A structure theorem for the unit group of the integral group ring of some finite groups

Transcription

1 A structure theorem for the unit group of the integral group ring of some finite groups Eric Jespers and Angel del Río Abstract We classify the finite groups G such that the unit group of the integral group ring ZG contains a subgroup of finite index which is the direct product of free products of abelian groups. The latter problem is also investigated for any order in an arbitrary semisimple finite dimensional Q-algebra. 1 Introduction For a finite abelian group A the structure of the unit group U(ZA) of the integral group ring ZA is well known. This result, due to Higman, can be considered as a generalization of Dirichlet s Unit Theorem on the unit group of the ring of integers in an algebraic number field. For a nonabelian finite group it follows from a general result of Borel and Harish-Chandra that the unit group U(ZG) is finitely generated (even finitely presented). For many finite groups G, Leal and Jespers explicitly described finitely many generators for a subgroup of finite index in U(ZG); this extends earlier results by Kleinert, Ritter and Sehgal. The groups G excluded in the result are those for which the rational group algebra QG has a simple component that is a non-commutative division algebra other than a totally definite quaternion algebra or that is a two-by-two matrix ring over the rationals, a quadratic imaginary field extension of the rationals or a non-commutative division algebra. We refer the reader to [5] for a survey on this matter. As mentioned by Kleinert in [10], what is still missing is a Unit Theorem, i.e., a basic structure theorem for the unit group U(ZG) for an arbitrary finite group G. According to Kleinert the minimum requirements for Research partially supported by the Onderzoeksraad of Vrije Universiteit Brussel, Fonds voor Wetenschappelijk Onderzoek (Belgium) and by DGES (PB C02-02) and Comunidad Autónoma de Murcia (project PB/16/FS/97 of Fundaçion Séneca) 1

2 a unit theorem is that it consists, in purely group theoretical terms, of a class of groups G such that almost all torsionfree subgroups of finite index in U(ZG) are members of G. Given a class of groups G, we say that a group U is virtually G if G contains a subgroup of finite index of U. For example a group is virtually abelian if it has an abelian subgroup of finite index. In this paper we consider the following type of structure theorem, called The Virtual Structure Problem for unit groups of integral group rings: for a class of groups G, classify the finite groups G, such that U(ZG) is virtually G. The Virtual Structure Problem has been solved for the class G consisting of the groups that are products of nonabelian free groups (see [9] and [11]). Earlier, in [4], the problem was solved for the class of nonabelian free groups. Leal and Jespers partially solved a question of Marciniak by showing that there are only finitely many possible groups G such that U(ZG) is virtually a nontrivial free product of abelian groups [7]. However it was not settled whether all the groups of the list obtained satisfy the required condition. Note that U(ZG) often contains subgroups that are a direct product of free products of free abelian groups of rank two (Proposition 2.3). However, as mentioned above, only few groups G are such that U(ZG) is virtually a free product of abelian groups. The reason for this is that if G has this property then the Wedderburn decomposition of QG only contains one simple component which is not a division algebra, and actually this component has to be a two-by-two matrix ring. So if QG has more than one simple component which is not a division ring, then not only are there many free products of abelian subgroups, but also U(ZG) contains subgroups that are direct products of free products of abelian groups. In this paper we therefore consider The Virtual Structure Problem for the class G of direct products of free products of abelian groups. Our solution in particular solves the above mentioned question of Marciniak (and thus completes the work started in [7]) and also deals with the case of direct products of arbitrary free groups, hence answering a question of Saorín. The first part of our solution can be stated for orders in arbitrary finite dimensional semisimple Q-algebras. All our orders are Z-orders; for the definition we refer to [14]. Proposition 1.1 Let O be an order in a finite dimensional semisimple Q- algebra A = n i=1 M n i (D i ), where each D i is a division algebra, and let O i be an order in D i. Then the following conditions are equivalent: 2

3 1. U(O) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. 2. n i=1 GL n(o i ) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. 3. For every i = 1,...,n, GL n (O i ) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. This result will be proved in section 3 and a similar proof shows that the results also holds for the class of direct products of free groups. So we are naturally led to deal with the virtual structure problem for linear groups over orders in finitely dimensional division Q-algebras. It is well known that GL 2 (Z) is virtually free nonabelian; actually, as shown in [10], M 2 (Q) is the only finitely dimensional simple Q-algebra A so that the unit group of some order in A is virtually nonabelian free. It is also well known that if D is a totally definite quaternion algebra, then the group of units of an order in D is virtually abelian [14, Lemma 21.3]. We show in Lemma 2.1 that if O is an order in a nonreal number field and n 2 then GL n (O) contains a free product of two free abelian groups of rank 2. However in section 4 we prove the following result. Proposition 1.2 Let O be an order in a finite dimensional division Q- algebra D. If GL n (O) is virtually G, then n 2. Moreover, if n = 2 then D = Q and if n = 1 and D is noncommutative, then the number of imaginary embeddings of the centre of D is at most 2. Finally in section 5 we prove the main theorem: Theorem 1.3 Let G be a finite group. The following conditions are equivalent: 1. U(ZG) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. 2. U(ZG) is virtually a (finite) direct product of (finitely generated) free groups. 3. Every nonabelian ) simple quotient of QG is isomorphic to either M 2 (Q), or H(K) with K = Q, Q( 2) or Q( 3). ( 1, 3 Q 4. G is either abelian or isomorphic to H C2 k, where H is one of the following groups: 3

