Irreducible Subgroups of GL_{1}(D) Satisfying Group Identities

R. Ebrahimian, D. Kiani, M. Mahdavi-Hezavehi ebrahimian@mehr.sharif.edu 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 finite dimensional F -central division algebra, and G be an irreducible subgroup of D^{∗} := GL_{1}(D). Here we investigate the structure of D under various group identities on G. In particular, it is shown that when [D : F] = p^{2}, p a prime, then D is cyclic if and only if D^{∗} contains a nonabelian subgroup satisfying a group identity.

1 Introduction

Let D be an F -central division algebra of degree n. We define D to be quasicrossed product (QCP for short) if it contains a maximal subfield L with a chain of fields F ⫋ K ⊆ L such that K/F is Galois and L/K is abelian Galois. If L = K, then D is crossed product. Also, D is said to be soluble QCP if Gal(K/F) is soluble. We also recall that a subgroup G of D^{∗} is irreducible if the F -linear hull of G, F [G] = D. Irreducible soluble subgroups of the multiplicative group of a divison ring were first studied by Suprunenko in [11]. Assume that D is a noncommutative finite dimensional F -central division algebra. It is clear that D^{∗} is an improper irreducible subgroup of D^{∗}. More generally, if N is a noncentral normal subgroup of D^{∗}, then, by Cartan-Brauer-Hua Theorem, N is irreducible. But N cannot satisfy a group identity since it is known that N contains a noncyclic free subgroup [3] . So, we are interested in proper irreducible subgroups of D^{∗} that are not normal. To construct such subgroups of D^{∗}, take a basis {g_{i}}_{1≤i≤n} of D/F and consider the subgroup G = ⟨g_{i}, 1 ≤ i ≤ n⟩. Since G is finitely generated it is not normal in D^{∗} (cf. [6]) and we clearly have F [G] = D. Therefore, we can always find nonnormal proper irreducible subgroups in D^{∗}. We shall see later on that if a nonnormal proper irreducible subgroup G of D^{∗} satisfies a group identity, then we may be able to determine some information about the structure of D. For example, it is shown that if D^{∗} contains an irreducible subgroup satisfying a group identity, then D is QCP. In particular, when [D : F] = p^{2}, p a prime, any nonabelian subgroup of D^{∗} is irreducible. In this case, it is proved that D is cyclic if and only if D^{∗} contains a nonabelian subgroup satisfying a group identity.

2 Notations and conventions

Let D be a division ring with center F and G be a subgroup of 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 shall say that G is irreducible if D = F [G]. For any group G we denote its center 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. H is called centerby-finite if H/Z(H) is finite. Let S be a subset of D, then the centralizer of S in D is denoted by C_{D}(S). For results related to central simple algebras see [8].

3 Irreducible Subgroups of D^{∗}

This section deals with a few results on division algebras whose multiplicative groups contain certain subgroups. Using these results we eventually prove our main theorem which asserts that if D^{∗} contains an irreducible soluble subgroup, then D is QCP. We then examine the structure of division algebras whose multiplicative groups contain irreducible abelian-by-finite subgroups. Using the results, it is shown that if D has a nonzero characteristic and D^{∗} contains an irreducible subgroup satisfying a group identity, then D is QCP. Finally, a criterion is given for a division algebra D of prime degree to be cyclic. To be more precise, it is shown that D is cyclic if and only if D^{∗} contains a nonabelian subgroup satisfying a group identity. We begin our study with the following lemma, for a proof see [2].

Lemma A. Let D be a finite dimensional F -central division algebra. If D is soluble crossed product, then D^{∗} contains an irreducible soluble subgroup.

We shall also need the following easy lemma from group theory.

Lemma 3.1. Let G be a group in which Z(G) is a maximal abelian normal subgroup. Then Z(G/Z(G)) is trivial.

