A E M S MaximalSubgroup (2)

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Journal of Algebra 259 (2003) 201225

www.elsevier.com/locate/jalgebra

Maximal subgroups ofGLn(D)

S. Akbari,a,b,R. Ebrahimian,a H. Momenaee Kermani,a and

A. Salehi Golsefidya

a Department of Mathematical Sciences, Sharif University of Technology, PO Box 11365-9415, Tehran, Iranb Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran

Received 16 January 2002

Communicated by Jan Saxl

Abstract

In this paper we study the structure of locally solvable, solvable, locally nilpotent, and nilpotent

maximal subgroups of skew linear groups. In [S. Akbari et al., J. Algebra 217 (1999) 422433] it

has been conjectured that ifD is a division ring andMa nilpotent maximal subgroup ofD, thenDis commutative. In connection with this conjecture we show that ifF[M]\Fcontains an algebraicelement over F, then Mis an abelian group. Also we show that C Cj is a solvable maximalsubgroup of real quaternions and so give a counterexample to Conjecture 3 of [S. Akbari et al.,

J. Algebra 217 (1999) 422433], which states that ifD is a division ring and Ma solvable maximal

subgroup of D, then D is commutative. Also we completely determine the structure of divisionrings with a non-abelian algebraic locally solvable maximal subgroup, which gives a full solution to

both cases given in Theorem 8 of [S. Akbari et al., J. Algebra 217 (1999) 422433]. Ultimately, we

extend our results to the general skew linear groups.

2002 Elsevier Science (USA). All rights reserved.

Keywords:Division rings; Skew linear groups; Nilpotent; Irreducible; Maximal subgroups

Introduction

The theory of linear groups over division rings of a fairly general type has been

intensively developedin recent decades. The most important results concerning skew linear

groups can be found in the interesting book of Shirvani and Wehrfritz (1986) [19].

* Corresponding author.E-mail addresses:[email protected] (S. Akbari), [email protected] (R. Ebrahimian),

[email protected] (H. Momenaee Kermani), [email protected] (A. Salehi Golsefidy).

0021-8693/02/$ see front matter 2002 Elsevier Science (USA). All rights reserved.

PII: S 0 0 2 1 -8 6 9 3 (0 2 ) 0 0 5 4 9 - 5


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202 S. Akbari et al. / Journal of Algebra 259 (2003) 201225

Since typically maximal subgroups are large, their algebraic properties carry over to

those of the whole group. In this article we are interested in investigating what happens

whenever a maximal subgroup of a general skew linear group admits some specific

algebraic property. There are many articles devoted to investigations on skew linear groups

include one-dimensional case, i.e.,GL1(D), the multiplicative group of a division ringD.

Recently the structure of maximal subgroups of division rings have been studied; for

instance, see [13]. In [3] the following two conjectures have appeared.

Conjecture A.LetD be a division ring andMa solvable maximal subgroup ofD. ThenDis commutative.

Conjecture B.LetD be a division ring andMa nilpotent maximal subgroup ofD . ThenDis commutative.

The special case of Conjecture B makes the following interesting conjecture.

Conjecture C.LetD be a division ring andMan abelian maximal subgroup ofD . ThenDis commutative.

The present article contains three sections. In Section 1 we establish that C Cj (C isthe field of complex numbers) is a solvable maximal subgroup ofH, whereH denotesreal quaternions. Thus this gives a counterexample to Conjecture A.

Also in Section 1 we show that ifMis a nilpotent maximal subgroup ofD , whereDis a division ring and M

\F contains an algebraic element over F, then M is an abelian

group. In Section 2 we construct a family of maximal subgroups ofGLn(D), n >1, and

completely classify the reducible maximal subgroups ofGLn(D). Moreover, Conjecture C

will be extended to skew linear groups and it will be shown that if D is an infinite-

dimensional division ring over its centerF andGLn(D) is algebraic overF, thenGLn(D)contains no abelian maximal subgroup. Finally, in Section 3 we generalize our results to

general skew linear groups and prove that for a locally finite-dimensional division ring D,

ifM is a locally nilpotent maximal subgroup ofGLn(D), thenMis abelian. Also we show

that locally finiteness of a maximal subgroup ofGLn(D) implies that GLn(D) is locally

finite.

Before stating our results, we fix some notations. Throughout, Dis a division ring with

center F, we shall denote their multiplicative groups by D, F, respectively. For anygroup G and any subset X ofG, CG(X), and NG(X) denote respectively the centralizer

and the normalizer ofXinG. In this articleGr denotes ther th derived subgroup ofG. Theset of primes which occur as the order of some element ofG will be denoted by (G).

Throughout this paperH < G means that His a not necessarily proper subgroup ofG.

