Crossed Product Conditions for Division Algebras of Prime

Power Degree

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 an F -central division algebra of degree p^{r}, p a prime. A set of criteria is given for D to be a crossed product in terms of irreducible soluble or abelian-by-finite subgroups of the multiplicative group D^{∗} of D. Using the Amitsur’s classification of finite subgroups of D^{∗} and the Tits Alternative, it is shown that D is a crossed product if and only if D^{∗} contains an irreducible soluble subgroup. Further criteria are also presented in terms of irreducible abelian-by-finite subgroups and irreducible subgroups satisfying a group identity. Using the above results, it is shown that if D^{∗} contains an irreducible finite subgroup, then D is a crossed product.

1 Introduction

Let D be an F -central division algebra of degree n. The algebra D is called a crossed product if it contains a maximal subfield K such that K/F is Galois. We shall say that D is a nilpotent crossed product if Gal(K/F) is nilpotent. A subgroup G of D^{∗} is said to be irreducible if F [G] = D. When n = p, a prime, it is shown in [6] that D is cyclic if and only if D^{∗} contains a nonabelian soluble subgroup. A criterion is also given in [3] for D to be a supersoluble (nilpotent) crossed product division algebra in terms of subgroups D^{∗}. More precisely, it is shown that D is a supersoluble (nilpotent) crossed product if and only if D^{∗} contains an abelian-by-supersoluble (abelian-by-nilpotent) irreducible subgroup. The aim of this paper is to generalize some of these results to a division algebra of a prime power degree p^{r}. In fact, we present a set of criteria for D to be a crossed product in terms of irreducible soluble or abelian-by-finite subgroups of D^{∗}. To be more precise, it is shown that D is a nilpotent crossed product if and only if D^{∗} contains an irreducible soluble subgroup. In addition, it is shown that, except for the case CharF = 0, p = 2 and r > 1, D is a crossed product if and only if either of the following conditions holds: (i) D^{∗} contains an irreducible abelian-by-finite subgroup, or (ii) D^{∗} contains an irreducible subgroup satisfying a group identity. Furthermore, it is proved that these conclusions also hold for the above excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}). Finally, given a non-commutative F -central division algebra D of index p^{r}, p a prime, using the above mentioned results, it is shown that if D^{∗} contains an irreducible finite subgroup G, then D is a crossed product. We note that soluble subgroups of the multiplicative group of a division ring were first studied by Suprunenko in [10].

2 Notations and conventions

We now recall some notations and conventions that are used throughout. Let D be an F -central division algebra and G be a subgroup of D^{∗}. The F -linear hull of G, i.e., the F -algebra generated by elements of G over F, is denoted by F [G]. G is called 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. Let S be a subset of D, then the centralizer of S in D is denoted by C_{D}(S). For notations and results, used in the text, on central simple algebras see [7].

3 Irreducible soluble subgroups

Let D be an F -central division algebra of degree p^{r}, p a prime. This section investigates the structure of D under the condition that D^{∗} has an irreducible soluble subgroup. To be more precise, it is shown that D is a crossed product if and only if D^{∗} contains an irreducible soluble subgroup. We begin our study with the following:

Lemma 1. Let D be a finite dimensional F -central division algebra. If D is a soluble crossed product, then D^{∗} contains an irreducible soluble subgroup.

Proof. Let K be a maximal subfield of D such that K/F is soluble Galois. By Skolem-Noether Theorem, for any σ ∈ Gal(K/F) there exists an element x ∈ N = N_{D^{∗}}(K^{∗}) such that σ(k) = xkx^{−1}, for all k ∈ K. Hence N_{D^{∗}}(K^{∗})/C_{D^{∗}}(K^{∗}) ≃ Gal(K/F). Since K is a maximal subfield of D, we have C_{D^{∗}}(K^{∗}) = K^{∗}. Therefore, N_{D^{∗}}(K^{∗}) is a soluble subgroup of D^{∗}. To complete the proof of the lemma, it is enough to show that N is irreducible, i.e., F [N] = D. Put D_{1} = F [N]. We have C_{D}(D_{1}) ⊆ C_{D}(K) = K, and hence C_{D}(D_{1}) is an intermediate field of the Galois extension K/F. By the fact that every element of Gal(K/F) is the restriction of an inner automorphism of N we conclude that C_{D}(D_{1}) ⊆ F ix(Gal(K/F)). Therefore C_{D}(D_{1}) = F. Now, by Centralizer Theorem, we obtain D = C_{D}(F) = C_{D}(C_{D}(D_{1})) = D_{1}, which completes the proof. □

The following lemma is used in many proofs below, its idea is due to Suprunenko [10].

Lemma 2. Let D be a finite dimensional F -central division algebra. Suppose that G is a subgroup of D^{∗} such that F^{∗} ⊆ Z(G). If K = Z(G) ∪ {0} is a subfield of D and G/K^{∗} is abelian, then we have [K[G] : K] = |G/K^{∗}| and hence G/K^{∗} is a finite group.

