ON THE NON-TRIVIALITY OF G(D) AND THE EXISTENCE OF MAXIMAL SUBGROUPS OF GL_{1}(D)

T. KESHAVARZIPOUR & M. MAHDAVI-HEZAVEHI

Abstract. Let D be an F -central division algebra of index n. Here we investigate a conjecture posed in [4] that if D is not a quaternion algebra, then the group G_{0}(D) = D^{∗}/F^{∗}D′ is nontrivial. Assume that either D is cyclic or F contains a primitive p-th root of unity for some prime p|n. Using Merkurjev-Suslin Theorem, it is essentially shown that if none of the primary components of D is a quaternion algebra, then G(D) = D^{∗}/RN_{D/F} (D^{∗})D′ ≠ 1. In this direction, we also study a conjecture posed in [1] or also [7] on the existence of maximal subgroups of D^{∗}. It is shown that if D is not a quaternion algebra with i(D) = p^{e}, then D^{∗} has a maximal subgroup if either of the following conditions holds: (i) F has characteristic zero, or (ii) F has characteristic p, or (iii) F contains a primitive p-th root of unity.

Let D be an F -central division algebra of index n. Denote by D′ the commutator subgroup of the multiplicative group D^{∗}. Given a subgroup G of D^{∗}, we shall say that G is maximal in D^{∗} if for any subgroup H of D^{∗} with G ⊂ H, one concludes that H = D^{∗}. We know, by Corollary 1 of [8], that G(D) := D^{∗}/RN(D^{∗})D′, where RN(D^{∗}) is the image of D^{∗} under the reduced norm of D to F, is an abelian torsion group of a bounded exponent dividing the index of D over F. This group is not trivial in general. For instance, if D is the algebra of real quaternions, then G(D) is trivial whereas for rational quaternions G(D) is isomorphic to a direct product of copies of Z_{2}, as it is easily checked. Assume that G(D) is not trivial, then by Prũfer-Baer Theorem (cf. [14], p. 105), we conclude that G(D) is isomorphic to a direct product of Z_{r_{i}}, where r_{i} divides the index of D over F. In this way, one may obtain normal maximal subgroups of finite index in D^{∗}. So, if G(D) is not trivial, then D^{∗} contains maximal subgroups. For some examples of non-normal maximal subgroups of D^{∗}, see [9]. It is shown in [9] that even for the case G(D) = 1, we may obtain maximal subgroups in D^{∗}. But, the question of whether D^{∗} contains a maximal subgroup for any noncommutative division ring D, is still open. In this note, we concentrate on the case where D is of finite dimension over its centre such that G(D) is trivial. When i(D) = p^{e}, p a prime, and G(D) = 1, it is shown in Theorem 1 and Theorem 3 that if either D is an F -central cyclic division algebra or F contains a primitive p-th root of unity, then D is a quaternion algebra. Also, in Proposition 1, it is proved that if one of the primary components of D is a p-algebra for some prime p|n, then G(D) ≠ 1. We then proceed to explore suitable conditions on D such that D^{∗} contains a maximal subgroup for an arbitrary division algebra of index n. It is essentially shown that when D is not a quaternion algebra with i(D) = p^{e}, then D^{∗} contains a maximal subgroup if either of the following conditions holds: (i) F has characteristic zero, or (ii) F has characteristic p, or (iii) F has a primitive p-th root of unity. We shall use the conventions and notations of [2] throughout. We begin our study with the following:

1991 Mathematics Subject Classification. 16K20.

Key words and phrases. division ring, maximal subgroup, splitting field.

Lemma 1. Let A be an F -central cyclic algebra of odd index n such that the skew field component of A is noncommutative. Then G_{0}(A) := A^{∗}/F^{∗}A′ ≠ 1, where A′ is the commutator subgroup of A^{∗}.

