Tits Alternative for Maximal Subgroups of GL_{n}(D) ^{∗}

M. Mahdavi-Hezavehi ^{†} mahdavih@sharif.edu Department of Mathematical Sciences Sharif University of Technology P. O. Box 11365-9415 Tehran, Iran

Abstract

Let D be a noncommutative division algebra of finite dimension over its centre F. Given a maximal subgroup M of GL_{n}(D) with n ≥ 1, it is proved that either M contains a noncyclic free subgroup or there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M for all 1 ≤ i ≤ r such that M/A is finite if CharF = 0 and M/A is locally finite if CharF = p > 0, where A = K_{1}^{∗} × ⋯ × K_{r}^{∗}.

1 Introduction

Let D be a division algebra of finite dimension over its centre F. Denote by M_{n}(D) the n×n matrix ring over D and SL_{n}(D) the commutator subgroup of the multiplicative group GL_{n}(D) = M_{n}(D)^{∗}. Given a subgroup G of GL_{n}(D), we shall say that G is maximal in GL_{n}(D) if for any subgroup H of GL_{n}(D) with G ⊂ H, one concludes that H = GL_{n}(D). We know, by Corollary 1 of [11], that G(A) := A^{∗}/RN(A^{∗})A′, where A := M_{n}(D), A′ = SL_{n}(D), and RN(A^{∗}) is the image of A^{∗} under the reduced norm of A to F, is an abelian torsion group of a bounded exponent dividing the degree of A over F. This group is not trivial in general. For example, if A is the algebra of real quaternions, then G(A) is trivial whereas for rational quaternions G(A) is isomorphic to a direct product of copies of Z_{2}, as it is easily checked. Assume that G(A) is not trivial, then by Prũfer-Baer Theorem (cf. [13, p. 105]), we conclude that G(A) is isomorphic to a direct product of Z_{r_{i}}, where r_{i} divides the index of A over F. In this way one may obtain normal maximal subgroups of finite index in GL_{n}(D). So, if G(A) is not trivial, then GL_{n}(D) contains maximal subgroups. For some examples of non-normal maximal subgroups of GL_{n}(D), see [12] or [3, p. 140]. It is shown in [12] that even for the case G(A) = 1 we may have maximal subgroups in GL_{n}(D). But the question of whether GL_{n}(D) has a maximal subgroup for any noncommutative division algebra D, is still open. Now, let D be a noncommutative division ring not necessarily of finite dimension over its centre F. The problem of whether GL_{n}(D) contains a noncyclic free subgroup seems to be posed first by Lichtman in [8]. Stronger versions of this problem which essentially deal with the existence of noncyclic free subgroups in normal or subnormal subgroups of GL_{n}(D) have been investigated in [4] and [5]. It is known so far that these problems have positive answers as long as we work with a division algebra of finite dimension over its centre. Further investigations for the infinite dimensional case are also dealt with in [4] and [5]. The study of maximal subgroups of GL_{n}(D) begins in [2] in relation with an investigation of the structure of finitely generated normal subgroups of GL_{n}(D), where D is of finite dimension over its centre F. In [2] and [10] we actually show that maximal subgroups arise naturally in GL_{n}(D), n ≥ 1, and finitely generated subnormal subgroups of GL_{n}(D), n ≥ 1, are central. This result is used to prove that a maximal subgroup of GL_{n}(D) can not be finitely generated for n ≥ 1. Therefore, we are not able to apply directly Tits’ result, that any finitely generated linear group either is soluble-by-finite or contains a noncyclic free group (cf. [19]), to a maximal subgroup M of GL_{n}(D) to explore the structure of M. In [2], it is also shown that there is a similarity between the behaviour of normal or subnormal subgroups of GL_{n}(D) and the maximal ones. So, it is natural to ask if there exists a noncyclic free group in a maximal subgroup of GL_{n}(D). In this direction, we observe that not every subgroup of GL_{n}(D) satisfies the Tits’ Theorem though any normal subgroup does so (cf. [16, p. 154]). We also mention that the soluble subgroups of the multiplicative group of a finite dimensional division algebra were studied in 1962 by Suprunenko [17] and the soluble subgroups of the multiplicative group of a finite dimensional simple algebra were considered by Zalesskii [22]. Inspired by Suprunenko’s results, it is shown in [12] that given a noncommutative maximal subgroup M of GL_{1}(D), then either M contains a noncyclic free subgroup or there exists a maximal subfield K of D which is Galois over F such that K^{∗} is normal in M and M/K^{∗} ≅ Gal(K/F). Using some results of algebraic groups and skew linear groups, in the present note our aim is to extend this result for n > 1, i.e., to prove a variation of Tits’ Theorem for maximal subgroups of GL_{n}(D). To be more precise, let D be a noncommutative division algebra of finite dimension over its centre F. Given a maximal subgroup M of GL_{n}(D), it is proved that either M contains a noncyclic free subgroup or there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M for all 1 ≤ i ≤ r such that M/A is finite if CharF = 0 and M/A is locally finite if CharF = p > 0, where A = K_{1}^{∗} × … × K_{r}^{∗}.

