Central units of integral group rings

We give an explicit description for a basis of a subgroup of finite index in the group of central units of the integral group ring $\Z G$ of a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in $G$. The basis elements turn out to be a natural product of conjugates of Bass units. This extends and generalizes a result of Jespers, Parmenter and Sehgal showing that the Bass units generate a subgroup of finite index in the center $\mathcal{Z} (\U (\Z G))$ of the unit group $\U (\Z G)$ in case $G$ is a finite nilpotent group. Next, we give a new construction of units that generate a subgroup of finite index in $\mathcal{Z}(\U(\Z G))$ for all finite strongly monomial groups $G$. We call these units generalized Bass units. Finally, we show that the commutator group $\U(\Z G)/\U(\Z G)'$ and $\mathcal{Z}(\U(\Z G))$ have the same rank if $G$ is a finite group such that $\Q G$ has no epimorphic image which is either a non-commutative division algebra other than a totally definite quaternion algebra, or a two-by-two matrix algebra over a division algebra with center either the rationals or a quadratic imaginary extension of $\Q$. This allows us to prove that in this case the natural images of the Bass units of $\Z G$ generate a subgroup of finite index in $\U(\Z G)/\U(\Z G)'$.


Introduction
Let Z(U(ZG)) denote the group of central units in the integral group ring ZG, for G a finite group. Then Z(U(ZG)) is equal to ±Z(G)×T , where T is a finitely generated free abelian subgroup of Z(U(ZG)) [PS02, Corollary 7.3.3].
Bass proved that if G is a finite cyclic group, then the so-called Bass units (also known as Bass cyclic units) generate a subgroup of finite index in U(ZG) [Bas66]. Moreover, he described an independent set of generators using the Bass Independence Lemma. In these investigations the cyclotomic units show up and therefore the Bass units are a natural choice. Next, Bass and Milnor proved this result for finite abelian groups reducing to cyclic groups and using K-theory. However, they did not describe an independent set of generators. Recently, these results were proved avoiding K-theory and the Bass Independence Lemma, and an independent set of generators is described [JdRVG11].
In this paper we construct a basis for a subgroup of finite index in the center of U(ZG) provided the finite group G is abelian-by-supersolvable and has the property that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. The basis elements are constructed as a (natural) product of conjugates of Bass cyclic units. For finite nilpotent groups G, constructions of central units of this type have earlier been considered (using K-theory) by Jespers, Parmenter, Sehgal [JPS96] in the context of finding finitely many generators for a subgroup of finite index in the center of U(ZG). Ferraz and Simón in [FS08] constructed a basis for the center of U(ZG) in case G is a metacyclic group of order pq, with p and q two distinct odd primes.
Next, for arbitrary finite strongly monomial groups G, we construct generalized Bass units and show that the group generated by these units contains a subgroup of finite index in Z(U(ZG)). This generalizes a result of Jespers and Parmenter [JP12] on metabelian groups.
For many finite groups G, it was proved that the group B G generated by the Bass units and the bicyclic units (of one type) has finite index in U(ZG) (see for example [RS89,RS91]). In [RS89] it was proved that if G is a finite nilpotent group such that QG does not have in its Wedderburn decomposition certain types of simple algebras, called exceptional components, then B G has finite index in U(ZG). Furthermore, Jespers and Leal have extended these results to a much larger class of groups [JL93]. It is proved that B G is of finite index in U(ZG) if G is a finite group such that QG has no exceptional components and G has no non-abelian homomorphic image which is fixed point free. In this paper we also show that for such groups the group generated by the bicyclic units is of finite index in the commutator group U(ZG) ′ . Furthermore, U(ZG)/U(ZG) ′ and Z(U(ZG)) (and K 1 (ZG)) have the same rank. This allows us to prove that in this case the natural images of the Bass units of ZG generate a subgroup of finite index in U(ZG)/U(ZG) ′ .

Preliminaries
We first recall the definition of Bass units, a classical construction of units in integral group rings.