Proof. We know that A ≃ ⊕_{i=0}^{n−1}Ka^{i}, where K/F is cyclic of degree n with a^{n} = α ∈ F. Thus, a is a root of the minimal polynomial x^{n} − α. Now, we have RN_{A/F} (a) = (−1)^{n+1}α. Assume on the contrary that G_{0}(A) = 1. Then there exist f ∈ F^{∗} and c ∈ A′ such that a = fc. Hence, RN_{A/F} (a) = f^{n} and therefore, f^{n} = (−1)^{n+1}α. Since n is odd, we obtain f^{n} = α and hence α ∈ N_{K/F} (K^{∗}). But this, by Theorem 14.7 of [6], contradicts the assumption A ≄ M_{r}(F) for any r, and so the result follows. □

The next result deals with F -central cyclic division algebras of degree a power of 2 such that G_{0}(D) is trivial. It is shown that in this case our cyclic division algebra takes a particular simple form.

Lemma 2. Let D be an F -central cyclic division algebra of index n = 2^{m} such that G_{0}(D) = 1. Then we have the following:

(i) There is an element a ∈ D^{∗} and a maximal subfield K such that D ≃ ⊕_{i=0}^{n−1}Ka^{i} with a^{n} = −1; where K/F is cyclic, Gal(K/F) =< σ >, and ax = σ(x)a, for all x ∈ K.

(ii) The left K-space D_{1} generated by even powers of a, i.e., D_{1} := ⊕_{i=0}^{n/2−1}Ka^{2i} is a cyclic division algebra with maximal subfield K and center E such that [E : F] = 2.

Furthermore, (i) is valid for any F -central cyclic algebra A with index n = 2^{m} and G_{0}(A) = 1.

Proof. (i) Since D is cyclic we have the representation D ≃ ⊕_{i=0}^{n−1}Ka^{i} for some a ∈ D^{∗} with a^{n} = α ∈ F. To end the proof, we claim that it is possible to take α = −1. It is clearly seen that a is a root of the minimal polynomial x^{n} − α. Therefore, RN_{D/F} (a) = (−1)^{n+1}α. Since G_{0}(D) = 1 we have a = fc for some f ∈ F^{∗} and c ∈ D′. Thus RN_{D/F} (a) = f^{n} and hence f^{n} = (−1)^{n+1}α. Since n is even we conclude that f^{n} = −α and so a^{n} = −f^{n}, i.e, (af^{−1})^{n} = −1. Therefore, we may replace a by af^{−1} to obtain the result. (ii) It is easily seen that the left K-space D_{1} is closed under addition and multiplication and so D_{1} is a ring. We claim that D_{1} is a division algebra. To see this, let x ∈ D_{1}. Then x^{−1} as an element of D, has the form x^{−1} = y + z where y ∈ D_{1} and the powers of a occurring in z are all odd. Therefore, xx^{−1} = x(y + z) = xy + xz = 1. Since xz = 1 − xy ∈ D_{1}, and the powers of a occurring in xz are odd, we conclude that xz = 0. i.e., x^{−1} ∈ D_{1} and the claim is established. It is now clear that K is a maximal subfield of D_{1}. For dimensional reasons we conclude that Z(D_{1}) = E ⊂ K such that [E : F] = 2. Therefore, we obtain D_{1} ≃ (−1, K/E, σ^{2}). Note that our Galois group here is Γ = {σ^{2}, σ^{4}, ⋯ }. □

In the next lemma, we show that for any F -central cyclic division algebra D of index a power of 2, the condition G_{0}(D) = 1 implies that D is a quaternion algebra.

Lemma 3. Let D be an F -central cyclic division algebra of index n = 2^{m}. If G_{0}(D) = 1, then D is a quaternion algebra.

