A note on solvable maximal subgroups in subnormal subgroups of GL n (D)
aa r X i v : . [ m a t h . R A ] F e b A NOTE ON SOLVABLE MAXIMAL SUBGROUPSIN SUBNORMAL SUBGROUPS OF GL n ( D ) HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI
Abstract.
Let D be a non-commutative division ring, and G be a subnormalsubgroup of GL n ( D ). Assume additionally that the center of D contains atleast five elements if n >
1. In this note, we show that if G contains a non-abelian solvable maximal subgroup, then n = 1 and D is a cyclic algebra ofprime degree over the center. Introduction
In the theory of skew linear groups, one of unsolved difficult problems is thatwhether the general skew linear group over a division ring contains maximal sub-groups. In [1], the authors conjectured that for n ≥ D , thegroup GL n ( D ) contains no solvable maximal subgroups. In [2], this conjecture wasshown to be true for non-abelian solvable maximal subgroups. In this paper, weconsider the following more general conjecture. Conjecture 1.
Let D be a division ring, and G be a non-central subnormal sub-group of GL n ( D ) . If n ≥ , then G contains no solvable maximal subgroups. We note that this conjecture is not true if n = 1. Indeed, it was proved in [1] thatthe subgroup C ∗ ∪ C ∗ j is a solvable maximal subgroup of the multiplicative group H ∗ of the division ring of real quaternions H . In this note, we show that Conjecture1 is true for non-abelian solvable maximal subgroups of G , that is, we prove that G contains no non-abelian solvable maximal subgroups. This fact generalizes themain result in [2] and it is a consequence of Theorem 3.7 in the text.Throughout this note, we denote by D a division ring with center F and by D ∗ the multiplicative group of D . For a positive integer n , the symbol M n ( D )stands for the matrix ring of degree n over D . We identify F with F I n via the ringisomorphism a a I n , where I n is the identity matrix of degree n . If S is a subsetof M n ( D ), then F [ S ] denotes the subring of M n ( D ) generated by the set S ∪ F .Also, if n = 1, i.e., if S ⊆ D , then F ( S ) is the division subring of D generated by S ∪ F . Recall that a division ring D is locally finite if for every finite subset S of D ,the division subring F ( S ) is a finite dimensional vector space over F . If H and K are two subgroups in a group G , then N K ( H ) denotes the set of all elements k ∈ K such that k − Hk ≤ H , i.e., N K ( H ) = K ∩ N G ( H ). If A is a ring or a group, then Z ( A ) denotes the center of A . Key words and phrases. division ring; maximal subgroup; solvable group; polycyclic-by-finitegroup.2010
Mathematics Subject Classification. HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI
Let V = D n = { ( d , d , . . . , d n ) | d i ∈ D } . If G is a subgroup of GL n ( D ), then V may be viewed as D - G bimodule. Recall that a subgroup G of GL n ( D ) is irreducible (resp. reducible, completely reducible ) if V is irreducible (resp. reducible,completely reducible) as D - G bimodule. If F [ G ] = M n ( D ), then G is absolutelyirreducible over D . An irreducible subgroup G is imprimitive if there exists aninteger m ≥ V = ⊕ mi =1 V i as left D -modules and for any g ∈ G themapping V i → V i g is a permutation of the set { V , · · · , V m } . If G is irreducible andnot imprimitive, then G is primitive .2. Auxiliary lemmas
Lemma 2.1.
Let D be a division ring with center F , and M be a subgroup of GL n ( D ) . If M/M ∩ F ∗ is a locally finite group, then F [ M ] is a locally finitedimensional vector space over F .Proof . Take any finite subset { x , x , . . . , x k } ⊆ F [ M ] and write x i = f i m i + f i m i + · · · + f i s m i s . Let G = (cid:10) m i j : 1 ≤ i ≤ k, ≤ j ≤ s (cid:11) be the subgroup of M generated by all m i j .Since M/M ∩ F ∗ ∼ = M F ∗ /F ∗ is locally finite, the group GF ∗ /F ∗ is finite. Let { y , y , . . . , y t } be a transversal of F ∗ in GF ∗ and set R = F y + F y + · · · + F y t . Then, R is a finite dimensional vector space over F containing { x , x , . . . , x k } . (cid:3) Lemma 2.2.