∗Key words: Free Subgroup, Division ring, maximal subgroup.

†AMS (1991) Subject classification : 15A33, 16K

2 Notations and conventions

Let D be a division ring with centre F. Given a subgroup G of GL_{n}(D), we denote by F [G] the F -algebra generated by elements of G over F. We shall say that G is absolutely irreducible if M_{n}(D) = F [G]. For any group G we denote its centre by Z(G). Given a subgroup H of G, N_{G}(H) means the normalizer of H in G, [G : H] denotes the index of H in G, and < H, K > the group generated by H and K, where K is a subgroup of G. We shall say that H is soluble-by-finite if there is a soluble normal subgroup K of H such that H/K is finite. Let S be a subset of M_{n}(D), then the centralizer of S in M_{n}(D) is denoted by C_{M_{n}(D)}(S). We shall identify the centre F I of M_{n}(D) with F. Some notations and conventions for linear groups and skew linear groups from [15], [16] and [18] are frequently used throughout.

3 Free subgroups in maximal subgroups

Given a division ring D with centre F, let M be a maximal subgroup of GL_{n}(D). This section essentially deals with maximal subgroups of GL_{n}(D) and how they sit in GL_{n}(D) with respect to F^{∗} and SL_{n}(D). We then present some commutativity theorems that enable us to prove our main result. To be more precise, let D be a noncommutative division ring not necessarily of finite dimension over its centre F. It is shown that there exists no maximal subgroup M of GL_{n}(D), n ≥ 1, containing F^{∗} such that [M : F^{∗}] < ∞. It is then proved that M is nilpotent if and only if M is the multiplicative group of a maximal subfield of M_{n}(D). We then use these results to show that given a maximal subgroup M of GL_{n}(D), n ≥ 1, if M is soluble, then there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M, 1 ≤ i ≤ r, such that M/A is finite, where A = K_{1}^{∗} × ⋯ × K_{r}^{∗}, and so M is abelian-by- finite. Using this fact, we then obtain the same conclusion when M ∩SL_{n}(D) is commutative or M/M ∩ F^{∗} is torsion for any maximal subgroup of GL_{n}(D). Finally, we apply these results to prove our main theorem that if M is a maximal subgroup of GL_{n}(D), n ≥ 1, then either M contains a noncyclic free subgroup or there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M for all 1 ≤ i ≤ r such that M/A is finite if CharF = 0 and M/A is locally finite if CharF = p > 0, where A = K_{1}^{∗} × … × K_{r}^{∗}. We begin our material with

Lemma 1. Let D be a division ring not necessarily of finite dimension over its centre F. If either n = 1 and D is noncommutative or n > 1 and D is infinite, then there exists no maximal subgroup M of GL_{n}(D), n ≥ 1, containing F^{∗} such that [M : F^{∗}] < ∞.

Proof. Assume on the contrary that there is a maximal subgroup M such that M/F^{∗} is finite and first suppose that [D : F] < ∞. Let x_{1}, …, x_{t} be the representatives for cosets of F^{∗} in M, i.e., M = F^{∗}x_{1} ∪ … ∪ F^{∗}x_{t}. Then, we have M =< x_{1}, …, x_{t} > F^{∗}, where < x_{1}, …, x_{t} > is the group generated by x_{1}, …, x_{t}. Take x ∈ GL_{n}(D)\M. By maximality of M, we obtain GL_{n}(D) =< x_{1}, …, x_{t}, x > F^{∗}. Put H =< x_{1}, …, x_{t}, x >. Thus, GL_{n}(D) = HF^{∗} and consequently we have SL_{n}(D) = H′ ⊂ H, i.e., H is normal in GL_{n}(D). Now, by Corollary 1 of [10], we conclude that H ⊂ F^{∗}, i.e., GL_{n}(D) = F^{∗} which means that n = 1 and D = F that is a contradiction. This takes care of the finite dimensional case.

Now consider the case [D : F] = ∞. As above, we may assume M = F^{∗}x_{1} ∪ … ∪ F^{∗}x_{t}. Put A = {Σ_{i=1}^{n}f_{i}x_{i}; f_{i} ∈ F }. It is clearly seen that A is a finite dimensional F -algebra and we have M ⊂ A^{∗}. Since A is of finite dimension over F we conclude that A ≠ M_{n}(D) and so M = A^{∗} by maximality of M in GL_{n}(D). If n = 1, then it is easily seen that A is a division algebra. Thus we have [A^{∗} : F^{∗}] < ∞. If A is infinite, then, by a result of Faith (cf. [7, p. 225]), A = F and so M = F^{∗} which is a contradiction. So, we may assume that A is finite. Now, Wedderburn’s Theorem implies that A is a finite field. So there exists an element a ∈ D^{∗} such that A^{∗} =< a >, i.e., a^{n} = 1 for some positive integer n. Since a is non-central in D, by Herstein’s Lemma (cf. [7]), there is an element b ∈ D^{∗} such that bab^{−1} = a^{i} ≠ a. Thus, b ∈ N_{D^{∗}}(A^{∗}) and so < M, b >⊂ N_{D^{∗}}(A^{∗}). Now, by maximality of M, we conclude that N_{D^{∗}}(A^{∗}) = D^{∗}. Therefore, by Cartan-Brauer-Hua’s Theorem, we have either A ⊂ F or A = D, and it is clear that none of these cases can occur. This completes the proof for the case n = 1 and [D : F] = ∞.

