Uniform (2,k)-generation of the 4-dimensional classical groups

0.33 MB·English

by  M. A. Pellegrini

Uniform (2,k)-generation of the 4-dimensional classical groups
M. A. Pellegrini
DipartimentodiMatematicaeApplicazioni,Universit`a degliStudidiMilano-Bicocca
ViaR.Cozzi,53,20125MilanoItaly
1
E-mail: [email protected]
1
0
M. C. Tamburini Bellani
2

n DipartimentodiMatematicaeFisica,Universit`aCattolicadelSacroCuore
a ViaMusei,41,25121BresciaItaly
J E-mail: [email protected]

2
M. A. Vsemirnov1
1

St.PetersburgDivisionofSteklovInstituteofMathematics
]
R 27,Fontanka,191023,St.Petersburg,Russia
E-mail: [email protected]
G
.
h
t
a Inthispaperwestudythe(2,k)-generationofthefiniteclassicalgroupsSL4(q),
m Sp4(q),SU4(q2)andtheirprojectiveimages. Herekistheorderofanarbitrary
[ element of SL2(q), subject to the necessary condition k ≥3. When q iseven
weallowalsok=4.

1
v
1. INTRODUCTION
8
5
We recall that a group is (2,k)-generated if it can be generated by two elements of respective
3
2 orders2andk. Ouraimistofinduniform(2,k)-generatorsofthe4-dimensionalclassicalgroups.
. Since two involutions generate a dihedral group, we assume k 3.
1 ≥
In this problema special case of a formula of L. Scott [20, Theorem1] plays a crucialrole. E.g.
0
1 it gives constraints for the similarity invariants of a (2,k)-generating pair of the groups under
1 consideration(see(1)). Moreoveritgivesarigiditycriterion,provedbyStrambachandVo¨lklein
: [21,Theorem2.3],whichisveryuseful(see(2)). WeacknowledgeA.Zalesskiiforhavingbrought
v
i our attention to this subject and to a systematic study of Scott’s result. Sections 2 and 3 are
X
dedicated to this study, aiming to develop more general techniques for linear groups.
r In Section 4 we fix the canonical forms of our uniform generators x,y. Having in mind the
a
(2,k)-generation of the projective images of the classical groups, we allow x2 = I. On the
±
other hand both y and its projective image have order k. The choice of the canonical form of y
is determined, for uniformity reasons, by the case k = 3 (see the beginning of Section 6). We
characterize the shapes of x and y, up to conjugation, subject to the condition that the group
x,y is absolutely irreducible (see (11)). The matrix x is uniquely determined by its order,
h i
y has an entry s which determines its order k and four indeterminate entries r ,...,r : their
1 4
values which still produce a reducible group are described by Lemma 4.1.
In Section 6, we are ready to prove a list of negative results. They show that, apart from
the groups Sp (q), which probably require a generator of order k 4 with similarity invariants
4 ≥
otherthany,ourpositiveresultsarethebestpossible. Inparticularthefollowinggroupsarenot
(2,3)-generated: SL (2), Sp (q) for all q, PSp (2a), PSp (3a), SU (9) and PSU (9). Moreover
4 4 4 4 4 4
1The third author was supported by the Cariplo Foundation, the Russian Foundation for Basic Research
(grant no.09-01-00784-a) andtheFederal TargetProgramme“Scientific andscientific-pedagogical personnel of
theinnovativeRussia2009-2013” (contract no.P265).
1
SL (3), SU (9) and PSU (9) are not (2,4)-generated: this last fact has requiredan unexpected
4 4 4
amount of details. The exception of symplectic groups were detected by Liebeck and Shalev in
[13]. Here we give an alternative proof.
InSection 7we specialize the values ofthe parametersr ,...,r , aiming to ourpositive results.
1 4
Let F be an algebraically closed field of characteristic p > 0. If (k,p) = 1, let us denote by ǫ
a primitive k-th root of unity in F. If k = p or k = 2p, we set respectively ǫ = 1 or ǫ = 1.
−
Writing s=ǫ+ǫ−1, our uniform generators have shapes:
0 0 1 0 1 0 0 r
2
0 0 0 1 0 1 0 r
x=d 0 0 0, d=±1, y = 0 0 0 41, r4 6=0.
−
 0 d 0 0  0 0 1 s 
   
   
By Corollary 7.1 the group H = x,y is absolutely irreducible provided r = ǫ±1r and
2 4
h i 6 −
r +r = (2 s)√d. We make the assumption that H is absolutely irreducible, which has
2 4
6 ± −
several consequences.
