Group algebras of Kleinian type and groups of units

0.24 MB·English

by  Gabriela Olteanu

GROUP ALGEBRAS OF KLEINIAN TYPE AND GROUPS OF UNITS
7
0 GABRIELAOLTEANUANDA´NGELDELR´IO
0
2
Abstract. ThealgebrasofKleiniantypearefinitedimensionalsemisimplerationalalgebrasAsuchthatthe
n groupofunitsofanorderinAiscommensurablewithadirectproductofKleiniangroups. Weclassifythe
a
SchuralgebrasofKleiniantypeandthegroupalgebrasofKleiniantype. Asanapplication,wecharacterize
J the group ringsRG, with Ran order ina number field and G afinite group, such that RG∗ is virtuallya

2 directproductoffree-by-freegroups.
1

]
T ThestudyofKleiniangroupsgoesbacktotheworksofPoincar´e[Poi]andBianchi[Bia]andithasbeenan
R active field of research ever since. In the last decades, it is strongly related to the Geometrization Program
. of Thurstonfor the classificationof 3-manifolds[EGM, MR, Mas, Thu]. The use of the methods of Kleinian
h
groups to the study of the groups of units of group rings was started in [PRR] and led to the notion of
t
a algebras of Kleinian type and finite groups of Kleinian type.
m
LetK be a numberfield, Aacentralsimple K-algebraandR aZ-orderinA. LetR1 denotethe groupof
[ units of R of reduced norm 1. Every embedding σ :K C induces an embedding σ :A M (C), where d
→ → d
is the degreeof A. Furthermore,σ(R1) SL (C). We say that A is of Kleinian type if either A=K or A is
1 d
⊆
v a quaternion algebra over K and σ(R1) is a discrete subgroup of SL2(C) for some embedding σ of K in C.
1 Moregenerally,analgebra of Kleinian type[PRR]isbydefinitionadirectsumofsimplealgebrasofKleinian
5 type. A finite group G is of Kleinian type if and only if the rational group algebra QG is of Kleinian type.
3
The finite groups of Kleinian type have been classified in [JPRRZ] where it has been also proved that
1
a finite group G is of Kleinian type if and only if the group of units ZG of its integral group ring ZG
0 ∗
7 is commensurable with a direct product of free-by-free groups. Recently, Alan Reid asked us in a private
0 communicationabout the consequencesof replacing the ring of rationalintegers by another ring of integers.
/ This leads to the following two problems:
h
t Problem 1. Classify the group algebras of Kleinian type of finite groups over number fields.
a
m Problem 2. Given a groupalgebra of Kleinian type KG, describe the structure of the group
of units of the group ring RG for R an order in K.
:
v
The simple factorsofKGare Schuralgebrasoverits center. So, in orderto solveProblem1, it is natural
i
X tostartclassifyingtheSchuralgebrasofKleiniantype. ThisisobtainedinSection1. Usingthisclassification
r andthatoffinitegroupsofKleiniantypegivenin[JPRRZ]weobtainthe classificationofthegroupalgebras
a
of Kleinian type in Section 2. In Section 3 we obtain a partial solution for Problem 2.
1. Schur algebras
All throughout K is a number field. We refer to the field homomorphisms K R as real embeddings
→
of K. By abuse of notation, a pair of complex embeddings of K is, by definition, a pair of conjugate field
homomorphismsK CwhoseimagesarenotembeddedinR. RecallthattheinfiniteplacesofK correspond
→
to the real embeddings of K and the pairs of complex embeddings of K. Furthermore, the finite places of
K correspond to the prime ideals of the ring of integers of K.
2000 Mathematics Subject Classification. 16S34, 20C05,16A26,16U60,11R27.
Keywords and phrases. Groupalgebras,Kleiniangroups,groupsofunits.
ResearchsupportedbyM.E.C.ofRomania(CEEX-ET47/2006), D.G.I.ofSpainandFundacio´nS´eneca ofMurcia.
1
2 GABRIELAOLTEANUANDA´NGELDELR´IO
A Schur algebra over K is a central simple K-algebra A which is generated over K by a finite subgroup
of the group of units A of A. Equivalently, a Schur algebra over K is a simple factor, with center K, of a
∗
group algebra of a finite group.
If L/K is a finite cyclic extension of degree n with Gal(L/K)= σ and a K then (L/K,σ,a) denotes
∗
h i ∈
the cyclic algebra L[uuxu 1 = σ(x),un = a]. Sometimes we abbreviate (L/K,σ,a) by writing (L/K,a), if
−
|
the generator σ is either clear from the context or not relevant. Recall that (L/K,a)is split if and only if a
is a norm of the extension L/K [Rei, Theorem 30.4].
A cyclic cyclotomic algebra is a cyclic algebra (L/K,a) with L/K is a cyclotomic extension and a is a
root of unity. A cyclic cyclotomic algebra (L/K,a) is a Schur algebra because it is generatedover K by the
finite metacyclic group u,ζ , where ζ is a root of unity of L such that L=K(ζ).
h i
Quaternion algebras are cyclic algebras of degree 2 and take the form a,b = K[i,j i2 = a,j2 = b,ji=
K |
−ij], for a,b ∈ K∗. We abbreviate H(K) = −1K,−1 . If A = aK,b and(cid:16)σ is(cid:17)a real embedding of K then
A is said to ramify at σ if R σ(K) A H((cid:0)R), or(cid:1)equivalent(cid:16)ly, i(cid:17)f σ(a),σ(b) < 0. Recall that a totally
⊗ ≃
definite quaternion algebra is a quaternion algebra A over a totally real field which is ramified at every real
embedding.