Every locally solvable periodic group is locally finite.Proof . Let G be a locally solvable periodic group, and H be a finitely generatedsubgroup of G . Then, H is solvable with derived series of length n ≥
1, say,1 = H ( n ) E H ( n − E · · · E H ′ E H. We shall prove that H is finite by induction on n . For if n = 1, then H is a finitelygenerated periodic abelian group, so it is finite. Suppose n >
1. It is clear that
H/H ′ is a finitely generated periodic abelian group, so it is finite. Hence, H ′ isfinitely generated. By induction hypothesis, H ′ is finite, and as a consequence, H is finite. (cid:3) Lemma 2.3.
Let D be a division ring with center F , and G be a subnormal subgroupof D ∗ . If G is solvable-by-finite, then G ⊆ F .Proof . Let A be a solvable normal subgroup of finite index in G . Since G is subnormal in G , so is A . By [13, 14.4.4], we have A ⊆ F . This implies that G/Z ( G ) is finite, so G ′ is finite too [12, Lemma 1.4, p. 115]. Therefore, G ′ is afinite subnormal subgroup of D ∗ . In view of [6, Theorem 8], it follows that G ′ ⊆ F ,hence G is solvable. Again by [13, 14.4.4], we conclude that G ⊆ F . (cid:3) For our further use, we also need one result of Wehrfritz which will be restatedin the following lemma for readers’ convenience.
Lemma 2.4. [16, Proposition 4.1]
Let D = E ( A ) be a division ring generated assuch by its metabelian subgroup A and its division subring E such that E ≤ C D ( A ) .Set H = N D ∗ ( A ) , B = C A ( A ′ ) , K = E ( Z ( B )) , H = N K ∗ ( A ) = H ∩ K ∗ , and let T be the maximal periodic normal subgroup of B . OLVABLE MAXIMAL SUBGROUPS (1) If T has a quaternion subgroup Q = h i, j i of order with A = QC A ( Q ) , then H = Q + AH , where Q + = h Q, j, − (1 + i + j + ij ) / i . Also, Q is normalin Q + and Q + / h− , i ∼ = Aut Q ∼ = Sym (4) .(2) If T is abelian and contains an element x of order not in the center of B ,then H = h x + 1 i AH .(3) In all other cases, H = AH . Maximal subgroups in subnormal subgroups of GL n ( D ) Proposition 3.1.
Let D be a division ring with center F , and G be a subnormalsubgroup of D ∗ . If M is a non-abelian solvable-by-finite maximal subgroup of G ,then M is abelian-by-finite and [ D : F ] < ∞ .Proof . Since M is maximal in G and M ⊆ F ( M ) ∗ ∩ G ⊆ G , either M = F ( M ) ∗ ∩ G or G ⊆ F ( M ) ∗ . The first case implies that M is a solvable-by-finite subnormalsubgroup of F ( M ) ∗ , which yields M is abelian by Lemma 2.3, a contradiction.Therefore, the second case must occur, i.e., G ⊆ F ( M ) ∗ . By Stuth’s theorem(see e.g. [13, 14.3.8]), we conclude that F ( M ) = D . Let N be a solvable normalsubgroup of finite index in M . First, we assume that N is abelian, so M is abelian-by-finite. In view of [17, Corollary 24], the ring F [ N ] is a Goldie ring, and hence itis an Ore domain whose skew field of fractions coincides with F ( N ). Consequently,any α ∈ F ( N ) may be written in the form α = pq − , where q, p ∈ F [ N ] and q = 0.The normality of N in M implies that F [ N ] is normalized by M . Thus, for any m ∈ M , we have mαm − = mpq − m − = ( mpm − )( m − qm ) − ∈ F ( N ) . In other words, L := F ( N ) is a subfield of D normalized by M . Let { x , x , . . . , x k } be a transversal of N in M and set∆ = Lx + Lx + · · · + Lx k . Then, ∆ is a domain with dim L ∆ ≤ k , so ∆ is a division ring that is finitedimensional over its center. It is clear that ∆ contains F and M , so D = ∆ and[ D : F ] < ∞ .Next, we suppose that N is a non-abelian solvable group with derived series oflength s ≥