Let G be a finite group, g an element of G of order n and k and m positive integers such that k m ≡ 1 mod n. Then the Bass unit based on g with parameters k and m is u k,m (g) = (1 + g + · · · + g k−1 ) m + 1 − k m n (1 + g + · · · + g n−1 ).
and hence we allow negative integers k with the obvious meaning.
u 1,m (g) = 1 and (4) for g ∈ G, n = |g| and k m ≡ k m 1 ≡ k m1 ≡ 1 mod n. By (2) we have for a non-negative integer i and from (1), (3) and (4), we have if kk 1 ≡ 1 mod n. Thus an integral power of a Bass unit is a Bass unit. Furthermore, from (1), (3) and (5) we deduce Let N be a normal subgroup of G. Using equations (1) and (6) together with the Chinese Remainder Theorem, it is easy to verify that a power of a Bass unit in Z(G/N ) is the natural image of a Bass unit in ZG.
If R is an associative ring and G is a group then R * α τ G denotes the crossed product with action α : G → Aut(R) and twisting τ : G × G → U(R) [Pas89], i.e. R * α τ G is the associative ring g∈G Ru g with multiplication given by the following rules: u g a = α g (a)u g and u g u h = τ (g, h)u gh , for a ∈ R and g, h ∈ G. Recall that a classical crossed product is a crossed product L * α τ G, where L/F is a finite Galois extension, G = Gal(L/F ) is the Galois group of L/F and α is the natural action of G on L. The classical crossed product L * α τ G is denoted by (L/F, τ ) [Rei75]. Our approach is making use of the description of the Wedderburn decomposition of the rational group algebra QG. We shortly recall the character-free method of Olivieri, del Río and Simón [OdRS04] for a certain class of groups, called strongly monomial groups.
Throughout, G will be a finite group. If H is a subgroup of G then N G (H) denotes the normalizer of H in G. We use the exponential notation for conjugation: where M(H/K) denotes the set of all minimal normal subgroups of H/K. We extend this notation by setting ε(H, H) = " H. Clearly ε(H, K) is an idempotent of the group algebra QG. Let e(G, H, K) be the sum of the distinct G-conjugates of ε( Clearly, e(G, H, K) is a central element of QG and if the G-conjugates of ε(H, K) are orthogonal, then e(G, H, K) is a central idempotent of QG.
A strong Shoda pair of G is a pair (H, K) of subgroups of G with the properties that K ≤ H N G (K), H/K is cyclic and a maximal abelian subgroup of N G (K)/K and the different conjugates of ε(H, K) are orthogonal. In this case C G (ε(H, K)) = N G (K) [OdRS04].
Let θ be a linear character of a subgroup H of G with kernel K. Then the induced character θ G is irreducible if and only if H/K is cyclic and [H, g] ∩ H ⊆ K for every g ∈ G \ H [Sho33]. A pair (H, K) of subgroups K H satisfying these conditions is called a Shoda pair. Therefore an irreducible character χ of G is monomial if and only if there is a Shoda pair (H, K) of G such that χ = θ G for a linear character θ of H with kernel K. If χ = θ G for θ as above with (H, K) a strong Shoda pair of G then we say that the character χ is strongly monomial. The group G is strongly monomial if every irreducible character of G is strongly monomial.
For finite strongly monomial groups, including abelian-by-supersolvable groups, all primitive central idempotents are realized by strong Shoda pairs, i.e. they are of the form e(G, H, K), with (H, K) a strong Shoda of G. However, different strong Shoda pairs can contribute to the same primitive central idempotent. Indeed, let (H 1 , K 1 ) and (H 2 , K 2 ) be two strong Shoda pairs of a finite group G. Then e(G, H 1 , K 1 ) = e(G, H 2 , K 2 ) if and only if there is a g ∈ G such that H g 1 ∩ K 2 = K g 1 ∩ H 2 [OdRS06]. In that case we say that (H 1 , K 1 ) and (H 2 , K 2 ) are equivalent as strong Shoda pairs of G. In particular, to calculate the primitive central idempotents of G if G is strongly monomial, it is enough to consider only one strong Shoda pair in each equivalence class. We express this by saying that we take a complete and non-redundant set of strong Shoda pairs.
The structure of the simple component QGe(G, H, K) is given in the following Theorem.  Note that the action α of the crossed product Q(ζ k ) * α τ N/H in Theorem 2.1 is faithful. Therefore the crossed product Q(ζ k ) * α τ N/H can be described as a classical crossed product (Q(ζ k )/F, τ ), where F is the center of the algebra, which is determined by the Galois action α.