First of all, by Remark 4.1, the triple (x,y,xy) is rigid and xy must have a unique similarity
invariant. It follows that, for any field automorphism σ, the matrices (xy)σ and (xy)−1 are
conjugate if and only if their characteristic polynomials are the same. They are respectively:
χ (t) = t4 drσt3 dsσt2 rσt+1,
(xy)σ − 4 − − 2
χ(xy)−1(t) = t4−r2t3−dst2−dr4t+1.
Suppose that r , r and s belong to F . If H Sp (q), then xy is conjugate to its inverse,
and we obtain t2he4necessary conditionqr = dr ≤. If d4= 1 and r = r , then H SO±(q) by
2 4 2 4
≤
Theorem 7.2(ii). This fact and Theorem 6.1(iv) explain why, in the symplectic case, we must
assume d = 1 and p odd. On the other hand, the conditions d = 1, p odd and r = r
2 4
sare suFffi.cieIfntH−to guSaUran(tqe2e),thtaatkiHng≤thSep4F(rqo)bbeyniTushemoraepmα7.2(vα).qNaesxtths−eupfipeoldseatuhtaotmro2r,pr4hi∈smF−qσ2,,
q 4
we∈obtain the n≤ecessary condition r = drq. On the other7→hand, this condition is sufficient to
2 4
guarantee that H is contained in SU (q2) by Theorem 3.1(ii).
4
Now suppose that the group H is contained in one of the classical groups under consideration.
If H is not the whole group, then it is contained in some maximal subgroup M. Following
Aschbacher’sstructureTheorem[1], the groupM belongsto oneofnine classes. Eightofthem,
denoted from to , correspond to natural subgroups. The remaining class results from
1 8
C C S
absolutely irreducible representations of finite simple groups. So, for fixed q and s,we have to
exclude all values of r and r for which H is contained in some M in the above classes. If
2 4
M , then it is reducible over F: thus Corollary 7.1 takes care of this case. If M ,
1 3 2
∈ C ∪C ∈ C
it stabilizes a direct sum decomposition: this possibility is considered in Section 8. If M ,
5
∈ C
it stabilizes a subfield: see Lemma 5.2. The case M , the class of classical subgroups,
8
∈ C
is studied in Theorem 7.1. In dimension 4, this analysis includes also the case M .
4 7
Indeed, up to conjugation, such an M is contained in GL (F) GL (F): hence it fi∈xesC, u∪pCto
2 2
⊗
a scalar, the matrix J J, where J = antidiag(1, 1). Finally, the cases M and M
6
⊗ − ∈ C ∈ S
are considered in Sections 9 and 10 respectively. The conditions obtained, except those given
by Lemma 5.2, are summarized in Table 4.
We are not aware of any published list of the maximal subgroups of the finite 4-dimensional
classical groups. So our reference was the Ph.D. thesis of Kleidman [11], whose list is based on
the work of several authors, namely [6], [9], [17], [18], [19], [22], [30].
ThelastSectioncontainsourpositiveresults. Theirprecisestatements,formulatedinTheorems
11.1, 11.2, 11.3 and 11.4, can be roughly summarized as follows. Up to a finite number of
exceptions, completely determined, we have that:
if d= 1, 0=r , 0=r F and F =F s,r2 , then x,y =SL (q);
• ± 2 6 4 ∈ q q p 4 h i 4
• if d=−1, r2 =−r4, p6=2, s6=2, 06=r4 ∈(cid:2)Fq an(cid:3)d Fq =Fp s,r42 , then hx,yi=Sp4(q);
2
(cid:2) (cid:3)
• if d=±1, s∈Fq, r2 =dr4q, 06=r4 ∈Fq2 and Fq2 =Fp r42 , then hx,yi=SU4(q2).
As a consequence we obtain that, for all k 3 such that k (q 1) or k (q+1) or k = p or
(cid:2) (cid:3)
≥ | − |
k =2p, the following simple groups are (2,k)-generated:
PSL (q), q >2, PSp (q), k =p, q odd, PSU (q2), q >3.
4 4 6 4
In particular they are (2,3)-generated,except PSp (2a) and PSp (3a).
4 4
Actually other papers establish explicitly the (2,3)-generationof many of them. To our knowl-
edge the groupsPSL (q) were consideredin [14], [23], [24] andthe groups PSp (q) were consid-
4 4
ered in [3].
MorerecentworkinarelatedareaisduetoMarion[15],whostudiesthegroupsPSL (q),n 3,
n
≤
which are epimorphic images of a given hyperbolic triangle group.
Finally we note that mostof our results are computer independent. Nevertheless MAGMA and
GAP have been of great help in the computational aspects of this paper.
