On subnormal subgroups in division rings containing a non-abelian solvable subgroup
aa r X i v : . [ m a t h . R A ] A ug ON SUBNORMAL SUBGROUPS IN DIVISION RINGSCONTAINING A NON-ABELIAN SOLVABLE SUBGROUP
BUI XUAN HAI AND MAI HOANG BIEN
Abstract.
Let D be a division ring with center F and N a subnormal sub-group of the multiplicative group D ∗ of D . Assume that N contains a non-abelian solvable subgroup. In this paper, we study the problem on the existenceof non-abelian free subgroups in N . In particular, we show that if either N is algebraic over F or F is uncountable, then N contains a non-abelian freesubgroup. Introduction
Let D be a division ring with center F and D ∗ denotes the multiplicative groupof D . Among the classical results on solvable subgroups in division rings, we notethe earliest theorem due to L. K. Hua [15] which states that if D ∗ is solvable,then D is commutative. Since then, there are many works devoted to this subject.In 1960, Huzubazar [16] showed that in division rings there are no non-centrallocally nilpotent subnormal subgroups. Later, Stuth [19] proved that if a subnormalsubgroup G of D ∗ is solvable, then G must be central, that is, G ⊆ F . Severaltheorems of such a kind of other authors one can find, for example in [10, 11, 12, 13,19] and in the references therein. This subject has close connection to the problemon the existence of non-abelian free subgroups in division rings. For instance, in[14, Theorem 2.2.1], it was proved that in the case when D is centrally finite, thatis, dim F D < ∞ , a subgroup G of D ∗ contains no non-abelian free subgroup if andonly if G contains a solvable normal subgroup of finite index. Our paper followsthis side. Namely, we establish some relations between the existence of non-abeliansolvable subgroups and non-abelian free subgroups in subnormal subgroups of D ∗ .Note that this subject is not quite strange. Indeed, in [18, Theorem 2], Lichtmanshowed that if a normal subgroup of D ∗ contains a non-abelian nilpotent-by-finitesubgroup, then it contains a non-abelian free subgroup. These results were extendedin [8] for a normal subgroup of D ∗ containing a non-abelian solvable subgroup withsome additional conditions. In this paper, we consider a subnormal subgroup of D ∗ containing a non-abelian solvable subgroup instead of a normal subgroup containinga non-abelian nilpotent-by-finite subgroup. In fact, assume that N is a subnormalsubgroup of D ∗ containing a non-abelian solvable subgroup. We will prove that if N is algebraic over F or the transcendence degree of F over its prime subfield P isinfinite, then N contains a non-abelian free subgroup (see theorems 4.1, 4.3 in thetext). Key words and phrases. subnormal subgroups, free subgroup, solvable subgroup, division ring,quotient division ring.2010
Mathematics Subject Classification.
The ideas of techniques of proofs in this paper are taken from [8], and they canbe shortly explained as the following: assume that N is a subnormal subgroup of D ∗ and G is a non-abelian solvable subgroup of N . Then, there exists a divisionsubring D of D which is isomorphic to the quotient division ring K ( t, σ ) of a skewpolynomial ring K [ t, σ ] in t over some subfield K with respect to an automorphismfield σ of K , and D ∩ N contains t . In some special cases, we may continueto choose a division subring D of D such that D is isomorphic to the quotientdivision ring L ( x, y ) of a polynomial ring L [ x, y ] in two commutating indeterminatesover a suitable division ring L . The construction of D and D will be presentedin Section 4 and with the above reason, in Section 2, we consider the existence ofnon-abelian free subgroups of subnormal subgroups in the quotient division ring ofa skew polynomial ring in one indeterminate over a field. In Section 3, our basedivision ring is the quotient division ring D ( x, y ) of a polynomial ring D [ x, y ] intwo commuting indeterminates over a division ring D .Remark that recently, the problem on the existence of free subobjects in thequotient division rings of skew polynomial rings have been received a considerableattention. For instance, the existence of free subalgebra in the quotient divisionrings of Ore polynomial rings was studied in [2, 3, 4, 8] while in [6], Gon¸calvesdealt with the existence of non-abelian free subgroups of a normal subgroup of thequotient division ring of a certain skew polynomial ring. Hence, the next section ofthe present paper is itself an interesting topic.2. The existence of free subgroups in K ( x, σ )Let G be a group. If a, b ∈ G , then we denote [ a, b ] = aba − b − and [ i +1 a, b ] =[ a, [ i a, b ]] for an integer i ≥
1. The following lemma will be used in the next.
Lemma 2.1.