4 (a) x,y x 4 = y 4 = [x 2,y] = [x,y 2 ] = [x,[x,y]] = [y,[x,y]] = 1, (b) x,y 1,...,y n x 4 = y 2 i = [y i,y j ] = [x 2,y i ] = [[x,y i ],y j ] = [[x,y i ],x] = 1, (c) x,y 1,...,y n x 4 = y 4 i = y2 i [x,y i] = [y i,y j ] = [x 2,y i ] = [y 2 i,x] = 1, (d) x,y 1,...,y n x 2 = y 2 i = [y i,y j ] = [[x,y i ],y j ] = [x,y i ] 2 = 1, (e) x,y 1,...,y n x 2 = y 4 i = y2 i [x,y i] = [y i,y j ] = [[x,y i ],x] = 1, (f) x,y 1,...,y n x 4 = yi 4 = x2 y1 2 = y2 i [x,y i] = [y i,y j ] = [yi 2,x] = 1, (g) x,y 1,...,y n x 4 = x 2 yi 4 = y2 i [x,y i] = [y i,y j ] = 1, (h) Z x where Z is an elementary abelian 3-group, x has order 2 or 4 and z x = x 1 zx = z 1 for every z Z, (i) Z H where Z is an elementary abelian 3-group, H = x,y = Q 8 and z x = x 1 zx = z 1 = y 1 zy = z y for every z Z. Note that as a corollary of the Theorem, and the results in [7] one obtains the following result. Corollary 1.4 The following are equivalent for a finite group G: 1. U(ZG) is virtually a free product of abelian groups. 2. U(ZG) is either virtually abelian or virtually nonabelian free. 3. QG is a direct product of fields, division rings of the form ( 1, 3 Q ) or H(K) with K = Q, Q( 2) or Q( 3) and at most one copy of M 2 (Q). 4. One of the following conditions hold: (a) G = Q 8 C2 n (in this case U(ZG) is finite), (b) G is abelian or (c) G is one of the following groups: D 6, D 8, Q 12 or P = a,b a 4 = b 4 = 1,[a,b] = a 2 (in this case U(ZG) is virtually nonabelian free). Throughout the paper G denotes the class of groups which are a finite direct product of finite free products of finitely generated abelian groups. Since for a finite group G the unit group U(ZG) is finitely presented, U(ZG) is virtually a direct product of free product of abelian groups if and only if U(ZG) is virtually G. 4

5 Further we use the following notation, where n is a positive integer, C n cyclic group of order n. ξ n n-th primitive root of unity. dihedral group of order 2n ) quaternion group of order 2n the generalised quaternion algebra over a ring R,i.e., D 2n Q 2n ( a,b R H(R) = = R(i,j ) i 2 = a,j 2 = b,ij = ji), where a,b R.. ( 1, 1 R [x,y] = xyx 1 y 1 for x and y in a group. Finally given 1 i,j n, E ij denotes the n n matrix having 1 at the (i,j) entry and 0 elsewhere. The size of E ij will be always clear from the context. For basic terminology and notation on group rings and their units we refer the reader to [14]. For the reader s convenience we recall here also some of the well known facts on free products of groups that will be needed in our proofs. Kurosh s Theorem ([12, Theorem 1.10]) states that any subgroup H of a free product G = A i of groups A i is itself a free product, H = F ( H j ) where F is a free group and each H j is the intersection of H with a conjugate of some factor A i of G. Commuting elements in a free product: ([13, Theorem 4.5]) if x and y are commuting elements in a free product G = A i of groups A i then they both belong to a conjugate of one of the factors A i of G. The Conjugacy Theorem for Free Products: ([12, Theorem 2.8]) Let G = A B be a free product of groups A and B. Assume that u = c 1 c n is a cyclically reduced element of G with n 2, that is, each c i belongs to one of the factors A or B, successive elements c i and c i+1 belong to different factors and also c 1 and c n belong to different factors. Then every cyclically reduced conjugate of u is a cyclic permutation of the c 1,...,c n. 2 ZG often contains direct products of nontrivial free products of free abelian groups of rank 2. It is well known that if a finite group G is not abelian or a Hamiltonian 2-group, then U(ZG) contains a nonabelian free subgroup. In section 2 we show that U(ZG) often contains direct products of nontrivial free products 5

6 free abelian groups of rank 2. Using a classical argument we first find such groups in linear groups. Lemma 2.1 Let K be a nonreal number field and O an order in K. If n > 1 then SL n (O) contains a free product of two free abelian groups of rank 2. Proof. Without loss of generality one may assume that n = 2. Since, by assumption, K is a nonreal number field, we may also assume that K is embedded in C so that K is not contained in R. Hence, it is easily seen that there exists and an additive subgroup I of rank two of O not contained in R, such that every nonzero element of I has norm at least 2. Let ) and X = Y = {( 1 a 0 1 {( 1 0 a 1 ) } : a I } : a I. Then X and Y are free abelian groups of rank 2. Let B be the set of complex numbers of length less than 1 and C the set of complex numbers of length larger than 1. Consider the classical action of GL 2 (C) on C, that is, the matrix A = (a ij ) acts on z C by mapping z to z A = a 11z+a 12 a 21 z+a 22. If z B and 1 x X then z x = z + a for some 0 a I. So 1 < a z z x 1+ a. Thus B x is properly z az+1 contained in C. If z C and y Y then z y = for some 0 a I. Then az z 1 > z and hence z y B. Thus C y B. By the Table-Tennis Argument, the group generated by X and Y is a free product of X and Y. We quote the following Lemma from [14, Lemma 4.6]. Lemma 2.2 If R S are two orders in a finite dimensional semisimple Q-algebra, then U(R) has finite index in U(S). Proposition 2.3 Let G be a finite group and write QG = i I M n i (D i ) where D i is a division ring. Let J = {j I : n j 1, and D i contains a nonreal number field}. Then U(ZG) contains a subgroup isomorphic to Z 2 Z 2. j J 6