2. SCOTT’S FORMULA AND RIGIDITY
We recall some basic consequences of Scott’s formula [20, Theorem 1]: for their background we
refer to [25]. Here X,Y denotes an absolutely irreducible sugbgroup of GL (L), where L is a
n
h i
field. Fora subset K ofM =Mat (L), let dK be the dimensionofC (K). Then, withrespect
n M M
to the conjugation action of X,Y on M, Scott’s formula gives the condition:
h i
dX +dY +dXY n2+2. (1)
M M M ≤
Moreover,if equality holds, namely if
dX +dY +dXY = n2+2 (2)
M M M
thetriple(X,Y,XY)isrigid[21,Theorem2.3]. Thismeansthat,foranyothertriple(X′,Y′,X′Y′)
with the same similarity invariants as (X,Y,XY), there exists g GL (L) such that X′ =Xg
n
∈
and Y′ =Yg. In particular the groups X,Y and X′,Y′ are conjugate.
h i h i
We mayidentify the space Ln Ln with M,via the linear extensionofthe mape e e eT.
⊗ i⊗ j 7→ i j
In particular the symmetric square S of Ln Ln is identified with the space of symmetric
⊗
matrices. Clearly, for any g K as above, the diagonal element g g acts as m gmgT, for
all m M. Now letus denot∈e respectivelyby dK anddˆK the dimen⊗sionofthe spa7→ceofK-fixed
∈ S S
points on S and on its dual. In this case, Scott’s formula gives the condition:
n(n+1)
dX +dY +dXY +dhX,Yi+dˆhX,Yi. (3)
S S S ≤ 2 S S
By [25, Lemma 1], setting K = X,Y , we have dK 1, dˆK 1. If charL=2, then dK =dˆK.
h i S ≤ S ≤ 6 S S
Moreover if dˆK =1, then dK =1 and K is contained in an orthogonalgroup.
S S
In characteristic 2 it may happen that dˆK = 0, and dK = 1. To exclude this possibility in
S S
certain situations, it may be useful the following:
Lemma 2.1. Let g GL (L), with charL=2. Assume that g is conjugate to its inverse. Then
n
dg n/2 if n is eve∈n, dg (n+1)/2 if n is odd.
S ≥ S ≥
Proof. Let V = m Mat (L) gmgT =m . Since g is conjugate to its inverse, there ex-
n
∈ |
ists h GL (L) such that gT = h−1g−1h. It follows that gmgT = m if and only if g
n
centrali∈zes mh−1,(cid:8)whence dim V = dg (cid:9)n. This inequality can be seen noting that,
M ≥
for each companion matrix c of the rational form of g, the algebra L[c] centralizes c. Now
consider the map from V to S: m m + mT. The image of this map lies in V S.
7→ ∩
3
Since charL = 2, the kernel is also in V S. Since at least one of the dimensions of the
∩
image and of the kernel is at least half of the dimension of V, this completes the proof.
3. GROUPS PRESERVING A FORM
The followingtheoreminthe unitarycaseappearedin[29, Lemma6.2]. When Lisa finite field
see also [27, Theorem 2.12].
Theorem 3.1. Let L be a field and σ Aut(L). Let X,Y GL (L). Suppose that Xσ X−1,
n
∈ ∈ ∼
Yσ Y−1, (XY)σ (XY)−1. Assume further that X,Y is absolutely irreducible and that
∼ ∼ h i
(2) holds.
(i) If σ =id, then X,Y fixes a non-degenerate symmetric or skew-symmetric form.
h i
(ii) If σ is an involution, then X,Y fixes a non-degenerate hermitian form.
h i
Proof. Let Mφ =Mat (L) be the X,Y -module equipped with the following action φ on it:
n
h i
φ(h).m=hσmhT,
for all h X,Y . Using the non-degeneratepairing (m ,m )=Tr(m m ) we canidentify Mφ
1 2 1 2
∈h i
with his dual. Via this identification, the dual representation φ∗ is equivalent to φˆ:
φˆ(h).m=(hT)−1m(hσ)−1.
Clearly hσ is conjugate to its transpose. If hσ h−1, then choose an invertible matrix g such
∼
that (hσ)T = gh−1g−1. Then hσmhT = m if and only if h(mTg)h−1 = mTg. In particular,
dh = dh . Now, Scott’s formula for the module Mφ together with the assumptions of the
Mφ M
Theorem and (2) imply that either dhX,Yi 1 or dˆhX,Yi 1.
Mφ ≥ Mφ ≥
First,weshowthatdhX,Yi 1yieldsdˆhX,Yi 1. Tothispurposeassumethat,forsomem=0,
Mφ ≥ Mφ ≥ 6
hσmhT =m (4)
for every h X,Y . Let V be the eigenspace of m relative to 0. In particular V = Ln, as
∈ h i 6
m = 0. For any h X,Y and any v V, we have mhTv = (hσ)−1mv = 0. Therefore,
6 ∈ h i ∈
V is XT,YT -invariant. By the absolute irreducibility of X,Y it follows V = 0 , i.e., m
h i h i { }
is non-degenerate. Inverting both sides of (4), we have (hT)−1m−1(hσ)−1 = m−1. Hence,
hTm−1hσ =m−1 and, since φˆ is equivalent to the dual representation φ∗, we have dˆhX,Yi 1,
Mφ ≥
as desired.