A subring O of a finite dimensional Q-algebra A is called an order if it is a finitely generated Z-module such that QO = A. For example, ZG is an order in QG when G is finite. It is known that the intersection of two orders in A is again an order in A and that, if O 1 ⊆ O 2 are orders in A, then the index of their unit groups [U(O 2 ) : U(O 1 )] is finite [Seh93]. Therefore, for two arbitrary are commensurable. Moreover, the group of units of an order in a finite dimensional semisimple algebra is finitely generated [Sie43,BHC62]. Hence the unit group of ZG is finitely generated if G is a finite group. Recall that in a finitely generated abelian group replacing generators by powers of themselves yields generators for a subgroup of finite index. We will use these properties several times in the proofs without explicitly mentioning.

A generalization of the Jespers-Parmenter-Sehgal Theorem
In this section we prove a generalization of a theorem of Jespers, Parmenter and Sehgal while also avoiding the use of K-theory. For a finite abelian-by-supersolvable group G such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G, the detailed description of the primitive central idempotents of QG and the Bass-Milnor Theorem allow us to show that the group generated by the set of Bass units of ZG contains a subgroup of finite index in Z(U(ZG)). Furthermore, one obtains a description for the generators of this subgroup.
In order to do this, we first need a new construction for central units based on Bass units in the integral group ring ZG.
The idea originates from [JPS96], in which the authors constructed central units in ZG based on Bass units b ∈ ZG for finite nilpotent groups G. We denote by Z i the i-th center, i.e. Z 0 = 1 and By induction, b (i) is independent of the order of the conjugates in the product expression and b (i) is central in Z Z i , g , since for every h ∈ Z i and for every i there exists x ∈ Z i−1 such that hg = xgh and Z i−1 , g Z i , g . In particular, b (n) ∈ Z(U(ZG)).
Note that the previous construction can be modified and improved by considering the subnormal series g Z 1 , g · · · Z n , g = G and taking in each step conjugates in a transversal for Z i in Z i−1 . Then, the two constructions differ by a power. The constructions remain valid when starting with an arbitrary unit u in ZG with support in an abelian subgroup.
We now generalize this construction to a bigger class of groups G. Throughout G will be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. It is clear that this class of groups contains the finite nilpotent groups, the dihedral groups D 2n = x, y | x n = 1 = y 2 , yxy = x −1 and the generalized quaternion groups Q 2n = x, y | x 2n = 1 = y 4 , x n = y 2 , y −1 xy = x −1 .
Let u ∈ U(Z g ), for g ∈ G of order not a divisor of 4 or 6. We consider a subnormal series N : where T i is a transversal for N i in N i−1 . We will prove that this construction is well defined by proving the following three properties.
Lemma 3.1. Let g ∈ G, u ∈ U(Z g ), N , N i and T i be as above. We have

Proof. It is easy to see that equation (II) implies (III). Hence it is sufficient to prove (I) and (II).
We prove these by induction on i. First assume i = 1. Then (I) and (II) are trivial since the support of u is contained in g = N 0 ⊳ N 1 . Now assume the formulas hold for Hence, by the induction hypothesis on (III), the latter equals c N i−1 (u) and we have proved (II).
By equations (I) and (II) we have that the construction is independent of the order of the conjugates in the product expression. Furthermore, c N m (u), the final step in our construction, is a central unit in ZG, which we will simply denote by c N (u).
Theorem 3.2. Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. Then the group generated by the Bass units of ZG contains a subgroup of finite index in Z(U(ZG)).
Proof. We argue by induction on the order of the group G. For G = 1 the result is clear. So assume now that the result holds for groups of order strictly less than the order of G.
Because of the Bass-Milnor Theorem [Bas66], we can assume that G is non-abelian. Write is a direct sum of non-commutative simple rings and QG G ′ ≃ Q(G/G ′ ) is a commutative group ring. Hence, each z ∈ Z(U(ZG)) can be written as z = z ′ + z ′′ , with z ′ ∈ Z(U(ZG(1 − G ′ ))) and z ′′ ∈ U(ZG G ′ ). Note that z ′ z ′′ = 0 = z ′′ z ′ . We will prove that some positive power of z is a product of Bass units. Since z is an arbitrary element of the finitely generated abelian group Z(U(ZG)), the result follows.