By a skew linear group we mean a subgroup of the general skew linear group GLn(D), for

some division ringD . LetG be a subgroup ofGLn(D) and setV= Dn , the space of rown-vectors over D . Then V is a D -G bimodule in the obvious way. We say that G is an

irreducible (respectively reducible, completely reducible) subgroup ofGLn(D), whenever

V is irreducible (respectively reducible, completely reducible) as a D -G bimodule. Also

Gis said to be absolutely irreducible ifF[G] = Mn(D). For a field F, we call a subgroup


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S. Akbari et al. / Journal of Algebra 259 (2003) 201225 203

GofGLn(F ) nonmodular ifG is torsion and Char F /(G). A field extension K /F issaid to be radical if for everyx Kthere exists a natural numbern(x)such thatx n(x) F.For a subset X ofMn(D) and a subring R ofD , R[X] denotes the subalgebra generatedbyR and X. Also ifX is a subset ofD, thenR(X)denotes the subdivision ring generated

byR and X. We write[V: K]r for the dimension of a right vector space Vover a fieldK .For a given ringR, the group of units ofR is denoted byU(R).

1. Maximal subgroups of division rings

Division rings have a more complicated structure than fields and much less is known

about them. It is easily seen that the multiplicative group of a field does not necessarilyhave a maximal subgroup. For example, ifM is a maximal subgroup of index n ofC,then for any xC, x n is contained in M, implyingM= C, since indeed any elementofCis annth power. It seems that the multiplicative group of a noncommutative divisionring has maximal subgroup and it was conjectured first in [2]. The structure of a maximal

subgroup of a group forces some properties in the group. For example, in [17] there is

a theorem which says that if a finite group contains an abelian maximal subgroup, then

it should be solvable. It is not true in general case, because there are some examples of

nonsolvable groups which have finite abelian maximal subgroups, see [15]. Since GLn(D)

is violently non-abelian it is intuitively see thatGLn(D)has no abelian maximal subgroup.

It is a natural question to ask what can be said about the structure of division rings whose

multiplicative subgroups have an abelian maximal group or a solvable maximal subgroup.

In what follows we shall give a counterexample to Conjecture A. Let us record the

following technical lemma.

Lemma 1. H= C,j + 1.

Proof. LetG= C,j+1. We claim that ifw, z C with|z| = |w| and z+jG thenw+jG. For z=0 it is clear, so assume that z=0. Since|z| = |w|, there exists someelementtCsuch thatwz1 =t /t. Now,(z +j)t=t(w+j) G. So,w+jG whichproves the claim.

Since each element ofH is of the formz1+ z2j where z1, z2C, to prove the lemma itsuffices to show that for any aC we havea+j G. Forr >0, let

u =1

r

1 + r +1 1 r1 + r2

i.

So|u| = 1 and using our claim, since 1+j G, we have u+j G. Thus (1+j)(u+j) G,which implies that(u 1)/(u + 1) +j G. On the other hand, we haveu 1u + 1

= r.Sincer > 0 was arbitrary, by our claim for all aC we havea+j G.


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The following theorem asserts a very interesting property of real quaternions, that there

is just one subgroup betweenCand H; actually it gives a counterexample to Conjecture3of [3].

Theorem 1. Let M be a proper subgroup ofH containingC then M= C orM=CCj. Therefore H has a solvable maximal subgroup.

Proof. It suffices to prove that for any z0C,H= C, z0+j. Let G= C, z0+jwhere|z0|2 =r0>0. By a similar method used in the proof of the previous lemma, sincez0+j G, bothr0 i +j andr0+j are in G. So,

r0+jr0 i +j= (r0i 1) + r0 r0 ij G.Let us put

z1=r0i 1

r0 r0 i.

Then|z1|2 = (r02 + 1)/(2r0) 1 and z1+jG. Again let|z1|2 =r1 1. Similarly(

r1+j)(r1 i +j) G. Now, if we define

z2=r1i 1

r1 r1 i

then

z2+j G and |z2|2 =r1

2 + 12r1

1.

Continuing this method we obtain the sequences {rn} and {zn} with the followingproperties:

rn=r2n1+1

2rn1 1, zn=

rn1i 1rn1 rn1 i

, zn+j G, and |zn|2 = rn.

Since the sequence{rn} is decreasing and bounded by 1, so when n tends to infinity rnconverges to 1. Consequently, there exists a natural number m such that zm

+j

G and

(t2 1)2/(4t ) < t where|zm| = t. Now let

u = (t2 1)2

4t+

t2

(t2 1)24t

2i.

Thus,(t+j)(u +j) Gwhich implies that

tu 1t+ u +j G and

t u 1t+ u= 1.


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So, 1+j G and by Lemma 1 we have G =H. On the other hand,Cis an abelian normalsubgroup of index 2 of the group CCj; so it is solvable. The proof is complete.

We note the following simple fact.

Lemma 2.LetG be a group andMa maximal subgroup ofG, then eitherZ(G) M orG M.

Proof. Assume thatZ(G) M, so by maximality ofMwe haveZ(G)M= G. ThereforeG= M M, which completes the proof.

Lemma 3.LetG be a group, such thatG/Z(G)is a locally finite group, thenG is locallyfinite.

Proof. It is sufficient to prove that for any finitely generated subgroup ofG, say N, N

is finite. By the assumption Z(G)N/Z(G) is finite and so is N/N Z(G). ThereforeN/Z(N)is finite, thusN is finite.