Let G be a group, and assume that N is a subnormal subgroup of G with a series of subgroups N = N n ⊳ N n − ⊳ · · · ⊳ N = G. If a ∈ N and b ∈ G , then [ i a, b ] ∈ N i for every positive integer i . In particular, [ n a, b ] ∈ N .Proof. The proof is simple by induction on i . (cid:3) Let R be a ring and σ : R → R an endomorphism of R . A skew polynomial ring in an indeterminate x over R corresponding to σ , denoted by R [ x, σ ], is defined asthe following:As a set, R [ x, σ ] consists of all polynomials in an indeterminate x with coefficientstaken from R , that is, R [ x, σ ] = { a + a x + · · · + a n x n | n ∈ N , a i ∈ R for every 1 ≤ i ≤ n } . The addition in R [ x, σ ] is usual while polynomials are multiplied formally with therule xa = σ ( a ) x . If R = K is a field and σ is an automorphism, then K [ x, σ ] is aPLID ( principal left ideal domain ) and also a PRID ( principal right ideal domain ),that is, every one-sided ideal in K [ x, σ ] is generated by one element. In this case, wedenote by K ( x, σ ) the quotient division ring of K [ x, σ ]. Every element of K ( x, σ )has the form f ( x ) g − ( x ), where f ( x ) , g ( x ) ∈ K [ x, σ ].In this section, we consider the existence of non-abelian free subgroup in a sub-normal subgroup of K ( x, σ ) containing x . The recent results on the existence offree subobject in K ( x, σ ) can be found in [2, 3, 4, 6]. N SUBNORMAL SUBGROUPS CONTAINING A SOLVABLE SUBGROUP 3
Lemma 2.2.
Let K be a field, and σ an automorphism of K with fixed subfield E = { a ∈ K | σ ( a ) = a } . If a ∈ K and E ( σ n ( a ) | n ∈ N ) is infinitely generated,then the set S = { σ i ( a ) σ i ( a ) · · · σ i n ( a ) | n ≥ , ≤ i < i , · · · < i n } is linear independent over E .Proof. Deny the conclusion, and let µ P j =1 α j σ i ,j ( a ) σ i ,j ( a ) · · · σ i n,j ( a ) = 0 for some µ ∈ N and α j ∈ E . Assume that µ is the smallest integer satisfying such a rela-tion, i.e., this relation is shortest. Put m = max { i n,j | ≤ j ≤ µ } . Then, thisdependence relation may be written in the form sσ m ( a ) + t = 0 , where s, t ∈ E ( σ i ( a ) | i < m ) . By the choice of µ , we have s = 0, which implies that σ m ( a ) = s − t ∈ E ( σ i ( a ) | i ≤ m − . As a corollary, σ m +1 ( a ) = σ ( σ m ( a )) ∈ E ( σ i ( a ) | i ≤ m )= E ( σ i ( a ) | i ≤ m − σ m ( a )) ⊆ E ( σ i ( a ) | i ≤ m − . Thus, E ( σ n ( a ) | n ∈ N ) is finitely generated that contradicts to the hypothesis. (cid:3) Let K be a field, σ an automorphism of K and D = K ( x, σ ). Assume that a ∈ K and N is a subnormal subgroup of D ∗ containing x with the following seriesof subgroups N = N n ⊳ N n − ⊳ · · · ⊳ N = D ∗ . Without loss of generality, we can suppose that n ≥ N = D ∗ , then wecan set N = N = D ∗ . For the convenience, we denote a ( i ) = σ i ( a ) for a ∈ D .Consider the sets { y i = a ( i ) x n | ≤ i < n } and { A i = 1 + y i | ≤ i < n } . Lemma 2.3.