Now consider the case dˆhX,Yi 1. Assume that m=0 is such that the equality
Mφ ≥ 6
(hT)−1m(hσ)−1 =m (5)
holds for any h X,Y . We show that m is invertible. Let V be the eigenspace of mT
∈ h i
relative to 0. For any h X,Y and any v V, we have (hv)Tm = vTm(hσ)−1 = 0, whence
∈ h i ∈
mT(hv) =0. Thus V is X,Y -invariant and V = 0 by the absolute irreducibility of X,Y .
h i { } h i
Therefore, m is non-degenerate. In particular, this implies that any two non-zero matrices in
Mat (L) satisfying (5) must be proportional.
n
Equation (5) shows that X,Y fixes a bilinear form m defined over L.
h i
Transpose both sides of (5) and apply σ. We have
(hT)−1(mT)σ(hσ)−1 =(mT)σ.
By what observed above,
(mT)σ =βm for some β L∗. (6)
∈
4
(i) Assume that σ = id. Repeating (6) twice, we have β = 1, i.e., m is either symmetric or
±
skew-symmetric.
(ii) Assume that σ is an involution. Let F = Inv(σ) be the subfield fixed pointwise by σ. Our
next aim is to find a suitable scalar α L∗ such that αm is hermitian, i.e., ((αm)σ)T = αm.
∈
Iterating (6) we have that ββσ = 1. By Hilbert’s Theorem 90 [8, page 297] for the extension
L/F, there is α such that β =α/ασ. Therefore,
((αm)σ)T =ασβm=αm,
as desired.
Corollary 3.1. Let L be a field and σ Aut(L). Let X,Y GL (L). Suppose that for some
n
∈ ∈
λ, µ L we have Xσ λλσX−1, Yσ µµσY−1, (XY)σ λµ(λµ)σ(XY)−1. Assume further
∈ ∼ ∼ ∼
that X,Y is absolutely irreducible and (2) holds.
h i
(i) If σ =id, then X,Y is contained in a conformal orthogonal or in the conformal symplectic
h i
group.
(ii) If σ is an involution, then X,Y is contained in a conformal unitary group. Moreover, if
h i
X, Y GL (F), where F =Inv(σ) is the subfield fixed pointwise by σ, then X,Y is contained
n
∈ h i
in a conformal orthogonal or in the conformal symplectic group defined over F.
Proof. Set X = λ−1X, Y = µ−1Y. Then Xσ = (λ−1)σXσ λX−1 = X−1, Yσ Y−1,
1 1 1 ∼ 1 1 ∼ 1
(X Y )σ (X Y )−1. By Theorem 3.1, X ,Y fixes a non-degenerate (symmetric or skew-
1 1 1 1 1 1
∼ h i
symmetric or, respectively, hermitian) form m. Hence
XσmXT =λλσm, YσmYT =µµσm,
i.e., X,Y is contained in the corresponding conformal group.
h i
Moreover,the proof of Theorem 3.1 shows that m is unique up to a scalar multiple. Therefore,
if σ is an involution and X, Y GL (F), then m = αm , where m GL (F). (Since
n 1 1 n
∈ ∈
λλσ, µµσ F, one can find the entries of m as a solution of a system of linear equa-
1
∈
tions defined over F). Since m is hermitian, we have mT = (ασ/α)m = βm . Applying
1 1 1
σ to the last relation, we find β =β−1, i.e., m is either symmetric or skew-symmetric.
1
Whenσ =id,Theorem3.1doesnotallowtodistinguishbetweensymmetricandskew-symmetric
forms, as both cases may arise. We give some conditions under which the symmetric case can
be detected.
Lemma 3.1. Let X,Y GL (L) and suppose that X,Y is absolutely irreducible. Assume
n
∈ h i
furtherthatdX+dY = n(n+1) andthatXY is conjugatetoitsinverse. Then X,Y iscontained
S S 2 h i
in an orthogonal group.
Proof.
SettingK = X,Y ,relation(3)givesdXY dK+dˆK. FromtheassumptionthatXY isconju-
h i S ≤ S S
gateto itsinverseitfollows1 dXY. Actually,whenp=2,wehavethe strongercondition2
≤ S ≤
dXY byLemma2.1. NowwemakerepeateduseofLemma1of[25]whichsays,first,thatdK 1
S S ≤
and dˆK 1. Moreover it says that, when p is odd, dK =dˆK. We conclude dK = dˆK = 1. Our
S ≤ S S S S
claim follows again from Lemma 1 of [25].
