Rational group algebras of finite groups: from idempotents to units of integral group rings

0.26 MB·English

by  E. Jespers

RATIONAL GROUP ALGEBRAS OF FINITE GROUPS: FROM
IDEMPOTENTS TO UNITS OF INTEGRAL GROUP RINGS
ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
0
1 Abstract. We give an explicit and character-free construction of a complete set of orthogonal
0 primitiveidempotentsofarationalgroupalgebraofafinitenilpotentgroupandafulldescription
2 of the Wedderburn decomposition of such algebras. An immediate consequence is a well-known
result of Roquette on the Schur indices of the simple components of group algebras of finite
n
nilpotentgroups. Asanapplication,weobtainthattheunitgroupoftheintegralgroupringZG
a
of afinite nilpotent group G has a subgroup of finite index that is generated by three nilpotent
J
groups for which we have an explicit description of their generators. Another application is a
8
newconstructionoffreesubgroupsintheunitgroup. Inalltheconstructionsdealtwith,pairsof

subgroups(H,K),calledstrongShodapairs,andexplicitconstructedcentralelementse(G,H,K)
]
A play a crucial role. For arbitrary finite groups we prove that the primitive central idempotents
of the rational group algebras are rational linear combinations of such e(G,H,K), with (H,K)
R strongShodapairsinsubgroupsofG.
.
h
t
a
m
1. Introduction
[
The investigationof the unit group (ZG) of the integralgroupring ZG of a finite groupG has
1 U
a long history and goes back to work of Higman [Hig] and Brauer [Bra]. One of the reasons for
v
the importance of the integral group ring ZG is that it is an algebraic tool that links group and
6
3 ring theory. It was anticipated for a long time that the defining group G would be determined by
2 its integral group ring, i.e. if ZG is isomorphic with ZH for some finite group H then G = H,
∼
1 the isomorphism problem. Roggenkamp and Scott [RS] showed that this indeed is the case if G is
.
1 a nilpotent group. Weiss proved a more general result [Wei], which also confirmed a Zassenhaus
0 conjecture. It was a surprise when Hertweck [Her] gave a counter example to the isomorphism
0 problem. In all these investigations the unit group (ZG) of ZG is of fundamental importance.
1 U
Thereisavastliteratureonthetopic. Forasurveyupto1994,thereaderisreferredtothebooksof
:
v PassmanandSehgal[Pas,Seh1,Seh2]. Amongstmanyothers,duringthepast15years,thefollowing
i
X problems have received a lot of attention (we include some guiding references): construction of
generatorsfor a subgroupof finite index in (ZG) [RS1, RS2, JL1], construction of free subgroups
r U
a [HP, MS2], structure theorems for (ZG) for some classes of groups G [JPdRRZ].
U
Essential in these investigations is to consider ZG as a Z-order in the rational group algebra
QG and to have a detailed understanding of the Wedderburn decomposition of QG. To do so, a
firstimportantstepis to calculatethe primitive centralidempotents ofQG. Aclassicalmethod for
this is to apply Galois descent on the primitive central idempotents of the complex group algebra
2000 Mathematics Subject Classification. 20C05,16S34,16U60.
Keywords and phrases. Idempotents, Groupalgebras,Grouprings,Units.
ResearchpartiallysupportedbyOnderzoeksraadofVrijeUniversiteitBrussel,FondsvoorWetenschappelijkOn-
derzoek(Flanders),thegrantPN-II-ID-PCE-2007-1projectID 532,contractno.29/28.09.2007,MinisteriodeCiencia
yTecnolog´ıaofSpainandFundaci´onS´eneca ofMurcia.
1
2 ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
CG. The latter idempotents are the elements of the form e(χ)= χ(1) χ(g−1)g, where χ runs
|G| g∈G
through the irreducible (complex) characters χ of G. Hence the primitive central idempotents of
P
QG are the elements of the form e(σ χ) (see for example [Yam]). Rather recently,
σ∈Gal(Q(χ)/Q) ◦
Olivieri, del R´ıo and Sim´on [OdRS1] obtained a character free method to describe the primitive
P
centralidempotentsofQGprovidedGisamonomialgroup,thatis,everyirreduciblecharacterofG
isinducedfromalinearcharacterofasubgroupofG. ThenewmethodreliesonatheoremofShoda
on pairs of subgroups (H,K) of G with K normal in H, H/K abelian and so that an irreducible
linear character of H with kernel K induces an irreducible character of G. Such pairs are called
Shoda pairs of G. In Section 2 we recall the necessary background and explain the description of
the primitive central idempotents of QG. It turns out that these idempotents can be built using
the centralelementse(G,H,K)(seeSection2forthedefinition)with(H,K)aShodapairofG. In
case the Shoda pair satisfies some extra conditions (one calls sucha pair a strong Shoda pair) then
one also obtains a detailed description of the Wedderburn component associated with the central
idempotent. This is an important second step towards a description of the simple components of
QG. This method is applicable to all abelian-by-supersolvable finite groups, in particular to finite
nilpotent groups.