So, we may assume that n ≥ 2 and [D : F] = ∞. We claim that A is simple. To see this, we first observe that the Jacobson radical J(A) = J of A is nilpotent since A is left Artinian. It is known that a multiplicative semi-group of nilpotent matrices over a division ring can be simultaneously triangularized (cf. [6, p. 135]). Thus, we may assume that each element of J is upper triangular. Now, denote by L the subring of all elements x in M_{n}(D) such that xJ ⊂ J. It is clear that A ⊂ L. For any d ∈ D we see that I + de_{1n} ∈ L. We claim that M contains only a finite number of these elementary matrices. To prove this, assume that I + d_{1}e_{1n}, I + d_{2}e_{1n} ∈ F^{∗}x_{i} for some d_{1}, d_{2} ∈ D^{∗}. Then, we have I + d_{1}e_{1n} = λx_{i} and I + d_{2}e_{1n} = μx_{i} for some λ, μ ∈ F^{∗}. This implies that d_{1} = d_{2} and λ = μ. Thus, at most t elementary matrices of the form I + de_{1n} belong to M. Since M = A^{∗} ⊂ L^{∗} and D is infinite we conclude that there is an element d ∈ D^{∗} such that I + de_{1n} ∈ L^{∗} \ M. By maximality of M, this implies that L = M_{n}(D), i.e., J is a left ideal of M_{n}(D). A similar argument applied to the subring R of all elements x ∈ M_{n}(D) such that Jx ⊂ J, we conclude that J is also a right ideal of M_{n}(D), i.e., J is a two sided ideal of M_{n}(D) which means that J = 0. So A is semisimple. Now, we observe that the centre Z(A) does not contain any nonzero zero-divisor. For otherwise, if 0 ≠ u ∈ Z(A) is a zero-divisor in A, consider l = l(u), the left annihilator of u in A. We have l ≠ 0 since u is a zero-divisor. Now, since u ∈ Z(A) we conclude that l is an ideal of A and we have lu = 0. Since entries of each element of l belong to a division ring we may reduce the j-th column of each element of l to zero. Now, the subring R of M_{n}(D) consisting of all elements x ∈ M_{n}(D) such that lx ⊂ l contains A and we have le_{ji} = 0. Thus, we conclude that I + de_{ji} ∈ R for all d ∈ D. Now, since F^{∗} is of finite index in M = A^{∗}, we conclude as above that there exists an element d ∈ D^{∗} such that I + de_{ji} ∈ R^{∗} \ M. Since M = A^{∗} ⊂ R^{∗} by maximality of M we obtain R = M_{n}(D), i. e., l is a right ideal of M_{n}(D). Since l is a two sided ideal of A, a similar argument applied to the subring L of M_{n}(D) consisting of all elements x ∈ M_{n}(D) such that xl ⊂ l, we may conclude that L is also a left ideal of M_{n}(D), i. e., l = 0 which is a contradiction and so no nonzero element of Z(A) is a zero-divisor. Now, since A is a semisimple Artinian ring we conclude that A is simple as claimed. Therefore, there is a positive integer n_{1} such that M ≅ GL_{n_{1}}(D_{1}), for some division ring D_{1} of finite dimension over its centre K, say. We know that F^{∗} is of finite index in M. Thus, the image of F^{∗} in GL_{n_{1}}(D_{1}), which is a subfield of K, is also of finite index in GL_{n_{1}}(D_{1}). We identify this image by F^{∗}. Thus, D_{1}^{∗}/K^{∗} is torsion and so, by a result of Kaplansky (cf. [7, 259]), we obtain D_{1} = K, i. e., A ≅ M_{n_{1}}(K), where K is of finite dimension over F. Now, since M/F^{∗} is finite, we conclude that GL_{n_{1}}(K)/F^{∗} is finite. Thus, we have GL_{n_{1}}(K) = ∪_{1}^{t}F^{∗}u_{i} for some u_{i} ∈ GL_{n_{1}}(K). Take an elementary matrix I + be_{ij} ∈ GL_{n_{1}}(K), where b ∈ K^{∗}. As observed above, for each b, I + be_{ij} may be contained in only one coset F^{∗}u_{i}. Therefore, there is a finite number of elements I + be_{ij} which occur in M ≅ GL_{n_{1}}(K). Thus, there are at most n^{2}t elementary matrices in GL_{n_{1}}(K). This means that SL_{n_{1}}(K), which is generated by I + be_{ij}, is finitely generated. Now, by Corollary 1 of [10], we conclude that SL_{n_{1}}(K) ⊂ K^{∗}. This in turn implies that n_{1} = 1 and consequently M = K^{∗}, i. e., K^{∗}/F^{∗} is finite. But it is known that this is not possible unless M = K^{∗} is finite. Now, by Theorem 4 of [1], that asserts that a normal subgroup of GL_{n}(D) does not contain a finite maximal subgroup, we arrive at a contradiction and so the proof is completes.