First we focus on the commutative component. Since G/G ′ is abelian, it follows from the Bass-Milnor Theorem that the Bass units of Z(G/G ′ ) generate a subgroup of finite index in U(Z(G/G ′ )). A power of each Bass unit of Z(G/G ′ ) is the natural image of a Bass unit of ZG. Hence, we get that z ′′ m = r i=1 b i for some positive integer m and some Bass units b i in ZG, where we denote the natural image of x ∈ ZG in Z(G/G ′ ) by x. It is well-known and easy to verify that u k,m (g) has finite order if and only if k ≡ ±1 mod |g|. In particular, there is a Bass unit based on g ∈ G of infinite order if and only if the order of g is not a divisor of 4 or 6. Hence we can assume that each b i is based on an element of order not a divisor of 4 or 6.
By the assumptions on G, we can construct central units in ZG which project to some power Since G is abelian-by-supersolvable and hence also strongly monomial, we know (see the Pre-  (ZHε(H, K)) is a finitely generated abelian group, it is easy to verify that { n∈NG(K) u n | u ∈ U(ZHε(H, K))} is of finite index in U((ZHε(H, K)) NG(K)/H ).
Next note that if α = 1 − ε(H, K) + βε(H, K) ∈ A, with β ∈ ZH, then α n = 1 − ε(H, K) + β n ε(H, K), for n ∈ N G (K). Hence, α and α n commute and thus the product n∈NG(K) α n is independent of the order of its factors. It follows from the previous that Let γ = 1 − ε(H, K) + δ ∈ B, with δ ∈ (ZHε(H, K)) NG(K)/H . Let T be a right transversal of N G (K) in G. Since ε(H, K) t ε(H, K) t ′ = 0 for different t, t ′ ∈ T , we get that γ t and γ t ′ commute and t∈T γ t = 1 − e(G, H, K) + t∈T δ t ∈ 1 − e(G, H, K) + ZGe(G, H, K). Clearly, t∈T γ t corresponds to a central matrix in QGe(G, H, K) with diagonal entry in (ZHε(H, K)) NG(K)/H . From the previous it follows that C = C(H, K) = { t∈T γ t | γ ∈ B} is a subgroup of finite index in Z (U (Z(1 − e(G, H, K)) + ZGe(G, H, K))). As each γ ∈ B is a product of Bass units in ZH, so is t∈T γ t a product of Bass units in ZG.
We can now finish the proof as follows. Write the central unit for each (H, K).
Remark 3.3. Only one argument in the proof of Theorem 3.2 makes use of the assumption that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. It is needed to produce a central unit as a product of conjugates of a Bass unit b. For that we use the construction c N (b). It is not clear to us whether an alternative construction exists for other classes of groups, even for metacyclic groups this is unknown.
At first sight, it looks like we do not use the properties of abelian-by-supersolvable groups, except for the fact that these groups are strongly monomial and hence we know an explicit description of the Wedderburn components. However, we can not generalize the proof to strongly monomial groups since we use an induction hypothesis on subgroups and, unlike the class of abelian-by-supersolvable groups, the class of strongly monomial groups is not closed under subgroups.
Corollary 3.4. Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. For each such cyclic subgroup g , fix a subnormal series N g from g to G. Then c Ng (b g ) | b g a Bass unit based on g, g ∈ G is of finite index in Z(U(ZG)).
Proof. Because Z(U(ZG)) is finitely generated and using Theorem 3.2, it is sufficient to show that if u = b 1 b 2 · · · b m ∈ Z(U(ZG)), with each b i a Bass unit based on g i ∈ G, then there exists a positive integer l so that u l is a product of c Ng (b g )'s, with b g a Bass unit based on g ∈ G. In order to prove this, for each primitive central idempotent e of QG, write for some positive integer m ′′ , i.e. u km ′ m ′′ ∈ c Ng (b g ) | b g a Bass unit based on g, g ∈ G .
For finite nilpotent groups of class n, we can always take the subnormal series N g : g Z 1 , g · · · Z n , g = G. Since both constructions c Ng (b) and b (n) only differ on a power, we can deduce the Jespers-Parmenter-Sehgal result.