1. Then we have such a series1 = N ( s ) E N ( s − E N ( s − E · · · E N ′ E N E M. If we set A = N ( s − , then A is a non-abelian metabelian normal subgroup of M . By the same arguments as above, we conclude that F ( A ) is normalized by M and we have M ⊆ N G ( F ( A ) ∗ ) ⊆ G . By the maximality of M in G , either N G ( F ( A ) ∗ ) = M or N G ( F ( A ) ∗ ) = G . If the first case occurs, then G ∩ F ( A ) ∗ isa subnormal subgroup of F ( A ) ∗ contained in M . Since M is solvable-by-finite, sois G ∩ F ( A ) ∗ . By Lemma 2.3, A ⊆ G ∩ F ( A ) ∗ is abelian, a contradiction. Wemay therefore assume that N G ( F ( A )) = G , which says that F ( A ) is normalized by G . In view of Stuth’s theorem, we have F ( A ) = D . From this we conclude that Z ( A ) = F ∗ ∩ A and F = C D ( A ). Set H = N D ∗ ( A ), B = C A ( A ′ ), K = F ( Z ( B )), H = H ∩ K ∗ , and T to be the maximal periodic normal subgroup of B . Then H is an abelian group, and T is a characteristic subgroup of B and hence of A . Inview of Lemma 2.4, we have three possible cases: Case 1: T is not abelian. HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI
Since T is normal in M , we conclude that M ⊆ N G ( F ( T ) ∗ ) ⊆ G . By themaximality of M in G , either M = N G ( F ( T ) ∗ ) or G = N G ( F ( T ) ∗ ). The firstcase implies that F ( T ) ∗ ∩ G is subnormal in F ( T ) ∗ contained in M . Again byLemma 2.3, it follows that T ⊆ F ( T ) ∩ G is abelian, a contradiction. Thus,we may assume that G = N G ( F ( T ) ∗ ), which implies that F ( T ) = D by Stuth’stheorem. By Lemma 2.2, T is locally finite. In view of Lemma 2.1, we concludethat D = F ( T ) = F [ T ] is a locally finite division ring. Since M is solvable-by-finite, it contains no non-cyclic free subgroups. In view of [5, Theorem 3.1], itfollows [ D : F ] < ∞ and M is abelian-by-finite. Case 2: T is abelian and contains an element x of order 4 not in the center of B = C A ( A ′ ).It is clear that x is not contained in F . Because x is of finite order, the field F ( x )is algebraic over F . Since h x i is a 2-primary component of T , it is a characteristicsubgroup of T (see the proof of [16, Theorem 1.1, p. 132]). Consequently, h x i is anormal subgroup of M . Thus, all elements of the set x M := { m − xm | m ∈ M } ⊆ F ( x ) have the same minimal polynomial over F . This implies | x M | < ∞ , so x is an F C -element, and consequently, [ M : C M ( x )] < ∞ . Setting C = Core M ( C M ( x )),then C E M and [ M : C ] is finite. Since M normalizes F ( C ), we have M ⊆ N G ( F ( C ) ∗ ) ⊆ G . By the maximality of M in G , either N G ( F ( C ) ∗ ) = M or N G ( F ( C ) ∗ ) = G . The last case implies that F ( C ) = D , and consequently, x ∈ F ,a contradiction. Thus, we may assume that N G ( F ( C ) ∗ ) = M . From this, weconclude that G ∩ F ( C ) ∗ is a subnormal subgroup of F ( C ) ∗ which is containedin M . Thus, C ⊆ G ∩ F ( C ) ∗ is contained in the center of F ( C ) by [13, 14.4.4].Therefore, C is an abelian normal subgroup of finite index in M . By the samearguments used in the first paragraph we conclude that [ D : F ] < ∞ . Case 3: H = AH .Since A ′ ⊆ H ∩ A , we have H/H ∼ = A/A ∩ H is abelian, and hence H ′ ⊆ H .Since H is abelian, H ′ is abelian too. Moreover, M ⊆ H , it follows that M ′ is alsoabelian. In other words, M is a metabelian group, and the conclusions follow from[4, Theorem 3.3]. (cid:3) Let D be a division ring, and G be a subnormal subgroup of D ∗ . It was showedin [4, Theorem 3.3] that if G contains a non-abelian metabelian maximal subgroup,then D is cyclic of prime degree. The following theorem generalizes this phenome-non. Theorem 3.2.
