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LONDON MATHEMATICAL SOCIETY LECTURE NOTE SERIES Managing Editor: Professor J.W.S. Cassels, Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, 16 Mill Lane, Cambridge CB2 1SB, England The books in the series listed below are available from booksellers, or, in case of difficulty, from Cambridge University Press.
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London Mathematical Society Lecture Note Series. 132
WHITEHEAD GROUPS OF FINITE GROUPS Robert Oliver Matematisk Institut, Aarhus University
1
ht f the f Cootbridge and sell wanl
d boks
y
;;IiIt: ll in 1534. ty has printed
d eonfinoously since 1584.
CAMBRIDGE UNNERSrTY PRESS Cambridge New York N ew Rochelle Melbourne Sydney
CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo
Cambridge University Press The Edinburgh Building, Cambridge C132 8RU, UK
Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521336468
© Cambridge University Press 1988
This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1988 Re-issued in this digitally printed version 2008
A catalogue record for this publication is available from the British Library
Library of Congress Cataloguing in Publication data Oliver, R. (Robert) Whitehead groups of finite groups / R. Oliver. p. cm. - (London Mathematical Society lecture note series: 132) Bibliography: p. Includes Index 1. Whitehead groups. 2. Finite groups. 3. Induction (Mathematics) I. Title II. Series QA171.044 1988 512'.22 - do 19
87 - 27725
ISBN 978-0-521-33646-8 paperback
PREFACE
This book
is written with
intention of
the
making more easily
accessible techniques for studying Whitehead groups of finite groups, as
well as a variety of related topics such as induction theory and p-adic It developed out of a realization that most of the recent
logarithms.
work in the field is scattered over a large number of papers, making it very difficult even for experts already working with K- and L-theory of finite groups to find and use them. but
experts,
application
also
at
The book is aimed, not only at such
nonspecialists
who
involving Whitehead groups,
overview of
the current
state
either
or who
of knowledge
in
need
some
just want the
to
subject.
specific
get an It
is
especially with the latter group in mind that the lengthy introduction as well as the separate introductions to Parts I, II, and III - have been They are designed to give a quick orientation to the contents of
written.
the book, and in particular to the techniques available for describing Whitehead groups.
I would like to thank several people, in particular Jim Davis, Erkki
Laitinen, Jim Schafer, Terry Wall, and Chuck Weibel,
for all of their
helpful suggestions during the preparation of the book. thanks
to
Ioan James for encouraging me
arranging its publication.
Also, my many
to write the book, and for
LIST OF NOTATION
The following is a list of some of the notation used throughout the book.
In many cases, these are defined again where used.
Groups: NG(H),
denote the normalizer and centralizer of
CG(H)
C
denotes a p-Sylow subgroup of
C
denotes a (multiplicative) cyclic group of order
D(2n),
n
Q(2n), SD(2n) denote the dihedral, quaternion, and semidihedral groups of order 2n
An denote the symmetric and alternating groups on n
Sn,
in G
(the abelianization) for any group G
Gab = C/[G,G] Sp(G)
H
letters
HA G denotes a semidirect product where H
is normal
GZCn and CSn
Gn 4 Cn and Gn A Sn
denote the wreath products
if C acts linearly on M (groups
MG
= {x E M: Cx = x}
M
= M/(gx-x: g EC, x EM) = H0 (G;M) } of invariants and coinvariants)
G
= H0(C;M)
Fields and rings:
Kp =
P
®Q K
if
K
is any number field and
p
a rational prime (so
Kp
is possibly a product of fields) Rp = 7Lp ®Z R
}K,
(}tK)P
if
(K
R
is the ring of integers in a number field
any field) denote the groups of roots of unity, and p-th
power roots of unity, in fn
(n Z 1)
K
denotes a primitive n-th root of unity
En
(n > 0),
when some prime
is fixed, denotes the root of unity
p
exp(2iri/pn) E C. (for any field
KCn
extension of J(R) (-)
K
and any
K containing
n>-
denotes the smallest field
1)
n
denotes the Jacobson radical of the ring R means "subgroup (or 1p module) generated by"
(-)R means "R-ideal or R-module generated by"
er j = ei j(r)
(where
10 j,
i, j > 1,
and
r E R)
matrix with single off-diagonal entry
r
denote the elementary in the (i,j)-position
K-theory:
for any 7- or Z-order
SK1(2l) = Ker[K1(2l) - K1(A)]
I
Ki (2I) = K1(2I)/SK1(2l)
simple Q-
C11(2l) = Ker[SK1(21) --» 19 pSK1(2Ip)] C(A) =
in a semi-
or s-algebra A
for any 7-order
21
A
for any semisimple I-algebra
SK1(21, I)
2l
and any 7-order
I
3 C A, where the limit is taken over all ideals of finite index (see Definition 3.7)
Cp(A)
denotes the p-power torsion in the finite group
Wh(R[G]) = K1(R[G])/(rg: r E WK, g E G)
whenever of
@ or
R
and
Wh'(R[G]) = Wh(R[G])/SK1(R[G])
is the ring of integers in any finite extension K
(and G
is any finite group)
Wh'(G) = Wh(G)/SK1(Z[G]) = Ki(7[G])/(±g) K2(R,I) = Ker[K2(R)
C(A)
K2(R/I)]
for any finite group C
for any ring
(see remarks in Section 3a)
R
and any ideal
I C R
Introduction Historical survey Algorithms for describing Wh(G) Survey of computations
Part I
1
3
4 14
General theory
20
Chapter 1. Basic algebraic background la. Orders in semisimple algebras lb. P-adic completion lc. Semilocal rings and the Jacobson radical ld. Bimodule-induced homomorphisms and Morita equivalence
21 21
Chapter 2. Structure theorems for K1 of orders 2a. Applications of the reduced norm 2b. Logarithmic and exponential maps in p-adic orders
40 40 50
Chapter 3. Continuous K2 and localization sequences 3a. Steinberg symbols in K2(R) 3b. Continuous K2 of p-adic orders and algebras 3c. Localization sequences for torsion in Whitehead groups
63 64 70 73
Chapter 4. The congruence subgroup problem 4a. Symbols in K2 of p-adic fields 4b. Continuous K2 of simple IQp-algebras 4c. The calculation of C(Q[G])
90 91 100 115
Chapter 5 First applications of the congruence subgroup problem an example 5a. Constructing and detecting elements in SK1: 5b. C11(R[G]) and the complex representation ring 5c. The standard involution on Whitehead groups
127 127 134 148
Chapter 6. The integral p-adic logarithm 6a. The integral logarithm for p-adic group rings 6b. Variants of the integral logarithm 6c. Logarithms defined on K2($P[G])
153 153 166 168
Part II
172
Group rings of p-groups
Chapter 7.
The torsion subgroup of Whitehead groups
Chapter S. The p-adic quotient of SK1(7L[G]): 8a. Detection of elements 8b. Establishing upper bounds Examples Sc.
27 32 37
p-groups
173 183 183 191
200
Chapter 9. Chapter 10.
Part III
205
C11(7L[G]) for p-groups
229
The torsion free part of Wh(G)
243
General finite groups
Chapter 11. A quick survey of induction theory 11a. Induction properties for Mackey functors and Green modules l1b. Splitting p-local Mackey functors
245 246 254
Chapter 12.
272
The p-adic quotient of SK1(7L[G]):
finite groups
Chapter 13. C11(7L[G]) for finite groups 13a. Reduction to p-elementary groups 13b. Reduction to p-groups 13c. Splitting the inclusion C11(7L[G]) C SK1(7L[G])
291 291
Chapter 14.
328
Examples
305 322
References
340
Index
348
INTRODUCTION
R
For any associative ring defined as follows.
For each
with unit, an abelian group
n >0,
invertible
nxn-matrices with entries in
subgroup of
GLn+l(R)
and set
CLn(R)
let
by identifying A E GLn(R)
GL(R) =
For each
Un=1GLn(R).
with
as a
GLn(R)
(A 01) E GLn+l(R);
En(R) C CLn(R)
let
n,
is
denote the group of
Regard
R.
K1(R)
be the
nxn-matrices - i. e., all those
subgroup generated by all elementary
which are the identity except for one nonzero off-diagonal entry - and Then by Whitehead's lemma (Theorem 1.13 below),
E(R) = Un-1En(R).
set
E(R) = [GL(R),GL(R)],
GL(R).
the commutator subgroup of
In particular,
and the quotient group
E(R) 4 GL(R);
K1(R) = GL(R)/E(R)
is an abelian group.
One family of rings to which this applies is that of group rings.
G
is any group, and if
is the free R-module with basis
R[G]
induced by the group product. units
u E R*,
If
is any commutative ring, then the group ring
R
G,
where ring multiplication is
In particular, group elements
can be regarded as invertible
and hence represent elements in
K1(R[G]).
g E G,
1xl-matrices over
The Whitehead group of
and R[G],
G
is
defined by setting
Wh(G) = K1(7L[G])/(±g: g E G).
By construction,
K1(R)
(or
Wh(G))
taking an arbitrary invertible matrix over it
to
the identity
operations.
(or
to
some
±g)
measures the obstruction to
R
(or
7L[G]),
and reducing
via a series of elementary
Here, an elementary operation is one of the familiar matrix
operations of adding a multiple of one row or column to another; and these
2
INTRODUCTION
elementary operations are very closely related to Whitehead's "elementary
deformations" of finite CW complexes.
This relationship leads to the
definition of the Whitehead torsion
T(f) E Wh(al(X))
f: X -- Y
of any homotopy equivalence
T(f) = 1
where
(i. e.,
between finite CW complexes;
the identity element) if and only if
f
X
induced by a series of elementary deformations which transform Y.
A homotopy equivalence
such that
f
is
into
T(f) = 1
is called a simple
only
studying homotopy
homotopy equivalence.
Whitehead
torsion plays a
not
role,
in
equivalences of finite CW complexes, but also when classifying manifolds.
The s-cobordtsm theorem (see Mazur [1]) says that if smooth closed n-dimensional manifolds, where
n
compact (n+l)-dimensional manifold such that
M ' > W
the inclusions then
W
and
is diffeomorphic to
N I
M
and
and if
5,
are
N
is a
W
and such that
8W = M II N,
are simple homotopy equivalences,
i W
M x [0,1].
In particular,
M
are
N
and
diffeomorphic in. this situation; and this theorem is one of the important
This theorem is
tools for proving that two manifolds are diffeomorphic.
also one of the reasons
for
the importance of Whitehead groups when
computing surgery obstructions.
When G
is a finite group, then
generated abelian groups,
K1(7L[G])
and
Wh(G)
are finitely
whose rank was described by Bass
(see
The main goal of this book
section on algorithms below, or Theorem 2.6).
is to develop techniques which allow a more complete description of for finite
G;
the
Wh(G)
and in particular which describe the subgroup
SK1(7[G]) = Ker[K1(7[G])
K1(D[G])].
This is a finite subgroup (Theorem 2.5), and is in fact by a theorem of
(Theorem 7.4 below)
Wall Wh(C).
SL(7L[G])
When
G
isomorphic
is abelian, then
to
the
full
torsion subgroup of
SK1(7L[G]) = SL(7L[G])/E(7L[G]),
denotes the group of matrices of determinant
Most of
where
1.
the general background results have been presented here
without proofs - especially when they can be referenced in standard textbooks such as Bass [2],
Curtis & Reiner [1],
Janusz [1],
Milnor
3
INTRODUCTION
Also, some of the longer and more technical proofs
[2], and Reiner [1].
have been omitted when they are well documented in the literature, or are
not needed for the central results.
for most
sketched,
Proofs are included, or at least
results which deal
directly with
the problem of
describing Whitehead groups.
Historical survey
Whitehead groups were first defined by Whitehead [1],
in order to
find an algebraic analog to his "elementary deformations" of finite CW equivalences
complexes,
and
complexes.
Whitehead also showed in [1] that
if
G = 7L;
to
simple
and that
homotopy
multilicative cyclic group of order
if
finite
CW
4
or
IGI
always denotes a
Cn
(Note that
Wh(C5) # 1.
between
Wh(G) = 1
n.)
A more systematic understanding of the structure of the groups
Bass' theorem [1,
came only with the development of algebraic K-theory.
Corollary 20.3],
showing that
the
Wh(G)
are finitely generated and
computing their rank, has already been mentioned.
focus attention on the torsion subgroup of and Wall [1] to be isomorphic to
Wh(G)
This made it natural to
shown by Higman [1]
Wh(G):
SK1(7L[G]).
Milnor, in [1, Appendix A], noted that if the "congruence subgroup problem" could be proven, then it would follow that all finite abelian groups
G.
SK1(7L[G]) = 1
for
This conjecture was shown by Bass, Milnor,
and Serre [1] to be false (see Section 4c below); but their results were still
sufficient
groups.
to show that
SK1(7L[G])
In particular, it was shown that
(Bass et al [1, Proposition 4.14]),
and any
n
if
(Lam, [1, Theorem 5.1.1]),
vanishes for many abelian
SK1(7L[G]) = 1
G = Cp x Cp
or if
if
G
is cyclic
for any prime
G = (C2)n
for some
p
n
(Bass et al [1, Corollary 4.13]).
The first examples of finite groups for which
constructed by Alperin, Dennis, and Stein [1].
SK1(7L[G]) # 1
were
They combined earlier
results from the solution to the congruence subgroup problem with theorems
about generators SK1(7L[G])
when
for
K2
G = (Cp)n,
of
finite n > 3,
rings,
and
p
to
explicitly describe
is an odd prime.
In
4
INTRODUCTION
particular,
is nonvanishing for all such
SK1(Z[G])
G.
Their methods
were then carried further, and used to show that for finite abelian if and only if either
SK1(7[C]) = 1 of
G has the form Cp or
G = (C2)n,
G,
or each Sylow subgroup
Cpn x Cp.
Later results of Obayashi [1,2], Keating [1,2], and Magurn [1,2], showed that
vanishes for many nonabelian metacyclic groups
SK1(7[G])
and in particular when G various
ad hoc
methods,
is any dihedral group.
which
did
give
not
generalizations to arbitrary finite groups. more
systematic
approach using
G,
These were proven using much
hope
for
having
To get general results, a
localization
sequences
is
needed -
extending the methods of Alperin, Dennis, and Stein - and it is that approach which is the main focus of this book.
Algorithms for describing Wh(G)
If
R
is any commutative ring, then the usual matrix determinant
induces a homomorphism
det
This
is
:
K1(R) = GL(R)/E(R) - R.
e -' K1(R)
split surjective - split by the homomorphism
induced by identifying R* = GL1(R).
Hence, in this case,
K1(R)
factors
as a product
K1(R) = R* x SK1(R),
where
SK1(R) = SL(R)/E(R)
If
R = 7L[G],
given earlier:
and
SL(R) = {A E GL(R)
:
det(A) =1).
then this coincides with the definition of Q[G]
is a product of fields, so
SK1(7L[G])
K1(D[G]) = (tD[G])*.
Determinants are not, in general, defined for noncommutative rings.
However, in the case of the group rings
7[G]
and
ID[G]
for finite
5
INTRODUCTION
groups
they can be replaced by certain analogous homomorphisms:
C,
One way to do this is to consider, for fixed
reduced norm homomorphisms. G,
the
the Wedderburn decomposition
k C[G] =
Mr. (C)
i=1
of the complex group ring as a product of matrix rings (see Theorem 1.1). For each
n,
the reduced norm on
GL(Q[G])
is then defined to be the
composite
nr
:
GLn(Q[G])
incl GLn(C[G]) -
[I
i=1
ri
(C)
III ]]*. i=1
These then factor through homomorphisms
nrZ[G]
: K1(Z[G]) - n Cw
and
nrO[G]
:
K1(Q[G]) --->
i=1
Also,
nro[G]
II C*.
i=1
is injective (Theorem 2.3), and so
(by definition)
SK1(7L[G]) = Ker[K1(7L[G]) --f Kl((Q[G])]
= Ker(nry[G]).
Note that when G
(1)
is commutative, then
Ker(nrl[G]) = Ker[det: K1(7L[G]) - (7L[G])*];
so that the two definitions of
SK1(7L[G])
coincide in this case.
For
more details about reduced norms (and in more generality), see Section 2a. Reduced norm homomorphisms are also the key to computing the ranks of
the finitely generated groups SK1(7L[G]) = Ker(nrZ[G])
K1(7L[G])
and
Wh(C).
Not only
is
finite, but - once the target group has been
restricted appropriately - Coker(nry[G])
is also finite (Theorem 2.5).
6
INTRODUCTION
A straightforward computation using Dirichlet's unit theorem then yields the formula
rk(K1(Z[C])) = rk(Wh(C))
= #(irreducible
R[G]-modules) - #(irreducible
O[G]-modules).
Furthermore, by the theorem of Higman [1] (for commutative
and Wall
C)
[1] (in the general case),
tors(K1(Z[G])) _ {±l} x Gab x SK1(Z[G])
(see Theorem 7.4 below). structure of
K1(Z[G])
Thus,
and
Wh(G)
structure of the finite group
as abstract groups,
at
least,
the
will be completely understood once the
SK1(Z[G])
is known.
A much more difficult problem arises
if
explicit generators for the torsion free group
one needs
to construct
Wh'(G) = Wh(G)/SK1(Z[G]).
One case where it is possible to get relatively good control of this is
G
when
is a p-group, for some regular prime In this case,
p = 2).
p
(including the case
logarithmic methods can be used to identify the
p-adic completion p 0 Wh'(G)
with a certain subgroup of
H0(G;2p[G])
(i. e., the free i module with basis the set of conjugacy classes in
G).
This is explained, and some applications are given, in Chapter 10; based on Oliver & Taylor [1, Section 4].
SK1(Z[G]):
When studying
SK1(Z[G]),
it is convenient to first
define a certain subgroup C11(Z[C]) C SK1(Z[G]). 2p [G]
For each prime
and P[G] denote the p-adic completions of
Section lb); and set
Z[G]
and
SK1(gp[G]) = Ker[K1(2p[G]) -4 K1(&[G])].
p,
D[G]
let
(see
Then set
C11(Z[C]) = Ker[SK1(Z[G]) - ® SK1(7LP[G])] P
The sum
19PSK1(2 P[G])
is,
in
fact,
a finite
sum - SK1( p[G])= 1
INTRODUCTION
whenever
p4IGI - and the localization homomorphism
Note that
3.9).
7
C11(7L[G]) = SK1(7L[G])
SK1(2p[G])x (Zp[G])*
G
if
is onto (Theorem
B
K1(8p[G])
is abelian:
in this case, and matrices over a 2p-algebra can
always be diagonalized using elementary row and column operations (see Theorem 1.14(1)). In particular,
SK1(g[C])
sits in an extension
1 ---> C11(7L[G]) --' SK1(7L[G]) a '' ®SKl(2p[G])
(2)
1.
p The groups
SK1(2 p[G])
and
C11(Z[G])
are described independently, using
very different methods, and it is difficult to find a way of handling them
both simultaneously. understand
the
In
extension
the remaining problems
one of
fact, (2)
in 2-torsion
is
to
have a natural
does
(it
splitting in odd torsion).
SK1( p[G]):
p-group for any prime
p
and any finite group
if the p-Sylow subgroup of
G
groups
for all
SK1(2p[GI) = 1
G,
If
G
SK1(Zp[G])
By a theorem of Wall [1, Theorem 2.5],
is abelian.
SK1(7Lp[G]) = 1
and
G,
is a
In fact, for most "familiar"
p.
is a p-group, then
(3)
SK1(2p[G]) - H2(G)/H2b(G);
where
H2b(C)
= Im [I {H2(K) : K C G, K abelian}
ind) H2(C)]
= (g-hEH2(G) : g,hEG, gh=hg) (see Section 8a).
Formula
(3)
is
shown
in Theorem
8.6,
and
the
isomorphism itself is described in Proposition 8.4. If
then set
G
is an arbitrary finite group, and if
Cr = {g E G:
homology group Let
H2(G;Zp(Gr)),
p
is a fixed prime,
(the "p-regular" elements).
where G
acts on
Zp(Gr)
Consider the
by conjugation.
8
INTRODUCTION
(D
: H2(G;2p(Gr)) ) H2(C;ap(Gr))
be induced by the endomorphism 4(Jrigi) = Jrigp
on 2 p(Gr);
H2(G;2 p(Gr)), =
be the group of 4-coinvariants.
H2(G;2p(Gr))/Im(1-4)
Then
SK1(2p[G]) -
(see Theorem 12.10).
and let
)),/H2b(G;Zp(Gr))
(4)
Here, in analogy with the p-group case:
H2b(G;7Lp(Gr)) = Im [
H2(K'2p(Kr)) i- H2(G.7Lp(Gr))41]
K C G
abelian = ((g^h)O k
:
g,h E G, k E Gr,
g,h,k commute pairwise).
The following alternative description of group
is often easier to use.
G,
two elements some
n.
g,h E G
for a non-p-
be "gyp conjugacy"
of order prime to
G
g
are Qp conjugate if
Zi = CG(gi)
Set
g1,...,gk E G
Let
class representatives for elements of
SK1(2 p[G]),
is conjugate to
p - where
hp
for
(the centralizer), and
xgix- 1
Ni = {x E G :
= gp
,
some
n}.
Then by Theorem 12.5 below,
k SK1(2p[G]) =
HO(Ni/Zi; H2(Zi)I 2b(Zi))(pY
(5)
i=1
C11(Z[C]): the part of
The subgroup
K1(7L[G])
C11(7L[G]) C SK1(7L[G])
can be thought of as
which is hit from behind in localization sequences.
One way to study this is to consider, for any ideal index, the relative exact sequence
I C 7L[G]
of finite
9
INTRODUCTION
K2(7L[G]/I) ---> SK1(7L[G],I) 4 SK1(7[G]) - K1(7L[G]/I)
After taking inverse limits
of Milnor [2, Lemma 4.1 and Theorem 6.2]. over all such
I,
this takes the form of a new exact sequence
® K2(7Lp[G]) -' m I SK1(7L[G],I) -L+ SK1(7[G]) e ' ® SK1(Zp[G])
(6)
p We now have another characterization of elements in 1
mod
which can be represented by matrices congruent to
SK1(7L[G])
for arbitrarily small ideals
I,
of finite index.
I C 7L[G]
The second term in (6) remains unchanged when any other 7L-order in
it is the set of
C11(7L[G]):
is replaced by
7L[G]
Hence, it is convenient to define
Q[G].
C(I[G]) = Jim SK1(7L[G],I)
(all
of finite index)
I C 7L[G]
I
- ® K2(®[G])].
Coker[K2(D[C])
This is a finite group; and
is a functor on the category of finite
C(-)
See Section 3c for more details.
dimensional semisimple Q@-algebras.
The computation of
C(I[G])
C(A)
Here,
D[G]
based on the solution
with center
1
if for some v: K
PK
otherwise.
to
the
IR,
K,
(some
Qt®vKA = Mr(IR)
r) (7)
jIl
}
of
is
In Theorem 4.13, it will be seen that for
congruence subgroup problem.
each simple summand A
(Theorem 3.12)
PP
P
denotes the group of roots of unity in
One convenient way
K.
to use this involves the complex representation ring RC(C).
Fix a group K = O(Cn)
G,
and fix any even
is a splitting field for
G,
n
such that
where
cn
exp(G) In.
root of unity, and we can identify the representation rings
The group
Cal(K/@) = (DJn)*
automorphisms) and on
7L/n
thus acts both on
(by multiplication).
Then
is a primitive n-th
RC(G)
Regard
RC(G) =R K(G)'
(via Galois
R1(G)
(the real
INTRODUCTION
10
representation ring) as a subgroup of
RC/IR(G) = RC(C)N(G)
Then we will see in Lemma 5.9 that
for short.
C(D[G]) = [RC,(G) 0 Z/n]
in the usual way, and set
RC(G)
(i. e.,
(Z/n)*
(Z/n)*-coinvariants) (8)
= RC/R(G)/([V]- a-[7a(V)]: V E RC(G), (a,n)= 1, Ta C Cal(K/Q))
This description, while somewhat complicated, has the advantage of being natural in that the induced epimorphisms
RC(G)
\ja
C(D[G])
C11(Z[G])
commute both with maps
induced by group homomorphisms and with maps
induced by restriction to subgroups (Proposition 5.2).
For example, one
immediate consequence
is generated by
of
this
is
induction from elementary subgroups of
groups with p-groups) - since
Cll(Z[G])
that
G
RC(G)
products of cyclic
(i.
e.,
is
generated by elementary
induction by Brauer's induction theorem.
Odd torsion in
C11(Z[C])
and
SK1(Z[G]):
For any finite group
G,
the short exact sequence (2) has a natural splitting in odd torsion, to give a direct sum decomposition
SK1(Z[G])[2] = C11(Z[G])[2] ® ®SKl(Zp[G])
(9)
P;'2
Furthermore, for odd K2(7Lp[G])
and
p,
there is a close relationship between the groups
H1(G;2p[C]) = H1(G;Z[G])(p)
(where
G
again acts by
conjugation) - close enough so that (6) can be replaced by an isomorphism
Cl1(Z[G])[2] = Coker[H1(G;Z[G]) - C(D[G])][2]
(10)
INTRODUCTION
G
When
is a p-group (and
11
is odd), this formula is shown in Theorem
p
9.5, and an explicit definition of
yG
is given in Definition 9.2.
is arbitrary, then the formula, as well as the definition of
*G'
G
If
are
derived in the discussion following Theorem 13.9.
G
When
C11(7L[G])(p),
Theorem 13.9.
is not a p-group, for any finite
G
then an alternative description of and any odd prime
is given in
p,
This takes the form
k C11(Z[G])(P)
al,...,°k C G
where
® H0(Ni/zi; i=1
5$(Zi)
(11)
(zi)
is a set of conjugacy class representatives for
cyclic subgroups of order prime to and
C11(Z[n]));
li
p;
and
Z. = CG(ai),
Ni = NG(ci),
is the set of p-subgroups.
2-torsion in when
G
then
SK1(Z[C]) = C11(Z[G])
Cl1(Z[G])(2) - even
The description of
SK1(Z[C]):
is a 2-group - is still rather mysterious.
G
If
is abelian,
can be described via formulas analogous to
(10) above (see Theorem 9.6 for
the case of an abelian 2-group, and
Theorem 13.13 for the general abelian case). 2-group, we conjecture that
SK1(7L[G])
When
C
is an arbitrary
can be (mostly) described via a
pushout square V
H1(G;7L[G])
H2(G)
J'PG
bC
C(Q[G])/CQ(Q[C])
Here,
CQ(Q[G]) C C(Q[G])
quaternionic summands:
SK1(Z[G])/Q(G)
denotes the subgroups of elements coming from
i. e., simple summands
matrix algebras over division algebras of the form fn = exp(2ai/2n) E C image of
CQ(Q[G])
(see Theorem 9.1).
under
Also,
of
A
Q[G]
Q(fn,j)
which are
(C Ii),
Q(G) C C11(Z[G])
8: C(Q[G]) - C11(Z[G]);
and
u
where is the
is defined
INTRODUCTION
12
by setting
u(g Oh) = g^h E H2(G)
for commuting g,h E G.
Note that
Coker(v) = H2(G)/b(G) = SK1(g2[G])
by (2).
The most interesting point here is the conjectured existence of a
lifting
©G
This is currently the only hope
of the isomorphism in (2).
for constructing examples where extension (1)
For more
is not split.
details, see Conjecture 9.7, as well as Theorems 9.6, 13.4, 13.12, and 13.14.
Induction
for odd
C11(7L[G])(p)
Each
theory:
the
of
SK1(2p[G]),
functors
has been given two descriptions above.
p,
and The
direct sum formulas (5) and (11) are based on a general decomposition
and are usually the easiest
formula in Theorem 11.8, computing
SK1(7Lp[G])
to apply when
as an abstract group.
C11(7L[G])(p)
or
The other
((4) and (10)) seem more natural, and are easier to use to
formulas
determine whether or not a given element vanishes. In both cases - SK1(2 p[G])
C11(7L[G])(p) - these formulas are
and
derived from those in the p-group case with the help of induction theory as formulated by Dress [2].
In the terminology of Chapter 11, these two
functors are "computable" with respect
to induction from p-elementary
subgroups (i. e., subgroups of the form
Cn x v
n
when
is a p-group).
See Chapter 11, and Theorems 12.4 and 13.5, for more details.
Detecting and constructing explicit elements: above algorithms have been stated so as to describe
as abstract groups.
But in fact,
For simplicity, Wh(G)
the
SK1(7L[G])
they can in many cases be used to
determine whether or not a given invertible matrix over Wh(G);
and
vanishes in
7L[G]
or to construct matrices representing given nonvanishing elements.
The procedures SK1(7L[G])
for
constructing explicit nontrivial
are fairly straightforward.
G = C4 X C2 X C21
G
in
is worked out in detail in Example 5.1; and essentially
the same procedure can be used to construct elements in any finite
elements
One example of this, for the group
(once the group
C11(7L[G])
C11(7L[G])
itself is known,
for
that is).
INTRODUCTION
Explicit elements in
can be constructed using Proposition
SK1(7Lp[G])
8.4, or Theorem 12.5 or
13
12.10; although Theorem 8.13 provides a much
simpler way of doing this in many cases.
The procedure for lifting an
element
can be found in the proof of
[A] E SK1(2 p[G])
note however
Theorem 3.9;
If
SK1(7L[C])
that
this depends on finding an explicit
A as a product of elementary matrices over
decomposition of
whether
to
is given, then the first step when determining
A E GL(7L[G]) it vanishes in
Wh(G)
C
is abelian, then
determining whether
to compute its reduced norm, and
is
determine (using (1)) whether or not if
OP[G].
SK1(7L[G]) = Cl1(7L[G]),
[A] = 1
Once this is done,
[A] E SK1(7L[G]).
and the procedure for
is fairly straightforward.
The details are
described in the proof of Example 5.1, and in the discussion afterwards.
If
is nonabelian, and if
C
[A]
is known to lie in
then one must next check whether or not it vanishes in primes
p1IGI.
8.4 when C
SK1(7L[G]),
The procedure for doing this is described in Proposition
is a p-group, and in Theorem 12.10 for general finite
both cases, this involves first choosing some group extension such that
SK1(2 p[G])
[A] E K1(2 p[G]),
maps trivially to
taking its
SK1(2p[G]);
G.
a: G - C A
then lifting
logarithm (more precisely,
In
to
its integral
r(W) E HO(G;7p[G])); and then composing that by a certain
logarithm
explicit homomorphism to
SK1(7Lp[G])
using formula (3) or (4) above.
The general procedure for detecting elements nonabelian
that at
for
SK1(71p[G])
G
the
Cl1(7L[G])
in
C11(7L[G])
for
is much less clear, although there are some remarks about
end of Section 5a.
The main problem (once
itself is understood) is to lift
[M] E Cll(7L[G])
the group to
C(D[G])
along the boundary map in sequence (6).
In some specialized cases, there are other ways of doing this.
The
proofs Propositions 16-18 in Oliver [1] give one example, and can be used to
detect
(certain)
nonabelian groups
G.
nonvanishing elements
in
C11(7L[G])
for
many
Another such example is given by the procedure in
Oliver [5] for detecting the Whitehead torsion of homotopy equivalences of S1-bundles.
INTRODUCTION
14
Survey of computations
The
examples
listed
here
both a survey
give
of
type
the
of
computations which can be made using the techniques sketched in the last section, as well as an idea of some of the patterns which arise from the
The first few examples give some conditions which are
computations.
necessary or sufficient for
Example
(Theorem
1
SK1(R[Cn]) = 1
5.6,
for any Finite cyclic group
Example 2
n),
or
Cn,
G
[3,
Theorem 3.3])
when R
is the ring of
al
Q.
(Theorem 14.2 and Example 14.4)
Cpn x Cp
or if
Alperin et
or
integers in any finite extension of
Cp,
to vanish.
SK1(7L[G])
(for any prime
and any
p
SK1(Z[G]) = 1 n),
C
if
G = (C2)n
if
(any
is any dihedral, quaternion, or semidihedral 2-group.
C
Conversely, if
is a p-group and
one of the above groups, or
p = 2 and Gab = (C2)n
G
then either
C11(7L[G]) = 1,
for some
is
n.
The next example (as well as Example 12) helps to illustrate the role
played by the p-Sylow subgroup
in determining the p-torsion in
Sp(G)
SK1(7[G])
Example 3 SK1(7[G])(P)
The SK1(7L[G])
=
next (or
(Theorem 1
if
14.2(i),
Sp(G) = Cpn
example Wh(G))
gives
or
Oliver
Cpn x p (any
or
completely
some
[1,
Theorem
2])
criteria
for
n).
different
This, together with the first three
to vanish.
examples, helps to show the hopelessness of finding general necessary and
sufficient conditions for particular that
Example 4
Wh(G) = 1
SK1(7L[C]) = if
(Theorem 14.1)
C
1
(or
Wh(G) = 1).
Note in
is any symmetric group.
Let
'C C b'
be the smallest classes of
finite groups which are closed under direct product and under wreath product with any symmetric group
S
n
;
and such that
T
contains the
INTRODUCTION
trivial group and contains
all
D(8),
G E 'C,
(Note that
also contains all dihedral groups.
'C'
as well as all symmetric groups.)
and SK1(Z[G]) = 1
Then
Wh(G) = 1
'e
for
for all G E V.
Note that the classes of finite groups SK1(Z[G]) = 1,
15
G
Wh(G) = 1,
for which
are not closed under products (see Example 6).
or
A slightly
stronger version of Example 4 is given in Theorem 14.1.
We now consider examples where that of abelian groups.
The easiest case is
SK1(Z[G]) # 1.
SKI(Z[G]) = C11(Z[G])
In fact, the exponent of
can be explicitly determined in this case.
Example 5
(Alperin et al [3, Theorem 4.8])
abelian group, and let dividing
for which
IGI
k(G) Sp(G)
Let
G
be a finite
be the product of the distinct primes is not cyclic.
p
Then
IGI
exp(SKI(Z[G])) =
k(G
(G)/;
where a=2 if (i)
G = (C2)n
for some
n Z 3,
(ii) S2(G) = C2,, x C2 for some (iii) S2(G) = C2. x C2 x C 2
and
e =1
or
n Z 3,
for some
or n Z 2;
otherwise.
We now consider some more precise computations of
SK1(Z[G])
in
cases where it is nonvanishing.
Example 6
(Example 9.8, and Alperin et al [3, Theorems 2.4, 5.1,
5.5, 5.6, and Corollary 5.9]) of
SK1(Z[G]) = C11(Z[G])
The following are examples of computations
for some abelian p-groups
G:
INTRODUCTION
16
(t)
If
p
is odd, then
SK1(ZL( p)k]) = R/P)N, where N = (ii)
p it -
(1/P)(P-1)(n-1)
(any prime
SK1(Z[Cp2 X CPn]) -
(LP)nP(P-1)/2
(iii) SK(Z[(C 1 )2 x C n ]) P P
i (Z/2)
n-1
x
(Z/P)p2-1
(iv)
SK1(Z[Cp3 x Cp3])
1
(v2)
P)
is odd
if
p
if
p = 2
(1p2)P-1
p
if
4
is odd
if p = 2 ®(Z/2s)
L(r)_(U2r-1)] ®r
(v) SKl(ZL(C2)kxC2n]) = [ r=2
J
We now look at some nonabelian p-groups: for
(P+P-1)
s=2
J
first for odd
p
and then
p = 2.
Example 7
(Example 9.9, and Oliver [7, Section 4])
odd prime, and let C11(Z[G]) =
C11(7L[G])
be a nonabelian p-group.
G
ICI = p3.
if
If
ICI
= p4,
be an
SK1(Z[G]) =
Then
(Z/p)P-1
p
Let
SK1(Z[G]) _
then
and:
(UP)
2(P-1)
if
CP X Cpl
if
(CP)3,
exp(G) = p
if
(C )3,
exp(G) = P2
(vp)(p2+3p-6)/2
(vP)(p2+p-2)/2
SK1(Z[G])
p
(UP)
3(p-1)/2
if
(1fP)P-1
Note that the
p-
and
if
p2-rank of
Cp x CP, Gab = C
C11(Z[G])
p
x CP,
( C
3 Cp x Cpl C G.
is a polynomial in
for each of the families listed in Examples 6 and 7 above.
p
Presumably,
this holds in general, and is a formal consequence of Theorem 9.5 below; but we know of no proof.
INTRODUCTION
Example 8
(Examples 9.9 and 9.10)
1
SK1(l[G]) = Cll(7L[G])
If
G
if
I 8/2 if
If
17
then
ICI = 16,
Gab = (C2)2 b
or
(C2)3
C4 xCZ.
Ga
is any (nonabelian) quaternion or semtdthedral 2-group, then for
all k>0: k
SKl(7LLG X (C2)k]) = Cl1(Z[G x (C2)k]) = (1/2)2
-k-1
We next give some examples of computations for three specific classes of non-p-groups.
Example 9
Assume C
(Example 14.4)
is a finite group whose 2-Sylow
subgroups are dihedral, quaternionic, or semidihedral.
Then
1(1LG])(2) = C11(7LG])(2) = (1/2)k
where
k
such that
and
is the number of conjugacy classes of cyclic subgroups (a)
lal
(b) CG(a)
is odd,
(c) there is no g E NG(a)
a C G
has nonabettan 2-Sylow subgroup,
with gxg 1 = x 1
for all
x E a.
Note, in the next two examples, the peculiar way in which 3-torsion (and only 3-torsion) appears.
Example 10
(Theorem 14.5)
3
1
SK1(1[PSL(2,pk)]) jl
1
For any prime
p and any k > 1,
if p=3, 24k, k>5 otherwise.
and
SK1(1LSL(2,pk)])
-1/3x1/3 if p=3, 24k, k>5 1
otherwise.
INTRODUCTION
18
Example 11
(Theorem 14.6)
alternating group on n
letters.
For any
be
the
and
r
m1 > m2 > ... > mr 2 0,
Fjni
odd
i=1
SK1(Z[An])
otherwise.
1
The next example involves the groups
G
An
let
1,
Then SK1(Z[G]) = C11(Z[G]),
if n = 13m` 2 27,
Z/3
group
n >
for which
Constructing a
is rather complicated (note that
SK1(2 p[G]) # 1
SK1(Z[G]) = C11(Z[G])
SK1(2 p[G]).
in all of the examples above); so instead of doing
that here we refer to Example 8.11 and the discussion after Theorem 14.1. For now, we just note the following condition for
Example 12 subgroup of
(Proposition 12.7)
SK1(Zp[G]) =
To end
we note
section,
the
(t)
Example 13
the p-Sylow
two
specific examples of
-13(1-g2)(1+h1)(1+h2)(3-g)
group of order
Then
and the nontrivial element is represented by the matrix
Z/2,
Set
concrete
C11(Z[G]).
G = C4 xC2 xC2 = (g) x (h1) x (h2).
Set
1 + 8(1-g2)(1+h1)(1+h2)(1-g)
(it)
if
1
G has an abelian normal subgroup with cyclic quotient.
matrices or units representing nontrivial elements in
SK1(Z[G])
to vanish.
SK1(2 p[G])
-(1-g2)(1+h1)(1+h2)(3+g) 1 + 8(1-g2)(1+h1)(1+h2)(1+gl )/
G = C3 x Q(8) = (g)x(a,b), S.
Then
SK1(Z[G]) = Z/2,
where
Q(8)
E GL (Z[G]) 2
is a quaternion
and the nontrivial element is
represented by the unit
1 + (2-g-g2)(1-a2)(3g +a + 4g2a + 4(g2-g)b + Sab) E (Z[G])*.
The matrix in (i) is constructed in Example 5.1.
is computed as a special
case of Example 9,
and
In (ii),
SK1(Z[G])
the explicit unit
representing its nontrivial element can be constructed using, the proof of
INTRODUCTION
Oliver [1, Proposition 17].
not a given element of
The general problem of determining whether or
Wh(G)
studied in Magurn et al
19
can be represented by a unit in
Z[G]
is
[1], and is discussed briefly in Chapter 10
(Theorems 10.6 to 10.8) below.
PART I: GENERAL THEORY
These first six chapters give a general introduction to the tools used when studying group rings.
K1
of 7L- and 2p orders, and in particular of integral
While some concrete examples of computations of
SK1(7L[G])
are given in Sections 5a and 5b, the systematic algorithms for making such computations are not developed until Parts II and III. The central chapters in Part I are Chapters 2, 3, and 4.
free part of
K1(21),
for any 7L- or 2p order 3,
In Chapter 3, the
using reduced norm homomorphisms and p-adic logarithms. continuous
K2
The torsion
is studied in Chapter 2
for p-adic algebras and orders is defined, and then used
to construct the localization sequences which will be used later to study SK1(7L[G])
for finite
subgroup problem:
C(@[G]) =
G.
Chapter 4 is centered around the congruence
the computation of one term
SK1(7L[G],n7L[G]) = Coker[K2(@[G]) ->
K2(® [G])]
in the localization sequence of Chapter 3.
In addition, Chapter 1 provides a survey of some general background material
on such subjects as semisimple algebras and orders,
theory, and K-theory of finite and semilocal rings.
number
Chapter 5 collects
some miscellaneous quick applications of the results in Chapter 4: example,
the results
that
dihedral, or quaternionic. K2(2 p[G]),
C11(7L[G])
=
1
G
whenever
for
is cyclic,
Also, the "standard involution" on
K1(7L[G]),
etc., studied in Section 5c, is the key to many of the later
results involving odd torsion in
C11(7L[G]) C SK1(7L[G]).
The integral
p-adic logarithm (Chapter 6), which at first glance seems useful only for getting an additive description of
K1(2 p[G])/torsion,
to play a central role in the computations of both C11(7L[G]).
will be seen later SK1(2 p[G])
and
BASIC ALGEBRAIC BACKGROUND
Chapter 1
By a Z-order (i. e.,
subring.
h
in a semisimple Q-algebra A
3
is a finitely generated
7L-module and
is meant a 7L-lattice
A =
which is a
One of the reasons why Whitehead groups are more easily studied
for finite groups
than for
theorems for semisimple
infinite groups
is
that strong structure
Q-algebras and their orders are available as
In fact, it is almost impossible to study the K-theory of group
tools.
71[G]
without considering some orders which are not themselves
group rings.
Furthermore, the use of localization sequences as a tool for
rings
studying
K1(21)
for
7-orders
2[
makes it also important to study the
K-theory of orders over the p-adic integers 2 p.
This chapter summarizes some of the basic background material about semisimple algebras,
orders, p-adic localization,
similar topics, which will be needed later on. mostly without proof.
semilocal
rings, and
The results are presented
The first two sections are independent of K-theory.
Section lc includes some results about
K1
of semilocal or finite rings,
as well as Quillen's localization sequence for a maximal order.
Section
ld contains a short discussion about bimodule-induced homomorphisms for Ki(-),
and in particular about Morita equivalences.
Recall that a number field is any finite field extension of ring of integers in a number field K:
over
la.
K
is the integral closure of
The
Q.
Z
in
i. e., the set of elements in K which are roots of monic polynomials Z.
Semisimple algebras and maximal orders
The definition of a semisimple algebra (or ring) varies somewhat; the
most standard is to define it to be a ring which is semisimple (i. e., a
direct sum of modules with no proper submodules) as a (left or right) module over itself.
Then a simple algebra is a semisimple algebra which
has no proper 2-sided ideals.
Throughout this book, whenever "semisimple
22
CHAPTER 1.
BASIC ALGEBRAIC BACKGROUND
algebra" is used, it is always assumed to mean finite dimensional over the
Our main references for this topic are Curtis & Reiner [1,
base field.
Section 3] and Reiner [1, Section 7].
For any field standard
K
of characteristic zero, and any finite group
representation
K-algebra.
theory
shows
that
K[G]
the structure of
In particular,
G,
a semisimple
is
as a semisimple
@[G]
D-algebra plays an important role when studying K1(Z[G]).
The center of any algebra A will be denoted
Theorem 1.1 K-algebra.
K
Let
be a field, and let
(Wedderburn theorem)
and numbers
over
K,
each
Mri(D1)
be any semtsimple
There are division algebras
rl,...,rk > 0,
such that
(D.)r'.
k
is simple if and only if
LOK A
(iii)
A
If
Z(A)
is a field.
is separable, then for any field
Z(A)/K
is a central simple K-algebra (i. e.,
K
Z(A).
In
K = Z(A)),
then
n E Z.
A is a central simple K-algebra,
If
is a simple subalgebra which contains
f: B - A
automorphtsm of
LO
K = Z(A).
(Sholem-Noether theorem)
B C A
ll Z(Di); i=1
is semtsimple with center
is simple if
for some
homomorphtsm
Proof
LOK A
L 2 K,
[A:K] = n2
(iv)
is simple and
A
If
particular,
Here,
k
[1 Z(Mri(Di)) = i=1
extension
ni-1Mri(D1).
Furthermore,
Z(A) -
(it)
A
D1,...,Dk
is a simple algebra, and has a unique irreducible module
isomorphic to
and if
A
Then the following hold:
(t)
and A
Z(A).
which fixes
K
K,
then any ring
is the restriction of an inner
A.
The Wedderburn theorem is shown, for example,
in Curtis &
Reiner [1, Theorems 3.22 and 3.28]; and the other statements in (i) are
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
easy consequences of that.
23
The other three points are shown in Reiner [1, 0
Theorem 7.18 and Corollary 7.8, Theorem 7.15, and Theorem 7.21].
Note in particular that a commutative semisimple K-algebra always factors as a product of fields.
In the case of group rings of abelian
groups, one can be more specific.
Recall that for any
primitive n-th root of unity; and that for any field
K
any field k K[G]
For any
n Z 1,
where for each
For any
induced by setting
More generally, for
Ki = K(in)
a homomorphism
n,
for some
[Kd:Q] = Bp(d)
a
elements of
0
which must be roots of unity.
As another example, consider group rings
is
C
and
C[G]
IR[G]
for a
C
itself (this is the case for any algebraically closed field); and
algebra).
IR
Note in particular that
not central in G,
C[G]
H.
are OI
is not a
and
(the quaternion
IH
C-algebra, since
is a product of matrix algebras over IR,
is a simple IQ-algebra with center
K,
Ot®K A
C,
IR,
C
is
The Wedderburn theorem thus implies that for any
a product of matrix algebras over
then
and the images of the
The only (finite dimensional) division algebra over
the only division algebras over
finite
a
is a product of fields by Theorem
K[G]
K
G.
is surjective.
(see Janusz [1, Theorem I.9.2]),
d
1.1, and each field component is generated by
finite group
Each
is an isomorphism.
The last point is clear:
G,
a
So
d.
is
Wd, and they are
Q[Cn]-representation
for each
dimension count shows that
g E C.
for some fixed generator
ad(g) = rd
G,
nilexp(G).
a-ilad:I[n])Rd Innd
distinct since n acts on Wd with order
Since
Kin denotes the
K,
D[Cn] =
i,
induces an irreducible
ad
denotes a
of characteristic zero and any finite abeltan group
[[i_1Ki,
Proof
n
K which contains the n-th roots of unity.
smallest field extension of
Example 1.2
n,
C,
and
HI.
and if
is a matrix algebra over either
K '- Ot IR
or
C,
and
IR[G]
Also, by (ii), if
IH.
is
A
is any embedding,
This last point
will play an important role later, for example when describing the image of the reduced norm in Theorem 2.3.
