Identities on Maximal Subgroups of GL_{n}(D)

D. Kiani, M. Mahdavi-Hezavehi

kiani@mehr.sharif.edu

mahdavih@sharif.edu

Department of Mathematical Sciences

Sharif University of Technology

P. O. Box 11365-9415

Tehran, Iran

Abstract

Let D be a division ring with centre F. Assume that M is a maximal subgroup of GL_{n}(D), n ≥ 1 such that Z(M) is algebraic over F. Group identities on M and polynomial identities on the F -linear hull F [M] are investigated. It is shown that if F [M] is a PI-algebra, then [D : F] < ∞. When D is noncommutative and F is infinite, it is also proved that if M satisfies a group identity and F [M] is algebraic over F, then we have either M = K^{∗}, where K is a field and [D : F] < ∞ or M is absolutely irreducible. For a finite dimensional division algebra D, assume that N is a subnormal subgroup of GL_{n}(D) and M is a maximal subgroup of N. If M satisfies a group identity, it is shown that M is abelian-by-finite.

1 Introduction

Let D be a division ring with centre F, and let n be a positive integer. Denote by A := M_{n}(D) the full n × n matrix ring over D and by A^{∗} := GL_{n}(D), the units of A. Given a subgroup M of A^{∗}, we shall say that M is maximal in A^{∗} if for any subgroup L of A^{∗} with M ⊂ L, one concludes that L = A^{∗}.

The study of maximal subgroups of A^{∗} begins in [1] and [9] 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 those papers we essentially show that maximal subgroups arise naturally in A^{∗}, and finitely generated subnormal subgroups of A^{∗}, are central. This result is used to prove that a maximal subgroup of A^{∗} can not be finitely generated. The reader may consult [7], and the references thereof for more recent results on multiplicative subgroups of A^{∗}. The object of this note is to investigate the algebraic structure of D when the F -linear hull F [M] satisfies a polynomial identity. We also study the structure of M whenever M satisfies a group identity. To be more precise, let M be a maximal subgroup of A^{∗} such that the centre of M, Z(M) is algebraic over F. It is shown that if F [M] is a PIalgebra, then [D : F] < ∞. As consequences of this, some results of [1] that are proved there for the case n = 1 may be generalized for n > 1. For examples, it is shown if [A : F] = ∞, then A^{∗} contains no abelian maximal subgroup which is algebraic over F. Also, if M satisfies a multilinear polynomial identity, then [D : F] < ∞. In this direction, it is proved that if D is noncommutative and M/F^{∗} is locally finite, then we have either M is absolutely irreducible and D is locally finite dimensional or [D : F] < ∞ and M = K^{∗}, where K is a subfield of A. We then turn to the case where M satisfies a group identity. Let D be a noncommutative division ring with infinite centre F and n ≥ 1. Assume that M is a maximal subgroup of A^{∗} such that F [M] is algebraic over F. It is shown that if M satisfies a group identity, then we have either M = K^{∗}, where K is a field and [D : F] < ∞ or M is absolutely irreducible. In particular, if M is a noncommutative soluble maximal subgroup of A^{∗} such that F [M] is algebraic over F, then M is abelian-by-locally finite. For a noncommutative finite dimensional F -central division algebra D, assume that N is a subnormal subgroup of A^{∗} and M is a maximal subgroup of N. It is proved that if M satisfies a group identity, then M is abelian-by-finite.

2 Notations and conventions

Let D be a division ring with centre F and G be a subgroup of A^{∗} = GL_{n}(D). We denote by F [G] the F -linear hull of G, i.e., the F -algebra generated by elements of G over F. We also denote by D^{n} the space of row n-vectors over D. Then D^{n} is a D − G bimodule in the obvious manner. G is said to be an irreducible (reducible) subgroup of GL_{n}(D) whenever D^{n} is irreducible (reducible) as D − G bimodule. Considering the elements of D^{n} as column vectors, we may regard D^{n} as a G − D bimodule. It is easily shown that D^{n} is irreducible (reducible) as a G − D bimodule precisely when it has the property as D − G bimodule. 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, and < H, K > the group generated by H and K, where K is a subgroup of G. We shall say that H is abelian-by-finite if there is an abelian 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. By a dilatation matrix D_{ii}(d), d ∈ D^{∗} we understand a diagonal n×n matrix whose diagonal entries are all 1 except the (i, i)-th entry which is d. Some notations and conventions for linear groups and skew linear groups from [11] and [12] are frequently used throughout.

