CROSSED PRODUCT CONDITIONS FOR CENTRAL SIMPLE ALGEBRAS IN TERMS OF IRREDUCIBLE

SUBGROUPS

T. KESHAVARZIPOUR & M. MAHDAVI-HEZAVEHI

Abstract. Let M_{m}(D) be a finite dimensional F -central simple algebra. It is shown that M_{m}(D) is a crossed product over a maximal subfield if and only if GL_{m}(D) has an irreducible subgroup G containing a normal abelian subgroup A such that C_{G}(A) = A and F [A] contains no zero divisor. Various other crossed product conditions on subgroups of D^{∗} are also investigated. In particular, it is shown that if D^{∗} contains either an irreducible finite subgroup or an irreducible soluble-by-finite subgroup that contains no element of order dividing deg(D)^{2}, then D is a crossed product over a maximal subfield.

1. Introduction

Let D be an F -central division algebra of index n. Assume that the algebra M_{m}(D) is generated by a subgroup G of GL_{m}(D) over a subring S which is normalized by G (written M_{m}(D) = S[G]). For H = G∩S⊴ G, we say that M_{m}(D) is a crossed product of S by G/H if M_{m}(D) = ⊕_{t∈T} tS, for some transversal T of H in G, and denote it by M_{m}(D) = S ∗ G/H. The classical crossed product simple algebra corresponds to the case where S is a maximal subfield of M_{m}(D) which is Galois over F. Also, we call a subgroup G of GL_{m}(D) irreducible if F [G] = M_{m}(D). In [4] and [10], the authors ask whether a division algebra generated by a soluble-by-finite irreducible subgroup G is necessarily a crossed product over a maximal subfield. For some special cases where D is of a prime power degree and G is soluble or finite, the answer to the above mentioned question is shown to be positive. In [8], this problem for arbitrary degree is investigated, and it is proved that when G is irreducible soluble-by-finite, then D is a crossed product over a (not necessarily maximal) subfield K. But for the case where G is torsion free, or metabelian, or the degree of D has the property that every finite group of order deg(D)^{2} is nilpotent, then K may be shown to be a maximal subfield. In this paper, we present a necessary and sufficient condition for which a central simple algebra is a crossed product over a maximal subfield in terms of its irreducible subgroups that contain a self-centralizing normal abelian subgroup. Various other crossed product conditions on subgroups of D^{∗} are also investigated. To be more precise, it is proved that D is a crossed product over a maximal subfield if D is spanned by an irreducible subgroup G such that G is either finite or soluble-by-finite containing no element of order dividing deg(D)^{2}, or locally nilpotent. Other special cases such as G being abelian-by-supersoluble or metabelian are also reviewed and it is shown that in these cases the irreducible subgroups involved all contain a self-centralizing normal abelian subgroup.

1991 Mathematics Subject Classification. 16K20.

Key words and phrases. Division ring, Crossed Product, Irreducible groups.

2. Notations and Conventions

Throughout, D is an F -central division algebra of index n and G is a subgroup of GL_{m}(D), the group of units of M_{m}(D). The F -linear hull of G, i.e., the F -subalgebra generated by elements of G over F in M_{m}(D), is denoted by F [G]. Given a subgroup H of G, N_{G}(H) means the normalizer of H in G and C_{G}(H) the centralizer of H in G. By Z(G) or Z(M_{m}(D)), we mean the center of the group G or the center of the central simple algebra M_{m}(D), respectively.

3. Crossed products in terms of irreducible subgroups

We begin this section by a lemma which gives us a useful tool to realize maximal Galois subfields of a central simple algebra in terms of irreducible subgroups containing a self-centralizing normal abelian subgroup.

Lemma 1. Let D be an F -central division algebra and G an irreducible subgroup of GL_{m}(D). If A is any abelian normal subgroup of G such that F [A] contains no zero divisor, then M_{m}(D) is isomorphic to the crossed product

M_{m}(D) = F [C_{G}(A)] ∗ G/C_{G}(A),

where F [C_{G}(A)] is a simple subalgebra of M_{m}(D). Moreover, we have [G : C_{G}(A)] = dim(Z(F [C_{G}(A)]) : F)