Let D be a division ring with center F , and G be a subnormalsubgroup of D ∗ . If M is a non-abelian solvable maximal subgroup of G , then thefollowing conditions hold:(i) There exists a maximal subfield K of D such that K/F is a finite Galoisextension with
Gal(
K/F ) ∼ = M/K ∗ ∩ G ∼ = Z p for some prime p , and [ D : F ] = p .(ii) The subgroup K ∗ ∩ G is the F C -center. Also, K ∗ ∩ G is the Fitting subgroupof M . Furthermore, for any x ∈ M \ K , we have x p ∈ F and D = F [ M ] = L pi =1 Kx i . OLVABLE MAXIMAL SUBGROUPS Proof . By Proposition 3.1, it follows that [ D : F ] < ∞ . Since M is solvable,it contains no non-cyclic free subgroups. In view of [4, Theorem 3.4], we have F [ M ] = D , there exists a maximal subfield K of D containing F such that K/F isa Galois extension, N G ( K ∗ ) = M , K ∗ ∩ G is the Fitting normal subgroup of M andit is the F C -center, and
M/K ∗ ∩ G ∼ = Gal( K/F ) is a finite simple group of order[ K : F ]. Since M/K ∗ ∩ G is solvable and simple, one has M/K ∗ ∩ G ∼ = Gal( K/F ) ∼ = Z p , for some prime number p . Therefore, [ K : F ] = p and [ D : F ] = p . Forany x ∈ M \ K , if x p F , then by the fact that F [ M ] = D , we conclude that C M ( x p ) = M . Moreover, since x p ∈ K ∗ ∩ G , it follows that h x, K ∗ ∩ G i ≤ C M ( x p ).In other words, C M ( x p ) is a subgroup of M strictly containing K ∗ ∩ G . Because M/K ∗ ∩ G is simple, we have C M ( x p ) = M , a contradiction. Therefore x p ∈ F .Furthermore, since x p ∈ K and [ D : K ] r = p , we conclude D = L p − i =1 Kx i . (cid:3) Also, the authors in [11] showed that if D is an infinite division ring, then D ∗ contains no polycyclic-by-finite maximal subgroups. In the following corollary,we will see that every subnormal subgroup of D ∗ does not contain non-abelianpolycyclic-by-finite maximal subgroups. Corollary 3.3.
Let D be a division ring with center F , G be a subnormal subgroupof D ∗ , and M be a non-abelian maximal subgroup of G . Then M cannot be finitelygenerated solvable-by-finite. In particular, M cannot be polycyclic-by-finite.Proof . Suppose that M is solvable-by-finite. Then by Proposition 3.1 , weconclude that [ D : F ] < ∞ . In view of [10, Corollary 3], it follows that M is notfinitely generated. The rest of the corollary is clear. (cid:3) Theorem 3.4.
Let D be a non-commutative locally finite division ring with center F , and G be a subnormal subgroup of GL n ( D ) , n ≥ . If M is a non-abeliansolvable maximal subgroup of G , then n = 1 and all conclusions of Theorem 3.2hold.Proof . By [5, Theorem 3.1], there exists a maximal subfield K of M n ( D ) con-taining F such that K ∗ ∩ G is a normal subgroup of M and M/K ∗ ∩ G is a finitesimple group of order [ K : F ]. Since M/K ∗ ∩ G is solvable and simple, we con-clude M/K ∗ ∩ G ∼ = Z p , for some prime number p . It follows that [ K : F ] = p and[ M n ( D ) : F ] = p , from which we have n = 1. Finally, all conclusions follow fromTheorem 3.2. (cid:3) Lemma 3.5.
Let R be a ring, and G be a subgroup of R ∗ . Assume that F isa central subfield of R and that A is a maximal abelian subgroup of G such that K = F [ A ] is normalized by G . Then F [ G ] = ⊕ g ∈ T Kg for every transversal T of A in G .Proof . For the proof of this lemma, we use the similar techniques as in the proofof [2, Lemma 3.1]. Since K is normalized by G , it follows that F [ G ] = P g ∈ T Kg for every transversal T of A in G . Therefore, it suffices to prove that every finitesubset { g , g , . . . , g n } ⊆ T is linearly independent over K . Assume by contrarythat there exists such a non-trivial relation k g + k g + · · · + k n g n = 0 . Clearly, we can suppose that all k i are non-zero, and that n is minimal. If n = 1,then there is nothing to prove, so we can suppose n >
1. Since the cosets Ag HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI and Ag are disjoint, we have g − g A = C G ( A ). So, there exists an element x ∈ A such that g − g x = xg − g . For each 1 ≤ i ≤ n , if we set x i = g i xg − i , then x = x . Since G normalizes K , it follows x i ∈ K for all 1 ≤ i ≤ n . Now, we have( k g + · · · + k n g n ) x − x ( k g + · · · + k n g n ) = 0 . By definition, we have x i g i = g i x , and x , x i ∈ K for all i . Recall that K = F [ A ] iscommutative, so from the last equality( k g x + k g x + · · · + k n g n x ) − ( x k g + x k g + · · · + x k n g n ) = 0 , it follows( k x g + k x g + · · · + k n x n g n ) − ( k x g + k x g + · · · + k n x g n ) = 0 . Consequently, we get( x − x ) k g + · · · + ( x n − x ) k n g n = 0 , which is a non-trivial relation with less than n summands because x = x , acontradiction. Therefore, T is linearly independent over K . (cid:3) Remark 1.