7 Proof. For every j J, let A j B j be a subgroup of GL nj (O j ) where O j is an order of a nonreal subfield of D j and A j and B j are free abelian of rank 2 (Lemma 2.1). Let X = j J A j B j. By Lemma 2.2, A j U(ZG) = B j U(ZG) = Z 2, for every j J and hence the subgroup generated by the subgroups A j U(ZG) and B j U(ZG) with j running in J is isomorphic to j J Z2 Z 2. 3 The class of virtually G groups is closed under subgroups of finite index and direct factors In this section we prove Proposition 1.1. Because the group of units of an order in a finite dimensional semisimple Q-algebra is finitely generated and because of Lemma 2.2, it is sufficient to show that the statements listed in the section title are satisfied. Lemma 3.1 Let G 1 and G 2 be groups. Then, G 1 G 2 is virtually G (resp. a finite direct product of finitely generated free groups) if and only if so are G 1 and G 2. Proof. One implication is obvious. For the converse, assume G = G 1 G 2 is virtually G. Let F = i I F i be a subgroup of finite index of G where each F i is a free product of finitely generated abelian groups and I is a finite set. Throughout the proof i denotes an element of I and j is either 1 or 2. Let π j : G G j and p i : F F i be the projection maps. Let H j = F G j and H = H 1 H 2. Also let I a = {i I : F i is abelian }, I b = I \ I a and A = i I a F i. Clearly we may assume that F i is infinite cyclic if i I a. Then H j has finite index in G j and hence H has finite index in F. Thus p i (H 1 ),p i (H 2 ) has finite index in F i. Let I j = {i I : p i (H j ) 1} and J j = {i I : p i (H j ) is nonabelian}. The inclusion J j I j I b is clear. Assume that i I 1 I b. Let 1 x p i (H 1 ). Then x is a nontrivial central element of C Fi (x), the centraliser of x in F i. By Kurosh s Theorem C Fi (x) itself is a free product of abelian groups. Since it contains a central element, it must be abelian. Since p i (H 2 ) C Fi (x), it follows that p i (H 2 ) is abelian and hence i J 2. This implies that J 1 J 2 =. Since p i (H 1 ),p i (H 2 ) has finite index in F i, p i (H 1 ),p i (H 2 ) is not abelian and hence p i (H 1 ) is not abelian. Thus i J 1. This proves that I 1 I b = J 1. By symmetry I 2 I b = J 2. Since I 1 I 2 = I we also get J 1 J 2 = I b. 7

8 Set p i (H 1 ) if i I a, X i = F i (H 1 (H 2 A)) if i J 1,. 1 if i J 2 We first prove that X i has finite index in F i for every i J 1. Since H has finite index in G and F i is normal in F i H, there are x 1,x 2,...,x n F i, such that F i H = n k=1 x kh. If x F i, then x = x k h, for some k = 1,2,...,n and h H. If i J 1 = I 1 I b, then i J 2 = I 2 I b and thus i I 2. So p i (π 2 (h)) = 1. If i J 2, we obtain similarly that p i (π 1 (h)) = 1. Hence, 1 = p i (x) = p i (π 2 (h)). Therefore π 2 (h) H 2 A and hence h X i. So {x 1,...,x n } contains a transversal for X i in F i. Note that by Kurosh s Theorem this implies furthermore that X i is a nontrivial free product of abelian groups. Next we prove that π 1 (X i ) is a free product of abelian groups, for any i I. This is trivial if i J 2 I a. If i J 1 then write X i = t T A t, where A t is abelian. If 1 = k l=1 π 1(a l ) where a l A tl, π 1 (a l ) 1 and t l t l+1 for every l = 1,2,...,k 1, then k l=1 a l H 2 X i A F i = {1}. This shows that π 1 (X i ) = t T π 1 (A t ). Let B be the group generated by all π 1 (X i ) with i I. By the previous paragraph and the fact that nontrivial free products have trivial centre, it is obvious that B is in G. It remains to show that B has finite index in G. Since we are assuming that F i = Z, if i Ia I 1, then X i has finite index in F i, if i I a I 1. Thus, for every i J 1 (I a I 1 ) there are finitely many a ij F i such that F i = j a ij X i. Let h H 1. Then for every i J 1 (I a I 1 ) there is j i and x i X i, such that p i (h) = a iji x i. Since p i (h) = 1 for every i J 1 (I a I 1 ), then h = i J 1 (I a I 1 ) p i(h) = i J 1 (I a I 1 ) a ij i x i. Therefore h = π 1 (h) = i J 1 (I a I 1 ) π 1(a iji )π 1 (x i ) i J 1 (I a I 1 ) π 1(a iji )B. Since H 1 has finite index in G 1 we obtain that B has finite index in G 1. The proof of the statement concerning direct products of free groups is completely similar. Lemma 3.2 The classes of virtually G groups and virtually finite products of free groups are closed under subgroups of finite index. Proof. Let F = n i=1 F i be a subgroup of finite index of a group G where F i is a free product of abelian groups. If H is a subgroup of finite index of G, then H F i has finite index in F i for every i. Thus n i=1 F i H has finite index in n i=1 F i and hence it also has finite index in G and in H. Finally 8

9 notice that F i H is a free product of abelian groups, by Kurosh s Theorem. 4 Virtually G linear groups In this section we prove Proposition 1.2. A crucial property of a virtually G group is the following. Lemma 4.1 If G is a virtually G group, then for every u,v G there exists a positive integer n, such that u n,v n is either abelian or contains a nonabelian free subgroup. Proof. Let H = m i=1 F i be a subgroup of finite index of G such that each F i is a finite free product of finitely generated abelian groups. For every i I, let π i : H F i be the natural projection. Then for every u,v G, there is n N such that u n,u n H. So to prove the lemma it is sufficient to show that if g,h H and [g,h] 1 then g,h contains a nonabelian free group. So let g,h H be two noncommuting elements in H. Then there exists i I with [x,y] 1, where x = π i (g) and y = π i (h). By Kurosh s Theorem x, y is a nontrivial free product of abelian groups. Let x,y = A 1 A 2 with 1 a 1 A 1 and 1 a 2 A 2. Then it is easily verified that u = a 1 a 2 a 1,v = a 2 a 1 a 2 is a free group of rank 2. Let u,v g,h be such that π i (u) = u and π i (v) = u. Then u,v is a free group of rank 2. This proves the lemma. Lemma 4.2 Let R be a domain of characteristic 0 with finitely many units. If G = G 1 G 2 is a subgroup of finite index of GL 2 (R), then either G 1 or G 2 is finite. Thus GL 2 (R) is virtually G if and only if it is virtually a free product of abelian groups. Proof. First note that if g GL 2 (R) commutes with 1 + ae 12 with 0 a R, then g is upper triangular. Since, by assumption, R has only finitely many units, it therefore follows that some power of g is of the form 1+xE 12. Let a 0 be such that g = 1 + ae 12 = g 1 g 2 with g i G i. Since g 1 and g 2 commute with g then they both are upper triangular and hence g n 1 and g n 2 are of the form 1 + xe 12 for some n, and either g 1 or g 2 is not periodic. Assume that g 1 is not periodic. Since every element of G 2 commutes with 9

