Supersoluble Crossed Product Criterion for Division Algebras

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. A criterion is given for D to be a supersoluble (nilpotent) crossed product division algebra in terms of subgroups of the multiplicative group D^{∗} of D. More precisely, it is shown that D is supersoluble (nilpotent) crossed product if and only if D^{∗} contains an irreducible abelian-by-supersoluble (nilpotent) subgroup.

1 Introduction

Let D be a division algebra with center F and degree n, (i.e. dim_{F} D = n^{2}). The algebra D is called crossed product if it contains a maximal subfield K such that K/F is Galois. D is said to be supersoluble crossed product if Gal(K/F) is supersoluble. We also recall that a subgroup G of D^{∗} is irreducible if F [G] = D. When n = p, a prime, it is shown in [1] that D is cyclic if and only if D^{∗} contains a nonabelian soluble subgroup. Here we generalize this result to a division algebra of arbitrary degree n. To be more precise, it is proved that D is supersoluble crossed product if and only if D^{∗} contains an irreducible abelianby-supersoluble subgroup. We then present a criterion for D to be nilpotent (abelian or cyclic) crossed product. In fact, it is shown that a noncommutative finite dimensional F -central division algebra D is nilpotent (abelian or cyclic) 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 (abelian or cyclic). We recall that soluble subgroups of the multiplicative group of a division ring were first studied by Suprunenko in [4].

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 (supersoluble) if there is an abelian normal subgroup K of H such that H/K is finite (supersoluble). 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 [3].

3 Supersoluble Crossed product Division Al-

gebras

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 a division algebra D is supersoluble crossed product if and only if D^{∗} contains an irreducible abelian-by-supersoluble subgroup. Further criteria are also given for when a division algebra is nilpotent (abelian or cyclic) crossed product. To be more precise, it is shown that a noncommutative finite dimensional F -central division algebra D is nilpotent ( abelian or cyclic) 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 (abelian or cyclic). we begin our study with the following:

Lemma 3.1. Let D be a finite dimensional F -central division algebra. If D is crossed product, then D^{∗} contains an irreducible abelian-by-finite subgroup.

Proof. Let K be a maximal subfield of D such that K/F is 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 an abelian-by-finite 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. □

Lemma 3.2. 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).

Proof. Consider the homomorphism σ : G ⟶ Gal(K/F) given by σ(x) = f_{x}, where f_{x}(k) = xkx^{−1}, for any k ∈ K. It is clear that ker σ = C_{G}(K). Now, we claim that F ix(imσ) = F. Choose an element a ∈ F ix(imσ). For any x ∈ G we have f_{x}(a) = a, and hence xa = ax. This shows that a ∈ C_{K}(G) = F since G is irreducible. So F ix(Gal(K/F)) ⊆ F ix(imσ) = F, which implies that K/F is a Galois extension and σ is surjective. Therefore we have G/C_{G}(K) ≃ Gal(K/F) and K/F is a Galois extension. □

Lemma 3.3. 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^{∗})].

Proof. Put D_{1} = C_{D}(K) and D_{2} = F [C_{G}(K^{∗})]. It is clear that D_{2} ⊆ D_{1}. By Centralizer Theorem, we have D ⊗_{F} K ≃ M_{t}(F)⊗_{F} C_{D}(K), where t = [K : F]. Therefore, by comparing dimensions of both sides of the last relation, we obtain [D_{1} : F][K : F] = [D : F] = [D : D_{1}][D_{1} : F]. Hence [D : D_{1}] = [K : F]. On the other hand any element of D can be written in the form ∑_{i=1}^{s} f_{i}g_{i}, where g_{i} ∈ G and f_{i} ∈ F, for any 1 ≤ i ≤ s. Now, let ℓ = [G : C_{G}(K^{∗})] and G = ⋃_{i=1}^{ℓ} C_{G}(K^{∗})x_{i}. Therefore, every element of D can be written in the form ∑_{i=1}^{ℓ} a_{i}x_{i}, where a_{i} ∈ D_{2}, for any 1 ≤ i ≤ ℓ. So [D : D_{2}] ≤ ℓ and we obtain [D : D_{1}] ≤ [D : D_{2}] ≤ [G : C_{G}(K^{∗})] = [K : F] = [D : D_{1}], and so D_{1} = D_{2}, as desired. □

The following theorem provides us a criterion for an F -central division algebra to be supersoluble crossed product:

Theorem 3.4. Let D be a noncommutative finite dimensional F -central division algebra. Then D is supersoluble 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 supersoluble.

