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On almost subnormal subgroups in division rings

Let D be a division ring with infinite center F , and G an almost subnormal subgroup of D. In this paper, we show that if G is locally solvable, then G ⊆ F . Also, assume that M is a maximal subgroup of G. It is shown that if M is non-abelian locally solvable, then [D : F ] = p for some prime number p. Moreover, if M is locally nilpotent then M is abelian.


Introduction
There has been recently a great deal of interest in algebraic structure of the multiplicative group D * of a division ring D. For a self-contained account of the development on this area of study, we refer to [14]. It is not much to say that the starting point is the Wedderburn Little Theorem stating that any finite division ring is a field. This famous theorem is the subject for many generalizations. Many authors paid attention on the question that how far D * , and more generally its subnormal subgroups, from being abelian. On this line, the well-known result of Hua states that if D * is solvable then D is a field. Some special types of subnormal subgroups such as nilpotent, solvable, and locally nilpotent have been examined. For instance, Stuth [26,Theorem 6 (iii)] proved that every solvable subnormal subgroup of D * is central, i.e, it is contained in the center F of D. In [17], Huzurbazar showed that this result remains true if the word 'solvable' is replaced by 'locally nilpotent'. In [10,Theorem 2.3], the authors extended the result of Hua by proving that if D * is locally solvable, then D is a field. It is clear that if G is locally nilpotent or solvable, then it is locally solvable. However, the converse does not hold: there exists a division ring which contains a locally solvable subgroup that is neither solvable nor locally nilpotent (see [25, 1.4.13]). In view of [26,Theorem 6 (iii)] and [10,Theorem 2.3], it is natural to ask whether every locally solvable subnormal subgroup of D * is central (see [11,Conjecture 1]). It was shown in [12] that the question has the positive answer in the case when D is algebraic over F . Recently, the positive answer to this question in general case have been obtained in [18]. One of our purposes in this paper is to extend this result for almost subnormal subgroups. We shall prove that in a division ring D with center F every locally solvable almost subnormal subgroup is central, provided F is infinite.
Let us recall the notion of almost subnormal subgroups. Following Harley [13], a subgroup H of a group G is called almost subnormal in G if there is a finite sequence of subgroups In all other cases H/GE * is abelian.
In the above lemma if we take E = F , where F is the center of D, then we have the following result.
Lemma 2.2. Let D be a division ring with center F , and G a locally nilpotent subgroup of D * such that F (G) = D. If H = N D * (G), then H/GF * is solvable. Theorem 3.10]). Let D be a division ring with infinite center F , and K a division subring of D. Assume that G is a non-central almost subnormal subgroup of D * . If g −1 Kg ⊆ K for any g ∈ G, then either K ⊆ F or K = D.
Proof. Trivially G satisfies a non-trivial group identity, and the result follows from [22,Theorem 2.2].
Proof. Let A = a 1 , a 2 , . . . , a n be a finitely generated subgroup of G ′ . Then we may write where x ij , y ij ∈ G for 1 ≤ i ≤ n. If N is the subgroup of G generated by all x ij , y ij , then A ⊆ N ′ . By hypothesis, we conclude that N Z(G)/Z(G) is finite. It follows by Schur's Theorem that N ′ = (N Z(G)) ′ is finite ([24, 10.1.3, p.278]); note that Z(G) ⊆ Z(N Z(G)). This implies that A is finite.
Lemma 2.6. If D is a divsision ring such that D (k) is locally nilpotent for some k ≥ 1, then D is commutative.
Proof. Assume by contradiction that G is not contained in F . Since G is an almost subnormal subgroup of D * , there is a finite chain of subgroups x s! y s! x −s! y −s! = 1 is a group identity of G 0 . It follows by [22,Theorem 2.2] that G = G 0 ⊆ F , a contradiction. Therefore C is not contained in F . By replacing G 0 by C if necessary, we can assume that G 0 is normal in G 1 .