24
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
A
If
is a simple algebra with center
division algebra
D,
then the index
A
of
A
field for
L ®K A
if
Proposition 1.3
A field
1/2
L 2 K
is a matrix algebra over
A
Let
for any splitting field
for
ind(A)I[L:K].
A,
L C A
division algebra, then any maximal subfield for
A and satisfies
Proof
domain with field of fractions A
generated
R-module and
A
K.
Then
A
is a
is a splitting field
See Reiner [1, Theorem 28.5, and Theorem 7.15].
K-algebra
If
[L:K] = [A:K]1/2
We now consider orders in semisimple algebras.
in
is called a splitting L.
be a simple algebra with center
L D K
for some
is defined by setting
ind(A) = ind(D) = [D:K]
(an integer by Theorem 1.1(iii)).
A = Mr(D)
and
K,
A)
an
K,
is defined to be an
If
R-order
21
R-lattice (i. e., which is a subring.
R
13
is a Dedekind
in a semisimple 21
is a finitely
A maximal R-order Our
is just an order which is not contained in any larger order.
main reference for orders and maximal orders is Reiner [1].
The most
important properties of maximal orders needed when studying Whitehead groups are listed in the next theorem (and Theorems 1.9 and 1.19 below).
Theorem 1.4
Fix a Dedekind domain R with field of fractions
characteristic zero, and let
A
be a semisimple K-algebra.
K
of
Then the
following hold.
A contains at least one maximal R-order, and any R-order in A
(i)
is contained in a maximal order.
(ii)
maximal i,
Dii
If
A = fl
R-order, then
where the
1Ai, 1)1
Ai
are simple and
splits as a product
is a maximal order in
Ai.
lB _ Ik=1Ni,
)1 C A
is a
where for all
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
A
(iii) If
is commutative, then there is a unique maximal R-order
A = fi_1Ki
I C A.
If
then
_ [[i_1Ri,
where the
where
R
ideals in
Ki
in
K.:
K,
i. e.,
Ki.
Any maximal R-order
all left (or right)
is hereditary:
Il C A
are projective as
V
are finite field extensions of
is the ring of R-integers in
Ri
the integral closure of
(iv)
25
1-modules, and all finitely generated
R-torsion free 1R-modules are projective.
(u)
C
If
containing
is any finite group, and if
R[G],
Proof
then
IGI'Th C R[G].
These are shown in Reiner [1]:
Theorem 10.5(1),
(1) in Corollary 10.4, (ii) in
in Theorem 10.5(iii),
(iii)
Corollary 21.5, and (v) in Theorem 41.1.
n > 2,
(iv)
in Theorem 21.4 and
o
Note that point (i) above is false if example, for
is a maximal order
Th C R[C]
A
is not semisimple.
the ring of upper triangular
nxn
matrices over
For
R
has no maximal Z-orders.
1.2 has already hinted at
Example
the
important role played by
cyclotomic extensions when working with group rings.
The
following
properties will be useful later.
Theorem 1.5 KCn Then
Fix a field
T(Cn) = Kn) a
(i)
integers.
for some unique
(K = @)
Gal(Q'n/tD)
K
n > 1
such that
each
(Z/n)*:
a E (Z/n)*.
=
(Brauer)
If
is
G
char(K).I'n.
Let
by a primitive n-th root of unity.
(ZJn)*,
Gal(KCn/K)
T E Cal(Kcn/K)
can be
has the form
Furthermore:
and
Zfn C QCn
In particular, under the identification
maximal Z-order in ([C]
(ii)
K
is an abelian Galois extension, and
KC /K
identified as a subgroup of
then
and
K,
denote a field extension of
D[Cn]
is the ring of
ndInnd'
the
ndInZCd-
is a finite group, and
is a splitting field for
G:
i. e.,
KCn[G]
char(K)4'exp(G)In,
is a product of
26
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
matrix algebras over
Kcn.
Gal(Kcn/K) C (U/n)*
The embedding
Proof
Gal(Kn//) = (7L/n)* 1.9.2]);
and
Exercise
2].
7L[n
The
is the ring of integers in statement
last
Corollary 15.18 and Theorem 17.1].
R
Note that when K, then
example,
in
QCn
then
(i)
Theorem
by Janusz [1, §1.9,
from Theorem
follows
Brauer's splitting theorem is shown in Curtis & Reiner [1,
1.4(iii).
field
(see Janusz [1,
[Q(n:'D] = v(n)
since
K
When
is clear.
O
is the ring of integers in an arbitrary number R
RCn need not be the integral closure of
Z[Z]
is the ring of integers in
f(v ),
in
7[v ,i]
but
For
Kcn.
is not
the ring of integers in D(v,i) = Q(C12)-
For any field K and any finite group order some
are
n
prime to
x E G
char(K)
and some
C-conjugate
if
are called
a E Gal(Kcn/K) C and
In-conjugate if and only if
only (g)
and
are
(h)
g,h E G
ga = xhx 1
For example,
(7LJn)*.
they
if
two elements
G,
K-conjugate if
and
conjugate;
g
and
they
are conjugate subgroups of
of
for
h are G.
The importance of K-conjugacy lies in the following theorem.
Theorem
1.6
characteristic zero, irreducible
(Witt-Berman
and for any
For
any
finite group
G.
theorem)
field
K[G]-modules - i. e., the number of simple summands of
- is equal to the number of K-conjugacy classes of elements in
Proof
K
The characters of the irreducible
for the vector space of all functions K-conjugacy classes.
of
the number of K[G]
G.
K[G]-modules form a basis which are constant on
(C ---> C)
This is shown, for example, in Curtis & Reiner [1,
Theorem 21.5] and Serre [2, §12.4, Corollary 2].
n
Note that there also is a version of the Witt-Berman theorem when char(K) > 0:
the number of distinct irreducible
the number of K-conjugacy classes in char(K).
G
K[G]-modules is equal to
of elements of order prime to
See Curtis & Reiner [1,Theorem 21.25] for details.
CHAPTER 1.
lb.
27
BASIC ALGEBRAIC BACKGROUND
P-adic completion
be the ring of integers in any number field
R
Let
maximal ideal
p C R,
the
p-adtc completion
For any
K.
R and K are defined
of
by setting
Rp =
im R/pn;
Kp =
p[p]
(p = char(R/p)).
n
Then Rp
is a local ring with unique maximal ideal
field of fractions. Rp
Furthermore,
is the integral closure of
pRp,
and Kp
is a finite extension of
Kp in
is its
gyp,
and
Kp.
1p Alternatively, valuation
and
RP
vp: K -+ 7 U -.
Kp
can be constructed using the p-adtc
This is defined by setting
vp(r) = max{n > 0 : r EP n}
for
r E R,
and
vp(r/s) = vp(r)- vp(s)
This induces a
in general. v px) (
K - based on the norm
topology on
- and
Ixlp = p
are the corresponding completions of
K
and
Kp
Note that
R.
and
RR
Rp
is
compact under this p-adic topology, since it is an inverse limit of finite groups.
If
p g R C K are as above, then for any semisimple K-algebra
any R-order
21 C A,
the p-adic completions of
A and
21
A
are defined by
setting
2[p =
Then
AP
is a
Z-order in
is a semisimple Ap.
P
%-algebra
(where
Note that if we regard
for any rational prime p and Ap =
Ap=Kp®KA=K 0R21p.
H/per = Rp ®R 21,
(i. e.,
p = char(R/p)),
A
p E 7), and any
and
21p
as a l-algebra, then
21 C A,
2(p = Z 0 21 p
®Q A.
The importance of using p-adic completions when studying
K1(21)
for
28
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
a 7-order
is due partly because it is far easier to identify the units
3
(and invertible matrices) in the
2p
than in
2[,
and partly because
analytic tools such as logarithms and exponents can be used when working in
For example, results in Chapters 6 and 8 illustrate how much
K1(21p).
Ki(2p[G])
more simply the groups groups
Ki(Z[G])
and
S C E
Then
E/F
Theorem 1.7
P p
q = pS,
Fix an algebraic number field A
Let
k
pip
(i. e.,
maximal ideals
K,
and
2I
and let
R 2I
be its be any
p
p,
Ap = pAp
RRp
pip
and
up = n21p pip
pip
p g R
which divide
(p = char(R/p))
is unramified for all but finitely many
p g R.
is a product of matrix rings over fields for all but
AP
finitely many
(iv)
be the
p 7 pR).
(ii) KpA
(tit)
q C S
R C F
and is totally
be any semisimple K-algebra, and let
Here, the products are taken over all maximal ideals p
let
A.
IIK, p
R
p C R and
is unramifted if
For any rational prime
(i)
Qp,
S/q = R/p.
ring of integers.
R-order in
are described than the
be any pair of finite extensions of
be the rings of integers, and let
maximal ideals. ramified if
SK1(2p[G])
SK1(7[G]).
and
E/F
Now let
and
p C R
2Cp
is a maximal
-order in
is a maximal R-order in
RR-order in AP
for all
A
Ap
for almost all
if and only if
21P
p
in
R;
is a maximal
p g R.
Proof The first two points are shown in Janusz [1]: (1) in Theorem 11.5.1, and (ii) in Theorem 1.7.3. Points (iii) and (iv) are shown in
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Reiner [1, Theorem 25.7 and Corollary 11.6].
29
0
The next proposition gives information about the groups
R
when
is the ring of integers in a finite extension
particular, the reciprocity map
F*/N(E*) = Gal(E/F)ab
the key to defining norm residue symbols in
Proposition 1.8 p C R C F
p.
&.
In
of
in (ii) below is
(Section 3a).
and let
be the maximal ideal and ring of integers.
µ C R*
Let
(i) to
R*,
be any Finite extension of
F
Let
K2(F)
F
F* and
be the group of all roots of unity of order prime
Then projection mod
for any generator n
of
}t
2-5'
(R/p)
and
p:
R* = g x (1+p)
(ii)
induces an isomorphism
p
F* = u x (l+p) x ('r)
and
For any Finite Galots extension
E/F,
there is a canonical
i somorpht sm
s
: F*/NE,,(E*) )
(the reciprocity map). where
Cal(E'/F),
contained in
(iii)
E'
If
E/F
Gal(E/F)ab
is not Galois,
then
F*/NE/F(E*)
denotes the maximal abelian Galois extension of
E.
If
E/F
is unramtfted, and if
q C S C E
are the maximal
ideal and ring of integers, then the norm and trace homomorphtsms
N = NS/R : S* are surjectiue.
Proof
R*
and
N R*
Also, for all
n
1,
Tr = TrS/R
:
S
%R
N(l+gn) = 1+pn and Tr(gn) = pn,
To see (i), just note that
_ Ln
F
(R/Pn)* =
n
((l+p)/(1+pn) x (R/p)*) - (l+p) x (R/p)*;
30
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
since
(l+p)/(l+pn)
is a p-group for each
ptI(R/p)*I.
and
n,
Point
(ii) is shown in Cassels & Frbhlich [1, §VI.2.2, and §VI.2.6, Proposition 4].
NS/R and TrS/R
The surjectivity of
by filtering e and
Section V.2]:
in (iii) is shown in Serre [1,
R
by the
and then using
pn,
analogous results about norms and traces for finite fields.
O
The following very powerful structure theorem for p-adic division algebras and their maximal orders is due to Hasse [1].
Theorem 1.9
D
ring of integers, and let n =
Q
of
n E D such that
and an element
S C E,
let
R C F
be a division algebra with center
Then there exists a maximal subfield
[D:F]1/2.
integers
F
Fix a finite extension
be the F.
Set
with ring of
E C D,
irEa 1 = E,
for which
the following hold: n-1
is unramified, and D =
E/F
(i)
i=0
n-1
®
A =
(ii)
is the unique maximal 2p -order in D
i
i=0
vA
(iii)
(iv)
is the unique maximal ideal in
is the maximal ideal in
irnR
Furthermore, for any is conjugate (in
Proof
end
Mr(D))
1
to
R.
and any maximal
1
order
]A
in
Mr(D),
7k
Mr(A).
See Hasse [1, Satze 10 & 47], or Reiner [1, Section 14 and
Theorem 17.3].
We
r >
A
0
the
section
by
noting
following
the
more
specialized
properties of p-adic group rings.
Theorem 1.10
Q, and let
R C F
Fix a prime
p,
let
F
be the ring of integers.
be any finite extension of
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
n
For any
(t)
such that
F[C
p4n,
31
where the
= ni-1Fi,
are
Fi
n] finite unramifted extensions of where
ni=1Ri,
is the ring of integers in
Ri
is unramified, and
R[Cn]
More precisely, then
i E F,
Proof
2
p = 2)
(if
is odd and
Cp E F,
p-1
or
(if
p = 2
or if
Since
1 E R,
is a maximal 2p order in
R[Cn]
F[C
n]
n Theorem 1.4(v). integers in
If
p C R
for each
R.
(ii)
K
For any field
K
and
F./F
for some
is a
R/p[Cn]
so
pRi
is unramified.
is
the
The last
i.
of characteristic zero, a cyclotomic algebra
is a twisted group ring of the form
finite cyclotomic extension of A
i,
Fi = FCn
statement follows since
over
p = char(R/p)Fn),
by
are the rings of
R.
is the maximal ideal, then
product of finite fields (since maximal ideal in
where the
R[Cn] = ni=1Ri
Hence,
Fi.
p and
is a product of matrix algebras over fields.
F[G]
(i)
p
if
is a product of matrix algebras
F[G]
G,
over division algebras of index dividing is odd).
Fcn/F
In particular,
F1.
is the ring of integers in F.
For any finite group
(ii)
R[Cn] _
Under this identification,
F.
K,
is a central simple K-algebra).
G = Gal(L/K),
and
L
where
A = LP[G]t,
is a
(3 E H2(G;pL)
(so
By the Brauer-Witt theorem (see Witt
[1], or Yamada [1, Theorem 3.9]), any simple summand similar to a cyclotomic algebra over its center.
A
of
F[G]
is
Then by another theorem
of Witt [1, Satz 12] (see also Yamada [1, Proposition 4.8 and Corollary 5.4]),
for any finite extension
(p = 2),
E 2 QP(Cp)
any cyclotomic algebra over
proves the last statement in (ii).
E
ind(A)I[Ep:E]jp-1
if
p
is odd, and
E 3 42(i) This
is a matrix algebra.
of
F[G]
with center
ind(A)j[E(i):E]j2
Many of the elementary properties of
K1(2I),
when H
if
E D F,
p = 2.
D
is a 2p order
-algebra, are special cases of results about semilocal
in a semisimple
Qp rings.
odd) or
The first statement then follows from
for any simple summand A
Proposition 1.3:
(p
These will be discussed in the next section.
32
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Semilocal rings and the Jacobson radical
lc.
For any ring
the Jacobson radical
R,
is defined to be the
J(R)
intersection of all maximal left ideals in
R;
intersection of all maximal right ideals in
R
or, equivalently,
the
(see Bass [2, Section
For example, the Jacobson radical of a local ring is its unique
111.2]).
maximal ideal, and the Jacobson radical of a semisimple ring is trivial.
I C R
An ideal
R
If
J(R).
is called a radical ideal if it is contained in
is finite, then
is a radical ideal if and only if it
I C R
is nilpotent (see Reiner [1, Theorem 6.9]). semisimple @p-algebra, then J(2I) is radical if and only if
p21
lim In = 0.
and
If
21
is a 1p order in a
J(21)/p21 = J(21/p21);
so I C 21
The next theorem helps to explain
n-wo
the importance of radical ideals when working in K-theory.
Theorem 1.11 any
n >1,
For any ring
a matrix
M E Mn(R)
invertible in Mn(R/J).
Proof
R
with Jacobson radical
J = J(R),
and
is invertible if and only if it becomes
In particular,
1+J C R*.
See Bass [2, Proposition 111.2.2 and Corollary 111.2.7].
The next example shows that p-adic group rings of p-groups are, in fact, local rings.
Example 1.12 of
R[G]
QP,
p C R
if
If
R
is the ring of integers in any finite extension
is the maximal ideal, and if
G
is any p-group, then
is a local ring with unique maximal ideal
J(R[G]) = {jrlgi
In particular,
Proof
ri E R,
gi E G,
Zr, E p}.
R[C]/J(R[G]) = R/p.
See Curtis & Reiner [1, Corollary 5.25].
We now recall some of the basic definitions and properties of
For any ring
R,
let
GL (R)
n
K1(-).
be the group of invertible nxn matrices
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
over
(any n> 1); and set
R
let
r E R,
GL(R,I)
and any
denote the elementary matrix which is
eij E GL(R)
r
in the (i,j)-position; and let
be the subgroup generated by the
ideal, then
For any i#j
GL(R) = Un-1GLn(R).
identity except for the entry GL(R)
33
eij.
If
the
E(R) C
is any (2-sided)
I C R
denotes the group of invertible matrices which are
congruent to the identity modulo normal subgroup of
GL(R)
K1(R) = GL(R)/E(R)
and
I;
and
containing all
E(R,I) ei.
denotes the smallest
for
r E I.
Finally, set
That these are, in
K1(R,I) = GL(R,I)/E(R,I).
fact, abelian groups will follow from the next theorem.
For the purposes of this chapter, we define
by setting
= H2(E(R)).
K2(R)
The usual definition
Steinberg group), as well as some of Steinberg symbols in
Theorem 1.13
K2(R),
K2(R),
for any ring
R,
(involving
the
the basic properties of,
e.
g.,
will be given in Section 3a.
For any ring
(Whitehead's lemma)
R,
and any ideal
I C R, E(R) = [GL(R),GL(R)] = [E(R),E(R)],
and
E(R,I) = [GL(R),GL(R,I)] = [E(R),E(R,I)].
and
For any
A,B E GLn(R,I),
K1(R,I).
Furthermore, there is an exact sequence
(0 A 1) E E2n(R,I),
[0 B]
in
K2(R) -+ K2(R/I) -+ Kl(R,I) - K1(R) -i K1(R/I).
Proof
The commutator relations are due to Whitehead and Bass, and
are shown in Milnor [2, Lemmas 3.1 and 4.3].
I)Q 01)(I -I
A
(0 A '/ = l0
The relation
I/(-A 1
OI)(0 I/(0
is clear from the definition of
I
E2n(R,I)
(and is part of the proof that as
[GL(R),GL(R,I)] C E(R,I));
and
an immediate consequence.
The exact sequence is constructed in Milnor [2,
[diag(A,B)]
34
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Lemma 4.1 and Theorem 6.2]; and can also be derived from the five term exact homology sequence (see Theorem 8.2 below).
A ring
R
is called semilocal
equivalently, if
0
R/J(R)
if
semisimple;
the functor
or
Thus, any
As one might guess, given
finite ring, and any ip order, are semilocal.
Theorem 1.11 above,
is
is artinian (see Bass [2, §111.2]).
R/J(R)
behaves particularly nicely for
K1
semilocal rings.
Theorem 1.14 The following hold for any semtlocal ring
(t)
Any element of
R.
is represented by a unit (t. e., by a
K1(R)
one-by-one matrix).
(it)
If
SK1(21) = 1
if
(iii) If
R 21
is commutative,
1)
Proof
R*.
a: R - S
is another semilocal ring, and
is an
K1(a): K1(R) - K1(S)
and
are all surjectiue.
These are all shown in Bass [2]:
(i) in Theorem V.9.1, (ii)
in Corollary V.9.2, and (iii) in Corollary 111.2.9.
The following relation in
Swan's presentation of
K2(R,I)
below
For any ring
are such that
the proofs
(when
be used in this book only in the case when
Theorem 1.15
I
then
in
later chapters.
is a radical ideal) will
I2 = 0.
R and any ideal
(l+rx) E R*,
0
due to Vaserstein [1], is often
K1(R,I),
useful, and helps to simplify some of
x E I
In particular,
the maps
GLn(a): CLn(R) - CLn(S)
(any n
K1(R)
is any commutative 2p order.
S
eptmorphtsm, then
then
I C R,
(l+xr) E R* and
(l+rx)(l+xr)-1 E E(R,I).
if
r E R
and
If
is a radical ideal (t. e.,
I
35
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
I C J(R)),
K1(R,I) = (1+I)/((l+rx)(l+xr)-1
then
r E R, x E I).
:
In particular, K1(R,I) = I/(rx - xr : rER, xE I) if I2 = 0. Proof
Recall that
E(R,I) = [GL(R),GL(R,I)]
(Theorem 1.13).
Using
this, the relation
shows
1/1x
particular, since again),
(0 u9 ) E E(R,I)
this shows that
1/
O1)l0
(l+rrx (1+0r)_1)
and that
(1)
E E(R,I).
u E (1+I)*
for any
In
(Theorem 1.13
(l+rx)(l+xr)-1 E E(R,I).
The presentation for [2,Theorem 2.1].
-
O1/10
(l+xr) E R*,
that
O1(x
(100rx
1+xr0
10
is due to Swan
I C J(R),
when
K1(R,I),
The last presentation (when
is a special case
I2 = 0)
of Swan's presentation, but is also an easy consequence of Vaserstein's O
identity.
When studying
for a
K1(21),
2p -order
get information about K1(21/I) and index.
K2(21/I)
The next result is a first step towards doing this.
Theorem 1.16 finite.
Let
Furthermore,
is a p-group if pj'IK1(R)I
Proof
if
R
R
R
be a finite ring.
(i)
K2(R) = 1
if
K2(R)
theorem,
is semisimple;
has p-power order (for any prime
By Theorem 1.14(i),
are both finite.
R = ni_1Mri(Di),
and hence fields.
R
and K2(R)
Then K1(R)
p);
R*
surjects onto
If
R
H2(E5(R))
K1(R).
onto
are
(it)
K2(R)
and
(iii)
is semisimple and has p-power order for some prime
Theorem 1], there is a surjection of and
it is often necessary to for ideals I C 21 of finite
21,
p.
By Dennis [1,
K2(R).
So
K1(R)
is semisimple, then by the Wedderburn
where the
D.
Then GL(R) = [[k=1CL(Di),
are finite division algebras E(R) = ((i=1E(D1);
and hence
36
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
k
H1=1Kn(Di)
Kn(R)
9.13]; and
n = 1,2.
for
p4IK1(R)I
R
if
By Theorem 1.11, if
So
(and hence the
J C R
by Milnor [2, Corollary
K2(R) = 1 Di)
have p-power order.
is the Jacobson radical, then the group
X = Ker[E(R) -v E(R/J)] S U (1+Mn(J)) n>1
is a union of finite p-groups.
Hence
H.(X)[p] =0 for all
and
i > 0;
the Hochschild-Serre spectral sequence (see Brown [1, Theorem VII.6.3]) applies to show that
K2(R)[p] = H2(E(R))[P] = H2(E(R/J))[p] = K2(R/J)[p]
Since
R/J
is semisimple,
K2(R)[p] = 1.
and hence
K2(R/J) = 1;
a
We end the section with some localization exact sequences which help to describe
M
when
Ki(1k)
is a maximal
g-
or
They are
1P order.
special cases of Quillen's localization sequences for regular rings (or
abelian categories). Theorem 1.17 in a semisimple
Qp-algebra
then there is for all
...
For any prime
(i)
n Z 0
A, and if
if
Li
J c I
7L-order
is a maximal
is the Jacobson radical,
an exact sequence
> Ki+l(IR) - K,+1(A) -+ Ki(Dt/J) -' Ki(T) - Ki(A) - ....
In particular,
p4'ISK1(R)I.
(ii) Fix a subring
Q-algebra ffi[
p,
A,
and let
Th[p:pE5'],
and let
A C Q and a maximal
A-order
I
in a semisimple
be any set of primes not invertible in (for
Jp c Dtp
p E f)
A.
Set
be the Jacobson radical.
Then there is an exact sequence
.. - Ki+1(Th) - Ki+l(M[1]) -
K1(ip/Jp) '--' Ki(l) pE#
Ki(Th[.i
]) - ....
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Proof
37
For example,
These follow from Quillen [1, Theorems 4 and 5].
in case (ii), if finitely generated
IR-modules,
respectively;
by
then
and
M(R[l]),
M(N),
denote the categories of
N'(-N)
R[!]-modules, and !O-torsion
Quillen
Theorem
[1,
5]
there
is
R-modules,
an
exact
localization sequence
Ki+l(M(1R[4])) -' Ki(Mt(R)) -' Ki(M(T)) -* K1(_(R[1])) -+ ...
Since
and
R
generated
or
R-
follows that
are hereditary (Theorem 1.4(iv)), all finitely
DI[!]
]R[!]-modules have finite projective resolutions.
Ki(M(IR)) = Ki(7R)
JR-module
M,
and
Ki(M(I[1])) = Ki(IR[l]).
M = ®(p), and each M(p)
It
For any
has a filtration by
So by devissage (Quillen [1, Theorem 4]),
R /Jp modules.
Jp).
K.(Mt(T)) = ® Ki(i P
In case semisimple,
K1(]IVJ)
p4'ISKI(m)I.
ld.
(i),
SKl(At)
=
-
Im[K1(Th/J)
has order prime to
p
Kl(1)].
Since
7R/J
is
by Theorem 1.16(iii), and so
o
Bimodule-induced homomorphisms and Morita equivalence
Define the category of "rings with bimodule morphisms" to be the category whose objects are rings; and where and
S,
isomorphism classes of (S,R)-bimodules
SMR
generated and projective as a left S-module. given by tensor product.
for any rings
R
where
such that
M
is finitely
Composition of morphisms is
The usual category of rings with homomorphisms
is mapped to this category by sending any SSR'
Mor(R,S),
is the Grothendieck group modulo short exact sequences of all
s3s2f(r).
f: R -> S to the bimodule
The importance of this category for
our purposes here follows from the following proposition.
38
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Proposition 1.18
For each
is an additive functor on the
K.
i,
category of rings with btmodule morphtsms.
M
Any (S,R)-bimodule
Proof
which
is
finitely generated and
projective as a left S-module induces a functor
M®R : E(R)
Z(-)
where modules.
denotes
E(S);
category of
the
finitely generated projective
So the proposition follows immediately from Quillen's definition
in [1] of
Ki(R)
using the Q-construction on
In the case of
and
K1(-)
directly.
Let
M ®P = Sk
of left S-modules.
SMR
K2(-),
$(R).
this can be seen much more
be any bimodule as above, and fix some isomorphism
For each n Z 1,
define homomorphisms
[M®R ]n: CLn(R) = AutR(Rn) - ) AutS(Snk) = GL
by setting [M®R]n
[NOR]n(a) = (M OR a) ®Id(Pn)
for each
(S)
a E Aut(Rn).
The
are easily seen to be (up to inner automorphism and stabilizing)
independent of the choice of isomorphism unique homomorphisms on K1(R) = H1(GL(R))
M ®P = Sk;
and
and hence induce
K2(R) t--- H2(E(R)).
13
As one example, transfer homomorphisms in K-theory can be defined in terms of bimodules.
R C S
If
is any pair of rings such that
S
is
projective and finitely generated as an R-module, then
trfR = [SOS], : K1(S) -- K1(R);
when
S
is regarded as an (R,S)-bimodule in the obvious way.
proposition
is
often
useful
when
verifying
the
The above
commutativity
of
K-theoretic diagrams which mix transfer homomorphisms, maps induced by ring homomorphisms, and others: isomorphisms of bimodules.
commutativity is checked by constructing
Examples of this can be seen in the proofs of
Proposition 5.2 and Theorem 12.3, as well as throughout Chapter 11.
39
BASIC ALGEBRAIC BACKGROUND
CHAPTER 1.
Another setting in which it is useful to regard
Ki(-)
as a functor
defined on rings with bimodule morphisms is that of Morita equivalence.
R
Morita equivalence between two rings
bimodule
and
i. e., for some bimodule 0S, MORN=S and
:
as bimodules.
In particular,
isomorphisms between
K1(R)
and
[NOS]*
and
[M®R]*
Rn
is invertible when regarded as an
In this case, the induced isomorphisms
-algebra
we saw
1.9,
Mn(D)
are
inverse
For any ring
mn
R
(Mn(R),R)-bimodule.
are precisely
Ki(Mn(R)) = Ki(R)
with CL
those induced by identifying GLm(S)
In Theorem
N ®S M=R
K1(S).
The simplest example of this is a matrix algebra. and any n > 1,
A
is an "invertible"
S
(R).
that any maximal 2P order
in a simple
a division algebra) is conjugate to a matrix
(D
algebra over the maximal order in
D.
This is not the case for maximal
Z-orders in simple Q-algebras; but a result which is almost as good can be stated in terms of Morita equivalence.
Theorem 1.19 Let
A
be any simple
V.
Let
maximal R-order in EndA(A)
an
for some
A C D A;
Write
K-algebra.
with field of fractions
and any maximal R-order in A
(T,A)-bimodule, and so
A
in
V.
and
V
Then
A
Furthermore,
Mn(A)
has the form A
Ki(m)
Proof
id incl
K.
is a
induce for all
i
N =
a commutative
Ki(D)
Ki(A) = Ki(Nn (D)).
See Reiner [1, Theorem 21.6 & Corollary 21.7].
is a
is inuertible as
square
Ki(A)
D
for some n-dimensional
be any maximal R-order.
A-lattice
where
A = Mn(D),
A = EndD(V)
division algebra, and identify D-module
R
Fix a Dedektnd domain
a
Chapter 2 STRUCTURE THEOREMS FOR K1 OF ORDERS
This chapter presents some of the basic applications of the reduced norm and logarithm homomorphisms to describe 2p order in a semisimple
7-order or
is a
example,
K1(21)
Q-
and
when
K1(A),
or @p algebra
A.
is shown to be finitely generated whenever
K1(21)
21
2[
For
is a
7L-order; and is shown to be a product of a finite group with a finitely
generated
7L-module in the
is determined.
K1(21)
In both cases, the rank of
7L-order case.
Also,
is shown (for both
SK1(21)
7L-
2p orders) to be the kernel of the "reduced norm" homomorphism from to units in the center of
and K1(21)
A.
The results about reduced norms are dealt with in Section 2a.
These
include all of the results about 7L-orders mentioned above, as well as some
properties of
applied to
2p orders.
Then,
for example,
show,
p-adically closed in
in Section 2b, p-adic logarithms are
that for any
(i. e.,
GL(21)
that
7l-order
K1(21)
21,
E(21)
is
is Hausdorff in the
p-adic topology).
Applications of the reduced norm
2a.
For any field
E D F
F,
and any central simple F-algebra
nrA.K: A* - F*
norm homomorphtsm
be any extension which splits Then for any a E A*,
E®FA -4-> Mn(E).
This is independent of the choice of by an inner automorphism of 1.1(iv) above). of
A
for
Cal(E/F)
Mn(E)
Furthermore,
E/F
if
splitting field
E.
A,
set gyp:
the reduced
and
Let
fix an isomorphism
nrA/(a) = detE(pp(l®a.)) E E*.
any two such isomorphisms differ
by the Skolem-Noether theorem (Theorem
nrA/F,(a) E F*
is Galois);
A,
is defined as follows.
(it is fixed by the action
and is independent of the choice of
For more details, see Reiner [1, Section 9a].
CHAPTER 2.
STRUCTURE THEOREMS FOR K1 OF ORDERS
As one easy example, consider the quaternion algebra R.
A C-linear ring isomorphism
gyp: CO
41
IH
with center
is defined by setting
Ui =+ M2(C)
IR
0
0
W(1®i) = (10 i )'
p(101) = (0 ').
W(10k) = li
p(10j) = (-l O)'
/.
Then, for any f = a+bi+cj+dk E 1H,
(g) = det(-c+di c+bi I - a2+b2+c2+d2.
nr
nrA/F: A* -> F*
It is immediate from the definition that For any
homomorphism.
algebra, and
n > 1,
is again a central simple
Mn(A)
is the extension to
GLn(A)
an isomorphism between
K1(A).
nrMn_1(A)/F'
of
the reduced norm extends to a homomorphism defined on factors through its abelianization
is a
So
and hence
GL(A),
nr.,
For example,
F-
induces
and the multiplicative group of positive
K1(H)
real numbers.
The first lemma lists some of the immediate properties of reduced norms.
Lemma 2.1 F-algebra.
(t)
Set
Fix a field n = [A:K]1"2.
and let
F,
Then the following hold.
detF(A ' A) = nrA/(u)n
(ii)
(iii)
nrA/(u) = un
If
for any
A = Mn(F),
be a central simple
A
then
for any
u E A*.
u E F*.
nrA/F: A* --> F*
is the determinant
homomorphism.
(iv)
tralizer of
Proof
E C A
If
E
in
A,
is a subfield containing then
F,
and if
nr ,T(u) = NE/(nrB/E(u))
B
is the cen-
for any
u E B*.
The first three points are shown in Reiner [1, Section 9a].
Point (iv) is shown in Draxl [1, Corollary 22.5]; and also follows easily
42
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
from the relations among (reduced) characteristic polynomials in Reiner [1, Theorems 9.5, 9.6a, and 9.10(iii)].
Now, for any semisimple A i
is simple with center
Q-
0
%-algebra A = ([i=1Ai,
or
and any Z-
Fi,
nrA: K1(A) --> fi(F.)* = Z(A)*
or 2p order
21,
where each we let
nr21: K1(21) -* Z(A) *
and
denote the homomorphisms induced by the product of the reduced norm maps
for the
Note that
Ai.
homomorphisms defined on
as a map on
nrA
and
nr2[
while
K1(-),
are used here to denote the
nrA,,F
denotes the reduced norm
* e.
The following lemma will be useful when computing Ker(nrA).
Lemma 2.2 A
and let
Let
be any
E
F-algebra.
Ker[K1(i):
(where
i(x) = 1®x)
Proof
be any finite field extension of degree
F
Then
:
is central in
trf o K1(i)
A
If
v
of
IR 0vK A
K
n.
By Proposition 1.18, the composite
Kl(E ®F A) ) K1(A)
K1(A)
is induced by tensoring with F
) K1(E ®F A)]
K1(A)
has exponent dividing
trf o K1(i)
Since
n,
A,
is multiplication by
E ®F A,
regarded as an
E ®F A = An n.
as
is a matrix algebra over
v: K y IR) Ui.
(A,A)-bimodules, and so
The result is now immediate.
is any simple ER-algebra with center
(i. e., an embedding
(A,A)-bimodule.
K,
0
then a real valuation
is called ramified in
A
if
Theorem 2.3 F = Z(A),
A
Let
be a simple
9 C A be a maximal
let
the ring of integers.
7L-
lor
Im(nrA)
and
F*
If
R C F
be
0 F*
are described as follows:
Im(nr,,,)
and
nrA(K1(A)) = FTM
(ii)
and let
pp-order,
is a %-algebra, then
A
If
(t)
Qp-algebra with center
or
Then
nrA : K1(A)
is injectiue; and
43
STRUCTURE THEOREMS FOR K, OF ORDERS
CHAPTER 2.
A
nrl(K1(IR)) = R*.
is a ID-algebra, then set
= {u E F* : v(u) > 0 for all ramified
v: F '- Itl;
R + = F+ fl R*.
Then
and
nrA(K1(A)) = F+
Proof
Recall that the index of
is defined by
Step 1 Set
nritK1(Th)) = R+.
A = Mr(D)
ind(A) = ind(D) =
We first consider the formulas for
n = ind(A).
Note first that
(Lemma 2.1(1)), and so
write
F
and
A C D a maximal order. E C D
and an element
u*
and
Im(nrN).
for any u E GLk(l),
Thk) E R*
nrA/F(u) E R*.
is a finite extension of
A = Mr(D)
a division algebra)
Im(nrA)
nrN(K1(M)) C R*:
A/F(U)n nr = detF(Ak u-> Ak) = detR(IRk
If
(D
[D:F]1/2.
Th = Mr(A),
then by Theorem 1.9, we can
Qp,
where
D
is a division algebra and
Also by Theorem 1.9, there is a maximal subfield a E D
such that
E/F
is unramified, such that
44
CHAPTER 2.
7Ea 1 = E,
and such that
then
generates the maximal ideal in
nn
nrDJE = NE/F
Lemma 2.1(iv), integers,
STRUCTURE THEOREMS FOR K1 OF ORDERS
the usual norm.
S C E
If
R.
By
is the ring of
NE',F(S*) = R* by Proposition 1.8(iii); and so
nr7(K1(71)) 2 (N
= F*.
(-1)n-lan)
,(S*), nrD/F(a)) = (R*,
Q,
then the formula for nr,,(A*)
is the Hasse-Schilling-Maass norm theorem
(see, e. g., Reiner [1, Theorem
If
33.15]).
F
is a finite extension of
To see that
M E GL(A)
nr(M) = u.
such that
distinct primes at which Rn =
nplnRp,
the injectivity of
(rk) E E(A
n
ikJk
(rj)... e
iljl
Choose elements t.
)
such that
GL(iiin
ikjk (rk) E
T1 , ... , Tk E DI[n]
So assuming
(shown in Step 2 below), there exist
e l.iji (r1),...,e
M -e
the
An = [[p'nAp,
nrAn([M]) E (Rn)* = nrin(K1(i n)).
nrl. = npinnrA P
elementary matrices
be the product of
n
Let
and choose
u E R+,
is not invertible; and set
M
Then
etc.
fix some
nr,N(K1(N)) = R+,
).
such that ft = rt
(mod
in) for all
Note that it suffices to do this on the individual coordinates (in
of the rt with respect to some fixed 7L-basis of
If we now set
fit.
then nrA,,F(M)=nrA,,F(M) = u and MEGL(IR). Step 2
The injectivity of
was first shown by Nakayama &
nrA
Matsushima [1] in the p-adic case, and for lu-algebras by Wang [1]. following combined proof, using induction on
[A:F],
The
is modelled on that
in Draxl [1].
Step 2a Assume first that
E C A
cyclic Galois extension of degree centralizer of
B*
such that
E
in
A.
is a subfield such that
n >
We claim that
1,
and let
[u] = 1
in
B
K1(A)
E/F
is a
denote the for any
u E
nrA/F(u) = 1.
Note first that
B
is a simple algebra with center
E
(see Reiner
CHAPTER 2.
[1, Theorem 7.11]),
45
STRUCTURE THEOREMS FOR K1 OF ORDERS
and that
the induction hypothesis.
nrB
So
[B:E] < [A:F].
is injective by
Furthermore, by Lemma 2.1(iv),
nrA/F(u) = NE, (nrB/E(u)) = 1. G = Gal(E/F) = 7Vn,
Set
and consider the exact sequence in cohomology fl-1
H 2(G;E*/nrB(B*)) --)
Here, if R-1
J+ E G
(2)
(G;nr B(e)) - H 1(G;E*).
G-module
is a generator, we identify for any
M:
(G;M) = {x E M: NC(x) = x+4,(x)+...+P-l(x) = 0}/{y(x)-x : x E M}.
In particular,
Appendix A]).
case since
fl- 1(G;E*) = 1
by Hilbert's Theorem 90 (see Janusz [1, fl-2(G; e/nr We))
Also, by Step 1,
E* = nrB(B*);
is a product of copies of
and in the
in the p-adic
Q-algebra case since
for certain real embeddings
{±l}
these real embeddings are permuted freely by Thus,
= 1:
E*/nrB(B*)
E
and
G.
So by (2), there is an element
H 1(G;nrB(B*)) = 1.
v E BTM
such that
,y(nrB/E(v)) (nrB/E(v))-1 = nrB,(u) . Furthermore, by the Skolem-Noether theorem (Theorem 1.1(iv)), there is an element (B
a E A
such that
is the centralizer of
axa E),
1
= 4'(x)
for all
x E E.
ava
Then
1
E B
and
nrB/E([a,v]) = nrB/E(ava 1)/nrB/E(v) = P(nrB/E(v))/(nrB/E(v)) = nrB/E(u)Thus I
1,
induction hypothesis; and hence
so
1
in
K1(B)
by the
[u] = 1 E K1(A).
Step 2b The rest of the proof consists of manipulations, using Lemma 2.2, to reduce the general case to that handled in Step 2a.
Fix a prime
46
STRUCTURE THEOREMS FOR K, OF ORDERS
CHAPTER 2.
and any
P,
Let
P
F
be a splitting field such that
[P:F']
and
is a p-power.
where D
is a division algebra,
Now let
E 2 F'(v)
Set B = KOF' [E:F']
D.
E
over
Let
Gal(E/F').
v,
K C E
and let
be the fixed
pl[K:F'];
Thus,
and
Calois extension, and
then
L C L
Step 2a applies to show that
such that L.
and
[K:F']
is a Calois extension
and so
[B:K] = [L:K]2 = 1,
centralizes
v
and
K C L C E,
L = K,
If
Then nrB/(v)
10v E B.
is a field, since
L
Also,
Otherwise, there is a subfield
[A:F],
F'.
and identify v E D with
are relatively prime;
of p-power degree.
nrD/F(v) = 1.
and
is a p-power by Proposition 1.3.
v E L = K ®F, E.
and
V E D*,
(3)
be any maximal subfield containing
field of some p-Sylow subgroup of ind(D)= [E:F']I[P:F']
Then
Cal(F/F).
[10u] = [v] E K1(D),
F' = Z(D),
be any normal closure of
=1,
is Calois, and let
P/F
Write
and
F' ®F A - Mr(D)
E Q E
p.
be the fixed field of a p-Sylow subgroup of
p4'[F':F],
show that
We will
nrAfF(u) = 1.
has finite order prime to
[u] E K1(A)
F' C P
such that
u E A
v = 1.
is a degree
L/K
In this case, since
p
[B:K] _
[v] = 1 E K1(B) = Kl(K ®F, D).
[v][K:F'] = 1
Lemma 2.2 now applies to show that
in
K1(D).
Hence
[10u][K:F'] = 1 E Kl(F' OF A)
by (3), and a second application of Lemma
2.2 shows that
in
and so
[u] [K:F] = 1
has order prime to
[u]
K1(A).
p
in
But
K1(A).
p4[K:F]
by construction,
0
The following lemma, due to Swan [2], makes it possible to compare with
K1(21)
algebra.
when
K1(B),
4 C 23
is any pair of orders in the same
It will also be used in the next chapter when constructing
localization sequences.
Lemma 2.4
Let
S-ideal contained in
R C S R.
be any pair of rings, and let
Then
E(S,I2) 9 E(R,I) 9 E(R).
I
be any
CHAPTER 2.
is ftntte, then the induced map
S/I2
If, furthermore,
STRUCTURE THEOREMS FOR K1 OF ORDERS
47
K1(R) --+ K1(S)
has finite kernel and cokernel.
Proof
If
diagonal entry r,s E I
denotes the elementary matrix with single off-
eij in
r
and any distinct
(i,j)-position, then
CL(S)
containing all such
for any
E(S, 12)
Hence, since by definition
i,j,k.
the smallest normal subgroup in
eij = [eik,ekj]
is
eij,
E(S, 12) C [E(S,I), E(S,I)] C [GL(S,I), CL(S,I)]
_ [CL(R,I), GL(R,I)] C [GL(R),CL(R,I)] = E(R,I)
(1)
(see Theorem 1.13).
Now consider the following diagram, with exact rows and column: E(S,I2)/E(R,12)
K1(R,I2) --> K1(R) --' K1(R/12)
K2(R/I2)
(2)
K2(S/I2) -) Kl(S,I2) -) K1(S) - ) K1(S/I2).
By Theorem 1.16,
Ki(R/I2)
E(S,I2)/E(R,I2)
Also,
and
Ki(S/I2)
(i = 1,2)
are all finite.
is finite since by (1),
E(S,I2)/E(R,I2) C (E(R) fl GL(R,I2))/E(R,I2) = Ker[K1(R,I2) -4 K1(R)].
This shows that three of the maps in square (2) have finite kernel and cokernel, and so the same holds for
K1(R) -' K1(S).
a
We are now ready to apply reduced norm homomorphisms to describe the structure of 7-
or
K1(2f,I) - modulo finite groups, at least - when
2 p order and
I C 2f
is an ideal of finite index.
2f
is a
48
Theorem 2.5 Let any
STRUCTURE THEOREMS FOR K1 OF ORDERS
QiAPTER 2.
A be a semtstmple
or2p -order in
Z-
A, and let
or %-algebra,
Q-
let
21
be
be an ideal of finite index.
I C 21
Then
SK1(21) = Ker(nr2)
(t)
nr2,I
(it)
:
and is finite.
K1(21,I) - R*
has finite kernel and cokernel, where
e is the product of the rings of integers in the field components of the center
Z(A).
(iii)
A
If
abeltan group.
is a Q-algebra, then
is a finitely generated
K1(21,I)
If
q = number of simple summands of
A,
r = number of simple summands of
fl
and
®Q A,
then rk2(K1(21,I)) = r - q. Proof
(i)
injectivity of that
The equality nrA.
If
sequences of Theorem 1.17, 1p -order.
Th _D 21
is a maximal order, then Lemma 2.4 shows
is finite if and only if
SK1(21)
is immediate from the
SK1(21) = Ker(nr2)
SK1(71t)
SK1(II)
is torsion,
is.
By the localization
and is finite if
R
is a
(A proof of this which does not use Quillen's localization
sequence is given by Swan in [3, Chapter 8].)
When
21
is a
Z-order, then by a theorem of Bass [1, Proposition
11.2], every element of Siegel [1] has shown that
K1(21)
GL2(21)
is represented by a 2x2 matrix. is finitely generated.
finitely generated, and hence finite, in this case. finiteness of
(ii)
Let
SKI(M)
M D 21
SK1(21)
is
Alternatively, the
follows from Theorem 4.16(i) below.
be a maximal order.
So
Also,
Then the maps
-)
K1(21,I)
--' K1(N) nr) K1(2l)
R*
all have finite kernel and cokernel: K1(2VI)
49
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
the first since
K2(2VI)
are finite (Theorem 1.16), the second by Lemma 2.4, and
nr,,
and
by
Theorem 2.3 and (i) above.
(iii) F.
If
A
is a D-algebra, then write
are fields, and let
F = Z(A) = [IF1,
be the ring of integers.
Ri C Fi
where the
By the
R* _ fl(R.)*
Dirichlet unit theorem (see Janusz [1, Theorem 1.11.19]),
is
finitely generated and
rk7(R*) _
[(no. field summands of IR 0
rk7(R*) _ i=1
= (no. field summands of
By (ii), the same holds for
IR ®Q F) - q = r - q.
o
K1(21,I).
In the case of an integral group ring, the formula for
can be given a still nicer form, defined in Section la.
g,h E G
are
IR-conjugate if
g (g)
is conjugate to and
Theorem 2.6 Fix a ftntte group
rk(K1(7L[G]))
using the concept of "K-conjugacy"
Note that in any finite group
ED-conjugate if the subgroups
Then
Fi) - 1]
i=1
1
(h)
G,
h
G,
or
two elements h-1;
and are
are conjugate.
and set
r = no. of
IR-conjugacy classes to
G,
q = no. of
D-conjugacy classes to
G.
rk(Wh(G)) = rk(K1(7[G])) = r - q.
Proof
By the Witt-Berman theorem (Theorem 1.6), for any K C C,
50
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
(no. K-conjugacy classes in
G) = (no. irred. K[G]-modules)
= (no. simple summands in
K[G]).
The result is now immediate from Theorem 2.5(iii).
o
Logarithmic and exponential maps in p-adic orders
2b.