Proof. The ”only if” part is clear from the proof of Lemma 3.1. Suppose that G is an irreducible subgroup of D^{∗} and A is an abelian normal subgroup of G such that G/A is supersoluble. Take A maximal. Therefore, we have a maximal abelian normal subgroup A of G such that G/A is supersoluble. Set G_{1} = K^{∗}G, where K = F (A). One may easily show that G_{1} is irreducible and K^{∗} is a maximal normal abelian subgroup of G_{1} such that G_{1}/K^{∗} is supersoluble. By Lemma 3.2, we conclude that K/F is Galois and G_{1}/C_{G_{1}}(K^{∗}) ≃ Gal(K/F). Therefore, K/F is supersoluble Galois. To complete the proof, it is enough to show that K is a maximal subfield of D. We know that G_{1}/K^{∗} is supersoluble, and hence there exists a normal series ⟨e⟩ = N_{t} ⊆ … N_{1} ⊆ N_{0} = G_{1}/K^{∗} such that N_{i}/N_{i+1} is cyclic. Set N = C_{G}(K^{∗})/K^{∗}. It is clearly seen that N is a normal subgroup of G_{1}/K^{∗}. We observe that there exists a minimal natural number s such that N ∩ N_{s} is nontrivial and so N ∩ N_{s+1} = ⟨e⟩. Therefore, we conclude that N ∩ N_{s} is a cyclic normal subgroup of G_{1}/K^{∗}. Thus, there exists x ∈ C_{G_{1}}(K^{∗})\K^{∗} such that ⟨K^{∗}, x⟩/K^{∗} is a normal subgroup of G_{1}/K^{∗}, and hence ⟨K^{∗}, x⟩ ≠ K^{∗} is an abelian normal subgroup of G_{1}. This contradicts the maximality of K^{∗} in G_{1}. Therefore, C_{G_{1}}(K^{∗}) = K^{∗} and the claim is established. Now, by Lemma 3.3, we obtain F [C_{G_{1}}(K^{∗})] = C_{D}(K), and hence C_{D}(K) = K. Thus, K is a maximal subfield of D and the proof is complete. □

The following theorem provides us a criterion for an F -central division algebra to be nilpotent crossed product:

Theorem 3.5. Let D be a noncommutative finite dimensional F -central division algebra. Then D is 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.

Proof. The ”only if” part is clear from the proof of Lemma 3.1. As in the proof of Theorem 3.4 D^{∗} contains an irreducible subgroup G_{1} and a maximal normal abelian subgroup K^{∗} such that G_{1}/K^{∗} is nilpotent. By Lemma 3.2, we conclude that K/F is Galois and G_{1}/C_{G_{1}}(K^{∗}) ≃ Gal(K/F). Therefore, K/F is nilpotent Galois. To complete the proof, it is enough to show that K is a maximal subfield of D. First, we claim that C_{G_{1}}(K^{∗}) = K^{∗}. Otherwise, since it is known that if G is a nilpotent group and 1 ≠ N ◁ G, then Z(G) ⋂ N ≠ 1, there is an element K^{∗} ≠ xK^{∗} ∈ C_{G_{1}}(K^{∗})/K^{∗} ⋂ Z(G_{1}/K^{∗}). This implies that ⟨K^{∗}, x⟩/K^{∗} is a normal subgroup of G_{1}/K^{∗}, and hence ⟨K^{∗}, x⟩ ≠ K^{∗} is an abelian normal subgroup of G_{1}. This contradicts the maximality of K^{∗} in G_{1}. Thus, C_{G_{1}}(K^{∗}) = K^{∗} and the claim is established. Now, by Lemma 3.3, we obtain F [C_{G_{1}}(K^{∗})] = C_{D}(K), and hence C_{D}(K) = K. Thus, K is a maximal subfield of D, and the proof is complete. □

Corollary 3.6. Let D be a noncommutative finite dimensional division algebra. If D^{∗} contains an irreducible locally nilpotent subgroup G, then D is nilpotent crossed product.