Lemma 4. LetD be a division ring with at least4 elements, andM a solvable maximal

subgroup of GLn(D). IfFis the center ofD, then F Munless whenn = 1andD= F.

In addition, ifn = 1then eitherMis abelian orF(M) = D, and so Z(M) = F.

Proof. By Lemma 2 we have F

MorSLn(D)

M; in the later case we conclude that

SLn(D) is solvable, but by [10, p. 138], we know that for n >1, (SLn(D))=SLn(D)which is a contradiction. So FM. If n=1 and F is not contained in M, then byLemma 2, D < M and so D is solvable and by Huas Theorem we have D=F. Toprove the second part of lemma, we note that since n=1, we have MF(M), whereF(M) is a division ring. IfM=(F(M)), then by Huas Theorem [12],Mis abelian. IfM= (F(M)), then by maximality ofMwe obtainF(M) = D.

Lemma 5. LetD be a division ring with centerF andMa nilpotent maximal subgroup

ofD, then Mis a metabelian group.

Proof. If M is abelian then M is a metabelian group. If not by Lemma 4 we have

Z(M)

=F. Since by assumption M /F is a nilpotent group then we have Z(M/F) is

nontrivial. Suppose thatFyZ(M/F)and y /F. By considering the homomorphism: MF, taken by the rule (x)=xy x1y1, we obtain that M/ CM(y) < F andso M < CM(y) . If F (M)=D, then we should have yF, which is a contradiction.Therefore F (M) =D. On the other hand, we have M < ND ((F(M)))and(F(M))0, then for every two elementsa, bM,a, b is a finite group and since characteristic of F is nonzero, this groupshould be cyclic and so M is an abelian group. IfMF, then M is nilpotent and byTheorem 4 we are done. So we can assume that M F. On the other hand, we have

M < ND ((F(M)))and, by maximality ofM, we find that M=ND ((F(M))). Thisyields (F(M)) < M. Therefore F (M) is radical over Fand, by Kaplanskys Lemma,F (M)/F is a purely inseparable extension or F is algebraic over its prime subfield. Butwe note thatM is torsion and so the first case does not occur. If the second case occurs,then since M /F and F are locally finite groups by [9, p. 154], we conclude that M islocally finite. Now, since Char F > 0,Mis an abelian group, a contradiction.

Thus we can assume that Char F= 0. We know that M is locally finite, and locallysolvable, and, by Theorem B, Mis solvable. Now consider a natural number r such that

Mr1 F andMr F. Consider the subdivision ring F (Mr1)ofD. IfF (Mr1) = D,then noting that M < ND ((F(M

r1))) and (F(Mr1)) < ND ((F(Mr1))), inthe view of CartanBrauerHua Theorem, we obtain (F(Mr1)) < M, and by HuasTheorem, F (Mr

1) is a field. Since M/F

is locally finite and (F(Mr

1))

< M, we

obtain that F (Mr1) is radical over F. Since Mr1 F, then F is a proper subfield ofF (Mr1) and, by Kaplanskys Lemma, CharF= p > 0, which is a contradiction. Thus wemay assume that F (Mr1)=D. Since Mr1 is locally finite,F[Mr1] is locally finite-dimensional overF. To see this, let x1, . . . , xk Mr1; thenG = x1, . . . , xkis finite, soF[G] is finite-dimensional over F. Therefore F (Mr1)=F[Mr1]. Now we have thatMr1 is locally finite andMr Z(Mr1), so we have that Mr1 is a nilpotent group. ByTheorem D, one may see that there exists an abelian normal subgroup L ofMr1 suchthat n= [Mr1 :L]


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this case we find that (F(N1)) < M and, by Huas Theorem, that N1 is abelian and

[M:(F(N1))]


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that proves the claim. Now, we have

x=

A B

E C

G, so

Ir 0

E Inr

G.

LetP andQ be two invertible matrices; we haveIr 0

0 P

Ir 0

E Inr

Q 0

0 Inr

=

Q 0

P EQ P

G.

Therefore, by the claim we obtain that

Ir 0

P EQ I nr

G.

If we findi, jsuch that Ir 0

eij Inr

G;

then we can choose invertible matrices P andQ such thatP eijQ=elm for any 1 l n r, 1 m r, and D. Therefore for every Dand 1 i, j n,In+ eijGso SLn(D)G (see [10, p. 137]) and, since diagonal matrices are contained in M, weobtainG = GLn(D). Now, we findi, jsuch that

Ir 0eij Inr

G.

By the theorem of [7, p. 380] there exist invertible matrices PandQ such that

P EQ =

Is 0

0 0

for some s 1. Ifs=1 we are done. Ifs 2 then there exists AGLnr (D) such thatAX= X+ e12whereX= P EQ. Now, we have

Ir 0

0 A Ir 0

X Inr =

Ir 0

X

+e12 A

G.

Using the claim we conclude that Ir 0

X+ e12 Inr

G.

On the other hand, Ir 0

X+ e12 Inr

Ir 0

0 Inr

Ir 0

X Inr

G.