LetGbe afinitegroup. Astrong Shoda pair ofGisapair(M,L)ofsubgroupsofGsuchthatL(cid:2)M(cid:2)G,
M/L is cyclic and M/L is maximal abelian in N (L)/L. (The definition in [ORS] is more general but for
G
our purposes we do not need such a generality.) If M = L (and hence M = G), then let ε(M,M) = M =
1 m QM; otherwise, let ε(M,L) = (L S), where S runs on the minimal subgroups of M
M m M ∈ −
c|on|taini∈ng L properly. Finally, let e(G,M,L) denote the sum of the different G-conjugates of ε(M,L).c
P Q
For every positive integer n, ζ denotes a complebx prbimitive n-th root of unity. We quote the following
n
result from [ORS].
Proposition1.1. LetGbeafinitegroupand(M,L)astrongShodapairofG. LetN =N (L),k =[M :L]
G
and n=[G:N]. Then e=e(G,M,L) is a primitive central idempotent of QG and QGe is isomorphic with
M (Q(ζ ) σN/M), an n n-matrix algebra over a crossed product of N/M over the cyclotomic field Q(ζ ),
n k ∗τ × k
with defining action and twisting given as follows: Let x be a generator of M/L and let γ :N/M N/L be
→
a left inverse of the natural epimorphism N/L N/M. Then for every a,b N/M one has
→ ∈
ζkσ(a) =ζki, if xγ(a) =xi and τ(a,b)=ζkj, if γ(ab)−1γ(a)γ(b)=xj.
In [ORS] it was proved that if G is a finite metabelian groups, then every primitive central idempotent
of QG is of the form e(G,M,L) for some strong Shoda pair (M,L) of G. This can be used to compute the
Wedderburn decompositionof QG for G metabelian. A method to compute the Wedderburn decomposition
of QG for G an arbitrary finite group is given in [Olt]. This method has been implemented in the GAP
packagewedderga[BKOOR]. Wewillmakeuseseveraltimes ofthismethodatdifferentstagesofthepaper.
We quote the following proposition from [JPRRZ].
Proposition 1.2. The following are equivalent for a central simple algebra A over a number field K.
(1) A is of Kleinian type.
(2) A is either a numberfield or a quaternion algebra which is not ramified at at most one infinite place.
(3) One of the following conditions holds:
(a) A=K.
(b) A is a totally definite quaternion algebra.
(c) A M (Q).
2
≃
(d) A M (Q(√d)), for d a square-free negative integer.
2
≃
(e) A is a division algebra, K is totally real and A ramifies at all but one real embeddings of K.
(f) A is a division algebra, K has exactly one pair of complex (non-real) embeddings and A ramifies
at all real embeddings of K.
GROUP ALGEBRAS OF KLEINIAN TYPE AND GROUPS OF UNITS 3
We need the following lemmas.
Lemma 1.3. If K =Q(√d) with d a square-free negative integer then
(1) H(K) is a division algebra if and only if d 1 mod 8.
(2) −1K,−3 is a division algebra if and only if≡d≡1 mod 3.
Proof. ((cid:0)1) Writ(cid:1)ing H(K) as (K(ζ4)/K, 1) one has that H(K) is a division algebra if and only if 1 is a
− −
sum of two squares in K. It is well known that this is equivalent to d 1 mod 8 [FGS].
(2) Assume first that A = −1K,−3 is not split. Then A is ramified≡at at least two finite places p1 and
p . Writing A as (K(ζ )/K, 1) and using [Rei, Theorem 14.1], one deduces that p and p are divisors
2 3 1 2
−(cid:0) (cid:1)
of 3. Thus 3 is totally ramified in K and this implies that D = 1, where D is the discriminant of K
3
[BS, Theorem 3.8.1]. Since D = d or D = 4d and p p mod 3, for each rational prime p, one has
3 ≡ (cid:0) (cid:1)
d= d = D 1 mod 3.
3 3 ≡ (cid:0) (cid:1)
Conversely, assume that d 1 mod 3. Then 3 is totally ramified in K. Let p be a prime divisor of 3 in
(cid:0) (cid:1) (cid:0) (cid:1) ≡
K. Then the residue field of K has order 3 and K (ζ )/K is the unique unramified extension of degree 2
p p 4 p
of K [Rei, Theorem 5.8]. Since v ( 3)= 1, we deduce from [Rei, Theorem 14.1] that 3 is not a norm of
p p
the extension K (ζ )/K . Thus K − A=(K (ζ )/K , 3) is a division algebra, henc−e so is A. (cid:3)
p 4 p p K p 4 p
⊗ −
Lemma 1.4. Let D be a division quaternion Schur algebra over a number field K. Then D is generated
over K by a metabelian subgroup of D .
∗
Proof. By means of contradictionwe assume that D is not generated over K by a metabelian group. Using
Amitsur’s classification of the finite subgroups of division rings (see [Ami] or [SW]) we deduce that D is
generatedby a groupG which is isomorphicto one ofthe following three groups: , the binary octahedral
∗
O
group of order 48; SL(2,5), the binary icosahedral group of order 120; or SL(2,3) M, where M is a
×
metacyclic group. Recall that = x,y,a,bx4 = x2y2 = x2b2 = a3 = 1,ab = a 1,xy = x 1,xb = y,xa =
∗ − −
O h |
x 1y,ya =x 1 .
− −
i
We may assume without loss of generality that G is one of these three groups. Let D be the rational
1
subalgebra of D generated by G. It is enough to show that D is generated over Q by a metabelian group.
1
So we may assume that D =Q(G) and so D is one of the factors of the Wedderburn decomposition of QG.
Computing the WedderburndecompositionofQ andQSL(2,5)andhavinginmind thatD hasdegree
∗
O
2 we obtain that D (Q(ζ )/Q(√2), 1), if G= and D (Q(ζ )/Q(√5), 1), if G=SL(2,5). In both
8 ∗ 5
≃ − O ≃ −
cases D is generated over its center by a finite metacyclic group.