For arbitraryfinite groups G, it remains an open problem to give a character free description of
the primitive central idempotents of QG. Only for very few groups that are not monomial, such a
description has been obtained (see for example [GJ] for alternating groups). In section 3 we show
that for arbitrary finite groups G the elements e(G,H,K) are building blocks for the construction
of the primitive central idempotents e of QG, i.e. every such e is a rational linear combination of
e(G,H,K), where (H,K) runs through strong Shoda pairs in subgroups of G. The proof makes
fundamental use of Brauer’s Theorem on Induced Characters. Presently we are unable to control
the rational coefficients in this linear combination.
In case G is an abelian-by-supersolvable finite group, then, as mentioned above, the primitive
central idempotents of QG are of the form e(G,H,K), with (H,K) a strong Shoda pair of G and
the simple component QGe(G,H,K) is described. For nilpotent groups G we will show to have a
much better and detailed control. Indeed, in Section 4 we describe a complete set of matrix units
(in particular, a complete set of orthogonal primitive idempotents) of QGe(G,H,K); a third step
in the description of QG. This allows us to give concrete representations of the projections ge of
the group elements g G as matrices over division rings. As a consequence, the recognition of the
∈
Z-order ZG in the Wedderburn description of QG is reduced to a linear algebra problem over the
integers. We include some examples to showthat the methodcannot be extended to, forexample,
finitemetacyclicgroups. Itremainsachallengetoconstructacompletesetofprimitiveidempotents
for such groups.
InSection5,wegiveseveralapplicationstotheunitgroup (ZG)forGafinite nilpotentgroup.
U
First we show that if G is a finite nilpotent group such that QG has no exceptional components
(see Section 5 for the definition) then (ZG) has a subgroup of finite index that is generated by
U
three nilpotent finitely generated groups of which we give explicit generators. The problem of
describing explicitly a finite set of generators for a subgroup of finite index in (ZG) has been
U
investigated in a long series of papers. Bass and Milnor did this for abelian groups [Bas], the
case of nilpotent groups so that their rational group algebra has no exceptional components was
done by Ritter and Sehgal [RS1, RS2], arbitrary finite groups so that their rational group algebra
has no exceptional components were dealt with by Jespers-Leal [JL1]. It was shown that the
Bass cyclic units together with the bicyclic units generate a subgroup of finite index. Some cases
with exceptional components have also been considered, see for example [GS, Jes, JL2, Seh3]). In
RATIONAL GROUP ALGEBRAS OF FINITE GROUPS 3
general,verylittleisknownonthestructureofthegroupgeneratedbytheBasscyclicunitsandthe
bicyclic units, except that “often” two of them generate a free group of rank two (see for example
[GP, GdR, Jes, JdRR, MS2, JL3]). In this paper, for G a finite nilpotent group, we not only give
new generators for a subgroup of finite index, but more importantly, the generating set is divided
into three subsets, one of them generating a subgroup of finite index in the central units and each
of the other two generates a nilpotent group. One other advantage of our method with respect to
the proofs and results given in [JL1, RS1, RS2] is that our proofs are (modulo the central units)
more direct and constructive to obtain an explicit set of generators for a subgroup of finite index
in the group of units of the integral group ring of a finite nilpotent group. Furthermore, we also
give new explicit constructions of free subgroups of rank two.
2. Preliminaries
We introduce some useful notation and results, mainly from[JLP] and [OdRS1]. ThroughoutG
is a finite group. If H is a subgroup of a group G, then let H = 1 h QG. For g G, let
|H| h∈H ∈ ∈
g = g and for non-trivial G, let ε(G)= (1 M), where“M runs tPhrough the set of all minimal
h i −
norm”al nontrivial subgroups of G. Clearly, H is an idempotent of QG which is central if and only
Q
ibf H is normal in G. If K⊳H G then let“ c
≤
ε(H,K)= (K M),
−
M/K∈YM(H/K) “
c
where (H/K) denotes the set of all minimal normal subgroups of H/K. We extend this nota-
M
tion by setting ε(K,K) = K. 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(H,K), that is, if T is a right transversal
“
of Cen (ε(H,K)) in G, then
G
e(G,H,K)= ε(H,K)t,
t∈T
X
where αg = g−1αg for α CG and g G. Clearly, e(G,H,K) is a central element of QG. If the
∈ ∈
G-conjugates of ε(H,K) are orthogonal, then e(G,H,K) is a central idempotent of QG.
A Shoda pair of a finite group G is a pair (H,K) of subgroups of G with the properties that
K EH, H/K is cyclic, and if g G and [H,g] H K then g H. A strong Shoda pair of
∈ ∩ ⊆ ∈
G is a Shoda pair (H,K) of G such that H EN (K) and the different conjugates of ε(H,K) are
G
orthogonal. We also have, in this case, that Cen (ε(H,K)) = N (K) and H/K is a maximal
G G
abelian subgroup of N (K)/K [OdRS1].
G
If χ is a monomial character of G then χ = ψG, the induced character of a linear character
ψ of a subgroup H of G. By a Theorem of Shoda, a monomial character χ = ψG as above is
irreducible if and only if (H,Ker ψ) is a Shoda pair (see [Sho] or [CR, Corollary 45.4]). A finite
group G is monomial if every irreducible character of G is monomial and it is strongly monomial
if every irreducible character of G is strongly monomial. It is well known that every abelian-by-
supersolvable group is monomial (see [Hup, Theorem 24.3]) and in [OdRS1] it is proved that it is
even strongly monomial. We will use these results in order to study the primitive idempotents of
groupalgebrasforsomeabelian-by-supersolvablegroups,including finite nilpotent groups. We will
also use the following description of the simple component associated to a strong Shoda pair.