Proof. To prove the corollary, it is enough to choose a basis for the vector space D over Z(D) from G, then consider the group generated by this basis and use Theorem 3.5. □

The multiplicative group of the real quaternion division algebra contains the quaternion group which is an irreducible 2-group. Therefore, by Corollary 3.5, it is nilpotent and even cyclic. The following result says that if the multiplicative group of a noncommutative division algebra D contains an irreducible p-group, then it is nilpotent crossed product with p = 2 and [D : F] = 2^{m} for some m ∈ ℕ.

Corollary 3.7. Let D be a noncommutative finite dimensional F -central division algebra. If D^{∗} contains an irreducible p-subgroup, then D is nilpotent crossed product with [D : F] = 2^{m}, for some m ∈ ℕ.

Proof. Let G be an irreducible p-subgroup of D^{∗}. Since G is locally nilpotent, by Corollary 3.6, we conclude that D is nilpotent crossed product. If p is odd, then by a theorem of [3, p. 45], G is abelian which contradicts the irreduciblity of G . If p = 2, then by the proof of Theorem 3.4, there exists a maximal subfield K of D such that GK^{∗}/K^{∗} ≃ Gal(K/F) and K/F is a Galois extension. Hence [K : F] is a power of 2 and the result follows. □

The next result gives us a criterion for when an F -central division algebra is abelian crossed product.

Theorem 3.8. 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.

Proof. The ”only if” part is clear from the proof of Lemma 3.1. Suppose that G is an irreducible subgroup of D^{∗} and A is an abelian normal subgroup of G such that G/A is abelian. Take A maximal and put K = F (A). Consider the group G_{1} = K^{∗}G. One can easily show that K^{∗} is a maximal abelian normal subgroup of G_{1} such that G_{1}/K^{∗} is abelian. Therefore, for the derived subgroup G′_{1} of G_{1} we have G1^{′} ⊆ K^{∗}. We claim that C_{G_{1}}(K^{∗}) = K^{∗}. On the contrary, if x ∈ C_{G_{1}}(K^{∗})\K^{∗}, then ⟨K^{∗}, x⟩ is an abelian subgroup of G. Since G1^{′} ⊆ ⟨K^{∗}, x⟩ we conclude that ⟨K^{∗}, x⟩ is also normal in G_{1}. This contradicts the maximality of K^{∗} in G_{1}, and hence the claim is established, i.e., C_{G_{1}}(K^{∗}) = K^{∗} . Now, by Lemma 3.2, we conclude that K/F is Galois and G_{1}/K^{∗} ≃ Gal(K/F) and so K/F is abelian Galois. To complete the proof, it is enough to show that K is a maximal subfield of D. To see this, by Lemma 3.3, we have F [C_{G_{1}}(K^{∗})] = C_{D}(K) which means that C_{D}(K) = K. Hence K is a maximal subfield of D and so the proof is complete. □

Let D be an F -central division algebra of prime degree. Suppose that D^{∗} contains an irreducible soluble subgroup. Using the fact that the degree of D is prime one may easily conclude that D^{∗} contains an irreducible metabelian subgroup. Now using Theorem 3.8, we obtain the following corollary, which is the main result of [1].

Corollary 3.9. 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.

Finally, we present a criterion for a finite dimensional F -central division algebra to be cyclic.

Theorem 3.10. A noncommutative finite dimensional F -central division algebra D is cyclic 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 cyclic.

Proof. The ”only if” part is clear from the proof of Lemma 3.1. So, assume that G is an irreducible subgroup of D^{∗} and A is an abelian normal subgroup of G such that G/A is cyclic. Put K = F (A) and consider the isomorphism K^{∗}G/K^{∗} ≃ G/K^{∗} ∩ G. This implies that K^{∗}G/K^{∗} is a cyclic group. Set H = K^{∗}G. Obviously H is an irreducible subgroup of D^{∗} and K^{∗} ◁ H. Therefore, by Lemma 3.2, we conclude that K/F is Galois and H/C_{H}(K^{∗}) ≃ Gal(K/F). Now, since H/K^{∗} is cyclic and K^{∗} ⊆ C_{H}(K^{∗}) we conclude that K/F is cyclic. To complete the proof of the theorem, it is enough to show that K is a maximal subfield of D. We know that C_{H}(K^{∗})/K^{∗} is cyclic and K^{∗} ⊆ Z(C_{H}(K^{∗})). Thus, we conclude that C_{H}(K^{∗})/Z(C_{H}(K^{∗})) is cyclic, i.e., C_{H}(K^{∗}) is abelian. Now, by Lemma 3.3, we obtain F [C_{H}(K^{∗})] = C_{D}(K) and hence C_{D}(K) is a field. Thus, K is a maximal subfield of D, and the proof is complete. □

The following example shows that working with certain subgroups of D^{∗} may be sometimes more useful than maximal subfields.