Proof. Let g_{1}, …, g_{t} be a set of linearly independent elements of G over K. It is clearly seen that g_{1}K^{∗}, …, g_{t}K^{∗} are distinct elements of G/K^{∗}. On the other hand, if g_{1}K^{∗}, …, g_{t}K^{∗} are distinct elements of G/K^{∗}, we shall show that g_{1}, …, g_{t} are linearly independent over K. To see this, since G/K^{∗} is abelian, for every g ∈ G we have gg_{i}g^{−1} = k_{i}g_{i} with 1 ≤ i ≤ t, where k_{i} ∈ K^{∗}. We claim that for each pair i ≠ j we can find an element g in G such that k_{i} ≠ k_{j}. For suppose that for each g in G we have gg_{i}g^{−1}g_{i}^{−1} = k_{i} = gg_{j}g^{−1}g_{j}^{−1} = k_{j}. Therefore, we conclude that [g, g_{j}^{−1}g_{i}] = 1, and hence g_{j}^{−1}g_{i} ∈ K^{∗}. This contradicts the choice of g′_{i}s, and so the claim is established. Now, suppose that g_{1}, …, g_{t} are linearly dependent over K and consider a relation

λ_{1}g_{1} + … + λ_{t}g_{t} = 0. (∗)

Of all relations of the form (∗), there must be at least one for which the number of nonzero terms is least. Let (∗) be such a relation. Now, we may assume that λ_{1} ≠ 0, λ_{2} ≠ 0 and choose g in G such that k_{1} = gg_{1}g^{−1}g_{1}^{−1} ≠ gg_{2}g^{−1}g_{2}^{−1} = k_{2}. From the relation (∗) we obtain k_{1}(λ_{1}g_{1} + … + λ_{t}g_{t}) − g(λ_{1}g_{1} + … + λ_{t}g_{t})g^{−1} = λ_{1}k_{1}g_{1} + … λ_{t}k_{1}g_{t} − (λ_{1}k_{1}g_{1} + … λ_{t}k_{t}g_{t}) = λ_{2}(k_{1} − k_{2})g_{2} + … + λ_{t}(k_{1} − k_{t})g_{t} = 0. Now, the last equation contradicts the choice of the relation (∗). Therefore, g_{1}, …, g_{t} are a linearly independent subset of G over K, and this completes the proof.

To prove our next lemma, we shall need the following results from [3].

Lemma A. 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 B. 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 C. Let D be a noncommutative finite dimensional F -central division algebra. Then D is a nilpotent 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 nilpotent.

Lemma 3. Let D be a finite dimensional F -central division algebra of index n. Assume that D^{∗} contains an irreducible soluble subgroup. Then we have the following:

(i) there is an irreducible soluble subgroup G and a maximal abelian normal subgroup K^{∗} of G such that K = K^{∗}∪{0} is a subfield of D and G/K^{∗} is finite. Furthermore, setting H := C_{G}(K^{∗}), then the derived group H′ of H is also finite. (ii) Assume the notation of (i). If H′ is abelian and n = q^{r}, q a prime, then G/K^{∗} is a q-group. (iii) Keep the notation of (i). If H′ is nonabelian and n = q^{r}, q a prime, then H^{t−2} is a q-group, where H^{i} denotes the i-th term of the derived series of H. (iv) If D is a non-crossed product with index i(D) = 2^{r}, then D^{∗} contains the finite quaternion subgroup Q_{8}, or SL_{2}(Z_{3}), or the binary octahedral group of order 48.

Proof. (i) Let G_{0} be an irreducible soluble subgroup of D^{∗}. By Lemma 3 of [5], we know that G_{0} is abelian-by-finite, i.e., there is an abelian normal subgroup A in G_{0} of finite index. Take A maximal in G_{0}, and set K = F (A). One may easily show that G_{0} ⊆ N_{D^{∗}}(A) and that K^{∗}G_{0} is an irreducible soluble subgroup of D^{∗}. Set G = K^{∗}G. Then, it is easily seen that K^{∗} is maximal abelian normal in G and G/K^{∗} is a finite group. Furthermore, we know that H/Z(H) is finite and hence, by Theorem 15.1.13 of [8], the derived group H′ is a finite group.

(ii) Because H′ ⊆ C_{G}(K^{∗}), K^{∗}H′ is an abelian normal subgroup of G. Hence, by maximality of K^{∗}, we have H′ ⊆ K^{∗} = Z(H). Therefore, H/K^{∗} is abelian. Now, by Lemma 2, we conclude that [K[H] : K] = |H/K^{∗}|. Since [D : F] = q^{2r} and F^{∗} ⊆ K^{∗} we conclude that [K[H] : K] divides q^{2r}, i.e., there exists a natural number s such that |H/K^{∗}| = [K[H] : K] = q^{s}. Now, by Lemma A, we have G/H ≃ Gal(K/F) and K/F is a Galois extension. Since i(D) = q^{r} there exists a natural number t such that |Gal(K/F)| = [K : F] = q^{t}. Thus, |G/H| = q^{t} and hence |G/K^{∗}| = q^{s+t}, i.e., G/K^{∗} is a q-group.