In the last section, reduced norms were used to compare
any 2P
-order
order.
K1(21),
for
with the group of units in the center of the maximal
21,
Now, p-adic logarithms will be used to get more information about
the structure of
Kl(2[).
Throughout this section,
will be a fixed prime, and the term
p
"p-adic order" will be used to mean any 8p algebra which is finitely
The results here are shown for
generated and free as a 2 p module. arbitrary p-adic orders,
to emphasize their
specialized properties of orders in semisimple &-algebras. order
R
K1(R)
is generated by units in
in
1+I
is semilocal, since
for any ideal
and
R;
2
R and any x E R,
So by Theorem 1.14(1),
is generated by units
- ...
define
2 3 Exp(x) = 1 + x + 2- + 3i + ...
3 +
2
K1(R,I)
Any p-adic
I C R.
For any p-adic order
Log(l+x) = x -
is finite.
R/J(R)
the more
independence of
and
3
whenever these series converge (in
case with the usual logarithm on
@ 01 R, IR,
at least).
Just as is the
p-adic logarithms can be used to
translate certain multiplicative problems involving units in a p-adic order to additive problems - which usually are much simpler to study.
The main results of this section are, for any p-adic order ideal
I C R,
that
Log
log,
:
induces a homomorphism
K1(R,I) -+ Q ®Z (I/[R,I]),
R
and any
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
that
is finite and
Ker(logl)
E(R,I)f1GLn(R,I)
Throughout this section
Log
between subgroups of e and group homomorphisms.
and b E 12.
for all
while
n.
are used to denote set maps
Exp
and
R,
and
log
For any pair of ideals
R generated by elements
the subgroup of
is a ap-lattice, and that
Im(logl)
is closed in GLn(R,I)
51
denote induced
exp
11,12 c R,
for all
Recall (Theorem 1.11) that for any radical ideal
every element in
a E I1
I c J(R),
The following lemma collects most
is invertible.
1+I
denotes
[11,12]
[a,b] = ab - ba
of the technical details which will be needed throughout the section.
Lemma 2.7
R
Let
be any p-adic order, let
Jacobson radical, and let
(t)
R0 = O ®Z R=R[ 1 ]
Set
and Log(v)
Log(u)
u,v E 1+ I,
and
fp E pfR.
Assume
I C fR
Then for all
converge in
(mod [R0,I0]).
Assume
(iii)
and
Exp
addition,
I c fR
Ip c pIJ.
and
Log
Then
for some Exp(x)
Proof of
Log(u)
f E Z(R)
such that 1 + I
and for any
I
Exp(x)
(2)
fp E pfR, and for all
and
x E I;
1 + I.
In
x,y E I:
(mod
E(R,I)).
The proof will be carried out in three steps. or
and
converges in
Exp(x + y) = Exp(x)'Exp(y)
such that
(mod [R,I]).
are inverse bijecttons between
Exp([R,I]) C_ E(R,I),
(1)
f E Z(R)
Log(u), Log(v) E I
Log(uv) = Log(u) + Log(v)
also that
and
IQ,
for some central element
u,v E 1+ I,
Then for all
IQ = @ ®g I = I[p].
Log(uv) = Log(u) + Log(v)
(it)
denote the
j = J(R)
be any radical ideal.
I C; J
(3)
The convergence
in all three cases will be shown in Step 1.
The
congruences (1) and (2) will then be shown in Step 2, and congruence (3) in Step 3.
Step 1
For any
n Z1,
J/pnR
is nilpotent in
R/pnR.
Hence, for
52
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
x E I C J,
any
lim(xn) = 0,
n-m
thus converges in
Log(l+x)
IQ.
IP C pI,
Under the hypotheses of (ii), n >1
p
(all rational primes except
any x E I,
xn/n E I
To see that
The series for
lim(xn/n) = 0.
and
n-m
for all
and hence
n,
converges when
Exp(x)
In C nI
and so
for all
1p C R).
are inverted in
So for
Log(l+x) E I.
Ip C pIJ,
note first that for
any n > 1, n!.P ([n/P]+[n/p2]+[n/p3]+...) E (2 )* P
where
For any n > p,
denotes greatest integer.
[.]
In C (IP)[n/P] C
Similarly, if
n > p2,
induction, for
any n > 1,
and by
In C
p([n/P]+[n/p2]+...+[n/pk]).I.Jk
In C
Thus,
then
= nl.IJk
_L.In C I
for all
n,
lim n-
in 1+I for any x E I. E x p o Log(l+x) = l+x, for x E I,
_L. In
= 0;
pk+l).
(if
pk < n
1
where
f E Z(R)
and
fP E pER,
then
and
m+n
\
U(I) = ([r, n s] : m,n> 1, fmrE lm, fnsE ln, fr,fsE I) C [R,I]. So congruences
(1) and (2) will both follow,
once we have shown the
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
53
relation
(mod
Log((l+x)(1+y)) = Log(l+x) + Log(l+y)
U(I))
(6)
for any I and any x,y E I. For ,each n> 1, length
n
let
w
C(w) = orbit of
be the set of formal (ordered) monomials of
W.
in two variables
a, b.
ab
k(w) = number of occurrences of
r(w) = coefficient of
k$w)
(-1)n-i-1
1
To see the formula for
w
in
Log(1+a+b+ab)
in
(k(w) )
r-i 1
i=O
``(k(w))
w
set
w E W,,,
under cyclic permatations
W.
in
For
i
/'
r(w), note that for each
ways as a product of
and
(ab)'s
i
w
i,
n-2i
a's
can be written in or
b's.
i
Fix an ideal ICJ and elements x,y E I. For any n > 1, any wEW,,, and any w' EC(w), w' is a cyclic permutation of w, and so (mod
w'(x,y) = w(x,y)
for some
i,j
such that
i+j =n.
[Ii,Ij]).
It follows that
00
Log(l+x+y+xy) =
I
I r(w)'w(x,y)
n=1 wEW, (7)
j
For fixed wEW,,, order
(mod
U(I)).
if
(i. e.,
IC(w)I = n/t
w
has cyclic symmetry of
t), and if k =
then
r(w')).w(x,y)
w'ECW
n=l
C(w)
contains
ma+(w')
:
w' E C(w)
k/t elements with
k-1
(ab)'s
(i. e., those of the
54
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
form b...a)
and
elements with k
(n-k)/t
(ab)'s.
So
k
k(")]
r(w') = w' CCw
kI
_
(_1)n-i-l,n11,r(n_k)(k) ((
1
k (-1
((
11
l1
+ (k-i)( )]
1=0
L
ni1
1=0
'l i /
if k=0 (so t=n).
n
Formula (7) now takes the form, for any
Log((1+x)(l+y)) a E
if k>O
0
(( k11
(_1}n-1,(nn +
x,y E I,
n (mod n /
n=l
U(I))
/
= Log(l+x) + Log(l+y);
and this finishes the proof of (6).
Step 3 Now assume that fp E pfR,
and that
I C fR
Ip C pIJ.
are inverse bijections between
I
and
1+ I.
Log(Exp(x)'Exp(y)) = x+ y
by (6).
f E R
for some central
Exp
In particular, by Step 1, So for any
(mod
and
Log
x,y E I,
U(I))
It follows that
Exp(x)'Exp(y) E Exp(x + y+ U(I))
for
such that
x,y E I;
(8)
and hence that
Exp(x)'Exp(y)'Exp(x+y)-1 E Exp(x+y+U(I))'Exp(-x-y) (9)
C Exp(U(I)) C Exp([R,I]). So it remains only to show that
rER and xE I,
Exp([R,I]) C E(R,I).
Note that for all
Exp(rx)-Exp(xr)-1
55
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
= l1+ r( I l n=1
n-1
x
00
x
))(1+ (I
rn!
() n-1 )r)-1 rn!
E E(R,I)
n=1
(10)
by Vaserstein's identity (Theorem 1.15).
Fix some 2p basis
and
v.i E I.
[rl,vl],...,[rm,v
for
[R,I],
where
ri ER
Define
y :
[R,I] -i Exp([R,I])
by setting, for any x = Gi-lai[ri,vi] E [R,I]
(ai E 71p):
m
(Exp(airivi) Exp(aiviri)
_
I. .
i=1
Im(y) C E(R,I)
Then
Exp(x)-Exp(y) ° Exp(x+y)
by (9).
Also, for any
(mod
k,e > 1
e > k > 1,
`1+(x) = Exp(x)
U(pkl) C p2kU(I) C p2k[R,I])
and any x E pk, y E peI,
Exp(x)-Exp(y) E Exp(y)'Exp(x)
So for any
(mod
[pkl,peI]
and any x E pk[R,I]
(mod
and any x,y E pkI,
For any k > 1
by (10).
and
pk+e[R,I]).
y E pk[R,I],
p2k[R,I])
pk+e[R,I]). ,P(x + y) = P(x)'J+(y) a
For arbitrary [R,I]
(mod
define a sequence
u E Exp(p[R,I]),
x0,xl,x2,...
by setting
x0 = Log(u) E p[R,I];
Log(,P(xi)-1.u).
xi+l = xi +
By (11), applied inductively for all
J+(xi) _ u,
xi+l = xi
i > 0,
(mod
p2+i[R,I]).
in
56
So
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
converges, and
{x.} 1
This shows that
u = P(lim x.). 1
i-
Exp(p[R,I]) C Im(y) C E(R,I).
Now define subgroups
Dk = (rx - xr : xE I, Recall the hypotheses on and
Ip C pIJ
U(IJk) =
k > 0,
by setting
rER, rx,xr E IJk) C [R,I] fl IJk. I C ER,
I:
In C nIJ
(so
for all
Dk,
(12)
for all
where n).
f E Z(R)
Then for all
Ep E pER,
and k > 0,
(by (5))
V
m,n>1
n C ([r, - s] : n >2, Er,Es E IJk,
Together with (8),
shows
this
(normal) subgroups of
k > 0
ri E R,
xi E I;
and any
(mod
x E Dk,
i
l(Exp(rixi)-Exp(xiri)-1)
(mod
for
k
x = Z(rixi - xiri)
(where
then
(mod
Exp(Dk+1))
(by (13))
(by (10))
E(R,I)).
Exp(Dk) C
In other words, Dk C p[R,I]
(13)
i
Exp(x) a
1
x,y E IJk,
Exp(U(IJk)) C Exp(Dk+1))
if we write
E IJk),
rixi, x r
are both
Exp(Dk) C Exp([R,I])
Also, by (9), for any
R*.
Exp(x)-Exp(y) a Exp(x+y)
For any
that
fnrs,,nsr E (IJk)n g nIJk+1\ C Dk.1
large enough
Exp([R,I]) = Exp(Do) C
for all (Dk C [R,I]fllJk);
k > O.
But
and so using (12):
C E(R,I).
0
Constructing a homomorphism induced by logarithms is now straightforward.
Theorem 2.8
For any p-adic order
and any 2-sided ideal x E If1J)
57
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
R
with Jacobson radical
the p-adic logarithm
I C R,
J C R,
Log(l+x)
(for
induces a unique homomorphism
logl
If, furthermore,
K1(R,I) -' ® (IILR,I])
I C R
for some central
f E Z(R)
such thatp E pER,
then the logarithm induces a homomorphism
K1(R,I) ---.+
log'
and
log'
Proof
Ip C pIJ.
is an isomorphism if
and I _for short, and let
Write R _
the Jacobson radical of
R.
Assume first that
I C J.
J be
By Lemma 2.7(i),
the composite
L:
1+I
Log)
I
r°
(1)
is a homomorphism.
For each n > 1,
let
Trn
IQ/[RI@]
be the homomorphism induced by the trace map. ideal
Ln
M(I) C M(R),
Then (1), applied to the
induces a homomorphism
1+ Mn(I) = GLn(R,I)
Log.
Trn.
For any n, and any u E 1 + M( I) and r E GLn(R) , Ln([r,u]) = Ln(rur 1) - L(u) =
and so L = U(L)
(2)
1)
factors through a homomorphism
- Trn(Log(u)) = 0;
58
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
log, : K1(R,I) = GL(R,I)/[GL(R),GL(R,I)] ) IQ/[RQ,IQ].
Now assume that
Consider the
Io = If1J.
is arbitrary, and set
I
relative exact sequence
K2(R/Io,I/Io) -' K1(R,Io) -' KI(R,I) --> K1(R/IO,I/Io) (see Milnor [2, Remark 6.6]).
isomorphically to summand of
The surjection
(R/J
is semisimple).
(Io C J)
I C gR
Lemma 2.7(ii),
and
f E R
logI
such that
fp E pER,
» I/[R,I]
The same argument as before then shows that
is bijective and
logI
defined on
Log 1([R,I]) C E(R,I)
K1(R,I).
If
L
Ip C pIJ,
by Lemma 2.7(iii); and
is an isomorphism.
The next result is based on a theorem of Carl Riehm [1].
says that for any p-adic order K1(R)
then by
and the composite
Log(1+I) C I,
factors through a homomorphism
so
is a
pl'IK1(R/Io,I/Io)I = IK1(I/I0)I.
L: 1+I -L'24 I
then Log
I/Io
: K1(R,I) -* IQ/[%,IQ].
for some central
is a homomorphism.
In particular,
extends uniquely to a homomorphism
log,
If
I/Io
R/I0, and by Theorem 1.16,
K2(R/Io,I/Io) = K2(I/Io) = 1
logo
sends
(I+J)/J, which is a 2-sided ideal and hence a ring
R/J
semisimple ring summand of
So
R/Io --% R/J
the p-adic topology on
R,
Roughly, it
R*
makes
into a Hausdorff group.
Theorem 2.9 Ker(log,)
For any p-adic order
is finite; and for all
n
R
and any 2-stded ideal
the group
I C R,
59
STRUCCURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
En(R,I) = GLn(R,I) fl E(R,I) = Ker[GLn(R,I) -+ K1(R,I)]
is closed (in GLn(R,I))
Proof
Set % = Q Oz R and
and Theorem
2.7(iii)
I,, = @ ®g I
as before, and write
: GL1(R,I) --» K1(R,I) -1 IQ/[RR,IQ].
L = log, o proj
By Lemma
in the p-adtc topology.
l+p21 -' p21
Log:
2.8,
is
a
homeomorphism, and factors through an isomorphism
logp2l : K1(R,p21) = (l+p21)rE1(R.P21) -_' P21/[R,P2I].
In particular,
E1(R,p2I) C E1(R,I) Now,
groups
[R,p2I] = Log(E1(R,p2I))
since
is compact:
GL1(R/pnR,(I+pnR)/pnR).
subgroup of
Ker(L)
[RO,I,4],
Ker(L).
are open subgroups of
GL1(R,I)
is open in
it is the inverse limit of the finite So
is compact, and any open
Ker(L)
has finite index.
It follows that
Ker(log,) = Ker(L)/ 1(R,I)
is finite.
Any open subgroup
of
a
topological
complement is a union of open cosets). in in
Ker(L) and hence also in GLn(R,I)
E1(Mn(R),Mn(I)).
for all
n,
GL1(R,I).
group
is
also closed
(its
In particular,
E1(R,I)
is closed
To see that
En(R,I)
is closed
just note that by definition,
En(R,I) _
0
The following description of the structure of consequence of Theorem 2.9.
K1(R,I)
is an easy
60
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
Theorem 2.10 For any p-adic order
R with Jacobson radical
J C R,
K1(R) - K1(R/J) ® K1(R,J),
where
K1(R/J)
is finite of order prime to
finitely generated p-module.
K1(R,I)
(i)
is
If
p,
and
is a
K1(R,J)
is any (2-sided) ideal, then
I C R
the product of a finite group with a finitely
generated 7LP module and
rk2 I/[R,I]; D
(ii) p)
if
K1(R,I)
is a -module (i. e., contains no torsion prime to
and
I C J;
(iii)
K1(R,I) =
Proof
Note first that
since
1+p21
im K1(R/pnR,(I+pnR)/p'1R) n
lattice.
by Lemma 2.7(iii).
Hence,
the image of
1+I,
K1(R,I)
logl
is a i
Log(1+p2I) = p2I
has finite index in
Since
Ker(logl)
is finite by Theorem 2.9,
K1(R,I)
is
now seen to be a product of a finite group with a 7Lp module, and
rk$ K1(R,I) = rk I/[R,I]. P
P
To prove (iii), note first that
GL1(R,I) =
im GL1(R/pnR,(I+pnR)/p'1R).
n
Since
E(R,I) fl GL1(R,I)
is closed in
GL1(R,I)
(Theorem 2.9), it is
also an inverse limit of groups of elementary matrices over.
R/p11R.
The
STRUCTURE THEOREMS FOR K1 OF ORDERS
CHAPTER 2.
description of
61
as an inverse limit then follows since
K1(R,I)
im
preserves exact sequences of finite groups. If
I C J,
then
= 1- (I+PnR)/PnR
GL1(R/p"R, (I+PnR)/prR)
is a p-group for all ip module, by (iii).
n.
K1(R,I)
So
Since
K2(R/J) = 1
(Theorem 1.16), the sequence
K1(R) -* K1(R/J) - 1
1 - K1(R,J)
is exact; and is split since finite of order prime to
is a pro-p-group, and hence a
is a
K1(R,J)
i module and
(Theorem 1.16 again).
p
K1(R/J)
is
a
Theorem 2.10 will be the most important application of these results
P-adic logarithms will again be used
needed in the next three chapters.
directly in Chapters 6 and 7, but in the form of "integral" logarithms for p-adic group rings, whose image is much more easily identified.
We end the chapter with the following theorem of Kuku [1], which applies results from both Sections 2a and 2b. D1
is a maximal i
and only if
A
(qn - 1)/(q - 1),
Proof
A
order, then
where
if
Then
order.
and where
n = ind(A),
is
SK1(Th)
and let
F,
cyclic
of
order
is the order of the
q
F.
it suffices to show this when
otherwise, if
]R = Mr(A)
A = Mr(D),
by Theorem 1.9.
be the ring of integers, and let By Hasse's description of
[DVJ : R/p] = n.
SKI(M) = 1
be a simple s-algebra with center
By Theorem 1.9,
division algebra:
ideals.
Let
be any maximal
residue field of
R C F
order in any semisimple Q-algebra, then
is a product of matrix algebras over fields.
Theorem 2.11 N C A
Note in particular that if
Also,
pj1SKI(IR) I
W
and
A C D
In particular,
p C R and
A
is a
is the maximal [A:F] = n2.
Let
J C DI
be the maximal
IVJ
is a field and
(Theorem 1.9),
by Theorem 1.17(1).
It follows that
62
CHAPTER 2.
STRUCTURE THEOREMS FOR K1 OF ORDERS
SK1(I) = Ker[nr,: K1(IR) --» K1(R)][P]
(Theorem 2.5)
Ker[(WJ)* - (R/p)*];
where the reduced norm is onto by Theorem 2.3(i). cyclic, this shows that
SK1(A1)
(Theorem 2.10)
Since
is cyclic of order
I('d/J)*III(R/p)*I = (qn-1)/(q-1).
13
(IVJ)*
is
Chanter 3 QONTINUOUS K2 AND LOCALIZATION SEQUENCES
So far, all we have shown about
is that it is finite.
SK1(7L[G])
In
order to learn more about its structure, except in the' simplest cases, exact sequences which connect the functors
The Mayer-Vietoris sequences of Milnor
K1
and
K2
are necessary.
Theorems 3.3 and 6.4] are
[2,
sufficient for doing this in some cases (see, e. g., the computation of But to get more systematic results, some
by Keating [2]).
SK1(7L[Q(8)])
kind of localization exact sequence is needed which compares the K-theory of
Z[G]
with that of
Q[G]
or a maximal order, and their p-adic
completions.
The results here on localization sequences are contained in Section 3c.
The principal sequence to be used (Theorem 3.9) takes the form
K2(iip) -> C(A)
for any Z-order
21
C(A) =
SK1(21) - ® SK1(up) -i 1
in a semisimple Q-algebra
4
A.
(1)
Here,
SK1(21,I) = Coker[K2(A) - ® K2(Ap)];
where the limit is taken over all ideals
I C 21
of finite index, and
where the last isomorphism is constructed in Theorem 3.12.
A specialized
version of (1) in the p-group case is derived in Theorem 3.15.
As can be seen above,
the continuous
K2
algebras plays an important role in these sequences.
of p-adic orders and These groups
K2(-)
are defined in Section 3b, and some of their basic properties are derived there.
This, in turn, requires some results about Steinberg symbols and
symbol generators for
K2(R):
results which are surveyed in Section 3a.
64
3a.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
Steinberg symbols in K2(R)
For any ring
the Steinberg group
R,
free group on generators
xij
for all
St(R)
i A j
is defined to be the
(i,j Z 1)
and all
r E R;
q(xij)
be the
modulo the relations
xij xij = x..
r
1
4: St(R) -b E(R)
An epimorphism
i # j
if i # 2, j = k i f iA B, jA k.
J Xi2
s
[xij,xk8] _
r,s E R and any
for any
is defined by letting
elementary matrix whose single nonzero off-diagonal entry is Then
(i,j)-position.
(in particular,
E(R)
in the
r
is the "universal central extension" of
St(R)
Ker(o) C Z(St(R))),
and
K2(R) = Ker(O) = H2(E(R)).
For details, see, e. g., Milnor [2, Chapter 5].
For any pair
u,v E Rw
of units, the Steinberg symbol
{u,v}
is
defined to be the commutator
{u,v} =
Since
Ker(o)
liftings.
where
[0-1(diag(u,u 1,1))
is central in
, m
l(diag(v,l,v 1))] E St(R).
this is independent of the choice of
St(R),
We are mostly interested in the case where
{u,v} E Ker(#) = K2(R).
to work with the
{u,v}
uv = uv,
and hence
However, it will occasionally be necessary
for noncommuting
u
and
v;
for example, in
Lemma 4.10 and Proposition 13.3 below.
The next theorem lists some of the basic relations between Steinberg symbols.
Theorem 3.1 or
St(R):
For any ring
R,
the following relations hold in
K2(R)
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
For any
(i)
u E R*,
For any u,v,w E R*
(ii)
For any
and
x,y,r,s E R
(iv)
X,Y E St(R)
If
O(Y) = diag(vl,...,vn),
and
0 = xy = xry = yx = ysx,
{l+sx,l+y} = {l+x, 1+ys}.
are such that
¢(X) = diag(ul,...,un)
ui,vi E R*,
where
1-u E R*.
if
{v,u} = {u,v}-l.
such that
{1+xr,l+y} = {1+x,l+ry}
{u,l-u} = 1
uv = vu and uw = wu,
such that
{u,vw} =
(iii)
and
{u,-u} = 1;
65
and
i
for each
= viui
u v
and
i
then [X,Y] = [[i_1{uivi}.
i > 2,
S C R
For any
(v)
projective as an
such that
R
is finitely generated and
S-module, and any commuting units
u E R* and v E
S*,
trfs({u,v}) = {trfs(u),v}.
trfs: Kn(R) --- Kn(S)
Here,
Proof
denotes the transfer homomorphism.
Point (i), and the relation
{v,u} = {u,v}-1,
are shown in
Milnor [2, Lemmas 9.8 and 8.2] and Silvester [1, Propositions 80 and 79] (it clearly suffices to prove these for commutative
R).
The relations in
(iii) are shown by Dennis & Stein [1, Lemma 1.4(b)] in the commutative and
case,
follow
in
using
Alternatively,
{l+x,l+y} = <x,y>
relation:
Propositions 96 and 97]).
symbols,
whenever
by
case
noncommutative
the
Dennis-Stein
same
proof.
from
the
(see Silvester [1,
xy = O = yx
The formula in (v)
the
follows
(iii)
is shown in Milnor [2,
Theorem 14.1].
When proving (ii) and (iv), it will be convenient to adopt Milnor's notation:
A * B = [0 1(A),
uniquely defined since any
u E
entries
R*
and any
u, u
1
l(B)] E St(R)
K2(R) = Ker(o) i x j,
in positions
following two points:
dij(u) i
for any
A,B E E(R).
is central in
St(R).
This is Also, for
will denote the diagonal matrix with
and
j
(and 1's elsewhere).
Note the
66
(1)
CHAPTER 3.
CONTINUOUS KZ AND LOCALIZATION SEQUENCES
A*B = 1
if A E EI(R),
disjoint subsets of
{1,2,3,...}.
[xi.,xkQ] = 1
relation:
(2)
MAM 1*MBM 1
[A,B] * M = 1;
whenever
B E EJ(R),
I
where
and
J are
This follows easily from the defining i # 2
= A*B
and
j # k.
for any
and in particular whenever
such that
A,B,M E E(R)
This is immediate
[A,B] = 1.
from the obvious relations among commutators.
Now, fix
X,Y
as in (iv), and set
B = (Y) = diag(vl,...,vn)
(where
A = ¢(X) = diag(ul,...,un)
[ui,vi] = 1
for all
i > 2).
[X,Y] = A *B = diag(A,A 1,1) * diag(B,1,B 1)
and
Then
(by (1))
(dl,n+l(u1) * d1,2n+1(v1))**.(dn2n-1(un)
dn,3n-l(vn)) (by (1))
=
(d12(u1) * d13(v1))...(d12(un}
d13(vn))
(by (2))
= {ul'v1}...{un,vn}.
To prove (ii), fix units Then, using the relation
{u,vw} = d12(u) *d 13(vw)
u,v,w E R*
such that
= d12(u)
we get
* (d
lv 1))
= (d12(u) * d13(v)).(d12(u) * diag(w,l,vw lv
=
(by (iv))
We now consider relative K2(R,I)
[u,v] = 1 = [u,w].
[a,bc] =
d13(v))
0
K2 groups.
Keune [1] has defined groups
which fit into a long exact sequence involving
(note that this is not the case with the [2, Section 6]).
In this book, however,
most convenient to take as definition
K2(R,I)
K3
K2
and
K3
defined by Milnor in
never appears; and it is
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
67
K2(R,I) = Ker[K2(R) - K2(R/I)],
R
for any ring
relations which hold in
will automatically hold in
K2(R)
The following lemma is frequently useful. the next section when defining continuous
Lemma 3.2
For any pair
In particular, symbol
I C R.
and any (2-sided) ideal
S C R
K2(R,I)
here.
It will also be needed in
K2.
of rings, and any R-ideal
I C S,
K2(R, 14) C Im[K2(S,I) --> K2(R)].
Proof
Let
D
P2
I C R
be any ideal, and consider the pullback square
R
Ji
(D = ((rl,r2) E RxR ; rl-r2 E I)). I
R -) R/II Ker(p2)
We identify
with
I.
By the Mayer-Vietoris sequence for the
above square (see Milnor [2, Theorem 6.4]), K2(D,I) Since
onto P2
K2(R,I).
Also,
pl
E(D,I) = E(R,I)
is split by the diagonal map
induces a surjection of by Milnor [2, Lemma 6.3].
A: R -i D, there is a split
extension
1 - E(R,I) > E(D) ) E(R) -4 1. The Hochschild-Serre spectral sequence for
(1)
(1)
(see Brown [1,
Theorem
VII.6.3]) then induces a surjection
H1(E(R);E(R,I) ab
) _»
Ker[H2(E(D)) -+ H2(E(R))] _ Im[H2(E(R,I)) -> H2(E(D))] (2)
Coker[H2(E(R,I)) -4 K2(D,I)]
-* Coker[H2(E(R,I))
This will now be applied to the ideals
E(R,I4) 9 [E(R,I2),E(R,I2)],
since
I4 C I2 C R.
E(R, 14)
is
-->
K2(R,I)].
Note first that
the smallest normal
68
subgroup in for
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
containing all elementary matrices
GL(R)
x,y E 12
and
trivially to
distinct indices E(R,I2)ab,
i,j,k.
Thus,
e.'' = [eik,ekj],
maps
E(R,I4)ab
and so by (2),
K2(R,I4) C Im[H2(E(R,I2)) - K2(R,I2)].
Furthermore,
by Lemma 2.4, and hence
E(R, 12) C E(S,I)
K2(R, 14) C Im[H2(E(R,I2)) -i K2(R)] C Im[H2(E(S,I)) --> K2(R)]
C Im[K2(S,I) -i K2(R)].
The next theorem lists some generating sets for the relative groups K2(R,I).
For
the purposes
in
simplest elements to use as generators.
Dennis-Stein symbols a,b E R
with
E K2(R)
l+ab E RM)
Steinberg symbols are
this book,
the
However, in many situations, the
(defined for any commuting pair
are the most useful.
We refer to Stein &
Dennis [1], and to Silvester [1, pp. 214-217], for their definition and relations.
Theorem 3.3
Fix a noetherian ring
Jacobson radical, and let that
[J,I] = 0.
I C R
R,
let
j = J(R)
be its
be a radical ideal of finite index such
Then
K2(R,I) = ({l+x,l+y} : xEJ, yE I) = (<x,y> : xEJ, yE I). Moreover, if (i)
(ii)
R
J = (a1,...,ak)R,
or
J = (p,al,...,ak)R
I C (a1,...,ak)R,
then
is finite, and if either
for some prime
p,
where
p
is odd or
CHAPTER 3.
K2(R,I) = ({1+ai,l+x}
Proof
Set
Im[K2(S,I)
Also, since
any symbol
[J, I] = 0,
to
x E I).
Then by Lemma 3.2,
-
K2(R/I',I/I')].
{l+x,l+y},
Since
K2(R,I).
x E j/14
for
R/I4
or
Case 1
R
is finite.
Assume
R = 7L+ I.
R
Then
generated by symbols
for
x E I.
and
rER
=
By Stein &
is commutative.
Dennis [1, Theorem 2.1] or Silvester [1, Corollary 104],
x,
and
finite by
is
this shows that we need prove the theorem only when either
assumption,
R = 7L+ I,
R.
K2(R)] Q K2(R,I') = Ker[K2(R,I)
can be lifted
y E I/I',
1
Point (i) is now an easy consequence of relation The refinement in (ii) is shown in Oliver [7, Lemma
1.1], using an argument involving Dennis-Stein symbols similar to that used in Case 1 above.
0
Even when Theorem 3.3 does not apply directly to often filter
I
by a sequence
I = IO D I1 D ...
Theorem 3.3 applies to each of the groups
K2(R,I),
one can
of ideals such that
K2(R/Ik'Ik-1/Ik)'
and obtain
70
CHAPTER 3.
generators for
CONTINUOUS KZ AND LOCALIZATION SEQUENCES
K2(R,I)
This technique is the basis of the
from that.
proofs of Theorem 4.11, Proposition 9.4, and Lemma 13.1 below.
The last part of Theorem 3.3 is especially useful in the case of group rings of p-groups.
Corollary 3.4
Fix a prime
and a p-group
p
G,
Q.
ring of integers in some finite unramtFied extension of any pair
Io C I C R[G]
R
and let
be the
Then, for
of ideals of finite index such that
gx - xg E Io
for all gEG and x E I, K2(R[G]/Io,I/Io) =({g,l+x}
if
p
is odd, or if p=2 and I C (4,g-1: gEG)RG
K2(R[G]/I0,I/Io) =
Proof
by
p,
g E C, x E I/Io)
:
otherwise.
{g,l+x} : gEG, x E I/Io)
By Example 1.12, the Jacobson radical
together with elements
immediate from Theorem 3.3.
g-1
for
J C R[G]
g E G.
is generated
So the result is
a
Other theorems giving sets of generators for
K2(R)
K2(R,I)
or
are
shown in, for example, Stein & Dennis [1] and Silvester [1].
There are
also some much deeper theorems, which give presentations for
K2(R)
terms of Steinberg symbols or Dennis-Stein symbols.
K2
was Matsumoto's presentation for Other examples of presentations of
in
The first such result
of a field (see Theorem 4.1 below).
K2(R)
or
K2(R,I)
have been given by
Maazen & Stienstra [1] and Keune [1] for radical ideals in commutative
rings, by Rehmann [1]
for division algebras, and by Kolster
[1]
for
noncommutative local rings.
3b.
21
Continuous
K2
of p-adic orders and algebras
As mentioned above, the goal is to describe
SK1(2l),
in terms of
K1(A),
in a semisimple Q-algebra
A,
for a Z-order Ki(2[p),
and
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
(for
K1(Ap)
However, the groups
i = 1,2).
for example,
F2&
(see Bass & Tate [1, Proposition 5.10]).
are
has uncountable rank for any finite extension
K2(F)
in Theorem 4.4 that
K2(A)
and
K2(2 p)
huge:
71
K2(F) - the continuous
In contrast, we will see
K2 - is finite for such
F.
Several different definitions have been used for a "continuous" for a topological ring
K2(R)
functor
algebra over
Zp
especially when
E(R)
have been used by Moore [1] and Rehmann [1,
in all dimensions as a
Kc(R)
limit of homotopy groups of certain simplicial complexes. here,
definition
following
the
is an
Definitions involving continuous universal
Section 5]; and Wagoner [1] has defined
purposes
R
Qp.
or
central extensions of
R,
is
But for the
simplest
the
and
most
convenient.
For any prime 2[
in
A,
p.
any semisimple
s-algebra
A.
and any p-order
set Coker[K2(2l, p-2l)
K2(21) =
K2(21)]
k and
K2(A)].
K2 (A) = 4Lm Coker[K2(21, p'-4)
k
By Lemma 3.2, for any pair Im[K2(B,p4kB)
So
K2(A)
2[ C B
I
of orders in A and any k> 0,
K2(A)] C Im[K2(B,pk2!)
--4 K2(A)].
is well defined, independently of the choice of order
3 C A.
Quillen's localization sequence for maximal 2p orders (Theorem 1.17)
can easily be reformulated in terms of K.
Theorem 3.5 semisimple
Fix a prime
%-algebra
p,
A, and let
there is an exact sequence
let
j C M
Th
be a maximal 2p -order in a be the Jacobson radical.
Then
72
CONTINUOUS KZ AND LOCALIZATION SEQUENCES
CHAPTER 3.
1 -' K ( ) ' K2(A) - K1(N/J) --> K1(Th) - Kl (A) --> ... . Proof
This is almost an immediate consequence of the localization
sequence in Theorem 1.17(1). K2(m/J) = 1
K2(A);
and
and so
p4IK1(m/J)I K2(V)
Since
IR/J
is semisimple of p-power order,
by Theorem 1.16.
injects into
K2(A)
Hence
K2(1t)
by definition of
injects into K2.
11
The formula in the next proposition could just as easily have been taken as the definition of
Recall that a pro-p-group is a group
K2(21).
which is the inverse limit of some system of finite p-groups.
Proposition 3.6 Fix a prime semisimple
Qp-algebra
A.
p,
and let
21
be a
2p order in some
Then
K2(21) =
im K2(21/p
)
k
In particular,
K2(21)
is a pro-p-group, and
Kc(A)
is the product of a
finite group and a pro-p-group.
Proof
By definition,
K2(21) =
im Coker[K2(21,pk21) -+ K2(21)]. k
The sequence
1 -4 Coker[K2(21,pkkti) ' K2(21)] - K2(Vpk21) - K1(2C,p is exact for all limits.
and hence is still exact after taking inverse
k,
But by Theorem 2.10(iii),
K1(21,p
)=
km
m
Kl(2/pn21,p
pN)
Kl(2/p' 1,pn21/pn21) = 1. n
)
In particular,
is a pro-p-group by Theorem 1.16(11).
K2(2I)
is any maximal order, then
IRCA
3.5, and so
K2(A)
K2(A)
If
is finite by Theorem
[K2(A) : K2(7)1)]
is pro-finite since
In fact, in Chapter 4,
3c.
73
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
is.
K2(A1)
will be shown to always be finite.
Localization sequences for torsion in Whitehead groups
We now want to describe semisimple Q-algebra
A,
when
SK1(21),
in terms of
K1
is a 77-order in a
21
K2
and
of
A,
21p
and
Ap
The usual way of doing this is via Mayer-Vietoris exact sequences based on
"arithmetic squares", and one example of such sequences is given at the end of the section (Theorem 3.16). convenient
to
make a different
sequences for ideals
I C 21
But for the purposes here, it has been approach,
using
of finite index.
the
relative
exact
This will be based on the
following definitions.
Definition 3.7
2ICA,
For any semtstmpte Q-algebra
A,
and any 7L-order
define ®SKl(2Ip)].
C11(21) = Ker[SKl(21)
p
More generally, for any (2-sided) ideal SK1(2I,I) = Ker[K1(21,I) -) K1(A)]
I C 21,
set
and
Cll(21,I) = Ker[SKl(21,I)®SKl(21p,Ip)]. p Then define
C(A) =
SK1(21,I)
I where the itmit is taken over all ideals
I C 21
of finite index.
74
CHAPTER 3.
The subgroup is
hit
can be thought of as the part of
C11(21)
from behind
sequences.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
detected by
e.,
(i.
Recall (Theorem 1.14(ii)) that
is commutative, so that
C11(21) = SK1(21)
As is suggested by the notation,
of the choice of order
21
in
K2)
in
localization
the
p
for all
SK1(21p) = 1
which
SK1(21)
if
21
in this case.
C(A) =
im SK1(21,I)
is independent
This is an easy consequence of Lemma
A.
2.4; and will be shown explicitly in Theorem 3.9.
The
C(A)
can be
characterized in several ways:
Cll(21,I)
C(A) = kim SK1(21,1) =
Cll(2I)
(Theorem 3.9)
(taken over all Z-orders
Coker[K2(A) - ® K2(Ap)].
21
in
A)
(Theorem 3.12)
p
It is the last description of
C(A),
in terms of
K2(-),
which will be
used to calculate these groups in Section 4c below.
The appearance of
C(A)
in the localization sequence for
helps to explain the close connection between computations of
the congruence subgroup problem. have implied that
SK1(R,I) = 1
number field K and of the groups
I C R
SK1(21)
SK1(21)
and
In fact, the original conjecture would whenever
R
is the ring of integers in a
is an ideal of finite index.
C(K) = j SK1(R,I)
The computation
follows as a special case of results
of Bass, Milnor, and Serre [1, Theorem 4.1 and Corollary 4.3] in their solution to the problem.
One difficulty which always occurs in localization sequences based on comparing
Ki(21)
with Ki(A)
(when
21 C A
is a 7L-order) is dealing with
the infinite products which arise in the p-adic completions of
The next lemma says that in the case of 0 SKI(21p)
and
21
and
0 K2(ip),
least, there is no problem - both of these are finite products.
A.
at
Lemma 3.8
Proof
be a maximal order.
It J 21
i
is a maximal order in
can assume that
Also, since
21 = V.
orders in the simple summands of Let and set
for such
for almost all
SK1(iip) = SK1(Rp) = 1
RP = [pIPRR,
and p C R (i)
A = K
containing p = char(R/p)
and
R = R.
P
P
K
p.
A
is simple.
p
for almost all
(Theorem
In particular,
p by Theorem 1.14(11). it suffices to
K2(Dip) = K2(Rp);
Recall that for each
p,
RP
= npiPRp
where the products are taken over all maximal ideals p.
We claim that
is odd, and
Then,
RP
is its maximal ideal by (ii). (1 + p2K
Furthermore,
So we
be the ring of integers,
K2(Rp) = 1 RP
(ii)
for any
is unramified over
is the ring of integers in Thus,
(R
pRp)*
(1 + p2x)1/P
Q-P.
Rp:
By
p.
K
and
has order prime to
consists of p-th powers in
so the binomial series for
such that
p
Theorem 1.7(ii), this is the case for all but finitely many
Fix such a
for all
= Ft
2
by Theorem 1.9.
p
It remains to consider the case of do this when
Then
factors as a product of maximal
Dt
be the center, let R 1p n = [A:K] 1/2. Then = Mn(Kp) Aip - Mn(Rp)
p.
we can assume that
A,
21 C A,
by Theorem 1.7(iv).
Ap
K = Z(A)
1.7(iii)); and
and any Z-order
for almost all primes
SK1(21p) = 1
Let and
pj'[1:21],
A
For any semtstmpie Q-algebra
and
K2(up) = 1
75
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
p
converges for any
pRp p.
is odd, and x E Rp.
In
particular, by Proposition 1.8(i),
(Kp)* _ (up, 1-pup, p : u E (Rp)*).
This,
together with identities of the form
(Theorem 3.1(1)),
shows that
{uP,v}
{p,p} = {-l,p} = {(-1)P,p}
K2(Kp)
{a,l-a} =
1
= {a,-a}
is generated by symbols of the form
(for
u E (Rp)*,
v E (Kp)*)
76
CHAPTER 3.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
{p,l-pup} = {up,l-pup}-1
(for
{l-pup,l-pvp} = {pvp(1-pup),l-pvp}
u E (k )*) p
u,v E (Rp)*)
(for
= {pvp(1-pup),(l-pvp)(1-pvp+p2upvp)-1} E { (Kp)*,1+p2Rp}.
In other words, every element of
is a p-th power.
K2(Kp)
But the
localization sequence of Theorem 3.5 takes the form c 1 -> K2 (Rp) - K2(Kp)
is a pro-p-group and
K2 (Bp)
We are now ready describing
is finite, and so
the main
to derive
defined, independently of the choice of order
Theorem 3.9
K2(Rp) = 1.
localization sequence
the same time, we show that
At
SK1(21).
(R/p)*
(R/p)
For any 71-order
21
C(A)
for
is well
21 C A.
in a semi simple Q-algebra
A,
there
is an exact sequence
® K2(2Ip)
where
2
C(A)
is induced by the inclusions
(21,I) C 21,
and
p
®SK1(3p) - 1;
SK1(21)
a
21 C 2[p,
(1)
by the inclusions
by the composites K2(2 [p) --> K2(2p/Ip) ---' K1(21,I).
Furthermore,
C(A) = 4Lm SK1(21,I) = im C11(21,I); I
and is independent of the choice of order
Proof
For each ideal
sequence for sequence
the pair
(I C 21
of finite index)
I
I C 21
(21,I)
21
in
A.
of finite index,
(Theorem 1.13)
the relative exact
restricts
to an exact
CONTINUOUS KZ AND LOCALIZATION SEQUENCES
CHAPTER 3.
77
K2(21/I) - ' SKl(21,I) -i SK1(21) - Kl(2VI). The first term is finite (Theorem 1.16), and so taking the inverse limit over all
gives a new exact sequence
I
Lm K2(2VI) - C(A) -f SKI(21) The first term in (2) is isomorphic to
2.10(iii).
I by Proposition 3.6 (and
$ K2(21p)
and the last term is isomorphic to
Lemma 3.8),
(2)
Kl(21/I).
I
npKl(21p)
by Theorem
This shows that sequence (1) is defined, and is exact except
possibly for the surjectivity of
For each prime 1im Kl(21/pk21).
we can choose
p,
SK1(21p)
Since
E.
is finite (Theorem 2.5(i)), and
SKi(21p) = 1
for almost all
such that for all primes
n > 1
p,
Kl(2p)
this shows that
p,
SK1(21p) }---' Kl(21 /-21p)
(3)
is injective.
Fix a prime
p
M E GL(2Ip) n E(Ap).
M =
and an element
In other words,
Write
(ri E Ap)
eiiji(rl)...e. (rk)'
a product of elementary matrices. ri,
[M] E SK1(21p).
choose a global approximation
Ti = ri
(mod
pa21p),
ri = 0
(mod
m21).
Write n = p .m, I. E 21[1]
where
pt.-
For each
such that
and
Note that it suffices to do this on the individual coordinates (in of ri with respect to some fixed 7L-basis of H. If we now set
p)
78
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
M
then ( = M
(mod
(rk) E GL(21) fl E(A),
ei13i(rl)...e.
=
MaI
pa21p),
(mod
and
m21),
[M] E SKl(21) = Ker[Kl(21) -i Kl(A)]. By (3),
the congruences guarantee that
pair of Z-orders in
For any
A.
So
e([M]) = [M].
It remains to prove the last statement.
First let
I C B
21-ideal
a
is onto.
B C 21
be any
of finite index, there
is a short exact sequence
1 --> E(21,I)/E(B,I) ) SK1(B,I) I 1 SKl(21,I) --> 1. By Lemma 2.4,
E(21,I2) C E(B,I)
trivially to
Ker(f1).
isomorphism
for any such
I,
In particular, the inclusion
4Lm SK1(B,I) = Jim SK1(21,I),
I
so that
Ker(f12)
so that
B C 21
C(A)
maps
induces an
is well defined.
I
Also, by definition of
C11(21,I),
1 -> Jim Cll(21,I)
there is an exact sequence
®Kl(2t ,Ip);
Jim SK1(21,I)
I
I
p
Note in particular that for any
p
Cll(21) = Im[C(A)
21,
the localization sequence above; and that
SK1(21)
SK1(21)]
in
sits in an extension
1 -* Cll(21) ' SKl(21) e + ® SKl(21p) - 1. p One easy consequence of Theorem 3.9 is the following:
Corollary 3.10 Q-algebras, and let
Then the Induced map
Let
f:
21 C A and
A - B B C B
be a surjectton of semtstmple be
Z-orders such that
f(21) C B.
CHAPTER 3.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
C11(f)
A = B
is surjectiue.
If
only for primes
PI[B:21]
:
79
C11(21) -» Cl1(B)
21 C B,
and.
then
has torsion
Ker(C11(f))
Consider the following diagram of localization sequences from
Proof Theorem 3.9:
p K 2p)
C(f)
K2(f) $ K2(BD) P Since
f: A -+ B
Hence
Cll(f)
so
whenever
pI[B:21],
Coker(K2(f))
and
C11(f)
-> C11(B) -> 1.
C(B)
is projection onto a direct summand,
is also onto.
Coker(K2(f))
Cll(21) ---+ 1
C(A)
If
surjects
onto
K2(Bp)
is
Ker(C11(f))
A = B,
then
Ker(C11(f)).
C(f)
But
is an isomorphism, K(B- p) = K2(up)
a pro-p-group for all
have torsion only for primes
We next want to prove an alternate description of K2(-).
is onto.
C(f)
and so
p,
PI[B:21].
C(A),
o
in terms of
The key problem when doing this is to define and compare certain
boundary maps for localization squares.
In fact, given any commutative
square
R a , R' f I S
of rings, inverse boundary maps
b: Ker(Kn(a) $ Kn(f)) 1 Coker(Kn+l(f') $ Kn+1(R))
can always be defined (and the problem is to determine when
6
is an
80
OONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
When n = 1,
isomorphism).
(i)
elements
there are two obvious ways of defining
For any M E CL(R) x E St(R')
and
such that
y E St(S)
6:
choose
[M] E Ker(K1(a) 0 K1(f)),
such that
+(x) = a(M),
+(y) _
Then
f(M).
61([M]) = [x 1y] E K2(S')
(ii)
(mod
Im(K2(R) $ K2(f')))
Define
BGL(R) + -) BGL(R')+),
K1(a) = ir1(homotopy fiber of
and similarly for
Consider the following diagram:
K1((3).
8
K2(a)
K2(R) - ) K2(R')
K K1(a)
1K2(f')
K2(f)
K2(S)
a+
1('a).
1K1(fo)
K2 R) ) K2(S')
K1
K1(13)
K (a)
K1(R) 1 -) K1(R') 1K1(f')
1K1(f)
ti P I , Kl(R)
fRI Kl-'' Kl(S');
where the rows are induced by the homotopy exact sequence for a fibration; and let
62
be the composite
Ker(K1(a) $ K1(f)) 3 Coker(K2(13) (D K2(f'))
62 =
Note
that any boundary homomorphisms constructed using Quillen's
localization sequences (Theorems 1.17 and 3.5) will be of type
Lemma 3.11
Let
R'
jfr
RR
S
be any commutatiue square of rings.