The previous Lemma is a special case of a more general fact (see Corollary 3.2 below), which
uses the following result, essentially proved in [28, Lemma 3.4].
Lemma 3.2. Let g GL (L) and let
n
∈
gTmg =m (7)
5
for some non-degenerate matrix m which is either symmetric or skew-symmetric. Let µ be the
g
minimal polynomial of g. Assume that either (i) degµ >2a or (ii) degµ =2a and the middle
g g
coefficient of µ is non-zero. Then dg a.
g S ≥
Proof. Define θ :Mat (L) S as follows: θ(u)=u+uT. Clearly, if
n
→
gTug =u, (8)
then gTuTg = uT and gTθ(u)g = θ(u). Notice that for any i the matrix u = mgi satisfies (8).
Let
a
U = c mgi :c L .
i i
( ∈ )
i=1
X
It follows from (7) that m−1(gi)Tm = g−i. Set λ = 1 if m is symmetric, λ = 1 if m is
−
skew-symmetric. Consider
a a a
gam−1θ c mgi = ga c gi+gam−1 c (gi)TmT
i i i
!
i=1 i=1 i=1
X X X
a a
= ga c gi+λga c m−1(gi)Tm
i i
i=1 i=1
X X
a a
= ga c gi+λga c g−i
i i
i=1 i=1
X X
a−1 2a
= λ c gi+ c gi.
a−i i−a
i=0 i=a+1
X X
Clearly, under the assumptions of the Lemma, this sum is zero only if all c vanish. Therefore,
i
thekerneloftherestrictionofθ toU istrivialand dim θ(U)= dim U =a. Inparticular,dg
S ≥
a.
Corollary 3.2. Let X,Y GL (L) be as in Theorem 3.1 with σ = id and assume further
n
∈
that:
n2+n degµ
dX +dY XY +2. (9)
S S ≥ 2 − 2
Then X,Y is contained in an orthogonal group.
h i
Proof. By Theorem 3.1(i), there exists a non-degenerate, symmetric or skew-symmetric, form
m which is preserved by X and Y. Clearly degµ > 2 degµg 1 . Thus, applying Lemma
g 2 −
3.2 to g = XY, we have dXY 1degµ 1. If L has(cid:16)characteris(cid:17)tic 2, we have the stronger
S ≥ 2 XY −
inequality dXY n 1degµ by Lemma 2.1. Thus, under assumption (9) we get
S ≥ 2 ≥ 2 XY
n2+n
dX +dY +dXY +1
S S S ≥ 2
and, when L has characteristic 2
n2+n
dX +dY +dXY +2.
S S S ≥ 2
Our claim follows by the same arguments used in the proof of Lemma 3.1.
6
4. CANONICAL FORMS AND SHAPES OF A (2,K)-GENERATING PAIR
Let F be an algebraically closed field of characteristic p 0. We fix an integer k 3. If p = 0
or (p,k) = 1, we denote by ǫ a primitive k-th root of≥unity in F. If (p,k) = p≥we suppose
k p,2p and, for k =p 3, we set ǫ=1, for k=2p 4 we set ǫ= 1.
Ne∈xt{we c}onsider two linea≥r transformations ξ,η of F4 w≥ith respective−similarity invariants:
t2 d, t2 d, d= 1,
• − − ±
t 1, t3 (1+s)t2+(1+s)t 1, s=ǫ+ǫ−1.
• − − −
The projective image of ξ has order 2. The Jordan form of η is respectively
1 1 1
1 1 0 0 1
 ,  ,   (10)
ǫ 1 1 0 1 0
−
 ǫ−1   0 1 1  1 1
     − 
     
according as (i) (k,p)=1, (ii) k =p or k =4 and p=2, (iii) k =2p 6. It follows that η and
≥
its projective image have order k.
We assume further that ξ,η acts irreducibly on F4. As the eigenspace V of η relativeto 1 has
dimension 2, and V ξ(hV) =i 0 by the irreducibility of ξ,η , we have V +ξ(V) = F4. Thus
= ξ(v ),ξ(v ),v ,∩v is basis of F4 whenever v ,v h isia basis of ξ(V). Considering the
1 2 1 2 1 2
rBation{alcanonicalformo}fthelineartransformation{induc}edbyη onF4/V,wemayassumethat
v and v are chosen so that the matrices of ξ and η, with respect to , have shapes:
1 2
B
0 0 1 0 1 0 r r
1 2
0 0 0 1 0 1 r r
x= , y = 3 4  (11)
d 0 0 0 0 0 0 1
−
 0 d 0 0  0 0 1 s 
   
   
for suitable r F with (r ,r )=(r ,r ) if k =p or k =4 and p=2 (i.e. if s=2).
i 1 3 2 4
∈ 6
For δ = √d, the corresponding eigenspace of x is:
±
(a,b,δa,δb)T a,b F . (12)
| ∈
For ǫ=1, the eigenspace of y relat(cid:8)ive to ǫj (j = 1) is gen(cid:9)erated by:
6 ±
u = (r ǫjr , r ǫjr , ǫj 1, ǫ2j+ǫj)T. (13)
ǫj 1− 2 3− 4 − −
Clearly when k=2p 6, i.e., ǫ= 1, the two vectors coincide.