Proof. Consider F [A], the F -linear hull of A in M_{m}(D). By assumption, F [A] is a commutative integral domain. Since D is of finite dimension over F, F [A] is algebraic over F. Thus, F [A] is a subfield of M_{m}(D). Now, applying Lemma 2.4 of [8], we obtain C_{M_{m}(D)}(F [A]) = F [C_{G}(A)] and that M_{m}(D) is a crossed product of C_{M_{m}(D)}(F [A]) by G/C_{G}(A). Now, by Centralizer Theorem ([2], p. 42), it is easily seen that C_{M_{m}(D)}(F [A]) is a simple subalgebra of M_{m}(D). Hence, its center is a subfield containing F. Also, the group G/C_{G}(A) acts as a group of automorphisms on the center of C_{M_{m}(D)}(F [A]), and F is the fixed field of this action. Therefore, we obtain

[G : C_{G}(A)] = dim(Z(F [C_{G}(A)]) : F),

as required. □

Now, we are prepared to prove the main theorem of this section as follows:

Theorem 1. Let D be a finite dimensional F -central division algebra. Then, M_{m}(D) is a crossed product over a maximal subfield if and only if there exists an irreducible subgroup G of M_{m}(D) and a normal abelian subgroup A of G such that C_{G}(A) = A, and F [A] contains no zero divisor.

Proof. Assume that M_{m}(D) is a crossed product over a maximal subfield K. Then, K/F is Galois and by a theorem of ([2], p.92), we can write M_{m}(D) = ⊕_{σ∈Gal(K/F)}Ke_{σ}, where e_{σ} ∈ GL_{m}(D) and for each x ∈ K and σ ∈ Gal(K/F) there exists σ(x) ∈ K such that e_{σ}x = σ(x)e_{σ}. Therefore, the elements e_{σ}’s as well as the group K^{∗} are contained in N_{GL_{m}(D)}(K^{∗}). This implies that N_{GL_{m}(D)}(K^{∗}) is an irreducible subgroup of M_{m}(D). Now, using the Skolem-Noether Theorem ([2], p.39) and the fact that C_{M_{m}(D)}(K) = K, we obtain the isomorphism N_{GL_{m}(D)}(K^{∗})/K^{∗} ≃ Gal(K/F). Hence, taking G := N_{GL_{m}(D)}(K^{∗}) and A := K^{∗}, one side of the proof is done. On the other hand, let G be an irreducible subgroup of M_{m}(D), and A a normal abelian subgroup of G such that C_{G}(A) = A, and F [A] contains no zero divisor. By Lemma 1, we conclude that M_{m}(D) is a crossed product of F [A] by G/A. Furthermore, F [A] is a subfield of M_{m}(D) such that [G : A] = dim(Z(F [A]) : F) = dim(F [A] : F). Therefore, F [A] is in fact a maximal Galois subfield of M_{m}(D) with Galois group isomorphic to G/A. Hence, the result follows. □

We observe that in the special case m = 1, the above result reduces to the following:

Theorem 2. Let D be a finite dimensional F -central division algebra. Then D is a crossed product over a maximal subfield if and only if there exists an irreducible subgroup G ⊆ D^{∗} and a normal abelian subgroup A of G such that C_{G}(A) = A.

4. irreducible subgroups which imply that D is a crossed product over a maximal subfield

In this section we deal with the special case where our central simple algebra is a non-commutative finite dimensional F -central division algebra D. In fact, we investigate some classes of irreducible subgroups G whose existence in D^{∗} give rise to a crossed product division algebra D over a maximal subfield. To do this, we consider the following cases and apply mainly Theorem 2 of the last section. In all cases below G will be an irreducible subgroup of D^{∗}. We also remark that given an irreducible subgroup G of an F -central division algebra D, if F̄ is the algebraic closure of F, then D ⊗_{F} F̄ = M_{n}(F̄) and we have F̄[G] = M_{n}(F̄). So, G is also an irreducible subgroup in the linear group sense.

Case 1. Abelian-by-supersoluble groups

Let G be an abelian-by-supersoluble subgroup of D^{∗}. We shall prove that D is a crossed product over a maximal subfield. By Theorem 2, it is enough to find an abelian normal subgroup in G such that C_{G}(A) = A. To do this, take A maximal abelian normal in G such that G/A is supersoluble. If C_{G}(A)/A ≠ 1, then there exists a normal subgroup H of C_{G}(A) such that H/A is a nontrivial normal cyclic subgroup of G/A. Now, take H/A to be the smallest nontrivial intersection of C_{G}(A)/A by the terms of the normal cyclic series of group G/A. It is easily seen that H is an abelian normal subgroup of G properly containing A. This contradicts the choice of A. Hence, we have C_{G}(A) = A.

