-generator group.

1 (ii) There exists a pair of generating sets {x1,y1} and {x2,y2} of G such that
v Σ(x ,y )∩Σ(x ,y )=1.
1 1 2 2
6
Then {x ,y } and {x ,y } are said to form a Beauville structure for G.
0 1 1 2 2
By using this characterization, one can study whether a given finite group is
9
6 a Beauville group. For example, Catanese [5] showed that a finite abelian group
0 is a Beauville group if and only if it is isomorphic to C ×C , with n > 1 and
n n
. gcd(n,6) = 1. On the other hand, a remarkable result, proved independently by
1
0 GuralnickandMalle[11]andbyFairbairn,MagaardandParker[6]in2012,is that
7 every non-abelian finite simple group other than A is a Beauville group.
5
1 If p is a prime, not much was known about Beauville p-groups until very re-
:
v cently (see [1] and [3]). In [7, Theorem 2.5], Ferna´ndez-Alcober and Gu¨l extended
i Catanese’s criterion in the case of p-groups from abelian groups to a much wider
X
family of groups, including all p-groups having a ‘good behaviour’ with respect to
r
a taking powers, and in particular groups of class <p.
Anon-cyclicfinitep-groupGissaidtobethin ifeverynormalsubgroupofGlies
betweentwoconsecutivetermsofthelowercentralseriesand|γ (G):γ (G)|≤p2
i i+1
foralli≥1. Indeed,thesegroupsare2-generator. Thusitisnaturaltoaskwhether
theyareBeauvilleornot. Furthermore,thestudyofthinp-groupsisalsomotivated
bythefactthattheyusuallygiveexamplesofgroupswhosepowerstructuresarenot
so well-behaved. Well-known examples of thin p-groups are p-groups of maximal
class and quotients of the Nottingham group. In [7], all Beauville quotients of the
Nottinghamgroupweredetermined. Thankstotheill-behavedpowerstructure,the
firstexplicitinfinitefamilyofBeauville3-groupswasgivenbyconsideringquotients
Keywords and phrases. Beauvillegroups;metabelianthinp-groups.
The second author is supported by the Spanish Government, grant MTM2014-53810-C2-2-P,
andtheBasqueGovernment, grantIT974-16.
1
2 N.GAVIOLI,S¸.GU¨L,ANDC.SCOPPOLA
of the Nottingham group over F . In [8], the existence of Beauville structures in
3
themostimportantfamiliesofgroupsofmaximalclass,inparticularinmetabelian
p-groups of maximal class, was studied.
The goal of this paper is to complete the study of Beauville structures in
metabelian thin p-groups. In this paper, we will exclude p-groups of maximal
class from our consideration of thin groups. This means, in particular, that the
prime 2 is excluded [12, TheoremIII.11.9]. Then according to TheoremA in [4], if
G is a metabelian thin p-group, then cl(G) ≤ p+1. If the class is < p, then the
existence of Beauville structures can be determined by using Corollary 2.6 in [7].
Thus we restrict to groups of class p or p+1. The main result of this paper is as
follows.
Theorem A. Let G be a metabelian thin p-group of class p or p+1 for p ≥ 5.
Then there are four cases in which there is a Beauville structure:
(i) cl(G)=p and |γ (G)|=p2.
p
(ii) cl(G)=p+1.
(iii) cl(G)=p, |γ (G)|=p and Gp =γ (G).
p p−1
(iv) cl(G) = p, |γ (G)| = p, Gp = γ (G) and G has at least three maximal
p p
subgroups of exponent p.
Notation. We use standard notation in group theory. If G is a group, H,K ≤ G
and N EG, then H ≡K (mod N) means HN/N = KN/N. If p is a prime, then
we write Gpi for the subgroup generated by all powers gpi as g runs over G, and
Ω (G) for the subgroup generated by the elements of G of order at most pi. The
i
exponent of G, denoted by expG, is the maximum of the orders of all elements of
G.
2. Proof of the main result
In this section, we give the proof of Theorem A. We begin by giving some prop-
erties of metabelian thin p-groups. Firstly, we recall the following more general
result of Meier-Wunderli. If G is a metabelian 2-generator p-group, then
(1) Gp ≥γ (G)
p
(see [13], Theorem 3).