3 Polynomial identities on F [M]

Given a maximal subgroup of A^{∗}, this section essentially deals with conditions on M that imply either the commutativity of M or [D : F] < ∞. The main result is Theorem 5 which asserts that if F [M] is a PI-algebra and Z(M) is algebraic over F, then [D : F] < ∞. Using this, it is shown that if M satisfies a multilinear polynomial identity and Z(M) is algebraic over F, then [D : F] < ∞. Furthermore, it is proved that if either n = 1 and D is noncommutative or n > 1 and D is infinite, then there exists no maximal subgroup M of A^{∗} containing F^{∗} such that [M : F^{∗}] < ∞. For a noncommutative division ring with centre F, it is also shown that if M/F^{∗} is locally finite, then we have either M is absolutely irreducible and D is locally finite dimensional or [D : F] < ∞ and M = K^{∗}, where K is a subfield of A. We begin our material with

Proposition 1. Let D be a division ring with centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D). Then M is either irreducible or it contains an isomorphic copy of D^{∗}.

Proof. The case n = 1 follows from Proposition 1 of [1]. So, we may assume that n ≥ 2. Now, consider the F -algebra F [M]. Since M is maximal we conclude that either GL_{n}(D) = F [M]^{∗} or F [M]^{∗} = M. The first case implies that M_{n}(D) = F [M], i.e., M is absolutely irreducible and so it is irreducible. Thus, we may assume that F [M]^{∗} = M. If M is not irreducible, then D^{n} has a nontrivial submodule as D − F [M] bimodule. Thus, by 1.1.1 of [12], there exists an invertible n × n matrix P over D such that P MP^{−1} ⊂ Σ,

where Σ = \{( \begin{matrix}A B \\ 0 C \end{matrix} ) | A ∈ GL_{s}(D), C ∈ GL_{n−s}(D), B ∈ M_{s×(n−s)}(D)\}.

It is clear that P MP^{−1} is also a maximal subgroup of GL_{n}(D) and we have P MP^{−1} ⊆ Σ ⊂ GL_{n}(D). Therefore, P MP^{−1} = Σ and since Σ contains a copy of D^{∗} we obtain the result.

We shall need the following lemmas to prove our main theorem.

Lemma 2. Let D be a division ring of infinite dimension over its centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D). If F [M] is a PI-algebra, then F [M] is a prime ring.

Proof. By Proposition 1, we know that either M contains a copy of D^{∗} or M is irreducible. If the first case happens, then D is a P I-algebra. This implies, by Kaplansky’s Theorem (cf. [11]), that [D : F] < ∞ which is a contradiction. So we may assume that M is irreducible. Now, using 1.1.14 of [12, p. 9], we conclude that F [M] is a prime ring.

Lemma 3. Let D be a division ring of infinite dimension over its centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D). If F [M] is a PI-algebra, then Z(F [M]) is a field.

Proof. Set A = F [M]. Assume that X ∈ Z(A). We consider two cases: Case 1. If X ∈ GL_{n}(D) and X^{−1} ∉ A, then we have < M, X >⊆

C_{M_{n}(D)}(Z(A)) and by maximality of M we conclude that C_{M_{n}(D)}(Z(A)) = M_{n}(D). This means that Z(A) = F which is a contradiction. Thus X^{−1} ∈ Z(A).