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Therefore Ir 0

e12 Inr

G,

and so we find that Ir 0

e12 Inr

G,

which completes the proof.

The following corollary completely determines the structure of reducible maximal

subgroups ofGLn(D), for a division ringD and a natural numbern.

Corollary 1.LetD be a division ring, n a natural number, andMa maximal subgroup of

GLn(D), then Mis irreducible or it is conjugate to one of the maximal subgroups given in

the previous theorem.

Lemma 8.LetDbe a division ring andG a subgroup of GLn(D) such thatG is irreducible

thenCMn(D)(G)is a division ring.

Proof. Let 0 = A CMn(D)(G). It is easily checked that ker Aand im Aare two invariantsubspaces ofDn underG. Now by irreducibility ofG we have im A

=Dn and ker A

=0,

which completes the proof.

By the above lemma we immediately obtain the following result.

Corollary 2. Let D be a division ring with center F and M be an abelian maximal

subgroup of GLn(D) such that n 2 or D= F; then K= M {0} is a field andK= CMn(D)(M). Furthermore, ifD is finite-dimensional overF then [K :F]2 = n2[D :F]and there is no field betweenF andK .

Proof. By Corollary 1 and Lemma 8, we have that CMn(D)(M) is a division ring; so

CMn(D)(M)=K . Now suppose that[D: F]


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Corollary 4. LetD be a division ring with centerFwith GLn(D) algebraic overF and

such that[D: F]=;then GLn(D) contains no abelian maximal subgroup.

Proof. IfGLn(D) has an abelian maximal subgroup, sayM, then by Corollary 2,M {0}is a field. Let aM\F. We have that F(a) is a field and m= [F(a):F]


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3. Solvable and locally nilpotent maximal subgroups ofGLn(D)

The maximal solvable, maximal nilpotent, and maximal locally nilpotent subgroups

of general linear groups were extensively studied by Suprunenko; the main results

are expounded in [20]. Our object here is to discuss the general skew linear groups

whose maximal subgroups are of some special types. For instance, abelian maximal,

solvable maximal, locally nilpotent maximal, torsion maximal, and locally finite maximal

subgroups of general skew linear groups will be investigated. Our first result in this

direction is as follows.

Lemma 10. LetD be a division ring with at least four elements and centerF. IfM is

a solvable maximal subgroup of GLn(D), whereD is noncommutative orn 3, thenMisirreducible.

Proof. By contradiction, suppose that M is reducible. So by Corollary 1 there exists

a suitable matrixP GLn(D) and a natural number msuch that 0 < m < nand

P MP1 =

A B

0 C

A GLm(D), C GLnm(D), B Mm(nm)(D)

.

On the other hand,Mis solvable, thereforeGLm(D) and GLnm(D) are solvable groups.ThusD is a field andm = n m = 1, son = 2, which is a contradiction.

Ifn = 2, the above lemma fails. To see this, consider the solvable maximal subgroup

Tr2(F ) =

a b

0 c

a, c F, b F

ofGL2(F ), where F is a field. The following theorem plays an important role in our proofs.

Theorem F [19, p. 7]. LetD be a locally finite-dimensional division ring with centerF

andG a subgroup of GLn(D). SetR= F[G].

(a) IfG is completely reducible, thenR is semisimple.

(b) IfG is irreducible, then R is simple.

Lemma 11. LetD be a locally finite-dimensional division ring over its centerFwith at

least four elements andMa non-abelian solvable maximal subgroup of GLn(D) such that

Dis noncommutative orn 3;thenMis absolutely irreducible.

Proof. By Lemma 10, M is irreducible. So by Theorem F, F[M] is a simple ring. Bymaximality ofMwe have two cases. IfU (F[M]) = GLn(D), there is nothing to prove; ifnot, thenU (F[M]) = M. Now, becauseF[M]is a simple ring andU (F[M])is a solvablegroup,Mis abelian, a contradiction.


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Theorem G [19, p. 102].LetD be a locally finite-dimensional division ring with center

F andG a locally nilpotent subgroup of GLn(D);thenG is solvable.

Corollary 5.LetD be a division ring which is locally finite-dimensional over its centerF,

with at least four elements, andM a non-abelian locally nilpotent maximal subgroup of

GLn(D);thenMis absolutely irreducible.

Proof. According to Lemma 11 and Theorem G, we can assume that F=D andn=2.Now, we claim thatMis irreducible. If not, there exists a suitable matrix P GL2(F )suchthatP MP1 is a subgroup of

Tr2(F ) =

a b0 c

a, c F, b F.By maximality ofMwe obtain that Tr2(F ) =P MP1; so it is a locally nilpotent group.But choose an element 1 = a F. Then the group

1 1

0 1

,

1 1

0 a

has trivial center, so it is not nilpotent, a contradiction. Therefore Mis irreducible. On the

other hand, using Theorem F we obtain thatF[M]is a simple ring. But, ifU (F[M]) = M,then M must be an abelian group, which is a contradiction. So, by maximality of M,

U (F[M

])=

GLn(D) and thereforeF[M

] =Mn(D), which completes the proof.