Finally, assume that G = SL(2,3) M, with M metacyclic. Then D is a simple factor of A A ,
1 Q 2
× ⊗
where A is a simple epimorphic image of QSL(2,3) and A is a simple epimorphic image of QM. If A
1 2 1
is commutative, then D is obviously generated by a metabelian group. Assume otherwise that A is not
1
commutative. Computing the Wedderburn decomposition of QSL(2,3) we have that A is isomorphic to
1
either H(Q) or M (Q(√3)). Since D is a divisionalgebra then the second optiondoes not hold. Since H(Q)
2
is generated over Q by the quaternion group of order 8, A A is generated over Q by a direct product
1 Q 2
of two metacyclic groups, hence so is D. ⊗ (cid:3)
For a positive integer n we set
η =ζ +ζ 1 and λ =ζ ζ 1.
n n n− n n− n−
Observe that η2 λ2 = 4 and hence Q(η2) = Q(λ2). Furthermore, if n 3, then λ2 is totally negative
n − n n n ≥ n
because if the integer i is relatively prime with n then ζn2i+ζn−2i ≤−1. Therefore, if λ2n ∈K then λ2nK,−1
ramifies at every real embedding of K. (cid:16) (cid:17)
We are ready to classify the Schur algebras of Kleinian type.
Theorem 1.5. Let K be a number field and let A be a non-commutative central simple K-algebra. Then A
is a Schur algebra of Kleinian type if and only if A is isomorphic to one of the following algebras:
4 GABRIELAOLTEANUANDA´NGELDELR´IO
(1) M (K), for K =Q or Q(√d) for d a square-free negative integer.
2
(2) H(Q(√d)), for d a square-free negative integer, such that d 1 mod 8.
≡
(3) −1,−3 , for d a square-free negative integer, such that d 1 mod 3.
Q(√d) ≡
(4) (cid:16)λ2nK,−1(cid:17), where n ≥ 3, ηn ∈ K and K has at least one real embedding and at most one pair of
(cid:16)complex(cid:17)(non-real) embeddings.
Proof. That the algebras listed are of Kleinian type follows at once from Proposition1.2.
Now we prove that the algebras listed are cyclic cyclotomic algebras and so they are Schur algebras.
Indeed, if A=M (K) as in (1), then A (K(ζ )/K,1) with n=3 if K =Q(ζ ) and n=4 otherwise. If A
2 n 4
≃
is a division algebra and either A=H(K) or A= −1K,−3 , for K =Q(√d), then K does not contain ζ4 nor
ζ3 (see Lemma 1.3) and hence H(K)=(K(ζ4)/K,(cid:0)−1) an(cid:1)d −1K,−3 =(K(ζ3)/K,−1). Finally, assume that
A = λ2nK,−1 . Let B = (K(ζn)/K,−1) = K(ζn)[u|uζnu−1(cid:0)= ζn−1(cid:1),u2 = −1], a cyclic cyclotomic algebra.
Then(cid:16)B =K(cid:17)[λn,u], and uλn =−λnu. Thus, B ≃ λ2nK,−1 =A.
Now we provethat if A is a SchuralgebraofKle(cid:16)inianty(cid:17)pe then one ofthe cases(1)-(4)holds. If A is not
a division algebra,Proposition1.2 implies that A=M (K) for K =Q or an imaginary quadratic extension
2
of Q, so (1) holds.
IntheremainderoftheproofweassumethatAisadivisionSchuralgebraofKleiniantype. ByLemma1.4,
A is generated over K by a finite metabelian group G. Then A=K B, where B is a simple epimorphic
L
⊗
image ofQGwith centerL and, by Proposition1.1,B is a cyclic cyclotomicalgebra(Q(ζ )/L,ζa) ofdegree
n n
2. Since A is of Kleinian type, so is B.
NowweprovethatListotallyreal. Otherwise,sinceLisaGaloisextensionofQ,Listotallycomplexand
therefore K is also totally complex. By Proposition 1.2, both L and K are imaginary quadratic extensions
of Q and so L = K and ϕ(n) = 4, where ϕ is the Euler function. Then either (a) n = 8 and K = Q(ζ ) or
4
K =Q(√ 2); or (b) n=12 and K =Q(ζ ) or K =Q(ζ ). If n=8, then B is generatedoverQ by a group
4 3
−
of order 16 containing an element of order 8. Since B is a division algebra,G=Q and so B =H(Q(√2)),
16
a contradiction. Thus n=12andhence B =(Q(ζ )/Q(ζ ),ζa), where d=6or 4. Since ζ is a normofthe
12 d d 6
extension Q(ζ12)/Q(ζ3), necessarily d = 4. So A = B = (Q(ζ12)/Q(ζ4),ζ4a) = ζQ4a(,ζ−4)3 . Since X = 1+ζ4,
Y =ζ is a solution of the equationζ X2 3Y2 =1, ζ is a norm of the extensi(cid:16)onQ(ζ(cid:17) )/Q(ζ ), and hence
4 4 4 12 4
−
so is ζa, yielding a contradiction.
4
So L is a totally real field of index 2 in Q(ζ ). Then L = Q(η ) and necessarily B is isomorphic to
n n
(Q(ζn)/Q(ηn),−1) ≃ λ2nL,−1 . This imply that A ≃ λ2nK,−1 . If K has some embedding in R, then (4)
holds. Otherwise K =(cid:16) Q(√d(cid:17)), for some square-free n(cid:16)egative(cid:17)integer d. This implies that L = Q. Then
n = 3, 4 or 6 and so A is isomorphic to either H(K) or −1K,−3 . Since A is a division algebra, Lemma 1.3
implies that, in the first case, d 1 mod 8 and condition (2) holds, and, in the second case, d 1 mod 3
and condition (3) holds. ≡ (cid:0) (cid:1) ≡ (cid:3)
2. Group algebras
In this section we classify the group algebras of Kleinian type KG for K a number field and G a finite
group. The classification for K =Q was given in [JPRRZ].