(iii) Suppose that H′ is nonabelian. Then, the soluble length of H is t = l(H) ≥ 3. Now, consider the derived chain ⟨e⟩ = H^{t} ⊂ H^{t−1} ⊂ … ⊂ H′ ⊂ H. It is clear that H^{t−1} is abelian and H^{t−2} is a nonabelian subgroup of H′. Now, we know that H^{t−1} ◁ G and H^{t−1} ⊂ C_{G}(K^{∗}). Thus, H^{t−1}K^{∗} is an abelian normal subgroup of G. Hence, by maximality of K^{∗}, we conclude that H^{t−1} ⊆ K^{∗}. Therefore, H^{t−2}K^{∗}/K^{∗} is an abelian subgroup of G/K^{∗}. Set N = H^{t−2}K^{∗}. We note that N is normal in G, and hence Z(N) is an abelian normal subgroup of G containing K^{∗}. By maximality of K^{∗}, we have Z(N) = K^{∗}. Now, N is a subgroup of D^{∗} such that N/K^{∗} is abelian and Z(N) = K^{∗}. By Lemma 2, we obtain [K[N] : K] = |N/K^{∗}|. By our assumption, we know that [K[N] : K] divides q^{2r}. Therefore, N/K^{∗} ≃ H^{t−2}/K^{∗} ∩ H^{t−2} is a qgroup. Now, by Lemma B, we have F [H] = K[H] = C_{D}(K). Put D_{1} = C_{D}(K). We know that Z(D_{1}) = K. We now claim that H^{t−2} is a q-group. To see this, let x ∈ H^{t−2}. Then, there exists a natural number s such that x^{q^{s}} ∈ K^{∗} ∩H^{t−2}. On the other hand, x^{q^{s}} ∈ H′ ⊂ D′_{1}, and hence RN_{D_{1}/K}(x^{q^{s}}) = 1. Since x^{q^{s}} ∈ K^{∗} = Z(D_{1})^{∗} we obtain RN_{D_{1}/K}(x^{q^{s}}) = (x^{q^{s}})^{i(D_{1})} = x^{q^{s+u}}, where i(D_{1}) = q^{u}. Therefore, x^{q^{s+u}} = 1 and so H^{t−2} is a finite q-group.

(iv) Let G be the irreducible soluble subgroup obtained by (i) and keep the notations of the above cases. If H′ is abelian, then by the case (ii), G/K^{∗} is a 2group and hence nilpotent and so G is an irreducible abelian-by-nilpotent subgroup of D^{∗}. Now, by Theorem C, we conclude that D is a nilpotent crossed product, which contradicts our assumption that D is a non-crossed product. Therefore, H′ is nonabelian. By (iii) we conclude that H^{t−2} is a finite 2-group. Now, by a result of [9, p.45], we conclude that H^{t−2} is cyclic or a (generalized) quaternion group. Since H^{t−2} is nonabelian we conclude that H^{t−2} is a (generalized) quaternion group. We recall that the (generalized) quaternion group of order 2^{u}, u ≥ 3, is defined with the presentation Q_{2^{u}} = ⟨ x, y | x^{2^{u−2}} = y^{2}, y^{4} = 1, yxy^{−1} = x^{−1}⟩. It is clear that ⟨x⟩ ◁ Q_{2^{u}} and Q_{2^{u}}^{(2)} = ⟨x^{2}⟩. Thus, ⟨x^{2}⟩ is a characteristic subgroup of H^{t−2} and hence ⟨x^{2}⟩ is an abelian normal subgroup of G. We note that ⟨x^{2}⟩ ⊂ C_{G}(K^{∗}) and so K^{∗}⟨x^{2}⟩ is an abelian normal subgroup of G. Therefore, by maximality of K^{∗}, we have ⟨x^{2}⟩ ⊂ K^{∗}. Thus, x^{2} ∈ Z(Q_{2^{u}}), and hence x^{−2} = yx^{2}y^{−1} = x^{2}. Therefore, x^{4} = 1. On the other hand, we have x^{2^{u−1}} = 1, and so u = 3, i.e., H^{t−2} ≃ Q_{8}. Now, assume that N is a maximal normal 2-subgroup of H′. For every g ∈ G, set N_{1} = gNg^{−1}. N_{1} is a normal 2-subgroup of H′. Hence NN_{1} is a normal 2-subgroup of H′. By maximality of N, we obtain N_{1} ⊆ N and so we have N ◁ G. We note that H′ is finite and hence N is a finite 2-group. Thus, as in the case of H^{t−2} above, we conclude that N ≃ Q_{8}. Therefore, by a result of [9, p.54], we have either H′ ≃ Q_{8} × M, where M is a group of odd order, or H′ ≃ SL_{2}(Z_{3}) × M, where M is a group of order m coprime to 6, or H′ is isomorphic to the binary octahedral group. In the first case, M is a characteristic subgroup of H′. Therefore, M is a normal subgroup of G. Let l be the soluble length of M. We know that M^{l−1} is a nontrivial abelian normal subgroup of G. Thus, M^{l−1} ⊆ K^{∗}, and hence for every x ∈ M^{l−1} we have x^{2^{α}} = RN_{C_{D}(K)/K}(x) = 1, which contradicts the fact that M is of odd order. Therefore, M is trivial and so H′ ≃ Q_{8}. One may easily show that other cases are also true by similar arguments and this proves the case (iv).