Observe that the only thin abelian p-group is the elementary abelian group of
order p2 and we refer to its lattice of normal subgroups as a diamond. It then
follows that if G is thin, then G/G′ is elementary abelian of order p2, and hence G
is 2-generator. Also, the lower and upper central series of a thin p-group coincide
[4, Corollary 2.2].
Note that if G is a thin p-group and g ∈γ (G)rγ (G), then
i i+1
(2) γ (G)=[g,G]γ (G)
i+1 i+2
(see [4], Lemma 2.1).
Now let G be a metabelian thin p-group. By Theorem A in [4], we have the
following:
(i) γ (G) is cyclic and γ (G)=1.
p+1 p+2
(ii) ThelatticeofnormalsubgroupsofGconsistsofadiamondontop,followed
byachainoflength1,atmostp−2diamonds,pluspossiblyanotherchain
of length 1.
As a consequence, cl(G) ≤ p + 1, and |G| ≤ p2p. We next recall the power
structure of a metabelian thin p-group.
METABELIAN THIN BEAUVILLE p-GROUPS 3
Lemma 2.1. Let G be a metabelian thin p-group, and l be the largest integer
such that Gp ≤ γ (G). Then 3 ≤ l ≤ p, γ (G) is cyclic, γ (G) = 1 and
l l+1 l+2
γ (G)p ≤γ (G).
2 l+1
Proof. See the proofs of Lemmas 1.2, 1.3 and 3.3 in [4]. (cid:3)
The next corollary follows directly from Lemma 2.1.
Corollary 2.2. Let G be a metabelian thin p-group. Then |Gp|≤p3.
Lemma 2.3. Let G be a metabelian thin p-group such that its lattice of normal
subgroups ends with a chain. Then the order of Gp cannot be p2.
Proof. If G is of class p+1, then Gp = γ (G), and hence |Gp| = p3. Thus we
p
assume that cl(G) = c ≤ p. Next observe that if M is a maximal subgroup of G,
then cl(M)<c≤p, and so M is regular.
Now suppose, on the contrary, that |Gp| = p2. Consider the quotient group
p
G = G/γ (G), which is regular. Then |G | = p, and hence |G : Ω (G)| = p.
c 1
Write Ω (G)=M/γ (G) for some maximal subgroup M of G. Since G is regular,
1 c
exp Ω (G)=p,andhenceMp ≤γ (G). Thisimpliesthat|Mp|=|M :Ω (M)|≤p
1 c 1
as M is regular. It then follows that G′ ≤Ω (M), and thus exp G′ =p.
1
Ontheotherhand,ifM isanarbitrarymaximalsubgroupofG,wehaveG′ ≤M
and since expG′ = p, we get G′ ≤ Ω (M) ≤ M. Then |Mp| = |M : Ω (M)| ≤
1 1
p. Since G is thin, this implies that Mp ≤ γ (G) for any maximal subgroup
c
M. But Gp = hMp | M maximal in Gi ≤ γ (G). Thus |Gp| ≤ p, which is a
c
contradiction. (cid:3)
We finally recall a commutator relation between the generators of G. More
specifically, if G is a metabelian thin p-group, then to every x ∈ G r G′ there
corresponds a y such that G=hx,yi and
(3) [y,x,x,x]≡[y,x,y,y]h (mod γ (G)).
5
where h is a quadratic non-residue modulo p [4, Theorem B].
BeforeweproceedtoproveTheoremA,wewilldeterminewhichmetabelianthin
3-groups are Beauville groups.
Theorem 2.4. A metabelian thin 3-group is a Beauville group if and only if it is
one of SmallGroup(35,3), SmallGroup(36,34), or SmallGroup(36,37).