≥ −
Lemma 4.1. Let x,y be defined as in (11). Then H = x,y is a reducible subgroup of SL (F)
4
h i
if and only if one the following conditions holds for some j = 1, and some δ = √d:
± ±
(i) r =r ǫjr +ǫ−jr ;
4 1 2 3
−
(ii) ∆=r r r r +δ−1((s 1)r r +r r )+(2 s)d=0.
1 4 2 3 1 2 3 4
− − − − −
Proof.
If (i) holds, taking v = ǫ−j,1,0,0 T, we get yxv = d(r ǫjr )v +ǫjxv. Thus v,xv is a
1 2
− − h i
2-dimensional H-module. On the other hand, if (ii) holds, there exists (a,b)= (0,0) such that
(cid:0) (cid:1) 6
w = (δa,δb,a,b)T is fixed by yT. Since w is an eigenvector of xT, the 1-dimensional space w
h i
is HT-invariant. It follows that H has a 3-dimensional submodule.
Viceversa, let W be a proper H-submodule.
Case 1. dim W =1. In this case W is generatedby a commoneigenvectoru of x and y. From
xW =W wegetu e ,e . Hence k =pifpisodd, k =4ifp=2and,uptoascalar,u=u
1 2 ǫj
6∈h i 6 6
for some j = 1. From xu=δu for some δ = √d, we get
± ±
r =ǫjr +δ−1(ǫj 1), r =ǫjr +δ−1(ǫj ǫ2j).
1 2 3 4
− −
7
These conditions imply condition (i).
Case 2. dim W = 3. In this case HT has a 1-dimensional invariant space. A generator must
have shape (δa,δb,a,b)T, in order to be an eigenvector of xT. And a non-zero vector w of this
shape is an eigenvector of yT only if yTw =w. This condition gives (ii).
Case 3. dim W = 2. Assume first that there is a non-zero v W such that yv = v. It
∈
follows that v and xv are linearly independent, hence generate W. By the shape of y, we may
assume v = (α,1,0,0)T for some α F. From yxv = λv +µxv we get µα = 1, µ = ǫj,
∈ −
and these conditions easily give that (i) must hold. Next suppose that yv = v for all non-zero
6
v W. It follows that ǫ = 1 and the characteristic polynomial of the linear transformation
η ∈, say, induced by y on F6 4/W must be (t 1)2. Considering the minimum polynomial of
0
−
y, we have (η I)(η ǫI)(η ǫ−1I) = 0. As the second and third factors are invertible,
0 0 0
we get η = I.−Thus −y must i−nduce the identity on F4/W. So we get that HT fixes the
0
space U of the fixed points of yT. Since U is fixed by xT, we can choose a non-zero vector
w U whichisaneigenvectorofxT. Hence w isHT-invariant,andcondition(ii)musthold.
∈ h i
The characteristic polynomials of xy and (xy)−1 are respectively:
χ (t) = t4 d(r +r )t3+(r r r r ds)t2+(r s r +r )t+1; (14)
xy 1 4 1 4 2 3 1 2 3
− − − −
χ(xy)−1(t) = t4+(r1s r2+r3)t3+(r1r4 r2r3 ds)t2 d(r1+r4)t+1. (15)
− − − −
Remark 4. 1. If x,y are as in (11), by a formula of Frobenius [8, Theorem 3.16, p. 207],
dim (CMat4(F)(x))=8, dim (CMat4(F)(y))=6. (16)
When H = hx,yi is absolutely irreducible, from (16) and (1) we get dim (CMat4(F)(xy)) = 4.
In particular equality holds in (1), i.e., the triple (x,y,xy) is rigid. Moreover xy has a unique
similarity invariant, equivalently its minimal and characteristic polynomials coincide.
5. FIELD OF DEFINITION
For lack of a reference, we sketch a proof of the following well known fact.
Lemma 5.1. Let Ω be the set of coefficients of the similarity invariants of h GL (F). If
n
hg GL (F ), with g GL (F) and F F, then Ω F . ∈
n 1 n 1 1
∈ ∈ ≤ ⊆
Proof. Call C and C the rational canonical forms of h and hg respectively in GL (F) and
1 n
GL (F ). FromC conjugate to C in GL (F), we have C =C. As all elements of Ω appear as
n 1 1 n 1
entries of C, we conclude that Ω F .