10 g 1, it now follows easily that every element of G 2 is periodic. Hence G 2 is finite. Lemma 4.3 Let U = A B be a nontrivial free product. Let a 1,a 2,b 1,b 2,c 1 = a 1 b 1,c 2 U be such that a 1 A and a 1,a 2, b 1,b 2 and c 1,c 2 are free abelian of rank 2. Then a 2,b 1,b 2,c 1,c 2 A. Proof. Because a 1,a 2, b 1,b 2 and c 1,c 2 are free abelian of rank 2 it follows (see Commuting Elements Theorem in the introduction) that each of these groups is contained in a conjugate of either A or B. In particular a 1,a 2 is contained in A. Moreover the length of the normal form of each b 1, b 2, c 1 and c 2 is odd. Write b 1 = g 1...g n hgn 1...g1 1, a product in normal form with h,g i in A B. If n 1, since c 1 = a 1 b 1 has odd length, then g 1 A. Thus the normal form of c 1 is (a 1 g 1 )... g n hgn 1 1. And therefore g1 1 c 1g 1 = (g1 1 a 1g 1 )... g n hgn 1...g2 1 is a cyclically reduced...g 1 word. However, as g1 1 c 1g 1 belongs to the conjugacy class of an element of length one, the Conjugacy Theorem for Free Products says that the cyclically reduced word (g1 1 a 1g 1 )... g n hgn 1...g 1 2 should be of length 1, a contradiction. So b 1 = h. Because c 1 = a 1 b 1 = a 1 h, has odd length, we get that b 1 = h A. Therefore, b 2,c 1,c 2 A. Although the following result is probably wellknown, we include a short proof. Lemma 4.4 Let d be a negative squarefree integer. Then the congruence subgroup of level 2 of PSL 2 (Z[ d]) is torsionfree. Proof. Let u = ( a b c d ) SL 2 (Z[ d]) be a periodic element in the congruence subgroup of level 2 of PSL 2 (Z[ d]). Then u is diagonalizable and the eigenvalues are roots of unity. If ξ is one of the eigenvalues, then ξ 1 is the other eigenvalue and ξ+ξ 1 Z[ d] R = Z. Therefore a + d = 0, ±1, ±2. Since a,d 1 mod 2, then a + d ±1. If a + d = 0, then 1 = ad bc = a 2 bc 1 mod 4, a contradiction. So a + d = ±2 and hence u = ±1. This proves the lemma. Now we can prove the first part of Proposition

11 Proposition 4.5 Let O be an order in a finite dimensional division Q- algebra D. If GL n (O) is virtually G, then n 2. Moreover, if n = 2 then D = Q. Proof. Assume that GL n (O) is virtually G and n 2. If n 3, then for any positive integer m, G = x = 1 + me 12 = (1 + E 12 ) m,y = 1 + me 23 = (1 + E 23 ) m is a nilpotent nonabelian group, contradicting Lemma 4.1. So n = 2. Now we prove that every unit of O is periodic and hence O has finitely many units. Let u be a unit of O and G = g = 1 + E 12,h = ue 11 + E 22. Then N = {1 + re 12 r O} G is a normal abelian subgroup of G and G/N is cyclic. Thus G is solvable. So G does not contain any nonabelian free subgroup and hence [g n,h n ] = 1 for some n. Therefore u must be periodic. Now we argue by contradiction. So assume that D Q and GL 2 (O) is virtually G. By Lemma 4.2, GL 2 (O) is virtually a free product of abelian groups. By [14, Lemma 21.3], D is either a quadratic imaginary extension of Q or a totally definite quaternion algebra over Q, so D contains Q( d) for a negative squarefree integer 1. By Kurosh s Theorem GL 2 (O) is virtually a free product of abelian groups, where O is the ring of integers of Q( d). Since PSL 2 (O) and GL 2 (O) have a common subgroup of finite index, PSL 2 (O) is virtually a free product of abelian groups. Let ω = d. Consider the following matrices in PSL 2 (O): a 1 = ( ),a 2 = ( 1 2ω 0 1 ) and C = ( and let L = a 1,a 2,b 1 = C 1 a 1 C,b 2 = C 1 a 2 C,c 1 = C 2 a 1 C 2,c 2 = C 2 a 2 C 2. Note that a 1,a 2, b 1,b 2 and c 1,c 2 are free abelian of rank 2 and also that a 1 c 1 = b 1. Since L is a subgroup of the congruence group of level 2 in PSL 2 (O), L is torsionfree (Lemma 4.4.) By Kurosh s Theorem L contains a subgroup of finite index K that is a free product of abelian groups. So K acts on a tree T with trivial edge stabilizers and abelian vertex stabilizers. By [1, Theorem IV.1.3], L acts on a tree T 1 with finite edge stabilizers and there is a K-map f from the set of vertices of T 1 to the set of vertices of T. Since L is torsionfree, the stabilizers of the edges are trivial and hence L is a free product of some of the stabilizers of the vertices. Moreover the intersection of K with the stabilizer L v of a vertex v of T 1 is embedded in the stabilizer of f(v) which is abelian. So L v is virtually abelian. Since L is not virtually abelian it is enough to ) 11

12 show that L is not a nontrivial free product. Assume the contrary and write L = A B with A 1 B. Since a 1,a 2 is a free abelian group of rank 2, then it is contained in a conjugate of either A or B. We may assume that it is contained in A. By Lemma 4.3, L = A, a contradiction. Corollary 4.6 Let G be a finite group such that U(ZG) is virtually G. If H is either a subgroup of G or an epimorphic image of G and e is a primitive central idempotent of QH, then QHe is either a division ring or isomorphic to M 2 (Q). In the latter case He is either D 8 or S 3. Proof. Let H be a subgroup of G and e a primitive central idempotent of QH so that QHe = M n (D) where D is a division algebra. Then QHe is a subring of a simple quotient S of QG. If n 2, then S is not a division ring and hence S = M 2 (Q) = QHe. In that case He = D 8 or S 3. If H is an epimorphic image of G then every simple component of QH is a simple quotient of QG. Hence the result follows. To finish the proof of Proposition 1.2 it only remains to show the following property. Proposition 4.7 Let O be an order in a finite dimensional division Q- algebra D. If U(O) is virtually G then the centre K of D has at most two imaginary embeddings. Proof. Let R = K O. Assume first that U(O) is virtually abelian and let A be an abelian subgroup of finite index of U(O). Since K(A) is a subfield of D and D is noncommutative, there is α O which is not in K(A). Then A 1 = A U(R[α]) has finite index in U(R[α]) and hence [U(R[α]) : U(R[A 1 ])] <. By [14, Lemma 21.2], K(A 1 ) is totally real and thus so is K. Assume know that U(O) is virtually G but it is not virtually abelian. Let U = A K 1... K n, a subgroup of finite index in U(O) with A abelian and each K i a nonabelian free product of abelian groups. Let D = K(A,K 1 ) and O = O D, an order in D. Since any element of O commutes with any element in K 2... K n, it follows that O (K 2... K n ) is abelian. Thus O U = A 1 K 1, where A 1 = A O (K 2... K n ) is abelian. Hence replacing D by D we may assume that U(O) contains a subgroup of finite index of the form A H, where A R and H = H 1 H 2 is a nontrivial free product. Let 1 h 1 H 1 and 1 h 2 H 2 and h = h 1 h 2. We claim that U(R[h])/U(R) has torsionfree rank at most 1. Otherwise there 12