Theorem 2.1. [OdRS1, Proposition 3.4] If (H,K) is a strong Shoda pair of G then
QGe(G,H,K)=M (QNε(H,K))=M (Q(ζ ) αN/H),
∼ r ∼ r m ∗τ
4 ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
where
m=[H :K], N =N (K), r =[G:N]
G
and the action α and twisting τ are given by
α(nH)(ζ )=ζi , τ(nH,n′H)=ζj ,
m m m
if n−1hnK =hiK and [n,n′]K =hjK, for hK a generator of H/K, n,n′ N and i,j Z.
∈ ∈
In the above theorem, ζ denotes a primitive m-th root of unity and we have used the notation
m
L αG, for L a field and G a group, to denote a crossed product with action α:G Aut(L) and
∗τ →
twisting τ : G G L∗ [Pas], i.e. L αG is the associative ring Lu with multiplication
× → ∗τ g∈G g
given by the following rules:
L
u a=α (a)u , u u =τ(g,h)u .
g g g g h gh
If the action of G on L is faithful then one may identify G with a group of automorphisms of L
and the center of L α G is the fixed subfield F = LG, so that G = Gal(L/F), and this crossed
∗τ
productisusuallydenotedby(L/F,τ)[Rei]. Werefertothesecrossedproductsasclassicalcrossed
products. This is the case for the crossed product QNε(H,K)) = Q(ζ ) αN/H in Theorem 2.1
∼ m ∗τ
which can be described as (Q(ζ )/F,τ), where F is the center of the algebra,which is determined
m
by the Galois action given in Theorem 2.1.
3. Primitive central idempotents
For an irreducible character χ of G and a field F of characteristic 0, e (χ) denotes the only
F
primitivecentralidempotentofFGsuchthatχ(e)=0. Inthis section,using Brauer’sTheoremon
6
InducedCharacters,wegive adescriptionofeveryprimitive centralidempotente (χ) ofarational
Q
group algebra QG corresponding to an irreducible character χ of a finite group G as a rational
linear combination of elements of the form e(G,H ,K ), with each (H ,K ) a strong Shoda pair in
i i i i
a subgroup of G, or equivalently, K is a normal subgroup of H with H /K cyclic.
i i i i
Theorem3.1(Brauer). [Bra]EverycomplexcharacterχofafinitegroupGisaZ–linear combina-
tion χ= a θG , a Z, of characters induced from linear characters θ of elementary subgroups
i i i i ∈ i
M of G, where by an elementary subgroup of G we mean one which is a direct product of a cyclic
i
P
group and a p-group for some prime p.
In particular, the M ’s are cyclic–by–p -groups for some primes p , hence by [OdRS1] each M
i i i i
is strongly monomial. As a consequence, every irreducible character of such a subgroup M is an
i
inducedcharacterθMi fromalinearcharacterθ ofasubgroupH ofM . So,θMi isirreducibleand
i i i i i
(H ,ker(θ )) is a strong Shoda pair of M .
i i i
We also will use the result [OdRS1, Theorem 2.1.] that describes the primitive central idempo-
tents e (ψG) of a rational group algebra QG associated to a monomial irreducible character ψG
Q
as
[Cen (ε(H,K)):H]
(1) e (ψG)= G e(G,H,K)
Q [Q(ψ):Q(ψG)]
where ψ is a linear character of the subgroup H of G and K is the kernel of ψ.
Proposition 3.2. Let G be a finite group of order n and χ an irreducible character of G. Then
the primitive central idempotent e (χ) of QG associated to χ is of the form
Q
1
e (χ)= a [Q(ζ ):Q(ψ )][C :H ]e(G,H ,K )
Q [Q(ζ ):Q(χ)] i n i i i i i
n
i
X
RATIONAL GROUP ALGEBRAS OF FINITE GROUPS 5
where a Z, (H ,K ) are strong Shoda pairs of subgroups of G (equivalently H /K is a cyclic
i i i i i
∈
section of G), C =Cen (ε(H ,K )) and ψ are linear characters of H with kernel K .
i G i i i i i
Proof. As it was mentioned in the introduction, for every χ Irr(G), we have
∈
e (χ)= e(χσ)= σ(e(χ))=tr (e(χ)),
Q Q(χ)/Q
σ∈Gal(Q(χ)/Q) σ∈Gal(Q(χ)/Q)
X X
where χσ is the character of G given by χσ(g) = σ(χ(g)), for g G. The interpretation of e (χ)
Q
∈
as a trace, suggests the following useful notation for the next arguments. For any finite Galois
extension F of Q containing Q(χ), let
eF = σ(e(χ))=tr (e(χ)).