Proof. Assume on the contrary that x ∈ G\Z(G) with Z(G)x ∈ Z(G/Z(G)). Using the fact that for any element y ∈ G we have Z(G)xy = Z(G)yx, we obtain yxy^{−1}x^{−1} ∈ Z(G). Thus, for any y ∈ G we have yxy^{−1} ∈ ⟨Z(G), x⟩. Hence, ⟨Z(G), x⟩ is an abelian normal subgroup of G properly containing Z(G). This contradiction completes the proof. □

We are now in a position to prove the following:

Theorem 3.2. Let D be a noncommutative finite dimensional F -central division algebra. Assume that D^{∗} contains an irreducible soluble subgroup. Then there exists an irreducible soluble subgroup S and a noncentral abelian normal subgroup A of S such that S/A is finite.

Proof. By Lemma 3 of [5], we know that any soluble subgroup S of D^{∗} is abelian-by-finite, i.e., there is an abelian normal subgroup A in S of finite index. We may take in our considerations A maximal when we deal with irreducible subgroups of D^{∗} which are clearly nonabelian. Hence, to prove the theorem we must show that there exists an irreducible soluble subgroup S that is not center-by-finite. On the contrary, assume that the set Σ of all irreducible soluble subgroups of D^{∗} are center-by-finite. Since D is of finite dimension over F we may view D^{∗} as a linear group in GL_{n}(F), where n = [D : F], and consequently each element of Σ is a linear group. Thus, by Huppert-Zassenhaus Theorem (cf. [1, p. 104]), the soluble length l of each element H ∈ Σ is at most 2n. Using this fact and Zorn’s Lemma, we conclude that every element of Σ is contained in a maximal element of Σ. Let G be a maximal element of Σ. It is clear that F^{∗} ⊆ G. By our assumption F^{∗} is a maximal abelian normal subgroup in G of finite index. We claim that if N is normal in G, then Z(F [N]) = F. Using the fact that G normalizes N we conclude that G normalizes F [N]. Obviously Z(F [N])^{∗}G is an irreducible soluble subgroup of D^{∗}. Also it is easily seen that Z(F [N])^{∗} is a normal abelian subgroup of Z(F [N])^{∗}G of finite index. Therefore, by our assumption we have Z(F [N]) = F. Now, G/F^{∗} as a soluble group must contain a nontrivial abelian normal subgroup. Let N/F^{∗} be a maximal abelian normal subgroup of G/F^{∗}. We claim that C_{G}(N) = F^{∗}. To see this, put M = NC_{G}(N). If M = N, then C_{G}(N) ⊆ N and hence C_{G}(N) = Z(N) = F^{∗}. Now suppose that M ≠ N. Considering the fact that M is normal in G and that N/F^{∗} is a maximal abelian normal subgroup of G/F^{∗}, we conclude that M/F^{∗} is not abelian. The group M/N is a nontrivial soluble subgroup of G/N. Thus, by looking at the derived series of M/N we can find a nontrivial abelian normal subgroup T/N of M/N such that T is normal in G. By the choice of N/F^{∗}, we obtain that T/F^{∗} is not abelian. Therefore, there exist z, t ∈ T such that ztz^{−1}t^{−1} ∉ F^{∗}. Using the definition of M, we can find n, m ∈ N and p, q ∈ C_{G}(N) such that z = np, t = mq. Therefore, we have ztz^{−1}t^{−1} = npmqp^{−1}n^{−1}q^{−1}m^{−1} = nmn^{−1}m^{−1}pqp^{−1}q^{−1}. On the other hand we have ztz^{−1}t^{−1} ∈ N since T/N is abelian. Thus, pqp^{−1}q^{−1} ∈ N and hence pqp^{−1}q^{−1} ∈ N ∩C_{G}(N) = Z(N) = F^{∗}. Now, using the fact that N/F^{∗} is abelian we obtain nmn^{−1}m^{−1} ∈ F^{∗} and so ztz^{−1}t^{−1} ∈ F^{∗}. This contradiction shows that if N/F^{∗} is a maximal abelian normal subgroup of G/F^{∗}, then C_{G}(N) = F^{∗} which establishes the claim. Now, set D_{1} = F [N] and, by the above observation, we note that Z(D_{1}) = F. Since G normalizes D_{1} we see that for any g ∈ G we may define a natural homomorphism f_{g} : D_{1} ⟶ D_{1}, given by f_{g}(x) = gxg^{−1}, for any x ∈ D_{1}. Hence, by Skolem-Noether Theorem there is an element a_{g} ∈ D_{1}^{∗} such that f_{g} = f_{a_{g}}. If u, v ∈ D_{1} satisfy f_{u} = f_{v}, then for any x ∈ D_{1} we have uxu^{−1} = vxv^{−1}. Therefore, u^{−1}v ∈ Z(D_{1}) = F, which shows that u, v are equal modulo F^{∗}, i.e., F^{∗}u = F^{∗}v. Now, for any x ∈ D_{1} we have gxg^{−1} = a_{g}xa_{g}^{−1}, and hence b_{g} = a_{g}^{−1}g ∈ C_{D}(D_{1}). The fact that b_{g} commutes with a_{g} implies that a_{g}, g, and b_{g} pairwise commute. Set A = ⋃_{g∈G} F^{∗}a_{g} and B = ⋃_{g∈G} F^{∗}b_{g}. We claim that both A, B are groups. To see this, it is enough to show that for any g, h ∈ G we have F^{∗}a_{g^{−1}} = F^{∗}a_{g}^{−1}, F^{∗}a_{h}a_{g} = F^{∗}a_{hg}, F^{∗}b_{g^{−1}} = F^{∗}b_{g}^{−1}, and F^{∗}b_{h}b_{g} = F^{∗}b_{hg}. We have b_{g} = a_{g}^{−1}g. For any x ∈ D_{1}, f_{a_{g^{−1}}}(x) = f_{g^{−1}}(x) = g^{−1}xg = (a_{g}b_{g})^{−1}x(a_{g}b_{g}) = a_{g}^{−1}xa_{g} = f_{a_{g}^{−1}}(x). Therefore, F^{∗}a_{g}^{−1} = F^{∗}a_{g^{−1}}. Also we have f_{a_{hg}}(x) = f_{hg}(x). Hence f_{a_{hg}}(x) = hgxg^{−1}h^{−1} = ha_{g}xa_{g}^{−1}h^{−1} = a_{h}a_{g}xa_{g}^{−1}a_{h}^{−1} = (a_{h}a_{g})x(a_{h}a_{g})^{−1} = f_{a_{h}a_{g}}(x). Therefore, F^{∗}a_{hg} = F^{∗}a_{h}a_{g} which shows that A is a group. Next considering the fact that a_{g} ∈ D_{1} and b_{h} ∈ C_{D}(D_{1}) we obtain b_{h}b_{g} = b_{h}a_{g}^{−1}g = a_{g}^{−1}b_{h}g = a_{g}^{−1}a_{h}^{−1}hg = (a_{h}a_{g})^{−1}hg. Thus, since A is a group we conclude that F^{∗}b_{h}b_{g} = F^{∗}(a_{h}a_{g})^{−1}hg = F^{∗}a_{hg}^{−1}hg = F^{∗}b_{hg}, F^{∗}b_{g}^{−1} = F^{∗}a_{g}g^{−1} = F^{∗}a_{g^{−1}}^{−1}g^{−1} = F^{∗}b_{g^{−1}}. Therefore, B is also a group. We claim that B is a soluble group that is normalized by G. To see this, consider the epimorphism θ : G ⟶ B/F^{∗} given by θ(g) = F^{∗}b_{g} for all g ∈ G. Hence B/F^{∗} as a homomorphic image of a soluble group is soluble, and so B is also soluble. Furthermore, since a_{g} ∈ D_{1} and B ⊆ C_{D}(D_{1}) for any g ∈ G we have gBg^{−1} = a_{g}b_{g}Bb_{g}^{−1}a_{g}^{−1} = a_{g}Ba_{g}^{−1} = B, which establishes our claim. Therefore, BG is an irreducible soluble subgroup of D^{∗}. By maximality of G we conclude that B ⊆ G. Now, we have B ⊆ C_{D}(D_{1}) ∩ G ⊆ C_{G}(D_{1}) = C_{G}(N) = F^{∗}. Since g = a_{g}b_{g} we obtain G ⊆ D_{1} and hence D_{1} = D, i.e., N is irreducible. By our assumption we conclude that F^{∗} is a maximal abelian normal subgroup of N. Therefore, by Lemma 3.1, Z(N/F^{∗}) is trivial, which is a contradiction and so the result follows. □

To prove our main theorems, we shall need the following results, for a proof see [2].