We start with some notation. The cyclic group of order n is usually denoted by C . To emphasize that
n
a C is a generatorof the group,we write C either as a or a . Recall that a group G is metabelian if
n n n
∈ h i h i
G has an abelian normal subgroup N such that A=G/N is abelian. We simply denote this information as
G=N :A. To give a concrete presentation of G we will write N and A as direct products of cyclic groups
and give the necessary extra information on the relations between these generators. By x we denote the
coset xN. For example, the dihedral group of order 2n and the quaternion group of order 4n can be given
GROUP ALGEBRAS OF KLEINIAN TYPE AND GROUPS OF UNITS 5
by
D = a : b , b2 =1,ab =a 1.
2n n 2 −
h i h i
Q = a : b , ab =a 1, b2 =an.
4n 2n 2 −
h i h i
If N has a complement in G then A can be identify with this complement and we write G = N ⋊A. For
example, the dihedral group also can be given by D = a ⋊ b with ab = a 1 and the semidihedral
2n n 2 −
h i h i
groups of order 2n+2 can be described as
DD22+−nn++22 == hhaaii22nn++11 ⋊⋊hhbbii22,, aabb ==aa22nn+−11..
Followingthenotationin[JPRRZ],forafinitegroupG,wedenoteby (G)thesetofisomorphismclasses
C
ofnoncommutativesimple quotientsofQG. Wegeneralizethisnotationand,forasemisimple groupalgebra
KG, we denote by (KG) the set of isomorphism classes of noncommutative simple quotients of KG. For
C
simplicity, we represent (G) by listing a set of representatives of its elements. For example, using the
C
isomorphisms
QD1−6 ∼=4Q⊕M2(Q)⊕M2(Q(√−2)) and QD1+6 ∼=4Q⊕2Q(i)⊕M2(Q(i))
one deduces that C(D1+6) = {M2(Q(i))} and C(D1−6) = {M2(Q),M2(Q(√−2))}. Notice that C(KG) =
KZ(A) A:A (G) . IfK isanumberfieldthenKGisofKleiniantypeifandonlyifeachelement
Z(A)
{ ⊗ ∈C }
of (KG) satisfies one of the conditions of Theorem 1.5.
C
The following groups play an important role in the classification of groups of Kleinian type.
= t x2 y2 :( x y ), with t=(y,x) and Z( )= x2,y2,t .
2 2 2 2 2
W h i ×h i ×h i h i ×h i W h i
n n
1n = (cid:0) ti 2 yi 2 (cid:1)⋊ x 4, with ti =(yi,x) and Z( 1n)= t1,...,tn,x2 .
W h i × h i h i W h i
(cid:18)i=1 i=1 (cid:19)
Qn Q
= y ⋊ x , with t =(y ,x)=y2 and Z( )= t ,...,t ,x2 .
W2n h ii4 h i4 i i i W2n h 1 n i
(cid:18)i=1 (cid:19)
= tQ2 x2 4 y2 4 :( x 2 y 2), with t=(y,x) and Z( )= x2,y2,t .
V h i ×h i ×h i h i ×h i W h i
n n
V1n = (cid:0) htii2× hyii4 (cid:1)⋊hxi8, with ti =(yi,x) and Z(V1n)=ht1,...,tn,y12,...,yn2,x2i.
(cid:18)i=1 i=1 (cid:19)
Qn Q
= y ⋊ x , with t =(y ,x)=y4 and Z( )= t ,x2 .
V2n h ii8 h i8 i i i V2n h i i
(cid:18)i=1 (cid:19)
Q 3 3
= t y2 : y , with t =(y ,y ) and
U1 1 i<j 3h iji2×k=1h ki2! (cid:18)k=1h ki2(cid:19) ij j i
≤ ≤
Z( Q)= t ,t ,tQ,y2,y2,y2 .Q
U1 h 12 13 23 1 2 3i
3
= t y2 y2 y2 : y , with t =(y ,y ), y4 =t , y4 =t and
U2 h 23i2×h 1i2×h 2i4×h 3i4 h ki2 ij j i 2 12 3 13
(cid:18)k=1 (cid:19)
(cid:0)Z(U2)=ht12,t13,t23,y12,y22,y32i(cid:1). Q
= ( t y ): x , with t=(y,x) and x2 =t2 =(x,t).
4 8 2
T h i ×h i h i
n n
= t y ⋊ x , with t =(y ,x), (t ,x)=t2 and Z( )= t2,...,t2,x2 .
T1n h ii4× h ii4 h i8 i i i i T1n h 1 n i
(cid:18)i=1 i=1 (cid:19)
T2n = Qn hyii8 ⋊Qhxi4, with ti =(yi,x)=yi−2 and Z(T2n)=ht21,...,t2n,x2i.
(cid:18)i=1 (cid:19)
Q n
= y2t y y : x , with t =(y ,x), (t ,x)=t2, x2 =t2,
T3n h 1 1i2×h 1i8× h ii4 h i2 i i i i 1
(cid:18) i=2 (cid:19)
Z(T3n)=ht21,y22,...Q,yn2,x2i and, if i≥2 then ti =yi2,
= Cn⋊P =(Cn Q): x , with Q a subgroup of index 2 in P and zx =z 1 for each z Cn.
Sn,P,Q 3 3 × h i2 − ∈ 3
We collect the following lemmas from [JPRRZ].
6 GABRIELAOLTEANUANDA´NGELDELR´IO
Lemma2.1. LetGbeafinitegroupandAanabelian subgroupofGsuchthateverysubgroupofAisnormal
in G. Let = H H is a subgroup of A with A/H cyclic and G H . Then (G)= (G/H).
′ H
H { | 6⊆ } C ∪ ∈HC
Lemma 2.2. Let A be a finite abelian group of exponent d and G an arbitrary group.
(1) If d2 then (A G)= (G).
| C × C
(2) If d|4 and C(G)⊆ M2(Q),H(Q), −1Q,−3 ,M2(Q(ζ4)) then C(A×G)⊆C(G)∪{M2(Q(ζ4))}.
(3) If d|6 and C(G)⊆nM2(Q),H(Q),(cid:16)−1Q,−3(cid:17),M2(Q(ζ3))o then C(A×G)⊆C(G)∪{M2(Q(ζ3))}.
n (cid:16) (cid:17) o
Lemma 2.3. (1) ( )= M (Q) .