In the next result we show how maximal subgroups of GL_{n}(D) sit in GL_{n}(D) with respect to F^{∗} and SL_{n}(D).

Proposition 2. Let D be a division ring not necessarily of finite dimension over its centre F. Assume that M is a maximal subgroup of GL_{n}(D), n ≥ 1. Then we have

(i) M contains F^{∗} or SL_{n}(D). (ii) Either M is (absolutely) irreducible or M is the group of units of a proper subring of M_{n}(D). (iii) Assume that D is noncommutative and [D : F] < ∞. Then M is nilpotent if and only if M is the multiplicative group of a maximal subfield of M_{n}(D).

Proof. (i) Assume that M does not contain F^{∗}. Then we must have GL_{n}(D) = F^{∗}M, and consequently SL_{n}(D) = M‘ ⊂ M.

(ii) Consider the F - algebra F [M] generated by M over F. By maximality of M, we have either GL_{n}(D) = F [M]^{∗} or M = F [M]^{∗}. In the first case we obtain M_{n}(D) = F [M], and the second case implies that M is the group of units of a proper subring of M_{n}(D).

(iii) One way is clear. So, assume that M is nilpotent. By (i), we have either F^{∗} ⊂ M or SL_{n}(D) ⊂ M. If the second case occurs, then SL_{n}(D) is nilpotent. Thus, in particular, the derived group D′ of D^{∗} is nilpotent and so D^{∗} is soluble. Therefore, by Hua’s Theorem (cf. [7, 223]), D is commutative which is a contradiction. So, we may assume that F^{∗} ⊂ M but SL_{n}(D) ⊄ M. The case n = 1 follows from Theorem 7 of [2]. Thus, we may suppose n > 1 and consider the F -algebra A := F [M] which is left Artinian since [D : F] < ∞. We may also view M as a linear group since [D : F] < ∞. By (ii), we conclude that either M is (absolutely) irreducible or A^{∗} = M. If the first case happens, i. e., M_{n}(D) = A, then it is clear that Z(M) = F^{∗} and M is an irreducible linear group (cf. [18, p. 100]). Now, it is known that if a linear group M is irreducible and nilpotent, then [M : Z(M)] < ∞ (cf. [17, p. 57]). But this is not possible by Lemma 1. Thus, we must have A^{∗} = F [M]^{∗} = M. If we prove that A is simple, then we obtain M ≅ GL_{m}(Δ) for some positive integer m and division ring Δ. Consequently, by Hua’s Theorem as above, we conclude that m = 1 and Δ = K is a field and therefore M is the multiplicative group of a field. Furthermore, we then have C_{M_{n}(D)}(A) = A by maximality of M, and this shows that M is the multiplicative group of a maximal subfield of M_{n}(D). So, it remains to prove that A is simple. To do this, we first observe that the Jacobson radical of A, J(A) = J, is nilpotent. As noted in the proof of Lemma 1, the elements of J may be assumed to be upper triangular. Now, let L be the subring of all elements x ∈ M_{n}(D) such that xJ ⊂ J. It is clear that A ⊂ L. For any d ∈ D^{∗} we see that the matrix D_{n}(d) = I + (d − 1)e_{nn} belongs to L. If M contains all matrices D_{n}(d) for d ∈ D^{∗}, then M contains a copy of D^{∗}. But this is not possible by Hua’s Theorem unless D is commutative which is a contradiction. Therefore, by maximality of M, we obtain L = M_{n}(D), i.e., J is a left ideal of M_{n}(D). A similar argument applied to the subring of all elements x in M_{n}(D) such that Jx ⊂ J, we conclude that J is a right ideal and thus J = 0. This means that J is semisimple. A similar argument as used in the proof of Lemma 1, one can show that Z(A) is a field and thus A is simple and the result follows.

To prove our next result, we need the following theorems:

Theorem A.(Rosenberg, [14]) If A is a simple ring with unit, the only subrings of M_{n}(A), n ≥ 2, invariant under all inner automorphisms of M_{n}(A), are subrings of the centre or M_{n}(A) itself.

Theorem B.(Snider, [16]) Let G be an absolutely irreducible subgroup of GL_{n}(D), let N be a normal subgroup of G and K a subring of M_{n}(D) normalized by G with F ⊂ K. If G/N is locally finite, then K[N] is semisimple Artinian.