13
, S' Then
62.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
81
S1 = S2 : Ker(K1(a) W K1(f)) ) Coker(K2((3) ® K2(f'))
Proof
Note first the following more direct description of
a: S1 -' BGL(R) +
any
a+oa and
such that
62([a])
:
:
D2 --) BGL(S)+.
S2 = D2 US1 D2 --' BG(S')+
as a CW complex whose 2-skeleton consists of one
BGL(S) +
for each element
(A)
relation among the elements in elementary matrix to some
of
is the homotopy class of the map
vertex, a 1-cell
A
and extend
[a] E Ker(wl(a+) ® irl(f+)),
) BCL(R')+,
(f+oaa) U (R+ af)
Regard
Fix
to maps
f+oa
as : D2
Then
62.
eS. E E(S).
X E St(S)
argument applies to
GL(S),
Then, given any
a 2-cell for each
and
BGL(S')+,
C(A)
C(A)
for each
[xS.]
a lifting of
A E E(S),
induces a null-homotopy of the loop BGL(R') +
We are now ready to reinterpret description of
A E GL(S),
and a 2-cell
(A).
and shows that
in terms of
The same
61 = 62.
K2(-).
0
The
in the following theorem will be the basis of its
computation in the next chapter.
Theorem 3.12
For any semisimple Q-algebra
A,
there is a natural
tsomorphism
C(A) = Coker[K2(A) - ®K2(Ap)]. p
Under this identification, in the localization sequence
® K2 (21 p)
O
C(A) a-> C11(21) -> 1
13
for a 7L-order
21 C A,
W
is described as follows.
is induced by the inclusions Given any M E CL(21)
21p C Ap,
such that
and
8
82
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
[M] E C11(21) = Ker[K1(21) -i Kl(A) ® I Kl(21p)], P M
lift
to
and to
x E St(A)
y = (yp) E II St(21p)
such that
x = yp
in
p St(Ap)
for almost all
X ly = yx
Proof
p
1
Then
[M] = 8([x 1y]) = 8([yx 1]),
Bp (x) E K2(Ap),
p
direct sum
For any
11 C A.
x E K2(A),
£ (x) E Im[K2(ip) -i K2(Ap)]
(this holds for each generator
for almost all
xr. E St(A)),
This shows that
by Lemma 3.8.
where
® K2 (Ap)] = C(A). P
E Coker[K2(A)
Fix a maximal Z-order
localizations all
p.
and
with
for almost K2(uip) = 1
maps into the
K2(A)
11 K2(Ap).
For each ideal
of finite index, consider the commutative
I C 1l
square
(1)
(1Ap, I
P
p
) -+p Ap
Recall (Milnor [2, Chapters 4 and 6]) that the
as direct summands of and similarly for
K.(D),
Ki('Aip,Ip).
where
can be regarded
D = {(r,s) E MxTh: r=s (mod I));
So Lemma 3.11 can also be applied to this
The inverse boundary maps
relative case.
Ki(111,I)
b3 =62
for (1) then take the
form
bI
C11(L1,I)
To see that
bl
)
Coker[K2(A) ® ®K2(Dip,Ip) P
P
actually maps into the direct sum (as opposed to the
direct product), note that for any decomposition of
- ® K2(Ap)]
M
[M] E C11(21,I),
and any explicit
as a product of elementary matrices over
A,
this
1p
decomposition will have coefficients in SI([M])
can be taken to be trivial in
For each of
K2(-).
p,
83
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
slim
for such
K2(AP)
p
by definition
for almost all
K2(1p) = 1
(so that
p).
Coker[K2(2P,Ip) -+ K2(Ap)] = K2(Ap)
Since, in addition,
inverse limit of the
for almost all
p,
the
takes the form
61
C(A) = 4Lm C11(LI,I) -j Coker[K2(A) - ® K2(Ap)].
Now consider the localization sequences
1 --' K2(T)
aA
K2(A)
® K1(%/Jp) --- SK1(Li) - 1 P =IId
I
I
®a p-)
1 -4 ® K2(T) -> ® K2(Ap) P
p/
K1(mp/JP) -> ® SK1(7AP) --> 1.
p
P
of Theorems 1.17 and 3.5 (where
P
Jp C Atp
is the Jacobson radical).
This
diagram, together with the localization sequence of Theorem 3.9, induces the following commutative diagram with exact rows:
P
Coker[K2(7R) -(1) K2(Di)]
- IId
a
a C(A) 18
(2a)
(2b)
C11(p) -> 1 1Id
(2)
1 -4 Coker[K2(7R) --> ® K2(AP)] -4 Coker[K2(A) --> ® K2Ap)] - Cl1(1) --+ 1 P
P
Square (2b) commutes since
(a')-1 = 62
in the notation of Lemma 3.11.
To see that (2a) commutes (up to sign), fix I
xr
let
in
xI E St(TR)
x by some
be a mod I mod I
x E 6 K2(fNp),
approximation to
approximation to
0(x) = ([#(xl)])1CM E -
r).
x
(i. e., replace each
Then
SK1(N,I) = C(A). I
and for each
84
Each
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
lifts to
(x1)
xI E St(A) -1
I
= x
go 'P(X) = (X-1 XI) -x I
It now follows from (2) that
b
The description of
The descriptions
is an isomorphism.
are immediate.
of b o V and 8 o b 1
o
0: C(A) -» Cll(21)
in Theorem 3.12 is in itself
of only limited use when working with concrete matrices. well
so that
x 1xI E 0) pSt(i p,Ip);
and
No matter how
is understood, it is difficult to deal with an element which
K2(Ap)
is presented only as a product of generators
xri E St(Ap).
In contrast,
the formula in the following proposition, while complicated to state, is easily applied in many concrete calculations.
Proposition 3.13
Let
be a Z-order in a semistmple Q-algebra
21
A,
and let
8 : Coker[K2(A) - ® K(Ap)] = C(A)
be the boundary map of Theorems 3.9 and 3.12. factorizations are
21[i]
Z[!]-orders.
= B x B',
Let
A = B x B';
(a d) E GL2(21)
» Cl (21)
Fix
where
n >
B g B
1,
and fix
and
B' C B'
be any matrix such that
ac = ca
and ad - cb =1, and such that c E B* X B" ,
Then
a E B x (B')and
[a a] = 8(X) E C11(21),
a E (2Ip)*
for all pin.
where
X = ({a,c}-1,1) E Im[( ® Kc(B ) x ® Kc(B')) , C(A)]; pin 2 p pin 2 p
and
X = ({a,c},1) E Im[( ® K°(B ) x ®Kc(B')) ) C(A)]. 2(.9p 2
Pjn
Proof
P
Pjn
Note first that these two definitions of
X are equivalent:
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
85
C(A) = Coker[K2(A) _- 4 ® K2(Ap)], P
and
({a,c},1) E K2(B)x K2(B') = K2(A).
Consider the matrix decompositions
c a
=
1
a 0 a0'
01
ca-'
b)
1 c-'d (a b1 (1 ac-' 0 -c-' ) c d) = (0 1 )(0 1 )1c 0
These give liftings of
ca-1. x21 h21(a)-1
x12-1.w21(c)-x121d
in
(pin)
E2(B).
in
St(B'),
in
b
and E2(2p)
to elements
(a d)
a12
E2(8 )
in
0 al
St(B),
St(8p)
St(Bp)
(pin)
and
St(2p)
(pin)
(P4n)
Here,
h21(a) = x21x12
(01
are liftings of
x21x21x12x21
and
a) Chapter 9] for more details).
(0
and
w21(c)
grespectively
The description of
from the following computation in
= x21x12
St(Bp)
for
(see Milnor [2,
0 1([c d]) pin,
x21
now follows
based on relations
in Milnor [2, Corollary 9.4 and Lemma 9.6]: ca-' .h_1(a) -1 a-'b ac-1 .w21(c),x12 c-'d/-l .x12 )(x12 (x21
x21-1.h21(a)-1.x12-1c-1.w21(c)-1.x12°-1
= _
= h21(a)
-1
(c ld - a lb = a 1c 1)
ac -a-'c-' ac -1 .x21.x12 x21 w21(c)
w21(ac).w21(c)-1.h21(a)-1
=
=
h21(ac)'h21(c)-1.h21(a)-1 = {c,a}
= {a,c}-1.
O
86
CHAPTER 3.
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
The use of this formula for detecting whether or not an explicit matrix vanishes in Note that when
3).
is commutative, any element of
2I
to a 2x2 matrix
reduced
will be illustrated in Example 5.1 (Step
Cl1(7L[C])
Proposition 11.2]).
(a d)
SK1(2C)
(see Bass
ad - be =1
where
can be [1,
So in principle Proposition 3.13 (or some variant)
can always be applied in this case.
Another example where this formula is
used can be seen in Oliver [5, Proposition 2.6].
We now focus attention on group rings. that
is a p-group for any finite
SK1(2 p[G])
p-group.
C,
Proof
Fix a prime
(Wall)
R C F
extension, and let SK1(R[C])
J C R[G]
Let
is a
Then for any finite
S
Then
So
it
will
suffice
E D F S
of
to
with ring of integers is a splitting field for
is a product of matrix rings over
M
SK1(R[G])
is
is
show
a
that
K1(R[G]/J).
[u] E SK1(R[G]).
E[G]-module, and if
be any finite
Ker[K1(R[G]) - K1(R[C]/J)]
and
Fix a finite extension that the residue field S[G]/J
F/&
let
be the Jacobson radical.
maps trivially to
be such that
p,
be the ring of integers.
pro-p-group by Theorem 2.10(ii).
such that
not only when G
is a p-group.
finite by Theorem 2.5(i),
SK1(R[G])
G,
This does not, of course, hold for arbitrary 2p orders.
Theorem 3.14
group
The following theorem shows
If
V
S.
V,
G;
such i. e.,
u E (R[C])*
finitely generated
is any
is any S[G]-lattice in
Let
S C E,
then
detS(M u-4 M) = detE(V - V) = 1
since
[u] = 1
in
K1(F[C]).
Hence, if we set
dets(M
M) = 1.
M = S OS M,
then
(1)
By the surjectivity of the decomposition map for modular representa-
tions (see Serre [2, §16.1, Theorem 33] or Curtis & Reiner [1, Corollary
87
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
18.14]), the representation ring
ules of the form M = SOS M. for any irreducible
RS(G) = KO(S[G]/J)
is generated by mod-
det5(T --+T)=1
So (1) extends to show that
S[C]/J-module
is a product of matrix algebras over
[u] E Ker[SK1(R[G])
Since
T.
and
S[G]/J =- §OR- (R[G]/J),
it follows that
S,
> K1(R[G]/J) )
i
K1(S[G]/J)].
o
The localization sequence of Theorem 3.9, in the case of group rings, at
least,
can now be split up in a very simple fashion into their
p-primary components.
For any semisimple (P-algebra
to be the p-localization of sum
C(A) = 0pCp(A) - since
K2(Ap) 3.6.
C(A).
C(A)
Note that
Cp(A)
splits as a direct
C(A)
is a quotient of
we define
A,
19 pK2(Ap).
and each
is a product of a finite group and a pro-p-group by Proposition In fact,
C(A)
Fix a number field
Theorem 3.15 integers.
will be seen in Chapter 4 to be finite for all
Then, for any finite group
K,
G
and let
R
A.
be its ring of
and each prime
p,
there are
exact sequences
) Cp(K[C]) G C11(R[G])(P) 8p
PP
K2(Rp[G])
(1) 1,
and
1 -> C11(R[G])(p)
,
SK1(R[G])(P)
0 SK1(Rp[G])
-i 1.
(2)
These sequences, together with the isomorphism
C(K[G])
are natural with respect
) Coker[K2(K[G]) -- ® K2(Kp[G])],
to homomorphisms of group rings, as well as
transfer (restriction) maps for inclusions of groups or of base rings.
Proof
The sequences follow immediately from Theorems 3.9 and 3.14;
since K2(p[G])
is a pro-p-group for each p by Proposition 3.6.
88
CONTINUOUS K2 AND LOCALIZATION SEQUENCES
CHAPTER 3.
Naturality with respect to homomorphisms of group rings is immediate.
For any inclusion
S[H] C R[G],
integers in a subfield of
where
H C G
(2) are all induced by some fixed inclusion with the usual isomorphisms
S
these
to
together
Sequence (2)
etc.
isomorphisms,
last
follow from the descriptions of
Theorems 3.9 and 3.12.
is the ring of
R[G] C Mk(S[H]),
Kj(Mk(S[H])) = Ki(S[H]),
is clearly natural with respect naturality of (1) and
and
the transfer maps for the terms in (1) and
K,
,p
and the
and
8
in
O
It has been simplest to derive the localization sequences used here by indirect means.
The usual way to regard localization sequences is as
Mayer-Vietoris sequences for certain "arithmetic" pullback squares.
We
end the chapter with an example of such sequences, due to Bak [2] in dimensions up to 2, and to Quillen (Grayson [1]) for arbitrary dimensions.
Theorem 3.16 and Fix a set
21[57']
Let
21
be any
7L-order in a semtsimpte
of (rational) primes.
= 21[!: p
295
Q-algebra
A,
Define
=pE5$ nup,
Then the pullback square
-) 21[
21
]
2(95 - A3, induces a Mayer-Vie torts exact sequence
... - K1(21) -*
K1(21[1])
W Ki (35,) - Ki %)
-' Ki-i (2) - .. . (1)
... - KO(H) -> Proof Let Pt(21,95) generated
f-torsion
21-
and
Pt(21,9s)
W KO(2195) -> KO(213,)
denote the categories of finitely
modules of projective dimension one.
and 2155-
CHAPTER 3.
CONTINUOUS KZ AND LOCALIZATION SEQUENCES
These categories are equivalent: generated the
for example,
note,
#-torsion module over either
2I
or
215$
89
that a finitely So
must be finite.
localization sequences of Quillen for nonabelian categories
(see
Grayson [1]) induce the following commutative diagram with exact rows:
... -> Ki( t(2I,55)) -* Ki(21) -* Ki(2I[1]) -i Ki-1(Pt(21,#)) =1
1
1
--' ... (2)
=1
--> Ki(Pt(2I9,41)) -' K1(2[q) -* Ki(A) -' Ki-1( The snake lemma applied to (2) now gives sequence (1), except for exactess at
KO(2I[!]) 0 KO(I );
and this last point is easily checked.
o
Chapter 4 THE CONGRUENCE SUBGROUP PROBLEM
The central result in this chapter is the computation in Theorem 4.13 of
C(A) = Jim SK1(2I,I) = Coker[K2(A) -b ® K2(Ap)]
-I
p
for a simple Q-algebra
A:
a complete computation when
A
is a summand
of any group ring
K[G]
the general case.
This computation is closely related to the solution of
for finite
but only up to a factor
G,
the congruence subgroup problem by Bass, Milnor, and Serre [1]. groups
in
(fl)
The
have already been seen (Theorems 3.9 and 3.15) to be
C(A)
important for computing
for 7L-orders
C11(21)
4.13 is needed when computing
SK1(7L[G])
In fact, Theorem
2[ C A.
in all but the most elementary
cases.
It is the second formula for
C(A)
(involving
K2(A)
which is used as the basis for the results here. originally taken by C. Moore in [1].
phisms between C(A)
and
K2(Ap)
and
K2(Ap))
This is the approach
The idea is to construct isomor-
and certain groups of roots of unity.
Norm residue symbols are defined in Section 4a, and applied there to prove Moore's theorem (Theorem 4.4) that
of unity in
F)
for any finite field extension
this is extended to the case of a simple of
K2(A)
(the group of roots
K2(F) = µF,
F Q
Qp -algebra
In Section 4b, A:
the computation
is not complete but does at least include all simple summands
of p-adic group rings.
The final computation of ®p K2(Ap)/Im(K2(A))
is then carried out in
Section 4c, based on Moore's reciprocity law (Theorem 4.12), and results
of Suslin needed to handle certain division algebras. applications are then listed:
for example, that
is a maximal 7-order, or an arbitrary
A few simple
Cl1(21) = 1
A-order when
7L 5 A C Q.
whenever
2[
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
4a.
91
Symbols in K2 of p-adic fields
By a symbol on a field X: F* x F* - G, for any
1 X u E F*.
from
comes
G
where
theorem of
Steinberg symbol with values in
Theorem 4.1
X(u,l-u) = 1
The importance of symbols when working in K-theory
following
the
is meant a bimultiplicative function
F
is any abelian group, such that
K2(F)
Matsumoto,
: F* 0 F*
{,}
that
is the "universal symbol" for
For any field
(Matsumoto)
which says
F,
the F.
the Steinberg symbol
) K2(F)
is surjectiue, and its kernel is the subgroup generated by all elements
u®(1-u) for units 1$uEF*.
In particular, any symbol
factors through a unique homomorphism
Proof
X:
F*xF* -> C
X: K2(F) -i C.
See, for example, Milnor [2, Theorem 11.1].
It is an easy exercise to show that the relations 1
(when
{u,l-u} = 1
F
and
follow as a formal consequence of the identity
K2(F)
in
{u,-u} = 1
is a field, at least).
When constructing symbols, the hardest part is usually to check the relation
X(u,l-u) =1.
The following general result is very often useful
when doing this.
Lemma 4.2 Fix a field
XE
F
and an abelian group
C.
Let
: E*xE*-4G
be bimultiplicatiue maps, defined for each finite extension
E 2 F,
which satisfy the relations
XE(u,v) = XF(NE/F(u),v)
for all
E.
Then, for any n Z 1
(all
and any
u E E*,
1 # u E F*,
v E FTM)
and
92
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
xF(u,l-u) E (xE(v,l-v)n : E/F finite extension,
In particular,
XF
G
is a symbol if
1 # v E E*).
contains no nontrivial infinitely
divisible elements.
Proof
Fix u E F*'-l,
and let
xn - u =
k [[ f(x)e' E F[x] i=1 1
be
the factorization as a product of powers of distinct irreducible
polynomials.
and set
In some algebraic closure of
Fi = F(ui).
Then ui = u
for all
i,
ui
of
fi,
and
k
k 1 - u =
fix roots
F,
ll fi(l)e' = Il NF i=l i=1
/F(1-ui)e'.
It follows that
k xF(u,l-u) =
k
II xF(u,NFj/F(1-ui)ei) =
xF:(ui, (1-ui)e')
D
i=1
i=1
k =
o
(ui,l-ui)ne'.
We now consider a more concrete example. Fix a prime and let
any finite extension of in
F.
For any W C µF,
extension such that
let
F
be
the norm residue symbol
(,)
is defined by setting
AF
p,
be the group of roots of unity
u
: F* ® F*
) W
(u,v)4 = s(v)(a)/a;
an = u
(n = IµI),
s : F*/NF(a)/F(F(a)*)
where
and where
) Gal(F(a)/F)
F(a)/F
is some
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
93
is the reciprocity map (see Proposition 1.8(11)).
Theorem 4.3 group µ C µF,
(t)
(,)
(ii)
If
symbol on E
Let
F
be any finite extension of
of roots of unity in
p
: F* x F*
E 7 F
Then
F.
is a symbol.
is any finite extension, and if
with values in
and fix some
@p,
(,)u,E
denotes the
then for any u E F* and any
µ,
v E E*,
(u,v)4,E = (u,NE/(v))u.
(iii)
For any
po C p,
*
and any
u,v E F ,
(u,v)W. = ((u,v)ul[1 1 o]
For any
(iv)
primes
there exists
q;.,
n-power torsion in
Proof
E(a)/E
(ii,iii)
u E F*
and any
njIWI,
v E F*
Set
n = Iµl
and
m = Iµol,
an = u.
E* E I Gal(E(a)/E)
INS
Is F* s > Gal(F(a)/F)
reciprocity maps, and where
(u,N E/F(V))
generates the
fix
u E F*,
and let
The diagrams
F* and
s) Gal(F(a)/F)
Id I res I
F*
commute by Serre [1, Section XI.3], where
(,)u,
(u,v) A
such that
for all
.i.
be an extension such that
definition of
u1/q f F
such that
res
So
Gal(F(an/m)/F) so
are the
denotes restriction maps.
By the
s,
sE'
and
for any v E E*,
= s(NE/F(v))(a)/a = 5E(v)(a)/a = (u'v)µ,E;
94
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
and the proof of (iii) is similar.
(u,l-u)µ =1
The relation
(i)
It suffices to show this when
(iv)
such that
u1/q f F,
roots of unity in
E = F(u1/q),
set
So for any
and let
(u,v) W generates the
Now, for any prime
of roots of unity of
F.
section, which says that
any finite extension
and
generates
F
F with values in
µq,
a
4p,
of µF,:
(,)F
the group
We can now prove the main theorem in this (')F
is the universal continuous symbol for
(C. Moore [1])
Q.
takes the form
E/F
q-power torsion in µ by (iii).
p
u E F*
Fix
µq be the group of q-th
(u,v)µQ = s(v)(u1/q)/ul/q
will denote the norm residue symbol for
Theorem 4.4
is prime.
Cal(E/F) = Z/q.
(E*)
v E F* - NE/F,(E*),
Finite extension of
n = q
Then the reciprocity map for
F.
F*/N
and
is immediate from (ii) and Lemma 4.2.
Let
p
be any prime, and let
Then the norm residue symbol
F
F.
be any
induces an
(')F
tsomorphism a K2(F)
R C F
Furthermore, if (µF,)P :
is the ring of integers, then
K2(R) = K2(F)(P)
the group of p-th power roots of unity.
Proof K2(F) (p)
µF.
Let is
p C R
be the maximal ideal.
clear from Theorem 3.5:
Proposition 3.6,
and
p4'IK1(R/p)I
The relation
K2(R)
K2 (R)
is a pro-p-group by
by Theorem 1.16.
By Matsumoto's theorem (Theorem 4.1), the norm residue symbol induces
a homomorphism
aF: K2(F) -1 PF.
If
n = IpF.I,
then
K2(R,p2nR) =
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
(Theorem 3.3).
{R*,1+p2nR} of
Set
{c,u}
So aF
x E R.
n = IgFI,
and let
u E F*
there exists
all elements (l+p2nx)l/n
factors through a : K2(F) --> µF. F
S E µF be a generator.
such that
has order at most
surjective.
{R*,1+p2nR} C Ker(aF):
Also,
n-th powers, since the Taylor series for
are
1+p2nR
converges for
95
aF({c,u}) (cn
n
= 1),
By Theorem 4.3(iv),
(c,u)F generates
=
aF
this shows that
Since
µF.
is split
Also, in the localization sequence
K2 (F) - Kl(R/P)
1 -' K2 (R)
is a pro-p-group, and
K2(R)
of Theorem 3.5, Proposition 1.8(i).
aF
Thus,
K1(R/p) = (µF)[p]
by
is an isomorphism of non-p-torsion; and we
will be done if we can show that
Z/p
if
pIiµFl
1
if
pjiµFi
K2(F) @ Z/p 25
(1)
1
.
Fix any it E p_p2. Then F*
= (n) x
by Proposition 1.8(i). pR = pe),
and set
Let
R*
e
eo = e/(p-1).
(l+irnr)p = 1 + pnrp
1 + pant +nrp = 1 + pirnr
= (ir) x µ x (1+p)
(2)
be the ramification index of
For any n > 1
and any
F
(i. e.,
r E R,
(mod
ppn+1)
if
n< eo
(mod
pn+e+l )
if
n = eo
(mod
pn+e+l)
if
n> eo.
(n+e = pn)
In particular,
(1 +
n) C (1 +
n+1).(F*)p
if
pin and n < peo,
if
n > peo.
(3) pn-e)p
(1 + pn) _ (1 +
96
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
(p-1)Ie
If
1 --i 1 +
eo E 7L), then consider the following diagram
(so
peo+l
1 + pep
peo+l) - 1
(1 + peo)/(1 +
at
lao lag
peo+l)
1 -4 1 + ppeo+l -+ 1 + ppeo
where
a; (u) = up
(and
(1 + ppeo)/(1 +
peo = eo + e).
the same order (both are isomorphic to
The domain and range of R/p),
a3
have
and so
IKer(a2)I = IKer(a3)I = ICoker(a3)I = ICoker(a2)I
IKer(a2)I = p
Also,
not
(p-1)Ie,
or
Cp E F.
depending on whether
1,
one of the following two cases holds:
(l+pm) C (F*)p
p1'IIFI,
and
(b)
pIIµF,I,
eo E Z.
and there exists
eo)p
and
(every element of
R
6 E 1 + ppep
or
such that
(1 + ppep) = (6).(1 + peo)p.
In case (b), for any u E R*,
6 a 1 - uxp
either
for any m Z peo;
(a)
b C (1 +
So whether or
(mod
there exists
x E pep
such that
(1+ppeo+l) C (F*)p)
is a p-th power mod
p).
{u,S) _ {u, l-uxp} =
Then
a 1
(mod
K2(F)p).
It follows that
{R*,6} C K2(F)p.
Now fix any write
1
maps, and set
µ = (c).
where
pn):
(modulo
pn > 2,
and let
N: K* - (aP)* be the trace and norm
and
p
C = exp(2iri/pn),
set
Then for any
u c 1+(1-z)2p[c],
p n.Tr(log u)
1
N ( u )
R
THE CONGRUENCE SUBGROUP PROBLEM
p
u)
if
p
if
p = 2
K,u)u =
R
is odd
(and n > 2).
See Serre [1, Proposition XIV.8 and Corollary] for the first
Proof
formula (the tame symbol). Hasse [1].
The formula for
(c,u)u
is due to Artin &
O
Note that Artin & Hasse in [1] also derive a formula for symbols of
(1-c,u),
the form
4b.
in the situation of (ii) above.
Continuous K2 of simple %-algebras
We now want to describe
with center
yA: K2(F) -+ K2(A),
A
whenever
by comparing it with
F,
homomorphism
K2 (A),
This will be based on a
K2 (F).
which
is a simple s-algebra
is
a
special
case
of
a
construction by Rehmann & Stuhler [1].
Proposition 4.8
If
A
is any simple
Qp algebra with center
F,
then there are unique homomorphisms
,PA
:
K2(F) - K2(A)
and
such that
4A({u,nrA/(v)}) = {u,v}
u E F* and
v E A*.
(i)
If
: K2(F) -4 K2(A)
(and similarly for
4A)
for any
Furthermore, the following naturality relations hold:
E C A
division algebra and triangle commutes:
PA
is any self-centralizing subfield (e. g., if
E
is a maximal subfield),
A
is a
then the following
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
K2(E)
101
K2(F)
yA incl
K2(A)
(it)
If
E D F
is any finite extension, then the following squares
commute: C
K2(F)
incl
c
c
trf
c
K2(E)
K2(F)
c
c
jE0A
+A
'PA
K2(E®FA)
10
trf
, K2(A).
K2 (A)
(iii)
for some
If
E 7 F
is any splitting field - i. e.,
E OF A = Mr(E)
r - then the following square commutes:
K2(F)
incl
K2(E)
j'PA 16
K2(A)
where
5
K2(EOFA)
is induced by the identification GLk(Mr(E)) = GLrk(E).
(iv)
For any
r > 1,
the triangle
K2 (F) - A K2 (A)
K2(Mr(A))
commutes; where
Proof
Let
S
is again induced by
y denote the composite
GLk(Mr(A)) = GLkr(A).
102
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
*
1 x(nrA)-1 = J+A
(nrA
F* x F*
PA
that
for all
'y(u,l-u) = 1
factors through
K2(F)
is equivalent to checking
u E
This will be done using Lemma 4.2. define
}
By Matsumoto's theorem (Theorem
is an isomorphism by Theorem 2.3).
4.1), showing that
{
x K1(A) ----) K2(A).
F
XE: E* x e --> K2(A)
For any finite extension
E/F,
by setting
XE(u+v) = trfA A(yEOA(u,v)).
For any u E E*, any v E F*, and any rl E A* such that nrA/F(n) = v, XE(u,v) = trfA®A({u,1o})
=
If
XF(NE/F(u),v).
n = [A:F]1/2;
XF(u,l-u)
=
then Lemma 4.2 now shows that for any
(u,l-u)
v E E*-{1}
EOA = trfA ({v,1-v}) = 1
(nrEOA/E(l-v) = (1-v)n
This shows that
u E F
is a product of elements
EOA
XE(v,l-v)n =
for
(Theorem 3.1(v))
= (NE/F(U)"')
PA
ring of integers, and if
by Lemma 2.1(ii)).
is well defined on ]R C A
K2(F).
If
R C F
is a maximal order, then for all
*A(K2(R.pkR)) = `YA({l+pkR,R*})
is the k > 1,
(Theorem 3.3)
= {l+pkR,ej C K2(p'p-j). So
yA
factors through
*A: K2(F) -> K2(A).
To prove (i), choose intermediate fields such that
K2(Fi) = {Fi,Fi_1}
for all
i
F = F0 C F1 C ... C Fk = E
(use Lemma 4.5).
For each
i,
CHAPTER 4.
THE CONGRUENCE SUBGROUP PROBLEM
Ai g A denote the centralizer of
let
in A
Fi
(so
103
Ak = E).
Consider
the following diagram:
K2(Fk) -> ... -> K2(Fi) t
c
K2(F1-1) -' ... -+ K2(FO)
ii1Y6
I*i
I'I"
c K2(Ai)
K2(Ak)
(where ¢i = 'PA. ) .
(1)
I_1 inc;
c 2(A,-1) -' ... -+ K2(AO)
For any u E Fi_1 and v E Ai , (Thm 3.1(v))
+i-lotrfi({u,nrA`/F`(v)}) = si-1({u,NF;/F:_1onrA;/F;(v)})
=
(Lemma 2.1(iv))
7-1((u,nrA._1/F,_1(v)))
_ {u,v} = incioPl({u,nrA,/Fi(v)}).
c(F ) = {F,F
Since
K2
by assumption (and
i i-1 }
i
nr
A; /F;
this shows that each square in (1) commutes.
2.3),
is onto by Theorem
In particular,
inc1E = *iotrfF : K2 (E) ) K2 (A).
By Lemma 4.5, it suffices to prove point (ii) when
K2(E)
A/F
And this follows easily upon noting that the reduced norm for restriction of the reduced norm for
EO
E
is the
(by definition).
The last two points are immediate, once one notes that for any and
r,
the standard isomorphism
for commuting
u,v E Aw,
(see Theorem 3.1(iv)).
S: K2(A)
) K2(Mr(A))
sends
A
{u,v},
{diag(u,...,u),diag(v,l,...,1)}
to the symbol o
The goal now throughout the rest of the section is to show, for as
many simple &-algebras as possible, that difficult
(and
injectivity.
still
not
completely
yA
solved)
problem
is
to
The next proposition will be used to do this when
odd, and in certain cases when p = 2.
The
is an isomorphism.
prove
p
is
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
104
Proposition 4.9 Fix a prime with center
p
Assume that
F.
for some
n >
extension
E D F
or that
2,
K2(F) --> K2 (E)
be a simple
p = 2 and
is odd, or that
C2n -[0
E F
and such that the induction map
A,
is injectiue.
of degree
n
restricts to a surjection of
µE
E D F
nlind(A),
there is a cyclotomic
such that the norm homomorphism onto
NE/F
It suffices to do this when
µF,.
is prime, and to show surjectivity onto the group
n= q
-algebra
Then there is a finite
is odd.
ind(A)
which splits
We first show that for any
Proof
extension
A
and let
p,
of q-power
(PF,)q
roots of unity.
Write
qr+1-st root of unity.
is a primitive E/F of
we may assume
'("F.)ql = qr;
Then
is unramified (Theorem 1.10(1)), and so (pE) q
onto
some
(up)q.
m > 3);
Now set
so
p = qr = 2,
C2m E E,
n = ind(A).
injects into
then
Let
E 7 F
F
trf
K2(F)
N
E
This commutes by the naturality of E®FE
then
by assumption (for
=
-1-
be any extension of degree n
such
[E:F] = n implies that
is a
E
To see that
consider the following diagram:
°E
pE
then
jCq
(see Reiner [1, Corollary 31.10]).
K2 (E),
q s p,
qr > 2,
C2m - X21 E F
and NE/F(CV)
The condition
K2(E)
with the bimodule
If
=
If
splitting field for
NE/F
If
where
induces a surjection
(S).(cl+gr).(c1+2gr)...(cl+(q-1)qr)
that NE/F(lE) = p F. K2 (F)
E = F(c),
[E:F] = q.
by Proposition 1.8(iii).
(lF)q
NB(C) =
generates
Set
r > 1.
incl K2
(E)
I °E
l1F
C
AE'
inclotrf
is induced by tensoring
(see Proposition 1.18);
and is hence the norm
aE:
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
homomorphism for the action of Cal(E/F) by Theorem 4.6, IK2(F)I.
and so
K2(F) = µF,;
I Im(incl) I
trf is onto
Also,
K2(E).
on
105
=
I Im(NE/F,) I
= IpFI =
El
The next lemma will be needed when showing that
any simple k-algebra A
Lemma 4.10
of index
maximal order.
F
Ker[K2(D)
F
for which
Then for any given
K2(D)] =
is injective for
2.
Fix a finite extension
division algebra with center
yA
Qp,
[D:F] = 4.
D
and let Let
IIi C D
be a be the
each element in
n,
i=
of
Im[K2(F,piIIi) -i K2(D)]
=1
{1+ pnx,1+ pny}
can be represented as a product of symbols pairs of elements
for commuting
x,y E V.
The proof is modelled on the proof by Rehmann & Stuhler [1,
Proof
Proposition 4.1] that for commuting u,v E D*.
is generated by Steinberg symbols
K2(D)
However, since we have to work modulo
{u,v}
pn'lfl,
the
proof is much more delicate in this setting.
Fix n > 2,
and define
Xn = ({u,v} : u,v E 1 + pnB,
We must show that
Step 1
Ker[K2(D) -+ K2(D)] C Xn.
Recall the symbols
any pair of units
uv = vu) C K2(D).
u,v E D*,
{u,v} E St(D),
and such that
defined in Section 3a for
({u,v}) = [u,v]
We are particularly interested here in the case where commute.
By Theorem 3.1(iv),
{u,v} = [x,y]
for any
u
and
(E GL1(D)).
v
x,y E St(D)
that
$(x) = diag(u,u2,...,uk)
and
0(y) = diag(v,v2,...,Vk),
do not such
106
CHAPTER 4.
and such that
ui = 1
THE CONGRUENCE SUBGROUP PROBLEM
or
for each
vi = 1
2 S i S k.
following relations among symbols, for arbitrary
Using this, the
u,v,x,y E D*,
follow
easily from corresponding relations among commutators:
{v,u} = {u,v}-1
(1)
1,vyv l} _ {u,vy};
{ux,v}
{uxu l,uvu
(2)
xvx
{ux
(3)
lv}.
NY l,YvY
(Y = v lxyx lv)
(4)
In particular, the relations in (2) show that for any u,v,x,y E l+pn1R such that [u,y] = [v,x] = 1, {u,v} a {u,vy} _- {ux,v}
Step 2 K2(-)
Set
(mod
Xn).
2I = 2p + p2nB C D,
a gp order in
D.
By definition of
(and Lemma 3.2),
Ker[K2(D)
Also, since
--> K2(D)] C Im[K2(2[,p2n-N) --) K2(D)].
is a local ring, results of Kolster [1] apply to show that
2l
each element of
K2(21)
is a product of
necessarily commuting) pairs of units
by
(5)
and
(2 p) *
1 +p 2n,,,
f E Ker[K2(D) -i K2(D)]
symbols
u,v E 2l*.
Since
relations (2) above
for
{u,v} 2f
(not
is generated
that
any
is a product of symbols {(gp)*, 1+pi-11},
and
show
has the form
f =
E
u;,v;
E
K2(2 p),
1+p
2,F-urthermore,
K2(Z/p-_ L),
(Theorem 3.3).
so
vanishes under projection to 'In fo E K9(2 p,p2 p) = {(gp)* , l+p p}
But for any a E (2p)* and any x E Lt,
CHAPTER 4.
THE CONGRUENCE SUBGROUP PROBLEM
{a,1+p2nx} _ {a(p-1)p" Here, the
(1+p2nx)1/(P'1)p"}
,
E
107
{l+pn+l p , l+pn5t1.
(p-1)pn-th root is taken using the binomial expansion.
We have now shown that
K2, (D)]
Ker[K2(D) -- +
Step 3 p
(a) C R
R
Let
C ({l+pnx,l+pny} : x,y E m)
be the ring of integers in
be the maximal ideal.
R/p-vector space.
For any a E R,
We regard
(C St(D)).
F = Z(D),
V/p7R
and let
as a 4-dimensional
denotes its image in
a
(6)
1/p0.
Define functions
A : (1 + pnB) - 1 -i (Th/pR) - 0
by setting, for any k
and
v
:
(1 + pnI.) - 1 --> Z>0,
0 and any a E IR N p1R:
µ(l+pnaka) = a E B/p)R
u(1+pn rrka) = k.
and
For a n y sequence u1 , ... ,uk E 1 + pnB,
set
i(ul,...,uk) = (W(u1),...,W(uk),l)R/p g 1R/pX;
R/p-vector subspace generated by these elements.
i. e., the
These
functions
manipulating symbols
(7)
For any
v(vo) = 0 and
will {u,v}.
(alternatively,
where
used
(mod
as
a
"bookkeeping
system"
when
The following two points will be needed.
u,v E 1 +pnR,
{uo,vo} E {u,v}
v = l+pnneb,
be
there exist
u(uo) = 0), Xn).
a,b E 1 N p1R.
uo,vo E 1 +pn1R
W(uo) = -µ(v),
To see this, write
such that
µ(vo) = W(u),
u = l+pnrrka
and
Then, by (5),
{u,v} a {l+pnirka,(l+pnveb)(l+pna)} = {l+pnrrka,l+pn(a+ireb+pnireba))
{(l+pnika)(l+pnak(a+aeb+pnieba))-1,1+pn(a+Teb+pnieba)1
=
(mod Xn)
{1-pn7 k+eb(l+pna)(l+pn7 k(a+rreb+pnneba))-l,l+pn(a+neb+pn\reba)I.
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
108
This proves the claim if
then a third such operation
2 = 0,
If
e > 0.
finishes the proof.
u,v E such that
For any
(8)
uo,vo E 1+pN
u = l+pn rra
{uo,vo} = {u,v}
Xn),
(mod
there exists and such that
we can do this with uo = u if and similarly for v. To see this, again write
3.
dimRp(µ(u)) = 2;
[u,v] # 1,
such that
1 +p' R
Furthermore,
and
v = l+pnireb where a,b E M'pM. The condition [u,v] # 1 implies that the elements l,u,v (and hence l,a,b) are F-linearly independent in D. If dimRp(u(u)) =2, so that a E V - (pJRUR), then
we can write b = a+pa+rrmbo,
a,(3 E R and bo4(a,l)R/p.
where
{u,v} = {1+pna a,(l+pn7r0b)(1+pnae(a+Ra))-1}
{u,1-pnV10+mbo(l+pnae(a+pa))-1}
=
and
µ(u,vo) = (a,bo,l)R/p
2-dimensional is similar.
u
analogous operation replaces
Step 4 that
µ(u,v)
elements
by
and
µ(x,y)
Xn);
such that
vo = x0,
or
u = l+pn rka,
a,b,c,d E and
Since
K
or
A
is
then an
u,v,x,y E 1 + p'
1
such
We will show that there are
{uo,vo} a {u,v}
and
and
{xo,yo}
}i(xo,yo) = u(x+y);
2(uo,vo) = li(xo,yo)
x = l+pnc,
Write
y = l+pnned,
dimRp(T/pl) = 4,
and since the sets
are linearly independent, there is a relation
d+T = 0 where
)
W(uo) it R/p.
v(v) = 0 = v(x).
v = l+pnb,
R - p7R.
{c,d,1}
xolvo
µ(v)
R/p
lie in
µ(v)
µ(uo,vo) = µ(u,v)
Using (7), we may assume that
{a,b,l}
The proof when
such that
uo
are 3-dimensional.
such that
and such that either
where
n
Now consider a n y 4-tuple of elements
uo,vo,xo,yo E 1 + pnf)1
(mod
{x,y}
and
}i(u)
(mod X
= {u,vo};
is 3-dimensional.
If both
So
is nonzero and
a
E R/p); or
R
is nonzero.
(9) Using (7) if
CHAPTER 4.
THE CONGRUENCE SUBGROUP PROBLEM
necessary to make some switches, we can arrange that
109
(3 x 0 X K.
For any a,(3,T E R such that (3 E R*,
{x,y} _ {1+pnc, (1+pnJred)(l+pn7rea3 1 c+pnirPi(3 1)} _ (1+ pnc
(mod
X )
n
1 + pnir8(d+a(31yc+T(3 1y) }
,
{(l+pnc)(1+pn((3d+ayc+"Ty)) _ < 1 + pn(c+Pxd+axyc+Txy)
1+pnne(d+aI3 lyc+TR 1y)}
,
, 1+ pnr0(d+a(3 1yc+10 1y)}
=
{xo,Yo}.
Note in particular that v(xo) = 0, and that A(xo) = Thus,
+(3
11(Yo) = d+«J3
µ(xo,Yo) = l(x,y).
Similarly, for any
uo = 1+pnlrk(a+XK 1ub) {uo,vo} a {u,v}
then
(mod
and
Xn),
l
1.
K E R* and X E R,
if
vo = 1+pn(b+iva+Xvub),
and µ(uo,vo) = 1i(u,v).
Now consider the equation
(10) By (9), we can find holds
a,(3,7,K,X E R,
pM = wV).
(mod
we can find a solution
If (10) holds
(mod
ire+11)
where
(3,K E R ,
aeM),
(mod
such that (10)
for some
e > 1,
then
unless
aer, and
r
R/p
11(xo1vo) and
1%(u,v)
=r
= µ(x,y)
= µ(u,v)+ µ(x,y).
If this ever happens, then
11(u,v)+Fl(x,Y) = A(uo,vo)+11(xo,Yo)+ since each has codimension one.
successive approximations yield
a,(3,1,K,X
Otherwise,
such that (10) holds, and
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
110
hence such that
vo = xo.
Step 5 We are now ready to prove the lemma.
E Ker[K2(D) --i K2(D)] 1 + pn%l.
is a product of symbols
symbols for commuting pairs
u,v E
for
it will suffice to show that
u,v E l+ pnB,
u,v,x,y E 1+ pnLl,
for
{u,v}
i. e., that f is a product of such
So to show that $ E Xn,
any product
By Step 2, any element
is congruent
(mod X
n
)
to another single symbol of the same form.
Fix such
u,v,x,y.
(using (8)) that there exist
uo,vo,xo,yo
such that either
We may assume that
}i(u,v)
vo = xo
and
[u,v] A 1 0 [x,y];
are 3-dimensional.
µ(x,y)
and hence
By Step 4, and
such that or
xolvo f V(uo,vo) = u(u,v) = l(x,y) = u(xo,yo)
In the first case, we are done by relation (3).
In the second case,
(mod Xn)
by
(4),
where
µ(x1)= µ(yo),
Then
Yo = volxoyoxolvo.
and µ(y1)= µ(xolvo).
µ(v1) = µ(vo)= µ(xo),
So by (11),
dimp(µ(xo,yo)) = 3. Step 4 (and (3)) can now be applied again, this time to to show that it is congruent
mod X
n
to a symbol of the same form.
The next theorem, due mostly to Bak & Rehmann [1], and Prasad & Raghunathan [1],
s-algebras.
shows
that
is an isomorphism for many simple
yA
Recall that the index of a central simple F-algebra
defined by setting
ind(A) = [D:F]
1/2
if
A
is
and
D
is a
with center
F.
Then
A = Mr(D)
division algebra.
Theorem 4.11
Fix a simple
there is a unique isomorphism
&-algebra
A
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
pF/T,
oA : K2 (A)
where
T C {i1},
111
and such that for any a E F* and any u E A*:
CA({a,u}) = (a, nrA/F,(u))F. T = 1
Furthermore,
(t)
p
is odd, or
44ind(A);
(it)
A
(tit)
if any of the following three conditions hold:
p = 2 and
-C21
E F
[2n
or
n > 2;
for some
or
is a simple summand of
for some finite group
K[G],
G
and some finite extension K Q Q.
Also, for any maximal order
K2(11) = K2(A)(p) = (PF,)p/T.
JR C A,
The last statement, that
Proof
K2(1R) = K2(A)(p),
By Theorem 4.4, it suffices to
the localization sequence of Theorem 3.5.
yA
show that most
2,
:
K2(F) - K2(A)
is surjective with kernel of order at
and an isomorphism if any of conditions (i) to (iii) hold.
proof will be carried out in three steps:
yA
will be handled in Step 3.
Ker(PA)
suffices to assume that Let
R C F
A
Set
The
will be
will be shown in Step 2, By Proposition 4.8(iv),
it
is a division algebra.
be the ring of integers,
the maximal ideals.
p
torsion prime to
dealt with in Step 1, the surjectivity of and
is immediate from
n = [A:F]1"2.
by a field E ? F and an element n
and let
A
p C R be
is generated
such that
(a)
E/F
(b)
there is a generator 11 E Gal(E/F) such that
is unramified,
and
j C 1R
By Theorem 1.9,
[E:F] = n,
and
irETr 1 = E
irxir
1
= n(x)
for
all xEE (c) ]R = S[a]
(where SCE is the ring of integers) ;
J = J(R) =,rJ,
CHAPTER 4.
112
An
THE CONGRUENCE SUBGROUP PROBLEM
and n generates the maximal ideal
E R,
Step
By Theorem 2.11,
1
ISK1(%I)I
p E R.
=
A
IK1(IVJ)I/IK1(R/p)I.
comparison of this with the localization sequence
K2 (A) --1 K1(m/J) -> SK1(M) - 1
i -> K2 (1l)
of Theorem 3.5 shows that IK2(A)[p]I = IK2(A)/2(h)I =
Since
E/F
1(M)I = IK1(R/p)I = IK2(F)[p]I
is unramified, the commutative diagram
c A
K 1 (R/p) E-- Kc(F)/Kc(R) -- K2 c(F)[1] 2 p
K°(A)L1] 2 p
incl I
Ithcl
I
K2(E)/K2(S) = K2(E)[ p] 1 -+ K2(E®FA)[-]
K1(S/pS)
(from Theorem 3.5 and Proposition 4.8(ii)) shows that injection of
K2(F)[p1]
into
K2(A)/[1p],
Step 2 We next show that via the subgroups K2(M) = K2(Th,J).
K2(M,Jk).
that in
by filtering
4A(K2(R)) = K2(T), By Theorem 1.16,
K2(7R/J) = 1,
K2(9t) and so
By Theorem 3.3, for each k > 1,
({l+lr,l+airk-1}
=
n1k,
induces an
and hence a bijection.
K2(E/Jk,Jk-1/Jk)
If
yA
then for any a,b E S,
: a E S).
the symbol relations in Theorem 3.1 show
K2(V/Jk,Jk-1/Jk)
(1+T, 1+abirk-1 } = { l+ir, l+baak-1 } = { 1+ab, l+airk-1 } = {l+rl(b)a,l+aak-1} = (1+7,
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
113
k(b).irk-1}.