In view of [9, Theorem 11], if D is a division ring with at least fiveelements and n ≥
2, then any non-central subnormal subgroup of GL n ( D ) containsSL n ( D ) and hence is normal. Theorem 3.6.
Let D be non-commutative division ring with center F , and G bea subnormal subgroup of GL n ( D ) , n ≥ . Assume additionally that F contains atleast five elements. If M is a solvable maximal subgroup of G , then M is abelian.Proof . If G ⊆ F , then there is nothing to prove. Thus, we may assume that G is non-central, hence SL n ( D ) ⊆ G and G is normal in GL n ( D ) by Remark 1.Setting R = F [ M ], then M ⊆ R ∗ ∩ G ⊆ G . By the maximality of M in G , either R ∗ ∩ G = M or G ⊆ R ∗ . We need to consider two possible cases: Case 1: R ∗ ∩ G = M .The normality of G implies that M is a normal subgroup of R ∗ . If M is reducible,then by [7, Lemma 1], it contains a copy of D ∗ . It follows that D ∗ is solvable, andhence it is commutative, a contradiction. We may therefore assume that M isirreducible. Then R is a prime ring by [14, 1.1.14]. So, in view of [8, Theorem 2],either M ⊆ Z ( R ) or R is a domain. If the first case occurs, then we are done. Now,suppose that R is a domain. By [17, Corollary 24], we conclude that R is a Goldiering, and thus R is an Ore domain. Let ∆ be the skew field of fractions of R ,which is contained in M n ( D ) by [14, 5.7.8]. Since M ⊆ ∆ ∗ ∩ G ⊆ G , either G ⊆ ∆ ∗ or M = ∆ ∗ ∩ G . The first case occurs implies that ∆ contains F [SL n ( D )]. Thus,if G ⊆ ∆ ∗ , then by the Cartan-Brauer-Hua Theorem for the matrix ring (see e.g.[2, Theorem D]), one has ∆ = M n ( D ), which is impossible since n ≥
2. Thus thesecond case must occur, i.e., M = ∆ ∩ G , which yields M is normal in ∆ ∗ . Since M is solvable, it is contained in Z (∆ ) by [13, 14.4.4], so M is abelian. Case 2: G ⊆ R ∗ .In this case, Remark 1 yields SL n ( D ) ⊆ R ∗ . Thus, by the Cartan-Brauer-Hua Theorem for the matrix ring, one has R = F [ M ] = M n ( D ). It follows by [15,Theorem A] that M is abelian-by-locally finite. Let A be a maximal abelian normal OLVABLE MAXIMAL SUBGROUPS subgroup of M such that M/A is locally finite. Then [14, 1.2.12] says that F [ A ] isa semisimple artinian ring. The Wedderburn-Artin Theorem implies that F [ A ] ∼ = M n ( D ) × M n ( D ) · · · × M n s ( D s ) , where D i are division F -algebras, 1 ≤ i ≤ s . Since F [ A ] is abelian, n i = 1 and K i := D i = Z ( D i ) are fields that contain F for all i . Therefore, F [ A ] ∼ = K × K · · · × K s . If M is imprimitive, then by [5, Lemma 2.6], we conclude that M contains a copy ofSL r ( D ) for some r ≥
1. This fact cannot occur: if r >
1, then SL r ( D ) is unsolvable;if r = 1, then D ′ is solvable and hence D is commutative, a contradiction. Thus, M is primitive, and [2, Proposition 3.3] implies that F [ A ] is an integral domain, so s = 1. It follows that K := F [ A ] is a subfield of M n ( D ) containing F . Again by [2,Proposition 3.3], we conclude that L := C M n ( D ) ( K ) = C M n ( D ) ( A ) ∼ = M m (∆ )for some division F -algebra ∆ . Since M normalizes K , it also normalizes L .Therefore, we have M ⊆ N G ( L ∗ ) ⊆ G . By the maximality of M in G , either M = N G ( L ∗ ) or G = N G ( L ∗ ). The last case implies that L ∗ ∩ G is normal inGL n ( D ). By Remark 1, either L ∗ ∩ G ⊆ F or SL n ( D ) ⊆ L ∗ ∩ G . If the first caseoccurs, then A ⊆ F because A is contained in L ∗ ∩ G . If the second case occurs,then by the Cartan-Brauer-Hua Theorem for the matrix, one has L = M n ( D ). Itfollows that K = F [ A ] ⊆ F , which also implies that A ⊆ F . Thus, in both case wehave A ⊆ F . Consequently, M/M ∩ F ∗ is locally finite, and hence D is a locallyfinite division ring by Lemma 2.1. If M is non-abelian, then by Theorem 3.4, weconclude that n = 1, a contradiction. Therefore M is abelian in this case. Now, weconsider the case M = N G ( L ∗ ), from which we have L ∗ ∩ G ⊆ M . In other words, L ∗ ∩ G is a solvable normal subgroup of GL m (∆ ). From this, we may concludethat L ∗ ∩ G is abelian: if m >