We are now prepared to prove the following.

Theorem 1. Let D be an F -central division algebra of index q^{r}, q a prime. If D^{∗} contains an irreducible soluble subgroup, then D is a crossed product.

Proof. We may consider the following two cases:

Case 1. CharF = p > 0. By Lemma 3, we know that there is an irreducible soluble subgroup G and abelian normal subgroup K^{∗} of G such that K = K^{∗} ∪ {0} is a subfield of D and G/K^{∗} as well as H′ is finite, where H = C_{G}(K^{∗}). Since CharF = p > 0, by a result of [4, p.215], we conclude that H′ is cyclic. Now, by Lemma 3, G/K^{∗} is a q-group and hence it is nilpotent. Thus, G is an irreducible abelian-by-nilpotent subgroup of D^{∗}. Now, by Theorem C, we conclude that D is a nilpotent crossed product, which completes the proof of this case.

Case 2. CharF = 0. We keep to the notations of the case 1. If H′ is abelian, then as in the above case we obtain the result. So, we may assume that H′ is nonabelian. By Lemma 3, we know that H^{t−2} is a finite q-group. If q is odd, then, by a result of [9, p.45], we conclude that H^{t−2} is cyclic, which contradicts the fact that H^{t−2} is nonabelian. So, we may assume that q = 2. We now proceed by induction on r. If r = 1, then it is clear that D is cyclic. Assume that the result holds for all n < r. Now, by a result of [9, p.45] again, we conclude that H^{t−2} is cyclic or a (generalized) quaternion. Since H^{t−2} is nonabelian we conclude that H^{t−2} is a (generalized) quaternion. As in the proof of Lemma 3, one may easily show that H^{t−2} ≃ Q_{8}. Therefore, H^{t−2} is normal in G. Set D_{1} = F [H^{t−2}]. It is clear that i(D_{1}) = 2 and Z(D_{1}) = F and D_{1} is a crossed product. Now, by the Double Centralizer Theorem, we have D ≃ D_{1} ⊗_{F} C_{D}(D_{1}). 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 the rule 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 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}, F^{∗}b_{h}b_{g} = F^{∗}b_{hg}. 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_{h}a_{g} = F^{∗}a_{hg} 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 soluble 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 is B. Set D_{2} = C_{D}(D_{1}). If we show that D_{2} = F [B], then by induction D_{2} is a crossed product. To prove this, put D_{3} = F [B]. Now, for all g ∈ G we have g = a_{g}b_{g} = (a_{g} ⊗ 1)(1 ⊗ b_{g}) = a_{g} ⊗ b_{g}. Therefore, we conclude that G ⊂ D_{1}⊗D_{3} and hence F [G] = D = D_{1}⊗D_{3} = D_{1}⊗D_{2}. Finally, one may easily see that [D_{3} : F] = [D_{2} : F], and so D_{3} ⊆ D_{2}, i.e., D_{3} = D_{2} and so the result follows.

Let D be an F -central division algebra of degree p^{r}, p a prime. Using the above result one may conclude that if D^{∗} contains an irreducible finite subgroup G, then D is a crossed product. To see this, by a result of [9, p.51, Thm 2.1.11], we know that either G is soluble or G ≃ SL_{2}(Z_{5}). If the first case happens, then the result follows from Theorem 1. If the second case occurs, then as in the course of the proof of Theorem 2.1.11 of [9, p.51], we have [ℚ(G) : ℚ] ≤ 8. Since ℚ ⊆ F we clearly have [F [G] : F] ≤ 8 and hence [D : F] = 4 because G is irreducible. Therefore, D is cyclic and so the result also follows for this case. Later on we shall present a different proof of this fact which may be of some interest. Now, combining Lemma 1 and Theorem 1, we are able to obtain one of our main results in the following form.

Corollary 1. Let D be an F -central division algebra of index p^{r}, p a prime. Then, D is a crossed product if and only if D^{∗} contains an irreducible soluble subgroup.