Q F/Q
σ∈Gal(F/Q)
X
Hence e (χ) = eQ(χ)(χ) = 1 (eF(χ)) for every finite Galois extension F of Q(χ). Using
Q Q [F:Q(χ)] Q
Brauer’sTheoremonInducedCharacters,wenowmaywriteχ= a ψG,withψ linearcharacters
i i i i
of elementary subgroups H with kernel K and a Z. Then
i i i
∈ P
eQ(ζn)(χ)= a eQ(ζn)(ψG)
Q i Q i
i
X
and, for every i, we will compute eQ(ζn)(ψG), as in the proof of [OdRS1, Proposition 2.1.]. (Note
Q i
that Q(ζ ) contains Q(χ), because it is an splitting field of G.)
n
Pute =e(ψ ). We knowthat =Aut(C)actsonthe leftandGactsonthe rightonψ andon
i i i
A
e (by composition and by conjugation respectively) and that their actions are compatible. Hence
i
one may consider G acting on the left on the set of irreducible characters of subgroups of G
A×
(and similarly on the e ’s) by (σ,g) ψ =σ ψ g−1.
i i i
· · ·
Let Gal(Q(ζ )/Q)= σ ,...,σ and T = g ,...,g a right transversal of H in G. Denote
n 1 l i 1 m i
by C =Cen (ε(H ,K ){). We have}that m {e g =e(}ψG), hence
i G i i k=1 i· k i
l l m m
eQ(ζn)(ψG) = σ e(ψG)= P σ e g = tr (e ) g
Q i j i j i· k Q(ζn)/Q i · k
j=1 j=1k=1 k=1
X XX X
m m
= [Q(ζn):Q(ψi)]trQ(ψi)/Q(ei)·gk = [Q(ζn):Q(ψi)]ε(Hi,Ki)gk
k=1 k=1
X X
= [Q(ζ ):Q(ψ )][C :H ]e(G,H ,K )
n i i i i i
The above computations now easily yield the desired formula for e (χ). (cid:3)
Q
Remark 3.3. Notice that the formula from Proposition 3.2 for the computation of the primitive
central idempotents e (χ) of QG associated to an irreducible character χ of G coincides with
Q
formula (1) in case χ is a monomial irreducible characterof G, that is χ is induced to G from only
one linear character ψ of a subgroup H, with kernel K such that (H,K) is a Shoda pair of G.
In general, as seen in Proposition 3.2, one has to consider all strong Shoda pairs in subgroups
of G that contribute to the description of a primitive central idempotent of QG. However, one
can reduce the search of the Shoda pairs that determine the primitive central idempotents of
QG to representatives given by a relation between such pairs of subgroups. Indeed, in [OdRS2,
Proposition 1.4], it is proved that if (H ,K ) and (H ,K ) are two Shoda pairs of a finite group
1 1 2 2
G and α ,α Q are such that e = α e(G,H ,K ) is a primitive central idempotent of QG for
1 2 i i i i
i=1,2, then e∈ =e if and only if there is g G such that Hg K =Kg H .
1 2 ∈ 1 ∩ 2 1 ∩ 2
6 ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
Remark 3.4. We would like to be able to give a bound for the integers a used in the previous
i
proposition and one would also like to give more information on the pairs of groups (H ,K ) that
i i
one has to consider in the description of e (χ).
Q
Notice that for monomial (respectively strongly monomial) groups, all primitive central idem-
potents are realized as elements of the form αe(G,H,K), with α Q, for some Shoda pair (H,K)
∈
(respectively strong Shoda pair and α = 1) in G. However, for the smallest non-monomial group,
which is SL(2,3), this is not true any more. Indeed, in [OdRS1, Example 5.7.], the two primi-
tive central idempotents corresponding to the non-monomial characters are e = 1e(G,B,A), and
1 2
e = 1e(G,B,1) 1e(G,B,A), withG= x,y ⋊ a ,asemidirectproductofthe quaterniongroup
2 4 −4 h i h i
x,y of order 8 by the cyclic group A= a of order 3, and with B = x2a . However, e can not
2
h i h i h i
be written as a rational linear multiple of some e(G,H,K) with (H,K) a pair of subgroups of G
such that KEH.
4. Primitive idempotents for finite nilpotent groups
We start this section by showing a method to produce a complete set of orthogonal primitive
idempotents of a classical crossed product with trivial twisting τ = 1, i.e. τ(g,h) = 1, for every
g,h G. Let L be a field of characteristiczero. Observe that (L/F,1) M (F), with n=[L:F],
n
∈ ≃
therefore a complete set of orthogonalprimitive idempotents of (L/F,1) contains n idempotents.
Lemma 4.1. Let A = (L/F,1) be a classical crossed product with trivial twisting and let G =
Gal(L/F) with n = G. Let e = 1 u and let x ,...,x be non-zero elements of L. Then
| | |G| g∈G g 1 n
the conjugates of e by x ,...,x form a complete set of orthogonal primitive idempotents of A if
1 n
and only if tr (x x−1)=0 for everPy i=j. (tr is the trace of L over F.)
L/F i j 6 L/F
Proof. As the twisting is trivial, u :g G is a subgroup of order G of the group of units of A
g
{ ∈ } | |
and hence e is an idempotent of A. Moreover u e = e for every g G. Therefore, if x L then
g
exe = 1 u xe = 1 xg−1u e = 1 xge = 1 tr∈ (x)e. Thus, if x ∈L then e
|G| g∈G g |G| g∈G g |G| g∈G |G| L/F ∈
and xex−1 are orthogonal if and only if tr (x)=0 and the lemma follows. (cid:3)
P P L/F P
Examples 4.2. (1) In the proof of Theorem 4.5 we will encounter some examples of classical
crossedproducts with trivial twisting with a list x ,...,x satisfying the conditions of the
1 n
previous lemma.