1
⊆
We denote the centre of GL (F) by F∗I.
4
Lemma 5.2. Forx,ydefinedasin (11),letH = x,y beconjugatetoasubgroupofGL (F )F∗I,
4 1
for some F F. Then F F s,(r +r )2 , whhereiF denotes the prime subfield.
1 1 p 1 4 p
≤ ≥
(cid:2) (cid:3)
Proof. Assume Hg GL (F )F∗I and write xg = x λ−1, yg = y ρ−1 with x ,y GL (F ),
4 1 1 1 1 1 4 1
λ,ρ F∗. The simila≤rity invariants of ρy are t ρ, t3 ρ(s+1)t2+ρ2(s+1)t ρ3.∈As (ρy)g =
ρyg =∈ y ,itfollowsfromthepreviousLemmath−atρan−dsareinF . Inthesame−way,asthesim-
1 1
ilarityinvariantsofλxaret2 dλ2,t2 dλ2,d= 1;wegetλ2 F . Asx y =(λxρy)g hastrace
1 1 1
− − ± ∈
dλρ(r +r ) F , it follows that (r +r )2 F .
1 4 1 1 4 1
∈ ∈
8
Lemma 5.3. Let q =pa be a prime power, with a>1. Denote by N be the number of non-zero
elements r F such that F r2 =F . Then:
q p q
∈ 6
(cid:2) (cid:3)
N 2(p 1) if a=2;
≤ − p(p⌊a/2⌋−1)
N (p 1,2) if a>2.
≤ − p−1
Proof.
For each α F∗ such that F [α] = F , there are at most 2 elements r F such that r2 = α.
∈ q p 6 q ∈ q
And, if p = 2 there is just 1. Thus our claim is clear for a = 2. For a > 2, considering the
possible orders of subfields of F , we have
q
p p⌊a/2⌋ 1
N (p 1,2) p+ +p⌊a/2⌋ (p 1,2) − .
≤ − ··· ≤ − p 1
(cid:16) (cid:17) (cid:0) − (cid:1)
6. NEGATIVE RESULTS
Let X,Y be elements of SL (F) whose projective images have orders 2 and 3. Then X2 =ihI,
4
for some h = 0,1,2,3, where i satisfies i2+1 = 0. Multiplying Y for a scalar, if necessary, we
may suppose that Y3 =I. Both X and Y have at least 2 similarity invariants, whence dX 8
M ≥
and dY 6. Now assume that X,Y is irreducible. If X or Y has more than 2 similarity
M ≥ h i
invariants, we get dX 10 or dY 10. From dXY 4, we get a contradiction with respect to
M ≥ M ≥ M ≥
(1). It follows that X and Y have 2 similarity invariants, which necessarily coincide with those
of x and y in (11), with s= 1 and some d= 1.
− ±
Theorem 6.1. Let q =pa be a prime power.
(i) Sp (q) is not (2,3)-generated.
4
(ii) If p=2,3, then PSp (q) is not (2,3)-generated.
4
(iii) SL (2) is not (2,3)-generated.
4
(iv) Sp (q) is not generated by elements having the same similarity invariants as x and y.
4
Proof. (i) and (ii). Let X,Y Sp (q) be preimages of a (2,3)-generating pair of PSp (q). By
∈ 4 4
the above considerations, we may assume X =x and Y =y as in (11), with d= 1, s= 1.
If d = 1, then dx = 6. Moreover dy = 4. Noting that xy is conjugate to its inv±erse, be−ing a
S S
symplecticmatrix,wehavethat x,y iscontainedinanorthogonalgroup,byLemma3.1. This
h i
contradiction proves (i) and also (ii) when p=2.
If d = 1 and p = 3, equating Tr(xy) and Tr((xy)−1) we get r = r r +r . As ǫ = 1, the
4 1 2 3
− −
group x,y is reducible by Lemma 4.1(i): a contradiction.
h i
(iii) Let X,Y be a (2,3) pair in SL (2), which generates an absolutely irreducible subgroup.
4
Up to conjugation X = x, Y = y as in (11), with s = 1. In all the cases in which x,y is
− h i
irreducible, namely (r ,r ,r ,r ) (1,0,0,0),(0,1,0,1),(0,0,1,1) , we have r = r +r . It
1 2 3 4 4 2 3
∈ { }
follows that xy and (xy)−1 have the same characteristic polynomial. Hence, by Remark 4.1,
they are conjugate. As above, case d=1, x,y is contained in an orthogonalgroup.
h i
(iv) If the claim is false, then Sp (q) could be generated by x,y of shape (11), with d = 1.