13 is k U(R[h]) such that h,k is free abelian of rank 2 and h,k U(R) = 1. We may assume that k = ax with a A and x H. Then x commutes with h and hence x = h m for some integer m. Then a = kh m U(R), a contradiction. Let s be the number of real embeddings of K and dim Q K = s + 2t. Let s 1 be the number of real embeddings of K[h] and dim Q F = s 1 +2t 1. By the previous paragraph and Dirichlet s Unit Theorem, s 1 + t 1 s t 1. Since s 1 +2t 1 2(s+2t) we get that t 1 s+2t s 1 2. Hence s 1 +t 1 s t t+ s 1 2 and thus t 1. So the number of imaginary embeddings of K is at most 2. 5 Proof of the Main Theorem In this section we prove Theorem 1.3. Note that (2) implies (1) is obvious and (3) implies (2) is a consequence of [14, Lemma 21.3] and the fact that GL 2 (Z) is virtually free nonabelian. In [9] and [11] it is shown that the groups listed in (4), (a)-(f) and (h) are precisely the finite groups G so that the noncommutative simple quotients of QG are of the type M 2 (Q), H(Q) or as in [9] and [11] one shows that ( 1, 3 Q ). Using similar arguments QG = 2 n+1 Q 2 n 1 (2 n 1)M 2 (Q) 2 2n 2 H(Q( 2)) when G is a group of type (g), and QG = 4Q H(Q) (3 k 1)M 2 (Q) 3k 1 H(Q( 3)) 2 when G is a group of type (i) such that Z has rank k. This proves (4) implies (3). Now we prove (1) implies (3). Let G be a finite group satisfying condition (1). By Corollary 4.6, for any subgroup or epimorphic image H of G, the simple quotients of QH are matrix rings of degree at most 2. Gow and Huppert have studied such groups. As an immediate consequence of their Theorem 6.1 and Theorem 6.2 in [3] and because of Corollary 4.6 we obtain the following property. Proposition 5.1 If H is either a subgroup of G or an epimorphic image of G then it is metabelian and has a nilpotent subgroup N of index 1 or 2. Furthermore H is supersolvable and the ordinary character degrees (that is, the character degree over a splitting field) are 1, 2 or 4. 13

14 The following Lemma follows by standard arguments. Lemma 5.2 If k 3, then M 2 (Q(i)) is a quotient of the rational group algebra of the group T 3 2 k = {x,y x 2k = y 3 = 1,xy = y 1 x} and hence U(ZT 3 2 k) is not virtually G. That (1) implies (3) now follows from Corollary 4.6 and the following Lemma. Lemma 5.3 Let e be a central idempotent of QG, such that QGe is a noncommutative division algebra. Then one of the following conditions hold: 1. QGe = H(Q) and Ge = Q QGe = H(Q( 2)) and Ge = Q QGe = H(Q( 3)) and Ge = Q QGe ) =, Ge = Q 12. ( 1, 3 Q Proof. We deal separately with the case that the group Ge is nilpotent and the case that it is not. So first suppose Ge is nilpotent. Then it follows from [6] that either QGe = H(Q(ξ 2 n 1 + ξ 1 2 n 1 )) and Ge = Q 2 n with n 3 or QGe = H(Q(ξ n )) and Ge = Q 8 C n with n an odd positive integer such that the order of 2 modulo n is odd. Because of Proposition 4.7 the latter does not occur. In the former case, it follows that D 2 n 1 is a homomorphic image of Q 2 n and thus also of G. But M 2 (Q(ξ 2 n 2 + ξ 1 2 n 2 )) is a simple component of QD 2 n 1. By Corollary 4.6, Q(ξ 2 n 2 + ξ 1 2 n 2 ) = Q and thus n = 3 or 4. So Ge = Q 16 and QGe = H(Q( 2)), or Ge = Q 8 and QGe = H(Q). Second assume Ge is not nilpotent. Hence by Proposition 5.1, G has a nilpotent subgroup N of index 2 so that Ne is a nilpotent subgroup of index 2 in Ge, and Ne is not a 2-group. Clearly QNe QGe, and thus QNe is either a field or a division ring. We now show that it has to be a field. Suppose the contrary, then since Ne is nilpotent, it follows again from [6] that Ne = Q 8 C n for some odd positive integer n such that the order of 2 modulo n is odd. Furthermore, QNe = H(Q(ξ n )). Write Ne = a,b,c with a,b = Q 8 and c a central element of order n. Now, let S be a Sylow 2- subgroup of Ge containing Q 8 = a,b. Then QSe is also a noncommutative 14