Proof. By Lemma 2, we may assume that D ≃ ⊕_{i=0}^{n−1}Ka^{i} with a^{n} = −1, where K/F is cyclic of degree n and for all x ∈ K, ax = σ(x)a with Gal(K/F) =< σ >. Thus, the characteristic of F is different from 2. Let D_{1} be the division subalgebra generated by the even powers of a. By Lemma 2, D_{1} is a cyclic division algebra with center E such that [E : F] = 2. It is clear that we have D = D_{1} ⊕ D_{1}a. If D_{1} is commutative, then we obtain Z(D_{1}) = D_{1} = K = E and so m = 1, which means that D is a quaternion division algebra. We now claim that D_{1} = E. i.e., n > 2 leads to a contradiction. To see this, set k = n/2 ≠ 1. Therefore, a^{k} ∈ D_{1} \ E, and so E and consequently F contains no square root of −1. Now, since G_{0}(D) = 1, for any x ∈ D^{∗} we have x = fc, for some f ∈ F^{∗} and c ∈ D′. By Skolem-Noether Theorem, we know that σ is inner. Thus, σ(x) = fσ(c) = fdcd^{−1} for some d ∈ D^{∗}. Hence, xσ(x) ∈ F^{∗}^{2}D′ for all x ∈ K^{∗}. Since CharF ≠ 2 and E/F is Galois of degree 2, we have N_{E/F} (−1) = 1. Therefore, by Hilbert’s ”Satz90”, there is an element b ∈ E such that bσ|_{E}(b)^{−1} = −1, where σ|_{E} is the restriction of σ to E. We also have bσ(b) ∈ F^{∗}^{2}D′. Hence b^{2} ∈ −F^{∗}^{2}D′, i.e., there is an element c ∈ D′ and a_{1} ∈ F^{∗} such that b^{2} = −a_{1}^{2}c. This implies that −a_{1}^{−2}b^{2} = c ∈ Z(D′) = F^{∗} ∩ D′ since b^{2} ∈ F^{∗}. Now, since F contains no square root of −1 and, by a result of [11], Z(D′) is a finite group of order dividing i(D) = 2^{m}, we conclude that Z(D′) = {−1, 1}. Therefore, we have either c = 1 or c = −1. If c = −1, then b^{2} = a_{1}^{2} and so b ∈ F. Now, from bσ(b)^{−1} = −1 we conclude that charF = 2 which is a contradiction. Thus, c = 1 and we obtain b^{2} = −a_{1}^{2}, i.e., (ba_{1}^{−1})^{2} = −1. This implies that E has a square root of -1, that is a contradiction. So we have k = 1, i.e, D_{1} = K = E and so the result follows. □

We are now able to prove one of our main results in the form of

Theorem 1. Let D be an F -central cyclic division algebra such that G_{0}(D) = 1, then D is a quaternion algebra.

Proof. By Corollary 15.3 of [13], we know that a central division algebra is cyclic if and only if its primary components are cyclic. Thus, if D ≃ ⊗_{i=1}^{k}D_{i} is the primary decomposition of D, then D_{i} is cyclic division algebra for each i. Now, by a result of [3], we know that G_{0}(D) ≃ G_{0}(D_{1}) × ⋯ × G_{0}(D_{k}). Hence, G_{0}(D_{i}) = 1 for all 1 ≤ i ≤ k. Finally, use Lemma 1 and Lemma 3 to obtain the result. □

To prove our next theorem we shall need the following:

Lemma 4. Let D be an F -central p-division algebra of index p^{e}, p a prime. Then D has a cyclic splitting field of degree p^{te}, for some positive integer t.

Proof. By Theorem 15.4 of [2], there are cyclic extensions L_{1}, …, L_{r} of degrees p^{e_{i}} over F and also elements a_{1}, …, a_{r} ∈ F^{∗} such that [D] = Σ_{i=1}^{r}[a_{i}, L_{i}/F, σ_{i}], where Gal(L_{i}/F) =< σ_{i} >. Set A_{i} := (a_{i}, L_{i}/F, σ_{i}). By Theorem 4.5.1 of [5], since the tensor product of A_{i}’s is also a cyclic p-algebra, we have ⊗_{i=1}^{r}(a_{i}, L_{i}/F, σ_{i}) = (a, L/F, σ) for some cyclic extension L/F. Hence, [L : F] = p^{s} for some integer s. Therefore, L is a cyclic splitting field for D of degree a power of p. Now, by a repeated use of Lemma 15.2 of [2], L can be chosen as a cyclic splitting field for D of degree p^{te} for some positive integer t. □

The next result essentially says that the multiplicative group of every F -central division p-algebra contains a normal maximal subgroup.

Theorem 2.Let D be an F -central division p-algebra of index p^{e}. Then we have G(D) ≠ 1.