Case 2. Assume that X ∉ GL_{n}(D). Then there exists an n × n matrix P in GL_{n}(D) such that the first row of P XP^{−1} is zero. We note that P MP^{−1} is a maximal subgroup of GL_{n}(D), F [P MP^{−1}] is a PI-algebra, and P XP^{−1} ∈ Z(F [P MP^{−1}]) = P Z(F [M])P^{−1}. Set J := XF [M]. Then J is an ideal in F [M] and P XP^{−1}(P F [M]P^{−1}) = P JP^{−1} = P XP^{−1}F [P MP^{−1}] is also an ideal in F [P MP^{−1}]. Put J′ = P JP^{−1} and B := {Y ∈ M_{n}(D)|Y J′ ⊂ J′}. It is clear that B is a ring and P AP^{−1} ⊆ B. On the other hand P XP^{−1} is a matrix whose first row is zero. Therefore, D_{11}(d) ∈ B for all d ∈ D. It is clear that P MP^{−1} ⊆ B^{∗}. Thus, B^{∗} = GL_{n}(D) or B^{∗} = P MP^{−1}. In the first case B = M_{n}(D) and so J′ is a right ideal and clearly J is a right ideal in M_{n}(D). In the second case D_{11}(d) ∈ P MP^{−1} for all d ∈ D. Therefore, F [P MP^{−1}] contains a copy of D^{∗} and so D is a P I-algebra which implies [D : F] < ∞ that is a contradiction. Similarly, there exists a matrix Q ∈ GL_{n}(D) such that QXQ^{−1} is a matrix whose first column is zero. Set C = {Y ∈ M_{n}(D)|(QJQ^{−1})Y ⊂ QJQ^{−1}}. As above, one may show that J is a left ideal in M_{n}(D). Consequently, J is an ideal in M_{n}(D). Since J ≠ 0 we obtain J = M_{n}(D). Therfore, F [M] = M_{n}(D) and so [D : F] < ∞ which is a contradiction. Thus, Z(A) is a field and the proof is complete.

Lemma 4. Let D be a division ring of infinite dimension over its centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D). If F [M] is a PI-algebra, then F [M] is simple and we have [F [M] : Z(F [M])] < ∞.

Proof. By Lemma 2-3, we conclude that F [M] is a prime ring whose centre is a field. Since F [M] is also a P I-algebra, by Theorem 7.5 of [3], we conclude that F [M] is simple. Finally, the rest of the result follows from Kaplansky’s Theorem.

Theorem 5. Let D be a division ring with centre Z(D) = F and n ≥ 1.

Assume that M is a maximal subgroup of GL_{n}(D) such that Z(M) is algebraic over F. If F [M] is a PI-algebra, then [D : F] < ∞.

Proof. We have M ⊆ F [M]^{∗}. By maximality of M we conclude that either F [M]^{∗} = GL_{n}(D) or F [M]^{∗} = M. The first case gives us F [M] = M_{n}(D). Now, use Kaplansky’s Theorem to obtain [D : F] < ∞. Therefore, we may assume that F [M]^{∗} = M. To complete the proof, we show that the assumption [D : F] = ∞ leads to a contradiction. Thus, suppose [D : F] = ∞. Then, by Lemma 4 and Artin-Wedderburn’s Theorem, we have F [M] ≅ M_{n_{1}}(D_{1}) for some positive integer n_{1} and division ring D_{1}, and so M ≅ GL_{n_{1}}(D_{1}). We claim that Z(M) = F^{∗}. For otherwise, since Z(M) is algebraic over F there exists a ∈ Z(M) \ F^{∗} such that [F (a) : F] < ∞. Now, we have F [M] ⊆ C_{M_{n}(D)}(F (a)) := A. If F [M] ≠ A, then, by the Centralizer Theorem, we conclude that A is a simple Artinian ring. Therefore, there exists a positive integer n_{2} and a division ring D_{2} such that A ≅ M_{n_{2}}(D_{2}) and so A^{∗} ≅ GL_{n_{2}}(D_{2}). We know that M ⊆ A^{∗}. If M = A^{∗}, then we clearly have F [M] = A which is a contradiction to our assumption. Thus, by maximality of M, one concludes that A^{∗} = GL_{n}(D), i.e., C_{M_{n}(D)}(F (a)) = A = M_{n}(D) which contradicts the fact that a ∈ Z(M) \ F. Therefore, we must have A = F [M] and so Z(F [M]) = Z(A) = F (a) by the Centralizer Theorem. Thus, by Lemma 4, this means that [F [M] : F] < ∞. Now, apply the Centralizer Theorem again to obtain M_{n}(D) ⊗ F (a)^{0p} ≅ M_{s}(F) ⊗ F [M] for some positive integer s. The last isomorphism implies that [D : F] < ∞ which contradicts our assumption. Therefore, we must have Z(M) = F^{∗} and the claim is established. Now, since F [M] is simple and F [M]^{∗} = M we obtain Z(F [M]) = F. Finally, consider the simple Artinian ring B := C_{M_{n}(D)}(F [M]). There exists a positive integer n_{3} and a division ring D_{3} such that B ≅ M_{n_{3}}(D_{3}). If F ≠ C_{M_{n}(D)}(F [M]), then F^{∗} ⊂ B^{∗} ≅ GL_{n_{3}}(D_{3}). Therefore. there is an X ∈ B^{∗} \ F^{∗} such that < M, X >⊆ C_{M_{n}(D)}(F (X)) and so M_{n}(D) = C_{M_{n}(D)}(F (X)) which is a contradiction to the fact that X ∉ F^{∗}. Thus, F = C_{M_{n}(D)}(F [M]). Now, using the Centralizer Theorem as above, one concludes that [D : F] < ∞ which is a contradiction and this completes the proof.