Case 2. Metabelian groups

Let G be a metabelian subgroup of D^{∗}. Take A maximal abelian normal in G such that G/A is also abelian. If C_{G}(A)/A ≠ 1, then for every element x ∈ C_{G}(A) \ A the group H =< A, x > is easily seen to be an abelian normal subgroup properly containing A. This contradicts the choice of A. So, we must have C_{G}(A) = A.

Case 3. Locally nilpotent groups

Assume that G is a locally nilpotent subgroup of D^{∗}. Then, by a theorem in [3], G is hypercentral. This reduces to the next item.

Case 4. Hypercentral groups If G is hypercentral, then by an exercise of ([7],p.354), we conclude that every maximal abelian normal subgroup of G is self centralizing. Hence, we get the result.

Case 5. Soluble-by-finite maximal subgroups of D^{∗}

If M is a maximal subgroup of D^{∗} that is soluble-by-finite, then M is not D^{∗}. Otherwise, D^{∗} satisfies a group identity which is not possible [1]. We now claim that M is an irreducible subgroup of D. Take F [M], the division subring of D generated by M over F. Since M is maximal two cases are possible, either F [M] = D or F [M]^{∗} = M. But the second case is not possible due to the fact that the multiplicative group of a division algebra does not satisfy a group identity. Therefore, D = F [M], i.e., M is an irreducible subgroup of D. Now, by Mal’cev Theorem ([11], p.44), M contains a normal abelian subgroup of finite index. Take A maximal. First, we prove that C_{D^{∗}}(A) ⊆ M. If not, then D^{∗} =< M, C_{D^{∗}}(A) >⊆ N_{D^{∗}}(A). This means that D = N_{D}(F [A]), which is a contradiction, by Cartan-Brauer-Hua Theorem [5]. Hence, C_{D^{∗}}(A) ⊆ M. Now, by Centralizer Theorem ([2], p.42), C_{D}(A) is a division subring of D with center F [A] whose multiplicative group is contained in M. Since M is a soluble by finite group, so is C_{D^{∗}}(A). It implies that C_{D}(A) = F [A]. Since F [A]^{∗} is a normal abelian subgroup of M containing A, we must have A = F [A]^{∗}. Therefore, C_{D}(A) = A, as desired.

We observe that the above result actually generalizes Corollary 4 of [6].

Case 6. Finite groups

First we prove the problem for the case where G is a finite soluble subgroup with a normal subgroup N such that G/N is supersolube and N has all its Sylow subgroups abelian. For such a group G, take A maximal abelian normal in G. If A ≠ C_{G}(A), then C_{G}(A)/A contains a nontrivial normal abelian subgroup L/A of G/A. Since G/A is soluble, we may take L/A to be the smallest nontrivial intersection of C_{G}(A)/A with the terms of the derived series of G/A. Clearly, A ⊆ Z(L), and because A is the maximal abelian normal subgroup of G, we necessarily have A = Z(L). Also, [L, L] = L′ ⊆ A ⊆ Z(L), So L is nilpotent of class two. Hence, we may write L ∩ N ≃ ∏ L_{p}, where L_{p}’s are the p-Sylow subgroups of L ∩ N. By assumption , all p-Sylow subgroups of N are abelian. Thus, we conclude that L∩N is an abelian normal subgroup of G. We claim that L∩N ⊆ A. If not, since L∩N ⊆ C_{G}(A), the group (L ∩ N)A is an abelian normal subgroup of G properly containing A. This contradicts the choice of A, so we must have L ∩ N ⊆ A. We also claim that AN ≠ LN. Otherwise, if LN = AN, then for all x ∈ L, we have x = an for some a ∈ A and n ∈ N. It follows that xa^{−1} = n ∈ L ∩ N ⊆ A, and so x ∈ A, a contradiction. Hence, we necessarily have AN/N ⊴ LN/N ⊴ G/N. Now, G/N is supersoluble and consequently G/AN is a supersoluble group. Therefore, the group LN/AN must contain a nontrivial normal cyclic subgroup in G/AN denoted by < x > AN/AN, where x ∈ L. Now, if we take A_{0} =< A, x >, then A_{0} is an abelian normal subgroup of G properly containing A, which is a contradiction to the choice of A. Therefore, we have C_{G}(A) = A and we obtain the result. Now, we consider the general case. By Amitsur’s Theorem ([9], p.46), if G is a finite subgroup of the multiplicative group of a division ring, then G is isomorphic to one of the following list of groups, and only the last one is insoluble.