1n 2
C W { }
(2) ( )= ( )= M (Q),H(Q) .
2n 2
C W C W { }
(3) ( ), ( ), ( ), ( ), ( ), ( ) M (Q),H(Q),M (Q(ζ )) .
1n 2n 1 2 1n 2 2 4
C V C V C V C U C U C T ⊆{ }
(4) ( ), ( ), ( ) M (Q),H(Q),M (Q(ζ )),H(Q(√2)),M (Q(√ 2)) .
2n 3n 2 2 4 2
C T C T C T ⊆{ − }
(5) Let G= .
n,P,Q
S
(a) If P = x is cyclic of order 2n then (G) = (G/ x2 ) ζ2n−1,−3 . In particular, if
h i C C h i ∪ Q(ζ2n−1)
P =C2 then C(G)={M2(Q)}, if P =C4 then C(G)= M2(Qn)(cid:16), −1Q,−3 (cid:17)oand if P =C8 then
C(G)= M2(Q), −1Q,−3 ,M2(Q(ζ4)) . n (cid:16) (cid:17)o
(b) If P =Wn1n and Q(cid:16) =hy1(cid:17),...,yn,t1,..o.,tn,x2i then C(G)= M2(Q), −1Q,−3 ,M2(Q(ζ3)) .
(c) If P = and Q= y2,x then (G)= M (Q),H(Q(√3))n,M (Q(ζ(cid:16))),M ((cid:17)Q(ζ )) . o
W21 h 1 i C { 2 2 4 2 3 }
We are ready to present our classification of the group algebras of Kleinian type.
Theorem 2.4. Let K be a number field and G a finite group. Then KG is of Kleinian type if and only if G
is either abelian or an epimorphic image of A H for Aan abelian group and one of the following conditions
×
holds:
(1) K =Q and one of the following conditions holds.
(a) A has exponent 6 and H is either , or , for some n, or H = with
Q= y ,...,y ,t ,...,t ,x2 , for soWme Wn a1nnd m.W2n Sm,W1n,Q
1 m 1 m
h i
(b) A has exponent 4 and H is either , , , , or , for some n.
U1 U2 V V1n V2n Sn,C8,C4
(c) A has exponent 2 and H is either , , , or with Q= y2,x , for some n.
(2) K =Q and has at most one pair of comTpleTx1n(noTn2-nreaTl3)nembeSdnd,Win2g1s,Q, A has expohne1nt i2 and H =Q .
8
6
(3) K is an imaginary quadratic extension of Q, A has exponent 2 and H is either , , or
1n 2n
W W W
, for some n.
Sn,C4,C2
(4) K =Q(ζ ), A has exponent 6 and H is either , or , for some n, or H = with
Q= y ,3...,y ,t ,...,t ,x2 , for some n andWm.W1n W2n Sm,W1n,Q
1 m 1 m
h i
(5) K =Q(ζ ), A has exponent 4 and H is either , , , , , or , for some n.
4 U1 U2 V V1n V2n T1n Sn,C8,C4
(6) K =Q(√−2), A has exponent 2 and H is either D1−6 or T2n, for some n.
Proof. To avoidtrivialities we assume that G is non-abelian. The main theoremof[JPRRZ]states that QG
is of Kleinian type if and only if G is an epimorphic image of A H for A abelian and A and H satisfy one
×
of the conditions (a)-(c) from (1). So, in the remainder of the proof, we assume that K =Q.
6
First we prove that if one of the conditions (2)-(6) holds, then KG is of Kleinian type. Clearly, if KG is
of Kleinian type and H is an epimorphic image of G, then KH is also of Kleinian type. So we may assume
that G = A H with A, H and K satisfying one of the conditions listed. We compute (KG) in all the
× C
cases.
If (2) holds, then (KG)= H(K) .
C { }
If (3) holds then, by Lemmas 2.2 and 2.3, one has C(G) ⊆ {M2(Q),H(Q), −1Q,−3 }, and so C(KG) ⊆
{M2(K),H(K), −1K,−3 }. (cid:16) (cid:17)
(cid:0) (cid:1)
GROUP ALGEBRAS OF KLEINIAN TYPE AND GROUPS OF UNITS 7
Similarly, if (4) holds then C(G) ⊆ C(H)∪{M2(Q(ζ3))} ⊆ {M2(Q),H(Q), −1Q,−3 ,M2(Q(ζ3))}. Hence
(KG)= M2(Q(ζ3)) , by Lemma 1.3. (cid:16) (cid:17)
C { }
Arguing similarly one deduces that if (5) holds then (KG)= M (Q(ζ )) .
2 4
C { }
Finally, assume that (6) holds. If H = D1−6 then C(G) = {M2(Q),M2(Q(√−2))} and so C(KG) =
M (Q(√ 2)) . Otherwise,thatisifH = ,forsomen,weshowthat (KG)= H(Q),M (Q),M (Q(√ 2)) .
2 2n 2 2
{ − } T C { − }
Let L be a proper subgroup of H such that H /L is cyclic. Using that (y,x)y2 = 1, for each y
′ ′
hy1,...,yni one has that T2n/L is an epimorphic image of T21 × C2n−1. Then Lemmas 2.1 and 2.2 im∈-
ply that (G) = ( ). So we may assume that G = . Now take B = Z( ) = t2,x2 C2 and L a
C C T21 T21 T21 h i ≃ 2
subgroup of B such that B/L is cyclic. If t2 L, then /L is an epimorphic image of and therefore
21
∈ T W
C(T21/L) ⊆ {M2(Q),H(Q)}. Otherwise L = hx2i or L = hx2t2i; hence T21/L ≃ D1−6 and so C(T21/L) ⊆
M (Q),M (Q(√ 2)) . Using Lemma 2.1 one deduces that ( )= H(Q),M (Q),M (Q(√ 2)) .