(2) Another situation where one can find always such elements correspond to the case when
L/F is a cyclic extension of order n and F contains an n-root of unity. Then L is the
splitting field over F of an irreducible polynomial of F[X] of the form Xn a. If α L
− ∈
with αn = a then x = 1,x = α,...,x = αn−1 satisfy the conditions of Lemma 4.1.
1 2 n
Indeed, the minimal polynomial of αi over F for 1 i < n is of the form Xn/d ai/d
≤ −
for d = gcd(n,i) and therefore tr (αi) = [L : F(αi)]tr (αi) = 0 and similarly
L/F F(αi)/F
tr (α−i)=0.
L/F
(3) We now construct an example where there are no elements x ,...,x satisfying the con-
1 n
ditions of Lemma 4.1. Consider the trivial cyclic algebra (L = Q(ζ )/F = Q(√ 7),1) of
7
−
degree3. Ifx ,x ,x satisfytheconditionsofLemma4.1thenα=x x−1andα−1 =x x−1
1 2 3 2 1 1 2
are non-zero elements of L with zero trace over F. This implies that the minimal polyno-
mial of α over F is of the form X3 a for some a F. But this implies that F contains a
− ∈
third root of unity, a contradiction.
RATIONAL GROUP ALGEBRAS OF FINITE GROUPS 7
Thegroupslistedinthefollowinglemmawillbe thebuildingblocksintheproofofTheorem4.5.
For n and p integers with p prime, we use v (n) to denote the valuation at p of n, i.e. pn is the
p
maximum p-th power dividing n.
Lemma 4.3. Let G be a finite p-group which has a maximal abelian subgroup which is cyclic and
normal in G. Then G is isomorphic to one of the groups given by the following presentations:
P = a,b apn =bpk =1, b−1ab=ar , with either v (r 1)=n k or p=2 and r 1 mod 4.
1 p
¨ | ∂ − − 6≡
P = a,b,c a2n =1, b2k =1, c2 =1, bc=cb, b−1ab=ar, c−1ac=a−1 , with r 1 mod 4.
2
P =¨a,b,c|a2n =1, b2k =1, c2 =a2n−1, bc=cb, b−1ab=ar, c−1ac=∂a−1 , wit≡h r 1 mod 4.
3
¨ | ∂ ≡
Note that if k = 0 (equivalently, if b = 1) then the first case correspond to the case when G
is abelian (and hence cyclic), the second case coincides with the first case with p = 2, k = 1 and
r = 1, and the third case is the quaternion group of order 2n+1.
−
Proof. Let A be a maximal abelian subgroup of G and assume that A is cyclic (generated by a)
andnormalinG. Put A =pn. ConsidertheactionofGonAbyinnerautomorphisms. SinceAis
| |
maximalabelianinG, the kernelofthis actionis A andtherefore G/Ais isomorphicto asubgroup
of the group of automorphisms of A.
Ifeitherpis oddorp=2andn 2thenAut(A)iscyclicandotherwiseAut(A)= φ φ ,
5 −1
≤ h i×h i
where φ is the automorphism of A given by φ (x)=xr.
r r
Assume that G/A is cyclic, so that G has a presentation of the form
(2) G= a,b apn =1,bpk =as,b−1ab=ar ,
¨ | ∂
with pn rpk 1 and pn s(r 1). If pi 3 then (1+xpi)p 1+xpi+1 mod pi+2 for every
| − | − ≥ ≡
i 1 and x Z. Using this, one deduces that if either p is odd or p = 2 and r 1 mod 4 then
v≥(rpi 1)=∈v (rpi−1 1)+1,foreveryi 1. Furthermore,fromtheassumptionth≡atAismaximal
p p
abelian−inG,onededu−cesthatn v (rpk≥ 1)=v (rpk−1 1)+1 nandhencev (rpk 1)=nand
p p p
≤ − − ≤ −
v (r 1)=n k. Therefore,v (s) n v (r 1)=k =v rpk−1 =v 1+r+r2+ +rpk−1
p − − p ≥ − p − p r−1 pÄ ··· ä
and hence there is an integer x such that x(1 +r + r2 +(cid:16) +(cid:17)rpk−1)+s 1 mod pn. Then
··· ≡
(axb)pk = 1 and, replacing b by axb in (2), we obtain the presentation of P . We have also proved
1
that v (r 1) = n k unless p = 2 and r 1 mod 4. Assume p = 2 and r 1 mod 4,
p
− − 6≡ 6≡
v (s) n v (r 1) = n 1 and v (1+r+r2+...r2k−1) = v r2k−1 n 1. If v (s) n
2 ≥ − 2 − − 2 2 r−1 ≥ − 2 ≥
then G = P . If v (s) = v (1+r+r2+...r2k−1) = n 1 then (ab(cid:16))2k = 1(cid:17). Replacing b by ab we
∼ 1 2 2 −
obtain again that G=P . Otherwise, v (s)=n 1 and v (1+r+r2+...r2k−1)=n. Therefore
v (r2k−1) > n > v (∼r2k−11 1) and by t2he first p−art of th2e proof (applied to a,b2 ) this implies
2 2
−
that k =2. Then G is the quaternion group of order 2n+1 which is P for k =0.