Proof. Assume on the contrary that G(D) = 1. By Lemma 4, we may choose a cyclic splitting field E for D such that [E : F] = p^{te} for some integer t. By Theorem 9.7 of [2], we can find an F -central cyclic algebra A such that E is a maximal subfield in A and also [A] = [D]. Consequently, A = M_{m}(D), where m = p^{(t−1)e}. Now, we claim that G(A) = 1. To prove this, by a theorem in [10], we know that G(A) = D^{∗}/RN(D)^{m}D′. Now, since G(D) = 1 we have D^{∗} = RN(D^{∗})D′. By taking reduced norm of both sides of the last relation we obtain RN(D^{∗}) = RN(D^{∗})^{p^{e}} and hence RN(D^{∗}) = RN(D^{∗})^{m}, i.e., G(A) = G(D) = 1, which establishes our claim. Thus, G_{0}(A) = 1. Now, by Lemma 1, we conclude that p = 2. Therefore, by Lemma 2, A can be written in the form A = (a, E/F, −1). Since −1 = 1, by Theorem 14.7 of [6], we will obtain the contradiction A ≃ M_{s}(F) and so the result follows. □

We shall need the following two lemmas to prove our next theorem.

Lemma 5. Let D be an F -central division algebra of index p^{e} such that F contains a primitive p-th root of unity and D has no non-cyclic Galois splitting field of degree a power of p over F. Then we have:

(i) If p = 2, then either D has a cyclic splitting field E of degree 2^{te} for some integer t such that −1 ∈ N_{E/F} (E^{∗}) or D has a cyclic splitting field E such that E is the splitting field of a minimal polynomial of the form x^{[E:F]} + 1 and F ⊆ E^{[E:F]}.

(ii) If p ≠ 2, then D has a cyclic splitting field of degree p^{te} for some positive integer t.

Proof. Since F has a primitive p-th root of unity we have (p, charF) = 1. Set L := F (ξ), where ξ is a primitive p^{e}-th root of unity and consider the L-algebra D ⊗_{F} L. By Theorem 17.1 of [2] which is a consequence of the Merkurjev-Suslin Theorem, D ⊗_{F} L has an abelian splitting field of the form K_{0} := L( \sqrt[p^{e}]{a_{1}}, …, \sqrt[p^{e}]{a_{t}}), for some a_{i} ∈ L. View L as a maximal subfield in M_{m}(F), where m := [L : F]. If σ_{i} ∈ Gal(L/F), by Skolem-Noether Theorem, there is an element A_{i} ∈ GL_{m}(F) such that

σ_{i}(x) = A_{i}xA_{i}^{−1} for all x ∈ L. Now, put E := L( \sqrt[p^{e}]{A_{i}a_{j}A_{i}^{−1}_{1}} : 1 ≤ j ≤ t, 1 ≤ i ≤ m). Since K_{0} ⊆ E, we conclude that E is a splitting field for D, and by Theorem 11.4 of [12], E/L is an abelian extension. We claim that |Gal(E/F)| = [E : F], i.e., E/F is also a Galois extension. To see this, for each i we may extend σ_{i} to E by the rule σ̄_{i}(x) = A_{i}xA_{i}^{−1}, for each x ∈ E, where A_{i}‘s and E may be viewed in M_{[E:F]}(F). We first show that σ̄_{i}(E) ⊆ E, which proves that σ̄_{i} is well defined. To see this, let α be a root of the polynomial x^{p^{e}} − A_{i′}a_{j}A_{i′}^{−1} in L[x]. Then, σ̄_{i}(α) = A_{i}αA_{i}^{−1} is also a root of x^{p^{e}} − A_{i}A_{i′}a_{j}A_{i′}^{−1}A_{i}^{−1}. Now, we have

A_{i}A_{i′}a_{j}A_{i′}^{−1}A_{i}^{−1} = σ_{i}σ_{i′}(a_{j}) = σ_{k}(a_{j}) = A_{k}a_{j}A_{k}^{−1},