Lemma B. Let D be a finite dimensional F -central division algebra. Suppose that K is a subfield of D containing F. If G is an irreducible subgroup of D^{∗} such that K^{∗} ◁ G, then K/F is Galois and G/C_{G}(K^{∗}) ≃ Gal(K/F).

Lemma C. Let D be a finite dimensional F -central division algebra and let G be an irreducible subgroup of D^{∗}. If K is a subfield of D containing F such that [G : C_{G}(K^{∗})] = [K : F], then C_{D}(K) = F [C_{G}(K^{∗})].

Theorem D. A noncommutative finite dimensional F -central division algebra D is abelian crossed product if and only if there exist an irreducible subgroup G of D^{∗} and an abelian normal subgroup A of G such that G/A is abelian. Equivalently D^{∗} contains an irreducible metabelian subgroup.

We are now prepared to prove one of our main results of this section in the following form:

Theorem 3.3. Let D be a noncommutative finite dimensional F -central division algebra. If D^{∗} contains an irreducible soluble subgroup, then D is soluble QCP.

Proof. As in Theorem 3.2, assume that Σ is the set of all irreducible soluble subgroups of D^{∗}. It is clear that every element of Σ is contained in a maximal element. Suppose that G is a maximal element of Σ. It is easily checked that F^{∗}G ∈ Σ, and by maximality of G we have F^{∗}G = G. Hence F^{∗} ⊆ G. Now, by Theorem 3.2, there exists an irreducible soluble subgroup S of D^{∗} containing a noncentral abelian normal subgroup B of finite index. Put T = F (B)^{∗}S. Since S normalizes B one can easily show that T is an irreducible soluble subgroup of D^{∗}. Therefore, T is contained in a maximal element G of Σ. Now, by Lemma 3 of [5], there exists a maximal abelian normal subgroup A in G of finite index. It is clear that [F (A)^{∗} : F^{∗}] is infinite since otherwise F would be a finite field (cf. [4, p. 213]), hence by Wedderburn’s Theorem, we conclude that D is a field, a contradiction. Therefore, G/F^{∗} is infinite. Since A is a maximal abelian normal subgroup of finite index in G we conclude that F^{∗} ≠ A. Set K = F (A). One may easily show that G ⊆ N_{D^{∗}}(K^{∗}) and that K^{∗}G is an irreducible soluble subgroup of D^{∗}. Therefore, by maximality of G we have K^{∗} ⊆ G. Thus, A ⊆ K^{∗} ◁ G, which shows that A = K^{∗}. Now, by Lemma B, K/F is Galois and G/C_{G}(K^{∗}) ≃ Gal(K/F). Therefore, K/F is a soluble Galois extension. If C_{D}(K) = K, one may easily conclude that K is a maximal subfield of D, hence L = K and so D is a soluble crossed product division algebra. So, we may assume that C_{D}(K) ≠ K. By Lemma C, we conclude that F [C_{G}(K^{∗})] = C_{D}(K). Put D_{1} = C_{D}(K). We have F [C_{G}(K^{∗})] = D_{1}, hence D_{1} contains an irreducible soluble subgroup. On the other hand by Centralizer Theorem we have Z(D_{1}) = K. Now, by Lemma 11 of [11] D_{1}^{∗} contains an irreducible metabelian subgroup. Therefore, by Lemma D, we conclude that D_{1} is an abelian crossed product division algebra. Thus, there exists a maximal subfield L of D_{1} such that L/K is abelian Galois. Now, K/F is Galois and Gal(K/F) is soluble. To complete the proof, it is enough to show that L is a maximal subfield of D. To see this, we have C_{D}(L) ⊆ C_{D}(K) = D_{1}, and hence C_{D}(L) ⊆ C_{D_{1}}(L) = L, which completes the proof of the theorem. □

As a special case of Theorem 3.3, we obtain the following corollary, which is the main result of [7].