4 Irreducible abelian-by-finite subgroups

This section turns to the case where the multiplicative group D^{∗} contains an irreducible abelian-by-finite subgroup. Let D be an F -central division algebra of index p^{r}, p a prime. It is proved that except when CharF = 0 and p = 2, r > 1, D is a crossed product if and only if D^{∗} contains an irreducible abelian-by-finite subgroup. Furthermore, the conclusion also holds for the excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}). Using the above result, and the Tits Alternative which asserts that a finitely generated linear group either contains a non-cyclic free subgroup or it is soluble-by-finite [11], we are able to show that D is a crossed product if and only if D^{∗} contains an irreducible subgroup satisfying a group identity. Furthermore, the conclusion also holds for the above excluded case provided that D^{∗} contains no finite subgroup isomorphic SL_{2}(Z_{5}). To prove our results, we shall need the following lemma.

Lemma 4. Given a field F of characteristic zero, let D be an F -central division algebra of index 2^{r}, r > 1. Assume that D^{∗} contains an irreducible abelian-by-finite subgroup. If D is a non-crossed product, then D^{∗} contains a copy of the finite group SL_{2}(Z_{5}).

Proof. Suppose that G is an irreducible abelian-by-finite subgroup of D^{∗} and A is a maximal abelian normal subgroup of G such that G/A is finite. Set K = F (A). It is clear that G ⊆ N_{D}^{∗}(K^{∗}), and hence G_{1} = GK^{∗} is an irreducible subgroup of D^{∗} so that G_{1}/K^{∗} is finite. One may easily show that K^{∗} is a maximal abelian normal subgroup of G_{1}. Put H = C_{G_{1}}(K^{∗}). By maximality of K^{∗}, we have Z(H) = K^{∗}. Now, we know that H/Z(H) is finite, and so by Theorem of [8, p.443, Thm. 15.1.13], the derived group H′ is a finite group. We claim that H′ is nonabelian. For if H′ is abelian, then H is soluble. Now, by Lemma A, we have G_{1}/H ≃ Gal(K/F) and K/F is a Galois extension. Thus, G_{1}/H is a 2-group and hence G_{1} is soluble. We note that G_{1} is an irreducible soluble subgroup of D^{∗}. By Theorem 1, we conclude that D is a crossed product which is a contradiction. Thus, H′ is nonabelian as claimed. Therefore, by a result of [9, p.51], this implies that either H′ is a soluble group or H′ ≃ SL_{2}(Z_{5}). If the first case occurs, then H is soluble and hence as above G is soluble. Therefore, by Theorem 1, we conclude that D is a crossed product which is a contradiction. So, we have H′ ≃ SL_{2}(Z_{5}), and the result follows.

Theorem 2. Let D be an F -central division algebra of index q^{r}, q a prime. If D^{∗} contains an irreducible abelian-by-finite subgroup, then, except for the case CharF = 0 and q = 2, r > 1, D is a crossed product. Furthermore, the conclusion also holds for the above excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}).

Proof. We consider the following three cases: Case 1. CharF = p > 0. Suppose that G is an irreducible abelian-by-finite subgroup of D^{∗} and A is a maximal abelian normal subgroup of G such that G/A is finite. Set K = F (A). It is clear that G ⊆ N_{D}^{∗}(K^{∗}), and hence G_{1} = GK^{∗} is an irreducible subgroup of D^{∗} so that G_{1}/K^{∗} is finite. One may easily show that K^{∗} is a maximal abelian normal subgroup of G_{1}. Put H = C_{G_{1}}(K^{∗}). By maximality of K^{∗}, we have Z(H) = K^{∗}. Now, we know that H/Z(H) is finite, and so by Theorem of [8, p.443], the derived group H′ is a finite group. Thus, by a result of [4, 4, Cor. 13.3], we conclude that H′ is cyclic. Therefore, H is a soluble group. Now, by Lemma A, we have G_{1}/H ≃ Gal(K/F) and K/F is a Galois extension. Thus, G_{1}/H is q-group and hence G_{1} is soluble. We note that G_{1} is an irreducible soluble subgroup of D^{∗}. By Theorem 1, we conclude that D is a crossed product, which completes the proof of this case. Case 2. CharF = 0. If q = 2 and r = 1, then it is clear that D is cyclic. So, we may assume that q is odd. Keeping to the notations of the above case, we know that H/Z(H) is finite as well as the derived group H′. Therefore, by a result of [9, p.51], we know that either H′ is a soluble group or H′ ≃ SL_{2}(Z_{5}). In the first case H is soluble and as in the above case we have that G_{1} is also a soluble subgroup of D^{∗}. Thus, by Theorem 1, we conclude that D is a crossed product. We claim that the second case leads to a contradiction. So, we may assume that H′ ≃ SL_{2}(Z_{5}). In the course of the proof of Theorem 2.1.11 of [9, p.51], that the only finite insoluble subgroup of a division ring is SL_{2}(Z_{5}) we obtain [ℚ(H′) : ℚ] ≤ 8. Since ℚ ⊆ K we conclude that [K[H′] : K] ≤ 8. On the other hand, we have K ⊆ Z(K[H′]). Set D_{1} = K[H′]. Now, we know that D_{1} is a division algebra with [D_{1} : Z(D_{1})] ≤ 8. Therefore, [D_{1} : Z(D_{1})] = 4 and hence 2 divides q^{r}, which contradicts our assumption that q is odd. Case 3. Assume that CharF = 0, p = 2, and r > 1. If D is not a crossed product, then, by Lemma 4, we conclude that D^{∗} contains a copy of the finite group SL_{2}(Z_{5}), which is a contradiction. This completes the proof of the theorem.