With the assumption and the symbols as above, we have(i) The element [ n x, A ] belongs to the intersection between N and the subgroup h A i | ≤ i < n i of D ∗ generated by all A i , ≤ i < n . Also, its first and last factorsare A n and A − n − respectively.(ii) The element [ n x, A − ] belongs to the intersection between N and the subgroup h A i | ≤ i < n i of D ∗ generated by all A i , ≤ i < n . Also, its first and last factorsare A − n and A n − respectively.Proof. We shall prove the Part (i). The proof of the Part (ii) is similar. In view ofLemma 2.1, [ i x, A ] ∈ N i for all i ; in particular, [ n x, A ] ∈ N n . Now, we prove byinduction on i > i x, A ] ∈ h A , · · · , A i i ; and(b) [ i x, A ] has A i and A − i − as its initial and last factor respectively.For any i >
0, we have xA i x − = x (1 + a ( i ) x n ) x − = 1 + a ( i +1) x n = A i +1 , and xA − i x − = ( xA i x − ) − = A − i +1 (1)which implies x h A , A , · · · , A i i x − ⊆ h A , A , · · · , A i +1 i . (2)We see that for i = 1, (a) and (b) are true in view of the following equalities:[ x, A ] = ( xA x − ) A − = A A − . BUI XUAN HAI AND MAI HOANG BIEN
Assume that (a) and (b) hold for 0 < i < n . We shall prove that they hold for i + 1. Indeed, by induction assumption and in view of (1), we have[ i +1 x, A ] = [ x, [ i x, A ]] = ( x [ i x, A ] x − )[ i x, A ] − ∈ h A , A , · · · , A i +1 i . Also, [ i x, A ] = A i BA i − . So, in view of (1), it follows[ i +1 x, A ] = x ( A i BA − i − ) x − ( A i BA − i − ) − = ( xA i x − ) xBA − i − x − A i − B − A − i = A i +1 xBA − i − x − A i − B − A − i . Thus, (a) and (b) both hold for i +1,and the proof is now complete. (cid:3) Let R be a ring and σ be an automorphism of R . The ring R [[ x, σ ]] = n α = ∞ X i = n a i x i | n ∈ Z , a i ∈ R o consisting of formal Laurent series over R with the usual addition and the multi-plication defined formally with the rule xa = σ ( a ) x is called a skew Laurent seriesring . Clearly, R [ x, σ ] is a subring of R [[ x, σ ]]. If R = K is a field, then K [[ x, σ ]] is adivision ring [17, Example 1.8] and K ( x, σ ) can be considered as a division subringof K [[ x, σ ]] via obvious injection. In particular, we have the following remark. Remark 2.4. If a ∈ K ∗ , then the element (1 + ax m ) − ∈ K ( x, σ ) can be writtenin the form: (1 + ax m ) − = ∞ X i =0 ( − i a i x mi = 1 − ax m + X i =2 b i x mi in K [[ x, σ ]] with b i = 0 for all i ≥ Theorem 2.5.
Let K be a field, and σ an automorphism of K with fixed subfield E = { a ∈ K | σ ( a ) = a } . Assume that D = K ( x, σ ) is the quotient division ring of K [ x, σ ] , and at least one of the following conditions holds:(i) There exists an element a ∈ K such that E ( a, σ ( a ) , σ ( a ) , · · · ) is infinitelygenerated.(ii) There exists an element a ∈ K \ E which is algebraic over E .If N is a subnormal subgroup of D ∗ containing x , then N contains a non-abelianfree subgroup.Proof. As in the proof of Lemma 2.3, for a ∈ K and i ∈ N , we denote a ( i ) = σ i ( a ).Consider a series of subgroups N = N n ⊳ N n − ⊳ · · · ⊳ N = D ∗ . Without loss of generality, we can suppose that n ≥ N = N = D ∗ .