Example 1. Let L/K be a cyclic field extension of degree n, and denote by σ the generator of Gal(L/K). Let D = L((T, σ)) be the division algebra of formal Laurent series. Although it is not hard to show that L((T^{n}))/K((T^{n})) is a cyclic extension and therefore D is a cyclic division algebra, we would like to show its cyclicity by using our criterion. It is known that Z(D) = K((T^{n})). If 1, t, …, t^{n−1} is a basis for the field extension L/K, then {t^{j}T^{i}}_{0≤i,j<n} is a basis for D over Z(D). Therefore, the group G = ⟨t^{j}T^{i}⟩_{0≤i,j<n} is an irreducible subgroup of D^{∗}. On the other hand, one can easily show that L ∩ G ◁ G and that G/L ∩ G is a cyclic group. Therefore G is an irreducible subgroup of D^{∗} which is abelian-by-cyclic. Thus, by Theorem 3.10, D is a cyclic division algebra. □

The rest of this section is devoted to the observation that one may not be able in general to replace ”division algebra” by ”central simple algebra” in the statements of some theorems above. First, we observe the following:

Theorem 3.11. Let D be an F -central crossed product division algebra, then for any natural number n, GL_{n}(D) contains an irreducible abelian-by-finite subgroup.

Proof. By Lemma 3.1, D^{∗} contains an irreducible abelian-by-finite subgroup G, say. We may view each element of D as a diagonal matrix in M_{n}(D). Suppose that M is the group of monomial matrices of GL_{n}(F) and E the group of diagonal matrices of GL_{n}(F). Set H = ⟨G, M⟩. We claim that H as a subgroup of GL_{n}(D) is irreducible. One can easily show that F [M] = M_{n}(F). Combining this equality with the fact that F [G] = D we conclude that F [H] = M_{n}(D). To complete the proof of the claim, it is enough to show that H is abelian-by-finite. It is easily seen that M/E ≃ S_{n}. Suppose that G/A is finite, where A is an abelian normal subgroup of G. Considering the fact that every element of G commutes with every elements of M we obtain H = GM and that AE is an abelian group. Using these facts, one may easily show that H/AE as a homomorphic image of G/A × M/E is a finite group. Therefore H is an irreducible abelian-by-finite subgroup of GL_{n}(D), as desired. □

We close the paper with the following remark which shows that in Theorems 3.8, and 3.10, we can not replace ”division algebra” by ”central simple algebra”.

Remark. Suppose that D is a crossed product division algebra and n a natural number. Using the notations of the proof of Theorem 3.10, we have M/E ≃ S_{n}. Therefore, M is an abelian-by-finite group. Hence, H is an irreducible abelian-by-finite subgroup of GL_{n}(D). Thus, if D is an abelian crossed product (cyclic) division algebra, then GL_{2}(D) contains an irreducible subgroup which is abelian-by-abelian (abelian-by-cyclic). Furthermore, if F is a field with more than 2 elements by applying to the group M, the group of monomial matrices, we may find an irreducible subgroup of GL_{2}(F) which is abelian-by-cyclic.

Example 2. We observe that there are some cyclic division algebras D, e.g., the division algebra of real quaternions ℍ, such that M_{n}(ℍ) is not crossed product for any n > 1. In fact it contains no subfield containing ℝ which is of degree 2n over ℝ. But for each natural number n, GL_{n}(ℍ) contains an irreducible abelian-by-finite subgroup, and it also contains an irreducible abelian-by-cyclic subgroup for n = 2 but it is not crossed product. Therefore, Theorems 3.8 and 3.10 are not valid if one replaces division algebras by central simple algebras.

The authors thank the Research Council of Sharif University of Technology for support. They are also indebted to Professor M. Lorenz for his constructive communications.

References

[1] M. Mahdavi-Hezavehi and J. P. Tignol, Cyclicity conditions for division algebras of prime degree, Proc. Amer. Math. Soc., to appear.

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

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

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