Since G 0 ⊆ F , it follows by Lemma 2.3 that F (G 0 ) = D. If we set H = N D * (G 0 ), then Lemma 2.2 says that H/G 0 F * is solvable. Since G 0 is assumed to be normal in G 1 , we conclude that G 1 ⊆ H. It follows that G 1 /G 1 ∩ G 0 F * ∼ = G 1 F * /G 0 F * is also solvable. Thus, there exists an integer k 1 for which G (k1) 1 ⊆ G 0 F * , which is a locally nilpotent group. If G (k1) 1 ⊆ F , then G 1 is solvable. According to Lemma 2.4, we conclude that G 1 ⊆ F , a contradiction. Therefore, we may assume that G is a non-central locally nilpotent normal subgroup of G 2 . By replacing G 1 and G 0 by G 2 and G (k1) 1 in the preceding paragraph, we conclude that G (k2) 2 is locally nilpotent, for some k 2 . By finite steps, we obtain that D (kn) is locally nilpotent for some k n . This fact together with Lemma 2.6 implies that D is commutative, a contradiction. Proof. Asumme that 0 = x ∈ F [G]. Then, x = f 1 g 1 + f 2 g 2 + · · · + f n g n for some f i ∈ F and g i ∈ G. The subgroup H = g 1 , g 2 , · · · , g n is finite, so F [H] is a domain which is a finite dimensional vector space over F . This implies that x where f ij ∈ F and g ij ∈ G. Let H = g ij : 1 ≤ i ≤ k, 1 ≤ j ≤ s be the subgroup of G generated by all g ij . Since G/G ∩ F * ∼ = GF * /F * is locally finite, the group HF * /F * is finite. Let {y 1 , y 2 , . . . , y t } be a transversal of F * in HF * and set Then, clearly S is a division ring that is finite dimensional over F containing Proof. Assume by contradiction that G ⊆ F . It follows by Lemma 2.3 that F (G) = D. In view of Lemma 2.9, we conclude that D is a locally finite division ring. According to [22,Theorem 4.3], we conclude that G contains a non-cyclic free subgroup. But this is impossible since G is locally finite.
Lemma 2.11. Every locally solvable periodic group is locally finite.
Proof. Let G be a locally solvable periodic group, and H a finitely generated subgroup of G. Then, H is solvable with derived series of length n ≥ 1, say, We shall prove that H is finite by induction on n. For if n = 1, then H is a finitely generated 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 ′ is finitely generated. By induction hypothesis, H ′ is finite, and as a consequence, H is finite.  Proof. By Lemma 2.3 we conclude that F (G) = D. If we set R = F [G], then R is an Ore domain by [30,Corollary 24]. Thus, the skew field of fractions of R is coincided with D. It follows by Lemma Proof. Assume by contradiction that G ⊆ F . Since G is an almost subnormal subgroup of D * , there is a finite chain of subgroups By the same argument used in the first paragraph of the proof of Proposition 2.7, we may assume that G 0 is normal in G 1 . It follows by Lemma 2.14 that N D * (G 0 ) = G 0 F * , which is a locally solvable group. Since G 1 ⊆ N D * (G 0 ), we conclude that G 1 is a locally solvable group. After finite steps, we obtain that D * is locally solvable. According to [10, Theorem 2.1], we conclude that D is commutative, a contradiction.

Locally solvable maximal subgroups
Let R be a ring, S a subring of R, and G a subgroup of the group of units of R normalizing S such that R = S[G]. Suppose that N = G ∩ S is a normal subgroup of G and R = t∈T tS, where T is some (and hence any) transversal T of N to G. Then, we say that R is a crossed product of S by G/N (see [31] or [25, p.23]). For the convenience of readers, we gather a number of theorems of B. A. F Wehrfritz, which will be used in the proofs of our results.   For a group G, we denote by Z 2 (G), which is defined by Z 2 (G)/Z(G) = Z(G/Z(G)), the second center of G. Remark 2. In the above lemma, the notation P, L A stands for a class of groups, in which A denotes the class of abelian groups and P and L the poly and local operators ( [31]). It is clear that the class of groups P, L A contains that of locally solvable groups.   Proof. For any ϕ ∈ Aut(G) and x ∈ C G (H), our task is to show that ϕ(x) ∈ C G (H). Take an arbitrary element h ∈ H. Since ϕ(H) = H, there exists h ′ ∈ H such that ϕ(h ′ ) = h. Now, we have which implies that ϕ(x) ∈ C G (H).