(i) Assume that there exists a ∈ K such that E ( a ( i ) | i ∈ N ) is infinitely gener-ated. We shall prove that the elements [ n x, A ] , [ n , A − ] defined as in Lemma 2.3generate a non-abelian free subgroup in N .Firstly, we claim that the subset { y i = a ( i ) x n | ≤ i ≤ n } generates a free E -algebra. Indeed, deny the conclusion, that is, there exists a non-zero polynomial F ( z , z , · · · , z n ) in the free E -algebra in n + 1 indeterminates z , z , · · · , z n suchthat F ( y , y , · · · , y n ) = 0 . We assume that F ( z , z , · · · , z n ) = F ( z , z , · · · , z n ) + F ( z , z , · · · , z n ) + · · · + F µ ( z , z , · · · , z n ) , N SUBNORMAL SUBGROUPS CONTAINING A SOLVABLE SUBGROUP 5 where F ℓ ( z , z , · · · , z n ) = P j + j + ··· + j ν = ℓ α j ,j , ··· ,j ν z j ℓ z j ℓ · · · z j ν ℓ ν is a homogeneouspolynomial of degree ℓ with coefficients in E for every 1 ≤ ℓ ≤ µ . Let s be a positiveinteger such that F s ( z , z , · · · , z n ) is non-zero (the existence of s is obvious since F is non-zero). Observe that the degree of F ℓ ( y , y , · · · , y n ) in the indeterminate x is 2 nℓ , so F ( y , y , · · · , y n ) = 0 if and only if F ℓ ( y , y , · · · , y n ) = 0 for every1 ≤ ℓ ≤ µ . In particular,0 = F s ( y , y , · · · , y n ) = X j + j + ··· + j ν = s α j ,j , ··· ,j ν y j ℓ y j ℓ · · · y j ν ℓ ν . After moving all coefficients to the left and the indeterminate x to the right, onehas F s ( y , y , · · · , y n ) = X j + j + ··· + j ν = s α j ,j , ··· ,j ν b j ,j , ··· ,j ν x ns , where b j ,j , ··· ,j ν = a ( j ) a ( j +2 n ) · · · a ( j +2( j − n ) a ( j +2 j n ) · · ·· · · a ( j +2 j n +2 n ) · · · a ( j ν +2( j ν − n +2 j ν − n + ··· j n ) .This implies that P j + j + ··· + j ν = s α j ,j , ··· ,j ν b j ,j , ··· ,j ν = 0. Since every coefficient b j ,j , ··· ,j ν belongs to S , by Lemma 2.2, α j ,j , ··· ,j ν = 0 which contradicts to theassumption that F s ( z , z , · · · , z n ) is non-zero.Next, we show that the set { A i = 1 + y i | ≤ i ≤ n } generates a free subgroupin ( K ( x, σ )) ∗ . Assume that there is a relation A n i A n i · · · A n t i t = 1 , where 0 ≤ i j < n and all n j = 0. We seek a contradiction by describing therepresentation of this relation in K [[ x, σ ]]. First of all, we need to know the rep-resentation of A ji in K [[ x, σ ]] with j ∈ Z ∗ . If j >
0, then A ji is a polynomial in K [ x, σ ] ⊆ K [[ x, σ ]], so we can write A ji = 1 + y i + A ′ i in which either A ′ i = 0 (incase j = 1) or the degree of A ′ i is greater than 2 n (in the indeterminate x ). If j <
0, then by Remark 2.4, one has A − i = 1 − y i + y i + · · · ∈ K [[ x, σ ]]. Hence, ifchar( D ) = 0, then A ji = 1 − y i + A ′ i where A ′ i is a series in K [[ x, σ ]] whose degreeof the first term is greater than 2 n . If char( K ) = p >
0, then we write j = p λ j with ( p, j ) = 1. Observe that A ji = (1 − y i + y i + · · · ) j = ((1 − y i + y i + · · · ) p α ) j = (1 − y p α i + y p α i + · · · ) j = 1 + j y p α i + A ′ i , where, A ′ i is a series in K [[ x, σ ]] whose degree of the first term is greater than 2 np α .Hence, in all cases, A ji can be written as A ji = 1 + b i y c i i + A ′ i where c i ∈ N , b i ∈ E and A ′ i is either 0 or a series in K [[ x, σ ]] whose degree of the first term is greaterthan 2 nc i . Therefore, the relation A n i A n i · · · A n t i t = 1 will be written in K [[ x, σ ]]as follows: (1 + b y c i + A ′ i )(1 + b y c i + A ′ i ) · · · (1 + b t y c t i t + A ′ i t ) = 1 , where c i > b i ∈ E ∗ . After expanding the left side, we see that there is aunique product b y c i b y c i · · · b t y c t i t = b b · · · b t y c i y c i · · · y c t i t , hence, this product is0 which is a contradiction since { y , y , · · · , y n } generates a free E -algebra. Thus, { A i = 1 + y i | ≤ i ≤ n } generates a free subgroup in D = K ( x, σ ). As a corollary,[ n x, A ] , [ n x, A − ] ∈ h A i | ≤ i ≤ n i generates a free subgroup in N by Lemma 2.3. BUI XUAN HAI AND MAI HOANG BIEN (ii) Assume that a ∈ K \ E and a is algebraic over E . Note that σ ( a ) is alsoalgebraic over E because E is the fixed field of σ . By induction, it is easy to seethat a ( i ) = σ i ( a ) is algebraic over E for any i ∈ N . In view of Part 1, we canassume that E ( a ( i ) | i ∈ N ) is finitely generated. This implies that K/E is a finiteextension, so the Galois group Gal(