for some A_{k} ∈ GL_{m}(F). This shows that σ̄_{i}(α) ∈ E, and hence σ̄_{i} ∈ Aut(E). Now, set G = {σ̄_{i}τ_{j} : σ_{i} ∈ Gal(L/F), τ_{j} ∈ Gal(E/L)}. It is clear that σ̄_{i}τ_{j} ∈ Gal(E/F) for all i, j. We claim that | G |= [E : F]. To see this, if for some i, i′, j, j′ we have σ̄_{i}τ_{j} = σ̄_{i′}τ_{j′}, then σ̄_{i} |_{L}= σ̄_{i′} |_{L} since τ_{j} |_{L}= τ_{j′} |_{L}. Hence, by Theorem 7.3 of [2], we obtain A_{i}A_{i′}^{−1} ∈ Z_{M_{m}(F)}(L) = L. Therefore, σ̄_{i} = σ̄_{i′} and hence τ_{j} = τ_{j′}, i.e, every two elements of G are distinct, and so the claim is established. Thus, E/F is a Galois extension of degree a power of p which is also cyclic by our assumption. We now show that F ⊆ E^{[E:F]}. To see this, we first claim that F ⊆ E^{p}. If b ∈ F \ E^{p}, since F contains a primitive p-th root of unity, then K = F (b^{1/p}) is a cyclic extension of degree p such that K ⊈ E. Therefore, E ⊗_{F} K is a non-cyclic Galois splitting field of degree a power of p over F that contradicts our assumption. So the claim is established. Now, consider the unique chain of all cyclic subfields in E: E_{0} = F ⊂ E_{1} ⊂ E_{2} ⊂ ⋯ ⊂ E_{k} = E. Because F ⊆ E^{p}, for each x ∈ F there exists y ∈ E such that x = y^{p}. From the uniqueness of the above chain we obtain F (y) = E_{1} or F (y) = F. This implies that F ⊆ E_{1}^{p}. Again, consider the skew field component of the E_{1}-central simple algebra D ⊗_{F} E_{1} with the same splitting field E. By taking b ∈ E_{1} \ E^{p} and using the same argument as above, we obtain E_{1} ⊆ E^{p} and hence E_{1} ⊆ E_{2}^{p}. Therefore, the repeated use of the argument implies that E_{i} ⊆ E_{i+1}^{p} and hence F ⊆ E^{[E:F]}, as required. Now, set Ω = {λ ∈ F : λ^{p^{r}} = 1, r ∈ ℕ}. We have τ ∈ Ω, where τ is a primitive p-th root of unity. Hence, Ω is a nontrivial group. If Ω is an infinite group, then τ ∈ N_{E/F} (E). Hence, by repeated use of Exercise 15.3 in [2], E can be extended to a cyclic extension of degree p^{te} for some t ∈ ℕ such that τ ∈ N_{E/F} (E) and the result follows. So assume that Ω is a finite cyclic group and consider ζ ≠ 1 as a generator of Ω. Since F ⊆ E^{[E:F]}, there exists η ∈ E such that η^{[E:F]} = ζ. If p^{s} is the minimum positive integer such that η^{p^{s}[E:F]} = ζ^{p^{s}} = τ, then η is a primitive p^{s+1}[E : F]-th root of unity. If not, we conclude that τ = 1, a contradiction. Now, we prove that E is a splitting field of the minimal polynomial x^{[E:F]} − ζ over F. To see this, take η_{0} = ζ and assume, by induction on i, that η_{i}, as a primitive p^{s+1+i}-th root of unity, be chosen such that E_{i} = E_{i−1}(η_{i}). Since E_{i} ⊆ E_{i+1}^{p}, there exists η_{i+1} ∈ E_{i+1} such that η_{i+1}^{p} = η_{i}. Hence, E_{i+1} = E_{i}(η_{i+1}), where η_{i+1} is a primitive p^{s+i+2}-th root of unity. Therefore, from our construction η, as a primitive p^{s+1}[E : F]-th root of unity, is not contained in E_{k−1}, i.e, F (η) = E. So, E is a splitting field of the minimal polynomial x^{[E:F]} − ζ over F. Now consider the following cases:

(i) If p ≠ 2, then N_{E/F} (η) = ζ and hence τ ∈ N_{E/F} (E). By Exercise 15.3 in [2], E can be extended to a cyclic extension E′ of degree p[E : F]. Now, by the repeated use of the construction above for E′ in place of E and using the fact that D has no noncyclic Galois splitting field of degree a power of p, we obtain a cyclic extension E of degree p^{te} for some integer t such that F ⊆ E^{[E:F]}.

(ii) If p = 2, suppose that ζ ≠ −1. Since −ζ = N_{E/F} (η) we have −1 ∈ N_{E/F} (E), and this is reduced to the above case. But, if ζ = −1, then we have a cyclic extension which is also the splitting field of the minimal polynomial x^{[E:F]} + 1 = 0, and also F ⊆ E^{[E:F]}.