= {1+ir,
So (17k
{ l+v, l+a(b-iik(b) # 1 since n1k).
)= 1, and b can be chosen so that =1
K2(M/Jk,Jk-1/Jk)
Hence
b-71 k(b) E
in this case.
then consider the relative exact sequence
nik,
If
)ak-1
K2(IR/Jk,Jk-1/Jk)
K2(m,Jk)
a + Kl(1R,Jk)
I
K1(M/Jk+1,Jk/Jk+l).
Since
nik,
[V,Jk] C Jk+l.
and hence
irk E R,
= V k. 11/j.
K1(DI/Jk+1,Jk/Jk+l) = Jk/Jk+1
For any
{1+ir,l+airk-1} E
K2(,R/Jk,
8'({1+n,l+ank-1})
Jk-1/Jk)
Then by Theorem 1.15,
(a E S),
[l+a,l+atrk-1]
= 1 +
_
k
and this vanishes if and only if a E R + pS. This shows that k > 1
and a E R.
4c
A
'n°l+ K2
(A)] = Im[K2(F(ir)) tr K2 (F)
is its own centralizer in
F(w)
{l+lr,l+aak},
for
In particular, using Proposition 4.8(i),
K2(%) C Im[K2(F(n))
(note that
is generated by symbols
K2(R)
A).
So
K2(A)]
K2(Th) C Im(4A),
and
is onto.
Step 3
and
If none of conditions (i) to (iii) are fulfilled, then p = 2
(}'F)2 = {fl};
injectivity of
yA
and so
IKer(4 ) I
, 2.
It thus remains to prove the
in p-torsion, when (i),
(ii), or (iii) holds.
By
Theorem 1.10(ii), any simple summand of a 2-adic group ring has index at most
2;
so it suffices to consider the first two conditions.
114
CHAPTER 4.
THE (X)NGRUENCE SUBGROUP PROBLEM
p
Assume first that
(i)
for some
n 2 2,
or that
is a splitting field K2 (F) -' K2 (E)
is odd, or that
ind(A)
E Q F
A
-C21
C2"
E F
Then by Proposition 4.9, there
is odd.
for
p = 2 and
such that the induced homomorphism
By Proposition 4.8(ii,iv), there is a
is injective.
commutative square K2(F)
incl
)
K_ (E)
K2 (A) -' K2(E®FA);
and so
is also injective.
PA
(ii)
Next assume that
trancendental extension
A and such that
F
E
p = 2, F
and that
There is a
ind(A) = 2.
(the "Brauer field") such that
is algebraically closed in
[2, Lemma 3 and Proposition 7]).
Then
theorem of Suslin [1, Theorem 3.6].
K2(F)
E
E
splits
(see, e. g., Roquette
injects into
by a
K2(E)
The following square commutes by
Proposition 4.8(iii):
K2(F)
'ncl
K2(E) Io
I 'PA
K2(A) --' K2(E®FA);
(note that we are using dtscrete On the other hand, [1, Theorem 4.3].
yA
Since
n > 1.
is finite, this implies that
follows that
here);
and so
PA
is injective.
is surjective by a theorem of Rehmann & Stuhler
By Lemma 4.10, any
n-th power for arbitrary K2 (F)
K2
r; E Ker[K2(A) -i K2(A)] yA
is an
is an isomorphism, and since
PA1(Tl) E Ker[K2(F) - K2(F)].
It
K2(F) = K2(A).
Now assume that
extension of degree
ind(A) = 2m, m.
Then
where
E®FA
m
is odd.
Let
E 2 F
be any
is a central simple E-algebra of
CHAPTER 4.
index
THE CONGRUENCE SUBGROUP PROBLEM
115
Consider
this follows from Reiner [1, Theorems 31.4 and 31.9].
2:
the following commutative diagram of Proposition 4.8(ii): incl) K2(E)
K2 (F)
c
yA
c
*A
yE®A
K2(A)
1- ) K (EOFA) tr K2(A).
m
in the top row is multiplication by
trf o incl
Proposition 1.18),
K (F)
c
'Pi
The composite
trf
and so
incl
(use
Hence
is injective in 2-power torsion.
is also injective in 2-power torsion; and this finishes the proof.
It is still unknown whether
A
Q2 algebra
with center
F.
K2(A) = K2(F)
K2).
*A: K2(F) - K2(A)
and
to show that
4c.
K2(-)
is always
But we have been unable to extend any of
these results to the case of continuous K2(-)
for an arbitrary simple
The argument in Step 3(ii) (based on
Suslin [1, Theorem 3.6]) shows that injective (using discrete
0
K
This difference between
2-
is the source of the (erroneous) claim by Rehmann [2]
K2(A) = K2(F) = µF
in general.
The calculation of C(Q[G])
If
R
is the ring of integers in a number field
congruence subgroup
of
SLn(R)
(for any
n Z 2)
K,
then a
is a subgroup of the
form
SLn(R,I) = (M E SLn(R) : M =-1
for any nonzero ideal
I C R.
(mod Mn(I))}.
The congruence subgroup problem
as
SLn(R)
of
En(R,mR):
by
originally stated was to determine whether every subgroup of finite index contains a congruence subgroup.
Any subgroup of
SLn(R)
of finite index
m
contains
definition, if
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
116
En(R,mR)
is generated by m-th powers in
the E(R,I)
n > 3,
all have finite index in
Conversely,
En(R).
SLn(R)
since
SK1(R,I) = SLn(R,I)/En(R,I)
(see Bass [2, Corollary V.4.5]) of nonzero ideals,
J C I C R
En(R,I) - any matrix in
for all
I C R;
Furthermore,
is generated by
SLn(R,I)
SLn(R/J,I/J)
n>3
conjecture holds for
is finite.
for any pair SLn(R,J)
can be diagonalized.
if and only if the groups
if and only if the group
Thus, the
SK1(R,I)
C(K) = 4im SK1(R,I)
and
vanish
vanishes.
For the original solution to the problem, where the use of Mennicke
symbols helps to maintain more clearly the connection with the groups
SLn(R, I) ,
we refer to the paper of Bass et al [1], as well as to the
treatment in Bass [2, Chapter VI].
The presentation here is based on the
approach of C. Moore, using the isomorphism
C(A) = 4Lm SK1(21,I) = Coker[K2(A) -e K2(Ap
shown in Theorem 3.12.
The groups
K2(Ap)
have already been described in
Theorem 4.11; and so it remains only to understand the image of
K2(A).
The key to doing this - for fields at least - is Moore's reciprocity law.
Norm residue symbols will again play a central role; and
description of
for a simple Q-algebra A
C(A)
be in terms of roots of unity in the center of
the
(Theorem 4.13 below) will A.
Recall that the valuations, or primes, in an algebraic number field K
consist of the prime ideals in the ring of integers (the "finite
primes"), and the real and complex embeddings of
Theorem 4.12
(Moore's reciprocity law)
number field, and let
A
be a simple
be the set of noncomplex valuations of
K.
Let
K
be an algebraic
Q-algebra with center
K
K.
Let
E
(i. e., the set of prime Ideals
and real embeddings) and set
2A = 2 - {v: K '-> f : Q2®vKA = Mr(W), some r}.
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
117
Then the sequence
® µK
K* 0 nrAIK(A*)
P
, µK --> 1
vE'A
is exact.
and
Here, p'K
for any
denote the groups of roots of unity, and
Kv E p k,):
C = (Cv)Vcl
p(U) _
µK
a)mV/m
[[
(mv = Iµj,I,
v
vE2
m =
KI)
1
This was proven (at least in the case A = K)
Proof
by C. Moore [1,
Theorem 7.4].
Note that
(v: K y It)
(K*,nrA/K(A*))v = 1
a E nrA/K(A*).
whenever
p ol(,) = 1
statement
when
v E 2
for any
A = K
2A:
since
v(a) > 0
It thus suffices to prove the 2A = B).
(so
This is just the
usual reciprocity law (see, e. g., Cassels & Frohlich [1, Exercise 2.9]).
For example, when relation
A = K = @,
p
1
and
p
and
are odd primes,
q
the
reduces to classical quadratic reciprocity
using the formula in Theorem 4.7(1).
A second, shorter proof of the relation case
A =
K,
is
in the
Ker(p)
given by Chase & Waterhouse
in
[1].
By
the
Hasse-Schilling-Maass norm theorem (Theorem 2.3(ii) above),
nrA/K(A*)
= {x E K : v(x) >0, all v E I- 7A} ;
and using this the proof in Chase & Waterhouse [1] of the relation Ker(p)
C
is easily extended to cover arbitrary
A.
O
We are now ready to present the description of the groups to a factor of unity.
{ill,
C(A) - up
at least - in terms of norm residue symbols and roots
This is due to Bass, Milnor, and Serre [1] in the case where A
is a field; and (mostly) to Bak & Rehmann [1] and Prasad & Raghunathan [1] in the general case.
CHAPTER 4.
118
Theorem 4.13 Let
THE CONGRUENCE SUBGROUP PROBLEM
A be a simple
@-atgebra with center
µK denote the group of roots of unity in
(i)
C(A) = 1
(ii)
if
C(A) = µK
if no embedding
and let
Then
K.
for some v: K I-- IR,
IROvKA = Mr(IR)
K,
v: K
splits
JR
A,
some
r
and if for
[2"-[21
each 2-adic valuation
v
of
either
K,
for some
E Kv
n Z 2,
or 41ind(Av) C(A) = µK or
(iii)
More precisely, if
IK/{tl}
otherwise.
C(A) = pK/T $ 1,
then there is an isomorphism
->
aA: C(A) = Coker[K2(A)
®K2(Ap)]
µK/T
p
such that for each (where
a E
(I{p)*
p,
each prime
and
u E (Ap)*),
p1p
of
K,
and each
{a,u} E K2(Ap)
aA({a,u}) = (a, nrA/K(u)) PK E PK.
In particular, each summand K2(Ap)
Proof
Let
surjects onto
C(A).
I be the set of all noncomplex valuations of
K
20 C 'A C 2:
(i. e.,
all finite primes and real embeddings).
Fix subsets
the set of finite primes of
e., prime ideals in the ring of
K
(i.
20
is
integers); and as in Theorem 4.12,
'A = 1 - {v: K '- IR For each (rational) prime 1.7(1)).
p,
:
IR ®vK A = Mr(N),
Kp = IVIPKv
In other words, we can identify
some
r).
and 1P = IvI pAv (1)
Consider the following commutative diagram:
(see Theorem
with ®vElK2(Av
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
K* ® nr
A/K
®}
(A*)
p
119
}-1
,
vE2o
(1)
Is
® K2(Av) ) C(A) -> 1.
K2(A)
Here,
is defined as in Theorem 4.12, and
p
is induced by the symbol
s
map
{,} : K* 0 K1(A)
(where
K1(A) = nr(A*) C K*
4.3(iii), the composite If
20
and so
= {±1}
p o[[(orv)
(K C IR),
uK
'A = 2o,
pK
C(A).
cA.
= }12 = {fl}
for
v E 'A-201
Hence
f
is onto, and
in this case.
A
i. e., if
both rows in (1) are exact. onto
satisfies the above formula for
is onto by Theorem 4.12.
"vE2o(')v
C(A) = Coker(f) = 1 If
Note that by Theorem
by Theorem 2.3).
then
2A,
I
) K2(A)
A
If
ramifies at all real places of
then
K,
In particular, (1) induces a surjection of
is a matrix algebra over a field, then
s
is
onto, and so C(A) = }. Otherwise, we use the
K2
reduced norm homomorphism of Suslin [1,
Corollary 5.7] to get control on
is any central simple nr2: K2(A) -> K2(F)
(2)
if
E 7 F
Coker(s).
F-algebra,
If
F
is any field and
then there is a unique homomorphism
which satisfies the naturality condition:
is any splitting field, then the square
K2(A)
1®
1nrA
K2(E ®F, A) = Io
incl
K2(F)
A
, K2(E)
CHAPTER 4.
120
commutes, where
Also,
THE CONGRUENCE SUBGROUP PROBLEM
is induced by the isomorphism E 0
6
F
there is a splitting field
algebraically closed in Proposition 7]).
(see,
i
such that
Roquette
g.,
e.
A
for
[2,
F
is
Lemma 3 and
So by a theorem of Suslin [1, Theorem 3.6]:
there exists a splitting field E
(3)
K2(F)
E
E D F
A = Mr(E).
K2(E)
F
such that the induced map
is injective.
Then (2) and (3) (and Proposition 4.8(ii)) combine to imply
for any u E F* and any a E A*,
(4)
nrA({u,a}) = {u,nrA/(a)}.
Assume now that condition (ii) holds; then
is an isomorphism for all
yA
K2(Kv) -_ KK(Av)
by Theorem 4.11.
v E 7o
Consider
the
following diagram:
(a^ )
K2(A) f - ® K2(Av)
"
vE2o
nrA
0(yA")-1
(5a)
f
(5b)
K
K2(Kv)
we define for convenience,
v E Y, - 2o,
if and only if
definition of the
al..
u,v < 0).
o
Ker[p
(
® pIC"/ vElo
: ® K2(Kv) -» vE2o
and it follows that
(and
If square (5a) also commutes, then a comparison
f) C Im(n(ak") o fK) fl =
K2(R) = µ, _ {±1}
Square (5b) commutes by the
of diagrams (5) and (1) shows that
Im(n(oA")
(5)
® Wkv vEl
vEy
aIR({u,v}) = -1
incl
ll(a^ )
K2(K)
Here, for
® I{" vE2o
C(A) = Coker(f) = pK.
µJ
CHAPTER 4.
THE CONGRUENCE SUBGROUP PROBLEM
121
It remains to check that (5a) commutes; we do this separately for each v E 2. This splits into three cases. Case 1
2 and
Assume first that
C2n C2n E Kv
for some
Then, by Proposition 4.9, splits
Av,
Kv ) E,,
where either
n > 2,
p = 2
or
p
and
ind(Av)
K2(E).
injects into
K2(Kv)
is odd.
E 2 Kv
there is a finite extension
and such that
p =
is odd, or
which
In the following
diagram:
K2(A) --' K2(Av) ) K2(E®KA) Ir1rA
1v)_1 (6b)
(6a)
6
(6)
K2(K) - K2(Kv) >--' K2(E) square (6b) commutes by Proposition 4.8(ii,iv), and (6a+6b) commutes by So
(2) above.
Case 2 odd
m.
degree m
(6a) also commutes.
Assume now that
Kv 2 O,
and that
ind(Av) = 2m
Using Proposition 4.9 again, choose an extension such that
K2(Kv)
injects into
K2(E).
for some
E 2 1{v
Then E @K A
of
has index
The same argument as in Case (see Reiner [1, Theorems 31.4 and 31.9]). 1v i. if it commutes for E OK A; 1 shows that square (5a) commutes for
2
e., that we are reduced to the case where If
ind(Av) = 2,
ind(Av) = 2.
then consider the following diagram:
K2(A) _--' K2(Av) InrA
K2(Av)
nr2
(7a)
K2(K) -- K2(Kv)
-> K2(Kv).
By Rehmann & Stuhler [1, Theorem 4.3], the form (4)
{a,u}
for
a E (Kv)*
(and the definition of
P).
and
(7)
(7b)
K2(Av)
is generated by symbols of
u E (Av)*;
and so (7b) commutes by
Square (7a) commutes by (2) and (3)
122
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
above; and so (5a) commutes in this case.
Finally, assume that v E 2 - 2o;
Case 3 for some
Then
r.
1. e. ,
K2(Av) _ {(Kv)*,(Av)}
that
Av = Mr(11)
by Rehmann & Stuhler [1,
x
K2(Av) n' K2(Kv) - K2(Kv)- K2(lt)- {fl}
The composite
Theorem 4.3].
is thus trivial (use (4) again); and so (5a) also commutes at such
v.
This finishes the proof of the theorem when (i) or (ii) holds. neither of these hold, then
is isomorphic to
C(A)
(''KK)2 = {±1},
The proof of
in odd torsion.
{
identical to that given above.
If
so we need only check that this
is
o
Theorem 4.13 immediately suggests the following conjecture.
Conjecture 4.14 For any simple Q-algebra A with center
(1
if
IRO
{l}11{
C(A)
vK
for some v: K '- -1
A = Mr(It)
It
K,
r
and some
otherwise.
By Theorems
whenever A
1.10(ii)
and 4.13,
Conjecture 4.14 holds at
is a simple summand of a group ring
and any number field
L.
L[G],
least
for any finite G
If Suslin's reduced norm homomorphism, when
applied to a simple s-algebra, could be shown always to factor through K2(-),
then the proof of Theorem 4.13 above could easily be modified to
prove the conjecture. We now consider some easy consequences of Theorem 4.13. theorems
depend,
in
fact,
not
Coker[K2(A) - ®pK2(Ap)], K2(Ap)
surjects onto
attention on
a-orders:
C(A).
on
the
full
The next two
description of
C(A) =
but only on the property that each factor The first explains why we focus so much
if any primes are inverted in a global order
3,
then C11(21) = 1. Theorem 4.15
is any
Let
A C ID
be any subrtng with
A-order in a semisimple
ID-algebra
A,
A
Z.
Cl1(21)
Then, if =
1.
21
More
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
precisely, if
123
denotes the set of primes not invertible in
9'
SK1(2I) =
then
A,
SK1(2Ip). pE#
Proof Let
be a maximal
IN D 2I
A-order in
and set n = [M:2I].
A.
The same construction as was used in the proof of Theorem 3.9 yields an exact sequence
Jim SKi(2I,I) I
® SKl(2[p) -> 1;
SKI(21)
where the limit is taken over all ideals
im Cll(T,I)
SK1(21,I) =
where
(1)
pE9'
I C 21
of finite index, and
for any maximal A-order
1R 2 21.
Furthermore, the same construction as that used in Theorem 3.12 (based on Quillen's localization sequence for a maximal order) shows that
Im C11(IR,I) = Coker[f9': K2(A) ->
K2(Ap)].
(2)
By Theorem 4.13, under the isomorphism
® K2(Ap)], P
C(A) = Coker[K2(A)
each factor
surjects onto
K2(Ap)
include all primes, im SK1(2I,I) = 1
the map
C(A). in
f51
(2)
in (1), and hence that
2
Hence, since is onto.
for a maximal i order
SK1(Th)
I
does not
It follows
is an isomorphism.
The next theorem allows us, among other things, description of
9'
that
o
to extend Kuku's (Theorem 2.11)
to
maximal Z-orders.
Theorem 4.16
(Bass et al [1]; Keating [3])
a semisimple D-algebra
for which iP
A,
then
C11(21)
is not a maximal order.
If
21
is any Z-order in
has p-torsion only at primes
In particular:
p
124
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
(i)
C1I(21) = 1,
and SK1(21) = ®pSKI(21p),
SK1(R) = 1
(ii)
R
if
if
is maximal;
21
is the ring of integers in any number field;
and
SK1(R[G])
(iii)
finite group and R
Proof
Let
has p-torsion only for primes
phIGI,
G
if
is a
is the ring of integers in any number field.
be any maximal order in
R J 21
and consider the
A;
localization exact sequence
C(A) -) SKI(') - ® SKI(i1p) - 1
® K2(fIp)
For each
of Theorem 3.9.
incl
K(Np)
K2(Ap)
p,
c1K2(Ip)
is the composite
pro j,, Coker[K2(A)
(D K2(Ap)
-> ® K2(Ap)]
K2(Ap)
and
surjects onto
by Theorem 4.11,
torsion in
C(A)).
now applies to show that
by Theorem 4.13.
C(A)
K2(1p)(p)
Hence
= C(A);
P
P
(P
and
Also,
(K2(9p)) = Cp(A)
so
C11(1) = 1.
is onto, and
Cll(21) = Ker[C11(21) -+ Cll('N)]
K2(ip)
=
(the p-power Corollary 3.10 has p-torsion
only for primes pl[ffi:21]. It remains only to prove point (iii). above,
Bp [G]
particular, 1.17(i).
pfl Cl1(R[G])I
for such
On the other hand, for each
Wall's theorem (Theorem 3.14).
So
and
p,
p.
IGI.
as In
p1'ISK1(Rp[G])I
and
by Theorem
is a p-group by
SKI(Rp[G])
C11(R[G])
have torsion only at primes dividing
R[G]
by Theorem 1.4(v).
For any group ring
is a maximal order for all
®pSKl(Rp[G])
both
0
Point (iii) above will be strengthened in Corollary 5.7 in the next chapter:
SKI(R[L])
p-Sylow subgroup
has p-torsion only for primes
Sp(G)
is noncyclic.
p
such that the
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
125
We end the section with a somewhat more technical application of
Theorem 4.13; one which often will be useful when working with group rings. G,
For example, it allows us to compare C(@[G]),
when K
with C(K[G])
Lemma 4.17 K-algebra.
K
Let
for a finite group
is a splitting field.
be any number field, and let
Then for any finite extension L 3 K,
A
be a semistmple
the transfer map
trfK : C(L ®K A) -» C(A)
is
surjecttue.
is a Galois extension,
L/K
If
then
induced
the
eptmorphtsm
trfo : HO(Gal(L/K); C(L ®K A)) - * C(A)
is an isomorphism in odd torsion; and is an isomorphism in 2-power torsion if either
(t)
K
has no real embedding and Conjecture 4.14 holds for
each simple summand of
Proof
or
A,
Note first that
simple summand of
A
(ii)
trfK
When proving the surjectivity of
L®K A.
C(A)
21IC(A)I.
is a sum of transfer maps, one for each
thus suffices to consider the case where the description of
is simple and
A
is simple and
trfK,
K = Z(A).
it By
in Theorem 3.12, it then suffices to show that
trf : K2(La OR Ap)
is surjective for any prime
p
in
K,
K2(A,)
and any
q I p
in
L.
And this
follows since the following square commutes by Proposition 4.8(11):
K2(Lq)
trf
N K2(Kp) 14;c
ISPc
K2(Lq OR A
where the transfer for
Lq 7 Kp
K2(Ap);
is onto by Theorem 4.6, and the two maps
126
*c
THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 4.
are onto by Theorem 4.11.
Now assume that
L/K
is Galois, and set
It will suffice to show that
if
A
C = Gal(L/K)
is simple,
isomorphism in odd torsion, and an isomorphism if L ®K Z(A) _ II
Write
NZ(A)).
extension of
then
Z(A);
factors transitively. which leave
L ®K A
Hence,
invariant (so
L1
iLi,
if
then
21IC(A)I
where each
is an C(A)
(so
is a finite Calois
Li
C
and
Ii=lLi®Z(A)A
for short. trfo
permutes the
is the subgroup of elements
C1 C C
G1 = Gal(L1/Z(A))),
then
HO(G; C(L ®K A)) = HO(G1; C(L1 ®Z(A) A)).
In other words, we are reduced to the case where
K = Z(A)
(and
G = C1,
L = L1).
In particular,
LOK A
is now a simple algebra with center
L.
By
Theorem 4.13, there are isomorphisms
µL/Tl
oLOA : C(L ®K A)
where
I.
{±l}.
and
oA : C(A)
PVTO
Furthermore, as abstract groups,
1K =
HO(G; pL) ;
since for any group action on a finite cyclic group,
coinvariants is isomorphic to the group of invariants. range of trfo
trfo
are thus isomorphic (in odd torsion if
is onto, it must be an isomorphism.
0
the group of
The domain and
C(A) $ pK).
Since
Chapter 5 FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
The results in this chapter are a rather miscellaneous mixture. Their main common feature is that they all are simple applications of the congruence subgroup problem (Theorem 4.13) to study
of group rings;
C11
applications which do not require any of the tools of the later chapters.
In Section 5a, the group
computation of
is used to illustrate the
G = C4 X C2 X C2
as well
(= Z/2);
SK1(Z[G])
as
the procedures
for
constructing and detecting explicit matrices representing elements of Several vanishing results are then proven in Section 5b:
SK1(Z[G]).
example, that
C11(R[G]) = 1
whenever
G
is cyclic and
R
is the ring
of integers in an algebraic number field (Theorem 5.6), that if
G
C11(Z[G]) = 1
is any dihedral, quaternion, or symmetric group (Example 5.8 and
Theorem 5.4),
and
natural epimorphisms
is generated by
C11(R[G])
that
elementary subgroups of
G
C11(R[G]);
$'R(;: RC(G)
C(ID[G])
and
epimorphisms which are
In Section 5c, the "standard involution"
on Whitehead groups is defined; and is shown, identity on
induction from
These are all based on certain
(Theorem 5.3).
constructed in Proposition 5.2.
5a.
for
C11(Z[G])
for example,
for any finite group
Constructing and detecting elements in SK1(Z[G]):
to be the
C.
an example
We first focus attention on one particular group abelian sketch the procedures for computing
SK1(Z[G]) (= C11(Z[G])),
G;
and
for con-
structing an explicit matrix to represent its nontrivial element, and for detecting whether a given matrix does or does not vanish in
Example 5.1
where
IgI
=4
Set
and
G = C4 X C2 X C2.
1h1I = 1h21 = 2.
Let
Then
g,hl,h2 E G
SK1(Z[G]).
be generators,
SK1(Z[G]) - Z/2,
and is
128 CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
generated by the element 11 + 8(1-gz)(1+h1)(l+hz)(l-g)
-(1-gz)(1+h1)(1+hz)(3+g)
-13(1-g2)(1+h1)(1+h2)(3-g)
1 + 8(1-gz)(l+hl)(1+hz)(l+g) 1
E SK (7L[G]).
This will be shown in three steps.
Proof
will be carried out in Step 1.
SK1(7L[G])
Then, in Step 3,
the procedure for is described.
SK1(7L[G])
the matrix just constructed is used to illustrate the
procedure for lifting it back to SK1(7[G]).
it vanishes in
The actual computation of
In Step 2,
constructing an explicit nontrivial element in
1
C(@[G])
and determining whether or not
This is, of course, redundant in the present
situation, but since the construction and detection procedures are very different, it seems important to give an example of each.
Step 1
An easy check shows that
D[G]
splits as a product
Q[G] = Q8 X Q(i)4
By Theorem 4.13, determine
C(@) = 1
ImkG
For each )(rs(g) = i,
:
Ars
Krs. G -+ (i)
let
Krs(hl) = (-1)r,
characters identifies one of the Let
) C(D[G]) = ((i))4]
K2(Z2[G])
r,s E {0,1},
denote the summand of
We must first
C(@(i)) = (i) = 7114.
and
Krs(h2) _ (-1)s. D(i)-sunvnands of
Q[G]
denote the character:
Each of these four D[G]
with
D(i) C C.
mapped isomorphically under
Krs'
so that
Q[G] = @[G/(gz)] X Aoo X Aol X Alo X A11.
Recall that the isomorphism induced by the norm residue symbol.
a({i,u}) =
i(N(u)-1)/4
a:
C(ID(i)) -_- (i)
is
For the purposes here, the formula
(N(a+bi) = a 2 + b2)
(1)
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 5.
of Theorem 4.7(ii) will be the most useful.
which (using (1)) lists values for
Consider the following table,
@(i)-summands, for some
at the
p,(x)
x E K2(Z2[G]).
chosen symbols
x
a(Xoo(x))
a(Xol(x))
a(Xlo(x))
a(X11(x))
{g,l+(l+hl)g}
i
i
1
1
{g,l+(l+h2)g}
i
1
i
1
{g,1+(1+hlhz)g}
i
1
1
i
{-h2,1+(1+h1)g}
-1
1
1
1
A quick inspection shows that
To see that
a
129
Im(cp(,)
has index at most 2.
is not onto, we define a homomorphism
WG
C(M[G])
{fl};
a(x) _
a(Xrs(x))2 r,s
In other words, Ker(a)
a
sends each
by the above table.
C(Q(i)) = (i)
To see that
onto
{±1);
2
and
Ker(a) = Im(,pG),
recall first
that by Corollary 3.4,
K2(7L2[G]) = ({-1,u}, {g,u}, {h1,u}, {h2,u}
The symbols squares in
,pG({hi,u})
C(Q[G]),
pC({-1,u})
and
and lie in
Ker(a).
: u E
(2[G])*).
have order at most
2,
are thus
Also, for any u E (7L2[G])
a({g,u}) = a({i, I Xrs(u)})2 E {fl}. r,s
Let
R: 7L2[G] - 22[i][C2 xC2] be induced by 13(g) = i, and write 13(u) = a + bh1 + ch2 + dh1h2
A direct calculation now gives
(a,b,c,d E 22[i]).
* ,
130 CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
R Xrs(u) = (a+b+c+d)(a+b-c-d)(a-b+c-d)(a-b-c+d) r,s = (a2+b2-c2-d2)2 - (2ab-2cd)2 = 1 Formula (1) now applies to show that This finishes the proof that
(mod
422[']).
a({g,u}) = 1.
Im(*G) = Ker(a).
So by Theorem 3.15,
SK1(Z[G]) = Coker(,pG) = Z/2.
Step 2
N C Q[G]
Let
N - (Z) 13 X (Z[i])4.
be the maximal order.
I = (16Z)e x (8Z[i])' = (16.1, 2 is in fact contained in generated (over
Then
and
N 7 Z[G],
Under this identification, the M-ideal
to see this,
Z[G]:
S N
2
just note that
is
EI
by the twelve idempotents
Z[G])
1)(12)
and
i6.(l+g2)(1±g)(1±h' )(lfh2)
Consider the following homomorphisms:
SK1(Z[G],I) a , SK1(Z[G])
SK1(T,I) - SK1(Z,16)8 x SK1(Z[i],8)4.
Here,
is an isomorphism by Alperin et al [2, Theorem 1.3].
f
SK1(Z[G])
is generated by
generates one of the
A E GL(Z[i],8)
regarding
GL(Z[i],S)
6o f- (x),
SK1(Z[i],8)
an explicit generator of matrix
1
which So
can be found by first constructing a generates
[A]
as a summand of
To find a generator of
x E SK1(1,I)
factors and vanishes in the others.
SK1(Z[G])
such that
for any
By Step 1,
SK1(Z[i],S),
and then
GL(1,I) = GL(Z[G],I) C CL(Z[G]).
SK1(Z[i],8),
consider the epimorphisms
CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM 8 s
(12[i]) --s K2(Z[i]/8)
(recall
SK1(Z[i])
that
K2(a2[i]),
=
and hence also
+: St(Z[i]/8) -" E(Z[i]/8)
SK1(Z[i],8)
By
(1)
above,
are generated by the symbol
K2(Z[i]/8),
{i,1+2i} = [+ 1(diag(i,l,i-1))
where
SK1(Z[i]+8)
by Theorem 4.16(11)).
1
131
+ 1(diag(1+2i,(1+2i)-1,1))];
,
is the canonical surjection.
Hence
is generated by the conmutator
8({i,1+2i}) = [diag(i,l,i-1), diag(M,1,M)] = [( 0), M] E CL(Z[i],8);
when
is any
M E GL2(7L[i])
(Recall that
approximation to
mod 8
by Theorem 1.13.)
diag(M,1,M 1) E E(Z[i])
To find
M,
diag(1+2i,(1+2i)-1).
we could take the usual decomposition _1 u12' e12. e211
diag(u,u 1) = e12* e21
u
then replace
by
u
and
1+2i
(1+2i)-1, and multiply it out.
e12 E E2(R),
*
1
that it is easier to use trial and error.
M = (
can be used; and shows that
1
65 - 641
1+2i
8
approximation to
mod 8 7L[i]
is small enough
For example,
8 13(1-2i)
SK1(7L[i],8)
A - (00 1 0 )(1+8i
by any
However, the ring
is generated by the matrix
i 0)(13(-821) 13(81-2i))( 0
1+2i/
-8(3+i)
E SL2(Z[i],8). -104(3-i)
(2)
65 + 641
Under the inclusion of
Q(i)
now lifts to the generator
as the simple summand
Aoo
of
@[G],
A
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
132 CHAPTER 5.
11 + 8(1-g2)(l+h1)(1+h2)(1-g)
-(1-g2)(l+hi)(1+h2)(3+g)
-13(1-g2)(1+h1)(1+h2)(3-g)
1 + 8(1-g2)(l+h,)(1+h2)(1+g)
E SK (7L[G]).
Step 3
1
We now reverse the process, and demonstrate how to detect
whether or not a given matrix vanishes in
We have seen in
SK1(Z[G]).
Step 1 that the two epimorphisms
C11(Z[G]) «-- C(D[G]) a » {+l)
So the idea is to first lift the matrix to an
have the same kernel.
X E C(D[G])
element
using Proposition 3.13, and then compute
a(X)
using the formula for the tame symbol in Theorem 4.7(i).
A = (a
Consider the matrix above.
Write
n = 130:
pin.
Aoo = Q(i)
the product of the primes at which
invertible.
a E B*
Q[G] = Aoo x B,
E SL2(7L[G])
d) where
Write
(a = 1
7L[n][G] = 2loo x B
in
where
By Proposition 3.13,
[A] = 8(X),
is as in Step 1.
Xoo(c) = -104(3-i) Boo C Aoo
c E (2[00)* = (7L[n][i])*,
B),
constructed in Step 2
and
and
Set
is not
Then
B C B.
for
a E (7L p[G])*
where
X = {Xoo(a),Xoo(c)} = (65-64i, -104(3-i))
E Im[ ®K2((Aoo)p) S ® K((Aoo)p)
p jn
proj),
C(Aoo) C C(Q[G])].
p
It remains to show that torsion only, and at odd primes
a(X) = -1. pJ130.
We are interested in 2-power
Hence, we can use the formula
(u,v)p = ((_l)P(u)P(v).up(v)/vp(u))(N(p)-1)/4 E (i) C (7[i]/p)*
(Theorem 4.7(i)) N(p) = I7L[i]/pl
(u,v)p = 1
if
for
and u
and
each prime p(-)
v
ideal
plpjn
in
denotes the p-adic valuation. are both units
mod
p.
7L[i]:
(3)
where
In particular,
Since
N(65-64i) = 8321 = 53.157,
we are left with only these two primes to consider.
Both split in
7L[i];
CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
and a direct computation shows that
133
is divisible only by the
65 - 641
prime ideals
pi = (7-2i):
i = 30,
c = -104(3-i) = -1
in
7[i]/p! = F53
P2 = (11-6i):
i = 28,
c = -104(3-i) = 88
in
7[i]/p2 = IF157
Formula (3), and the definition of
a
in Step 1, are now used to compute
a(X) = [(65-64i,-104(3-i))pj.(65-64i,-104(3-i))p2]2
= 153/.(
571
0
(+1).(-1) = -1.
=
The above method for computing
is not very
C C(Q[G])
practical for large groups; and much of the rest of the book (Chapters 9 and 13, in particular) is devoted to finding more effective ways of doing this.
is known, however, the construction and detection
Once
procedures in Steps 2 and 3 above can be directly applied to for an arbitrary finite abelian group
Note in particular that any
G.
can be reduced using elementary operations to a 2x2 matrix
M E GL(7[G]) a b c d
SK1(7L[G])
with
ad - be = 1
Also, when
(see Bass [1, Proposition 11.2]).
constructing matrices, it is most convenient to take as ideal
(in
I C Lt
Step 2) the conductor
I = {xET : x1RC7L[G]} (i. e., the largest
JR-ideal contained in
Theorem 1.3] applies to show that iR
Z[G]).
Then Alperin et al [2,
SK1(7L[G],I) = SK1(Lt,I).
I
and
D[G],
and
Also,
both factor as products, one for each simple component of
the rest of the procedures are carried out exactly as above.
When
G
is nonabelian,
elements is similar.
be isomorphic to I2
the procedure for constructing explicit
The main difference is that
SK1(H,I);
SK1(7L[G],I)
need not
so it might be necessary to replace
I
by
(see Lemma 2.4); or to use the description in Bass et al [1, Theorem
4.1 and Corollary 4.3]
to determine whether
SK1(IR.I)
is large enough.
134 CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
Theorem 4.11 can then be used to represent elements of symbols, which are lifted to
exactly as above.
SK1(bt,I)
The procedure for detecting a given
lies in
[A]
[A] E C11(Z[G])
but also why it lies there.
C11(7[G]),
A
One way of doing this (sometimes) is to first replace
A' a 1
such that
(mod E(Z[C]))
is much harder
The main problem is that one must
in general in the nonabelian case. know, not only that
by
K 2('1/1
where
(mod 12);
denotes the conductor from the maximal order
by some
A' a A again
I C 7L[G]
This is probably the
IN.
hardest part of the procedure - the descriptions of
SK, (2 [G])
in
p
Chapters 8 and 12 are unfortunately too indirect to be of much use for this - but Z[G]/12 is after all a finite ring. Then A' can be split up and analyzed in the individual components, and in most cases reduced to
elements in
SL2(R,I)
for some ring of integers
R.
For some nonabelian groups, there are alternate ways of detecting elements in
Examples of such techniques can be extracted from
C11(7L[G]).
the proofs of Propositions 16, 17, and 18 in Oliver [1].
5b.
and the complex representation ring
C11(R[G])
By Theorem 4.13, for any number field C(K[G])
K
and any finite group
G,
is isomorphic to a product of roots of unity in certain field
components of the center this description how
Z(D[G]).
C(K[G])
However, it is not always clear from
acts with respect to,
for example, group
homomorphisms and transfer maps.
One way of doing this is to use the
complex representation ring
for "bookkeeping" in C(K[G]).
RC(G)
Throughout this section, all number fields will be assumed to be subfields of
C.
In particular, for any number field
identified as a subgroup of
splitting field for rings over
K).
G
RC(G);
and
RK(G) = RC(G)
(i. e., whenever
For any number field
K
norm residue symbol defines an isomorphism 4.13). if
We fix a generator
IpKI = n
and
K
cK E C(K)
K[G]
K,
RK(G)
whenever
can be
K
is a
is a product of matrix
with no real embeddings, the
aK: C(K) _ pK C C*
by setting
has no real embeddings, and
(Theorem
cK = cK = 1
otherwise.
By
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 5.
L/K
Theorem 4.6, for any pair
135
of number fields,
trfK(cL) = cK E C(K).
G
If
is any finite group,
then we regard
as a subgroup of
C(K)
C(K[C]) - the subgroup corresponding to the sunenand
K and
usual way.
consider
G,
In particular, K[G]-module
dimensional
V,
of
cK as an element of
trivial action - and in this way regard For fixed
K
multiplication by
with
C(K[G]).
RK(G)-module in the
as an
C(K[G])
K[G]
[V],
for any finite
is the endomorphism induced by the functor
V ®K : K[G]-mmd ) K[G]-mMd.
Alternatively, in terms of Proposition 1.18, multiplication by induced by the
(K[G],K[G])-bimodule V ®K K[G],
ture is induced by setting
R C K
Similarly, if
necessarily projective)
GO(R[G])
Cl1(R[G])
into a G0(R[G])-module; where
There are surjections
R[G]-modules.
and
K0(K[G]) = RK(G)
(K-
C(K[G]) a a Cll(R[C]);
GO(R[G])-linear by the description of
is
this way,
C11(R[G])
Now, if
K
by setting
is a splitting field for
'KG(v) =
L 2 K
a
in Theorem 3.12.
In
can, in fact, be regarded as an RK(G)-module.
'+KG
and if
g,h,k E G and v E V.
for
is the Grothendieck group on all finitely generated (but not
G0(R[G])
6
is
is the ring of integers, then tensor product
R by R[G]-modules makes
over
and
gv®ghk
[V]
where the bimodule struc-
we define a homomorphism
G,
: RC(G) = RK(G) --' C(K[G])
v E RK(G).
is a splitting field for
YKG = trf o fLC : RC(G)
G,
If
K
we let
) C(L[G])
and C are arbitrary, EKG be the composite
trf ) C(K[G]).
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
136 CHAPTER S.
Finally, if
is the ring of integers, we write
R. C K
6RGoEKG
$RG =
YRG will be given in Lemma 5.9(ii).
A more explicit formula for
R C K
K
Fix a number field
Proposition 5.2 let
: RC(G) - C11(R[G])
be the ring of integers.
and a finite group
Then
and.
$KG
defined, independently of the choice of splitting field.
G,
and
are well
$RG
Furthermore:
JKG and YRG are both surjecttue.
(L)
(ii) for any number field
L D K
with ring of integers
S C L,
the
following two triangles commute:
RC(G)
I
C(L[G]) trf
(1)
and
RC(G)
7
(la)
For any
trf
\ 1\
C(K[G])
(iii)
C11(S[G])
C11(R[G]).
and any group homomorphism
H C G
f: G' -' G,
the
following diagrams commute:
RC(G') --L- Cll(R[G'I)
RC(G') - ' C(K[C']) f*
RC(G)
t
1Res
IC(f)
(2)
-- C(K[G])
(3)
If*
and
(2a)
I1(1 )
RC(G) --I- Cl,(R[C]) IRes
Itrf
(3a)
' .. Cl1(R[H]) It1.1
RC(H)
(iv) &G
RC(H)
C(K[H])
: RA(G)
are both RK(G)-linear.
> C(K[C])
and
.
$RG : RC(G) --+ C11(R[G])
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 5.
If
(u)
Proof
K C It,
137
R.(G) C Ker(&G).
then
8RG
By Theorem 3.15, the boundary maps
are surjective, and
So it
are natural with respect to all of the induced maps used above. suffices to prove the claims for
Note first that for any
: C(L[G]) ) C(K[G])
trfKG
is
RK(G)-linear
the transfer homomorphism
L D K,
(RK(G) C RL(G)).
In other words,
x E C(L[G])).
(v E RK(G),
This amounts to showing, for any
K[G]-module
(4)
the commutativity of
V,
the following square
[L®KV]. C(L[C])
C(L[G]) ItrfKG
ItrfKG
[V] C(K[G])
C(K[G]).
This in turn follows from Proposition 1.18, since each side is induced by (K[G],L[G])-bimodule
the
V ®K L[G]
multiplication on both factors, and
L[G]
(where
acts
K[G]
by
left
by right multiplication on the
second factor).
L ? K
If
RL(G).
are both splitting fields for
So using (4), for any
trfLG(f
KG LG (v)) =
KG
G,
then
v E RC(G),
L
KG L
arbitrary
for any splitting field
L ? K.
K ? K.
&G(v)-
K
In other words, triangle (1) commutes in this case. $KG = trfKGojKG
RC(G) = RK(G) _
But by definition,
and so (1) commutes for
This proves (ii), and also shows that
defined, independently of the choice of splitting field.
YKG
is well
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
138 CHAPTER 5.
To prove the surjectivity of field for
EKG'
Then for each simple summand A
G.
V
tensoring by
of
V,
A-m$d;
and hence an isomorphism from
RL(G)-module structure on C(L[G])
trfKG
fLC
is onto.
Also,
with irreducible
L[G]
C(L)
to
L-MW
to
In other words,
C(A).
restricts to an isomorphism
C(L[G]);
RC(G) 0 C(L) = RL(G) 0 C(L) C RL(G) 0 C(L[G])
and so
be a splitting
is a Morita equivalence from
module
the
L 2 K
again let
trf G is onto by Lemma 4.17, and so
EKG =
is onto.
o $LG
We next check point
(iii).
Using the commutativity of
suffices to show that squares (2) and (3) commute when K field for
G',
G,
and
(1),
it
is a splitting
This amounts to showing that the following
H.
diagram commutes:
ResH
RK(f) R __K(G') (5a)
C(K[G'])
For any
) itRK(H)
' RK(G)
(5b)
C(f)
C(K[G])
K[G']-representation
V,
trf
(5)
I.cK , C(K[H]).
Proposition 1.18 again applies to show
that
V ®K,-(-K)
RK(f)([V])'cK = [K[G] ®K[G,]
= [K[G] ®K[G,]]* c [V ®K]*(cK) = C(f)([VI-cK).
So (5a) commutes, and the proof for (5b) is similar.
To prove (iv), w E RK(G)
and
let
L Q K
v E RL(G) = RC(G).
be a splitting field for Then by definition of
w'$'LC(v);
and so
JKG
is
RK(G)-linear.
G.
(and (4)),
Fix
CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
Finally, assume
is such that
K C IR
be a splitting field for
Then
K.
and let
RK(G) = R1k(G),
(Theorem 4.13), so
C(K) = 1
139
L
K
cK = 1;
and using (4):
G(cL) =
YKG(FIR(G)) = trfKG(RK(G).cL) =
R.(G) C Ker(fKG)
Thus,
in this case; and the commutativity of (1) allows
us to extend this to arbitrary K C R.
0
YZG: RC(G) --» C(l[G])
These strong naturality properties of the make
1.
into an excellent bookkeeping device for comparing, for example,
C(®[G]) H C G.
or
with
C11(7L[G])
C(l[H])
or
for subgroups
C11(7[H])
The next few results present some applications of this, and more
will be seen in later chapters. For any prime Cn x v,
where
n
p,
a p-elementary group is a finite group of the form
is a p-group.
According to Brauer's induction theorem
(see Serre [2, §10, Theorems 18 and 19], or Theorem 11.2 below), for any finite group from
G,
elementary
is generated by elements which are induced up
RC(G) subgroups
- i.
G
of
p-elementary for some prime
e.,
subgroups
p - and for each prime
generated by induction from p-elementary subgroups.
p,
which are
RC(G) (p)
is
So Proposition 5.2
has as an immediate corollary:
Theorem 5.3
Let
R
Then for any finite group
be the ring of integers in any number field G,
C(K[G])
induction from elementary subgroups of and of
Cl1(R[G])(p) G.
and G.
C11(R[G])
K.
are generated by
For each prime
p,
Cp(K[G])
are generated by induction from p-elementary subgroups
0
The naturality properties of used to show that
SK1(7[C])
fRG
in Proposition 5.2 can also be
vanishes in many concrete cases.
We start
with a very simple result, one which also could be shown directly using Theorem 4.13(1).
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
140 CHAPTER 5.
Theorem 5.4
G
Let
of matrix rings over
Proof
be a finite group such that
Then
R.
C11(7L[G]) = 1.
RC(G) = R.(G).
By hypothesis,
is a product
IR[G]
By Proposition 5.2(v),
RR(G) C Ker[RC(G) - * C(D[G])];
and so
a
C(D[G]) = C11(7L[G]) = 1.
Note in particular that
Theorem 5.4 applies to elementary abelian
Sn
2-groups, to all dihedral groups, and to any symmetric group is a product of matrix algebras over 2.1.12]).
@:
see James & Kerber [1, Theorem
This result will be sharpened in Theorem 14.1, with the help of
later results about
SK1(2 p[G])
and
Wh'(G).
We next consider cyclic groups, and show that R
(Q[Sn]
is the ring of integers in any number field.
information about
K2(Rp[Cn])
SK1(R[Cn]) = 1
when
Clearly, to do this, some
is needed, and this is provided by the
following technical lemma.
Lemma 5.5 let
R C F
p
Fix a prime
and a finite extension
be the ring of integers.
F
of
Qp,
and
G,
the
Then for any cyclic p-group
transfer homomorphism
trfRG : K2(R[G])
) K2(R)
is surjectiue.
Proof
E D F
Let
ring of integers.
Then
particular, if assumption that
Sten 1
Let
o trfRGRC, trf RR o trfSS = trf RG R
This shows that
by Theorem 4.6.
pk = IGI, T
Spk+1 ,p
p g R
S C E
be any finite extension, and let
trfRG
be the
trf RR is onto
and
is onto if trfSG
is.