K/E ) is finite. If d is the order of σ , then wehave x d b = σ d ( b ) x d = bx d for any b ∈ K. Hence, x d belongs to the center Z of K ( x, σ ). It follows that every element of K ( x, σ ) can be written in the form α + α x + · · · + α d − x d − , with α i ∈ Z . This means that K ( x, σ ) is a finite dimensional vector space over Z .By [7, Theorem 2.1], N contains a non-abelian free subgroup. (cid:3) The existence of free subgroups in D ( x, y ) ∗ Let D be a division ring with center F . Assume that x, y are two commutingindeterminates and they also commute with all elements from D . We denote by D ( x, y ) the division ring of fractions of the polynomial ring D [ x, y ] = D [ x ][ y ] in x, y over D ; by R = D [ x ][[ y ]] the power series ring in y over the polynomial ring D [ x ]. Let N be a subnormal subgroup of D ( x, y ) ∗ and assume that there existnon-commuting elements a, b ∈ N ∩ D such that c = b − ab commutes with a .Let D [[ y ]] and x act on R by left multiplication and via right multiplication by a respectively, that is f ( y ) · α = f ( y ) α and d · x = xa for any f ( y ) ∈ D [[ y ]] , α ∈ R, d ∈ D . With this left action, R is a left R -module. Assume that K is a commutativesubring of R containing a . Then, it is shown in [5] that R is an ( R, K )-bimodule(left R -module and right K -module). If M = 1 K + bK is the right K -submoduleof R generated by { , b } and S = { r ∈ R | r · M ⊆ M } , then we can show that(or see [8, Section 5]) M is a free right K -module with the basis { , b } and S is asubring of R containing K [[ y ]]. This implies that the map φ : S → End( M K ) ∼ = M ( K ) , r φ r , where φ r : M → M, φ r ( m ) = rm , for every m ∈ M , is a ring homomorphism.Moreover, the image φ a of a via φ is the matrix (cid:20) a c (cid:21) . Now consider fourelements d = b ( ab − ba ) − ( bab − − x ); d = ( ab − ba ) − ( a − x ); d = − b ( ab − ba ) − ( bab − − x );and d = b ( ab − ba ) − ( a − x ) . Then, we have d , d , d , d ∈ S and φ d = (cid:20) (cid:21) ; φ d = (cid:20) (cid:21) ; φ d = (cid:20) (cid:21) ; and φ d = (cid:20) (cid:21) . We also conclude (or see [8, Section 5]) φ ( y ) = (cid:20) y y (cid:21) , hence, both 1 + yd and 1 + yd are invertible in S and the images of 1 + yd and 1 + yd via φ are φ d = (cid:20) y (cid:21) and φ d = (cid:20) y (cid:21) respectively. N SUBNORMAL SUBGROUPS CONTAINING A SOLVABLE SUBGROUP 7
Lemma 3.1.
For any n ≥ , there exists u ∈ S such that u is invertible in S andthe image of u via φ is φ ( u ) = (cid:20) − a − c ) n +1 y (1 − a − c ) n +1 y − (1 − a − c ) n +1 y − (1 − a − c ) n +1 y (cid:21) . Proof.
The element u is constructed by modifying the construction of w in [8,Proposition 5.8]. Let K = F [ a, a − , c, c − ] be the subring of D generated by { a, a − , c, c − } and K = K [[ y ]] be the subring of D [[ y ]][ x ] generated by all seriesin indeterminate y whose coefficients are in K . We have (cid:20) − a − c ) n +1 y (1 − a − c ) n +1 y − (1 − a − c ) n +1 y − (1 − a − c ) n +1 y (cid:21) = I + ( a − c ) n +1 y (cid:20) − − (cid:21) = I + φ ( y )(1 + a − c ) n +1 ( φ ( d ) + φ ( d ) − φ ( d ) − φ ( d )) . According to [8, Lemma 5.7], for every 1 ≤ i, j ≤ f ∈ F , there existsan element f ′ ∈ S ∩ D [ x ] such that φ ( f ′ ) = f φ ( d ij ), which implies that there existsan inverse element u ∈ S of I + φ ( y )(1+ a − c ) n +1 ( φ ( d )+ φ ( d ) − φ ( d ) − φ ( d )).Thus, φ ( u ) = (cid:20) − a − c ) n +1 y (1 − a − c ) n +1 y − (1 − a − c ) n +1 y − (1 − a − c ) n +1 y (cid:21) . (cid:3) The following lemma is the key result in this section.