2 2 2n 2 2
{ − } C T { − }
Now it is clear that in all the cases the elements of (KG) are of one of the types (1)-(4) of Theorem 1.5
C
and so KG is of Kleinian type.
For the remainder of the proof we assume that KG is of Kleinian type, G is non-abelian (and K = Q).
6
Then QG is of Kleinian type, that is G is an epimorphic image of A H for A and H satisfying one of the
×
conditions (a)-(c) from (1). Furthermore, K has at most one pair of complex embeddings, by Theorem 1.5.
We have to show that K and G satisfy one of the conditions (2)-(6). We consider several cases.
Case 1. Every element of (G) is a division algebra.
C
This implies that G is Hamiltonian and so G Q E F with E elementary abelian 2-group and F
8
≃ × ×
abelian of odd order [Rob, 5.3.7]. If F =1, then (2) holds. Otherwise, (KG) contains H(K(ζ )), where n
n
C
is the exponent of F. Thereforen=3 andK =Q(ζ ), by Theorem1.5. Since Q is an epimorphic image of
3 8
, condition (4) holds.
W
In the remainder of the proof we assume that (G) contains a non-division algebra B. Then B =M (L)
2
C
for some field L and therefore M (KL) (KG). Since K =Q, KL is an imaginary quadratic extension of
2
∈C 6
Q and L K. Let E be the center of an element of (G). Then KE is the center of an element of (KG).
⊆ C C
If KE =K,then the two complex embeddings of K extends to more than two complex embeddings ofKE,
6
yielding a contradiction. This shows that K contains the center of each element of (G).
C
Case 2. The center of each element of (G) is Q.
C
ThenLemmas2.2and2.3implythatC(G)⊆{M2(Q),H(Q), −1Q,−3 }. Usingthis andthe maintheorem
of [LR] one has that G=A H, where A is an elementary abe(cid:16)lian 2-g(cid:17)roup and H is an epimorphic image
×
of , , or , for some n. So G satisfies (3).
W W1n W2n Sn,C4,C2
Case 3. At least one element of (G) has center different from Q.
C
Then the center of each element of (G) is either Q or K. Using Lemmas 2.2 and 2.3 one has: If A and
C
H satisfy condition (1.a) then K = Q(ζ ), hence (4) holds. If either A and H satisfy (1.b) or they satisfy
3
(1.c) with H = , for some n, then K =Q(ζ ) and condition (5) holds.
1n 4
T
Otherwise, A has exponent 2 and H is either , , , for some n, or , for Q= y2,x . Since
T T2n T3n Sn,W21,Q h 1 i
A has exponent 2, one may assume that G = A H , for H an epimorphic image of H and H is not an
1 1 1
×
epimorphic image of any of the groups considered above. We use the standard bar notation for the images
of QH in QH .
1
Assume first that H = . Then (M = y,t ,L = ty 2 ) is a strong Shoda pair of and, by using
−
Proposition 1.1, one deduceTs that if e = e(Th,M,iL) = Lh1−2y4i, then QGe ≃ H(Q(√2)). SiTnce H(Q(√2)) is
not of Kleinian type, we have that e=0, or equivalently y4 L. Hence either y4 =1, t2 =1 or t=y 2. So
−
∈
H is an epimorphic image of either / y4 , / t2 or b/ ty2 . In the first case H is an epimorphic image
1 1
T h i T h i T h i
of , in the second case H is an epimorphic image of , which contradicts the hypothesis that H is not
11 1 1
T V
an epimorphic image of the groups considered above. Thus H1 is an epimorphic image of T/hty2i ≃ D1−6.
In fact H1 = D1−6, because every proper non-abelian quotient of D1−6 is an epimorphic image of W. Then
C(G)=C(D1−6)={M2(Q),M2(Q(√−2))} and so K =Q(√−2). Hence condition (6) holds.
8 GABRIELAOLTEANUANDA´NGELDELR´IO
Assume nowthat H = . By the firstpartofthe proofwe have (H)= H(Q),M (Q),M (Q(√ 2)) .
2n 2 2
T C { − }
Since we are assuming that one element of (G) has center different from Q, then M (Q(√ 2)) (H )
2 1
C − ∈ C
and so K =Q(√ 2). Hence condition (6) holds.
−
In the two remaining cases we are going to obtain some contradiction.
SupposethatH = . WemayassumethatnistheminimalpositiveintegersuchthatGisanepimorphic
3n
T
imageof AforAanelementaryabelian2-group. Thisimpliesthat t2,t ,...,t iselementaryabelianof
T3n× h 1 2 ni
order 2n and hence hy12,y2,...,yni≃C4n. Let M =hy1,y2,...,yni and L1 =ht1y1−2,y2,y3,...,yni). Then
(M,L )isastrongShodapairofH. ByusingProposition1.1,weobtainthatQHe(H,M,L ) H(Q(√2)).
1 1
≃
This implies that e(H,M,L ) = 0, or equivalently t2 = y4 L . Since t2 has order 2 and t2 t ,...,t
1 1 1 ∈ 1 1 1 6∈ h 2 ni
one has t21 = t1y1−2tα22···tαnn for some α1,...,αn ∈ {0,1}. Since, by assumption, H1 is not an epimorphic
image of , we have α = 0 for some i 2. After changing generators one may assume that α = 1
2n i 2
T 6 ≥
and αi = 0 for i ≥ 3. Thus t1 = y1−2t2. Let now L2 = ht1y1−2,y2y1−2,y3,...,yni. Then (M,L2) is also a
strong Shoda pair of H and QGe(H,M,L ) H(Q(√2)). The same argument shows that y 4 =t2 L =
2 ≃ 1 1 ∈ 2
ht1y1−2,y2y1−2,y3,...,yni. This yields a contradiction because t1y1−2 = (y2y1−2)2 ∈ hy2y1−2,y3,...,yni and
y14 6∈hy2y1−2,y3,...,yni.