(a) A group that all of whose Sylow subgroups are cyclic.

(b) C_{m}]Q, where Q is a quaternion of degree 2^{t}, and C_{m} a cyclic group of order m. By C_{m}]Q, we mean the split extension of C_{m} by Q in which Q acts on C_{m}.

(c) Q × M, where Q is quaternion of degree 8, and M a group of type (a).

(d) The binary octahedral group of order 48.

(e) SL(2, 3) × M, where M is a group of type (a).

(f) SL(2, 5).

The groups of (a), (b) and (c) are special cases of the soluble groups that are dealt with previously. Now, we examine the case (d). The binary octahedral group G of order 48, which has a normal quaternion subgroup Q of order 8. If D = F [G] is a division algebra generated by the octahedral subgroup G over the center F, then the division subring F [Q] has degree 2 since Q is a nonabelian group with a center containing two elements. It implies that Z(F [Q]) = F. Now, by Theorem 2.6 of [8], we can write D as F [X_{1}] ⊗_{F} F [X_{2}], where F [X_{1}] = F [Q] and F [X_{2}] = C_{D}(F [Q]). We also have F^{∗} ⊆ X_{2}, and [G : C_{G}(X_{2})] = [X_{2} : F^{∗}] in which C_{G}(X_{2}) = G ∩ F [X_{1}]. This means that [X_{2} : F^{∗}] is at most 6, and then it follows that X_{2} is abelianby-supersoluble. Hence, both F [X_{1}] and F [X_{2}] are crossed product algebras over their maximal subfields by part 1, and so is D as desired. To deal with the case (e), assume that D = F [G], where G is as in (e) and let Q be the normal quaternion subgroup of SL(2, 3) of order 8. We first claim that F [Q] = F [SL(2, 3)]. To do this, consider SL(2, 3) as a semidirect product of Q by C_{3}, written by Q]C_{3}, where C_{3} is the cyclic group of order 3 with generator c and Q =< x, y|x^{2} = y^{2}, y^{4} = 1, x^{y} = x^{−1} >. c acts on Q by cyclically permuting y, x, xy. But conjugation by the element t = −(1 + x + y + xy)/2 ∈ F [Q] also has the same effect. Thus ct^{−1} centralizes F [Q], and hence so does t. Therefore, c commutes with t. But t and c are both cube roots of unity in the field F [t, c]. It follows that c ∈< t >. Thus, F [Q] = F [SL(2, 3)] and hence Z(F [Q]) = Z(F [SL(2, 3)]) = F. Now, again by Theorem 2.6 of [8], we may write D = F [SL(2, 3)] ⊗_{F} F [X_{2}]. Repeating the same argument as used in the proof of Theorem 2.6 of [8], we may take X_{2} = M. So, one sees that both factors of the decomposition of D are crossed product division algebras over their maximal subfields. Hence, D is also a crossed product over a maximal subfield, as required. To prove the final case (f), let D be the division algebra generated by G = SL(2, 5). It is known from Theorem 2.1.11 of [[9], p. 51] that D can be generated by the quaternion subgroup of order 8. Hence, D is a crossed product over a maximal subfield.

We remark that the last result generalizes the case where the degree of D is a prime power, which is dealt with in [4].