In
it will suffice to prove the lemma under the l/ p
k-1
E F.
be the maximal ideal, and let
v:
F* -1 7
be
CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
Fix a primitive
the valuation.
the ramification index;
i. e.,
p-th root of unity
Let
C.
141
e = v(p)
By assumption,
pR = pe.
e > e(o(cpk+1)) = Pk(P-1)
Choose any x
be
(1)
such that
e/(Pk-l(P-1))
- 1 > 0,
v(x) =
(2)
k
u = 1- xP
and set
E R*
v(x2pk)
v(px2pk-1)
v(PXpk)
2 k
k-1
k
k-1
- pe/(p-1) = v(P(l-C));
Z
and so x , pxp , px
(i+xP
From (1) and (2) we get inequalities
(x E p).
E p(1-c)R. k-1
k
)p
It follows that
-1-(xk-1)P
k (-p)1/p-1.
where
x and
xk-1 =
v(xk-1) =
k-1
(mod
e/pk-1 +v(x)
p(1-C)R)
= e/pk-2(p-1) - 1.
Upon repeating this procedure, we get sequences
x = xk , xk_l , ... , x0 E R
and
u = uk
' uk-l ' ... , uO E R*;
where for each 0 < i < k-1,
v(xi) =
xi = (-P)1/p 'xi+1'
e/(Pi-1(p-1))
ui = (1 + (xi+l)P')P.ui+1 = 1 - (x,)1
In particular,
uo - 1-xo
(mod
p(1-C)R),
is a p-th power, then there exists
uo
(mod
Then
v(y) < e/(p-1),
= v(yP).
But
p4'v(xo)
In other words,
so
v(py) >
and
(3)
p(1-C)R).
and
y E p
uo - 1-x0 = (1+y)p = l+py+...+yp
- 1;
p4v(xo) < v(p(1-c)).
If
such that
(mod
v(yp).
P(1-c)R
=
pe/(p-1)).
It follows that
v(xo)
by (3), and this is a contradiction. uo
is not a p-th power in
F.
The same argument
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
142 CHAPTER 5.
shows that
is not a p-th power in any unramified extension of
uo
since the valuations remain unchanged. over
So
F(uo
F,
is ramified
) = F(ul/p)
F.
Step 2 integers.
Now set
Since
E/F
by Proposition 1.8(ii).
For any
* ),
if
(,) p
denotes the
(v,u) p #i by definition.
then
(cp),
theorem (Theorem 4.4),
g E G
Furthermore, if
(E*) - Z/p
v E R* -N (S E/F
norm residue symbol with values in
by Moore's
be the ring of
is ramified,
R*/NE/F(S*) = F*/N
Hence,
S C E
and let
E = F(u1/p),
generates
{v,u}
K2(R).
is any generator, then k
{v,u} = {V,l-xpk} = SV, tl ll
(1-
1X)
i=1
= Sv,trfRG(l-gx)I = trfRG({v,l-gx});
and so
trfR
:
K2(R[G]) -* K2(R)
For any cyclic p-group
and any
Proposition 5.2(iv) can be used to make quotient ring of the local ring it
suffices
to
o
is onto.
G,
RC(G)(p).
find any element
(Theorem 3.1(v))
R 9 K
such that
K
splits
C11(R[G]) = SK1(R[G]) So to show that
x E Ker[jtiRG: RC(G)(p)
which is not contained in the unique maximal ideal of
G,
into a
SK1(R[G])=1,
-4
SK1(R[G])]
RC(G)(p).
This is
the idea behind the proof of the following theorem.
Theorem 5.6
Let
R
be the ring of integers in any number field
Then, for any finite cyclic group
Proof
Set
G = Cn,
any finite extension SK1(R[G])
Cn,
for short.
L D K,
K.
SK1(R[Cn]) = 1.
If
S
is the ring of integers in
then the transfer map from
is surjective by Proposition 5.2(i,ii).
SK1(S[G])
to
It therefore suffices
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
CHAPTER 5.
G
Assume first that
is a p-group for some prime
1trf2
j trfl
Theorem 4.13
Z/pe[G*],
by Example
then 1.12.
Ker[8: Cp(K[G])
is an ideal in Vootrf1
-s
K[Cm] = Km
an inclusion
by
Z/pe
n
n = IGI
with
is
(RC(G) = RK(G)),
and so
But
-
p(K[G])]
Cp(K[G]) = Im(v)+ Ker(trf2),
SK1(R[G])(p) = Coker(q,) = 1.
Fix a prime
is arbitrary.
is p-torsion free.
Write
E K by assumption);
R[Cm] C Rm
Cp(K[G])
JRG: RC(G) - SK1(R[G])
SK1(R[G])(P)] = Im[w: K2(RP[G])
is surjective,
SK1(R[G])
If we identify
since
Cp(K[G]) = Zlpe[G*].
Now assume that
Then
and
is contained in the unique maximal
Ker(trf2) Also,
RC(G)-linear by Proposition 5.2(iv)
show that
Z/pe[e]
is a splitting field by assumption); and the right-hand
(K
square commutes by Proposition 5.2(iii). the ring
by Theorem
Coker(wo) = Cl (R) = 1
and K induce isomorphisms on
fKG
4.16(ii),
RC(l)/PQ = Ups.
is onto by Lemma 5.5,
trf1
(GM = Hom(G,C*))
1Res
o p(K)
0
K2C (Rp)
since
pe =
Cp(K[G]) A-' R_((;)/PQ = LPQ[C ]
K (RP[G])
ideal
Set
p.
Consider the following commutative diagram:
'(PK)P 1.
Here,
when R contains the n-th roots of unity.
SK1(R[G]) = 1
to prove that
143
of index prime to
n =
pin;
we will
where
pjm.
and by Theorem 1.4(v) there is p.
So by Corollary 3.10,
m SK1(R[G])(P)
If
C
= SK1(R[Cm X Cpk]) = ®SK1(R[Cpk]) = 1.
is a finite group, and if
Sp(G)
cyclic, then any p-elementary subgroup of can be combined with Theorem 5.6 to give:
G
0
(the p-Sylow subgroup) is is cyclic.
So Theorem 5.3
144 CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
Corollary 5.7
R
Let
be the ring of integers in any number field.
Then for any finite group
G,
C11(R[G])(p) = 1.
O
cyclic,
By RG'
Theorem
and any prime
together
5.6,
RC(G) -+ Cll(R[G])
with
p
such that
naturality
the
in Proposition 5.2,
Sp(G)
properties
is
of
factors through the
$RG
complex Artin cokernel
AC(G) =
Coker[ ® RC(H)
Ind) RC(G)]
= R C(G)/(
HCG cyclic
In fact, for any fixed for
R
I
lnd (R (H))).
HCG cyclic
$RG: AC(G) - Cl1(R[G])
G,
is an isomorphism
large enough (see Oliver [7, Theorem 5.4]);
represents the "upper bound" on the size of
C11(R[G])
so that
as
AC(G)
R varies.
The next example deals with some other familiar classes of finite groups, and illustrates the use of the Artin cokernel to get upper bounds on the order of
C11(R[G]).
Example 5.8
G
Let
be any finite dihedral, quaternionic, or semi-
R
dihedral group (not necessarily of 2-power order), and let
of integers in any number field C11(R[G]) = 1
Proof
K.
Then
1C11(R[G])j
if either R has a real embedding, or if
Gab
be the ring
C
E Z(Q[G]),
vopri(x) = a'Tr A(lox) = vF,
E C.
But for any g E G, x(') = XV(g); and so v o pri(E) = v o pri(x).
n
The above results will now be applied to describe the involution on C(f[G])
and
torsion in
C11(Z[G]).
C11(Z[G])
involution acts on
K
in
K
in Chapters 9 and 13.
(Bak [1]) For any finite group
Theorem 5.12
generally, if
This will be important when describing the odd
and
C(@[G])
via the identity.
Cll(Z[G])
is an algebraic number field such that
for all primes
and if
phIGI,
R C K
is unramtfted
Cll(R[G]),
and on
be the set of all primes if
K = Q,
pIIGI.
for
For convenience, let
Proof
p
and the set of primes
pj1GI
has p-torsion only for
9f
otherwise.
p E
Let
So by Theorem 3.15
and Proposition
Ff1K = ID
and F'
is trivial
p E 1.
A
be a simple summand of
a1[G],
and set
and
cn = exp(2iri/n).
So the assumption on
under any embeddings into
A' = K®IDA
C.
Hence,
is simple (Theorem 1.1).
has the same p-th power roots of unity as
F = Z(A).
F C D(cn),
Brauer's splitting theorem (Theorem 1.5(11)), exp(G)
C11(R[G])
By Theorem 4.16(111),
5.11(i), it suffices to show that the involution on C(K[G]) for all
More
is the ring of integers,
then the standard involution acts via the identity on Cp(K[G])
the standard
G,
K
F' = KO F Q
where
n =
implies that
is a field
Furthermore, for any F.
Then by
p E 9$,
152 CHAPTER 5.
FIRST APPLICATIONS OF THE CONGRUENCE SUBGROUP PROBLEM
p E ', and let (p,) p be the groups of p-th power roots of
Fix unity in
F
(and
nothing to prove).
C(A') # 1
Assume that
F').
(otherwise there is
Let
a : Cp(A') = Coker[K2(A') --> ®K2(Aq)](P) q
be the isomorphism (p-locally) of Theorem 4.13.
Then for each (finite) prime
q
in
(PF)P
Set
µ = (AF)p = (p'F')p'
and for any
F',
a E (F')*
and any
a({a,u}) = (a,nrA'/F'(u))µ,F'' q
By Proposition 5.11(ii), invariant, and acts on
the involution leaves
µ = (iF,)p
via
(and hence
F
(f I+ f-1).
F')
So by Lemma 5.10,
a({a,u}) = a({a,u}-1) =
(by naturality:
(a,
_
1 for f E }i)
= a({a,u}). Thus,
{a,u} = {a,u}
in
bols, the involution on
Cp(A).
Since
is trivial.
Cp(A)
is generated by such sym-
Cp(A)
o
Note that Theorem 5.12 does not hold for arbitrary
the above restrictions, standard involution on invariant.
negative
it
K[G]
is
easy
equivariant with respect is
R[G].
Without
construct examples where
the
does not even leave all simple components
It is not hard to show that
involution on C(K)
to
(-1),
to
JKG: RC(G) -» C(K[G]) the
involution
(note
is always that
the
by Lemma 5.10(1)).
In Chapter 7 (Corollary 7.5), we will see that the involution also acts via the identity on Wh'(G)
(= Wh(G)/SK1(Z[G]))
for any finite
G.
Chapter 6 THE INTEGRAL P-ADIC LOGARITHM
In Chapter 2, p-adic logarithms were used to get information about
the structure of
for a g-order
Kl(21)
U.
When
there is also an "integral p-adic logarithm":
is a group ring,
21
defined by composing the
usual p-adic logarithm with a linear endomorphism to make it integral valued.
This yields a simple additive description of
p-group
G
for any
(Theorems 6.6 and 6.7); and in later chapters will play a key
role in studying Integral
groups
Ki(Zp[G])
and
Wh'(G)
as well as
SKl(7L[G]),
Ki(2p[G])
itself.
logarithms have also been important when studying class for finite
D(7L[G]) C K0(7L[G])
G.
One example of this is Martin
Taylor's proof in [2] of the Frohlich conjecture, which identifies the [R] E KO(Z[G]),
class L,
when R
is the ring of integers in a number field
is a tamely ramified Galois extension, and
L/K
Another application is the logarithmic description of
G = Gal(L/K).
D(Z[G]),
when
G
is a p-group and p a regular prime, in Oliver & Taylor [1]. Throughout this chapter,
p
will denote a fixed prime.
working with group rings of the form and
R
R[G],
G
where
is a finite group
is the ring of integers in a finite extension
from Example 1.12 that if
G
F
of
p g R
is a p-group, and if
We will be
Recall
is the maximal
ideal, then
J(R[G]) = (p,r(l-g): g E G, r E R) = {Irigi E R[G]
6a.
:
Jri E p}.
The integral logarithm for p-adic group rings
In Theorem 2.8, a logarithm homomorphism
log,: Kl(2I,I) --> Q®Z (1/[21, 1]) was constructed, for any ideal
I
in a 2 p order
H.
When applying this
154
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
to a p-adic group ring extension
F
R[G],
(for any ideal
I C R[G]),
and
I/[R[G],I] - HO(G;I)
and
where the homology in both cases is taken with
the conjugation action of
to
HO(G;F[G])
is the ring of integers in any
it is convenient to identify
of
F[G]/[F[G],F[G]] - HO(G;F[C])
respect
R
where
HO(G;R[G])
G
on
In particular,
F[G].
can be regarded as the free
R-modules with basis the set of conjugacy classes of elements of
Mostly, we will be working with
Q;
over
so that
R C F
for which
is a field and
R/pR
extension
F
Finite group
G.
is unramified
F
V E Gal(F/&) - the
p(r)E rp (mod pR)
for any
r E R.
(Compare M. Taylor [1, Section 1] and Oliver [2,
Definition 6.1 Section 2]).
and
Gal((R/pR)/ffP).
Hence, in this case, there is a unique generator Frobentus automorphism - such that
F-
Let
R
of
@p.
be the ring of integers to any ftntte unramifted Define
4: HO(G;F[G]) --> HO(C;F[G]),
for any
by setting
G,
k
k
aigi) i=1
p(ai)gi.
(ai E F,
gi E G)
i=1
Define
r = rRG
by setting
:
K1(R[G]) - HO(G;F[G])
r(u) = log(u) -
for
u E K1(R[G]).
To help motivate this construction, consider the case where
G
abelian. Then 4 is a ring endomorphism, and so r([u]) = log(u) -
for
u E 1+J(R[G])
(J = Jacobson radical).
But
up a $(u) (mod pR[C]),
is
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
155
log(up/$(u)) E log(1+pR[G]) C pR[C],
and so
r([u]) E R[G].
The proof that
rRG
also is integral valued in the nonabelian case
is more complicated.
Theorem 6.2
R
Let
unramifted extension F
be
ring of
the
in some
integers
Then for any finite group
of
finite
G,
rRG(Kl(R[G])) S HO(G;R[G])
The map
r
is natural with respect to maps induced by group homomorphisms
and Galois automorphisms of
F.
finite and unramtfied over i),
For any G and any extension K/F
(both
if SK and R C F denote the rings of
integers, then the following squares commute: r
K1(R[G])
HO(G;R[G])
incl
incl
K1(S[G])
HO(C;S[G])
Proof
Let
J = J(R[G])
K1(S[G]) r HO(G;S[G1) Tr
trf
K1(R[G]) r - HO(G;R[G])
denote the Jacobson radical.
For any x E J,
3
2
-I -L.[xpk - D(xk)]
HO(G;R[C])).
(mod
k =l
So it suffices to show that primes other than
pkl[xpk - 4(xk)]
p are invertible in
R)
for all
k;
or (since all
that
4,(xpn-1)]
pnl[xpn -
(in HO(G;R[G]))
for all
Write x = jrigi,
set
n > 1
and all
q = pn,
x E R[G].
and consider a typical term in
xq
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
156
ril...riq.gi1...giq.
Z/pn
Let
of p -t
act by cyclically permuting the conjugate terms, where
leaving each term invariant.
pt
so that we get a total
gi's,
is the number of cyclic permutations
is a pt-th power, and the
Then q
sum of the conjugate terms has the form
1-t -t
t pt
pn-t.Tp g_
E H0(G;R[G])
pn-t ri,
(r =
n gi,). j=l J
S =
,
j=1
If t = 0, then this is a multiple of pn. t-1 t-i
t > 0, then there is a n-1
If
in the expansion of
corresponding term
xp
It
remains only to show that
pn-t rp t ,g
But
t
= pn-t.41(rp t-1.
pn-tP(rp t-1 ) =
t
(mod
).
pn).
since pI[rp-'V(r)]
pt-1
pt I[r pt
t-1
gp
Naturality with respect to group homomorphisms is immediate from the
definitions, and naturality with respect to Galois automorphisms holds since they all commute with the Frobenius automorphism
Gal(F/&)
is cyclic since
F
is unramified).
commutes with the inclusion maps since %JR =
ip
S D R,
If
(note that then
r
vR'
To see naturality with respect to the trace and transfer maps, first
note that
4'
commutes with the trace (since it commutes with Galois It suffices therefore to show that
automorphisms). For
s
E
Tr(s)
S,
multiplication by
s
= TrS/R(s) as an
is
the
trace
logotrf = Trolog. of
the matrix for
R-linear map (see Reiner [1, Section la]).
Hence, for any x E S[G] and any n > 0, log(trf(l + pnx)) = log(1 +
Tr(log(l + pnx))
(mod
p2n-1).
k
For any u E 1+J(R[G]),
up
E 1+pR[G]
for some k
(u
has p-power order
k+n in
(R/p[G]) *).
Then up
E 1+pn+1R[G]
for all
n > 0,
and so if
n k:
CHAPTER 6.
THE INTEGRAL P-ADIC LOGARITHM
157
k+n n-k. log(trf(up
log(trf(u)) = p
=_
))
pk-n,Tr(log(upk+n)) = Tr(log(u))
Since this holds for any n > k,
In fact,
"RG
(mod
p2n+l/pn+k = pn-k+l).
the congruence is an equality.
13
is also natural with respect to transfer homomor-
phisms for inclusions of groups, although in this case the corresponding restriction map on H0(G;R[G])
This will be shown,
is much less obvious.
for p-groups at least, in Theorem 6.8 below.
The next lemma collects some miscellaneous relations which will be needed.
Lemma 6.3
(t)
For any group C and any any element
(tt)
Let
K
P(1-g)2Z[G])
(mod
(l-g)P = (l-gp) - p(1-g)
g E C.
Then the
be any finite field of p-power order.
sequence
0,IFp is exact, where
Proof
(i)
Tr
-ino1)K 1p K
Tr
>ffp -->0
(1)
denotes the trace map.
Just note that
gP = [1 - (1-g)]P = 1 - P(1-g) + (-1)P(1-g)P
1 - p0-g) -
(ii)
Ker(1-gyp) = Fp
P(l-g)21[G])
since
Tr o (1-p) = 0
the
image
of
rG
(or
to p-groups. of
its
generates
'p
A counting argument then shows that (1) is exact.
Attention will now be restricted identifying
(mod
The trace map is onto by Proposition 1.8(iii),
by definition of the trace, and Gal(K/Fp).
(1-g)P
13
Both here,
restrictions
to
when
certain
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
158
subgroups),
and
later
when
studying
SK1(R[G]),
techniques is to work inductively by comparing for
z E Z(G)
central of order
a commutator - i. e.,
with
K1(R[G])
for some
main
the
of
K1(R[G/z])
In particular, the case where
p.
z = [g,h]
one
z
is
(as opposed to a
g,h E C
product of such elements) plays a key role when doing this.
The reason
for this is (in part) seen in the next proposition.
Proposition 6.4
extension
F
let
of
be the ring of integers in any finite
R
Let
be the maximal ideal, and let
pCR
T
denote the composite
T : R -»R/p Tr ,Fp. G
Then for any p-group
and any central element
of order
z E G
p,
there is an exact sequence
1 -, (z) ---, K1(R[G],(l-z)R[G])
where
w((1-z)lrigi) = T(jri)
lg. H0(G;(1-z)R[G]) W-,
ffp -4 0;
ri E R and gi E G.
for any
(1)
If F/& is
unramified, and if we set
H0(G;(l-z)R[G]) = Im[HO(G;(1-z)R[G]) - HO(G;R[G])]
= Ker[H0(C;R[C])
then
rRG(1+(1-z)f) = log(1+(1-z)f)
-+ in
HO(G/z;R[G/z])];
HO(G;(1-z)R[C])
for all
f ER[G]
and
[HO(G;(1-z)R[G])
Proof
Set
:
I = (1-z)R[G],
the Jacobson radical.
is a commutator
z
1
if
p
otherwise.
rG(l+(1-z)R[G])] =
for short, and let
Note that
J = J(R[G])
(1-z)p E p(1-z)R[G]
denote
by Lemma 6.3(1).
So Theorem 2.8 applies to show that the p-adic logarithm induces a homomorphism
log'
and an isomorphism
log
Ii ,
which sit in the following
CHAPTER 6.
THE INTEGRAL P-ADIC LOGARITHM
159
commutative diagram with exact rows:
lloglJ
1-zRG l--a
RG
Kl(R[G],(1-z)J) --> Kl(R[G],(1-z)R[G])
Kl ( (1-z )J
(1-z)J
'
logo
Ilogl
0 -> HO(G;(1-z)J) --> HO(G;(1-z)R[G]) ) HO G ,
(2)
(l-z)
-0.
Also, by Theorem 1.15 and Example 1.12, there are isomorphisms (1z)RrG
(1z
Kl((R z)J '
where
- R[G]/J =- R/p =- HO(G .
-z)JG)
a(l+(1-z)g) = f
);
f E R[G]/J.
for
Now consider the following diagram I
log ) HO(G;(l-z)R[G])
K1(R[G],(1-z)R[G])
a"
I
0 --1 Fp ` ' R/p where
p
a'(1+(1-z)lrigi) = Iri
denotes the reduction of
and
r ER.
a'((1-z)lrigi) = jr-,.
->0
Here,
F E R/p
The bottom row is exact by Lemma 6.3(ii);
and square (3) commutes since for r E R and g E G,
a"(Log(l+(1-z)rg)) = a'((1-z)(rg-rpgp)) = r-p(r) E R/p. Then (3) is a pullback square by diagram (2), and so isomorphic kernel and cokernel. w = Tr o a", If
and since
F/Op
a'
maps
is unramified (so
log(1+(1-z)g) _ (1-z)i
for some
I'G(1+(1-z)f) = log(1+(1-z)f)
:
and
1-p have
The exactness of (1) now follows since (z)
isomorphically to
FP = Ker(l-ip).
TG
is defined), then for any
17,
and
E R[G],
4((l-z)ti) = (1-zp)$(ij) = 0.
So
By the exactness of (1),
in this case.
1
[HO(G;(1-z)R[G])
logI
if
(1-z)g = 0 E HO(C;R[G])
some g E C
rG(l+(1-z)R[G])] =
p
otherwise.
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
160
In other words, the index is for some
if and only if
g,
The next lemma,
g
if and only if
1
z = [h,g]
for some
is conjugate to
zg
0
h,g E G.
on the existence of central commutators, will be
needed to apply Proposition 6.4.
Lemma 6.5
C
Let
normal subgroup generated by commutators in commutator
C
of order
z E Z(G)
p.
Fix any commutator
Proof
Since
C
Then
G.
be a nontrivial
H
p.
xo E H-1.
go E C not commuting with
If
is not central, then
xo
x, = [xo,go] E HW1.
and set
xo,
is nilpotent, this procedure can be continued, setting
Xk = [g,h] has order
=
until
0
p.
The main result of this chapter can now be shown.
It gives a very
simple description of the image of the integral logarithm on
Theorem 6.6
Fix a p-group
C,
of Q with ring of integers R C F.
:
x;
xk E H-1 is central for some k > 0. Then, if n-1 n-1 and n for some n > 1, = [g,hp ] and has order p xpk
[x;_1,g;_1] E HW1,
w _ WRC
contains a
In particular, any nonabelian p-group
contains a central commutator of order
choose any
H a C
be a p-group, and let
H0(G;R[G]) --) (E) x Gab
K1(R[G]).
and a finite unramifted extension Set
F
e = (-1)p-1, and define
by
w(Jaigi)
=
11(egi)Tr(a,)
Then the sequence
1 -4 K1(R[G])/torsion
T) HO(G;R[G])
(E) xCab ---> 1
(1)
is exact. Proof
Assume first that
Log(1 + pR) = pR p = 2,
then
if
p
G = 1,
is odd, and
Log(1+2r) a 2(r-r2) a
It follows that
the trivial group.
Log(1 +4R) = 4R (mod
if
4R)
By Theorem 2.8, p = 2.
for any
Also, if r E R.
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
161
p
if
Log(R*) = Log(1+ pR)
PR
is odd
=
p = 2.
Furthermore, since
Log(1+pR)
is c-invariant,
'-Log(e)
r(R*) _ ( 1 -
( ( 1
-
pic i(r)). i=1
So
and (1) is exact in this case.
Im(F) = Ker(w),
Now assume that WGoFG = 1;
G
We first show that
is a nontrivial p-group.
it suffices by naturality to do this when
R = 2p.
Let
I = {jrigi E R[G] : Zri = 0}
For any
u = l+Zri(1-ai)gi E 1+I, where riEZp
G
is abelian and
denote the augmentation ideal.
up = 1 + pZri(l-ai)gi + lrp(1-ai)pgi
(mod
1 + plri(1-ai)gi + Zri[(1-ap) - p(1-ai)]gi
(Lemma 6.3(i))
4(u) + pjri(1-ai)(gi-gp) a 'D(u)
(p(ri) = ri)
This shows that
up/4(u) E 1+p12,
and hence that
F(u) = log(u) - 1.0(log(u)) = I.log(up/4(u)) E
On the other hand, for any
r E 2p
and any
T(1+I) c I2 C Ker((j),
12.
a,b,g E G,
(Eg)r(.ag)-r(ebg)-r(eabg)r
w(r(1-a)(1-b)g) =
Thus,
pI2)
= 1 E (e) x Cab.
and so
r(K1(R[G])) = r(R*x (1+I)) = (I'(R*) , T(1+I)) C Ker((j). Now fix some central element is a commutator if
G
z E Z(G)
of order
is nonabelian (Lemma 6.5).
inductively that the theorem holds for
d,
Set
p,
(3) such that
G = C/z,
z
assume
and consider the following
THE INTEGRAL P-ADIC LOGARITHM
162
CHAPTER 6.
diagram (where
a: G -* G denotes the projection): 1
1
1
1
1
1
1 -i K1(R[G],(1-z)R[G])/tore -r2-+ 9O(G;(1-z)R[G])
I
r
1 -+ K1(R[G])/tors
Ker(aab)
+ 1
1
C
G HO(G;R[G])
C
. (_).dab
1
laab I' K1(R[G]))/tors G - HO(G;R[G])
Since
K(a)
wo
is clearly onto,
Im(To) C Ker(wo)
To
The
Also, the top row is
injective by Proposition 6.4,
is
by (3); and using Proposition 6.4 again: z
is a commutator
1
if
p
otherwise (i. e., if
} = ICoker(To)I
IKer(aao)I =
Since
, (e),Gab . 1 j
is onto (Theorem 1.14(111)), the columns are all exact.
bottom row is exact by the induction hypothesis. exact:
-C
(iGorG = 1
G
is abelian) J
by (3), the middle row is exact by the 3x3 lemma.
One simple application of Theorem 6.6 is to the following question of Wall, which arises when computing surgery groups.
2-group, and set
problem is to describe the cohomology group 7/2
Let
G be an arbitrary
Wh'(22[G]) = K1(7l2[C])/({tl}x dab x SK1(22[G])).
acts via the standard involution
Assuming Theorem 7.3 below,
Wh'(22[G])
H1(Z/2;Wh'(7L2[G])),
(g N g 1)
The
where
(see Section 5c).
is torsion free, and so the exact
sequence of Theorem 6.6 takes the form
1 ---1 Wh' (22[G]) - HO(G;7L2[G]) -- {tl} xGab _ 1; with the obvious involution on each term.
Also,
H1(7L/2;HO(G;7L2[G])) = 1,
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
since the involution permutes a a2 basis of
163
We thus get an
HO(G;Z2[G]).
exact sequence
HO(7L/2;HO(G;Z2[G])) 0 HO(7L/2;{tl}xCab)
H1(Z/2;Wh'(12[G])) -i 1;
and this yields the simple formula
{[g]EGab H1 (l//2;Wh'(72[G]))
:
[g2]=l}
([g]: g conjugate g-1)
Theorem 6.6 gives a very simple description of and the torsion subgroup of chapter (Theorem 7.3).
will be identified in the next
K1(2p[G])
This suffices for many applications; for example,
to prove the results on
C11(7[G])
and
in Chapters 8 and 9
SK1(7L[G])
But sometimes, a description of
below.
K1(2p[G])/torsion,
Ki(R[G])
(= K1(R[G])/SK1(R[G]))
The following version of the
up to extension only is not sufficient.
logarithmic exact sequence helps take care of this problem.
Theorem 6.7 Let
extension F
of
Qp.
R
be the ring of integers in any finite unramifted
For any p-group
(v,A): Ki(R[G])
((j,O): HO(C;R[C])
by setting, for
G,
define
) (Gab ® R) W (R/2)
and
(Gab ® R) ® (R/2)
gi E C and a, ai E R
(with reductions
(u,e)((1+pa)(l+ l al(gl-l))) = (I gi® ai, a),
a, ai E R/2),
and
(u,0)(Zaigi) = (I gi®ai, 1a1).
Then
v,
B,
and
r
are all well defined on
Ki(R[G]),
and the sequence
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
164
t u
1 - K1(R[G])(p)
8
HO(G;R[G]) ®
® R) ® (R/2)
0
10 (gyp-1)
(wA
(Gab
1®
0
(Gab
® R) ® (R/2) --> 0
is exact.
The main step is to show that the composite of the above two
Proof
The injectivity
homomorphisms is zero, and this is a direct calculation.
is a consequence of Theorem 7.3 below, which says that
of
Gab.
Ker(rRG) = tors(Ki(R[G])) = tors(R*) x
The exactness of the whole sequence then follows easily from Theorem 6.6. See Oliver [8, Theorem 1.2] for more details.
G
When
o
is an arbitrary finite group, then
TRG
sits in exact
sequences analogous to, but more complicated than, those in Theorems 6.6 and 6.7.
Since their construction depends on induction theory, we wait
until Chapter 12 (Theorem 12.9) to state them. Several naturality properties for
r
were shown in Theorem 6.2.
One
more property, describing its behavior with respect to transfer maps for To state this, we define, for
inclusions of groups, is also often useful.
any prime
p and any pair
ResH
as follows.
Fix
:
g E G,
H C G
of p-groups, a homomorphism
HO(G;7lp[G]) - HO(H;gp[H])
let
x1,...,xk
be double coset representatives
for H\G/(g), and set ni = min In >0 : gn E xi1Hxi} for 1 < i K k. Then define ResH(g) _
xignlxil E H0(H;R[H]) i=1
For example, if
G and H are p-groups and
[C:H] = p,
then
CHAPTER 6.
THE INTEGRAL P-ADIC LOGARITHM
165
p L-1
ResH(g) =
xlgx 1 i=O
(any
x E G'H)
if
if g f H .
gp Theorem 6.8 For any pair H C C
of
p-groups, the diagram
rG
Ki(Lp[G]) --=- HO(G;Zp[G]) trf
Ki(2p[H])
Here,
as subgroups of
(6)
10C
N WG
e = (-1)p-l, and
(a) x
jib
1
(e) x e are identified
(a) x Gab and Ki(R[H]),
-' 1
c
HO(H;Zp[H])
Ki(R[G])
x Gab
IResH
(1) rH
commutes.
g E H
and RH
is the restriction of
the transfer map.
Proof
The easiest way to prove the commutativity of (1) is to split
it up into two squares:
Kl(2p[G]) Itrf
Kill p[H])
Here, if
log (la)
HO(G;QP[G]) 1ResH
(lb)
Is
log ) HO(H;Q,[H])
1-
1
p
)
al,.... am denote right coset representatives for
res(g) = I{aigail for any
p
HO(G;[G])
g E G.
H C G,
1 S i m, aigail E H) E
The commutativity of (lb)
is straightforward, and the
commutativity of (la) follows from the relations res(upn
log(u) = lim -L. (up - 1);
n-w pn
then
trf(u) = lim (1+
- 1))l/'pn .
n-
See Oliver & Taylor [1, Theorem 1.4] for details.
0
166
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
ResH
In Oliver & Taylor [1, Theorem 1.4],
H C G
arbitrary pair
is in fact defined for an
above formulas can also be extended to include the case of
R
Z p
unramified over
R[G],
for any
in this case the Frobenius automorphism for
;
appears in the formula for
6b.
The
of finite groups (not only for p-groups).
R
ResH.
Variants of the integral logarithm
We list here, mostly without proof,
some useful variants of
the
The
integral logarithm, and of the exact sequence describing its image. first theorem is a generalization of Proposition 6.4 and Theorem 6.6.
It
is used in Oliver [4] to detect elements in K2(R[G]).
Theorem 6.9 Let
extension F
of
R be the ring of integers in any ftntte unramtfled
Qp,
a: G ---i C be any surjectton of p-groups, and
let
set Ia = Ker[R[G]
Set
3
R[G]].
Then there is an exact sequence
K = Ker(a) C G.
r H0(G;Ia) a K/[G,K] -i 1.
K1(R[G],Ia)
Here, for any
Proof
r E R,
g E G,
and w E K,
wa(r(l-w)g) = w
r(r).
This is an easy consequence of results in Oliver [4]; but
since it was not stated explicitly there we sketch the proof here.
a p-group d and a Z-order
21
to be the pullbacks
b,
Pt G
+G
Ia
1132
a
a
G
2[
lba
R[G] IRa
R[G] R R[G].
Define
CHAPTER 6.
Set
I;
= Ker[R(3i
THE INTEGRAL P-ADIC LOGARITHM
167
: R[G] 0 R[G]] (i = 1,2), and let p: R[G] -4 21
be
the obvious projection. Then Ker(,y) = I1 fl I2 = III, (see Oliver [4, Lemma 2.4]);
so
21
!--' R[d]/1112
and
K1(21) - K1(R[G])/(l+I1I2)
(1)
Also, by Oliver [4, Theorem 1.1] (and this is the difficult point):
rRG(l + I1I2) = Im[I1I2 - HO(G;R[G])]
together with the exact sequence of Theorem 6.6
Formulas (1) and (2), (applied to
R[d]),
now combine to give an exact sequence
21
K1(21)
(where
e = (-1)p-l,
(2)
(e)xGab
HO(G;21)
as usual).
We thus get a commutative diagram
-> (e) x C
H0(G;21)
K1(21)
Ib2)
IKl2)
1
rC) HO(G;R[G])
-- -
1
132b
R
(E)
xG^ab
-+ 1,
where the vertical maps are split surjective (split by the diagonal map
G)GCGxG). Then Ker(K1(b2)) = K1(R[6],Ia),
Ker(H(b2)) n-' HO(G;Ia),
and this proves the theorem.
O
Logarithms can be used to study, (R[G])*,
but also its center.
to Theorem 6.6.
Ker((3zb) = KI[G,K]
not only the abelianization of
The following theorem is in a sense dual
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
168
Theorem 6.10
R
Let
unramified extension
F
be
Q.
of
the
ring of
integers in any finite
Then for any p-group
G,
there is an
exact sequence
1 -+ (e) x Z(G)
lncl 1+ J(Z(R[G])) r Z(R[G]) ' (e) x Z(G) --+ 1.
Proof See Oliver [9].
o
The last result described here involves polynomial extensions of the base ring.
Theorem 6.11 polynomial algebra
Let
7[s]p- denote
7L[s].
For any p-group
the p-adic completion of C,
Let
I C 7L[s]p[G]
the
denote
the augmentation ideal, and define
Ki(R[S]p[G]+I) = Im[K1(Z[s]p[G],I)
-'
K1(Z[S]p[p][G])].
Then there is a short exact sequence
1 ---* Ki(7L[s]p[G].I)
T
HO(G;I) ® (Z[s]p®Gab) u 4-1
Proof
See Milgram & Oliver [1].
those in Theorem 6.7 above.
6c.
(7L[s]p ® Gab) - 1.
The homomorphisms are analogous to
0
Logarithms defined on K( p[G])
The "logarithm" homomorphisms discussed here are not needed describing the odd torsion in
C11(7L[G]),
for
but they could be important in
describing the 2-power torsion, and do help to motivate the conjectures in Chapter
9.
description of
In any
case,
K2(7Lp[G])
Theorem 6.12 below does give a complete when
p
is any prime and
G
is an abelian
p-group; and Conjecture 6.13 would give an analogous description (though
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
169
only up to extension, in general) for arbitrary p-groups.
The natural target group for Connes' cyclic homology group
G
K2
integral logarithms turns out to be
HC1(-).
If
is any commutative ring and
is any finite group, then
HC1(R[G]) = H1(G;R[G])/(g®rg: g E G, r E R)
where
C on
We identify
R[G].
is abelian; and for arbitrary
g®rh E H1(G;R[G])
Theorem 6.12
and
HC0(R[G]) = HO(G;R[G]);
is as usual defined with respect to the conjugation
Hn(C;R[G])
action of G
R
C
for any abeltan p-group
R C F
denote the trace map.
Then,
F2.
of
Q,
HCl(R[G]).
g®h+h®g -' O
and such that
G,
F
there is a short exact sequence
G,
1 -> K2(R[G])/{±G,±G} which is natural in
Tr: R - 2p
g,h E G.
and let
Fix an unramified extension Let
whenever
this allows us to define elements
r E R and any commuting pair
for any
be the ring of integers.
with G ® R[G]
H1(G;R[G])
r2
and w2
(1)
satisfy the following
two formulas:
(t)
For any g E G and any u E (R[G])*,
(ii) For any
Proof
g,h E G and any
r E R,
F2({g,u}) = g ®F(u).
G2(g ® ah) =
See Oliver [6, Theorems 3.7 and 3.9].
Other explicit formulas for example, formulas for [6, Theorem 4.3]).
F2({a,u})
Also,
F2
F2
when
O g 1h.
o
are also given in Oliver [6]:
a E R* and
u E (R[G])*
for
(Oliver
has been shown (Oliver [6, Theorem 4.8]) to
be natural with respect to transfer maps.
The obvious hope now is that similar natural exact sequences exist for nonabelian p-groups.
Some more definitions are needed before a
precise conjecture can be stated.
For an arbitrary group
G,
Dennis has defined an abelian group
170
THE INTEGRAL P-ADIC LOGARITHM
CHAPTER 6.
H2(G),
which sits in a short exact sequence
0>7V2®Gab -->H2(G) ---- H2(G)--->0, and such that H2(G) = (G ® G)/(g®h+h®g) whenever easiest way to define
for nonabelian G
H2(G)
H2(G)
C
is abelian.
The
is as the pullback
(Gab ®Gab)/(gOh+h®g)
I
I
H2(G) ---> H2(Gab) = (Gab ® Gab)/(gog) For any commuting pair denote the images of ((g,h)
g,h E C,
and
g^h E H2(G)
g®h E H2((g,h))
and g®h E H2((g,h)),
G.
respectively
Loday [1] has defined a natural homomorphism
XG
:
H2(G) _+ K2(7L[G])/{-1,G},
which will be considered in more detail in Section 13b. note that
AG(g^h) = {g,h}
{g,g} = {-l,g}). Qp,
will
being an abelian group).
For any
of
g^h E H2(C)
If
R
for any commuting pair
For now, we just
g,h E G
(recall that
is the ring of integers in any finite extension
then for the purposes here we set
Wh2(R[G]) = Coker[H2(G) ' + K2(7L[G])/{-1,iG)
Note that when G
is abelian,
Z'-4R) K2(R[G])/{-l,:G}].
Wh2(R[G]) = K2(R[G])/{±G,EG}.
The obvious conjecture is now:
Conjecture 6.13 of
P
For any p-group
with ring of integers
R,
G,
and any unramtfted extension F
there is an exact sequence
CHAPTER 6.
THE INTEGRAL P-ADIC LOGARITHM T
HC2(R[G])
(03
H3(G)
r? ,.
- Wh2(R[C])
W2
HC1(R[G])
Wh(R[G]) - HCO(R[G])
natural in
171
H2(G)
H1(G) --' 0,
and satisfying the formulas:
G,
(t)
r2({g,u}) = g®fRH(u)
if gEG,
(ii)
tu2(g ® rh) =
lh
H=CG(g),
and uER[H]*
for commuting g,h E G.
Note that the existence and exactness of the last half of the above
sequence follows as a consequence of Theorems 6.6, 7.3, and 8.6 (and is
included here only to show how it connects with the first half).
For
by Theorem 8.6.
See
example,
Coker(cw2) = H2(G)/H2b(G) = SKl(R[G])
Oliver [6, Conjectures 0.1 and 5.1] for some more detailed conjectures.
The results in Oliver [4] also help to motivate Conjecture 6.13.
In
particular, by Oliver [4, Theorem 3.6], there is an exact sequence
H3(G) ---> Wh2(R[G])
where
Wh2(R[C])
2) IH(R[G])
is a certain quotient of
- 4 H2(G);
Wh2(R[G]),
and where
°i(R[G]) = H1(G;R[G])/(g®rgn: gEG, r E R, n>1) (recall that
HC1(R[G]) = H1(G;R[G])/(g®rg)).
conjectured contribution of
H3(G)
to
This helps to motivate the
Ker(r2),
at least defined to this quotient group
N!(R[G])
and shows that of
HC1(R[G]).
r2
is
This
sequence can also be combined with Theorem 6.12 to prove Conjecture 6.13
for some nonabelian groups, including some cases - such as where
ii # 1.
G = Q(8) -
But presumably completely different methods will be needed
to do this in general.
PART II: GROUP RINGS OF P-GROUPS
We are now ready to study the more detailed structure of for finite
G.
For various reasons, both the results themselves (e. g.,
the formulas for
C11(Z[G])
as well as the methods
SK1(2p[G])),
and
used to obtain them, are simplest when G
is a p-group.
some of the induction proofs, it is important that is a local ring.
2p [G]
(Theorem 6.6),
when G
K1(Z[G])
C
For example, in
is nilpotent and
Also, the image of the integral logarithm
and the structure of
(Theorem 9.1),
@[G]
rG
are simpler
is a p-group.
The central chapters, Chapters 8 and 9, deal with the computations of SK1(2 p[G])
and
C11(Z[G]),
Theorem 8.6, where
The most important results are
respectively.
is described in terms of
SK1(7Lp[G])
Theorems 9.5 and 9.6, where formulas for
C11(Z[G])
H2(G);
are derived.
and Some
examples are also worked out at the end of each of these chapters.
Chapter 7 SK1(Z[G]) G.
is
centered around Wall's
theorem
is the full torsion subgroup of
Wh(G)
In contrast, the torsion free part of
10, mostly using logarithmic methods. arbitrary elements of
Wh'(G)
Wh(G)
that
is studied in Chapter
Also, the problem of representing
(= Wh(G)/SK1(Z[G]))
discussed at the end of Chapter 10.
(Theorem 7.4)
for any finite group
by units in
Z[G]
is
Note that Chapters 7 and 10, while
dealing predominantly with p-groups, are not completely limited to this case.
These four chapters are mostly independent of each other.
The main
exception is Theorem 7.1 (and Corollary 7.2), which give upper bounds on the torsion in
Wh(7Lp[G])
for a p-group
in Chapter 7 when showing that
G.
Wh'(2 p[G])
These are used, both later
is torsion free, and in
Section 8b when establishing upper bounds on the size of
SK1(2 p[G]).
Chanter 7
If
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
G
finite extension K1(R[G])
of
F
or
Ill
is the ring of integers in any
then obvious torsion elements in
Qp,
include roots of unity in
SKI(R[L]).
R
is any finite group, and if
elements of
F,
and elements in
G,
These elements generate a subgroup of the form
µFx Gab x SK1(R[G]) C- K1(R[C])
To see that this is, in fact, a subgroup, note that
pF x Gab C K1(R[G])
µF,x Gab
injects into
(F[Gab])* = K1(F[Gab]) - and hence
K1(F[G]) - since it is a subgroup of
that
(1)
(pF, x Gab) fl SK1(R[G]) = 1.
and
In particular, if we define the Whitehead group Wh(R[G])
by setting
Wh(R[G]) = Kl(R[G])/(pF x Gab),
then
SKI(R[L])
can also be regarded as a subgroup of
Wh(R[G]),
and
Wh'(R[C]) = Wh(R[G])/SK1(R[G]) = Ki(R[G])/(UF x Gab).
Note that when R g 7L,
this notation is far from standard (sometimes one
divides out by all units in
When G
that the only torsion in In particular,
R).
is abelian, then
Wh'(G)
Ki(7L[G]) = (7L[G])*;
(7L[G])*
is given by the units
is torsion free in this case.
motivation for Wall [1] to show that finite group torsion Wh'(R[G])
G;
subgroup
i.
of
and Higman [1] showed
e.,
Wh'(G)
tg
for
g E G.
This provided the
is torsion free for any
that the subgroup in (1) above is the full
KI(7L[G]).
is torsion free whenever
More
generally,
showed
is a number field and
F
finite (Theorem 7.4 below), or whenever
Wall
F
G
that is
is a finite extension of 1
CHAPTER 7.
174
and G
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
is a p-group (Theorem 7.3)
If one only is interested in the results on torsion in
Wh'(R[G]),
then Theorem 7.1 and Corollary 7.2 below can be skipped.
These are
directed towards showing that
R
is torsion free when
Wh(R[G])
ring of integers in a finite extension of
a normal abelian subgroup of
index
Wall's proof
p.
is the
is a p-group with in
[1]
that
is torsion free in this situation is simpler than the proof
Wh'(R[G])
But the additional information in Corollary
given here in Corollary 7.2.
7.2 (and in Theorem 7.1 as well) about
Wh(R[C])
Chapter 8 to get upper bounds on the size of
itself will be needed in
SK1(R[G]).
Recall the exact sequence of Proposition 6.4:
&,
integers in any finite extension of z E G
G
and
Qp,
is central of order
1 -> (z)
p,
if
G
if
R
is the ring of
is any p-group, and if
then there is an exact sequence
) K1(R[G],(1-z)R[G])
loo -> H0(G;(l-z)R[G])
)
ffp -+ 0.
In particular, this gives a precise description of the torsion subgroup of K1(R[G],(1-z)R[G]).
The next theorem gives an upper bound for the number
of those torsion elements which survive in
Fix a prime
Theorem 7.1 central of order
p.
p
K1(R[G]).
and a p-group
G,
- be the equivalence relation on 0
if
g ^- h
r g
is conjugate to
R
h E G),
generated by:
or
for any
i
prime to
p.
is the ring of integers in any finite extension of %,
Ker[tors Wh(R[C])
where
h,
some
!l [g,h] = zl
Then, if
z C C
Set
0 = {g E C : g conjugate zg} = {g E G : [g,h] = z,
and let
and let
- tors
Wh(R[C/z])] = (Ulp)N,
be
CHAPTER 7.
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
N=0
if 0 = 0
N < 10/ - 1
if
More precisely, if
0 0 0
0 $ 0,
(t. e., if
be as in Proposition 6.4, and fix any are
Tr+ Ip
r E R
Ker[tors Wh(R[G])
Proof
with
T(r) g 0.
--equivalence class representatives in
--> tors
By Proposition 6.4,
Then if
the elements
0,
(for 1 0,
[g lh,l-r(g-h)k] = 1 - (r(z ig-z ih)k - r(g-h)
1 + (1-z
(g-h)k + r2(g-h)2k + r3(g-h)3k + ...)
1 -
Since
poi,
(mod (1-z)2).