Combining Lemma 1 and Theorem 2, we obtain the following.

Corollary 2. Let D be an F -central division algebra of index p^{r}, p a prime. Then, except when CharF = 0 and p = 2, r > 1, D is a crossed product if and only if D^{∗} contains an irreducible abelian-by-finite subgroup. Furthermore, the conclusion also holds for the above excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}).

Using the above result, and the Tits Alternative which asserts that a finitely generated linear group either contains a non-cyclic free subgroup or it is soluble-by-finite [11], we are able to prove the following criterion.

Corollary 3. Let D be an F -central division algebra of index p^{r}, p a prime. Then, except when CharF = 0 and p = 2, r > 1, D is a crossed product if and only if D^{∗} contains an irreducible subgroup satisfying a group identity. Furthermore, the conclusion also holds for the above excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}).

Proof. The “only if “ part is clear by Lemma 1. Assume that G is an irreducible subgroup of D^{∗} satisfying a group identity. Since [D : F] < ∞ we may view G as a linear group. Let G_{1} be a subgroup of G generated by the elements of a basis of D over F. Thus, by Tits Alternative, we know that G_{1} is soluble-by-finite, i.e., there is a soluble normal subgroup N of G_{1} such that G_{1}/N is finite. Now, by Lemma 3 of [5], N is abelian-by-finite. Thus, G_{1} is abelian-by-finite. Therefore, by Theorem 2, D is crossed product.

Now, one may apply the above results to prove the following criterion for D to be cyclic. This is one of the main results of [2].

Corollary 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 subgroup satisfying a group identity.

Proof. The “only if“ part is clear by Lemma 1. If p = 2, then D is cyclic. Let p be an odd prime. Now, by Corollary 3, one can easily show that D is cyclic. □

Let D be an F -central division algebra of finite index i(D) = n and G be an irreducible subgroup of D^{∗}. Assume that A is a maximal abelian normal subgroup of G. We conclude this section with some remarks concerning the relation between the cardinal of G/A and the dimension of D/F.

Remark 1. Let D be a finite dimensional F -central division algebra. If G is an irreducible subgroup of D^{∗} with maximal abelian normal subgroup A such that G/A is nilpotent, then |G/A| = i(D). To see this, set G_{1} = K^{∗}G, where K = F [A]. It is easily seen that G_{1} is irreducible with maximal abelian normal subgroup K^{∗} such that G_{1}/K^{∗} ≃ G/A and so G_{1}/K^{∗} is nilpotent. As in the proof of Theorem 3.4 of [3], one may easily check that K/F is Galois and K is a maximal subfield of D. Therefore, we have C_{G_{1}}(K^{∗}) = K^{∗}. Now, by Lemma B, we have G_{1}/K^{∗} ≃ Gal(K/F), i.e., |G_{1}/K^{∗}| = |G/A| = [K : F] = i(D).

Remark 2. Let D be an F -central division algebra of index p^{r}, p a prime. Assume that G is an irreducible subgroup of D^{∗} with maximal abelian subgroup A such that G/A is finite. Then, except when CharF = 0 and p = 2, r > 1, we have |G/A| = i(D). Furthermore, the conclusion also holds for the excluded case provided that D^{∗} contains no finite subgroup isomorphic to SL_{2}(Z_{5}). To prove this, we may use Theorem 1, Theorem 2, and the Remark 1 to obtain the result.

5 Irreducible finite subgroups

Let D be an F -central division algebra of degree p^{r}, p a prime. This section studies the structure of D under the condition that D^{∗} has an irreducible finite subgroup. Using Amitsur’s classification of finite multiplicative subgroups of a division ring, it is proved that if D^{∗} contains an irreducible finite subgroup, then D is a crossed product.

Theorem 3. Let D be an F -central division algebra of index q^{r}, q a prime. If D^{∗} contains an irreducible finite subgroup G, then D is a crossed product.