Proposition 3.2.
Let D be a division ring, D ( x, y ) the quotient division ring of thepolynomial ring D [ x, y ] in two commuting indeterminates x, y and N a subnormalsubgroup of D ( x, y ) ∗ with a series of subgroups N = N n ⊳ N n − ⊳ · · · ⊳ N = D ∗ . Assume that a ∈ N ∩ D and b, c ∈ D with ab = bc , a = c and ac = ca . Let u be theelement in Lemma 3.1. Put v = [ n a, y ( ab − ba ) − ( a − x )] and w = [ n a, − yb ( ab − ba ) − ( bab − − x )] . (1) If char( D ) = 0 , then v, w generates a free subgroup in N . (2) If char( D ) > , then [ n v, u ] and [ n w, u ] generates a free subgroup in N .In particular, N contains a non-abelian free subgroup.Proof. Let F be the center of D , K = F [ a, a − , c − ] the subring of D generated by { a, a − , c, c − } and K = K [[ y ]] F = F [[ y ]] the subring of D [[ y ]][ x ] generated byall series in indeterminate y whose coefficients are in K . Let d = ( ab − ba ) − ( a − x )and d = − b ( ab − ba ) − ( bab − − x ). Then, 1 + yd , yd is invertible in S and v = [ n a, yd ] and w = [ n a, yd ].1. Assume that char( D ) = 0. By Lemma 2.1, v and w belong to N . We may showby inductive on n that the image of v and w via φ are φ v = (cid:20) − ac − ) n y (cid:21) and φ w = (cid:20) − ac − ) n y (cid:21) respectively. Observe that ( φ v − = ( φ w − = 0 and( φ v − φ w −
1) = (cid:20) (1 − ac − ) n y
00 0 (cid:21) is not algebraic over Q , so that accordingto [8, Lemma 5.3], φ v = (cid:20) − ac − ) n y (cid:21) and φ w = (cid:20) − ac − ) n y (cid:21) generates a free subgroup in GL ( K ). Hence, v are w , the inverse images of thesematrices, generates a free subgroup in N . BUI XUAN HAI AND MAI HOANG BIEN
2. Assume that char( D ) = p >
0. Put α = (cid:20) − ac − ) n (cid:21) and β = (cid:20) (cid:21) . Then, α = β = 0, and βα = (cid:20) − ac − ) n (cid:21) is not nilpotent.Hence, by [8, Lemma 5.2], φ v = 1 + yα, φ w = 1 + yβαβ and φ u = 1 + y (1 − β ) αβα (1 + β )generates the subgroup of GL ( K ) isomorphic to the free product C p ∗ C p ∗ C p , where C p is the cyclic group of order p . Therefore, { u, v, w } , the set of inverse images ofthese matrices, also generates the subgroup h u, v, w i of S ∗ which is isomorphic tothe free product C p ∗ C p ∗ C p . Observe that both [ n v, u ] and [ n w, u ] are in N since v, w ∈ N (Lemma 2.1 and Part 1). Applying Lemma 2.1 (3) and (4), we concludethat [ n v, u ] and [ n w, u ] generates a free subgroup in N . (cid:3) A subnormal subgroup containing a solvable subgroup
Theorem 4.1.