Before proving the next theorem, we need the following remark which is an easy

consequence of [5, p. 165].

Remark 1.Let D be a division ring with at least four elements and center F, R a semi-

simple subring ofMn(D), and U(R) a normal subgroupofGLn(D); then Ris either central

or equalsMn(D).

Lemma 12.LetMbe a solvable maximal subgroup of GLn(D), whereD is a division ring

with at least four elements, andn 5. ThenMis primitive.

Proof. By contradiction suppose that M is imprimitive; then by maximality of M we

obtain thatMGLr (D) Sk , the wreath product ofGLr (D) and Sk , wherer k=n. SinceMis solvable, then GLr (D) is solvable and so r=1 and D is a field. Thus k=n and soSn is solvable, a contradiction. ThereforeMis a primitive subgroup ofGLn(D).

Theorem 8. LetFbe an infinite field andM a non-abelian solvable maximal subgroup

of GLn(F),n 5; then n is a prime number, there is a maximal subfieldK of Mn(F )such that[M: K ] =n and K/F is a cyclic extension,[K: F] =n, and the groupsGal(K/F) andM /K are isomorphic toZn. Furthermore, there exists xM such thatM=ni=1 Kxi andMn(F ) =ni=1 Kx i .


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Proof. By Lemma 11,Mis absolutely irreducible. Now, using Theorem 6 of [20, p. 135],

Mcontains a maximal abelian normal subgroup, say N, such that|M/N| 2. Now suppose that[G:M] =3. We know that thenumber of 2-Sylow subgroups of G is 1 or 3. But M is not a normal subgroup of G.

Hence we may assume that G has three abelian maximal subgroups, say M1, M2, M3.

For any i=j, we have G= Mi , Mj< CG(MiMj) and Z(G)=MiMj= {I}.


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This implies that|M1M2M3| =44 and that G should have exactly two 3-Sylowsubgroups, a contradiction. Now suppose that[G:M] =3. Since M is not a normalsubgroup of G, there exists an element x G such that xM x1 =M. We know thatZ(G)= {I}< M. If aMx Mx1, then M and xM x1 are contained in CG(a).It follows thatM xM x1 = Z(G) = {I}. If 3|M|, then by the first Sylows Theorem[G:M] =3. So we can assume that 3| |M|, and so 6| |M|. If|M| =6, then we have|M| = |xM x1| 12, and so G has at least 72 elements, a contradiction. Hence assumethat|M| =6. It is well known that[G :G] =2. We know that{I}< M G. SinceMis maximal, we have thatM G, so|MG| = 72, which is a contradiction.

The following conjecture can be viewed as a variation of Conjecture A appeared in the

introduction.

Conjecture 1.LetD be a noncommutative division ring, then forn >1, GLn(D) has no

solvable maximal subgroup. Furthermore, ifF is a field, then forn 5, GLn(F ) has no

solvable maximal subgroup.

It seems that the above conjecture remains true if one replaces solvable by locally

solvable. In [3] authors try to prove that if the multiplicative group of a division ring has

an abelian maximal subgroup, then the division ring is a field. In what follows we want to

pose a similar question for general skew linear groups.

Conjecture 2.LetD be a division ring;then forn > 1, GLn(D) has no abelian maximal

subgroup, and forn = 1, ifD has an abelian maximal subgroup, thenD is a field.

The following lemma, apart from being useful in field theory, has a key role in the proof

of Lemma 14.

Lemma 13.LetK /Fbe a field extension such thatK is infinite andK /F is a torsiongroup, thenK /Fis a purely inseparable extension if and only if (K/F)is a finite set.

Proof. One side is clear. To prove the other side by contradiction, suppose that K /F is

not a purely inseparable extension. SinceK /F is a torsion group, then by KaplanskysLemma, CharK= p > 0 and Kis algebraic overZp. Now, for any element x K\F thereexistsyZp(x) such that (Zp(x))=yand o(y) = |(Zp(x))| =pl 1. But we knowthatx K\F, soy K\F. Now, let (K/F) = {p1, . . . , pr}; then there exist nonneg-

ative integers 1, . . . , r such that z=yp

11 ...p

rr

F; thusZp(z) is a proper subfield ofZp(x). Let|Zp(z)| =pk , thus k|l , k=l . Thereforey (pk1)p11 ...p

rr =1 and by the fact

thato(y) =pl 1 we conclude that (pl 1)|(pk 1)p11 . . . prr ; so there exist nonnega-tive integers 1, . . . , r such that (p

l 1)/(pk 1) = p11 . . . prr . On the other hand, K\F

is an infinite set and by the fact that any nonzero polynomial has finitely many roots we

can choose a suitable element x such that l is as large as we need. Now, by the theorem

of [8], p l 1 has a prime factor, say q , such that for every i|l , qpi 1, and hencefor anyi < l , q pi 1 and so q=pjfor some j, 1 j r . Now, choose y1, . . . , yr+1such thato(yi ) = pli 1,yi K\F, andl1< < lr+1. Thus, for everyi , 1 i r+ 1,


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there exists a natural numberji , 1 ji r , such thatpji is a prime factor ofplj 1 and

pjips 1, for everys < lj. So pji s are distinct numbers, which is a contradiction.