The next result generalizes Theorem 4.1 of [1].

Corollary 6. Let D be a division ring with centre Z(D) = F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D) such that Z(M) is algebraic over F. If M satisfies a multi-linear polynomial identity, then [D : F] < ∞.

The following result is also a generalization of Corollary 4.2 of [1].

Corollary 7. Let D be a division ring of infinite dimension over its centre Z(D) = F and n ≥ 1. Then GL_{n}(D) contains no abelian maximal subgroup which is algebraic over F.

Corollary 8. Let D be a division ring with 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 that there is a maximal subgroup M such that [M : F^{∗}] < ∞. Then, we have [F [M] : F] < ∞, i.e., F [M] is a P I-algebra and M is algebraic over F. Thus, by Theorem 5, we obtain [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 Theorem 5 of [2], we conclude that H ⊂ F^{∗}, i. e., GL_{n}(D) = F^{∗} which means that n = 1 and D = F that is a contradiction and so the result follows.

Corollary 9. Let R be a semisimple Artinian F -algebra with [Z(R) : F] < ∞. Assume that M is a maximal subgroup of R^{∗} such that Z(M) is algebraic over F. If F [M] satisfies a polynomial identity, then we have [R : F] < ∞.

Now, assume that M is a maximal subgroup of GL_{n}(D) containing F^{∗} such that M/F^{∗} is locally finite. The next result gives some information about the algebraic structure of M.

Theorem 10. Let D be a noncommutative division ring with centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D) containing F^{∗} such that M/F^{∗} is locally finite. Then we have either M is absolutely irreducible and D is locally finite dimensional or [D : F] < ∞ and M = K^{∗}, where K is a subfield of M_{n}(D).