Theorem C. Let D be a division ring that is not a locally finite field and let n > 1 be an integer. If N is any non-central normal subgroup of GL_{n}(D), then N contains a noncyclic free subgroup(cf. [16, p.154]).

Theorem D. Every irreducible soluble subgroup of GL_{n}(F) has an abelian normal subgroup of finite index(cf. [18, p. 135]).

The next result gives a criterion for when a maximal subgroup M of GL_{n}(D) is soluble, i. e., M is soluble if and only if there exists an abelian normal subgroup A such that M/A is finite and soluble.

Theorem 3. Let D be a noncommutative division algebra of finite dimension over its centre F. Suppose M is a maximal subgroup of GL_{n}(D), n ≥ 1. If M is soluble, then there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M, 1 ≤ i ≤ r, such that M/A is finite, where A = K_{1}^{∗} × ⋯ × K_{r}^{∗}, and so M is abelian-by-finite.

Proof. The case n = 1 follows from Corollary 4 of [12]. So, we may assume that n ≥ 2. By Proposition 2, we know that either F^{∗} ⊂ M or SL_{n}(D) ⊂ M. If SL_{n}(D) ⊂ M, then SL_{n}(D) is soluble. But, by Theorem C, we know that SL_{n}(D) contains a noncyclic free subgroup which is a contradiction. So, we may assume that F^{∗} ⊂ M but SL_{n}(D) is not contained in M. Now, consider the F -algebra A := F [M]. By Proposition 2, we have either A = F [M] = M_{n}(D) which implies that M is (absolutely) irreducible or A^{∗} = M. If the first case occurs, then since [D : F] < ∞ we conclude that M is an irreducible linear group (cf. [18, p. 100]). Therefore, by Theorem D, M contains an abelian normal subgroup B, say, of finite index. Now, consider K := F [B]. By Theorem B, we conclude that F [B] is semisimple Artinian. If F [B]^{∗} ⊄ M, then by maximality of M we have GL_{n}(D) =< F [B]^{∗}, M >⊂ N_{GL_{n}(D)}(F [B]^{∗}).

This implies, by Theorem A, that either F [B] = M_{n}(D) or F [B] ⊂ F. The first case can not happen since n > 1 and the second case implies that [M : F^{∗}] < ∞ which contradicts Lemma 1. Thus, we must have F [B]^{∗} ⊂ M. Now, since F [B] is commutative we have K ≅ K_{1} ×⋯×K_{r} for some fields K_{i}. It is now clearly seen that M/K^{∗} is finite and the result follows in this case.

Therefore, suppose A^{∗} = M. Since [D : F] < ∞, A is left Artinian. We claim that A is semisimple. To see this, we first note that the Jacobson radical J(A) = J of A is nilpotent. As observed in the proof of Lemma 1, we may assume that each element of J is upper triangular. Now, denote by L the subring of all elements x ∈ M_{n}(D) such that xJ ⊂ J. It is clear that A ⊂ L. For any d ∈ D^{∗} we see that the dilatation matrix D_{n}(d) = I + (d − 1)e_{nn} ∈ L. If M contains D_{n}(d) for all d ∈ D^{∗}, then M contains a copy of D^{∗}. Now, by Hua’s Theorem (cf.[7, p. 223]), this reduces to D = F which is a contradiction. Therefore, there is an element d ∈ D^{∗} such that D_{n}(d) ∈ L^{∗} \ M. By maximality of M, this implies that L = M_{n}(D), i. e., J is a left ideal of M_{n}(D). A similar argument applied to the subring R of all elements x ∈ M_{n}(D) such that Jx ⊂ J, one concludes that J is also a right ideal of M_{n}(D). Thus, J is two-sided and we have J = 0. Therefore, A is semisimple, i. e., there exist positive integers n_{i} such that A ≅ M_{n_{1}}(D_{1}) × ⋯ × M_{n_{r}}(D_{r}) for some division rings D_{i}, 1 ≤ i ≤ r. This means that M = A^{∗} ≅ GL_{n_{1}}(D_{1}) × ⋯ × GL_{n_{r}}(D_{r}). But this is not possible since M is soluble whereas GL_{n_{i}}(D_{i}), by Theorem C, contains a noncyclic free subgroup unless n_{i} = 1 for all 1 ≤ i ≤ r. Now, use Hua’s Theorem to conclude that D_{i} = K_{i} for some fields K_{i}. This implies that M ≅ K_{1}^{∗} × ⋯ × K_{r}^{∗} and so the result follows.

In the following result we show that if M ∩ SL_{n}(D) is commutative, then M is abelian-by-finite. This is used later on to prove that if M/M ∩ F^{∗} is torsion, then M is abelian-by-finite.