Reducing to a basis of products of Bass units for Z(U(ZG))
In this section, we obtain a basis formed by products of Bass units of a free abelian subgroup of finite index in Z(U(ZG)), for G a finite abelian-by-supersolvable group G such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G.
First, we need some properties of our construction of central units.
Lemma 4.1. Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. Let u, v be units in Z g for g ∈ G and let N be a subnormal series N 0 = g ⊳ N 1 ⊳ · · · ⊳ N m = G. Assume h ∈ G and denote by N h the h-conjugate of the series N , i.e. N h : Proof. Let u, v ∈ Z g . Then clearly c N 0 (uv) = uv = c N 0 (u)c N 0 (v). By an induction argument on i, we now get that x commute by properties (I) and (II). This proves (A). Let u ∈ Z g and h ∈ G. We prove that

then by the induction hypothesis
Let G be a group. If g ∈ G, we denote by C g the conjugacy class of g in G. R-classes and Qclasses are a generalization of this. For a given element g in a group G of exponent e, the R-class of g is defined as the union C g ∪ C g −1 , and the Q-class of g is defined as the union gcd(r,e)=1 C g r . The number of Q-classes of a group coincides with the number of conjugacy classes of cyclic subgroups of G and, by a result of Artin, this number coincides with the number of irreducible rational characters of G, i.e. the number of simple components of QG [CR62, Cor. 39.5, Th. 42.8].
Let g ∈ G and define S g = {l ∈ U(Z |g| ) : g is conjugate with g l in G}.
In other words, S g is the image of the homomorphism where l h is the unique element of U(Z |g| ) such that g h = g l h . The kernel of this homomorphism is Cen G (g). We denote S g = S g , −1 and we always assume that transversals of S g in U(Z |g| ) contain the identity 1.
Theorem 4.2. Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. Let R denote a set of representatives of Q-classes of G. For g ∈ R choose a transversal T g of S g in U(Z |g| ) containing 1 and for every k ∈ T g \ {1} choose an integer m k,g with k m k,g ≡ 1 mod |g|. For every g ∈ R of order not a divisor of 4 or 6, choose a subnormal series N g from g to G. Then is a basis for a free abelian subgroup of finite index in Z(U(ZG)).
Proof. For every g ∈ R of order not a divisor of 4 or 6, we choose a subnormal series N g from g to G. Then for each h ∈ G of order not a divisor of 4 or 6, we agree to choose the subnormal series N h to be the x-conjugate of N g when h = x −1 g i x with g ∈ R and i coprime to the order of g. By Corollary 3.4, the set generates a subgroup of finite index in Z(U(ZG)). Let t = ϕ(|G|). We first prove that generates a subgroup of finite index in Z(U(ZG)). To do so we sieve gradually the list of units in B 1 , keeping the property that the remaining units still generate a subgroup of finite index in Z(U(ZG)), until the remaining units are the elements of B 2 . By equation (1), to generate B 1 it is enough to use the Bass units of the form u k,m (h) with h ∈ G, 1 ≤ k < |h| and k m ≡ 1 mod |h|. Hence one can assume that k ∈ U(Z |h| ). By (6), for every Bass unit u k,m (h) we have u k,m (h) i = u k,t (h) j for some positive integers i and j. Thus, by (A), units of the form c N h (u k,t (h)) with k ∈ U(Z |h| ) generate a subgroup of finite index in Z(U(ZG)).
By the definition of a Q-class, we know that each h ∈ G is conjugate to some g i , for g ∈ R and (i, |g|) = 1. Hence, by (3), (A) and (B), we can reduce further the list of generators by taking only Bass units based on elements of R.
By (4), we can exclude k = 1 and still generate a subgroup of finite index in Z(U(ZG)).
Let g ∈ G be of order n. We claim that if l ∈ S g and k ∈ U(Z n ) then c Ng (u l,t (g k )) has finite order. As u n−l,t (g k ) = u l,t (g k )g −lkt , by (8), we may assume without loss of generality that l ∈ S g . By (3), (A) and (B) we have c Ng (u l i+1 ,t (g k )) = c Ng (u l,t (g k ))c Ng (u l i ,t (g kl )) = c Ng (u l,t (g k ))c Ng (u l i ,t (g k )).