□

Lemma 6. Let G be a finite non-cyclic p-group. Then G has at least two distinct normal subgroups of index p.

Proof. If G is an abelian group, then the conclusion is clear. So assume that G ≠ Z(G) and consider the group G/Z(G). From group theory we know that G/Z(G) is also a non-cyclic p-group. Now, use induction on the order of G to obtain the result. □

Now, we are able to prove the following interesting result.

Theorem 3. Let D be an F -central division algebra of index p^{e} such that F contains a primitive p-th root of unity and G(D) = 1. Then D is a quaternion algebra.

Proof. First assume that D has a non-cyclic Galois splitting field E of degree a power of p. Since G(D) = 1, by corollary 4.19 of [10], we have N(D^{∗}) = RN(D^{∗}), i.e., F^{∗}^{p^{e}} = F^{∗}^{p^{2e}}. By Lemma 6, G := Gal(E/F) has at least 2 distinct normal subgroups H_{1}, H_{2} of index p in G. If M_{1}, M_{2} are the fixed fields of H_{1}, H_{2} in E , respectively, then from Galois theory both M_{1}, M_{2} are cyclic extensions of degree p in E over F. Therefore, by Hilbert’s ”Satz90”, for i = 1, 2 there is b_{i} ∈ M_{i} such that b_{i}^{−1}σ_{i}(b_{i}) = τ, where Gal(M_{i}/F) =< σ_{i} >, and τ here is a primitive p-th root of unity in F. From the relation F^{∗}^{p^{e}} = F^{∗}^{p^{2e}}, since b_{i}^{p} ∈ F^{∗}, there are also c_{1}, c_{2} ∈ F^{∗} such that (b_{i}^{p})^{p^{e}} = c_{i}^{p^{2e}}, and hence (b_{i}^{p}(c_{i}^{−1})^{p^{e}})^{p^{e}} = 1. Let Ω denote the group of p^{e}-th roots of unity in F. Since b_{i} ∉ F, then b_{i}^{p}(c_{i}^{−1})^{p^{e}} for i = 1, 2 are generators of Ω. But, this is not possible since M_{1} ≠ M_{2}, and both M_{1}, M_{2} lie in E. Thus, we may assume that D has no non-cyclic Galois splitting field of degree a power of p. Now, by Lemma 5, we consider two following cases:

(i) If p ≠ 2, by Lemma 5, D has a cyclic splitting field E of degree p^{te} for some integer t. From the proof of Theorem 2 with m = p^{(t−1)e}, E can be embedded in the cyclic algebra A = M_{m}(D) as a maximal subfield such that G(A) = 1. But, by Lemma 1, we obtain M_{m}(D) = M_{r}(F) for some r ∈ ℕ, which is not possible.

(ii) If p = 2, by Lemma 5, suppose that D has a cyclic splitting field E of degree 2^{te} such that −1 ∈ N_{E/F} (E), then the cyclic algebra defined in (i), by Lemma 2, can be written in the form M_{m}(D) = ⊕_{i=0}^{[E:F]−1}Ea^{i} such that a^{[E:F]} = −1. But, −1 ∈ N_{E/F} (E). Therefore, by the proof of Lemma 14.7 of [6], we obtain M_{m}(D) = M_{r}(F) for some r ∈ ℕ, that contradicts our assumption. So, D has a cyclic splitting field E in which the minimal polynomial x^{[E:F]} + 1 splits. If η is an element in E such that its minimal polynomial over F is x^{[E:F]} + 1, then −η^{2^{k}} = N_{E/F} (η) = 1, where [E : F] = 2^{k}. On the other hand, since 1 + N_{E/F} (η) = N_{E/F} (η + 1) = RN_{M_{m}(D)/F} (η + 1) ∈ F^{2^{k}}, it follows that \sqrt{2} ∈ F. Thus, if k > 1, then η^{2^{k}} + 1 = (η^{2^{k−1}} + 1)^{2} − 2η^{2^{k−1}} can be decomposed further which leads to a contradiction that the minimal polynomial of η has degree less than [E : F]. Therefore, we have k = 1 which means that D is a quaternion algebra.

□

Finally, we shall need the following lemmas to prove our last result.