Lemma 4. Let D be a noncommutative division algebra of finite dimension over its centre F and n ≥ 1. Suppose M is a maximal subgroup of GL_{n}(D). If M ∩ SL_{n}(D) is commutative, then there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M, 1 ≤ i ≤ r such that M/A is finite, where A = K_{1}^{∗} × ⋯ × K_{r}^{∗}, and so M is abelian-by- finite.

Proof. Assume that M ∩ SL_{n}(D) is commutative. It is clear that SL_{n}(D) ⊄ M. We have M′ ⊂ M ∩ SL_{n}(D) and so M is soluble. Now, by Theorem 3, the result follows.

Theorem 5. Let D be a noncommutative division algebra of finite dimension over its centre F. Assume that M is a maximal subgroup of GL_{n}(D) with n ≥ 1. If M/M ∩ F^{∗} is torsion, then there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M, 1 ≤ i ≤ r such that M/A is finite, where A = K_{1}^{∗} × ⋯ × K_{r}^{∗}, and so M is abelian-by-finite.

Proof. The case n = 1 follows from Theorem 6 of [12]. So, we may assume that n ≥ 2. By Proposition 2, we know that either F^{∗} ⊂ M or SL_{n}(D) ⊂ M. If SL_{n}(D) ⊂ M, then P SL_{n}(D) = SL_{n}(D)/Z(SL_{n}(D)) is torsion. Thus, by Corollary 2 of [9], we obtain D = F which is a contradiction. So, we may assume that F^{∗} ⊂ M but SL_{n}(D) ⊄ M. Consider the F -algebra A := F [M]. Since M is maximal in GL_{n}(D) we conclude that either A^{∗} = M or A^{∗} = GL_{n}(D). We now deal with these cases separately:

Case 1. If A^{∗} = GL_{n}(D), then we clearly have A = F [M] = M_{n}(D) which means that M is an irreducible linear group since [D : F] < ∞ (cf. [18, p.100]). We now consider two subcases:

Subcase 1. Char F = 0. By Theorem 1 of [19], either M contains a noncyclic free subgroup or it is soluble-by-finite. The first case can not occur since M/F^{∗} is torsion. Thus, there is a soluble normal subgroup S in M such that [M : S] < ∞. Now, consider the F -algebra F [S]. We know that < F [S]^{∗}, M >⊂ N_{GL_{n}(D)}(F [S]^{∗}).

If F [S]^{∗} ⊄ M, then, by Theorem A, we conclude that either F [S]^{∗} ⊂ F^{∗} or F [S] = M_{n}(D). If F [S]^{∗} ⊂ F^{∗}, then S is central and so [M : F^{∗}] < ∞. Thus, by Lemma 1, we obtain a contradiction. Now, if F [S] = M_{n}(D), then S is an irreducible linear group (cf. [18, p. 100]). Therefore, by Theorem D, we conclude that S contains an abelian normal subgroup of finite index, consequently, M contains an abelian normal subgroup B, say, of finite index.

Put K = F [B]. Then, we have F [B]^{∗} ≠ GL_{n}(D) since n > 1. Thus, K^{∗} = F [B]^{∗} ⊂ M. If K^{∗} ⊂ F^{∗}, then [M : F^{∗}] < ∞ which contradicts Lemma 1. So, assume that F^{∗} ⊂ K^{∗} ⊂ M. By 1.2.5 of [16, p.11], we conclude that K is semiprime and since [D : F] < ∞ we conclude that K is semiprime Artinian. Therefore, by Theorem 10.24 of [7, p.173], K is semisimple. Thus, there exist fields K_{i}, 1 ≤ i ≤ r, and positive integers n_{i} such that K^{∗} ≅ GL_{n_{1}}(K_{1}) × ⋯ × GL_{n_{i}}(K_{i}), where for each i, K_{i} contains a copy of F. Since K^{∗} ⊂ M and M/F^{∗} is torsion we conclude that K_{i} is radical over F. By Kaplansky’s Lemma (cf. [7]), we conclude that char F = p > 0 which is a contradiction.

So, we may suppose F [S]^{∗} ⊂ M. Now, by 1.2.5 of [16, p.11] again, we conclude that F [S] is semiprime and as above F [S] is semisimple Artinian since [D : F] < ∞. Therefore, F [S]^{∗} ≅ GL_{m_{1}}(D_{1}) × ⋯ × GL_{m_{t}}(D_{t}) for some positive inegers m_{i} and division rings D_{i}. This case, via a theorem of Kaplansky (cf. [7]), also leads to a contradiction. This takes care of subcase 1.