Before proving the next result, we need the following three theorems.

Theorem H [19, p. 4].A subnormal subgroup of a completely reducible skew linear group

is completely reducible.

Theorem I [19, p. 215]. LetH be a locally nilpotent normal subgroup of the absolutely

irreducible subgroup G of GLn(D); then H is center by locally finite andG/ CG(H ) is

periodic.

Theorem J [6].LetGbe a locally finite group andMan abelian maximal subgroup ofG.

ThenG is solvable.

Now we are in a position to prove the following lemma.

Lemma 14. LetFbe an infinite field that is algebraic overZp, then forn 2, GLn(F )has no locally nilpotent maximal subgroup.

Proof. First we claim that ifMis a locally nilpotent maximal subgroup ofGLn(F ), then

M is an abelian group. By contradiction, suppose that M is not an abelian group. By

Corollary 5 and Lemma 4, M is absolutely irreducible andF

Mand so Z(M)

=F.

Now, by the theorem of [20, p. 135], there exists an abelian normal subgroup N of Msuch that|M/N|


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we note that F has no element of order 9, because 922r 1 for any natural number r ;

soF3 = F. ThusFis not divisible and hence has a maximal subgroup.

Theorem 9. Let D be a division ring with center F and M a locally finite maximal

subgroup of GLn(D);thenD= F,Char F= p > 0, andFis algebraic overZp .

Proof. By the method used in the proof of Lemma 9, it is not hard to see that we can

assume that M is irreducible. Now, by Theorem E we obtain that F[M] is a primeand semisimple ring, so it is a simple ring. On the other hand, by Lemma 2 we

conclude that FM or SLn(D)M. But, if SLn(D)M then M GLn(D), andso

[GLn(D)

:M

] 0 andFis algebraic over Zp.

Now, by the fact that F[M] is a simple ring, one concludes that Z(F[M]) is a field.In fact, we claim that F=Z(M). Let xZ(M)\F. By the fact that x is torsion andF[x] is a field, we can use SkolemNoether Theorem and find y GLn(D) such thatyx y1 =x i =x . Thereforey, M NGLn(D)(x). Thus, maximality ofM yields thatx GLn(D), which is a contradiction, because if n=1, then (F[x]) is a normalsubgroup of D which contradicts CartanBrauerHua Theorem, and if n > 1, thenSLn(D) must be a subgroup ofx, and it is obviously a contradiction. So, F= Z(M).

Now, we have two cases. If U (F[M])=GLn(D), then F[M] =Mn(D). So for anya GLn(D), there existmi s inMsuch that,a= m1 + + mr . Now, sinceMis a locallyfinite group,Zp[m1, . . . , mr] is a finite ring. Thus a must be torsion, which completes

the proof. In the other case, we haveU (F

[M

])=

M. Since

F[M

] is a simple ring, thereexists a division ringD1such that,F[M] Mm(D1)asF-algebra and, sinceF= Z(M),we have F= Z(D1). Now, by the fact that Mis torsion and Jacobsons Theorem [13,p. 219], we obtainD1= F. So F[M]is a finite-dimensional simple ring with centerF. ByCentralizer Theorem, we have that

Mn(D) F (F[M])op CMn(D)(F[M]) FMm2 (F ) ()

andCMn(D)(F[M])is a simple ring. Since Mis a maximal subgroup which is not normalin GLn(D), NGLn(D)(M)=M. So U (CMn(D)(F[M]))M, therefore[CMn(D)(F[M]):F] < . Now, by()we can conclude thatD is a finite-dimensional division ring overF.ButFis algebraic over Zp , so again by Jacobsons TheoremD= F, which completes theproof.

According to previous result, we pose the following conjecture.

Conjecture 3. LetD be a division ring andMa torsion maximal subgroup of GLn(D);

thenD is a field and is algebraic overZp .

Theorem K [19, p. 14]. Let G be an absolutely irreducible subgroup of GLn(D), N

a normal subgroup of G, and K a subring of Mn(D) normalized by G. IfZ(D)KandG/Nis locally finite, thenK[N]is semisimple.


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Maximal locally nilpotent linear groups has been described in details by D.A.

Suprunenko [20]. The following theorem shows that locally nilpotent maximal subgroups

ofGLn(D), whereD is a finite-dimensional division ring, should be abelian.

Theorem 10.LetD be a finite-dimensional division ring with centerF, which is infinite,

andMa locally nilpotent maximal subgroup of GLn(D);thenMis an abelian group.

Proof. By contradiction, suppose that Mis non-abelian. So by Corollary 5 and Lemma 4,

Mis absolutely irreducible andFM, soZ(M) =F. SinceD is a finite-dimensionaldivision ring, we can assume GLn(D) as a subgroup ofGLk (F ) for some suitable k . By

Theorem I we conclude that M/F is a locally finite group; so by Lemma 3,M is locally

finite. We divide the proof into three steps.