3 (cid:10) (cid:11)
Assume now that G/A is not cyclic, so p = 2 and G/ a = b c with c acting by inversion
h i h i×h i
on a . This provides a presentation of G of the form
h i
(3) G= a,b,c a2n =1,b2k =as,ca=a−1c,cb=aibc,b−1ab=ar,c2 =1 or c2 =a2n−1 .
h | i
Replacing b by bc if needed, one may assume that v (r 1) = n k 2 and v (r2k 1) = n.
2 2
− − ≥ −
So,applying the firstpartofthe proofto a,b , we mayassumethat b2k =1. Then c=b2kcb−2k =
h i
a−i(1+r+...+r2k−1)c and so 2n i(1+r +...+r2k−1) = ir2k−1. As v r2k−1 = k, we have
| − − r−1 2 r−1
(cid:16) (cid:17)
8 ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
v (i) n k = v (r 1). Hence, there exists an integer j so that j(r 1) i 0 mod 2n. It
2 2
≥ − − − − ≡
is easy to verify that the commutator of b and ajc is 1. So, replacing c by ajc if needed, we may
assume thatb andc commute andwe obtainthe presentationofP , ifc2 =1, andthe presentation
2
of P , if c2 =a2n−1. (cid:3)
3
WealsoneedthefollowingresultonsplittingofaHamiltonianquaternionalgebraH(F)=F[i,j
|
i2 =j2 = 1,ji+ji=0].
−
Lemma 4.4. Let F be a field of characteristic zero. Then the quaternion algebra H(F) splits if
and only if x2+y2 = 1 for some x,y F. In that case 1(1+xi+yj) and 1(1 xi yj) form a
− ∈ 2 2 − −
complete set of primitive idempotents of H(F).
Furthermore, if F =Q(ζ ,ζ +ζ−1) with m odd then 1 is the sum of two squares of F if and
m 2n 2n −
only if m=1 and either n 3 or the multiplicative order of 2 modulo m is even.
6 ≥
Proof. The first part can be found in [Seh1, Proposition 1.13]. Now assume that F =Q(ζ ,ζ +
m 2n
ζ−1) with m odd. If m=1 then F is totally realand therefore 1 is notthe sum oftwo squaresof
2n −
F. So assume that m=1. If n 2, then F =Q(ζ ) and the result is well known (see for example
m
6 ≤
[Mos, FGS] or [Lam, pages 307–308]). Finally assume that m=1 and n 3. Then F contains √2
6 ≥
and, as 2 is not a square in Q , the duadic completion of Q [Lam, Corollary 2.24], we deduce that
2
[F :Q ] is even. Then 1 is a sum of squares in F for every place p of F and hence 1 is a sum
2 2 p
− −
of squares in F (see [Lam, page 304]). (cid:3)
Nowwearereadytoshowaneffectivemethodtocalculateacompletesetoforthogonalprimitive
idempotents of QG for G a finite nilpotent group. Since G is abelian-by-supersolvable and hence
strongly monomial, it follows from [OdRS1, Theorem 4.4] that every primitive central idempotent
ofQGis ofthe forme(G,H,K)with(H,K)a strongShoda pairofG andthereforeitis enoughto
obtainacomplete setoforthogonalprimitiveidempotents ofQGe(G,H,K)foreverystrongShoda
pair (H,K) of G. This is described in our main result that we state now.
Theorem 4.5. Let G be a finite nilpotent group and (H,K) a strong Shoda pair of G. Set e =
e(G,H,K), ε = ε(H,K), H/K = a , N = N (K) and let N /K and H /K = a (respectively
G 2 2 2
h i h i
N2′/K andH2′/K = a2′ )denotethe2-parts(respectively,2′-parts)ofN/K andH/K respectively.
h i
Then a2′ has a cyclic complement b2′ in N2′/K.
h i h i
Acomplete setoforthogonal primitive idempotents of QGeconsists oftheconjugates of b2′β2εby
the elements of T2′T2TG/N, where T2′ ={1,a2′,a22′,...,a[2N′2′:H2′]−1}, TG/N denotes a left transver-
sal of N in G and β and T are given according to the cases below. c
2 2
(1) If H /K has a complement M /K in N /K then β = M . Moreover, if M /K is cyclic
2 2 2 2 2 2
then there exists b N such that N /K is given by the following presentation
2 2 2 ”
∈
a ,b a 2n =b 2k =1, a b2 =a r ,
≠ 2 2 | 2 2 2 2 ∑
and if M /K is not cyclic, there exist b ,c N such that N /K is given by the following
2 2 2 2 2
∈
presentation
a ,b ,c a 2n =b 2k =1, c 2 =1, a b2 =a r, a c2 =a −1, [b ,c ]=1 ,
≠ 2 2 2 | 2 2 2 2 2 2 2 2 2 ∑
with r 1 mod 4 (or equivalently, a 2n−2 is central in N /K). Then
2 2
(i) T≡= 1,a ,a2,...,a2k−1 , if a 2n−2 is central in N /K and M /K is cyclic; and
2 { 2 2 2 } 2 2 2
RATIONAL GROUP ALGEBRAS OF FINITE GROUPS 9
(ii) T = 1,a ,a2,...,a2k−1−1,a2n−2,a2n−2+1,...,a2n−2+2k−1−1 , otherwise.