15 division ring (since it is contained in QGe). So, again by [6], Se = Q 16. Hence we may assume that there exists g Ge \Ne with g,b = Q 16. Since c is normal in Ge (as it consists of the elements of odd order in Ne) we get that gc = c i g for some 1 < i < n and n i 2 1. Since i 1, there is an (odd) prime divisor p of n that divides i + 1. Let n = pq. It follows that g,b,c = Ge and H = g 2,b,c p = a,b,c p is a normal subgroup of Ge. Furthermore Ge/H = x = g, y = c with x 2 = 1, y p = 1 and xy = y 1 x. Thus Ge/H = D 2p. Hence M 2 (Q(ξ 2p + ξ2p 1 )) is a simple component of Q(D 2p ), and thus also of QG. Since Q(ξ 2p + ξ2p 1 ) is a real field and the Q-dimension of Q(ξ p + ξp 1 ) is more than 1 (note that p 3 as the order of 2 modulo p is odd), this contradicts with Corollary 4.6. So QNe is a field and thus Ne = c c n = 1 for n that is not a power of 2. Hence Ge = g,c g 2 = c j, c n = 1, gc = c i g for some g Ge, 0 j < n, 1 < i < n and i 2 1 ( mod n). Write n = 2 k m for some k and (m,2) = 1. Replacing g by g m we also may assume that m j. Write < c j >=< c j 1 >, with j 1 n. So c j = (c k ) j 1 for some k with (k,n) = 1. Replacing c by c k we therefore also may assume that m j and j n. Since Ge is contained in a division ring and ge is not central, we have that j 0. As Ge is not nilpotent and g 2 is central, H = Ge/ g 2 = x,y x 2 = y j = 1,xy = y i x (where i is the remainder of i modulo j) is not abelian and hence j does not divide i 1. Because of Theorem 2.2 in [8], M 2 (Q j ) is a simple component of QH, where Q j = Q(ξ j + ξj i,ξ 2 + ξj 2i,...). By Corollary 4.6, Q j = Q. Thus, by Lemma 3.2 in [8] j = 3 or 6 (because j is not a power of 2) and hence m = 3. Note that j does not divide i 1, and thus j does not divide i 1. Furthermore, as n = 2 k m (i 1)(i + 1), we get that i is odd. Let N 2 = c 3 the Sylow 2-subgroup of N, and let H = g,c 3. Then H is a 2-group and QHe is a field or a division ring. If it is a field, then H is cyclic (so all its subgroups form a chain). Since g < c 3 > it follows that c 3 g. Now gc 2k g 1 = c 2ki. Since c 2k is an element of order 3 and Ge is not nilpotent, this element c 2k cannot be central, for otherwise Ge = H C 3 is nilpotent. Thus g is an element of order 2 k+1, y = c 2k has order 3 and gx = x 1 g. So Ge = T 3 2 k+1 and by Lemma 5.2, k = 1. That is Ge = Q 12 and QGe ) = H 1 =. ( 1, 3 Q Now assume that QHe is a noncommutative division ring. Since H is a 2- group it follows from [6] that He = H = Q 2 k+1. But, D 2 k is a homomorphic image of Q 2 k+1. By Corollary 4.6 and [8], k = 2; that is H = Q 8 and Ge = 24. So gc 3 g 1 = c 3. But we also know that c 2k is not central. Hence, gc 2k g 1 c 2k. But c 2k is an element of order 3. So gc 2k g 1 = c 2k. 15

16 As 2 k and 3 are relatively prime we get gcg 1 = c 1. Consequently, and QGe = H(Q( 3)). Ge = g,c g 2 = c 12 = 1,gcg 1 = c 1 = Q 24 In the remainder of the paper we prove (3) implies (4). First we prove a few technical lemmas. Lemma 5.4 The class of finite groups G such that U(ZG) is virtually G is closed under quotients and subgroups. Proof. Let N be a normal subgroup of G. Since Q(G/N) is a quotient of QG, it follows from Proposition 1.1 that U(Z(G/N)) is virtually G. Assume know that H is a subgroup of G and let S be a noncommutative simple quotient of QH. By Lemma 2.2 we have to prove that U(R) if virtually G for some order R in S. Note that S is a subring of a simple quotient S of QG. If S is a division ring, then by Lemma 5.3, U(R ) is virtually abelian, where R is an order in S. Therefore U(R) is virtually abelian as well, where R = R S is an order in S. Otherwise S = M 2 (Q) and U(R ) is virtually free nonabelian, where R is an order in S (Lemma 4.2). Thus S R is an order in S and U(S R ) is virtually free nonabelian. Lemma 5.5 If G = G 1 G 2 is nonabelian and U(ZG) is virtually G, then either G 1 or G 2 is an elementary abelian 2-group. Proof. Assume that G 1 is nonabelian and g G 2 has order n 1. Then there exists a simple quotient S of Q(G 1 g ) that contains Q(ξ n ) in its centre. By Corollary 4.6 and Lemma 5.3 Q(ξ n ) is real, and hence n = 2. For a subgroup H of G we denote by H 1 the idempotent QG. H h H h in Lemma 5.6 If G is nonabelian and U(ZG) is virtually G, then the centre of G has exponent 2 and the exponent of G/Z(G) is a divisor of 12. If furthermore G is nilpotent then it is a 2-group. Proof. By Lemma 5.3 and Corollary 4.6, for every primitive central idempotent e of QG, the group Ge is either abelian or isomorphic to Q 8, D 8, S 3, D 12, Q 16 or Q 24. Since the centre of the latter groups is of exponent at most 2, we get that g 2 (1 G ) = 1 G and hence (g 2 1) G = g 2 1, for 16

17 g Z(G). Since G 1, a support argument therefore easily implies that g 2 G. Thus g 2 = 1. So we have proved that Z(G) has exponent 2. Furthermore, since the previously mentioned groups have exponent a divisor of 12 modulo their respective centre, we get that G/Z(G) has exponent a divisor of 12. The same argument shows that if G is nilpotent and nonabelian, then G is a 2-group. Frequently we will use that if G is a finite group so that ) the nonabelian simple quotients of QG are either H(Q), M 2 (Q) or (such groups we ( 1, 3 Q will call admissable), then G does not contain elements of order 8 and G is one of the groups listed in (a)-(f), (h) in the Theorem (see [9]). Note that, because of Lemma 5.3 and Corollary 4.6, if G is a 2-group (and U(ZG) is virtually G), then G is admissable precisely when Q 16 is not an epimorphic image of G. Lemma 5.7 If G is nilpotent and U(ZG) is virtually G, then 1. Q 8 is not an epimorphic image of G/Z(G). 2. G/Z(G) = H C2 n for some H which is either trivial or is as in the cases (b), (d) or (e) as stated in Theorem If x is an element of order 8 in G and y G with [x,y] 1, then < x,y > = Q 16. Proof. Let E be the set of primitive central idempotents of QG. Note that, because of Lemma 5.3 and Corollary 4.6, for every e E, Ge is either abelian or Q 8, D 8 or Q 16. Therefore, the group Ge/Z(Ge) is either trivial, an elementary abelian 2-group or D 8. Moreover, if x G has order at most 4, then xe has order at most 4 and since Ge is either Q 8, D 8 or Q 16, then x 2 e Z(Ge). Thus x 2 Z(G). We conclude that if x 2 Z(G) then x has exponent 8. Moreover if t = [x,y] 1 and x has order 8, then [t,x],t 2 Z(G) and hence [t,x 2 ] = 1. Thus x 2 is a central element of order 4 of t,x. By Lemma 5.6, this group is abelian. We conclude that if x has order 8 then [x,[x,y]] = 1 for every y G. Assume that Q 8 is an epimorphic image of G/Z(G). Thus there are x,y G such that t = [x,y],x 2,y 2 Z(G). This implies that x and y have order 8. Therefore [t,x] = [y,t] = 1. Since t is noncentral, there is e E such that te Z(Ge). This implies that Ge = Q 16 and Ge = xe,ye. Then either 17