Then, arguing inductively we deduce that c Ng (u l,t (g k )) i = c Ng (u l i ,t (g k )), and in particular c Ng (u l,t (g k )) t = c Ng (u l t ,t (g k )) = c Ng (u 1,t (g k )) = 1, by (1) and (4). This proves the claim.
With g and n as above, every element of U(Z n ) is of the form kl with k ∈ T g and l ∈ S g . Using (3) again we have u kl,t (g) = u k,t (g)u l,t (g k ). By the previous paragraph, c Ng (u l,t (g k )) has finite order. Hence we can reduce the generating system and take only k ∈ T g \ {1}.
The remaining units are exactly the elements of B 2 . Thus B 2 has finite index in Z(U(ZG)), as desired.
To finish the proof we need to prove that the elements of B are multiplicatively independent. To do so, it is enough to show that the rank of Z(U(ZG)) coincides with the cardinality of B. It is easy to see that |B| = g∈R |T g | − |R| and |R| equals the number of Q-classes. By construction, [U(Z |g| ) : S g ] equals the number of conjugacy classes contained in the Q-class of g. Furthermore, [S g : S g ] = 1 when g is conjugated to g −1 and [S g : S g ] = 2 when g is not conjugated to g −1 . Therefore |T g | = [U(Z |g| ) : S g ] is exactly the number of R-classes contained in the Q-class of g. Hence |B| equals the number of R-classes minus the number of Q-classes in G. By a result in [RS05,Fer04], this number coincides with the rank of Z(U(ZG)) and the proof is finished.

Generalization to strongly monomial groups
As mentioned in Remark 3.3, it is not known whether Theorem 3.2 remains valid for other classes of groups, including metacyclic groups. In this section we construct generalized Bass units and show that the group they generate contains a subgroup of finite index in the central units of the integral group ring ZG for finite strongly monomial groups G. This generalizes Corollary 2.3 in [JP12] on generators for central units of the integral group ring of a finite metabelian group.
Let R be an associative ring with identity. Let x be a torsion unit of order n. Let C n = g , a cyclic group of order n. Then the map g → x induces a ring homomorphism Z g → R. If k and m are positive integers with k m ≡ 1 mod n, then the element is a unit in R since it is the image of a Bass unit in Z g . In particular, if G is a finite group, M a normal subgroup of G, g ∈ G and k and m positive integers such that gcd(k, |g|) = 1 and k m ≡ 1 mod |g|. Then we have Since As in the proof of Theorem 3.2, one obtains that H, K)) + ZGe(G, H, K))) . The products involved in the definition of B do not depend on the order of multiplication because the factors α g belong to Z(1 − ε(H, K)) + ZHε(H, K) and this is a commutative ring. To prove that the product in the definition of C is independent on the order of the product observe that if γ ∈ B and t 1 , t 2 ∈ G then γ = 1 − ε(H, K) + γ 1 ε(H, K) for some γ 1 ∈ ZH. If t 1 = t 2 then ε(H, K) t1 ε(H, K) t2 = 0, because (H, K) is a strong Shoda pair. Using this it is easy to see that γ t1 and γ t2 commute.
Take now an arbitrary central unit u in Z(U(ZG)). Then we can write this element as follows

Finite groups without exceptional components
All the results of the previous sections refer to strongly monomial finite groups. In this section we drop this assumption. Instead we suppose that the Wedderburn decomposition of the rational group algebra does not contain some simple algebras which we call exceptional components. An exceptional component of QG is an epimorphic image of QG which is either a non-commutative division algebra other than a totally definite quaternion algebra, or a two-by-two matrix algebra over a division algebra with center either the rationals or a quadratic imaginary extension of Q. Note that our notion of exceptional is less restrictive than in the references [RS89,Seh93,JL93]. This is possible by using stronger results on the congruence subgroup problem than those used in the mentioned references (see Theorem 6.2).