(3) shows that
kr(g-h) k + kr2(g-h)2k + kr3(g-h)3k + ... E C'
for any
k > 1.
For
k
large enough,
r(g-h) k E
C C'
by (a).
A
downwards induction on k now shows that
for all
r(g-h) k E C'
(when
plk
particular,
this holds since
C'
r(g-h) E C' n R(0) = C.
k > 0
contains all p-th powers).
In
CHAPTER 7.
178
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
To prove (c), fix
This proves (d).
such that
j
and
[hj,g] = z,
consider the comnutator
[hj,l-r(1-g)] = 1 - r((1-zg) -
1 - (1-Z) (rg + r2g(1-g) + r3g(1-g)2 + ...).
By (3),
rg - r2g(1-g) + r3 g(1-g)2 - ... E C'.
By (d),
rkge = rkh = rkg
(mod
C)
k and any
for any
(4)
a
prime to
p;
and so (4) reduces to give
0 E rg + (-1)p-lrpg(1-g)p-l = (r-rp)g + rpgp = (r-rp)g
It follows that (r-rp)g E C' fl R(fl) = C.
(mod
C').
0
Later, in Section 8b, Theorem 7.1 will play a key role when obtaining
upper bounds for the size of
SK1(R[G]).
But for now, its main interest
lies in the following corollary.
Fix a prime
Corollary 7.2
p,
in any finite extension of
Let
abeltan normal subgroup H a G is torsion free.
Proof choose
z E Hf1Z(G)
is torsion free.
of order
G
R
be the ring of integers
be any p-group which contains an
such that
In particular,
This is clear if
and let
G/H
Then Wh(R[G])
is cyclic.
SK1(R[G]) = 1.
G = 1.
Otherwise, we may assume
p, and assume inductively that
H # 1,
Wh(R[G/z])
Define
0= {gEG : [g,h] = z, some hEG}, and let
- be the equivalence relation on
fl
defined in Theorem 7.1.
Theorem 7.1, we will be done upon showing that
If fl # 0, G/H.
Choose
then f ix any
h E 11
such that
g E fl,
and any
[g,h) = z.
-
is transitive on
xEG-H
Either
ghi
By
0.
which generates or
gih
lies in
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
179
(G/H being cyclic); we may assume by symmetry that
ghi =
CHAPTER 7.
H
for some
a E H.
i
If we write h = bxj
for some
b E H,
then
z = [g,h] = [ghi,h] = [a,bxf] = [a,x3] = [ax,xj]
= [ax,xj(ax)-3] = xi(ax)-3
the last step since
g '- h in
0;
It follows that
E H.
-
ghi
'x
(ax)- j
= a - xf " a x ~ xi
and hence that the relation is transitive.
We are now ready to describe the torsion in
o
Wh(R[G])
in the p-adic
case.
Theorem 7.3
(Wall [1])
Fix a prime
integers in any finite extension Wh'(R[G])
is torsion free.
F
of
p,
and let
4lp.
R be the ring of
Then for any p-group
G,
In other words,
tors(K1(R[G])) = ILF x Gab x SK1(R[G]);
where
µF, C R*
Proof
If
is the group of roots of unity in
G
F.
is abelian, then the theorem holds by Corollary 7.2.
So the result is equivalent to showing, for arbitrary
G,
that
pr* : tors Ki(R[G]) --> tors K1(R[Gab])
is injective on torsion.
Fix
G,
and assume inductively that the theorem holds for all of its
proper subgroups and quotients.
If
G
is cyclic, dihedral, quaternionic,
or semidihedral, then the theorem holds by Corollary 7.2. simple summands of
F[G]
Otherwise, all
are detected by restriction to proper subgroups
and projection to proper quotients (see Roquette [1], Oliver & Taylor [1, Proposition 2.5], or Theorem 9.1 below).
In other words, the restriction
maps and quotient maps define a monomorphism
CHAPTER 7.
180
ZResH $ ZProj
GIN
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
: K1(F[G]) >-->
® K1(F[H]) ® ® K1(F[G/N]) HC_G
NAG
[G:H]=p
INI=p
So the corresponding homomorphism for
For any
H C G
of index
Ki(R[G])
(1)
is also injective.
consider the following commutative
p,
diagram:
tors Ki(R[G]) p> tors Ki(R[G/[H,H]]) >pr,, tors K1(R[Gab])
It2
Iti tors Ki(R[H]) . pea
Here, the pre
t;
Ki(R[Hab]).
tors
are transfer maps and the
pr;
are induced by projection;
is injective by the induction assumption, and
pr3
by Corollary 7.2
contains an abelian subgroup of index
p).
Hence, for any
(G/[H,H]
u E Ker(pr3oprl),
tl(u) = 1 E Ki(R[H]).
Thus, for any p.
.
Also,
u E Ker(pr*),
Proj(u) = 1
hypothesis again);
and so
for all u = 1
TrfH(u) = 1 N A G
for all
p
of order
Ki(R[G])
of index
(by the induction
by (1).
Note that Theorem 7.3 only holds for p-groups. the torsion in
H C G
Formulas describing
in the non-p-group case are given in Theorems
12.5 and 12.9 below.
In order to prove the corresponding theorem for global group rings (in particular, for
Wh(G)),
some induction theory is needed.
For this
reason, the next theorem might technically fit better after Chapter 11, but organizationally it seems more appropriate to include it here.
Theorem 7.4 free. K,
(Wall [1]) For any finite group
More generally, if
and if
R
G,
Wh'(G)
is torsion
is the ring of integers in any number field
pK C R* denotes the group of roots of unity in
tors(K1(R[G])) = pK x
Gab
x SK1(R[G]).
K,
then
(1)
CHAPTER 7.
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
Fix a prime
Proof
The proof of (1) for p-power torsion will be
p.
carried out in three steps: p-hyperelementary, p g R,
ideal
and when
181
G
when
C
is a p-group, when
C
is
Note that for any prime
is arbitrary.
the completion homomorphisms
K1(F[G]) - K1(Fp[G]),
Ki(R[G]) >----% Ki(Rp[G])
are injective (the reduced norm maps are injective by Theorem 2.3).
Assume G
Step 1
pK x
For any prime p1p
is a p-group.
in
R,
Gab C tors Ki(R[G]) -i tors Ki(Rp[G]) = µ(Kp)x
Since the inclusion
Ki(R[G]) >---> Ki(Rp[G])
G
Assume
(R* -- (Rp)*)
tors Ki(R[G]) = w K-
is p-hyperelementary - i. e.,
G
normal cyclic subgroup of p-power index - but not a p-group. q A p
prime
dividing
IGI,
contains a Fix some
H a G be the q-Sylow subgroup.
and let
may assume inductively that the theorem holds for Let
as
Gab.
a direct summand, this shows that
Step 2
contains
Gab.
q g; R be any prime ideal dividing
We
G/H.
and set
q,
I = Ker[Rq[G] ) Rq[G/H]].
Then
I
is a radical ideal, since q D Zq
follows from Example 1.12).
Hence
and
K1(Rq[G],I)
H
is a q-group (this
is a pro-q-group (Theorem
2.10(11)), and so
torspKj(Rq[G]) = torspKj(Rq[G/H])
(p 0 q).
But
torspKj(R[G/H])
hypothesis, and so
=
(}
x (C/H)ab)(p)
by
the
induction
torspKi(R[G]) = (µK x Gab)(py
Step 3 By standard induction theory (see Lam [1, Chapter 4], Bass [2,
Chapter XI],
or Theorem 11.2 below),
for any finite group
G,
CHAPTER 7.
182
torspKl(R[G]) So
THE TORSION SUBGROUP OF WHITEHEAD GROUPS
is generated by induction from p-hyperelementary subgroups.
torspKj(R[G])
= (WK
x Gab)(p)
by Step 2.
0
As an easy consequence of Theorem 7.4, we now get:
Corollary 7.5
(Wall)
involution acts on Wh'(G)
Proof R. C F.
each
Fi
Write
For any finite group
G,
the standard
by the identity.
Z(D[G]) = HFi,
where the
denote the rings of integers.
are fields; and let
F.
The involution on
D[G]
acts on
via complex conjugation (Proposition 5.11(11)), and the reduced
norm homomorphism
nrZ[G] : K1(Z[G]) --' R(R1)-
commutes with the involutions by Lemma 5.10(11). SK1(Z[G])
is finite; and for each
conjugation. involution.
So O
Wh'(G)
i,
Also,
(Rj)*/torsion
= K1(Z[G])/torsion
Ker(nrZ[G]) =
is fixed by complex
is also fixed by the
Chapter 8 THE P-ADIC QUOTIENT OF SK1(Z[C]):
The
central
result
of
this
chapter
P-GROUPS
is
the
construction
of
an
isomorphism
0G : SK1(2p[G]) = SK1(7L[G])/C11(7z[G]) -'' H2(G)/12b(G)
for any prime
p
and any p-group
C.
Here,
is the
H2b(G) C H2(G)
subgroup generated by elements induced up from abelian subgroups of In fact,
in Theorem 8.7, we will
whenever
R
see that
is the ring of integers in any finite extension of
In Section 8a, the homomorphisms
The definition of
be surjective. SK1(R[G])
0RC
are constructed, and shown to
involves lifting elements of
for some appropriate
K1(R[G]),
to
ORG
C
surjecting onto
and then taking their integral logarithms (see Proposition 8.4).
8b then deals mostly with the proof that Section
G.
SK1(R[C]) = H2(C)/H2b(G)
some
8c,
SK1(7Lp[G]) = 1,
examples
are
given,
0RG both
Section
is an isomorphism. of
groups
and of groups for which it is nonvanishing.
C;
for
In
which
The last
result, Theorem 8.13, gives one way of constructing explicit nonvanishing elements of
SK1(2 p[C])
in certain cases.
Throughout this chapter,
8a.
p
will denote a fixed prime.
Detection of elements
The following proposition is the basis for detecting all elements in SK1(2 p[G]).
Proposition 8.1
extension
F
of Qp.
Let
R
be the ring of integers in any unramtfted
Then, for any extension
1 - K -- G
C -4 1
CHAPTER 8.
184
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
of p-groups,
Kfl [G,G]
Coker[SK1(Ra) : SK1(R[G]) -' SK,(R[G])] =
More precisely, u E KI(R[G]),
z. K;
and u
for any then
TRG(u) = Jri(zi-1)gi
Tr: R - 2p
Proof
for some
g1E G,
u
to
ri E R,
and
corresponds under (1) to the element
nzir(r:)
Here,
and any lifting of
u E SK1(R[G]),
(1)
([g,h] E K: g,hE G)
E Kfl
is the trace map.
For convenience, set
Ko = ([g,h] E K: g,h E G)
and
Ia = Ker[R[G] - 4 R[G]].
The snake lemma applied to the diagram
1 -i SKI(R[G])
xµF,xO^ab
-+ Ki(R[G]) -i Wh'(R[G]) -+ 1
1SK1(Ra) x a b
K(a)
I
1 - SKI (R[G]) x uF x dab - K1(R[G])
!Th,(a)
- Wh'(R[G]) -i 1
induces an exact sequence
K1(R[G],Ia) - Ker(Wh'(a)) - ) Coker(SKi(Ra)) -1 1.
Also, the following diagram with exact rows
1 --> Wh'(R[G]) - HO(G;R[G]) Wh'(a)
1 -* Wh'(R[C])
I
-4 (e)xGab -> 1 laab
H(a)
HO(G; R[G])
(e)
x}IGab
1
(2)
CHAPTER S.
THE P-ADIC QUOTIENT OF SK1(Z[G]):
P-GROUPS
(see Theorems 6.6 and 7.3) induces a short exact sequence
K
r 1 -i Ker(Wh'(a)) - ) HO(G;I a)
185
of kernels
-* 1
(3)
Kfl[G,G]
where
HO(G;Ia) = Ker(H(a)) = Im[HO(G;Ia) 0 HO(G;R[G])]. It remains to describe
r6(K1(R[6],Ia)).
This could be done using
the exact sequence of Theorem 6.9, but we take an alternate approach here to emphasize that the difficult part of that theorem is not needed.
We first check that there is a well defined homomorphism
HO(G;Ia) ) K/Ko U(Jri(zi-1)gi) = nzTr(r')
such that
It suffices to check this when contains no commutators.
Ko = 1;
be conjugate.
And
defined on
itself (and
Ia
G
K. = Ker((3i) (. K)
Then
/32
O^ab
in this
K
cannot
since it is well
HO(G;Ia)'
HO(G;K) = K/[G,K] = K).
so that
Set
and
Ii = Ker[R(3i: R[G] -i+ R[G]]
is split by the diagonal map from G
Gab =
gi E G.
is central and
in the same coset of
is well defined on
R[G] = R[G] W I2,
and
K
and
HO(G;Ia) = Ho(G;Ia)
Now define a = { (g,h) E G x a: a(g) = a(h) } ,
is a pullback square.
zi E K,
1. e., when
In particular,
case, since two distinct elements of
U
rl E R,
for
to
G.
(i = 1,2).
In particular,
K1(R[G]) - K1(R[G]) ® Kl(R[G]+I2)+
x (K2/[G,K2])
Consider the following diagram:
186
rG)
K1(R[G]+I2)
HO(G;I2)'
If1
G K2/[G,K2]
I
(4)
I
HO(G;Ia) ) K/K- ----- 1;
r
are induced by
f;
- 1
f,
f2
K1(R[G],Ia)
where the
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[C]):
CHAPTER S.
The top row is a direct
(31: 6 - G.
summand of the exact sequence of Theorem 6.6 applied to hence is exact.
Furthermore,
K2 = K,
and so
Ker(f3)
K1(R[6]),
and
is generated by
elements of the form
([g,h],l) = (ghg l,h)'(h,h)-1 = -6(r'((gghg 1,h)-(h,h))) E w6(Ker(f2))
for
g,h E G
such that
[g,h] E K
w6(Ker(f2)) = Ker(f3),
words,
Tr(r) = 1).
(and where
In other
and so the bottom row in (4) is exact.
It now follows that
Coker(SK1(Ra)) = Coker[K1(R[G],Ia) - 4 Ker(Wh'(a))]
(by (2))
rG(Ker(Wh'(a)))/rG(K1(R[G],Ia))
(by (3))
io rG(Ker(Wh'(a)))
(by (4))
_ (Kfl[G,G])/Ko =
Kfl [G,G]
(by (3))
([g,h]EK: g,hEG)
The description of the isomorphism follows from the definition of
Proposition 8.1 shows that elements in the difference between commutators in of commutators in
K.
"universal group" for onto
The functor Coker(SK1(a)),
K
H2(G)
SK1(R[G])
(when
ii.
o
are detected by
G = G/K),
and products
will now be used to provide a
for all surjections
a
of p-groups
G. If
G
is any group, and
G = F/R
where
F
is free, then by a
formula of Hopf (see, e. g., Hilton & Stammbach [1, Section VI.9]),
CHAPTER S.
THE P-ADIC QUOTIENT OF SK1(R[G]):
P-GROUPS
187
H2(G) - (Rfl[F,F])/[R,F].
If
g, h
is any pair of commuting elements
in
we let
G,
g-h E H2(G)
denote the element corresponding to
g
h
and
to
g,h E F
.
If
generated by such elements.
for any liftings of
[g,h] E Rfl[F,F]
is abelian, then H2(G) - A2(G)
G
So for arbitrary
H2b(G) = Im[[{H2(H): H C G, H abelian}
is
G,
2Ind
H2(G)]
= (g-h: g,h E G, gh = hg) C H2(G). Theorem 8.2
Let
1 1 K -4 G -4 G - 1
Then for any
groups.
Z[G]-module
M,
be any extension of
there is a "five term homology
exact sequence"
H2(G;M)
* H2(G;M)
M Kab
In particular, when M = 7L,
H2(G)
Ha
H2(G)
3M -+ H1(G;M)
this takes the form
3- K/[G,K] ---> b a
where for any commuting pair
K
(mod
- 1; g,h E
[G,K]).
is central, then this can be extended to a 6-term exact sequence
K ® G - H2(G) H a where
, dab
g,h E G and any ltftings to
ba(g^h) = [g,h]
If
H1(G;M) -> 0.
q(h ®g) = h-g E H2b(G)
a H2(G) b > K
0
ab aab
for any h E K and g E
Gab
-- 1,
CHAPTER 8.
188
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
The 5-term sequences are shown in Hilton & Stammbach [1,
Proof
Theorem VI.8.1 and Corollary VI.8.2]; and the formula for g^h and naturality.
from the definition of formula for
6"
are shown in Stammbach [1, V.2.2 and V.2.1].
7,
G a G-* 1
When 1-4 K
follows
The 6-term sequence, and the 13
is a central extension, then
can be regarded as the image of the extension
[a] E H2(G;K)
da
under the
epimorphism
H2(G;K) - Hom(H2(G),K)
in the universal coefficient
So
theorem.
it
is not surprising that
central extensions can be constructed to realize any given homomorphism
H2(G) - + K. Lemma 8.3
For any finite group G and any subgroup T C H2(G),
(i)
there is a central extension Sa: H2(G) -- b K
(ii)
is surjectiue with kernel
For any pair H C G
G -4 1
K -1 G
1
such that
T.
of finite groups, there is an extension
1 -4K--'G-G-+ 1 of finite groups, such that if we set H2(ao) = 0.
then
and if
G
(iii)
H a G,
If
is a p-group then G
M,
and ao = all H - H.
can be chosen such that
K C Z(H);
can also be taken to be a p-group.
For any finite group
or 1P [G]-module
H = a-1(H)
then
G,
and any finitely generated
there is an extension 1 -+ K -* G
Z[C]-
G --> 1 of
finite groups such that
H2(a;M) = 0 : H2(G;M) - H2(G;M).
Proof
(i)
Write
G = F/R,
there is an exact sequence
where
F
is free.
By Theorem 8.2,
THE P-ADIC QUOTIENT OF SKI(Z[G]):
CHAPTER S.
F
0 = H2(F) - H2(G)
Furthermore, so
R/[F,R]
P-GROUPS
189
Fab _ Gab
) R/[F,R]
1.
and all of its subgroups are free abelian groups, and
Fab
splits as a product
R/[F,R] = Ro/[F,R] x J(H2(G))
Ro a F
for some
a: G -» G
let
where
be the projection, then
morphism.
So for any
that
is surjective with kernel
6
T C H2(G),
ba: H2(G)
and
is an iso-
has the property
T.
where
is free and finitely
F
By Theorem 8.2, there is an exact sequence
generated.
0 = H2(F;M) --> H2(G;M) --> Rab®Z[G]M Here,
G = F/Ro, R/Ro
aT: 6/6a(T) --3+ G
C = F/R,
Again write
(iii)
If we now set
[F,R] C Ro C R.
is a finite p-group and
H2(G;M)
H1(F;M).
M
Rab
is a finitely gener-
0Z[G]
ated
Z-
or 1p module.
index such that into
(R/T)
So there is a normal subgroup
and such that
[R,R] C T C R,
If we now set
M.
G = F/T,
T a F
H2(G;M)
K = R/T,
of finite
still injects
and let
a: G - G
®Z[G]
be the surjection, then
H2(G;M)
So
6a
H
is injective in the exact sequence
H2(G;M)
K ®Z[G]M.
H2(a;M) = 0.
(ii)
and
H2(G;M) = H2(a
extension and then
Now fix H C C, H)
and set
for any
M = Z(G/H).
a: G -» G.
Then
H2(G;M) - H2(H);
So by (iii), there is an
1 -- K -+ G a G -+ 1, such that if we set
ao = alH, [H,K] 4 G,
then
H2(ao) = 0
and
and we can replace
6ao a
by
is injective. G/[H,K]
(so
H = a l(H) If H d C, K 9 Z(ff))
THE P-ADIC QUOTIENT OF SK1(E[G]):
CHAPTER S.
190
without changing the injectivity of
bao.
If
can be replaced by any p-Sylow subgroup.
Now, for any extension
G
P-GROUPS
is a p-group, then
G
O
1 - K -> G
G - 4 1
of p-groups, the
5-term exact sequence of Theorem 8.2 induces an exact sequence
K fl [G,G]
b (L)
([g,h]EK: g,hEG) Coker(SKl(Ra))
By Lemma 8.3(i), for any So
H2(G)/H2b(G)
among all
G,
there exists G -* G
such that
represents the largest possible group
a: G --» G.
Proposition 8.4
H2(a) = 0.
Coker(SK1(Ra)),
This is the basis of the following proposition:
R
Let
unramified extension of
Qp.
be
the ring of integers in any finite
Then for any p-group
G,
there is a natural
surjection
0RG :
SKI(R[L]) _ H2((;)/&b(G),
characterized by the following property.
For any extension
1 ->KC-> 1 of p-groups, for any
of
u,
gi E 6),
u E SK1(R[G]),
and for any lifting
if we write FRG(u) = jri(zi-l)gi
(where
ri ER,
u E KI(R[G])
zi EK,
and
then
Sa(ORG(u)) =
Furthermore,
p-group G
Proof
IIzr(r.) E K/([g,h] E K: g,h E G).
0RC
is an isomorphism if
surjecting onto
ORG
(ba: H2(C) '-'' K/[G,K])
is an isomorphism for any
G.
The only thing left to check is the last statement.
By the
ORG
above discussion,
8b.
of p-groups.
191
SK1(Ra) = 1
is an isomorphism if and only if
G a G
some surjection G
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(g[G]):
CHAPTER S.
for
Clearly, this property holds for
if it holds for any p-group surjecting onto
0
G.
Establishing upper bounds
0RG
It remains to show that the epimorphism an isomorphism.
While lower bounds for
Coker(SK1(Ra))
for surjections
SK1(R[G])
a: G - G,
established by studying
Coker(SK1(f))
of a subgroup of index
P.
when
of Proposition 8.4 is were found by studying
the upper bounds will be
G
f: H
is an inclusion
The following lemma provides the main induc-
tion step.
Lemur 8.5 extension of if
Let Qp.
SK1(R[H]) = 1,
Proof
be the ring of integers in any finite unramifted
Then for any patr HG of p-groups with then 0RG
[G:H] = p,
is an tsomorphism.
For the purposes of induction, the following stronger state-
ment will be shown: 0RG
R
G D H
for any pair
of p-groups with
[G:H] = p,
factors through an isomorphism
Oo
:
Coker[SK1(R[H]) -> SK1(R[G])] (1)
_- Coker[H2(H)I 2b(H) -'-2(G)I2b(C)].
Note that
Oo
is onto by Proposition 8.4.
Let
f: H -4 G
denote the
inclusion, and let
SK1(f): SK1(R[H]) -i SK1(R[G]),
H2(f): H2(H) - H2(G),
and
Wh(f): Wh(R[H]) -4 Wh(R[C'),
H2IH2b(f): H2(H)I 2b(H)
denote the induced homomorphisms.
Fix some
x E G'-H;
-
H2(C)/H
and fix
r E R
(C)
such
192
CHAPTER S.
Tr(r) =1E 2p
that
(Proposition 1.8(111)).
f E SKI(R[L])
Choose any must show that
ABL(E) E I
such that
f E Im(SK1(f)).
(f)).
We
This will be done in three steps.
In
Step 1, Theorem 7.1 will be used to show that fo E Wh(R[H])
h1 EH
such that
F
Step
2,
with a certain subquotient of
H;
we
3, this is used to show that
Step 1
in G
p
and where the first
identify
Then, in Step
f E Im(SK1(f)).
We can assume that
by Theorem 1.14(11).
for some
and then show that
corresponds under this identification to
Oo([f])
order
In
(
Wh(f)(fo),
= Ji=1r(h1 - xh1x 1),
( f o )
[h1,x] E [H,H].
satisfy b(f))
1
P-GROUPS
THE P-ADIC QUOTIENT OF SKI(Z[G]):
H
is nonabelian:
z E [H,H]
Fix
(Lemma 6.5),
set
otherwise
SK1(R[G]) =
which is a central commutator of
fl = H/z,
G = G/z,
and let
Hf0G
f
H
be the induced maps.
G
Consider the following commutative diagram:
Coker(SK1(f))
O
) Coker(H2/H b(f))
IS(a) 1H(a)
Coker(SK1(f)) Coker(H
Here,
0o
morphism);
0RG
is induced by
and
S(a)
and
(f)).
(and is assumed inductively to be an isoH(a)
are induced by
a.
In particular,
[f] E Ker(Oo) C Ker(S(a)).
Consider the following homomorphisms:
SK1(R[H])
SK
f
SK
pr
) Coker(SK1(f)) -> 1
ISKI(a)
18K) SK1(R[H])
SK1(R[G])
1S(a) f
SK1(R[G]) - Coker(SK1(f)) -' 1
THE P-ADIC QUOTIENT OF SK1(Z[C]):
CHAPTER S.
Since
S(a)([f]) = 1,
there exists
By Proposition 8.1, we can lift
SK1(a)(f).
In particular, since go
z
is a commutator in
is conjugate in
C
such that
to some
)
193
SK1(f)(ui) =
11 E Wh(R[H])
for some a E Z and any desired
such that rRH(rf) = ra(1-z)go, that
E SK1(R[H])
P-GROUPS
to
G,
we may choose
go E H. such
go
zgo.
Now set
0 = {g E C : g conjugate zg} = {g E C : [g,h] = z, some h E C) $ 0,
- be the relation on
and let either
g E H,
or
[g,h] = z
each --equivalence class of
from Theorem 7.1.
n
for some
n
h E H
G/H
(since
includes elements of
g E fl,
For any
is cyclic).
So
By Theorem 7.1,
H.
Ker(SK1(a)) = (Exp(r(1-z)(g-h)) : g,h E H fl 0)
c (Wh(f)(r1(r(g-zg))) : g E H, g conj. zg
Since
Wh(f)(i) a f
(mod
f = Wh(f)(fo),
and where
is conjugate in C
gi
x E G
Recall that some
ri S p-l
particular,
such that
rRH(fo) = r(1-z). gi,
zgi
to
generates
G/H.
xr'gix ri
for all
i.
Hence, for each
i,
there is
H
to
zgi.
is conjugate in
rRH(Eo) = Jr(gi - xr'gix r') E HO(H;R[H]).
can find elements
hi,...,hn E H
rRH(fo) _ Z r(hi-xhix 1),
By relabeling,
In
we
such that
and
[hi,x] E [H,H]
(all
i=1
Step 2 Now set
K = Coker(H,,/
C).
this shows that we can write
Ker(SK1(a))),
where
in
b(f)) = H2(G)I(
b(G), Im(H2(f))),
i).
(2)
CHAPTER 8.
194
THE P-ADIC QUOTIENT OF SK1(7L[G]):
P-GROUPS
1 -> K --> G -L G - 1
for short, and fix a central extension
such
that
b: is
H2(C) --» K = Coker(H2/H2b(f))
the projection (use Lemma 8.3(i)).
(where
Kfl[H,H] = 1
Im(H2(f)) C Ker(b).
since
H = (3 1(H)),
In particular,
The Hochschild-Serre spectral sequence for
1 - H -L G -' Cp -4 1
(see Brown [1, Theorem VII.6.3]) induces an exact sequence a3
C) H1(Cp;Hab)
H3(Cp)
, Coker(H2(f)) --' 0.
H1(Cp;-)
The usual identification of
with invariant elements modulo
norms takes here the form
{h E Hab
H1 (Cp ;Hab) _
(h.xhx
: xhx
1
=h
in Hab}
1...xp-lhxl-p:
Under this identification,
- {hEH: [h,x] E [H,H]} hEH)
h E H)
8(H3(Cp)) = (xp)
with the corresponding sequence for
by naturality (compare this
1 -4 (xp) --> (x) --' Cp --> 1).
So
there is an isomorphism
{hEH : [h,x]E[H,H]) at
Coker(H2(f)).
.
hEH) Furthermore,
aiI(Uab(G))
and so
a1
_ (hEH : h conj. xhx l in H) Q ((hx)p
: h E H);
factors through an isomorphism
{hEH: [h,x] E [H,H]} a :
E H: h conj. xhxl in
By construction, for each
h,
H)
Coker (H2Hab 2
f ))
(
= K.
(3)
CHAPTER S.
a(h) = [(3 1x,(3 lh]
By (2), i.
rRH(fo) =
zi E K
Theorem 6.6).
Then
h
a 1(Oo(f))
can choose
(mod 2)
(5)
r(hi - ix + (1-z.))
rltA(u. ) =
(3),
1fll) =
SK1(R[H])).
(mod
Also,
= ni=l hi'
we can write hJ
ai=l hi =
is conjugate in
(otherwise just take
fl E Wh(R[H])
(use
II z.l E K = Coker(H2/H2b(f)) i=1 1
where each
assume n a m
there
and so by the formula in Proposition 8.4,
Oo([f]) _
Now by
for all
Since K(1[U] = 1,
Wh((3IH)(ni=lui) - fo
Then by (4) and (5),
[hi,x] E [H,H]
(mod
such that
rRG(Wh(f)(ui)) = r(1-z1);
Step 3
where
such that
[hi,x] = zi
ui E Wh(R[H])
(4)
hi,x E G.
of
195
(mod
Zi-lr(hi-xhix l),
hi,z E G
Choose liftings
are unique elements
Fix
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
such that
H
to xhx 1.
6.6, we
hm+l =1).
rR(f1) = r(?i=1hi
We may
- L j= lh') j
Then
n r(hi - xhix l) - r(fl) +
1
i=1
r(h'-xhix 1) = 0 E HO(H;R[H]); i=1
and
fo - [fl,x]
Im(SK1(f)),
(mod
SK1(R[H])).
It follows that
and this finishes the proof.
The proof that
f = Wh(f)(fo) E
n
ORG always is an isomorphism is now just a matter of
choosing the right induction argument.
196
THE P-ADIC QUOTIENT OF SK1(Z[G]):
CHAPTER S.
Theorem 8.6 Let extension of Q .
R be the ring of integers in any finite unramifted
Then for any p-group
G,
0RG : SK1(R[G]) - ) H2(
is an isomorphism.
on SK1(R[G])
Proof
surjection
This will be shown by induction on (SK1(R[G]) = 1
if
subgroup such that
a1: G1 -» G/Z(G) and
H = a 1(H).
a
and so
1 0j@
and
IG/Z(G)I,
H2(a) = 0,
Also,
8.5 now applies to show that 0RG
H1 = a,1(H/Z(G)),
G be the pullback
1
This shows that
and so
H a C be
G
112b(H) 2 Ker(H2(a)) = H2(H)
of Theorem 8.2.
G,
Let
1 - K -+ H
Then
IH/Z(H)I
induction hypothesis.
onto
acts
By Lemma 8.3(ii), there is a
of p-groups, with
1
ailHi;
1)
Fix any non-
IC/Z(C)I.
is abelian); and let
H 2 Z(C).
G1 a -
extension, so
C
H2(a1IH1) = 0.
a
and set
(g '-4
by negation.
Ker(aj) C Z(H1)
that
(G)
Furthermore, the standard involution
abelian p-group G
any index p
P-GROUPS
is a central
is an isomorphism by the
since
a
factors through
by the 6-term exact sequence
SK1(R[H]) = H2(H)/H2b(H) = 0;
ORG
is an isomorphism.
is an isomorphism by the
But
and Lemma
G
surjects
last statement in
Proposition 8.4.
By
the
description
[u] E SK1(R[C]),
shows that
of
ORG
0RG([u]) = -0([u]).
SK1(R[G])
in
Since
Proposition
ORG
for
8.4,
any
is an isomorphism, this
is negated by the standard involution.
o
Theorem 8.6 can in fact be extended to include group rings over arbitrary finite extensions of
Qp.
This does not have the same import-
THE P-ADIC QUOTIENT OF SK1(Z[G]):
CHAPTER S.
ance when studying
SK1(Z[G])
P-GROUPS
197
as does the case of unramified extensions;
but we include the next theorem for the sake of completeness.
Theorem 8.7 F
of
R
Let
be the ring of integers to any finite extension
Then for any p-group
Q.
G,
SK1(R[G]) = H2(G)
If
E 7 F
is a finite extension, and if
is the ring of integers,
S C E
then
(i)
i*
SK1(R[G])
:
E/F
isomorphism if
(it)
trf
isomorphism if
Proof
:
E/F
(induced by inclusion) is an
SK1(S[G])
)
is totally ramified;
SK1(S[G])
'
and
SK1(R[G])
(the
totally ramified.
is
an
is unramified.
Note first that for any finite extension E
a unique subfield
transfer)
F C E
such that
To see this, let
F/Qp
of
Qp,
is unramified and
p C S C E
there is
E/F
is
be the maximal ideal and
I(S/p)*I.
ring of integers, and set
roots of unity in Theorem 1.10, Also,
E/F
F/Op
E,
um
be the group of m-th
and set F = Qp(µm) C E and R = 2p[µm]. is unramified, and
R C F
By
is the ring of integers.
IR/pRI = IS/pl = m+l.
this shows that it suffices to prove (i) and (ii)
under the assumption that
OSG
Let
is totally ramified since
In particular,
unramified,
m =
F
is unramified over
Qp.
If
E
is also
then the following triangle commutes by the description of
and ORG
in Proposition 8.4:
SK1(S[G])
OSG -
trf)
SK1(R[G])
ORG
H2(G)1 -2b(G).
198
THE P-ADIC QUOTIENT OF SK1(Z[G]):
CHAPTER S.
P-GROUPS
So the transfer is an isomorphism in this case.
Let
is totally ramified (and F/& is unramified).
E/F
Now assume that
p C S be the maximal ideal.
Ker[i*: SK1(R[G])
-+
Then
SK1(S[G])]
Ker[SK1(R[G]) -' K1(R/p[G]) = K1(S/p[G])]
C torspIm[K1(R[G],p) -4 K1(R[G])].
log: K1(R[G],p) -9 HO(G;pR[G])
Using the logarithm homomorphism Theorem 2.8, one checks easily that
K1(R[G],p)
is odd, and that the only torsion is
{t1)
Thus, in either case,
now shown by induction on
IGI,
see Oliver [2, Proposition 15].
p
p = 2.
if
The surjectivity of
is injective.
i*
is p-torsion free if
of
using Theorem 7.1 again.
i*
is
For details,
o
We end the section by showing that natural, not only with respect
the
isomorphisms
0RG
are
to group homomorphisms, but also with
respect to transfer homomorphisms induced by inclusions of p-groups.
Proposition 8.8
Let
R
be
the ring of integers in any finite
Then for any pair H C G
extension of
of p-groups, the square
P.
SK1(R[G])
0G
trfSK
H2(G)/ b(G) ItrfH
SK1(R[H])
commutes.
Here,
homomorphisms for
trfSK K1
and
and H2,
trfH
are induced by the usual transfer
respectively.
CHAPTER 8.
Proof
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
199
By Theorem 8.7, it suffices to show this when R = Zp.
Using
Lemma 8.3(11), choose an extension
1 -K-1G-c-+ G- 1, of p-groups, such that is injective, and
0
K C 2(H)
H=a 1(H),
Then
H2(ao) = 0.
and
SK1(2 p[ao]) = 1
ao =alH Sao: H2(H) '- K
By the description of
by Theorem 8.6.
in Proposition 8.4, it will suffice to show that the following squares
all commute:
SK1(2p[G]) ` trf
trfK
(1)
SK
1
SKI(2p[H]) `
G , HO(G;$p[G])
' K1(2p[G])
3 K1(2p[G])
trfj
(2)
ao
' Kl(gp[H]) t
ResH
(3) r
*
K1(Lp[H]) -H) HO(H; p[H])
HO(G;Ia) a K/([g,h] E K: g,h E G)
H2(G)
NG/H
2b (G)
trfH
(5)
IResH HO(H;Iao)
Here,
ResH
ao
so H2(
' K/([g,h] E K: g,h E H)
is the homomorphism of Theorem 6.8;
NO (G;Ia) = Ker[HO(G;Zp[G]) -" HO(G;ap[G])]I
r
wa(I ri(1-a.)gi) =
(and similarly for
HO(H;Iao)
the conjugation action of
(ri E Zp,
[Iaii
G/H
ai E K,
gi E G),
p
and w
ao
on
K.
and
);
N .1H
is the norm map for
The commutativity of (1) and (2)
is clear, (3) commutes by Theorem 6.8, and (4) by definition of The commutativity of (5) follows since
trfH
Res
splits as a composite
200
CHAPTER S.
THE P-ADIC QUOTIENT OF SK1(7[G]):
H2(G) ) H2(G;7L(G/H)) ---
H2(H)
(see Brown [1, Section 111.9]); and similarly for f2
are induced by the inclusions
IgEG/fig
and
i2(1) = 1
il,i2:
(note that
P-GROUPS
Here,
NG/H .
where
7L(G/H),
7L
is only 7[H]-linear).
i2
and
f1
il(l) _ n
Examples
Sc.
It turns out that
describe
H2(G)
need not be computed completely in order to
In practice, the following formula
H2(G)/1 b(G) = SK1(7Lp[G]).
provides the easiest way to make computations and to construct examples.
Lemma 8.9
Fix a central extension
1 -* K -4 G
G
1
of
p-groups, and define
: g,h E G, gh = hg} C H2(G)
A(G) = {g-h E H2(G)
(a subset of
H2(G)).
Let
Sa: H2(G) -+ K
5-term homology exact sequence (Theorem 8.2).
be the boundary map in the Then, if
R
is the ring of
integers in any finite extension of
SK1(R[G]) = H2(G)/H2b(G) = Ker(Sa)/(A(G) fl Ker(Sa)).
In particular,
Proof
SK1(R[G]) = 1
if
H2(a) = 0.
Consider again the 6-term homology exact sequence for a
central extension (Theorem 8.2):
K ® Gab
Here,
H2(G)
Ha
T(x®g) = x^g E H2b(G)
H2 (G)
S
a
for any
Ker(H2(a)) C H2b(G). Furthermore,
K
xEK
Gab
I
Gab
and any
1.
g E G.
So
CHAPTER 8.
THE P-ADIC QUOTIENT OF SKI(Z[C]):
201
P--GROUPS
H2(a)(H2b(G)) = (g-h E H2(G) : g,h lift to commuting elements in G)
= (A(G) fl Ker(ba)); and the result follows.
o
As a first, simple application of Lemma 8.9, we note the following conditions for
SK1(2 p[G])
Theorem 8.10 of
Q.
Then
to vanish.
R
Let
be the ring of integers in any finite extension
SK1(R[G]) = 1
if
c
is a p-group satifying any of the
following conditions:
(t)
there exists
H a G
such that
H
is abeltan and
G/H
is
cyclic, or
(it)
[G,G] is central and cyclic, or
(itt)
G/Z(G)
Proof
See Corollary 7.2 and Oliver [2, Proposition 23].
is abeltan of rank < 3.
The smallest p-groups p = 2,
or
p5
if
p
G
with
SK1(lLp[G]) A 1
a
have order 64 if
is odd (see Oliver [2, Proposition 24]).
following examples are larger, but are easier to describe.
Example 8.11
Fix n > 1,
and set cpn
G = (a,b,c,d
:
[G,[G,C]] = 1 = apn = bpn =
= dpn = [a,b][c,d]).
Then SK1(2p[C]) = z/p.
Proof
By construction,
1
(Cpn)5
G
sits in a central extension
G , Gab = (Cpn)4 _ 1;
The
202
CHAPTER 8.
where
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
ba: H2(Gab) --> (C Pn)5
is surjective with kernel
Ker(Sa) = (a^b + c^d) = 7/pn.
Then
in the notation of Lemma 8.9, and the result
A(Gab)f1Ker(ba) = 1,
follows.
O
G
Recall that for any
and
n,
GZCn
denotes the wreath product
The next proposition describes how H2((;)/b(G)
Gn M Cn.
and
SK1(2p[G])
act with respect to products and wreath products.
Proposition 8.12
= H2(G)/-ab(G) 0 H2(H)/112b(H),
H2(G x H) b(G x H)
In particular, if
and if
SK1(R[G]) = 1
R
factor of
and
is the ring of integers in any finite extension of
if
C
and
See Oliver [2, Proposition 25].
1 < i 1,
G and H are p-groups, then
SK1(R[G x H]) = SK1(R[G]) ® SK1(R[H])
Proof
H,
Cn) = H2(G)/H2b(G)
H2(G2
Also,
G and
For any finite groups
The only point that is at all
G2 n.
let f; : G ---> C C n be the inclusion into the i-th
Fix x E (G2Cn)'Gn
such that
x(91,....gn)x 1 = (g2,...,gn+g1)
xn = 1,
(for all
and such that
(g,,...,gn) E GP).
Define
T = (fi(g)^f;(h)
: g,h E G, i0J) 9 H2"(G2 n)
CHAPTER 8.
P-GROUPS
THE P-AI)IC QUOTIENT OF SK1(Z[G]):
203
A straightforward argument using the Hochschild-Serre spectral sequence (see Brown [1, Theorem VII.6.3]) shows that
H2(GZCn)/T.
fl* : H2((;)
Then
is generated by
H2b(GZCn)/T
for commuting
g,h E Gn);
as well as all
gx^h
h =
g-h
(i. e., elements
fl*(H2b(G))
For elements of the last type,
[gx,h] = 1.
induces an isomorphism
f1
g,h E Gn
for
such that
g =
if
and
then a direct computation shows that
gx"h = fl*((g,...g,)^h1) E fl*(H2b(G));
fl*
and so
induces an isomorphism H2(G)/H2b(G) = H2(G'Cn)
a G -- 1
By Theorem 7.1, if 1 -> (z) --> G extension of p-groups such that Izi = p, then g,hE0`\).
Ker(SK1(2 pa)) = (Exp((1-z)(g-h)): Also, for a n y
[Exp(r(l-z)(g-h))]
r E 2p,
is any central
(ft = {gEG: g conj. zg})
depends only on
r
p2p),
(mod
g and h modulo a certain equivalence relation
and on the classes of in
b(G2 n).
ti
It is natural now to check where these elements are sent under the
(1.
isomorphism
EG.
This is done in following theorem, which describes one
case where elements directly
in
(in contrast
can be constructed or detected
SK1(2 p[G]) to
the
6G
in
z E G
of
very indirect definition of
Proposition 8.4).
Theorem 8.13 order
p,
and let
Fix a p-group
C
and a central commutator
a: C - C/z denote the projection.
ring of integers in any finite extension of maximal ideal, and let by
a and by
a: R[C] -s ffp[G/z]
r: R -» R/p
Tr»
Fp.
Qp,
let
Let
p g R
R
be the
be the
be the epimorphism induced
Set
0 = {gEG : g conjugate to zg} = {gEG : [g,h] = z,
some
h E G}.
204
P-GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
CHAPTER S.
Define functions
Wp(n/z)
0 H2(G/z)/(Ker(ba) fl A(C/z)) < * { H2(G)/H2b(G)
where
X(g) = a(g)-a(h)
where
a*
any
is induced by
for any H2(a)
g,h E 0
such that
[g,h] = z;
[u] = [1 + (1-z)f] E Ker(SK1(Ra)),
if we set
and
Then for
(an injection by Lemma 8.9).
Log(u) = (1-z)D,
then
a(n) = a(g - fp) E Fp(fl/z) and -KG([u]) = a*1 o X(a(ii)) = a*1 o X(a(I; -Ep))Proof
See Oliver [2, Proposition 26].
equivalence relation defined in Theorem 7.1, 1p(0/-).
Note that if then
X
-
This then gives a new interpretation of the inequality
rk,,(Ker(SK1(Ra)))
of Theorem 7.1.
0
In/-I-1
is the
factors through
Chanter 9 C11(Z[G]) FOR P--GROUPS
We now turn to the problem of describing p
p-group and
odd
p
is any prime.
C11(Z[G]),
G
when
is a
This question is completely answered for
in Theorem 9.5, and partly answered in the case of 2-groups in
Theorem 9.6.
Conjecture 9.7 then suggests results which would go further
towards describing the structure of 2-group case.
Cll(Z[G])
Some examples of computations of
(and
SK1(Z[G]))
Cll(Z[G])
in the
are given at
the end of the chapter, in Examples 9.8 and 9.9.
All of these results are based on the localization sequence
> Cp(Q[G]) - G ' Cll(Z[G]) -' 1
K2(Zp[C])
The group
of Theorem 3.15.
there are
Theorem 4.13.
So
Cll(Z[G])
be computed:
can
Cp(I[G])
has already been described in
two remaining problems
to
solve before
a set of generators must be found for
and a simple algorithm is needed for describing their images in
C(D[G]).
The first problem is solved (in part) in Proposition 9.4,
and the second in Proposition 9.3. If
and the
])k=lCC(Ai),
unity in
where the
l[G] = ni=1A
Z(A.).
Cp(A,)
Ai
are simple, then
Cp(D[G]) =
have been described in terms of roots of
The following theorem helps to make this more explicit,
by listing all of the possible "representation types" which can occur in a
group ring of a p-group: summands.
i. e., all of the isomorphism types of simple
As usual, when p
root of unity
is fixed, then fn
(any n2 0)
denotes the
fn = exp(2ai/pn) E C.
Theorem 9.1
simple summand of
Fix a prime @[G].
If
A
be any
p
and a p-group
C,
and let
p
is odd, then
A
is isomorphic to a
206
CHAPTER 9.
matrix algebra over Cp(A).
p = 2,
If
algebras D
Q(fn)
then A
C11(7L[G]) FOR P-GROUPS
For some
n > 0;
and
K2(Ap)(p) = (fn)
is a matrix algebra over one of the division
in the following table:
D
KZ(A2)
C(A) = C2(A)
{fl}
1
Q
Q(fn)
(n > 2)
Q(fn+fn-l )
(n>_ 3)
{fl}
1
Q(fnfnl)
(n > 3)
{fl}
{fl}
Q(fn, j) (S Di)
(n > 2)
{fl}
{fl}
Proof
See Roquette [1].
to
subquotient of
that G;
Section 2 of [1], Roquette shows that
In
the division algebra for any isomorphic
irreducible representation of
a primitive,
of
(fn)
(f n)
faithful
G
is
representation of
some
and in Section 3 he shows that the only p-groups with
primitive faithful representations are the cyclic groups; and (if
For each such
the dihedral, quaternion, and semidihedral groups. Q[G]
has a unique faithful summand
G
Q(fn)
Q(2n+1)
4.11 and 4.13.
K2(Ap)
and
Cp(A)
M2(Q(fnfnl) )
follow immediately from Theorems
a
We next turn to the problem of describing
certain Steinberg H1(G;Z[G]),
SD(2n+1)
Q(fn, j)
M2(Q(fn+fn1) )
The computations of
G,
given by the following table:
A,
D(2n+1)
A
p = 2)
where
symbols
G
{g,u}
acts on
7L[G]
bookkeeping device for doing this.
,p({g,u}) E Cp(Q[G]),
E K2(2p[G]).
The homology group
via conjugation, provides a useful
Note that for any
are conjugacy class representatives for elements of
k
C,
G,
if
then
k
H1(C;7L[G]) = ®Hl(CG(gi))®Z(g=) = ®CG(g;)ab ®Z(gi) i=1
for
i=1
g1,...,gk
CHAPTER 9.
In particular,
H1(G;Z[C])
207
C11(71[G]) FOR P-GROUPS
is generated by elements g@ h,
for commuting
g,h E G.
Let
o(;
Cp(ID[G]) I iEI (P{ ')p
:
be the isomorphism of Theorem 4.13: and
I C {1,...,k}
where
@[G] = fi=1Ai,
Ki = Z(A1),
is an appropriate subset.