Finally assume that H = , with Q = y2,x and set y = y . Since, by assumption, G does not
Sn,W21,Q h 1 i 1
satisfy(1.a),onehast=t =1. Moreover,asinthepreviouscaseonemayassumethatnisminimal(forGto
1
6
beaquotientofH A,withAelementaryabelian2-group). LetM = Cn,x,t andL= Z ,tx2 ,whereZ
× h 3 i h 1 i 1
isamaximalsubgroupofZ =Cn. Then(M,L)isastrongShodapairofH andQHe(H,M,L) H(Q(√3)).
3 ≃
Thus 0 = e(H,M,L) = L(1 z)(1 t), where z Z Z . Comparing coefficients and using the fact that
1
− − ∈ \
t=z, for each z Z, we have L(1 z)=0, that is L=Z and this contradicts the minimality of n. (cid:3)
6 ∈ b b − b
b b 3. Groups of units
InthissectionwestudythevirtualstructureofRG forGafinitegroupandRanorderinanumberfield
∗
K. Moreprecisely,wecharacterizethefinitegroupsGandnumberfieldsK forwhichRG isfinite,virtually
∗
abelian, virtually a direct product of free groupsor virtually a directproduct of free-by-freegroups. We say
thatagroupvirtually satisfiesa grouptheoreticalconditionifithas asubgroupoffinite index satisfyingthe
given condition. Notice that the virtual structure of RG does not depends on the order R and, in fact, if
∗
S is any order in KG, then S and RG are commensurable (see e.g. [Seh, Lemma 4.6]). Recall that two
∗ ∗
subgroups of a given group are said to be commensurable if their intersection has finite index in both. It is
easy to show that a group commensurable with a free group (respectively, a free-by-free, a direct product
of free groups, a direct product of free-by-free groups) it is virtually free (respectively, free-by-free, a direct
product of free groups, a direct product of free-by-free groups).
One important tool is the following lemma.
Lemma 3.1. Let A= n A be a finite dimensional rational algebra with A simple for every i. Let S be
i=1 i i
an order in A and S an order in A .
i i
Q
(1) S is finite if and only if for each i, A is either Q, an imaginary quadratic extension of Q or a
∗ i
totally definite quaternion algebra over Q.
(2) S is virtually abelian if and only if for each i, A is either a number field or a totally definite
∗ i
quaternion algebra.
(3) S is virtually a direct product of free groups if and only if for each i, A is either a number field, a
∗ i
totally definite quaternion algebra or M (Q).
2
(4) S is virtually a direct product of free-by-free groups if and only if for each i, S1 is virtually free-by-
∗ i
free.
Proof. We are going to use the following facts:
GROUP ALGEBRAS OF KLEINIAN TYPE AND GROUPS OF UNITS 9
(a) S is commensurable with n S and S is commensurable with Z(S ) S1. (This is because S
∗ n i=1 i∗ i∗ i ∗× i
and S are both orders in A.)
i=1 i Q
(b) S1 isfinite ifandonlyifA is eitherafieldoratotallydefinite quaternionalgebra(see[Seh,Lemma
i Q i
21.3] or [Kle1]).
(c) If A is neither a field nor a totally definite quaternion algebra then if S1 is commensurable with a
i i
direct product of groups G and G then either G or G is finite [KR].
1 2 1 2
(d) S1 is infinite and virtually free if and only if A M (Q) (see e.g. [Kle2, page 233]).
i i ≃ 2
(1) By (a), S is finite if and only if S is finite for each i if and only if Z(S ) and S1 are finite for each
∗ i∗ i ∗ i
i. By the Dirichlet Unit Theorem, if A is a number field, then S is finite if and only if A is either Q or an
i i∗ i
imaginary quadratic extension of Q. Using this and (b), one deduces that if A is not a number field then
i
Z(S ) and S1 are finite if and only if A is a totally definite quaternion algebra over Q.
i ∗ i i
(2)Clearly S is virtually abelianif andonly if S1 is virtually abelianfor eachi. If A is either a number
∗ i i
field or a totally definitive quaternion then S is virtually abelian, by (b). Assume otherwise that A is
i∗ i
neither commutative nor a totally definite quaternion algebra. Using (c) and the fact that S1 is finitely
i
generated one deduces that S1 is virtually abelian, then it is virtually cyclic and so A = M (Q), by (d).
i i 2
This yields a contradiction because SL (Z) contains a non-abelian free group.
2
(3) By (a) and [JR, Lemma 3.1], S is virtually a direct product of free groups if and only if so is S1 for
∗ i
eachi. As in the previous proof,if A is neither commutative nor a totally definite quaternionalgebra,then
i
S1 is virtually a direct product of free groups if and only if S1 is virtually free if and only if A M (Q).
i (4) Is proved in [JPRRZ, Theorem 2.1]. i i ≃ 2 (cid:3)
The characterization of when RG is finite or virtually abelian are easy generalizations of known results
∗
for integral group rings.
Theorem 3.2. Let R be an order in a number field K and G a finite group. Then RG is finite if and only
∗
if one of the following conditions holds:
(1) K = Q and G is either abelian of exponent dividing 4 or 6, or isomorphic to Q A, for A an
8
×
elementary abelian 2-group.
(2) K is an imaginary quadratic extension of Q and G is an elementary abelian 2-group.
(3) K =Q(ζ ) and G is abelian of exponent 3 or 6.
3
(4) K =Q(ζ ) and G is abelian of exponent 4.
4
Proof. If K =Q, then R=Z and it is well known that ZG is finite if and only if G is abelian of exponent
∗
dividing 4 or 6 or it is isomorphic to Q A, for A an elementary abelian 2-group.
8
×
If one of the conditions (1)-(4) holds, then KG is a direct product of copies of Q, imaginary quadratic
extensions of Q and H(Q). Then RG is finite by Lemma 3.1.