Proof. We first observe that CharF = 0. Since otherwise G is cyclic and since G is irreducible we obtain D = F which is a contradiction. If q is odd, then the result follows from Corollary 3. So we may assume that q = 2. By a result of [9, p.51], we know that either G is soluble or G ≃ SL_{2}(Z_{5}). If the second case occurs, then as in the course of the proof of Theorem 2.1.11 of [9, p.51], that the only finite insoluble subgroup of a division ring is SL_{2}(Z_{5}), we may obtain [ℚ(G) : ℚ] ≤ 8. Since ℚ ⊆ F we clearly have [F [G] : F] ≤ 8 and hence [D : F] = 4 because G is irreducible. Therefore, D is cyclic and so the result follows for this case. It remains to consider the case where G is soluble and q = 2. By Lemma 3 of [5], we know that G is abelianby-finite, i.e., there is an abelian normal subgroup A in G of finite index. Take A maximal in G, and set K = F (A). One may easily show that G ⊆ N_{D^{∗}}(A) and that G_{1} = K^{∗}G is an irreducible soluble subgroup of D^{∗} with maximal abelian normal subgroup K^{∗}. Set H = C_{G}(A), H_{1} = C_{G_{1}}(K^{∗}). It is clearly seen that H_{1} = HK^{∗}. Since elements of K^{∗} and H pairwise commute we conclude that H′_{1} = H′. Now, by Lemma B, K/F is Galois with G_{1}/H_{1} ≅ Gal(K/F). Therefore, G_{1}/H_{1} is a 2-group and hence it is nilpotent. Now, one may easily show that G ∩ H_{1} = H and G_{1} = GK^{∗} = GH_{1}. Thus, we have G_{1}/H_{1} ≅ G/H and hence G/H is a 2-group. If H′_{1} is abelian, then as in the proof of Lemma 3, one may easily show that H_{1}/K^{∗} is a 2-group. Now, since A = H ∩ K^{∗} and H_{1} = HK^{∗} we conclude that H/A is a 2-group. This means that G/A is also a 2-group and hence G is abelian-by-nilpotent. Therefore, by Theorem C, we conclude that D is crossed product. Thus, we may assume that H′_{1} = H′ is nonabelian. Let l(H) = t be the derived length of H. As in the proof of Lemma 3, one may easily show that H^{t−2} is a nonabelian 2-group and it is isomorphic to the quaternion group Q_{8}. This means that H contains a normal subgroup isomorphic to Q_{8}. Now, assume that T = O_{2}(H) is a maximal normal 2-subgroup of H. As in Lemma 3, it is easily seen that O_{2}(H) ≃ Q_{8}. Now, by a result of [9, p.54], we have either H ≃ Q_{8} × M, where M is a group of odd order, or H ≃ SL_{2}(Z_{3}) × M, where M is a group of order m coprime to 6, or H is isomorphic to the binary octahedral group. We deal with these cases separately as follows:

Case 1. H ≃ Q_{8} × M, where M is a group of odd order. We claim that M is normal in G. Since H is normal in G for each g ∈ G and m ∈ M we have gmg^{−1} = (q, m_{1}) ∈ H. Comparing the orders of both sides of the last relation, one may easily conclude that q = 1 and so the claim is established. Now, we show that M is abelian. Otherwise, M′ is nontrivial. If l is the soluble length of M, then l ≥ 2. Thus, M^{l−1} ⊆ M′ is a nontrivial abelian subgroup. This implies that AM^{l−1} is an abelian normal subgroup of G and hence by the choice of A we obtain M^{l−1} ⊆ A. By Lemma B, we know that F [H_{1}] = C_{D}(K). Since M ⊆ H ⊆ H_{1} we obtain M′ ⊆ C_{D}(K)′. Take an element x ∈ M^{l−1} ⊆ A ⊆ K^{∗}. We have 1 = RN_{C_{D}(K)/K}(x) = x^{2^{s}}, where i(C_{D}(K)) = 2^{s}. This shows that the order of x is a power of 2 which contradicts the fact that M has odd order. Hence M′ must be trivial and so M is abelian. It is clear that H/M ≃ Q_{8} and G_{1}/H_{1} ≃ G/H is a 2-group. Since M is normal in G we conclude that G/M is also a 2-group. This says that G is abelian-by-nilpotent and hence, by Theorem C, D is a crossed product.

Case 2. H ≃ SL_{2}(Z_{3}) × M. Since the order of M is prime to 6 and |SL_{2}(Z_{3})| = 24, as in the case 1, we conclude that M is an abelian normal subgroup of G. Now, M as an abelian normal subgroup of D^{∗} is cyclic. Set M =< m > such that for each natural number s with (s, 6) = 1 we have m^{s} = 1. Since SL_{2}(Z_{3}) ⊆ G we have 2||G| and hence there exists g ∈ G such that g^{2} = 1, i.e., −1 ∈ G.