Step 1 (M F). By contradiction, suppose M < F. Obviously, for any a M\Fwe have that Fa Mand, in particular, that M /Fa is locally finite. On the otherhand, M is absolutely irreducible, so, by Theorem K, F[a] is a semisimple ring. ThusF[a] F1 Fr (as F-algebra) for suitable fields, Fi s. But F[a] is a noncentraland commutative subring ofMn(D), and M is a subgroup ofNGLn(D)(U(F[a])); so byRemark 1 it is concluded that U (F[a]) < M. Now, we prove that F[a] is a field. If not,let b=(1, . . . , 1, x) for an arbitrary xF. Now, since M /F is a torsion group, thereexists a natural numbermsuch thatbm F, and therefore xm = 1. SoFis algebraic overZp and, by Jacobsons Theorem, D=F, which contradicts Lemma 14. So we obtain thatF[a]is a field, in particular,a+ 1 M. Buta / F, so there existsc M\CM(a). On the

other hand, we have:

cac1a1 1 =

cac1a1 c(a+ 1)c1(a+ 1)1

(a+ 1).

Therefore,a+ 1 must be an element ofF, a contradiction.

Step 2(Mis finite). First we claim that ifNis a normal subgroup ofMsuch thatN < M

and there exists an elementz N\Fsuch that[N: CN(z)] < , thenNis a finite group.

By finiteness of[N:CN(z)], there exists a subgroup K ofCN(z) such that K Nand[N: K]< . Letm = |N/K |and G = xm, x N. We prove thatG F. SupposeG F. By the fact that Nis a normal subgroup ofM, we conclude thatG M; thereforeM

NGLn(D)(U(F

[G

])). Now, since Mis completely reducible and G

M, by TheoremH

we obtain that G is also completely reducible. On the other hand, G is locally finite,so by Theorem E it is concluded that F[G] is a semisimple ring. Now, by maximalityof M we have two cases. If NGLn(D)(U(F[G]))=GLn(D) then U (F[G])GLn(D).But G F, so by Remark 1 we have F[G] =Mn(D); but zCGLn(D)(G)\F, whichis a contradiction. Therefore NGLn(D)(U(F[G]))=M. So U (F[G]) is locally solvableand, by Zassenhaus Theorem [20, p. 137], it is a solvable group. Now, by the fact that

F[G] is a semisimple ring, we conclude that there exist suitable field extensions ofF,Fi s, such thatF[G] F1 Fr (asF-algebra). As we noted in the first step, we haver= 1. SoF[G]is a field and (F[G])/Fis a torsion group. Now, by Kaplanskys Lemma


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and Lemma 14, F[G]/F must be a purely inseparable extension. But we know thatG isa torsion group, soG F. Letx N; so x m G F, therefore det(xm) = xmk, wherek= n[D: F]. ButN M, thus det(x) = 1. So it is concluded thatx mk = 1. On the otherhand, N is a normal subgroup ofM, and Mis completely reducible, so by Theorem H,

Nis also completely reducible. Now, by Burnsides Theorem we conclude thatNis a finite

group, which proves the claim.

Now, letr be the largest number such thatMr F. Thus for anyxMr+1 we havexk = det(x) = 1. Therefore, sinceMr+1 M,Mr+1 is completely reducible and soMr+1is finite, which implies that Mr is an FC group. Now, by the above claim and Step 1, we

obtain thatMr is finite. By continuing this method, we find that M is finite.

Step 3.By finiteness ofM, we obtain that Mis an FC group and, on the other hand, it isa linear group; so by Theorem 4 of [20, p. 180] we conclude that M/F is a finite group,which contradicts Lemma 9.

In [3] it is shown that ifDis a division ring with center FandMis a maximal subgroup

ofD containingF, thenM/F cannot be a finite group. In relation toGLn(D), we havefollowing result.

Theorem 11.LetD be a finite-dimensional division ring with centerF andChar F=0,n a natural number, andMa maximal subgroup of GLn(D) such thatF

M;then M/Fcannot be a locally finite group.

Proof. By contradiction, suppose that M/F is locally finite. If M is reducible, thenby Corollary 1 we find that M is conjugate to one of the maximal subgroups given

in Theorem 7. So, there exists a natural number r such that (In+ e1n)r F, whichcontradicts Char F= 0. Thus M is irreducible. It implies that F[M] is a simple ring,henceF[M] Ms (D1), whereD1is a division ring. Now maximality ofMyields thatMis absolutely irreducible. Since otherwise GLs (D1)=U (F[M]) =M, thus GLs (D1)/Fis a torsion group and, by Kaplanskys Lemma and the fact that Char F= 0, we obtainthat D1=F and s=1. So M is abelian and, by Corollary 2, we have that M {0} isa field but M/F is torsion; so Kaplanskys Lemma implies that Char F=0, which isa contradiction. ThusMis absolutely irreducible, and Z(M) =F. Now, we claim that ifNMthen F[N] = Mn(D)orN F. We know that Mis completely reducible, soF[N]is semisimple. On the other hand, M