Proof. LetG be a metabelianthin 3-group. Then|G|≤36. Note that the smallest
Beauville3-groupS =SmallGroup(35,3)isoforder35,anditisthe onlyBeauville
3-group of that order [1]. Furthermore, by using the computer algebra system
GAP [10], it can be seen that S is metabelian thin. Thus, if |G| = 35, then G is
a Beauvile group if and only if G ∼= S. We next assume that |G| = 36. It has
been shown in [1] that there are only three Beauville 3-groups of order 36, namely
S = SmallGroup(36,n) for n = 34,37,40. However, if n = 40 then S is not thin
since |Z(S)| = 9 and thus Z(S) 6= γ (S). On the other hand, if n = 34 or 37
4
then again by using the computer algebra system GAP, one can show that S is a
metabelianthin3-group. Consequently,GisaBeauvillegroupifandonlyifG∼=S
for n=34 or 37. (cid:3)
Thus we assume thatp≥5. LetG be a metabelian thin p-groupwith cl(G)=p
orp+1. Thenwehavethreecases: Gisofclassp+1,orisofclasspand|γ (G)|=p2,
p
orisofclassp and|γ (G)|=p. Inthe firsttwocases,wehaveGp ≤γ (G). Itthen
p p
follows from (1) that Gp = γ (G). Note that we also have γ (G)p ≤ γ (G), by
p 2 p+1
Lemma 2.1. On the other hand, in the last case if l is the largestinteger satisfying
Gp ≤γ (G), then l =p−1 or p and hence γ (G)≤Gp ≤γ (G).
l p p−1
4 N.GAVIOLI,S¸.GU¨L,ANDC.SCOPPOLA
Ourfirststepistocalculatethepthpowersofxtymoduloγ (G)forall0≤t≤
p+1
p−1 if G= hx,yi and γ (G)p ≤ γ (G). To this purpose, we need the following
2 p+1
lemma.
Lemma 2.5. [14, Lemma 6] Let G be a metabelian p-group and x,y ∈ G. Set
σ =y and σ =[σ ,x] for i≥2. Then
1 i i−1
(p) (p)
(xy)p =xpypσ 2 ...σ p z,
2 p
where
p−1p−1
z = [σ , σ ]C(i,j),
i+1 j 1
i=1j=1
Y Y
and
p−1
k k
C(i,j)= .
i j
k=1(cid:18) (cid:19)(cid:18) (cid:19)
X
Lemma 2.6. Let G be a metabelian thin p-group such that γ (G)p ≤ γ (G). If
2 p+1
x and y are the generators of G satisfying (3), then for all 0≤t≤p−1
(4) (xty)p ≡(xp)typ[y,x,p−2y]1−−h2tt2[y,x,p−3y,x]1−2th2t2 (mod γp+1(G)).
Proof. By Lemma 2.5, we have
p−1p−1
(xty)p =(xp)typ[y,xt](p2)...[y,p−1xt](pp) [y,ixt,jy]C(i,j).
i=1j=1
Y Y
Since γ (G)p ≤γ (G), it then follows that
2 p+1
(xty)p ≡(xp)typ[y, xt] [y, xt, y]C(i,j) (mod γ (G)).
p−1 i j p+1
1≤i,j
Y
i+j≤p−1
Notethat fori+j >0, C(i,j)is the coefficientofuivj in p−1(1+u)k(1+v)k,
k=0
where
p−1 P
(1+u)k(1+v)k ≡ (u+v)+uv p−1 (mod p).
k=0
X (cid:0) (cid:1)
Inthe previousexpressionthe monomialsoftotaldegreelessthanpappearonly
in (u+v)p−1 ≡ p−1(−1)rurvp−r−1 (mod p), and hence
r=0
P 0 (mod p) if i+j <p−1,
C(i,j)≡
((−1)i (mod p) if i+j =p−1.
Thus the condition γ (G)p ≤γ (G) implies that
2 p+1
p−1
(xty)p ≡(xp)typ [y, xt, y](−1)i (mod γ (G)).
i p−i−1 p+1
i=1
Y
On the other hand, notice that for 1≤t≤p−1
(5) [y,xt,xt,xt]≡[y,xt,y,y]ht2 (mod γ (G)).
5
Since G is metabelian, for every a,b∈G and c∈G′ we have [c,a,b]=[c,b,a], and
this, together with (5), yields that
[y,xt, y]−(ht2)s−1 (mod γ (G)) if i=2s−1 ,
[y, xt, y](−1)i ≡ p−2 p+1
i p−i−1 ([y,xt,p−3y,xt](ht2)s−1 (mod γp+1(G)) if i=2s ,
METABELIAN THIN BEAUVILLE p-GROUPS 5
and hence
P(p−1)/2(ht2)s−1
(xty)p ≡(xp)typ [y,x, y]−t[y,x, y,x]t2 s=1 (mod γ (G)).