18 xe or ye has order 8. If xe has order 8, then e = [te,ye] = [x 2 e,ye] = x 4 e e, a contradiction. This proves (1). To prove (2) note that, by Lemma 5.6 G/Z(G) has exponent a divisor of 4. If G/Z(G) is abelian, then Q 16 is not an epimorphic image of G and thus G is admissable, so that G/Z(G) is an elementary 2-group. If G/Z(G) is not abelian then, since G/Z(G) does not have Q 16 as an epimorphic image, we obtain from [9] G/Z(G) is of the form H C2 n, where H is one of the groups of cases (a)-(f) in the Theorem 1.3. Since the groups of cases (a), (c) and (f) have Q 8 as an epimorphic image, H is of one of the remaining cases, (b), (d) or (e). To prove (3) we may assume that G = x,y, where x has order 8 and [x,y] 1. Note that since G has an element of order 8, the group Q 16 is an epimorphic image of G and G/Z(G) is nonabelian of rank 2. By (2), G/Z(G) is one of the groups in cases (b), (d) or (e) with n = 1. Note that for n = 1, the groups of (d) and (e) are equal to D 8. So G/Z(G) = D 8 = x 1,y 1 x 4 1 = y2 1 = x2 1 [x 1,y 1 ] = 1 or G/Z(G) = x 1,y 1 x 4 1 = y2 1 = [x 1,[x 1,y 1 ]] = [y 1,[x 1,y 1 ]] = [y 1,x 2 1 ] = 1. In both cases we may assume that xz(g) = x 1 and yz(g) = y 1. Then y 2 Z(G). Set t = [y,x]. By the first part of the proof, [t,x] = 1 and t 2 Z(G). Therefore yx n = t n x n y. Thus yt = y 2 xy 1 x 1 = x 5 yx 3 = x 5 t 3 x 3 y = t 3 y. This implies that tx 2 is central and hence [x 1,y 1 ]x 2 1 = 1 or equivalently [x 1,y 1 ] = x 2 1. Thus G/Z(G) = D 8. By Lemma 5.6, (tx 2 ) 2 = 1, or equivalently t 2 = x 4. Assume that e = (1 t 2 )ỹ2 tx 2 0. Since t 2 G, we know that e(1 G ) = e and thus QGe is a sum of noncommutative simple quotients of QG. Note that the centre of QGe contains Q(x+tx)f = Q( 2), for some primitive central idempotent f of QG with fe = f. Therefore QGf is a noncommutative division ring. But (1+y)(1 y)f = 0 and hence either (1 y)f = 0 or (1+y)f = 0. So yf = ±f and thus Gf = x,y f is abelian, a contradiction. Thus (1 t 2 )ỹ2tx 2 = 0. A similar argument shows that (1 t 2 )ỹ2 (1 tx 2 ) = 0 and hence (1 t 2 )ỹ2 = 0. Comparing coefficients one obtains that y 2 t 2. Since t 2 = x 4 1, it follows that y 2 = t 2. Assume know that e = (1 t 2 )(1 tx 2 ) 0 and set X = xe and Y = xye. Then Ge = X,Y X 8 = Y 2 = 1,Y X = X 3 Y. However this is impossible as Q(Ge) has a simple component of the type M 2 (Q( 2)). Therefore (1 t 2 )(1 tx 2 ) = 0 and comparing coefficients one deduces that tx 2 t 2. Since t x 2, it follows that tx 2 = 1 and hence t = x 6. So G = Q

19 We now prove that (3) implies (4) when G is nilpotent. Theorem 5.8 Let G be a nilpotent nonabelian finite group such that U(ZG) is virtually G. Then G is isomorphic to one of the groups (a)-(g) listed in Theorem 1.3. Proof. Because of Lemma 5.6 and the assumption, G is a nonabelian 2- group. If G has no elements of order 8 then Q 16 is not an epimorphic image of G and hence G is admissable. Thus G is one of the groups (a)-(f). So assume that G has an element of order 8. By Lemma 5.6, G contains a subgroup isomorphic to Q 16 and hence G/Z(G) is nonabelian. By Lemma 5.7, G/Z(G) = H K, where K = z 1 Z(G),...,z m Z(G) = C m 2 and H is one of the groups of types (b), (d) or (e) as listed in Theorem 1.3. Assume first that H is as in (e), that is, H = x,y 1,...,y n x 2 = y 4 i = y 2 i [x,y i ] = [y i,y j ] = [[x,y i ],x] = 1 Write x = uz(g) and y i = v i Z(G). We may assume that v 1 has order 8. Since [v 1,v i ] Z(G), then v 1,v i = Q16. By Lemma 5.7, [v 1,v i ] = 1. Thus v 1 is a central element of order 8 in v 1,...,v n. By Lemma 5.6, this group is therefore abelian. Hence we may assume that each v i has order 8. Since [v i,u] 1, then, by Lemma 5.7, [u,v 2 i ] = u2. Thus [u,v 2 1 v2 n ] = 1, so y2 1 y2 n = 1. We conclude that n = 1. Since the groups of type (d) and (e) are isomorphic for n = 1, the previous paragraph shows that we may assume that H is of type either (b) or (d), that is, H is either or H 1 = u,v 1,...,v n u 4 = v 2 i = [v i,v j ] = [u 2,v i ] = [[u,v i ],v j ] = [[u,v i ],u] = 1, H 2 = u,v 1,...,v n u 2 = v 2 i = [v i,v j ] = [[u,v i ],v j ] = [u,v i ] 2 = 1. If H = H 1, we may assume that u = xz(g), where x has order 8. Thus [x,z i ] = 1. Set v 1 = yz(g). By Lemma 5.7, [x,y] = x 2 and hence [u,v 1 ] = u 2, which is not true. So H = H 2. We may assume that uv 1 = yz(g) where y has order 8. Set v i = x i Z(G). By Lemma 5.7, [y,x i ] = y 2. Therefore [u,v i ] = [uv 1,v i ] = [uv 1,v j ] = [u,v j ] for every i j and hence n = 1. Set x = x 1. Then y,x = Q 16. Let 1 zz(g) K. Since [y,z] Z(G), by Lemma 5.6, [y,z] 2 = 1. Hence, by Lemma 5.7, [y,z] = 1. 19