1, then in view of Remark 1, one has L ∗ ∩ G ⊆ Z (∆ )or SL m (∆ ) ⊆ L ∗ ∩ G , but the latter cannot happen since SL m (∆ ) is unsolvable; if m = 1 then L = ∆ , and according to [13, 14.4.4], we conclude that L ∗ ∩ G ⊆ Z (∆ ).In short, we have L ∗ ∩ G is an abelian normal subgroup of M and M/L ∗ ∩ G islocally finite. By the maximality of A in M , one has A = L ∗ ∩ G . Because we arein the case L ∗ ∩ G ⊆ M , it follows that L ∗ ∩ G = L ∗ ∩ M . Consequently, A = L ∗ ∩ M = C GL n ( D ) ( A ) ∩ M = C M ( A ) , which means A is a maximal abelian subgroup of M .By Lemma 3.5, F [ M ] = ⊕ m ∈ T Km for some transversal T of A in M . Thus,for any x ∈ L , there exist k , k , . . . , k t ∈ K and m , m , . . . , m t ∈ T such that x = k m + k m + · · · + k t m t . Take an arbitrary element a ∈ A , since xa = ax , itfollows that k m a + k m a + · · · + k t m t a = ak m + ak m + · · · + ak t m t . By the normality of A in M , there exist a i ∈ A such that m i a = a i m i for all1 ≤ i ≤ t . Moreover, we have a and a i ’s are in K which is a field, the equalityimplies k a m + k a m + · · · + k t a t m t = k am + k am + · · · + k t am t , from which it follows that( k a − k a ) m + ( k a − k a ) m + · · · + ( k t a t − k t a ) m t = 0 . HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI
Since { m , m , . . . , m t } is linearly independent over K , one has a = a = · · · = a t .Consequently, m i a = am i for all a ∈ A , and thus m i ∈ C M ( A ) = A for all 1 ≤ i ≤ t .This means x ∈ K , and hence L = K and K is a maximal subfield of M n ( D ).Next, we prove that M/A is simple. Suppose that N is an arbitrary normalsubgroup of M properly containing A . Note that by the maximality of A in M , weconclude that N is non-abelian. We claim that Q := F [ N ] = M n ( D ). Indeed, since N is normal in M , we have M ⊆ N G ( Q ∗ ) ⊆ G , and hence either N G ( Q ∗ ) = M or N G ( Q ∗ ) = G . First, we suppose the former case occurs. Then Q ∗ ∩ G ⊆ M , hence Q ∗ ∩ G is a solvable normal subgroup of Q ∗ . In view of [2, Proposition 3.3], Q is aprime ring. It follows by [8, Theorem 2] that either Q ∗ ∩ G ⊆ Z ( Q ) or Q is a domain.If the first case occurs, then N ⊆ Q ∗ ∩ G is abelian, which contradicts to the choiceof N . If Q is a domain, then by Goldie’s theorem, it is an Ore domain. Let ∆ bethe skew field of fractions of Q , which is contained in M n ( D ) by [14, 5.7.8]. Because M normalizes Q , it also normalizes ∆ , from which we have M ⊆ N G (∆ ∗ ) ⊆ G .Again by the maximality of M in G , either N G (∆ ∗ ) = M or N G (∆ ∗ ) = G . Thefirst case implies that ∆ ∗ ∩ G is a solvable normal subgroup of ∆ ∗ . Consequently, N ⊆ ∆ ∗ ∩ G is abelian by [13, 14.4.4], a contradiction. If N G (∆ ∗ ) = G , then∆ = M n ( D ) by the Cartan-Brauer-Hua Theorem for the matrix ring, which isimpossible since n ≥