Lemma 7. Let D be an F -central division algebra of index p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}}. Suppose that D = D_{1} ⊗_{F} … ⊗_{F} D_{k} is the primary decomposition of D with i(D_{i}) = p_{i}^{e_{i}}. If G(D) = 1, then G(D_{i}) = 1 for all 1 ≤ i ≤ k.

Proof. It is enough to prove the result for the case D = A ⊗_{F} B, where A, B are two division algebras such that (i(A), i(B)) = 1 and also G(A ⊗_{F} B) = 1. Consider the following embeddings:

A →^{i} A ⊗_{F} B →^{i_{1}} A ⊗_{F} B ⊗_{F} B^{op} →^{i_{2}} A ⊗_{F} M_{m}(F) →^{j} M_{m}(A),

where m = i(B), and set φ = j ∘ i_{2} ∘ i_{1}. Thanks to Dieudonne determinent, we then obtain the following homomorphisms

A→\frac{A ⊗_{F} B}{(A ⊗_{F} B)′} →^{det∘φ} \frac{A}{RN_{A/F} (A^{∗})A′} = G(A).

By Corollary 2.4 of [3], since the exponent of G(A) divides i(A) and (i(A), i(B)) = 1, we conclude that the image of A under det ∘ φ is G(A). Now, we claim that for each y ∈ RN_{A⊗_{F} B/F} (A ⊗_{F} B), we have det∘φ(y) = 1. By the Reduced Tower formula [2], for each x ∈ A⊗_{F} B we have

RN_{A/F} (det(x)) = RN_{M_{m}(A)/F} (x) = RN_{A⊗_{F} B/F} (x)^{m}.

If y = RN_{A⊗_{F} B/F} (x) ∈ F, then

det∘φ(y) = det∘φ(RN_{A⊗_{F} B/F} (x)) = RN_{A⊗_{F} B/F} (x)^{m} = RN_{A/F} (det(x)),

i.e, the image of det ∘ φ(y) in G(A) is identity, and so the claim is established. Hence, we obtain the following embeddings

A→G(A ⊗_{F} B) →^{det∘φ} G(A).

Therefore, since the domain of det ∘φ is identity, and also det ∘ φ is surjective, we obtain G(A) = 1, and similarly G(B) = 1. □

Proposition 1.Let D be an F -central division algebra of index p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}}. If either of the following conditions holds, then we have G(D) ≠ 1.

(i) One of the primary components of D is a p_{i}-algebra.

(ii) F contains a primitive p_{i}-th root of unity for at least one i, and none of the primary components of D is a quaternion algebra.

Proof. Assume on the contrary that G(D) = 1. If D_{i} is an i-th primary component of D that satisfies (i) or (ii), then by Lemma 7, we have G(D_{i}) = 1. By Theorem 2, D_{i} is not a p_{i}-algebra, i.e., D_{i} does not satisfy (i). Therefore, by Theorem 3, we conclude that D_{i} is a quaternion division algebra which contradicts our assumption. □

Corollary 1. Let D be an F -central division algebra that satisfies the conditions of Proposition 1. Then D^{∗} has a maximal subgroup.

Proof. Since G(D) ≠ 1 the result follows. □

Corollary 2. Let D be an F -central division algebra of index p^{e} such that D is not a quaternion algebra. Then D^{∗} has a maximal subgroup if either of the following conditions holds.

(i) F has characteristic zero.

(ii) F has characteristic p.

(iii) F has a primitive p-th root of unity.

Proof. (i) Assume that F has characteristic zero. If G(D) ≠ 1, then the result follows. So, assume that G(D) = 1. If Z(D′) ≠ 1, then D′ contains a primitive p-th root of unity. Therefore, the proof is reduced to (iii). But, when Z(D′) = 1 we have D^{∗} = F^{∗} × D′. Hence, by Theorem 6 of [1], F^{∗} has a normal maximal subgroup. So, D^{∗} has also a normal maximal subgroup.

(ii) If F has characteristic p, then by Theorem 2, we have G(D) ≠ 1

and so the result follows. (iii) Assume that F has a primitive p-th root of unity. If G(D) ≠ 1,

the result follows. So, assume that G(D) = 1. By Theorem 3, D is a quaternion algebra that is a contradiction.

□

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

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

E-mail address: keshavarzi pour@mehr.sharif.edu &, mahdavih@sharif.edu