2 { 2 2 2 2 2 2 }
(2) if H /K has no complement in N /K then there exist b ,c N such that N /K is given
2 2 2 2 2 2
∈
by the following presentation
a ,b ,c a 2n =b 2k =1, c 2 =a 2n−1, a b2 =a r, a c2 =a −1, [b ,c ]=1 ,
≠ 2 2 2 | 2 2 2 2 2 2 2 2 2 2 ∑
with r 1 mod 4 and we set m=[H2′ :K]/[N2′ :H2′]. Then
≡
(i) β2 =b2 and T2 ={1,a2,a22,...,a22k−1}, if either H2′ =K or the order of 2 modulo m
is odd and n k 2 and
“
(ii) β = b 1+xa−22n−2+≤ya22n−2c2 and T = 1,a ,a2,...,a2k−1,c ,a c ,a2c ,...,a2k−1c
2 2 2 2 { 2 2 2 2 2 2 2 2 2 2}
with
“
x,y Q a[N2′:H2′],a2k +a−2k ,
∈ î 2′ 2 2 ó
satisfying (1+x2+y2)ε = 0, if H2′ = K and either the order of 2 modulo m is even
6
or n k >2.
−
Proof. We start the proof by making some useful reductions. Taking T = T a left transversal
G/N
of N in G, the conjugates of ε by elements of T are the “diagonal” elements in the matrix algebra
QGe=M (QNε). Hence, followingthe proofof[OdRS1, Proposition3.4],one cansee that itis
G/N
sufficient to compute a complete set of orthogonal primitive idempotents for QNε= QHε N/H
∗
and then add their T-conjugatesin order to obtainthe primitive idempotents of QGe. So one may
assume that N = G, i.e. K is normal in G and hence e = ε and T = 1 . Then the natural
{ }
isomorphism QGK Q(G/K) maps ε to ε(H/K). So, from now on we assume that K = 1 and
≃
hence H = a is a cyclic maximal abelian subgroupof G, which is normalin G and e=ε=ε(H).
“
h i
If G=H then QGe is a field, T2 =T2′ = 1 and b2′ =β2 =1; hence the result follows. So, in the
{ }
remainder of the proof we assume that G=H.
6
The map aε ζ induces an isomorphism f : QHε Q(ζ), where ζ is a primitive H -root of
7→ → | |
unity. Using the description of QGe given in Theorem 2.1, one obtains a description of QGe as a
classical crossed product (Q(ζ)/F,τ), where F is the image under f of the center of QGe.
We first consider the case when G is a p-group. Then G and H = a satisfy the conditions
h i
of Lemma 4.3 and therefore G is isomorphic to one of the three groups of this lemma. Moreover,
H has a complement in G if and only if G = P or G = P and, in these cases, τ is trivial. We
∼ 1 ∼ 2
claim that in these cases it is possible to give a list of elements x ,...,x of Q(ζ) (pk = [G : H])
1 pk
satisfying the conditions of Lemma 4.1 and the elements f−1(x ),...,f−1(x ) correspond to the
1 pk
conjugating elements in G given in the statement of the theorem in the different cases. To prove
this we will use the following fact: if K is a subfield of Q(ζ) such that ζ K, ζi K (with
p
∈ 6∈
i = 1,...,pk 1) and, moreover, ζ K if p = 2 then tr (ζi) = 0. To see this notice that
4 Q(ζ)/K
− ∈
if d is the minimum integer such that ζipd K then Q(ζi)/K is cyclic of degree pd and ζi is a
∈
root of Xpd ζipd K[X]. Then Xpd ζipd is the minimal polynomial of ζi over K. Hence
− ∈ −
tr (ζi)=0 and thus tr (ζi)=0.
Q(ζi)/K Q(ζ)/K
AssumefirstthatG=P andv (r 1)=n k(equivalentlyapn−k Z(G)),thatis,eitherpisodd
1 p − − p ∈
or p=2 and r 1 mod 4). Then F is the unique subfield of index [G:H]=pk in Q(ζ) and such
≡
thatifp=2thenζ4 ∈F. NamelyF =Q(ζpn−k)=Q(ζpk). Ifwesetxi =ζi,fori=0,1,...,pk−1,
then x x−1 = ζi−j. If i = j then ζi−j F and hence tr (x x−1)= tr (ζi−j)= 0. Thus,
i j 6 6∈ Q(ζ)/F i j Q(ζ)/F
byLemma4.1,the conjugatesofbby1,ζ,ζ2,...,ζpk−1 formacompletesetoforthogonalprimitive
b
10 ERICJESPERS,GABRIELAOLTEANU,ANDA´NGELDELR´IO
idempotents of (Q(ζ)/F,1). Then the elements f−1(xi) form the elements of T2′ if p is odd or the
elements of T , in case (1.i).