4
Again,by Lemma 3.1,the group x,y is containedinanorthogonalgroup. Acontradiction.
h i
The first three points of the previous Theorem give a unified proof of known results. Indeed it
hadbeenshownbyLiebeckandShalev[13,Proposition6.2]thatPSp (q)isnot(2,3)-generated
4
for p=2,3. And SL (2)=Alt(8) is not (2,3) generated by a result of Miller [16].
4 ∼
We recall the presentations of certain groups that will be used. Some of them are well known.
9
Lemma 6.1. Let G be a non trivial group. In (i)-(iv) suppose that G= S,T .
h i
(i) If S2 =T3 =(ST)5 =1 then G=Alt(5);
∼
(ii) if S2 =T3 =(ST)7 =[S,T]4 =1, then G=PSL (7);
∼ 2
(iii) if S2 =T7 =(TS)3 =(T4S)4 =1, then G=PSL (7);
∼ 2
(iv) if S2 =T4 =(ST)7 =(ST2)5 =(T(ST)3)7 =T(ST)3T2(ST)3S(T(T(ST)3)2TST)2 =1
then G=PSL (4);
∼ 3
(v) If G = T ,T ,T and T2 = T2 = T2 = (T T )3 = (T T )3 = (T T )4 = (T T T )5 = 1,
h 1 2 3i 1 2 3 1 2 2 3 1 3 1 2 3
then G=Alt(6);
∼
(vi) If G= T ,T ,T ,T ,T , where T ,...,T satisfy the following conditions
1 2 3 4 5 1 5
h i
T3 =1, i=1,...,5,
i
(T T )2 =1, i=1,...,4
 i i+1 ,
[T ,T ]=1 i j >2,
 TiT−j1T T−1T−1 =1, i| =−1,|2,3.
i i+1 i+2 i i+2
then G∼=Alt(7). 
Proof. See [5] and [4] for (i)-(iii), [2] for (iv) and [26, Theorem 1] for (v), (vi).
Lemma 6.2. SU (4) and PSU (9) are not (2,3)-generated.
4 4
Proof. SU (4) = PSp (3) is not (2,3)-generated by Theorem 6.1(ii). By contradiction, let
4 ∼ 4
X,Y be thepreimageinSU (9)ofa(2,3)-generatingpairofPSU (9). Bywhatobservedatthe
4 4
beginningofthisSection,wemayassumethatX,Y areasx,y in(11),forsomed= 1ands=
±
1. Since (xy)3 mustbe conjugateto(xy)−1,we obtainthe conditionsr3+r3 =d(r +r r )
a−nd r r r r 2 F . By Lemma 4.1, we have also to impose that1r 4r +r1 = r2−an3d
1 4 2 3 3 1 2 3 4
− − ∈ − 6
that r r r r i(r r +r r ) = 0. Finally, not all the r ’s belong to the prime field.
1 4 2 3 1 2 3 4 i
We listth−e possib±le4-tu−ples satisf−yinga6llthese conditions,denoting by ξ anelementofF such
9
that ξ2 ξ 1=0.
− −
Case d=1. There are 48 such 4-tuples. Namely:
A) (1,ξ,ξ7,1) (ξ,1,ξ7,ξ6) (ξ,ξ3, 1,ξ6)
± ± ± −
(0,0,ξ,ξ7) (0,ξ,0,ξ3) (ξ2,ξ5,ξ, 1) (ξ2,ξ6,ξ5,ξ) (ξ2,ξ,ξ2,ξ)
B) ± ± ± − ± ±
(ξ,0,ξ6,0) (ξ,ξ2,0,0) (ξ,ξ6,ξ7, 1) (ξ,ξ3,ξ2, 1)
(cid:26) ± ± ± − ± −
and their images under the field automorphism ξ ξ3.
7→
Case d= 1. There are 54 such 4-tuples. Namely:
−
A) (0,0,ξ2, ξ2) (0,ξ2,0,ξ2) (ξ2,0,0,0) (ξ,ξ7,ξ2,1) (ξ,ξ6,ξ3,1) (ξ2,ξ5,ξ7,ξ2)
± − ± ± ± ± ±
(0,0,ξ,ξ3) (ξ,0,1,0) (ξ,1,ξ3,ξ2) (ξ,ξ7, 1,ξ2) (1, 1,ξ,ξ5)
B) ± ± ± ± − ± −
(0,ξ,0,ξ7) (ξ, 1,0,0) (1,ξ,ξ5,ξ6) (1,ξ5,1,ξ5)
(cid:26) ± ± − ± ±
and their images under the field automorphism ξ ξ3.
7→
In both cases A) the matrix (xy)5 is scalar, hence H/Z(H)=Alt(5) by Lemma 6.1(i).
∼
In both cases B), both (xy)7 and [x,y]4 are scalar and so H/Z(H) = PSL (7) by Lemma
∼ 2
6.1(ii).
Lemma 6.3. SL (3) and SU (9) are not (2,4)-generated.
4 4
10