Subcase 2. Char F = p > 0. Consider the group G := SL_{n}(D)∩M which is normal in M. If G ⊂ F^{∗}, then by Lemma 4, the result follows. So, we may assume that G is not central. Now, take x ∈ G. We know that x^{n(x)} = a ∈ F^{∗}. Taking the redaced norm RN of M_{n}(D) to F from both sides of the last equation, we conclude that 1 = RN(x)^{n(x)} = a^{m}, where m = \sqrt{dim_{F} M_{n}(D)}. This means that G is a torsion group. Since [D : F] < ∞, we conclude that G is a torsion linear group. Thus, by Schur’s Theorem (cf.[7, p. 154]), G is locally finite. Now, consider the P -algebra S = P [G], where P is the prime subfield of F. By 1.1.14 of [17, p. 9], we conclude that S is a semisimple Artinian ring. Therefore, S = S_{1} × ⋯ × S_{r}, where S_{i} is simple Artinian. Suppose that r = 1. Then S is a matrix ring, S = M_{t}(K), say, for some locally finite field K. If S^{∗} ⊄ M, then GL_{n}(D) =< S^{∗}, M >⊂ N_{GL_{n}(D)}(S^{∗}). This means, by Theorem A, that either S^{∗} ⊂ F or GL_{n}(D) = S^{∗}. If S^{∗} ⊂ F, then G ⊂ F^{∗} and so, by Lemma 4, the result follows. Otherwise, S = M_{t}(K) ≅ M_{n}(D), for some positive integer t and locally finite field K. This implies that D is algebraic over its prime subfield and so D = F, by a result of Jacobson (cf. [7]), which is a contradiction. Therefore, we may assume that r > 1 and set T = ∩_{1}^{r}N_{M}(S_{i}). Consider the F -algebra F [T] = B. Then B is semiprime by 1.2.5 of [16, p.11], and it is also left Artinian since [D : F] < ∞. Thus, B is semisimple Arinian. The central idempotents of S are central in B, and so B is not simple. Therefore, we have in particular B ≠ M_{n}(D). Thus, B ≅ B_{1} × ⋯ × B_{t}, where B_{i} is simple Artinian. Now, consider B^{∗} = GL_{n_{1}}(D_{1}) × ⋯ × GL_{n_{t}}(D_{t}) for some positive integers n_{i} and division rings D_{i}. If B^{∗} ⊄ M, then GL_{n}(D) =< B^{∗}, M >⊂ N_{GL_{n}(D)}(B^{∗}). This means, by Theorem A, that either B = M_{n}(D) which is impossible or B^{∗} ⊂ F^{∗} which is nonsense since B is not simple. Therefore, F^{∗} ⊂ B^{∗} ⊂ M. Since M/F^{∗} is torsion, by Kaplansky’s Theorem (cf. [7]), we conclude that D_{i} = K_{i} is commutative for all i. If for some i, n_{i} > 1, then the matrix D_{1}(k) = I + (k − 1)e_{11}, k ∈ K_{i} is torsion modulo F^{∗}. Thus, K_{i} is algebraic over the prime subfield and since K_{i} contains a copy of F we conclude that F is algebraic over the prime subfield, i. e., D is algebraic over P and so D = F by Jacobson Theorem. This contradiction allows us to assume n_{i} = 1 for all i and D_{i} = K_{i} for some fields, where K_{i} contains a copy of F for all i. Thus, B is commutative and this in turn implies that G is commutative. Now, by Lemma 4, the result follows. This establishes Case 1.

Case 2. Assume that A^{∗} = M. Then F [M] is left Artinian since [D : F] < ∞. We claim that F [M] is simple. To prove this, we first observe that the Jacobson radical of A, J(A) = J, is nilpotent. As we observed in the proof of Lemma 1, the elements of J may be assumed to be upper triangular. Now, let L be the subring of all elements x ∈ M_{n}(D) such that xJ ⊂ J. It is clear that A ⊂ L. For any d ∈ D^{∗} we see that the matrix D_{n}(d) belongs to L. If M contains all D_{n}(d) for d ∈ D^{∗}, then D^{∗}/F^{∗} is torsion and by Kaplansky’s Theorem, we conclude that D = F which is a contradiction. Otherwise, by maximality of M, we obtain L = M_{n}(D), i. e., J is a left ideal of M_{n}(D). A similar argument applied to the subring of all elements x in M_{n}(D) such that Jx ⊂ J, we conclude that J is a right ideal and thus J = 0. This means that A is semisimple. A similar argument as used in the proof of Lemma 1, one can show that Z(A) is a field and thus A is simple, i. e., A ≅ M_{r}(Δ), for some positive integer r and some division ring Δ. As in the Case 1, we conclude that Δ is commutative by Kaplansky’s Theorem since M/F^{∗} is torsion. So, M ≅ GL_{r}(K) for some field K. If r > 1, then D_{1}(k) = I + (k − 1)e_{11}, k ∈ K^{∗} must be torsion modulo F^{∗}. This implies, as before, that D is algebraic over its prime subfield and so D = F which is a contradiction. Thus, we must have r = 1 and so M ≅ K^{∗} is commutative and this completes the proof of the theorem.

We are now in a position to prove our main result as

Theorem 6. Let D be a noncommutative division algebra of finite dimension over its centre F. Assume that M is a maximal subgroup of GL_{n}(D), n ≥ 1. Then either M contains a noncyclic free subgroup or there exists a finite family {K_{i}}_{1}^{r} of fields properly containing F with K_{i}^{∗} ⊂ M for all 1 ≤ i ≤ r such that M/A is finite if CharF = 0 and M/A is locally finite if CharF = p > 0, where A = K_{1}^{∗} × … × K_{r}^{∗}.