2. Therefore, the case N G ( Q ∗ ) = M cannot occur. Next, weconsider the case N G ( Q ∗ ) = G . In this case we have Q ∗ ∩ G E GL n ( D ), and henceeither Q ∗ ∩ G ⊆ F or SL n ( D ) ⊆ Q ∗ ∩ G by Remark 1. The first case cannot occursince Q ∗ ∩ G contains N , which is non-abelian. Therefore, we have SL n ( D ) ⊆ Q ∗ .By the Cartan-Brauer-Hua theorem for the matrix ring, we conclude Q = M n ( D )as claimed. In other words, we have F [ N ] = F [ M ] = M n ( D ).For any m ∈ M ⊆ F [ N ], there exist f , f , . . . , f s ∈ F and n , n , . . . , n s ∈ N such that m = f n + f n + · · · + f s n s . Let H = h n , n , . . . , n s i be the subgroup of N generated by n , n , . . . , n s . Set B = AH and S = F [ B ]. Recall that A is a maximal abelian subgroup of M . Thus,if B is abelian, then A = B and hence H ⊆ A . Consequently, m ∈ F [ A ] = K , fromwhich it follows that m ∈ K ∗ ∩ M = A ⊆ N . Now, assume that B is non-abelian,and we will prove that m also belongs to N in this case. Since M/A is locally finite,
B/A is finite. Let { x , . . . , x k } be a transversal of A in B . The maximality of A in M implies that A is a maximal abelian subgroup of B , and that A is also normalin B . By Lemma 3.5, S = Kx ⊕ Kx ⊕ · · · ⊕ Kx k , which says that S is an artinian ring. Since C M n ( D ) ( A ) = C M n ( D ) ( K ) = L is afield, in view of [2, Proposition 3.3], we conclude that A is irreducible. Because B contains A , by definition, it is irreducible too. It follows by [14, 1.1.14] that S is aprime ring. Now, S is both prime and artinian, so it is simple and S ∼ = M n (∆ )for some division F -algebra ∆ . If we set F = Z (∆ ), then Z ( S ) = F . Since B is abelian-by-finite, the group ring F B is a PI-ring by [12, Lemma 11, p. 176].Thus, as a hommomorphic image of
F B , the ring S = F [ B ] is also a PI-ring. ByKaplansky’s theorem ([12, Theorem 3.4, p. 193]), we conclude that [ S : F ] < ∞ .Since K is a maximal subfield of M n ( D ), it is also maximal in S . From this, weconclude that F ⊆ C S ( K ) = K , and that K is a finite extension field over F .Recall that A is normal in B , so for any b ∈ B , the mapping θ b : K → K givenby θ b ( x ) = bxb − is well defined. It is clear that θ b is an F -automorphism of K . OLVABLE MAXIMAL SUBGROUPS Thus, the mapping ψ : B → Gal(
K/F )defined by ψ ( b ) = θ b is a group homomorphism withker ψ = C S ∗ ( K ∗ ) ∩ B = C S ∗ ( A ) ∩ B = C B ( A ) = A. Since F [ B ] = S , it follows that C S ( B ) = F . Therefore, the fixed field of ψ ( B ) is F , and hence K/F is a Galois extension. By the fundamental theorem of Galoistheory, one has ψ is a surjective homomorphism. Hence, B/A ∼ = Gal( K/F ).Setting M = M ∩ S ∗ , then m ∈ M , B ⊆ M , and F [ M ] = F [ B ] = S .The conditions F ⊆ K and F ⊆ F implies that K = F [ A ] = F [ A ]. It is clearthat A is a maximal abelian subgroup of M , and that A is also normal in M .If M /A is infinite, then there exists an infinite transversal T of A in M suchthat S = F [ M ] = ⊕ m ∈ T Km by Lemma 3.5. It follows that [ S : K ] = ∞ ,a contradiction. Therefore, M /A must be finite. Replacing B by M in thepreceding paragraph, we also conclude that M /A ∼ = Gal( K/F ). Consequently, B/A ∼ = Gal( K/F ) ∼ = M /A . The conditions B ⊆ M and B/A ∼ = M /A imply B = M . Hence, m ∈ M = B ⊆ N . Since m was chosen arbitrarily, it followsthat M = N , which implies the simplicity of M/A . Since
M/A is simple andsolvable, one has
M/A ∼ = Z p , for some prime number p . By Lemma 3.5, it followsdim K M n ( D ) = | M/A | = p , which forces n = 1, a contradiction. (cid:3) Now, we are ready to get the main result of this note which gives in particular,the positive answer to Conjecture 1 for non-abelian case.