NGLn (D)(U(F

[N

])). IfU (F

[N

]) GLn(D) then

by Remark 1 we obtain thatN F orF[N] = Mn(D). IfU (F[N]) M, then by the factthatF[N]is semisimple andU (F[N])/Fis locally finite and using Kaplanskys Lemmawe conclude that F[N] =F or Char F= p. This completes proof of the claim. Now,by Theorem 10, M is not nilpotent; so M F, thus F[M] =Mn(D). Since M/F isa locally finite group, then by Lemma 3, Mis locally finite. On the other hand, Char F= 0,so Mis nonmodular and, by a theorem of [9,p. 161], it contains a normal abelian subgroupof finite index, say N. Let[M :N] = m and G = xm, x Mwhich is a normal subgroupof M. Two cases may occur. If GF then for every x M we have xm F; thusdet(xm)=x mk =1 where k=n[D:F] and, by Burnsides Theorem, we conclude that


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M is finite, soMis an FC group; thereforeM/Fis finite which contradicts Lemma 9. IfF[G] = Mn(D), thenMn(D) is commutative, a contradiction.

If in Theorem 10 we omit finite dimensionality ofD and add absolutely irreducibility

toM, then we obtain the following result.

Theorem 12. LetD be an infinite division ring with centerF andM a locally nilpotent

maximal subgroup of GLn(D) which is absolutely irreducible;then D=F andn=1; inparticular,M is abelian.

Proof. Since Mis locally nilpotent and absolutely irreducible, by Theorem I, M/F is

locally finite. So by Lemma 3, M is locally finite. Now, we divide the proof into twosteps.

Step 1. Suppose that M F. Thuswe obtainthat, for any y M\F,U (F[y]) is a normalsubgroup of M. Therefore, by Theorem H, U (F[y]) is completely reducible and, byTheorem E,F[y]is semiprime. Buty is algebraic over F, soF[y]is an artinian ring. Thusit is a semisimple ring. So there exist suitable fieldsFi s such thatF[y] F1 Fk .Now, byU (F[y]) Mand the fact that any polynomial with coefficients in Fhas finitelymany roots in its extensions, it is concluded that y has finitely many conjugates in M. So

CM(y) is finite index in M. Now, since M /CM(y) is finite, M is absolutely irreducible.

Furthermore, since U (F[y]) M, we conclude that CM(y) M and, by Theorem K,F[CM(y)] is semisimple. On the other hand, MNGLn(D)(U(F[CM(y)])). So, by thefact that y /

F and Remark 1, we obtain U (F

[C

M(y)

])

M. We know that M is

locally nilpotent andF[CM(y)]is semisimple, so there exist suitable fields Ei s such thatF[CM(y)] E1 Et. Now, we claim that t=1. If not, for any a F, by the factsthat M /F is torsion and U (F[CM(y)])M, we can find a natural number m such that(1, . . . , 1, am)F, which implies that F /Zp is algebraic. ThereforeM /F and F arelocally finite and so Mis also locally finite, which by Theorem 9 and Lemma 14 means

that we are done. SoF[CM(y)]is a field. Now, by the fact that CM(y) is finite index inM,we obtain thatF[M] = Mn(D) is of finite dimension overF[CM(y)]. Thus, by Lemma 6,Dis finite-dimensional over Fand Theorem 10 completes the proof.

Let r be a natural number such that Mr F. Now, we claim that F[Mr] =Mn(D).Since M/FMr is locally finite and M is absolutely irreducible, by Theorem K weconclude thatF

[Mr

]is a semisimple ring. On the other hand, M

NGLn(D)(U(F

[Mr

])),

so ifF[Mr] =Mn(D), then it is concluded that U (F[Mr])M. Thus, by the fact thatM is locally nilpotent, it is concluded that there exist suitable fields, Fi s, such that

F[Mr] F1 Fk . With the same argument as we did before, k=1. So F[Mr]is a field. Thus by the fact that M/F is locally finite, we obtain thatF[Mr]/F is radicaland so by Kaplanskys Lemma we find that F is algebraic overZp or that F[Mr]/F isa purely inseparable extension. But as we noted before, Fcannot be algebraic overZp .So F[Mr]/F is a purely inseparable extension. Now, since Mr is torsion and F[Mr]/Fis a purely inseparable extension, we conclude that Mr F, a contradiction. ThereforeF[Mr] = Mn(D) and soZ(Mr ) F.


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Step 2. Assume that M F. Thus F[M] =Mn(D). We show that Mn(D) is locallyfinite-dimensional overF. To see this, let x1, . . . , xs M; thenH= x1, . . . , xsis finite,thus[F[H] :F]


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Acknowledgments

The first author is indebted to the Research Council of the Sharif University of

Technology for support. The authors are indebted to the Institute for Studies in Theoretical

Physics and Mathematics for support. Also the authors wish to thank K. Mallahi Karai

for useful comment on the proof of Lemma 13. We wish to express our indebtedness and

gratitude to the referee for valuable suggestions.

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A E M S MaximalSubgroup (2)

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