p−2 p−3 p+1
(cid:16) (cid:17)
Notethatsincehisaquadraticnon-residue,wehave (p−1)/2(ht2)s−1 = 2 .
s=1 1−ht2
Consequently, we get
P
(xty)p ≡(xp)typ[y,x,p−2y]1−−h2tt2[y,x,p−3y,x]1−2th2t2 (mod γp+1(G)),
for 0≤t≤p−1, as desired. (cid:3)
Lemma 2.7. Let G be a metabelian thin p-group such that |γ (G)|≥p2 and let x
p
and y be the generators of G satisfying (3). Then for every t ∈ {0,1...,p−1}
0
there exist at most three distinct t∈{0,1...,p−1} such that
(6) h(xt0y)pi≡h(xty)pi (mod γ (G)).
p+1
Proof. Since |γ (G)|≥p2, we have Gp =γ (G) and γ (G)p ≤γ (G), by Lemma
p p 2 p+1
2.1. Also note that |γ (G) : γ (G)| = p2. Notice that, as a consequence of (2),
p p+1
l = [y,x, y] and m = [y,x, y,x] are linearly independent modulo γ (G).
p−2 p−3 p+1
Thus(l,m)isabasisofγ (G)moduloγ (G). Ifwesetxp ≡lαmβ (mod γ (G))
p p+1 p+1
and yp ≡lγmδ (mod γ (G)) for some α,β,γ,δ ∈F , then by (4), we have
p+1 p
(7) (xty)p ≡lγ+αt−1−2htt2 mδ+βt+1−2th2t2 (mod γp+1(G)).
Observe that as rational functions in t, neither f(t) = γ +αt− 2t nor g(t) =
1−ht2
δ+βt+ 2t2 are zero.
1−ht2
Wenowfixt ∈{0,1...,p−1}. Then(6)holdsifandonlyifthereexistsλ∈F∗
0 p
such that
f(t)=λf(t ) and g(t)=λg(t ).
0 0
If f(t ) = 0 or g(t ) = 0, then we have f(t) = 0 or g(t) = 0, that is (1−ht2)(γ +
0 0
αt)−2t=0 or (1−ht2)(δ+βt)+2t2 =0. Otherwise,we have f(t) = g(t) . Then
f(t0) g(t0)
g(t )f(t)−f(t )g(t)=0, that is
0 0
g(t ) (1−ht2)(γ+αt)−2t −f(t ) (1−ht2)(δ+βt)+2t2 =0,
0 0
whichisapol(cid:0)ynomialintofdegree≤(cid:1) 3. Thus(cid:0)ineverycase,thereare(cid:1)atmostthree
distinct t∈{0,1...,p−1} such that h(xt0y)pi≡h(xty)pi (mod γp+1(G)). (cid:3)
Lemma 2.8. Let G be a metabelian thin p-group such that γ (G)p ≤ γ (G). If
2 p+1
M is a maximal subgroupof Gand a,b∈MrG′, then haip ≡hbip (mod γ (G)).
p+1
Proof. If we write b = aic for some c ∈ G′ and for some integer i not divisible by
p, then by the Hall-Petrescu collection formula (see [12, III.9.4]), we have
(p) (p)
(aic)p =apicpc 2 c 3 ...c ,
2 3 p
where c ∈ γ (ha,ci) ≤ γ (G). Thus (aic)p ≡ api (mod γ (G)), and hence
j j j+1 p+1
haip ≡hbip (mod γ (G)). (cid:3)
p+1
Remark 2.9. If we replace x with x∗, where x∗ ∈ GrG′ is not a power of x,
there exists a corresponding y∗ satisfying (3). Then x ∈ h(x∗)t0y∗,G′irG′ for
some 0 ≤ t0 ≤ p−1, and according to Lemma 2.8, we have hxpi ≡ h((x∗)t0y∗)pi
(mod γ (G)). It then follows from Lemma 2.7 that there exist at most three
p+1
distinct t∈{0,1...,p−1} such that hxpi≡h(xty)pi (mod γ (G)).
p+1
The following corollary is an immediate consequence of Lemmas 2.7 and 2.8.