Corollary 3.4. Let D be an F -central division algebra of prime degree p. Then D is cyclic if and only if D^{∗} contains a nonabelian soluble subgroup.

Proof. the ”only if” part is clear with Lemma A. By Theorem 3.3, D is soluble QCP. Thus, there exists a noncentral subfield K such that K/F is Galois and Gal(K/F) is soluble. Now, for dimensional reasons, the proof is complete. □

The next result establishes the structure of division algebras whose multiplicative subgroups contain irreducible abelian-by-finite subgroups.

Proposition 3.5. Let D be a noncommutative finite dimensional F - central division algebra. Assume that D^{∗} contains an irreducible abelian-by-finite subgroup G. If CharF = p > 0, then D is QCP.

Proof. By our assumption, there exists an abelian normal subgroup A in G of finite index. Since G is nonabelian we may take A maximal. We now divide the proof into two cases:

Case 1: Suppose that A is noncentral in G. Therefore, A is also noncentral in D^{∗}. Set H = F (A)^{∗}G. Since G normalizes A one can easily see that H is an irreducible abelian-by-finite subgroup of D^{∗}. Let K = F (A). It is clear that H/K^{∗} is finite and K ≠ F. By Lemma B, we conclude that K/F is Galois and H/C_{H}(K^{∗}) ≃ Gal(K/F). If C_{D}(K) = K, we conclude that K is a maximal subfield of D and L = K, hence D is QCP. So, we may assume that C_{D}(K) ≠ K. Now, using Lemma C, we obtain F [C_{H}(K^{∗})] = C_{D}(K). Set D_{1} = C_{D}(K). We have that C_{H}(K^{∗})/K^{∗} is a finite group. On the other hand by the Centralizer Theorem, we have Z(D_{1}) = K^{∗}. Hence Z(C_{H}(K^{∗})) = K^{∗}. Now, by group theory, we know that the derived group C_{H}(K^{∗})′ is a finite group (cf. [9, p. 443]). On the other hand CharF = p > 0. Now, by a theorem of [4, p. 204], we conclude that C_{H}(K^{∗})′ is cyclic. Therefore, D_{1}^{∗} contains an irreducible metabelian subgroup. Hence, by Theorem D, we conclude that D_{1} is an abelian crossed product division algebra. Thus, there exists a maximal subfield L of D_{1} such that L/K is Galois and Gal(L/K) is abelian. Now, we have C_{D}(L) ⊆ C_{D}(K) = D_{1}, and so C_{D}(L) ⊆ C_{D_{1}}(L) = L, which shows that L is a maximal subfield of D. Therefore, D is QCP.

Case 2: Assume that A = Z(G). Since G is irreducible we conclude that Z(G) is central in D^{∗}. Because A is maximal in G we conclude that every abelian normal subgroup of G is contained in F^{∗}. Now, we know that G/Z(G) is finite and hence the derived group G′ is a finite group (cf. [9, p. 443]). On the other hand, charF = p > 0, by a theorem of [4, p. 204], we conclude that G′ is cyclic. Therefore, G is an irreducible soluble subgroup of D^{∗} and by Theorem 3.3, we obtain that D is QCP, which completes the proof. □

Corollary 3.6. Let D be a noncommutative finite dimensional division algebra of nonzero characteristic. If D^{∗} contains an irreducible subgroup satisfying a group identity, then D is QCP.