Let D be a division ring with center F , and assume that N is a sub-normal subgroup of D ∗ containing a non-abelian solvable subgroup. If transcendencedegree of F over the prime subfield P is infinite, then N contains a non-abelian freesubgroup.Proof. Assume that H is a non-abelian solvable subgroup of N , so H ( n ) = 1 forsome n >
1. Without loss of generality, we can assume n = 2 because if n >
2, wecan consider the subgroup H ( n − instead of H . Therefore, the derived subgroup H ′ is abelian. Put S = n H ′ ≤ A ≤ H | A abelian o . Let G be a maximal element in S (there exists such a G by Zorn’s Lemma). Then, G is an abelian normal subgroup of H . Take a ∈ H \ G and K = P ( G ). Then, K isa field and aKa − = K . It follows that the subring K [ a ] of D generated by K ∪ { a } can be written in the form K [ a ] = n n X i =0 a + a a + · · · + a n a n | n ∈ N , a ∈ K o . Also, the map σ : K → K defined by σ ( b ) = aba − for every b ∈ K is an automor-phism of K . Consider the homomorphismΦ : K [ t, σ ] → K [ a ] , n X i =0 a i t i n X i =0 a i a i , where K [ t, σ ] is the skew polynomial ring in an indeterminate t over K correspond-ing to σ . It is obvious that Φ is a surjection. If ker (Φ) = 0, then ker (Φ) = h f ( t ) i is generated by polynomial f ( t ) ∈ K [ t, σ ]. Therefore, K [ a ] ∼ = K [ t, σ ] / h f ( t ) i isa finite dimensional vector space over K . This implies that K [ a ] is a divisionsubring and by [1, Lemma 1], it is centrally finite. Note that N ∩ ( K [ a ]) ∗ is anon-central subnormal subgroup of ( K [ a ]) ∗ because a ∈ N ∩ ( K [ a ]) ∗ . Hence, by[7, Theorem 2.1], N ∩ ( K [ a ]) ∗ contains a non-abelian free subgroup. Now, assumethat ker (Φ) = 0. Then, Φ is isomorphism, so Φ can be extended naturally tothe isomorphism Φ : K ( t, σ ) → K ( a ). Let K be the subfield of K generated by { σ i ( a ) | i ∈ Z } . Then, the restriction of σ on K is the isomorphism of K . There-fore, K ( t, σ ) is a division subring of K ( t, σ ). Put M = Φ( N ) ∩ K ( t, σ ). Then, M is a subnormal subgroup of K ( t, σ ) ∗ containing t . N SUBNORMAL SUBGROUPS CONTAINING A SOLVABLE SUBGROUP 9 If K /P is infinitely generated, then by Theorem 2.5, M contains a non-abelianfree subgroup. If K /P is finitely generated, then trdeg ( K /P ) is finite. Buttrdeg ( F/P ) = ∞ implies the existence of elements x, y ∈ F such that x, y arealgebraically independent over P . By [5, Lemma 7.1], x, y are also algebraicallyindependent over K ( a ). Then, M = N ∩ K ( a )( x, y ) ∗ is a non-central normalsubgroup of K ( a )( x, y ) ∗ because a ∈ M . By Proposition 3.2, M contains a non-abelian free subgroup. The proof of the theorem is now complete. (cid:3) It is well known that every uncountable field has infinite transcendence degreeover its prime subfield, so the proof of following corollary is elementary.
Corollary 4.2.
In a division ring D with uncountable center, every subnormalsubgroup of D ∗ containing a non-abelian solvable subgroup contains a non-abelianfree subgroup. Theorem 4.3.
Let D be a division ring with center F . Assume that N is asubnormal subgroup of D ∗ containing a non-abelian solvable subgroup H . If H is algebraic over F , then N is contains a non-abelian free subgroup.Proof. Let n be the derived length of H , that is, n is the smallest positive integersuch that the n -th derived subgroup H ( n ) = 1. Since H is non-abelian, n ≥
2. Put G = H ( n − . Then G is a non-abelian metabelian subgroup. Set ℜ = { A | G ′ ≤ A ≤ G and A is abelian } . By Zorn’s Lemma, there exists a maximal element in ℜ , say M . Then, M is anabelian normal subgroup of G . Put K = F [ M ] and fix an element a ∈ G \ M .Clearly, a normalizes K , that is, aKa − = K and the set L = P i ∈ N Ka i is asubspace of the left vector space D over a field K . By the fact that a is algebraicover F , it is easy to see that L ⊆ P mi =0 Ka i for some positive integer m . Moreover,it is easy to that aKa − = K , which implies that L is a division subring of D andfinite dimensional left vector space over K . By Zorn’s Lemma, K is contained insome maximal subfield K of L , so L is a finite dimensional left vector space over K , and in view of [17, (15.8)], L is centrally finite. Let N = N ∩ L . Then N isnon-abelian subnormal in L ∗ . Therefore, N contains a non-abelian free subgroupby [7, Theorem 2.1]. (cid:3) References [1] S. Akbari, R. Ebrahimian, H. M. Kermani, and A. S. Golsefidy, Maximal subgroups of GL n ( D ), J. Algebra (2003) 201–225.[2] J. P. Bell and J. Z. Goncalves, Free algebras and free groups in Ore extensions and free groupalgebras in division rings,
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