If m ∈ F^{∗}, then m ∈ Z(G). Therefore, 1, m, ⋯, m^{s−1}, −1, −m, ⋯, −m^{s−1} are distinct elements of Z(G) for if m^{i} = −m^{j} with 0 ≤ i, j ≤ s − 1, then raising to the power of s we obtain 1 = −1 which is a contradiction to the fact that CharF = 0. Thus, |Z(G)| > 2s. Now, G as an irreducible subgroup of D^{∗} contains a basis g_{1}, g_{2}, ⋯, g_{t} with t = 2^{2r}. Since g_{1}, g_{2}, ⋯, g_{t} are linearly independent over F we conclude that g_{1}Z(G), g_{2}Z(G), ⋯, g_{t}Z(G) are distinct elements of G/Z(G) and hence |G/Z(G)| ≥ t. Therefore, we have |G| ≥ 2^{2r} × 2s. On the other hand, we have |M| = s and so |H| = 2^{3} × 3 × s and also G/H ≃ Gal(K/F), where K/F is Galois. If [K : F] = 2^{r}, then K is a maximal subfield of D and hence D is crossed product. So, we may assume that [K : F] ≤ 2^{r−1}. In this case we obtain |G| ≤ 2^{3} × 3 × s × 2^{r−1}. Therefore, 2^{2r+1} × s ≤ 2^{r+2} × 3 × s which implies that 2^{r−1} ≤ 3, i.e., r = 1 or r = 2. If r = 1, then it is clear that D is cyclic. If r = 2, then, by a result of [7, p. 183], D is a crossed product.

If m ∉ F^{∗}, then MA is an abelian normal subgroup of G. By maximality of A, we conclude that M ⊆ A ⊆ K^{∗}. Since m is not in F we obtain |Gal(K/F)| = 2^{u} with u ≥ 1. Since Z(H) is an abelian normal subgroup of G, by maximality of A , we have Z(H) = A. Therefore, A =< −1 > ×M and hence |A| = 2s. Since Q_{8} is normal in G let O_{2}(G) = Q_{2^{l}}. It is clearly seen that (Q_{2^{l}})^{2} =< x^{2} >, where Q_{2^{l}} =< x, y|x^{2^{l−1}} = y^{4} = 1, yxy^{−1} = x^{−1} >. Now, one may easily show that N =< x^{2} > is normal in G. Since the orders of M and N are coprime we have M ∩ N = 1. Therefore, each element of M commutes with each element of N, i.e., MN is abelian. Since −1 ∈ N we obtain A ⊆ MN. But this contradicts the choice of A unless < x^{2} >=< −1 >, i.e., x^{4} = 1 and l = 3. Thus, O_{2}(G) = Q_{8}. Now, by a result of [9, p.54] again we have three subcases to consider as follows:

Subcase 1. G ≃ Q_{8} × M_{1}, where the order of M_{1} is odd. If |M_{1}| = 2n + 1, then |G| = 2^{3} × (2n + 1). Now, we have |H| = 2^{3} × 3 × s, where s is odd, and |G/H| = |Gal(K/F)| = 2^{u} with K ≠ F. Therefore, (2n + 1) = 3s × 2^{u} which is not possible.

Subcase 2. G ≃ SL_{2}(Z_{3}) × M_{1}, where the order of M_{1} is prime to 6. Since the order of M_{1} is odd, as in the Subcase 1, we obtain a contradiction.

Subcase 3. G is isomorphic to the binary octahedral group of 48 elements. Then, |G| = 2^{4} × 3. Since −1 ∈ Z(G) we obtain |Z(G)| ≥ 2. As before, because G is irreducible we have |G/Z(G)| ≥ 2^{2r}. Therefore, 2^{2r+1} ≤ 2^{4} × 3. This means that either r = 1 or r = 2, and as above we conclude that D is a crossed product.

Case 3. H is isomorphic to the binary octahedral group of 48 elements. Then, |H| = 2^{4} × 3. As in the Subcase 3, we conclude that |G| ≥ 2^{2r+1}. In addition, as in the previous cases, we have |G/H| < 2^{r−1}, and hence |G| ≤ 2^{r−1} × 2^{4} × 3, i.e., 2^{2r+1} ≤ 2^{r+3} × 3. This implies that either r = 1 or r = 2 or r = 3. For the cases r = 1 or r = 2, as before, we conclude that D is crossed product. Assume that r = 3. If D is not a crossed product, then, by Lemma 3, D^{∗} contains a copy of Q_{8}. It is clear that [F [Q_{8}] : F] = 4. Set B = F [Q_{8}]. Then, by Centralizer Theorem, we have D ≃ B ⊗ C_{D}(B). Since i(C_{D}(B)) = 4, by a result of [7, p. 183], C_{D}(B) is a crossed product. Therefore, D which is a tensor product of crossed products is a crossed product division algebra. This completes the proof of the theorem.

The authors thank the Research Council of Sharif University of Technology for support.

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