Proof. Consider the F -algebra F [M]. If F [M]^{∗} = GL_{n}(D), then we have F [M] = M_{n}(D). Since M/F^{∗} is locally finite we conclude that D is locally finite dimenisional and so the result follows. Thus, we may assume throughout that F [M]^{∗} ≠ GL_{n}(D). By maximality of M we conclude that F [M]^{∗} = M. We observe that since M/F^{∗} is locally finite then F [M] is locally finite dimensional. Therefore, for each finite set of elemnts m_{1} ⋯ m_{r} ∈ M we have [F [m_{1} ⋯ m_{r}] : F] < ∞. We next claim that F [M] is a P I-algebra satisfying P (X, Y) = (XY − Y X)^{n}. To do this, let x, y ∈ F [M]. Then, there exist elements m_{i}, n_{j} ∈ M with 1 ≤ i ≤ t, 1 ≤ j ≤ s such that x = m_{1} + ⋯ m_{t} and y = n_{1} + ⋯ n_{s}, and we have [F [m_{1}, ⋯ m_{t}, n_{1}, ⋯ n_{s}] : F] < ∞. Therefore, A = F [m_{1}, ⋯ m_{t}, n_{1}, ⋯ n_{s}] is an Artinian P I -ring and so the Jacobson radical J = J(A) of A is nilpotent. Therefore, by 1.3.9 of [12], we conclude that J^{n} = 0. Now, by Wedderburn-Artin Theorem , there exist positive integers n_{1}, ⋯, n_{k} such that B := \frac{A}{J(A)} ≅ M_{n_{1}}(D_{1}) × ⋯ × M_{n_{k}}(D_{k}) as F -algebras , for some division rings D_{i}, where 1 ≤ i ≤ k. Now, we have B^{∗} = {m + J|m ∈ M ∩ A} ≅ GL_{n_{1}}(D_{1}) × ⋯ GL_{n_{k}}(D_{k}). Since B^{∗} is torsion over F we conclude that each D_{i} is torsion over F where F ⊆ Z(D_{i}). Thus D_{i} is torsion over Z(D_{i}) and by a result of Kaplansky (cf. [5]), we obtain D_{i} = Z(D_{i}) := F_{i}. Therefore, we obtain B ≅ M_{n_{1}}(F_{1}) × ⋯ × M_{n_{k}}(F_{k}). If there exists i such that n_{i} ≠ 1, then the dilatation n_{i} × n_{i} matrix D_{11}(f), where f ∈ F_{i}, is torsion over F. In particular, this implies that F^{∗} is a torsion group. Now, F^{∗} and M/F^{∗} are locally finite. Therefore, by a well-known result of group theory, M is locally finite. By Proposition 1, either M contains an isomorphic copy of D^{∗} or M is irreducible. The first case says that D^{∗} is locally finite. This implies, by a result of Jacobson (cf. [5]), that D = F which is a contradiction.

Thus, we may assume that M is irreducible and so it is completely reducible. Therefore, by 1.1.14 of [12], we conclude that F [M] is semisimple Artinian. Thus, as in the above case, there exist positive integers m_{1}, ⋯, m_{k} such that F [M] ≅ M_{m_{1}}(D_{1})×⋯×M_{m_{k}}(D_{k}), for some division rings D_{i}, where 1 ≤ i ≤ k. Now, we have F [M]^{∗} = M ≅ GL_{m_{1}}(D_{1}) × ⋯ GL_{m_{k}}(D_{k}). Since M is locally finite, as above, we conclude that D_{i} = F_{i} = Z(D_{i}). Therefore, F [M] ≅ M_{m_{1}}(F_{1})×⋯×M_{m_{k}}(F_{k}). This means that F [M] satisfies a polynomial identity and so by Theorem 5 we conclude that [D : F] < ∞. Since F^{∗} is torsion we obtain D = F which is a contradiction. Therefore, in the decomposition of B for all i we have n_{i} = 1, i.e., B ≅ F_{1} × ⋯ × F_{k}. This implies that for each x, y ∈ A we have xy − yx ∈ J and therefore (xy − yx)^{n} = 0 since J^{n} = 0. Thus, F [M] satisfies P (X, Y) = (XY − Y X)^{n}. Therefore, by Theorem 5, we obtain [D : F] < ∞. Since M is irreducible and [D : F] < ∞, by 1.1.12 of [12], we conclude that F [M] is simple Artinian, i.e., F [M] ≅ M_{t}(D_{1}) for some t and division ring D_{1}. Now, we may use a similar argument as above to deduce that D_{1} = F_{1} = Z(D_{1}). If t ≠ 1, we may conclude that F is torsion and since [D : F] < ∞ we obtain D = F which is a contradiction. Therefore, t = 1 and we have F [M]^{∗} = M ≅ F_{1}^{∗} which completes the proof.

Corollary 11. Let D be a division ring with centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D). If M is locally finite, then we have D = F.

Proof. If F^{∗} ⊈ M, then SL_{n}(D) ⊂ M. Since M is torsion, we conclude that SL_{n}(D) is torsion. Thus, by Corollary 2 of [8], we conclude that D = F. So, we may assume that F^{∗} ⊆ M. This implies that F^{∗} is torsion and so M/F^{∗} is torsion. Now, Theorem 10 implies that D is algebraic over F. Therefore, D is algebraic over its prime subfield and so by a result of Jacobson we conclude that D = F.