Proof. If M is commutative, the result follows from (iii) of Proposition 2. Thus, we may suppose that M is noncommutative. Now, the case n = 1 follows from Theorem 8 of [12]. So, we may assume that n > 1. Suppose M is a noncommutative maximal subgroup of GL_{n}(D). We know, by Proposition 2, that either SL_{n}(D) ⊂ M or F^{∗} ⊂ M. If SL_{n}(D) ⊂ M, then M contain a noncyclic free subgroup by Theorem C. Thus, we may assume that F^{∗} ⊂ M but SL_{n}(D) ⊄ M. Since [D : F] < ∞ we may view M as a linear group over F. Now, consider the F -algebra F [M] generated by M over F. Since M is maximal we have either F [M]^{∗} = M or F [M] = M_{n}(D), i.e., M is absolutely irreducible. We consider these cases separately and assume that M does not contain a noncyclic free subgroup.

Case 1. Assume that M is absolutely irreducible. If Char F = 0 and M does not contain a noncyclic free subgroup we conclude, by Theorem 1 of [19], that M contains a soluble normal subgroup T of finite index, i.e., [M : T] < ∞. If T ⊂ F^{∗}, then by Lemma 1 we obtain a contradiction. Now, by 1.1.7 of [16], T is completely reducible, and hence it contains an abelian normal subgroup A, say, of finite index. Set K = F [A]. Then, we have K^{∗} = F [A]^{∗} ≠ GL_{n}(D) since n > 1. If K^{∗} ⊂ F^{∗}, then [M : F^{∗}] < ∞ and we arrive at a contradiction by Lemma 1. So, assume that F^{∗} ⊂ K^{∗} ⊂ M. Now, by 1.2.5 of [16, p. 11], we conclude that K is semiprime. Since we are in the finite dimensional case K is semiprime Artinian, i.e., K is semisimple. Therefore, there exist fields K_{i}, such that K ≅ K_{1} × K_{2} × ⋯ × K_{r} and we clearly have [M : K^{∗}] < ∞ and so the result follows.

Now let Char F = p > 0. If M does not contain a noncyclic free subgroup, then every finitely generated subgroup of M does not contain a noncyclic free subgroup. By Tit’s Theorem (cf. [19]), we conclude that every finitely generated subgroup of M contains a soluble normal subgroup of finite index. Therefore, by a result of Wehrfritz (cf. [20]), M/R(M) is a torsion linear group, where R(M) is the unique maximal soluble normal subgroup obtained by Zassenhaus-Maltsev Theorem (cf. [21]). Thus, by Schur’s Theorem, M/R(M) is locally finite. Set S = R(M). If S = F^{∗}, then M/F^{∗} is torsion. Thus, by Theorem 5, the result follows. So, we may assume that F^{∗} ⊂ S ⊂ M. Since S is completely reducible it has an abelian normal subgroup V of finite index. We can assume in fact that it is the unique maximal abelian normal subgroup and because of this it is normal in M. If V ⊂ F^{∗}, as above we obtain a contradiction. Now, the quotient group M/V is locally finite. By a similar argument used above, we see that the linear envelope F [V] is a direct sum of fields. This completes the proof of the case F [M] = M_{n}(D).

Case 2. Assume that F [M]^{∗} = M and M does not contain a noncyclic free subgroup. We claim that A := F [M] is simple Artinian. Since the proof is more or less similar to the proof of Theorem 5, we only give the outlines. We first observe that J(A) = J is nilpotent and we may take the elements of J to be upper triangular. Denote by L the subring of all elements x ∈ M_{n}(D) such that xJ ⊂ J. If for all d ∈ D^{∗} the matrices D_{n}(d) = I + (d − 1)e_{nn} ∈ M, then M contains a copy of D^{∗} and this contradicts the assumption that M does not contain a noncyclic free subgroup by Goncalves’ Theorem as above. Thus, there is an element d ∈ D^{∗} such that D_{n}(d) ∈ L^{∗} \M. By maximality of M, this implies that J is a left ideal of A. A similar argument, as used above, shows that J is also a right ideal and so J = 0. Therefore, A is semisimple Artinian since [D : F] < ∞. One may easily check as in the proof of Lemma 1 that Z(A) is a field and so A ≅ M_{r}(Δ) for some division ring Δ and positive integer r. Since, by our assumption M = A^{∗} ≅ GL_{r}(Δ) does not contain a noncyclic free subgroup we obtain r = 1 and Δ = K for some field K. Thus, M = K^{∗} which contradicts our assumption that M is noncommutative and so the proof is complete.

The author thanks the referee for his constructive comments. He is also indebted to the Research Council of Sharif University of Technology for support.

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