6 N.GAVIOLI,S¸.GU¨L,ANDC.SCOPPOLA
Corollary 2.10. Let G be a metabelian thin p-group such that |γ (G)|≥p2. If M
p
is a maximal subgroup of G, then there exist at most two maximal subgroups M ,
1
M different from M such that Mp ≡Mp ≡Mp (mod γ (G)).
2 1 2 p+1
Before we present the main result, we also need the following remark.
Remark 2.11. Let G be a finite 2-generator p-group. Then we can always find
elements x,y ∈ GrΦ(G) such that x,y and xy fall into the given three maximal
subgroups of G. Let M , M and M be three maximal subgroups of G. Choose
1 2 3
x∈ M rΦ(G) and y ∈M rΦ(G). Since each element in the set {xyj | 1 ≤j ≤
1 2
p−1} falls into different maximal subgroups, there exists 1≤j ≤p−1 such that
xyj ∈M rΦ(G). Thus if we put x∗=x and y∗=yj, then elements in the triple
3
{x∗,y∗,x∗y∗} fall into the given three maximal subgroups.
WearenowreadytoproveTheoremA.Wedealseparatelywiththe casesinthe
theorem.
Theorem 2.12. Let G be a metabelian thin p-group with cl(G) = p such that
|γ (G)| = p2, where p ≥ 5. Then G has a Beauville structure in which one of the
p
two triples has all elements of order p2.
Proof. We divide our proof into three cases depending on the number of maximal
subgroups whose pth powers coincide, and in every case, we take into account
Corollary 2.10 and Remark 2.11. First of all, note that since G has at most three
maximal subgroups of exponent p and p ≥ 5, there are at least three maximal
subgroups of exponent p2.
Case 1: Assumethatthereisa1-1correspondencebetweenmaximalsubgroups
M of exponent p2 and Mp. Choose a set of generators {x ,y } such that o(x )=
i i 1 1 1
o(y )=o(x y )=p2.
1 1 1
Case 2: Assume that there exist three maximal subgroups of exponent p2 such
that their pth power subgroups coincide. Then choose a set of generators {x ,y }
1 1
such that x ,y and x y fall into those maximal subgroups.
1 1 1 1
InbothCase1and2,sincep≥5,wecanchooseanothersetofgenerators{x ,y }
2 2
so that each pair of elements in {x ,y ,x y | i = 1,2} is linearly independent
i i i i
modulo G′ by Remark 2.11.
Case 3: Assume that we are not in the first two cases. Then there exist two
maximalsubgroupsM , M ofexponentp2 suchthat Mp =Mp andMp 6=Mp for
1 2 1 2 1
all other maximal subgroups M.
Let us first deal with p ≥ 7. We start by choosing a set of generators {x ,y }
1 1
where x ∈ M and y ∈ M are such that o(x y ) = p2, say x y ∈ M . Then
1 1 1 2 1 1 1 1 3
there might be a maximalsubgroupM suchthat Mp =Mp (note thatthere is no
4 3 4
other i 6= 3,4 satisfying Mp = Mp, otherwise we are in Case 2). Since p ≥ 7, we
i 3
can choose another set of generators {x ,y } so that x ,y ,x y ∈/ M and each
2 2 2 2 2 2 4
pair of elements in {x ,y ,x y | i = 1,2} is linearly independent modulo G′, by
i i i i
Remark 2.11.
If p = 5 then by using the construction of metabelian thin p-groups in [4] and
the computer algebra system GAP, one can show that there is no metabelian thin
5-group of class 5 such that |γ (G)| = 52 and 5th powers of maximal subgroups
5
coincide in pairs. Thus in Case 3, there exists a maximal subgroup, say M , of
3
exponent 52, where all other M5 are different from M5. Then choose sets of gen-
3
erators {x ,y } and {x ,y } so that x ∈ M , y ∈ M and x y ∈ M and each
1 1 2 2 1 1 1 2 1 1 3
pair of elements in {x ,y ,x y |i=1,2} is linearly independent modulo G′.
i i i i
We claim that, in every case, {x ,y } and {x ,y } form a Beauville structure
1 1 2 2
for G. If A={x ,y ,x y } and B ={x ,y ,x y }, then we need to show that
1 1 1 1 2 2 2 2
(8) hagi∩hbhi=1,
METABELIAN THIN BEAUVILLE p-GROUPS 7
for all a ∈ A, b ∈ B and g,h ∈ G. Note that o(a) = p2 for every a ∈ A. Assume
firstthato(b)=p. Ifhagi∩hbhi=6 1forsomeg,h∈G, thenhbhi⊆hagi, andhence
haG′i = hbG′i, which is a contradiction, since a and b are linearly independent
modulo G′. Thus we assume that o(b) = p2. If (8) does not hold, then h(ag)pi =
h(bh)pi, which contradicts the choice of b. (cid:3)
In order to deal with the case cl(G)=p+1, we need the following lemma.