Kx i . It remains to prove that K * ∩ G is the Hirsch-Plotkin radical of M . Note that we are in the case that K * ∩ G ⊆ M , we conclude that K * ∩ G = K * ∩ M . Thus, we have M/K * ∩ M ∼ = M K * /K * ∼ = Z p . Let H be the Hirsch-Plotkin radical of M . Then HK * /K * ≤ M K * /K * , thus either H ⊆ K * or HK * = M K * . The first case implies that H ⊆ K * ∩ G, so H = K * ∩ G; we are done. If the second case occurs, then F [HK * ] = F [M ] = D. It follows that HK * is a locally nilpotent absolutely irreducible subgroup of D * , hence it is center-by-locally finite by [28,Theorem 1]. It is clear that the center of HK * is contained in F * . This yields that HK * /F * is locally finite, from which it follows that K/F is a non-trivial radical Galois extension. According to [21, 15.13], we conclude that D is algebraic over a finite subfield. But then Jacobson's Theorem ( [21, 13.11]) says that D is commutative, a contradiction. Proof. Assume by contradiction that M is non-abelian. We claim that [D : F ] < ∞. Let T = τ (M ) be the unique maximal periodic (hence locally finite) normal subgroup of M , and Z/T be the center of M/T . Since T is normal in M , it follows by Lemma 3.6 that either F (T ) = D or T is abelian. If the first case occurs, then Lemma 2.9 says that D is a locally finite division ring. According to [9, Theorem 3.1], we have [D : F ] < ∞; we are done. Now assume that T is abelian. This fact implies that Z is a solvable group with the derived series length of s ≥ 1, say. Since T and Z/T is a characteristic subgroup of M and M/T respectively, we conclude that Z is a characteristic subgroup of M . Again in view of Lemma 3.6, either F (Z) = D or Z is abelian. Let us consider the following two possible cases: Since D is non-commutative, it follows that Z is non-abelian. Thus, we may suppose that it is solvable with derived length s ≥ 2. Therefore, there exists such a series Let N be the maximal subgroup of M with respect to the property: N is an abelian normal subgroup of M containing Z. We shall show that M/N is a simple group. For, let P be a normal subgroup of M properly containing N . Clearly P is non-abelian. By Lemma 3.6, we have F (P ) = D. Since P is locally nilpotent, it follows by [30,Corollary 24] that F [P ] is an Ore domain whose skew field of fractions is coincided with D. In view of Lemma 3.1, we conclude that F [M ] is a crossed product of F [Z] by M/Z. It follows by Lemma 3.3 that M = P Z = P ; recall that Z ⊆ P . This fact shows that M/Z is simple. Note that M = Z since M is non-abelian. Because M/Z is locally nilpotent, we conclude that M/Z is finite of prime order ([24, 12.5.2, p.367]). If we set L = F (Z), then L is a subfield of D, over which D is finite dimensional. This fact yields [D : F ] < ∞.
By what we have proved, we conclude that n := [D : F ] < ∞. We know that D ⊗ F D op ∼ = M n (D). Therefore, we may view M as a subgroup of GL n (F ) to conclude that it is a solvable group. According to Theorem 3.9, we conclude that M is distinguish from its Hirsch-Plotkin radical, which contrasts to the fact that M is locally nilpotent.  Let B = B(M ) be the subgroup of M such that B/T is the Hirsch-Plotkin radical of the group G/T . As we see in the proof of Lemma 2.13, the group B is locally nilpotent. It follows that B is the the Hirsch-Plotkin radical of M (Remark 1). If M = B, then M is locally nilpotent and thus it is abelian by Theorem 3.10, a contradiction. We may therefore assume that M = B. We claim that M/B is a simple group. In fact, let C be a normal subgroup of M properly containing C. It is clear that C is non-abelian. It follows by Lemma 3.6 that F (C) = D. According to Lemma 3.2, we conclude that F [M ] is a crossed product of F [B] by M/B. Moreover, by [30,Corollary 24], it follows that F [C] is an Ore domain whose skew field of fractions is coincided with D. Therefore, we may apply Lemma 3.3 to conclude that M = BC, which implies that M = C. Thus, the group M/B is simple, as claimed. Again by [24, 12.5.2, p.367], we have M/B is a finite group of prime order. In view of Lemma 3.6, either B is abelian or F (B) = D. Let us consider the following two subcases.  By what we have proved, we conclude that n := [D : F ] < ∞. We know that D ⊗ F D op ∼ = M n (F ). Thus, we may view M as a subgroup of GL n (F ), to conclude that M is solvable. The results follow by Theorem 3.9.