2
Assume now that G is still P , but with p = 2 and r 1 mod 4 (equivalently, a2n−2 is not
1 6≡ 2
central). In this case ζ F and F(ζ ) is the unique subextension of Q(ζ)/Q(ζ ) of index [G :
4 4 4
H]/2 = 2k−1. That is F(ζ6∈4) = Q(ζ2n−k+1) = Q(ζ2k−1). We take xi = ζi and x2k−1+i = ζ2n−2+i =
ζ ζi, for 0 i < 2k−1. Hence, if i = j then x x−1 is either ζ±1 or ζ±i or ζ±1ζ±i, with i =
4 ≤ 6 i j 4 4
1,2,...,2k−1 1. As ζi F(ζ ), we have tr (ζi)=0. Since
− 6∈ 4 Q(ζ)/F(ζ4)
tr (ζ )=tr tr (ζ )=[Q(ζ):Q(ζ )]tr (ζ )=0,
Q(ζ)/F 4 F(ζ4)/F Q(ζ)/F(ζ4) 4 4 F(ζ4)/F 4
tr (ζi)=tr tr (ζi)=0
Q(ζ)/F F(ζ4)/F Q(ζ)/F(ζ4)
and
tr (ζ ζi)=tr tr (ζ ζi)=tr (ζ tr (ζi))=0,
Q(ζ)/F 4 F(ζ4)/F Q(ζ)/F(ζ4) 4 F(ζ4)/F 4 Q(ζ)/F(ζ4)
we deduce that tr(x x−1)=0 for everyi=j. Thenf−1 maps these elements to the elements ofT
i j 6 2
for case (1.ii).
Now assume that G = P2. Then F = Q(ζ2n−k +ζ2−n1−k). Since r ≡ 1 mod 4, n−k ≥ 2. Then
the same argument as in the previous case shows that the 2k+1 elements of the form x = ζi and
i
x = ζ2n/2+i =ζ ζi, for 0 i< 2k satisfy the conditions of Lemma 4.1. The elements f−1(x )
2k+i 4 ≤ i
form now the set T of case (1.ii).
2
Now we consider the non-splitting case, i.e. G = P . Then the center of QGe is isomorphic to
3
F =Q(ζ)hb,ci =Q(ζ2n−k +ζ2−n1−k) and bQGεb=bQha,ciεbF +Fa2n−2 +Fc+F(a2n−2c)∼=H(F),
which is a division algebra, as F is a real field. Then bε is a primitive idempotent of QGe. Hence
QGe M (H(F))andfromthefirstcbaseonbecanbprovidebthe2k orthogonalprimitiveidempotents
≃ 2k
needed in this case by taking the conjugates of b by 1b,a,a2,...,a2k−1, and this agrees with case
(2.i). This finishes the p-group case.
Letusnowconsiderthegeneralcase,whereGibsnotnecessarilyap-group. ThenG=G G
2× p1×
···×Gpr =G2×G2′,withpi anoddprimeforeveryi=1,...,r. Then(H,1)isastrongShodapair
ofGifandonlyif(H ,1)isastrongShodapairofG ,foreveryi=0,1,...,m,(withp =2)and
pi pi 0
ε(H) = ε(H ). Using this and a dimension argument it easily follows that the simple algebra
i pi
QGε(H) is the tensor product over Q of the simple algebras QG ε(H ). On the other hand, we
have seeQn that for i≥1, QGpiε(Hi)≃Mpki(Q(ζpni−ki)), for pnii =pi|Hip|iand pkii =[Ni :Ki]. Then
i i
QG2′ε(H2′) ≃ M[G2′:H2′](Q(ζm)), with m =| H2′ | /[G2′ : H2′] (= [H2′ : K]/[G2′ : H2′]) and then
a complete set of orthogonal primitive idempotents of QG2′ε(H2′) can be obtained by multiplying
thedifferentsetsofidempotentsobtainedforeachtensorfactor. ObservethateachG ,withi 1,
pi ≥
takes the form ai ⋊ bi and so G2′ = a ⋊ b , with a = a1...ar and b = b1...br. Having in
h i h i h i h i
mind that apkii is centralone can easily deduce, with the help of the Chinese Remainder Theorem,
that the product of the different primitive idempotents of the factors from the odd part (i.e. the
conjugates of bi by 1,ai,a2i,...,aipkii−1 are the conjugates of bε by 1,a,a2,...,a[G2′:H2′]−1. In the
notation of the statement of the theorem, a=a2′ and T2′ = 1,a,a2,....,a[G2′:H2′]−1 as wanted.
{
If G is oddbthen the proof is finished. Otherwise we shobuld combine the odd and even parts
| |
of G. If H has a complement in G then QG ε(H ) is split over its center and hence we can take
2 2 2 2
T as in the 2-group case. However, if H does not have a complement in G then QG ε(H ) =
2 2 2 2 2
M[G2:H2]/2(H(Q(ζ2n−k+ζ2−n1−k)))andhenceQGε=M[G:H]/2(H(F)),withF =Q(ζm,ζ2n−k+ζ2−n1−k)).
If H(F) is not split (equivalently the conditions of (2.i) hold) then we can also take T as in the
2
2-group case. However, if H(F) is split then one should duplicate the number of idempotents, or