Lemma 2.13. [9,Lemma4.2] LetG bea finitegroup and let {x ,y }and {x ,y }
1 1 2 2
be two sets of generators of G. Assumethat, for a given N EG, the following hold:
(i) {x N,y N} and {x N,y N} is a Beauville structure for G/N,
1 1 2 2
(ii) o(u)=o(uN) for every u∈{x ,y ,x y }.
1 1 1 1
Then {x ,y } and {x ,y } is a Beauville structure for G.
1 1 2 2
Theorem 2.14. Let G be a metabelian thin p-group with cl(G) = p+1, where
p≥5. Then G has a Beauville structure.
Proof. By Theorem 2.12, G = G/γ (G) has a Beauville structure in which one
p+1
ofthetwotripleshasallelementsoforderp2,i.e. theyhavethesameorderinboth
G and G. Then we can apply Lemma 2.13 and thus G is a Beauville group. (cid:3)
We next analyze the case cl(G) = p and |γ (G)| = p. Recall that we have
p
γ (G)≤Gp ≤γ (G), and thus there are two possibilities:
p p−1
(i) Gp =γ (G),
p−1
(ii) Gp =γ (G).
p
Observe that by Lemma 2.3, Gp cannot be a proper subgroup of γ (G) of
p−1
order p2.
Theorem 2.15. Let G be a group in case (i). Then G has a Beauville structure .
Proof. Firstofall,noticethatthereexistsapairofgeneratorsaandbofGsuchthat
ap and bp are linearly independent modulo γ (G). By the Hall-Petrescu formula,
p
we have
(p) (p)
(atb)p =atpbpc 2 ...c p ,
2 p
where c ∈γ (hat,bi). Since γ (G)p ≤γ (G), by Lemma 2.1, we get
j j 2 p
(atb)p ≡atpbp (mod γ (G))
p
for1≤t≤p−1. Observethat,similarlytoLemma2.8,foreverymaximalsubgroup
M, m ∈ M and c ∈ G′ , we have (mc)p ≡ mp (mod γ (G)). It then follows that
p
the power subgroups Mp are all different modulo γ (G).
p
On the other hand, since G = G/γ (G) is of class p−1, it is a regular p-group
p
such that |Gp| = p2. According to Corollary 2.6 in [7], G is a Beauville group
since p≥5. From the observationabove, all elements outside G′ are of order p2 in
both G and G. Then we can apply Lemma 2.13 to conclude that G is a Beauville
group. (cid:3)
Theorem 2.16. Let G be a group in case (ii). Then G has a Beauville structure
if and only if it has at least three maximal subgroups of exponent p.
Proof. If the number of maximal subgroups of exponent p is less than three, then
Ω (G)iscontainedintheunionofatmosttwomaximalsubgroups. Since|Gp|=p,
1
it then follows from Proposition 2.4 in [7] that G has no Beauville structure.
On the other hand, if at least three maximal subgroups have exponent p, then
we choose a triple in which all elements have order p. Since p ≥ 5, we can choose
another triple such that each pair of elements in the union of the two triples is
linearly independent modulo G′. Then G has a Beauville structure. (cid:3)
8 N.GAVIOLI,S¸.GU¨L,ANDC.SCOPPOLA
Acknowledgments
We would like to thank G.Ferna´ndez-Alcober for helpful comments andsugges-
tions. Also,the secondauthorwouldliketothankthe DepartmentofMathematics
at the University of L’Aquila for its hospitality while this researchwas conducted.
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Universita` degliStudidell’Aquila, L’Aquila, Italy
E-mail address: [email protected]
DepartmentofMathematics,MiddleEastTechnicalUniversity,06800Ankara,Turkey
E-mail address: [email protected]
Universita` degliStudidell’Aquila, L’Aquila, Italy
E-mail address: [email protected]