4 Group identities on M

Let D be a noncommutative division ring with infinite centre F and n be a positive integer. Given a maximal subgroup M of A^{∗} such that F [M] is algebraic over F, the key result of this section is to show that if M satisfies a group identity, then we have either M = K^{∗}, where K is a field and [D : F] < ∞ or M is absolutely irreducible. Particularly, if M is a noncommutative soluble maximal subgroup of A^{∗} such that F [M] is algebraic over F, it is proved that M is abelian-by-locally finite. When D is of finite dimension over F, assume that N is a subnormal subgroup of A^{∗} and M is a maximal subgroup of N. It is proved that if M satisfies a group identity, then M is abelian-by-finite.

Theorem 1. Let D be a noncommutative division ring with infinite centre F and n ≥ 1. Assume that M is a maximal subgroup of GL_{n}(D) such that F [M] is algebraic over F. If M satisfies a group identity, then we have either M = K^{∗}, where K is a field and [D : F] < ∞ or M is absolutely irreducible.

Proof. If F^{∗} ⊈ M, then M is normal in GL_{n}(D). This means that F [M] is normal in M_{n}(D). Thus, by a result of [10], we conclude that either F [M] ⊂ F or F [M] = M_{n}(D). If the first case occurs, then we have M ⊆ F^{∗} which is a contradiction to the fact that M is a maximal subgroup of GL_{n}(D). The second case says that M is absolutely irreducible. So, let F^{∗} ⊂ M, we have two cases to consider. If F [M]^{∗} = GL_{n}(D), then M is absolutely irreducible. So, assume that F [M]^{∗} = M. By Proposition 3.1, we know that either M contains an isomorphic copy of D^{∗} or M is irreducible. If the first case occurs, then D^{∗} satisfies a group identity. But since F is infinite this is impossible by a result of Amitsur (cf. [11]). Therefore, M must be irreducible. Thus, by 1.1.14 of [12, p. 9], we conclude that F [M] is a prime ring. Now, by Theorem 5.5 of [6], we have F [M] ≅ M_{r}(K) for some positive integer r, where K is an extension field of F. This shows that F [M] satisfies a polynomial identity. Now, by Theorem 3.5, we conclude that [D : F] < ∞. Finally, we have F [M]^{∗} = M ≅ GL_{r}(K). If r ≠ 1, then M contains a copy of SL_{r}(K), i.e., M contains a free subgroup (cf. [12]). This contradicts the fact that M satisfies a group identity. Therefore, we have r = 1 and this completes the proof.

Corollary 2. Let D be a division ring with infinite centre F and n ≥ 1. Assume that M is a noncommutative maximal subgroup of GL_{n}(D) such that F [M] is algebraic over F. If M satisfies a group identity, then M_{n}(D) is algebraic over F.

Corollary 3. Let D be a division ring with infinite centre F and n ≥ 1. Assume that M is a noncommutative nilpotent maximal subgroup of GL_{n}(D) such that F [M] is algebraic over F. Then M is centre-by-locally finite. Therefore, M/F^{∗} is locally finite and so D is locally finite dimensional.

proof. M satisfies a group identity since M is nilpotent. Thus, by Theorem 1, we conclude that M is absolutely irreducible. Now, by a result of [12, p. 213], this implies that M is centre-by-locally finite and so the proof is complete.

Corollary 4. Let D be a division ring with infinite centre F and n ≥ 1. Assume that M is a noncommutative soluble maximal subgroup of GL_{n}(D) such that F [M] is algebraic over F. Then M is abelian-by-locally finite.

proof. M satisfies a group identity since M is soluble. By Theorem 1, we conclude that M is absolutely irreducible. Now, by a result of [13], this implies that M is abelian-by-locally finite and so the proof is complete.