Let O be an order in a division ring D. For an ideal Q of O we denote by E n (Q) the subgroup of SL n (O) generated by all Q-elementary matrices, that is where I is the identity matrix in M n (O) and E ij denotes the matrix unit having 1 at the (i, j)-th entry and zeroes elsewhere. Theorem 6.3. Let G be a finite group such that QG has no exceptional components and denote by π the natural projection π : U(ZG) → U(ZG)/U(ZG) ′ . Then for every torsion-free complement T of ±Z(G) in Z(U(ZG)), we have (1) T ∩ U(ZG) ′ = 1 and (2) π(T ) has finite index in U(ZG)/U(ZG) ′ .
Proof. Let t ∈ T ∩ U(ZG) ′ . Since t is a central element with reduced norm 1 in all components, t is torsion. Since T is torsion-free, t = 1. Hence T ∩ U(ZG) ′ is trivial and π(T ) ≃ T .
Denote QG = e M ne (D e ), where each D e is a division ring and e runs through the primitive central idempotents of QG. Let O e be an order in D e . By Lemma 6.1, there exists a non-zero ideal Q e of O e such that 1 − e + E ne (Q e ) ⊆ U(ZG) ′ for all primitive central idempotents e of QG. Now define S = e (1 − e + E ne (Q e )) ⊆ U(ZG) ′ .
It is not hard to prove, using properties of the reduced norm, that SL ne (O e ) together with the center of GL ne (O e ) generate a subgroup of finite index in GL ne (O e ). Because of the assumption on QG and Theorem 6.2, it follows that also E ne (Q e ) together with the center of GL ne (O e ) generate a subgroup of finite index in GL ne (O e ) for n e ≥ 2. This is also true when n e = 1, since the only allowed non-commutative division ring is a totally definite quaternion algebra and hence SL 1 (O e ) is finite by a result of Kleinert [Seh93, Lemma 21.3].
Since T is a free abelian group of finite index in Z(U(ZG)), which is commensurable with e Z(GL ne (O e )), T is isomorphic to a subgroup of finite index in e Z(GL ne (O e )).
Combining these results, one gets that T is isomorphic to a subgroup of finite index in Since T is a free abelian group, it follows that the rank of U(ZG)/U(ZG) ′ equals the rank of Z(U(ZG)).
The next theorem can be proved making use of K-theory.
Theorem 6.4. Let G be a finite group such that QG has no exceptional components. Then the natural images of the Bass units of ZG generate a subgroup of finite index in U(ZG)/U(ZG) ′ .
Proof. Let B = B(G) be the group generated by Bass units in ZG and consider the natural homomorphism B → U(ZG)/U(ZG) ′ → K 1 (ZG).
It is well known that the rank of K 1 (ZG) equals the rank of Z(U(ZG)), which equals the rank of U(ZG)/U(ZG) ′ by Theorem 6.3. Let k be this common rank. Since the image of B in K 1 (ZG) has rank k [Bas66], the image of B in U(ZG)/U(ZG) ′ must have rank k too. Hence this image has finite index in U(ZG)/U(ZG) ′ .
Remark 6.5. The proof of Theorem 6.3 together with the results in [JL93] show that the bicyclic units generate a subgroup of finite index in U(ZG) ′ if G is a finite group such that G has no non-abelian homomorphic image which is fixed point free and QG has no exceptional components.
Corollary 6.6. Let G be a finite group. Then the natural images of the Bass units of ZG generate a subgroup of finite index in GL 3 (ZG)/GL 3 (ZG) ′ .
Proof. Note that M 3 (QG) has no exceptional components. Therefore we can adapt the proof of Theorem 6.4.
The Bass units of Z ζ |G| G are all the units of the form: u k,m (ζ i |G| g) = (1 + ζ i |G| g + · · · + (ζ i |G| g) k−1 ) m + (1 − k m ) ' ζ i |G| g, where g ∈ G, ζ i |G| g has order n, 1 ≤ j ≤ |G|, k is a positive integer with k m ≡ 1 mod n and ' ζ i |G| g = 1 n n−1 j=0 (ζ i |G| g) j . Corollary 6.7. Let G be a finite group of order different from 1, 2, 3, 4 and 6. Then the natural images of the Bass units of Z ζ |G| G generate a subgroup of finite index in U Z ζ |G| G /U Z ζ |G| G ′ .
Proof. When ζ |G| is a primitive |G|-th root of unity, Q ζ |G| G has no exceptional components since Q ζ |G| is a splitting field of G [JL93]. Therefore we can adapt the proof of Theorem 6.4.