Theorem 3.7.
Let D be a non-commutative division ring with center F , G asubnormal subgroup of GL n ( D ) . Assume additionally that F contains at least fiveelements if n > . If M is a non-abelian solvable maximal subgroup of G , then n = 1 and the following conditions hold:(i) There exists a maximal subfield K of D such that K/F is a finite Galoisextension with
Gal(
K/F ) ∼ = M/K ∗ ∩ G ∼ = Z p for some prime p , and [ D : F ] = p .(ii) The subgroup K ∗ ∩ G is the F C -center. Also, K ∗ ∩ G is the Fitting subgroupof M . Furthermore, for any x ∈ M \ K , we have x p ∈ F and D = F [ M ] = L pi =1 Kx i .Proof . Combining Theorem 3.2 and Theorem 3.6, we get the result. (cid:3) References [1] S. Akbari, R. Ebrahimian, H. Momenaei Kermani, and A. Salehi Golsefidy. Maximal subgroupsof GL n ( D ). J. Algebra (2003), 201-225.[2] H. R. Dorbidi, R. Fallah-Moghaddam, and M. Mahdavi-Hezavehi. Soluble maximal subgroupsin GL n ( D ). J. Algebra Appl. (6) (2011), 1371-1382.[3] P. Draxl. Skew Fields . London Math. Soc. Lecture Note Ser. (Cambridge University Press,1983).[4] B. X. Hai and N. A. Tu. On multiplicative subgroups in division rings. J. Algebra Appl. (3) (2016), 1650050 (16 pages).[5] B. X. Hai and H. V. Khanh. On free subgroups in maximal subgroups of skew linear groups. arXiv:1808.08453v1 [math.RA]. [6] I. N. Herstein. Multiplicative commutators in division rings. Israel J. Math. (1978), 180-188.[7] D. Kiani. Polynomial identities and maximal subgroups of skew linear groups. ManuscriptaMath. (2007), 269-274.[8] C. Lanski. Solvable subgroups in prime rings.
Proc. Amer. Math. Soc. (1981), 533-537. HUỲNH VIÊ. T KHÁNH AND BÙI XUÂN HẢI [9] M. Mahdavi-Hezavehi and S. Akbari. Some special subgroups of GL n ( D ). Algebra Colloq. (4) (1998), 361-370.[10] M. Mahdavi-Hezavehi, M. G. Mahmudi, and S. Yasamin. Finitely generated subnormal sub-groups of GL n ( D ) are central. J. Algebra (2000), 517-521.[11] M. Ramezan-Nassab and D. Kiani. Nilpotent and polycyclic-by-finite maximal subgroups ofskew linear groups.
J. Algebra (2014), 269-276.[12] D. S. Passman.
The algebraic structure of group rings . (New York: Wiley- Interscience Pub-lication, 1977).[13] R. W. Scott.
Group Theory . (Dover Publication, INC, 1987).[14] M. Shirvani and B. A. F. Wehrfritz.
Skew Linear Groups . (Cambridge Univ. Press, 1986).[15] B. A. F. Wehrfritz. Soluble normal subgroups of skew linear groups.
J. Pure Appl. Algebra (1986), 95-107.[16] B. A. Wehrfritz. Normalizers of nilpotent subgroups of division rings. Q. J. Math. (2007),127-135.[17] B. A. F. Wehrfritz. Goldie subrings of Artinian rings generated by groups. Q. J. Math. Oxford (1989), 501-512. Faculty of Mathematics and Computer Science, VNUHCM - University of Science,227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam.
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