20 We furthermore prove that [x,z] = z 2. For this consider the admissable group N = y 2,x,z. Note that N = y 4,[x,z]. If [x,z] = 1, then z is central in N and, z 2 = 1. So assume that [x,z] 1. Note that also [x,z] y 4, because otherwise y 2 z is a central of order 4 in N. So N has rank 2 and therefore by [9, Lemma 9], z 2 [x,z] and y 4 z 2 = (y 2 z) 2 [y 2 z,x] = y 4 [z,x]. This implies that z 2 1 and hence z 2 = [x,z]. Since z has order either 2 or 4, we get xzx 1 = z 1. In particular z1 1 z 1 2 = xz 1 z 2 x 1 = (z 1 z 2 ) 1 and hence [z 1,z 2 ] = 1. Since z i is not central, zi 2 = [x,z i] 1. Setting y 1 = y and y i = yz i 1 for i 2, these elements satisfy the relations of the group of type (g) in the Theorem. Since Z(G) is an elementary abelian 2-group, G = G 1 C2 k, for some k, where G 1 = x,y 1,...,y n, n = m + 1. It is easy to show that Z(G 1 ) = y 4,z2 2,...,z2 m. Moreover y 4k z 2l zm 2lm = [x,y 2k z l zm lm ] 1. Therefore Z(G 1 ) has order 2 n. Since G 1 /Z(G 1 ) D 8 C2 m, the group G 1 has order 2 2n+2. Since the group of type (g) in the Theorem has the same order, it follows that G 1 is of type (g). Let H and Z be two groups and ϕ : H Aut(Z) a group homomorphism. The semidirect product defined by this action is denoted by Z ϕ H. To deal with the nonnilpotent case we need the following Lemma. Lemma 5.9 If G is a nonabelian finite group such that U(ZG) is virtually G then G = Z ϕ H, where Z is an elementary abelian 3 group, H is a 2-group and Im ϕ = σ with σ(z) = z 1 for every z Z. Proof. By Lemma 5.6, we may assume that G is not nilpotent. By Proposition 5.1, G has a nilpotent subgroup N of index 2. Then, again by Lemma 5.6, N has to be abelian. Let N = N 2 N 3 where N 2 is a 2-group and N 3 is a 3-group. Let x G \ N. One may assume that x 2 N 2. So G = N 3 ϕ N 2,x, H = N 2,x is a 2-group and Im ϕ = σ = ϕ(x) has order 2. Since Z(G) is an elementary abelian 2-group, then Z(G) N 3 = 1 and for every y N 3, yσ(y) Z(G). So σ(y) = y 1. Thus, x,y / x 2 = D 2n, where n is the order of y. Therefore n is either 1 or 3, that is, N 3 is elementary abelian. The next result finishes the proof of the Theorem. Theorem 5.10 Let G be a nonnilpotent finite group such that U(ZG) is virtually G. Then G = (Z ϕ H) C2 m where Z is an elementary abelian 3-group and one of the following condition holds: 20

21 1. H = x is cyclic of order 2 or 4 and z x = x 1 zx = z 1 for every z Z. 2. H = Q 8 = x,y x 4 = x 2 y 2 = x 2 [x,y] and z x = xzx 1 = z 1 = yzy 1 = z y, for every z Z. Proof. We know G = Z ϕ K where Z is an elementary abelian 3-group, K is a 2-group and ϕ is as in Lemma 5.9. First we deal with the case K abelian. Let x be an element of maximal order in K. If x acts trivially on Z, then x is central and hence it has order 2. So K is an elementary abelian 2-group. Let g G \Ker ϕ. Then K = g Ker ϕ and G = (Z ϕ g ) Ker ϕ as desired. On the other hand, if x does not act trivially on Z, then x,y = T 3n, where n is a the order of x. By Lemma 5.2, n 4. Again K = x H where H is an elementary abelian group embedded in Ker ϕ and hence G = (Z ϕ x ) H, again as desired. Next assume that K is nonabelian, so write K = H H 1, where H is of one of types (a)-(g) in the Theorem and H 1 is an elementary abelian 2-group. Note that H 1 Ker ϕ, because if h is an element of order 4 of H and h 1 H 1 \ Ker ϕ, then for every 1 y Z, either h or hh 1 is a central element of order 4 in h,h 1,y. So G = (Z ϕ H) H 1. We claim that H is not isomorphic to D 8 = x,y x 2 = y 2 = (xy) 4 = 1. To prove this we may assume that Z = z is cyclic of order 3. There are two possibilities, x Ker ϕ (and so y Ker ϕ) or x,y Ker ϕ. Let e = (1 (xy) 2 )(1 z). In the first case the centre of QGe is Q(a) where a = (1 + 2z)xe and a 2 = 3. So QGe = M 2 (Q( 3)), in contradiction with Corollary 4.6. In the second case the centre of QGe is Q(a) where a = (1 + 2z)xye and a 2 = 3. So QGe = M 2 (Q( 3)), again in contradiction with Corollary 4.6. As a consequence H is not one of the groups of type (a)-(e) because Z,x,y 1 / x 2,y1 2 = Z D 8. Similarly H is not of type (g) because Z,x,y 1 / x 2,y1 4 Z D 8. So H is of type (f) and because Z,x,y 2 / x 2 = Z D 8, it follows that n = 1. This finishes the proof. References [1] W. Dicks and M.J. Dunwoody, Groups acting on graphs, Cambridge University Press,

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