∗
Conversely, assume that RG is finite and K = Q. Then ZG is finite and therefore G is either abelian
∗ ∗
6
of exponent dividing 4 or 6 or isomorphic to Q A, for A an elementary abelian 2-group. Moreover,R is
8 ∗
×
finite andsoK =Q(√d)for da square-freenegativeinteger. If the exponentofG is 2then (2)holds. If the
exponent of G is 4 then one of the simple components of KG is Q(√d,ζ ) and therefore d= 1, that is (4)
4
−
holds. If the exponent of G is 3 or 6, then one of the simple components of KG is Q(√d,ζ ), hence d= 3,
3
and therefore (3) holds. −(cid:3)
Theorem 3.3. Let R be an order in a number field K and G a finite group. Then RG is virtually abelian
∗
if and only if either G is abelian or K is totally real and G Q A, for A an elementary abelian 2-group.
8
≃ ×
Proof. As in the proof of Theorem 3.2, the sufficient condition is a direct consequence of Lemma 3.1.
Conversely, assume that RG is virtually abelian. Then ZG is virtually abelian and therefore it does
∗ ∗
not contain a non-abelian free group. Then G is either abelian or isomorphic to G Q A, for A an
8
≃ ×
elementaryabelian2-group[HP]. In the latter caseone ofthe simple components ofKGis H(K)andhence
K is totally real, by Lemma 3.1. (cid:3)
10 GABRIELAOLTEANUANDA´NGELDELR´IO
Theorem 3.4. Let R be an order in a number field K and G a finite group. Then RG is virtually a direct
∗
product of free groups if and only if either G is abelian or one of the following conditions holds:
(1) K =Q and G H A, for A an elementary abelian 2-group and H is either , , / x2 ,
1n 1n
≃ × W W W h i
, / x2 , / x2t , or H = , for some n and t=1,2 or 4.
W2n W2n h i W2n h 1i T3n Sn,C2t,Ct
(2) K is totally real and G Q A, for A an elementary abelian 2-group.
8
≃ ×
Proof. The finite groups G such that ZG is virtually a direct product of free groups were classified in [JR]
∗
and coincides with the groupsabelian orsatisfying condition (1). So, in the remainder ofthe proofone may
assume that R= Z, or equivalently K =Q, and we have to show that RG is virtually a direct product of
∗
6 6
free groups if and only if either G is abelian or (2) holds.
If either G is abelian or (2) holds then RG is virtually abelian, hence RG is virtually a direct product
∗ ∗
of free groups, because it is finitely generated.
Conversely, assume that RG is virtually a direct product of free groups and G is non-abelian. Since
∗
K = Q, M (K) is not a simple quotient of KG, hence Lemma 3.1 implies that every simple quotient of
2
6
KG is either a number field or a totally definite quaternion algebra. In particular, G is Hamiltonian, that
is G = Q A F, where A is an elementary abelian 2-group and F is abelian of odd order. If n is the
8
× ×
exponent of F then H(K(ζ )) is a simple quotient of KG and this implies that n = 1 and K is totally
n
real. (cid:3)
Theorem 3.5. Let R be an order in a number field K and G a finite group. Then RG is virtually a direct
∗
product of free-by-free groups if and only if either G is abelian or one of the following conditions holds:
(1) G is an epimorphic image of A H with A abelian and K, A and H satisfy one of the conditions
×
(1), (4), (5) or (6) of Theorem 2.4.
(2) K is totally real and G A Q , for A an elementary abelian 2-group.
8
≃ ×
(3) K =Q(√d),fordasquare-freenegativeinteger,SL (Z(√d))isvirtuallyfree-by-free,andG A H
2
≃ ×
where A is an elementary abelian 2-group and one of the following holds:
(a) H is either , / x2 , / x2 or , for some n.
W1n W1n h i W2n h i Sn,C2,1
(b) H is either , or / x2t , for some n and d 1 mod 8.
2n 2n 1
W W W h i 6≡
(c) H = for some n and d 1 mod 3.
Sn,C4,C2 6≡
Proof. To avoid trivialities we assume that G is non-abelian.
We first show that if K and G satisfy one of the listed conditions then RG is virtually a direct product
∗
of free-by-free groups. By Lemma 3.1 this is equivalent to show that if B (KG) and S is an order in B
∈ C
thenS1 isvirtuallyfree-by-free. ByusingLemmas2.2and2.3,itiseasytoshowthatifK andGsatisfyone
of the conditions (1) or (2), then B is either a totally definite quaternion algebra or isomorphic to M (K)
2
for K =Q(√d), with d=0, 1, 2or 3. In the firstcase S1 is finite andin the secondcase S1 is virtually
− − −
free-by-free (see Lemma 3.1 and [JPRRZ, Lemma 3.1] or alternatively [Kle1], [MR, page 137] and [WZ]).
If K and G satisfy condition (3) then Lemmas 1.3, 2.2 and 2.3 imply that B = M (Q(√d)). Since S1 and
2
SL (Z[√d]) are commensurable and, by assumption, the latter is virtually free-by-free, we have that S1 is
2
virtually free-by-free.
Conversely,assumethatRG isvirtuallyadirectproductoffree-by-freegroups. LetB beasimplefactor
∗
of KGandS anorderin B. By Lemma 3.1, S1 is virtually free-by-freeandhence the virtualcohomological
dimension of S1 is at most 2. Then B is of Kleinian type by [JPRRZ, Corollary 3.4]. This proves that KG
is of Kleiniantype. Furthermore,B is of one of the types (a)-(f) from Proposition1.2. However,the virtual
cohomological dimension of S1 is 0, if B is of type (a) or (b); 1 if B is of type (c); 2 if it is of type (d) or
(e); and 3 if B is of type (f) [JPRRZ, Remark 3.5]. Thus B is not of type (f). Since every simple factor of
KG contains K, either K is totally real or K is an imaginary quadratic extension of Q and KG is split.
ByTheorem2.4,GisanepimorphicimageofA H withAabelianandK,AandH satisfyingoneofthe
×
conditions (1)-(6) of Theorem 2.4. If they satisfy one of the conditions (1), (4), (5) or (6) of Theorem 2.4,