To prove our final result, we need the following

Lemma 5. Let D be a division algebra of finite dimension over its centre F. Assume that G is a subgroup of D^{∗}. If G satisfies a group identity, then G is abelian-by-finite.

proof. Assume that G satisfies a group identity. Since [D : F] < ∞ we conclude that G is a linear group. By a theorem of Platonov (cf. [14. p. 149]), G is soluble-by-finite, i.e., there exists a soluble normal subgroup N of G such that G/N is finite. Since [D : F] < ∞ we conclude that F [N] is a division ring, and therefore F [N] is semisimple. Thus, N as a linear group over F is completely reducible. Therefore, N is a completely reducible soluble linear group. So, by a theorem of [4, p. 111], N is abelian-by-finite and consequently, G is abelian-by-finite which completes the proof.

Theorem 6. Let D be a noncommutative division algebra of finite dimension over its centre F and n ≥ 1. Assume that N is a subnormal subgroup of GL_{n}(D) and M is a maximal subgroup of N. If M satisfies a group identity, then M is abelian-by-finite.

proof. The case n = 1 follows from Lemma 4. So, we may assume that n ≥ 2 and N is a subnormal subgroup of GL_{n}(D). By Theorem 11 of [8], we have either N ⊂ F^{∗} or SL_{n}(D) ⊂ N, i.e., N is normal in GL_{n}(D). We now claim that M is irreducible. For otherwise assume that M is reducible. Therefore, D^{n} has a nontrivial submodule as D −F [M] bimodule. Thus, there exists an invertible n × n matrix P over D such that P MP^{−1} ⊂ Σ, where

Σ = \{( \begin{matrix}A B \\ 0 C \end{matrix} ) | A ∈ GL_{s}(D), C ∈ GL_{n−s}(D), B ∈ M_{s×(n−s)}(D)\} ∩ N. It is

clear that P MP^{−1} is also a maximal subgroup of N. Set

T = ( \begin{matrix}SL_{s}(D) & o \\ o & SL_{n−s}(D) \end{matrix} ) .

Then, we have T ⊂ SL_{n}(D) and so T ⊂ Σ. Now, we clearly have P MP^{−1} ⊆ Σ ⊆ N. By maximality of P MP^{−1} we conclude that either P MP^{−1} = Σ or N = Σ. If the second case occurs, then I + e_{n1} ∈ SL_{n}(D) whereas I + e_{n1} ∉ Σ = N and this contradicts the fact that SL_{n}(D) ⊆ N. Therefore, P MP^{−1} = Σ which implies that Σ satisfies a group identity, and so T ⊂ Σ satisfies a group identity. Now, by 4.5.1 of [12], we conclude that D = F which is a contradiction. Thus, M is irreducible as claimed. Therefore, by 1.1.12 of [12], F [M] is simple Artinian. So, there exists a positive integer t and a division ring D_{1} such that F [M] ≅ M_{t}(D_{1}) as F -algebras. Thus, F [M]^{∗} ≅ GL_{t}(D_{1}) and (F [M]^{∗})′ ≅ SL_{t}(D_{1}). Since N is normal in GL_{n}(D) we have (F [M]^{∗})′ ⊂ N. If F [M]^{∗} ∩ N = M, then we have SL_{t}(D_{1}) ≅ (F [M]^{∗})′ ⊆ F [M]^{∗} ∩ N = M. Since M satisfies a group identity, by 4.5.1 of [12] again, we conclude that either D_{1} is a locally finite field or t = 1 and D_{1} = Z(D_{1}). In the first case F is also a locally finite field and since [D : F] < ∞ we conclude that D = F which is a contradiction. The second case implies that M is abelian and the result follows. Finally, if N ∩ F [M]^{∗} = N, then N ⊆ F [M]^{∗}. Since N is normal in GL_{n}(D) we conclude that F [M] = M_{n}(D), i.e., M is absolutely irreducible. By a result of [14, p. 149], M is soluble-by-finite, i.e., there is a soluble normal subgroup N_{1} such that M/N_{1} is finite. Now, by 1.2.12 of [12], F [N_{1}] is semisimple Artinian. Therefore, N_{1} is a soluble completely reducible linear group. By a theorem of [4, p. 111], N_{1} is abelian-by-finite and therefore M is abelian-by-finite which completes the proof.

The authors thank the Research Council of Sharif University of Technology for support. The second author is also indebted to the Abdus Salam International Centre for Theoretical Physics for its support during his stay in 2002.

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