Proof. Suppose that G is an irreducible subgroup of D^{∗} which satisfies a group identity. Let {g_{1}, …, g_{n}} ⊆ G be a basis of the vector space D over the field Z(D) and set G_{1} = ⟨g_{1} … g_{n}⟩. Then G_{1} is an irreducible subgroup of D^{∗}, which is finitely generated and satisfies a group identity. By Tits’ Alternative (cf. [12, p. 150]), we conclude that G_{1} is soluble-by-finite. Now, Lemma 3 of [5] implies that G_{1} is abelian-by-finite. Therefore, by Theorem 3.5, the result follows. □

Finally, using above results, a criterion is given for cyclicity of a division algebra of prime index in terms of subgroups of D^{∗}. This criterion includes those given in [7] and provides us with various conditions on D to be cyclic.

Theorem 3.7. Let D be a finite dimensional F -central division algebra of prime degree p. Then D is cyclic if and only if D^{∗} contains a nonabelian subgroup satisfying a group identity.

Proof. One way is clear from Lemma A. On the other hand, assume that D^{∗} contains a nonabelian subgroup G satisfying a group identity. We may view G as a linear group over F. By a result of Platonov (cf. [12, p. 149]), we may conclude that there is a soluble normal subgroup S of finite index in G. If S is nonabelian, the result follows from Corollary 3.4. So we may assume that S is abelian. Set T = F^{∗}G and A = F^{∗}S. It is easily seen that A is a normal abelian subgroup of finite index in T. We may take A maximal in T. Two cases may occur:

Case 1: If A = F^{∗}, then T/F^{∗} is a nontrivial finite subgroup of D^{∗}/F^{∗} and the result follows from the Theorem of [7].

Case 2: If A ≠ F^{∗}, then K = F [A] is a maximal subfield of D. Since A is normal in T, by Lemma B, we conclude that K/F is Galois and it is clear that K/F is cyclic which completes the proof.

□

The above result provides us with various cyclicity conditions on subgroups of D^{∗}. There are some conditions on subgroups of D^{∗} that imply those subgroups satisfy a group identity. We collect some well-known ones in the following:

Corollary 3.8. Let D be an F -centeral division algebra of prime degree p. Then D is cyclic if either of the following conditions holds:

(a) D^{∗} contains a nonabelian soluble subgroup.

(b) D^{∗} contains a nonabelian subgroup of a bounded exponent.

(c) D^{∗} contains a nonabelian finite subgroup.

The authors thank the referee for his constructive comments. They are also indebted to the Research Council of Sharif University of Technology for support.

References

[1] D. Dixon, The structure of linear groups, Van Nostrand, London, 1971.

[2] R. Ebrahimian, D. Kiani, M. Mahdavi-Hezavehi, Supersoluble crossed product criterion for division algebras , Preprint.

[3] J. Z. Goncalves, Free groups in subnormal subgroups and the residual nilpotence of the group of units of group rings, Canad. Math. Bull. 27(1984), 365-370.

[4] T. Y. Lam, A first course in noncommutative rings, Second edition,GTM, No. 131, Springer-Verlag, (2001).

[5] M. Mahdavi-Hezavehi, Free subgroups in maximal subgroups of GL_{1}(D), J. Alg. 241 (2001) 720-730.

[6] M. Mahdavi-Hezavehi, M. G. Mahmudi, S. Yasamin, Finitely generated subnormal subgroups of GL_{n}(D) are central, Journal of Algebra 225, 517-521 (2000).

[7] M. Mahdavi-Hezavehi and J. P. Tignol, Cyclicity conditions for division algebras of prime degree, Proc. Amer. Math. Soc. 131(2003), 3673-3676.

[8] L. Rowen, Ring theory, Volume II, Academic Press, INC, (1988).

[9] W. R. Scott, Group theory, Dover, New York 1987.

[10] M. Shirvani and B. A. F. Wehrfritz, Skew linear group, LMS Lecture Note Series, No. 118, 1986.

[11] D. A. Suprunenko, On solvable subgroups of multiplicative groups of a field, English Transl., Amer. Math. Soc. Transl., (2)46(1965), p. 153-161.

[12] A. E. Zalesskii, Linear group, in “Algebra IV”, part II( A. I. Kostrikin I. R. Shafarevich, Eds.), Encyclopedia of Mathematics and Science. Springer Verlag, Berlin/New York, 1993.