Case 7. Soluble subgroups containing no element of order dividing deg(D)^{2}

Let G be an irreducible soluble subgroup of D^{∗}. By Mal’cev Theorem ([11], p.44), G contains an abelian normal subgroup A of finite index . As before, take A maximal abelian normal in G. If C_{G}(A)/A ≠ 1, then due to solublity of G/A, the group C_{G}(A)/A contains a nontrivial abelian subgroup L/A which is normal in G/A. We have also Z(L) = A. For every a, b, c ∈ L, it is easily seen that for (a, b) = aba^{−1}b^{−1} ∈ A we have (a, b)(a, c) = (a, bc). Now, for every a ∈ L, define the group homomorphism ϕ_{a} : L → A by ϕ_{a}(x) = (a, x), where x ∈ L. It is clear that A ⊆ ker ϕ_{a}. Therefore, the image of ϕ_{a} is a finite subgroup of A whose elements are of order dividing [L : A]. We now claim that [L : A]|(D : F). To do this, take a transversal l_{1}, ..., l_{t} of A in L, and consider F [L], the F -linear hull of L in D. We want to prove that l_{i}’s are a linearly independent generating set of F [L] over F [A]. That F [L] is generated by l_{i}’s over F [A] is evident. But for the latter, take the relation ∑_{i=1}^{s} a_{i}l_{i} = 0, where a_{i} ∈ F [A], and s is chosen minimal. For each l_{j}, we have ∑_{i=1}^{s} a_{i}l_{j}l_{i}l_{j}^{−1} = 0, and by subtracting two relations we obtain ∑_{i=1}^{s} a_{i}(l_{j}l_{i}l_{j}^{−1}l_{i}^{−1} − 1)l_{i} = 0. Since s is minimal for each i we obtain l_{j}l_{i}l_{j}^{−1}l_{i}^{−1} = 1. In other words, L is an abelian normal subgroup of G properly containing A. This is not possible by choice of A, hence l_{i}’s are linearly independent over F [A]. Therefore, [L : A] = dim(F [L] : F [A]) divides deg(D)^{2}. Because each element of G is not of order dividing deg(D)^{2}, we conclude that ϕ_{a}(L) = {1}. Since a is chosen arbitrary in L, we conclude that L is an abelain normal subgroup of G properly containing A. This is again a contradiction. Therefore, we have C_{G}(A) = A, as desired.

Case 8. G is soluble-by-finite satisfying the condition of part 7, and G does not contain any normal subgroup isomorphic to SL(2,5).

Take A maximal abelian normal in G of finite index. If C_{G}(A)/A ≠ 1, then consider S, the maximal soluble normal subgroup of C_{G}(A) which clearly contains A in itself. We have A ≠ S for otherwise from Lemma 5.5 in [8] we obtain A = C_{G}(A), a contradiction. Hence, S/A is a nontrivial normal soluble subgroup of G/A which is contained in C_{G}(A)/A. Therefore, we can choose a nontrivial abelian normal subgroup L/A of G/A that is contained in C_{G}(A)/A. Now, apply a similar argument as in the previous case to obtain an abelian normal subgroup in G which is self-centralizing.

We observe that Cases 7 and 8 in fact generalize the last theorem of [8] in which the torsion free case is investigated. Finally, we remark that the counterexample given in the last section of [8] provides us with an irreducible subgroup G whose maximal abelian normal subgroup A does not satisfy the condition C_{G}(A) = A. Therefore, in general it is not true that every maximal abelian normal subgroup of an irreducible subgroup G of D^{∗} is a self-centralizing abelian normal subgroup.

Acknowledgment. The authors are indebted to the Research Council of Sharif University of Technology for support.

References

[1] S. A. Amitsur, Rational identities and applications to algebra and geometry, J. Algebra 3 1966 304–359.

[2] P. K. Draxl, Skew Fields, London Math. Soc. Lecture Notes Series. Vol 81, Cambridge, Univ. Press Cambridge(1983).

[3] M. S. Garascuk, On the theory of generalized nilpotent linear groups, Doki. Akad. Nauk BSSR 4, 276-277(1960).

[4] D. Kiani, M. Mahdavi-Hezavehi. Crossed product conditions for division algebras of prime power degree, J. of Algeb., 283 (2005) 222231.

[5] T. Y. Lam, A first course in noncommutative rings, Springer-Verlag, (1991).

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

[7] D. J. S. Robinson, A Course in the Theory of Groups, Graduate Text in Mathematics, No. 80, Springer-Verlag, (1982).

[8] M. Shirvani, On soluble-by-finite subgroups of division algebras , J. of Algeb., 294 (2005) 255277.

[9] M. Shivani & A. F. Wehrfritz, Skew linear groups, Cambridge University Press, (1986).

[10] J.-P. Tignol, M. Mahdavi-Hezavehi, Cyclicity conditions for division algebras of prime degree, Proc. Amer. Math. Soc. 131 (2003) 36733676.

[11] A. F. Wehrfritz, Infinite linear groups, Berlin. Springer, (1973).

Department of Mathematical Sciences, Sharif University of Technology, P. O. Box 11365-9415. Tehran, Iran

E-mail address: keshavarzipour@math.sharif.edu &, mahdavih@sharif.edu