Definition 9.2
p
Fix a prime
and a p-group
and define a
G,
homomorphtsm
yG : H1(G;Z[G]) = H1(G;2 p[G]) 0 Cp(ID[G])
as follows.
Write
irreducible module all
each
V.
such that
i
i E I,
(Q[G]
_ Hk
where each
Ai,
and center
K
I C {1,...,k}
Let
i. e., such that
CC(A.) X 1;
2r-1+1
if
1
otherwise.
and
p = 2
Then, for any commuting pair
Ki
Di = EndAi (V,) !9 R.
For
g,h E C,
= Q(Er)
set
,PG(g ®h) = a l((detK (g,Vh)E'/iEI/ E
G,
PKi
(Vi = {x E Vi: hx = x})
Cp(tD[G]).
Note in particular the form taken by
Xi: G - *
be the set of
set
ei
such a
is simple with
A.
when
PG
where the
K1
be the corresponding character.
Let
write
Q[G] = [Ik 1Ki
Definition 9.2, and set
I = {i: K. ¢ IR}.
aGo4G(g®h)
C
II
Fix
are fields, and let ei
Then
= (yi(g®h))iEI E
is abelian.
iEI (PK:)p
be defined as in
208
Cl1(Z[G]) FOR P-GROUPS
CHAPTER 9.
r Xi(g)6'
yi (g ®h) _ {l
where
1
When p
yG
is odd, the
if
Xi(h) = 1
if
Xi(h) # 1.
are easily seen to be natural with respect
This is not the case for 2-groups:
to homomorphisms between p-groups.
naturality fails even for the inclusion C2 '- C4. We now focus attention on certain Steinberg symbols in symbols of the form
each term in yG
u
g E G and
where
{g,u}
commutes with
CG (g)])*
[
(i. e.,
The next proposition describes how
g).
is used to compute the images in C(@[G])
Proposition 9.4 will show that when
u E (P
K2(2p[G]):
p
of the
{g,u}.
is generated by
Im(8C)
is odd,
Afterwards,
the images of such symbols.
Proposition 9.3 let
p and a p-group
Fix a prime
If
C.
be the distinct quaterntonic simple summands of
All ...'A1
t. e., those simple summands which are matrix algebras over some
m.
Cp(@[G]) C Cp(O[G])
Define
H C C
any
ID(fm,j)
for
is odd
if
p
if
p = 2.
2
Hi=lCp(Ai)
g E G,
Q[G];
by setting
11 Cp(Q[C]) =
Then, for any
then
p = 2,
such that
[g,H] = 1,
and any
u E (Zp[H])*,
tiPG({g'u}) = 'PG(g ®rH(u))
Proof
(mod
Cp(Q[G]))
This is a direct application of the symbol formulas of Artin
and Hasse (see Theorem 4.7(11)).
Fix a simple summand projection, and let
V
A
D[G],
be the irreducible
the center, and assume that
K
K = Q(fr)
(pr > 2)
or Q(fr frl)
as usual.
Let
and
"A
of
let
A-module.
Let
be the
K = Z(A)
be
has no real imbeddings.
In other words,
where
fr = exp(2ai/pr)
(p =2,
r Z 3) ;
denote the composites 'PA
X: @[G] -> A
CHAPTER 9.
Cl1(Z[G]) FOR P-GROUPS
209
C1 fA : K (Zp[G])
We must show that Set
EK =
p(A).
PA({g,u}) = PA(g or H(u))
pn = exp(G),
1
Cps
-±G-. Cp(l[G])
: H1(G;Z[G])
'PA
and
p(A),
CC(Q[G])
for any g,u as above. Define
L = K(fn) = Q(fn).
and let
1
if
p > 2, or p=2 and
K5--= @(f r-fr 1
1+2r-1
if
p = 2 and K = @(f )
(r
and similarly for
Set W = L ®K V,
EL.
g
the eigenspace for
nj.
Then, for each
on W
Wj
leaves each
W.
(r
3)
2)
E (fn)
and let
distinct eigenvalues of
on
)
Write W = W1 ®... ®Wm, h E H C CG(g),
where
be the Wj
is
the action of
h
invariant.
Write
k
k
Log(u) _
aihi,
ai(hi -
F(u) _
i=1
(hi E H,
ai E 4P)
i=1
Then by definition of
y,
CA0'A(g®rH(u)) _ fl (1)T'EK,
(1)
j=1
where
aA: Cp(A)
for each
WK
is the norm residue symbol isomorphism, and where
j,
k T. = i
Note that
Ti E 2
Now let
L
,pA
k ai.[dimL((W.)hi) -
i=1
for all
j,
since
denote the composite
r1(u) E 2 [H]
(2)
(modulo conjugacy).
C11(g[G]) FOR P-GROUPS
CHAPTER 9.
210
SPA
: K2(Zp[G])
- p(A)
10
-'' Cp(L®KA).
Then
aL®Ao*A({g,u}) =
ll (Tij,detL(u,Wj))L. j=1
The Artin-Hasse formula (Theorem 4.7(11)) takes here the form
(T)j)S'E`'(3)
°L®A°'PA((g,u)) = fl
j=1
where for each
j,
Si =
k
k (4)
=
Trj: EndL(Wi) - L
Here, (in
L)
of
h
on
is the trace map, and
(h)
is the character
W..
C E (pL)p = (gn),
For each
1-p if c=1 (c) _
- P 10
In particular, for each
j
if
Ici = p
if
ICI
p2.
and each h E CG(g),
Substituting this into (4) and comparing with (2) now gives
k S. _ i
T.. i_1
CHAPTER 9.
C11(7L[G]) FOR P-GROUPS
211
Now consider the diagram
' Cp(A) A -' (PK)p
K2( p[G])
Cp(K)
I
(5a)
incl
(5b)
It,
(5)
1 10
UT
(90p E
CP(L OK A)
Here,
(5a+5b)
homomorphism
by
commutes
Proposition
and
(nj)TJeK
II
is
a
j=1
for all
So the relation
J.
follow, once we show that
) = TEL b
i(c
It suffices to do this when K = O(fn-l)
there
so
By (1) and (3),
coaAo i (g®rH(u)) = II (Rj)SJeL
j=1 S. = I.
and
4.8(11),
which makes each square commute.
c
UAOPA({g,u}) =
and
Cp(L).
or Q(fn fnl).
PA({g,u})
=
sA(g ®FH(u))
E (µK)
for any
P.
i. e., when L = @(fn),
[L:K] = p;
will
and
Consider the following diagram:
-L
CP(L)
incl, Cp(L) CP(K)
°L
= °K
(li)p l (11L)p;
(µL)P
where
T(fn) = (fn)q
(L,L)-bimodule
L ®K L
q = [(µL)P:(1K)P].
if
commutes by Theorem 4.6.
°L
The composite
The left-hand square
inclo trf
homomorphism for the action of
Gal(L/K).
If
K
then
(pII1(fn)1+1p°")EK
1(fn-1)EK = LOT(fn)6K =
=
(f n1
e6)P+(Pi)Pf"1
= ((En)
i=-0
-
is induced by the
(see Proposition 1.18), and is hence the norm
- fn-1 - (fn-1) EL
if
p
if
p = 2.
(fn)(l+2^"2)(2+2^"1)
= fn-i =
is odd
212
CHAPTER 9.
If, on the other hand,
Cl1(7L[G]) FOR P-GROUPS
K = l(En fnl),
then
e< = 1,
1(-I) Eg = LOT(En) = (fin)'(-fnl) _ -1 = (-1)
and this finishes the proof.
13
The remaining problem is to determine to what extent generated by symbols
G
When
{-1,-1}
p = 2).
if
I C R,
K2(p[G])
is generated by
The next proposition gives some
partial answers to this in the nonabelian case.
R and any ideal
is
of the type dealt with in Proposition 9.3.
{g,u}
is abelian, then by Corollary 3.4,
such symbols (and
K2(!p[C])
Recall that for any ring
we have defined K2(R,I) = Ker[K2(R) -+ K2(R/I)]
(and similarly for K2).
Proposition 9.4
Fix a prime
p
and a p-group
the ring of integers in some Finite unramifted extension
(i)
For any central element
and let
G,
R
be
F 2 %. Then
z E Z(G), \
K2(R[G],(l-z)R[G]) = ({g,l-r(1-z)ih} : g,h E G, gh = hg, r E R, i > 1).
(it)
For any
H a G
such that
denotes the projection, and if
Hfl[G,G] = 1,
a: G --b G/H
if
Ia = Ker[[p[G] --v 2p[G/H]]1
then
K2(Ip[G],Ia) = ({g,l-r(1-z)h} : g,h E G, gh = hg, r E R, z E H).
(tit)
If
p
is odd, then
K,(R[CG(g)])+\.
K2(R[G])+ = ({g,u} : g E G, u E
Here,
K2(R[G])+
is the group of elements in
standard involution.
K2(R[G])
fixed under the
CHAPTER 9.
Proof
The most important result here is the first point; all of the
others are easy consequences generators for
of
K2(R[G],(1-z)R[G])
W flk=llk = 0;
The main
that.
I2 2 ...,
and such that for each
give generators for
Corollary 3.4 applies to
k,
K2(2p[G]/Ik,Ik_l/Ik).
lifted in several stages to
idea when finding
is to construct a filtration
(1-z)R[C] = IO 2 I1
where
213
C11(7L[G]) FOR P-GROUPS
These generators are then
The exact sequences for pairs of
K2(2p[C]).
ideals are used to show at each stage that all elements which can be lifted are products of liftable Steinberg symbols; and that the given symbols are the only ones which survive.
The complete proof is given in
Oliver [7, Theorem 1.4].
(ii)
H a G and
If
commutes in G
then a pair of elements g,h E G
Hfl[G,G] = 1,
if and only if it commutes in G/H.
Hence, if
1 =HOCH1 C ... CHk=H H. a G
is any sequence such that
and
for all
IHiI = pi
i;
then all
of the symbol generators given by (i) for each group
Ker[K2(2p[G/Hi])
lift to symbols in
(iii) define
K2(ap[G]).
Now assume
p
is odd.
For each
such that
u(rh) E K1(R[(h)])+
Theorem 6.6).
- 2($p[i+l])J
hEG
T(h)(u(rh)) =
Note that this element is unique, since
and each 1)
r E R, (see is
torsion free by Theorem 7.3.
Recall the formula for the action of the standard involution on a symbol in Lemma 5.10(1).
g E G
and
g,h E G,
u E (R[G]) *.
In particular,
So
and the homomorphism
{g,u) = {g,u}
{g,u(rh)} E K2(R[G])+
for any commuting
for any commuting
214
C11(Z[G]) FOR P-GROUPS
CHAPTER 9.
) K2(R[G])+,
AG : H1(G;R[G]+)
setting
defined by y
G(g
We claim that since
1
order
AG
µp ¢ R. set
p,
2
) = {g,u(rh)}
for any commuting
is uniquely defined and natural in
r E R,
andany
A
Iz = (1-z)R[G],
If
> 1,
IGI
G = G/z,
G.
This is clear if
is surjective.
g,h E G
C = 1:
then fix a central element
and assume inductively that
K2 (R) _
z E Z(C)
AG
is onto.
of
Set
and consider the following diagram:
a H1(G;R[G]+)
) H1(G;R[G]+) H + HO(G;Iz)
IG
(lb)
ti
a
(la)
led
K2(R[G],Iz)+
L
where
is
((Izp) C Piz).
-) the
K2(R[G])+ -- K2(R[G])+
K1(R[G],Iz)
logarithm homomorphism constructed in Theorem 2.8
Square (la) commutes by the naturality of
row is part of the relative exact sequence for the ideal 1.13).
L(1)
The bottom
A.
Iz
(see Theorem
The upper row is part of the homology sequence induced by the
conjugation G-action on the short exact sequence
(and note that
To see
R[G] --) R[G] -> 0
1z
0
H1(G;R[G])
that
surjects onto
square
together with liftings
g,h E G.
LoaKoee(g 0
Set
H = (z,h),
fix any commuting
Then, for any
r E R.
LoaK({g,u(rh)}) = Log([g,u(rh)]).
an abelian group.
Log([g,u(rh)]) =
=
commutes,
(lb)
Hl(a;R[G])).
Then, in
1)
1
(1-z)R[H],
- Log(u(rh))
- Log(u(rh))
g,h E a
CHAPTER 9.
(1 -
1
=
l
215
C11(Z[G]) FOR P-GROUPS
- Log(u(rh)))
l-zp = 0)
- F(u(rh))
.g l - r.h+2-1
=
r.h+21).
= 6.(g ®
The surjectivity of
AG
together with the fact that
now follows from the commutativity of (1), K2(R[G],Iz)+ C Im(OC)
(by (1)).
13
The proof of Proposition 9.4(i) can also be adapted to show that for
any prime p,
any p-group
G,
and any i> 1,
if p'>2
J ({g,l+plh} : g,hEG, gh=hg) l {g,1-2h} : g,hEG, gh=hg)
K2(Zp[G],p i ) = 1
The description of
C11(Z[C])
when
C
if pi=2.
is an odd p-group is now
immediate.
Theorem 9.5
For any odd prime
p and any p-group
H1(G,Z[G]) G Cp(Q[G]) ) C11(Z[G])
is exact.
with center
In other words, if Ki,
D[G] = Rk=1Ai,
the sequence
G,
- 1
where each
then
A.
k C11(Z[G])
=
Coker[oGoyG
:
H1(G;Z[G])
(}1Ki)p]
i=1
k [111 (UKI)p]/(UGo'G(
Proof
h)
: g,h E G, gh = hg).
Consider the localization sequence
K2(Zp[G]) G Cp(l[G]) G + CI1(Z[G]) - 1
is simple
216
C11(Z[G]) FOR P-GROUPS
CHAPTER 9.
of Theorem 3.15.
By Proposition 5.11(1),
P(@[G])
the standard involution, and Theorem 5.12.
and aG both commute with
vG
is fixed by the involution by
Hence
Im(cC) = V,(y( p[G])+
= (v,({g,u)) : g E G, u E K1(2 p[CC(g)])+)
(Prop. 9.4(i1i))
= ($G(X) : g E G, x E r(K1(2 p[CG(g)])+))
(Prop. 9.3)
(Theorem 6.6)
= 1G(Hi(G; p[G]+)).
4i(g®h) = JG(g®h 1)
But
by definition, and so
0
Im(f.) = Im(SPG).
Note in particular that by Theorem 9.5, for any odd prime p and any p-group
G,
the kernel of
8G: Cp(f[G]) -» C11(Z[G])
elements which come from rank 2 abelian subgroups of if
4
denotes the set of rank 2 abelian subgroups
is generated by In other words,
G.
H C G,
then there is
a pushout square
® C (Q[H]) s ® Cl (Z[H])
H" 1
HEdp 1
1
s Cll(Z[G])
Cp(D[G])
If, furthermore,
C11(Z[H]) = 1
this is the case whenever
for all
Cpl x Cpl ¢ G),
C11(Z[G]) = Coker[ Cp(D[H])
In the case of 2-groups,
H E A
(and by Example 9.8 below
then
Cp(Q[G])].
the situation is more complicated.
following theorem gives an algorithm which completely describes
when case.
G
The
C11(Z[G])
is abelian, but which only gives a lower bound in the nonabelian
CHAPTER 9.
C11(Z[G]) FOR P-GROUPS
Theorem 9.6 Fix a 2-group G,
217
and set
X = (2g, (1-g)(1-h): g,h E G) C Z[G].
Write
where the
[Gab] = ni-1Ki
Ki
are fields, set
9 = (1Si5k : Ki ¢IR) = {1SiSk : Ki *Q}; and f or each projection.
let
i
G
Then if
*C
C ox
ki: G -
("K )2
be the character induced by
is abelian, the sequence aG
' C('U[G])
) C11(Z[G]) ----> 1
is exact; and so "K` .]/(cGo+C(g®x): g E G, x E X).
SK1(Z[G]) = Cl1(Z[G]) - [ iE9
Otherwise,
8G
induces a surjection
C11(Z[G])
»
Coker[H1(C;Z[G]) G' C(@[G])
Coker[cub o*
Proof
If
G
augmentation ideal
K2(1[G]) = K2(
is abelian,
proj- C(D[Cb])]
: Hl(G;Z[G]) - iII'K]
then by Corollary 3.4, applied to the
I = (g-1: g E G) C Z2[G],
) ®
K(22[G],I)
Also, by Theorem 6.6, since
6 ({g,u} : g E G, u E 1+I).
(1-g)+ (1-h)=(1-gh)
(mod I2)
r(1+I) = Ker[w: HO(G;I) -, Gab, = { aigi: Iai = 0, fgi` =
Hence, by Proposition 9.3,
for
4
g,h E G.
= I2
218
CHAPTER 9.
C11(7L[G]) FOR P-GROUPS
Im[,pG: K2(Z2[G]) -0 C(Q[G] )] = (J+G(g ®r(u)) : g E G, U E I+I) = +G(G 0 12).
The relations
PG(g ®g) =
and
1
and hence that
*G(G® I2) = 4G(G®X);
Cll(7[G]) = Coker(,pG) = C(l[G])Iq'G(G®X) = [ n (l iE9
If
set
G
show that
PG(g ®h) = PG(g O h 1)
a: G -' Gab
is nonabelian, let
Ia = Ker(22[G]
l
)]Ia o\G(G®X).
`
denote the projection, and
Consider the following homomorphisms:
7L2[Cab]).
K(a)
.
K2(7L2[Gab] ) a K1(g2[G] , Ia) r
H1(Gab;22[Gab])
Both rows are exact, and g E Gab and
ra
u E 22[Gab],
H
I
HO(GIa)
is the homomorphism of Theorem 6.9.
and any liftings to
g E G and u E 22[G],
ra(aa({g,u})) = ra(Lg,u])
1
- rG(u)
(by definition of
Ia
= %(g®r(u)). Hence, for any x E q(2 2[G])'
if we write
k K2(a)(x) = {-1,-1}r ll {gi,ui}
(gi E Gab,
ui E
i=1
(using Corollary 3.4), then
k
k g®r(ui))
aH( i=1
= ra oaa(ll {gi,ui}) = ra as K2(a)(x) = 1. i=1
For any
C1,(Z[G]) FOR P-GROUPS
CHAPTER 9.
219
In other words,
k wO
(Proposition 9.3)
b(K2(a)(x)) _
E *dab(Ker(aH)) = C(a)opp(H1(G;Z2[G]))
So there is a surjection
C a
C11(Z[G]) --s Coker[K2(22[G]) -C) C2(Q[G])
and this finishes the proof.
Recall Conjecture 6.13:
C2(D[Cab])]
o
that for any p-group
G,
there should be an
exact sequence r2
H3(G) -) W2(Zp[G])
Hl(G;Zp[G])/(g®g)
which is natural with respect to group homomorphisms.
not be enough to give a general formula for
H2(G),
This would still
in the 2-group
C11(Z[G])
case (for reasons discussed below), but it does at least suggest the following approximation formula:
Conjecture 9.7 each Ki.
Ai
Fix a 2-group
and write
Q[G] = nk=1Ai,
is a matrix algebra over a division algebra
Di
i E 1-1
if and only if
D.
and
$=(i
:K1¢IR)
is a quaternion algebra).
where
with center
Set
9={i : D1gIR} (so
G,
Define
220
Cl (Z[G]) FOR P-GROUPS
CHAPTER 9.
C2(I[GJ) _
C1Q(z[C]) = aG(C2(QCGJ)) S C11(Z[C]);
C2(Ai) S C2(QLGJ);
11
iEl-$
and let
G - C11(z[G])/Cll(Z[GJ)
G > C2(RCGJ)/C2(QLGJ)
H1(G;Z[GJ)
be the homomorphisms induced by
yG
and
8G,
Then there
respectively.
are homomorphisms gab: Uab(G)
--> C11(Z[G])/ClQ(Z[G]),
9: H2(G) -p SK1(Z[G])/C1Q(Z[G]),
such that the following are pushout squares: _ vab G
W2
H1(G;Z[C])
.
( )
H2 (
r
)
19ab llpG 1®
a'
G
Cl1(Z[G])/Cl1(Z[G]) -> SK1(Z[G])/C1Q(Z[G]).
C2(Q[G])/C2(Q[G])
To see the connection between Conjectures 9.7 and 6.13, assume that Conjecture 6.13 holds, and consider the following diagram:
Kj(Z2CGJ)
r2
02
) H1(G;z2[G])/(g®g)
b(G) -> 0
=lid
I
I Gb 1
61
(22CGJ) G C2(Q[G])/C2(D[G]) G C11(z[GJ)/C1Q(Z[C]) . 1.
Both rows are exact; and the left-hand square computes on symbols
when
u E (22[CG(g)])*,
go r(u)). (see
If
Oliver
pG = +G o r2
semidihedral;
r2 [6,
r2({g,u}) =
is also natural with respect to transfer homomorphisms Conjecture
can be and
by Proposition 9.3 (note that
{g,u},
5.1]
reduced
this
is
to
easily
for the
details),
case where
checked.
The
then
G
relation
the is
cyclic
first part
of
or the
conjecture would then follow immediately.
The second part of the conjecture (the existence of
6
defined on
C11(Z[G]) FOR P-GROUPS
CHAPTER 9.
221
H2(G)) is motivated partly by the isomorphism
SK1(Z[G])/C11(Z[G]) = H2(G)/H2b(G)
of Theorem 8.6; and partly by the existence of homomorphisms
H2(G) --) Lp(Z[G]) E- H1(Zl2;SK1(Z[G]))
defined via surgery.
There is some reason to think that this surgery
defined map can be used to show that
This conjecture seems at present
at least, is well defined.
Aab,
to be the best chance
for getting
information about the extension
1 - 1 Cll(Z[G]) - SK1(Z[G]) -'' SKi(2[G]) -' 1
when
G
is a 2-group.
In fact,
if the conjecture can be proven,
should then be easy to construct examples of
it will be shown in Section 13c that
In contrast,
not split.
extension always splits when G There
seems
C11(Z[C])
or
be
to
no
SK1(Z[G])
is a p-group and p
the element
property that
r2(x)= 0,
but
For example, if
central and cyclic,
then
this,
the image of
G
K2($2[G])
{g,u}
has the
K2(7L2[C])
in
be handled with the
is a 2-group such that
[G,G]
is
can be shown to be generated by
g E G
for
principal, at least - also when @[G]
and
C2(Q[G])
u E (I2[CC(g)])*.
Using
can be described - in
contains quaternionic components.
Another class of nonabelian 2-groups for which computed using Proposition 9.4 is that of products
abelian and C11(Z[G])
G= (a,b) =Q(8),
pG(x)#l.
There are, however, some other cases which can
and symbols
describe
x = (a2,r-1(l+a+b+ab)) E K2 (22[G])
present techniques.
{-1,-1},
would
The problem with including
quaternionic components in the above diagram is that when
for example,
this
is odd.
conjecture which
obvious
completely.
it
G where this extension does
is already known.
C11(Z[G]) G x H,
Fix such G and
where
can be
H
H, and set
is
222
C11(Z[G]) FOR P-GROUPS
QllPTER 9.
I = Ker[Z[G x H] --4 Z[G]]
IQ = Ker[@[G x H] - + @[G]].
and
Then
C11(Z[G x H]) = C11(Z[G]) 0 Cl1(Z[C x H], I);
and using Proposition 9.4(11):
C11(Z[G x H], I) = Coker[K2(Z2[G x H],I2) -+ C(R[G x
C(@[GxH],IQ)/(v0({g,l+(1-z)h}): zEH, g,hEGxH, gh=hg). A special case of this will be shown in Example 9.10 below. We now look at some more specific examples of computations.
The case
of abelian p-groups will first be considered.
It will sometimes be convenient to describe elements in using the epimorphism
its projection
$G: RC(G)
p(@[G])
fG'p: RC(G)
description of G (but adapted to any irreducible C[G]-representation summand of such that If
CP(A)
and let
Q[C],
V
to p-torsion.
A
let
a: K = Z(A) '-1 C
is the irreducible C @ K A-module. 1,
if
isomorphism, and if
(})p
GA: Cp(A) pn = I(
}P
be the unique simple
be the unique embedding,
Then $G,P([V]) E Cp(A).
is the norm residue symbol
$G,P([V]) = aAloa 1(En) E CP(A).
(t)
SK1(Z[CPn x CP]) = 1
(it)
SK1(Z[CP2 X CP2]) -
For
then
)
Example 9.8 Ftx any prime
Recall the
given in Lemma 5.9(11).
G,P) V,
C(I[G])
of Section 5b - or rather
)1 C(ID[G])
p.
Then
for any n 2 0,
(Z/p)P-1.
and
(fn = exp(2ni/Pn))
CHAPTER 9.
C11(Z[G]) FOR P-GROUPS
The two computations will be carried out separately.
Proof
simplify the notation, the groups
Step 1 and
223
For each
SK1(Z[Go]) = 1
p
Gn-1 = n/(gp- n-1) Also, if
for each
p = 2,
then
Then
n.
SK1(Z[Gi]) = 1
p
If
Igl = pn
is odd,
then
is easily seen to be generated by the elements
,P(h®ghi) n Z 2,
where
p
(C('D[G1]) = C(Q[C2xC2]) = 1).
Cp(Q[Gi]) = (Z/p)p1
Now fix
n
by Theorem 5.6.
by Theorem 5.4
are written additively here.
write G = C n x C = (g,h),
We identify
= p.
IhI
n,
Cp(D[G])
To
and
(OS i Sp-l)
P(g®h). SK1(Z[n-1]) = 1.
and assume inductively that
(2a, (1-a)(1-b): a,b E G)
Z[G
if
p = 2
if
p
Set
X = Z[Gn ]
so that
SK1(Z[Gn]) = Cp(D[Gn])/y(Gn®X)
@[Gn] = Q[Gn-1] x A,
where
is odd;
by Theorem 9.5 or 9.6.
Write
is the product of those simple summands
A
upon which
g
denote the
C[Gn]-representation with character
acts with order
For each
pn.
r = 0,...,p-1, XV,(g) _fin,
Vr
let
)(V, (h) =
1
n-1
_ (En)rp elements
(En = exp(2wi/pn)). f(Vr),
Then
each of which has order
Since SK1(Z[n-1]) = 1,
CP(A)
is generated by
the
pn.
we have
CC(Q[Gn]) = (Cp(A), *(Gn ®X))
Also, a direct computation shows that for each O S r
(1)
p-1,
224
C11(7L[G]) FOR P-GROUPS
CHAPTER 9.
(PZlgipn-2)(1 + (p-1)g 1-0
'plg ®
rpn-1h
-
Plgipn_t)) i-0
if
Y((p-l)-Vr) + P Cp(I[Gn-1])
p
is odd
E
n
r
if p = 2.
-1])
Together with (1), this shows that for each
r,
there is some
'q r E Gn ® X
such that
y(11r) E 3'(Vr) +
The elements that
y
y(,qr)
then generate
is onto, and hence that
Step 2
The proof that
Cp(A).
Together with (1), this shows
SK1(Z[Gn]) = 1.
SK1(7[C4xC4]) = 1/2
is very similar to the
proof of Example 5.1, and we leave this as an exercise. odd.
Write C = Cpl x Cpl
for short, fix generators
So assume
g,h E G,
p
is
and set
H
_ (gp,hp).
Let
1
and
denote the sets of irreducible
512
tions upon which C acts with order
p and
p2,
C[G]-representa-
respectively.
Define
a . "2 - "1 by letting
a(V),
satisfies a,b E C,
for any V E 312,
Xa(V) = ()(V)p.
Then by Definition 9.2, for any generating pair
p(a®b) = g(V O a(V)),
tion such that
be the representation whose character
XV(a) = f2 and
where
V E
2
is the unique representa-
XV(b) = 1.
Now define an epimorphism
(Z/p)p+l
p : Cp(@[G])
by setting
s p(Q[G/H]) = Cp(Q[ pxCp]) t--
(V)
for
V E1; p(g(V)) = 4(a(V))
for
V E2.
CHAPTER 9.
We have just seen that (a,b) = G.
for all
Ker(13) C y(C ®Z[G]);
/3oy(a®b) = 1
Also,
p);
exponent
Cl1(Z[G]) FOR P-GROUPS
and
poy(a®b) = 1
and that
a E H
if
225
if b E H.
j3oy(a®b) = (3oy(a®1)
has
Cp(@[G/H])
(since
Since
if
y(a®a) = 1
it now follows that
a,
(Z/P)P-1.
O
SK1(Z[G]) = Cp(@[G/H])/(Roy(S®l), P-,P(hOl)) =
Some more complicated examples of computations of abelian p-groups
G
SK1(Z[G])
for
Some
can be found in Alperin et al [3, Section 5].
of these are listed in Example 6 at the end of the introduction.
The next example illustrates some of the techniques for computing for nonabelian p-groups
Cll(Z[G])
G
already have seen one example of this: quaternion,
or semidihedral
C11(Z[G]) = 1
for any dihedral,
(2-)group by Example 5.8.
Note that for
when G
groups of the same size, it is often easier to compute C11(Z[G])
is nonabelian - C(@[G])
is smaller in this case, and computations can
G
frequently be carried out via comparison with proper subgroups H
which C11(Z[H])
Example 9.9
We
using Theorems 9.5 and 9.6.
for
is already known.
Ftx a prime
Then C11(Z[G]) A 1,
p,
G
and let
be a nonabelian p-group.
p = 2 and
unless (possibly)
Gab has exponent
2.
Also,
SK1(Z[G]) = Cl1(Z[G]) = (Z/P)p-1
(i)
if
p
is odd and
jr,
= p3;
and
(it)
if
p = 2 and
IGI = 16,
11 SK (Z[G]) = C1 (Z[G]) 1
Proof
p
1
Z/2
then
if
Gab . (C2) 2
if
Gab
or
(C2)3
C, X C2 .
The proof will be split into two cases, depending on whether
is odd or
p = 2.
Note first that all of the groups
(ii) have abelian subgroups of index
p.
Hence
C
SK,(! [C]) = 1
in (i) and for these
226
C11(Z[G]) FOR P-GROUPS
CHAPTER 9.
G by Corollary 7.2, and
Case 1
p
Assume
Ho = [G,G] a G.
is odd, and fix a nonabelian p-group
Then
[G,Ho]
So
is also nonabelian.
H2(C) - H2(Gab)
ordered so that
Ho
(G
being nilpotent); and
Gab
Ho/[G,Ho] -
-'
Set
G.
G/[G,Ho]
in the five term homology exact sequence
b 0 1
Write
of Theorem 8.2.
SK1(Z[G]) = C11(Z[G]).
(G/Ho)ab
C/Ho = Gab = (g1) x... X(gk),
and let
S(g1^g2)0 1;
o=((gl)°+(g2)',go,.... gk)4G/Ho
where the
g;
are
be the subgroup such that
H a G
Then
- 1
has the property that any
H
commuting pair g,h E C generates a cyclic subgroup in G/H. Now consider the composite
y ,P'
: H1(G;Z[G]) G i Cp(f[G])
s Cp(D[G/H])
(1)
CP(D[CP x p]) = (Z/P) By the construction of
and
y'(g201)
H,
for all
is generated by y'(g1@l)
Im(y')
Hence, there is a surjection
a E G).
) Coker(*(,) a » Coker(*') = (Z/p)p-1;
C11(Z[G])
and
we see that
(y(a®a) = 1
(2)
C11(Z[G]) s 1. If
=
IGI
p3,
so that
H = [G,G] = Cp,
then all nonabelian
K C G,
C[G] -representations are induced up from proper subgroups
which C11(Z[K]) = 1.
So the
Ker(a*) C Im(pp)
in (1) above,
for
and a
is
an isomorphism in (2).
Case 2 Now assume that such that
dab
1, and write e.,
g1,g2
arrange the
and that
is not elementary abelian.
C/Ho = Gab = (gi) x ... x (gk)
gi
so that
surjects onto
C/H.
Igil 2 4.
Then
G
Set
lift to noncommuting elements of
((g1)41(g2)2,g3,...,gk) G
p = 2,
Let
G/H = C4 X C21
is a nonabelian p-group Ho = [G,C],
such that
S(gl^g2) # 1
G/[G,Ho]);
H C C
as in Case (i.
but this time
be such that
H/Ho =
and no abelian subgroup of
CHAPTER 9.
C11(Z[G]) FOR P-GROUPS
227
Define
' C(Q[G]) -a C(D[s])
: H1(G;Z[G])
y'
C(ID[C4 xC2]) = (L4)2, as before; so that there by Theorem 9.6 is a surjection
C11(Z[G])
» Coker(+').
Then in this case,
IM(V) = (`V'(g10 1), i'(g20 1)=`V'(g2g192),','(gi0 g2)),
and this has index 2 in C(Q[G/H]). G
If
is any nonabelian group of order 16, then
K
all proper subgroups
G
C11(Z[K]) = 1
for
(see Examples 5.8 and 9.8, and Theorems 5.4
So by Proposition 5.2, there is a commutative square
and 5.6).
RC(K) -5 ®C11(Z[K]) = 1 KEG
Nr-u
G
RC,,R(G) G - C1 l (Z[G]) ; IC11(Z[G])I < jCoker(f)j.
and hence order that
2
G
Gab = C4 X C2,
if
1
always has an abelian subgroup
nonabelian
irreducible
C[K]-representations).
2
and order
if GabC4xC2,
is easily checked to have
Coker(f)
otherwise
K
of index
C[G]-representations
We have seen that
are
C11(Z[G])
(note, for example, 2,
and that all
induced
up
from
has order at least
and this completes the computation.
o
As has been mentioned above, Proposition 9.4 can be used to calculate C11(Z[C x H]),
Cll(Z[C]) of this.
for any abelian 2-group
is already known.
H.
and any 2-group
G
for which
The last example illustrates a special case
228
CHAPTER 9.
Example 9.10 Let
C
C11(Z[G]) FOR P-GROUPS
be any 2-group.
Then, for any
C11(Z[C X (C2)k]) =
k,
®li/'C11(Z[G],21);
i=0
where for each
i Z1,
C11(Z[G],21) - C2(D[G])/(p({g,l+2ih}): g,h E G, gh = hg).
In particular, if
G
is any quaternton or semidihedrat 2-group, then
C11(Z[G X (C2)
Proof
For abelian
1.10 and 1.11].
G,
(U2) 2 k -k-1
k])
this is shown in Alperin et al [3, Theorems
The proof in the nonabelian case is almost identical;
except that Proposition 9.4(i) is now used to construct generators for p(K2(Z2[G],21)) C C2(@[G]).
The last formula (when
semidihedral) is an easy exercise.
D
G
is quaternion or
Chapter 10 THE TORSION FREE PART OF Wh(C)
So far, all of the results on
K1(Z[G])
and
presented here
Wh(G)
have dealt with either their torsion subgroups or their ranks; and that
suffices when trying to detect whether or not any given vanishes.
x E Wh(G)
For many problems, however, it is necessary to know specific
generators for
Wh'(C) = Wh(G)/SK1(Z[G]);
for some prime
p.
or to know generators p-locally
In general, this problem seems quite difficult, since
it depends on knowing generators for the units in rings of integers in global cyclotomic fields, and this is in turn closely related to class
numbers. One case which partly avoids these problems is that of p-groups for p
regular primes
(including the case
p = 2).
For such
Wh'(Zp[C])
G,
is a free 2p module by Theorems 2.10(i) and 7.3; and so the inclusion 7L[G] C 2p [G]
induces a homomorphism 1p @Wh'(G) -+ Wh'( p[G]).
This is
a monomorphism (Theorem 10.3 below); and the image of the composite
TG :
p 0 Wh' (G) }---i Wh' (Zp[C]) Y-G HO(G; p[C] )
will be described in Theorems 10.3 and 10.4.
One consequence of these
results (Theorem 10.5) is a description of the behavior of surjections, and under induction from cyclic subgroups of the
In
last
part
of
determining which elements of by units.
the
chapter,
Wh'(C)
we
(or of
turn Wh(C))
to
Wh'(C) under
C.
the problem of are representable
Theorem 10.7 gives some applications of logarithmic methods to
this problem in the case of 2-groups. all elements in Wh'(Q(32)x C2 X C2)
For example, it is shown that not
are represented by units in the group
In addition, some of the results in Magurn et al [1] are listed:
ring.
these include examples (Theorem 10.8) Wh'(G)
of quaternion groups for which
is or is not generated by units.
The first step towards obtaining these results is to establish an upper bound for the image of
fC
in
HO(G;p[G]).
This is based on a
230
CHAPTER 10.
THE TORSION FREE PART OF Wh(G)
simple symmetry argument, and applies in fact to arbitrary finite
Fix a prime
Lemma 10.1
p,
(g+g-1 -gn-g n,
be any finite group.
Let
be defined as above, and set
FG: 2p @ Wh' (G) ---+ HO(C;p[C]) Y(C) =
G
and let
G.
h-hm : g,h E G, h conj. h 1,
(n,Igl)=1, (m,Ihl)=1) C HO(G;p[G]). Then
fG(2p O Wh'(G)) C Y(G).
Set
Proof
R
By Theorem 1.5, field for
K = Wn (In = exp(2ni/n)),
n = exp(G),
is the ring of integers in
and K
K,
and
R = 7Lcn.
is a splitting
In particular, we can write
G.
K[G]
k n Mm.(K) i=1
for some
Consider the following commutative diagram:
mi.
K1(7LCG])
Z "R
K1(R[G]) f[ det o pr;
[I R*
iInlog =1 Ilog
I log
fl.Tr o pri
[C])
HO(C;
)
HO(G;KpCG])
k fi Kp
i=1
where
pri:
K[G] --» M(K) m,
denotes
the projection onto
the
i-th
component.
By Theorem 1.5(1) again, for any rya E Gal(K//) = Ga1(QC n//)
acts on
HO(G;p[G])
17a(ri).gi.
HO(G;Kp[G]) g E G,
Then and
ikp
and fTropri
a E (Uln)*,
such that HO(G;Kp[G])
Ta(Cn) = (cn)a via the action
commutes with
(note that each matrix
can be diagonalized).
there is an element
the
Also,
(7/n)*
ya(jrigi)
=
(Un)*-actions on
pri(g) E Mri(K),
for any
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
Write T = (Z/n)for short. u/u = i_1(u)/u E (±Cn)
For any u E R*,
and
H Ta(u) = NK, (u) E Z* _ (tl). aET
So by the commutativity of (1), for any
T_1(Tropriolog(u)) = Troprlolog(u)
a
u E K1(Z[G])
and any 1Sis k,
I Ta(Tropriolog(u)) = 0.
and
T-linear isomorphism,
7_1(log(x)) = log(x)
and
for all
x E K1(Z[C]).
the
(O(Zrigi) = jrigp).
Ta
231
Also,
HO(G;6p[G]))
17 a(log(x)) = 0 aET
(in
rG = (1 -
and $
It follows that
rG(K1(Z[G])) S HO(G;Zp[G]) n x E
x,
ll
We now restrict to the case where TG(Zp ®Wh'(G)) = Y(G)
and to describe
Im(TG)
when
C
whenever p = 2.
for some
ring of integers.
p
The goal is to
is an odd regular prime,
and to Hilbert, and
p = 2,
p.
Proposition 10.2 K C ((En)
is a p-group.
The key to these results is the
following proposition, due to Weber for Iwasawa for odd
I Ta(x) = 01 aET
a
= Y(G).
show that
commutes with
Fix a prime
n Z 1
p
and a number field
(gn = exp(2ai/pn)).
Let
K
such that
R = Z[gn]f1K
be the
Then the homomorphism
LK
induced by the inclusion
: Zp®ZR ) (Rp) R C Rp
(and by the
Zp-module structure on
232
CHAPTER 10.
THE TORSION FREE PART OF Wh(G)
is injectiue. Coker(LK)
p
If
is p-torsion free.
p = 2),
is regular (possibly p = 2,
If
then (and
K = Q(fn+fn
and if
then
R = 7[in+(nl]),
{u E R* : v(u) > 0,
v: K '4
all
{u2 : u E R*}.
IR}
(1)
=
The injectivity of
Proof conjecture.
For a proof, see, e. g., Washington [1,Corollary 5.32].
We next show that regular.
L C K
If
Coker(cK);
if = fn,
is a special case of Leopoldt's
cK
Coker(cK)
is any subfield, then
thus suffices
it
torsion free whenever
is
for short, and let
to consider
p = (1-f)R C R
p
is
Coker(cL)
is a subgroup of
the case
K = D(fn).
Set
denote the maximal ideal.
Define indexing sets
11.
and by
We include here a construction
due to Hambleton & Milgram [1].
n > 3
(otherwise
1)
J = 0).
Set
1)R =
(so AR =
p2).
(1+ X)(1 -X) = -(1 +f 2 +E-2); and an easy induction shows that NKo/Q (1+A) = -1.
Let
a set
p = 2,
If
this shows that there is a 7L-basis for
Janusz [1, Theorem 1.11.19]),
Coker(cK)
=
j E J,
1+(l_E)j(l_f-1)j_(1_E)2j
= 1 + (1-02i(-if -1) a 1 + (1-02j+1
In either case, since
for
by assumption, and hence that
(1+(1-f)j(1-f-1)j)(1+(1-f)j)-2 =
2=
(3)
bi E Z'p7L)
x2j = uj
{xi}iEI
(Bp)*/(tif)
233
THE TORSION FREE PART OF Wh(G)
g E Cal(K/a)
so for all
i > 1,
be the element
(Ko = Q($)niR = Q(A))
g(g) = if3.
Then
g(A) = A3 - 3A;
(4)
and
234
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
g(Ai) -
Ai
(A3 =
R.
-4X)(
g(x)
X
i-R-1
p21+4.
) E Ai+2R =
(5)
B=0
Set u1=1+ A for all odd
(u1E R* by (4)); and define inductively J
3.
If congruence (3) holds for
(l+Aj-2+a)-1.(1+g(Aj-2)+g(a))
uj =
Xj-2
= 1 + (g(Aj-2) -
+ g(a) - a) (1 +
Xj-2)
(A3-3A)i-2
= 1 +
then
u.-2,
aEp2j-3(IIR=Aj-1R)
(some AJ-2 + a)-1
Aj+1R + A2j 2R C Aj+1R)
(mod
1 + (g(AJ-2)-
uj = uj12-g(uj-2)
_ Aj-2 = 1 + NJ
Aj+1R = p2j+2).
(mod
In other words, congruences (3) are satisfied for all finishes the proof that
Coker(cK)
It remains to prove (1):
K = @( n+gn1).
in
places of
K;
Set
the description of strictly positive units
G = Cal(K/1D).
Let
be the set of all real
V
v: K y R.
Define
if v(u) > 0
r0 ® 7L/2:
X (u) =
where
jl
vEV
Regard M = 0vEVDJ2
is a
M.
7L/2[G]-module of rank
By (4),
unique maximal proper submodule of
M;
rk7(R) = [K:Q] - 1 = IVI - 1 theorem
(Janusz
and so
and
[1,
1,
v(u) < 0.
so that
Im(X)
and hence
NK/Q(l+L+f-1) = -1;
Then by Example 1.12,
by Dirichlet's unit
if
1
as a free
7L/2[G]-submodule of
and this
j;
is torsion free.
i. e., all embeddings
A=$X
(by (5))
the
Im(X) ¢
A
RM =
is surjective.
ZIVl-1
Also,
x Zf2
Theorem 1.11.19]);
and
implies that
{u E R*: v(u) > 0,
all
v: K y IR} = Ker(X) = {u2 : u E R*}.
o
this
CHAPTER 10.
THE TORSION FREE PART OF Wh(C)
235
Proposition 10.2 will now be combined with a Mayer-Vietoris sequence to give information about Wh'(G).
For any prime
p and any p-group
Yo(G) _ (g+g
-gn-g n :
G,
define
g E C, pin) S HO(G;2p[G]);
Y(G) = Yo(C) + (g - gn : g E G, g conj. g 1, pin) S HO(G;2p[G])
Note that
C Yo(G),
abelian, and that
Y(G) = Yo(C)
that
if
p
is odd or if
is torsion free.
HO(G;7Lp[G])/Y(G)
G
As before,
is
rG
denotes the composite r
rG :
7Lp ®Wh'(G) - Wh'(7Lp[G]) - HO(G;ap[GI),
where the first map is induced by the inclusion 2p module structure on
Theorem 10.3 maximal order
1
For any prime
Al C Q[G],
ap ® Wh' (G)
By Lemma 10.1,
Wh'(2 p[G]).
Im(P.) C Y(G).
for any p-group
p,
and the
7L[G] C 2 p[G]
G.
and for any
there is an exact sequence
- tors
Y(G)
Coker[gp ®K' (T)
- K'( 'Aip)1.
Furthermore,
(t) (Q[G]
Im(IG) = Y(G)
p
is an odd regular prime, or if
p = 2
and
is a product of matrix algebras over fields; and
(ii)
YO(G) C IM(TG) C Y(C)
Proof Ki,
if
and let
Write
D(fi)
G
is an arbitrary 2-group.
where each
D[G] = ni-lAi,
R. C Ki
is contained in
if
be the ring of integers. for some
i.
Ai
is simple with center
By Theorem 9.1, each
In the following diagrams
Ki
236
CHAPTER 10.
THE TORSION FREE PART OF Wh(G)
7Lp®K1(Di) --) K1(fRp)
7Lp®Wh'(G) --- Wh'(7Lp[G])
(la)
[[nr
fli,
7Lp® (Ri)* 11
K.
and
I am _
ll
*
(j71p0 (Ri)
n(Ri)p
(lb)
Inn!.
*
(1)
I (lnr 1 ^* Il(Ri)p
Ki
the reduced norm homomorphisms have finite kernel and cokernel (Theorem 2.3 and Lemma 2.4), and the is finite.
Ker(t)
But
tKl
Wh'(G)
are both injective.
rG = rGot
are injective by Proposition 10.2.
is torsion free, and
Also,
hence
rk,(Wh(G)) = rk2(Y(G))
So
and
t
by Theorem
P
2.6, and so
[Y(G):Im(fG)]
is finite.
By Milnor [2, Theorem 3.3], for each there is a Mayer-Vietoris exact sequence
n>1
such that
pnM C 7L[G],
K1(7[G]) -) Kl (DI) ®Kl (7[G]/per) ) K1(R/p') The group
Ker[K1(7L[G]) -i Kl(71)] C SK1(7L[G])
is finite, by Theorem
2.5(i), and so this sequence remains exact after taking the inverse limit over
n.
Since
and
K1(Z[G]/p M) =
K1(V/p
i)
=- Ki(ip)
nm
by Theorem 2.10(iii), this shows that the sequence
K1(7[G]) ) K1(N) ® K1(2p[G]) - K1('Np)
is exact; and remains exact after tensoring by surjects onto
SK1(Uip)
7Lp.
Also,
SK1(1A)
(Theorem 3.9), and so the top row in the following
diagram is exact:
1 -, Ip ®Wh' (G) JrG Y(G) `
Wh' (2 p[G]) - Coker[8p ®K' (M) -' K' (ip)] JrG 0 HO(C;7Lp[G])
(p) (2)
CHAPTER 10.
By Theorem 6.6,
with
237
THE TORSION FREE PART OF Wh(G)
rGI(Y(G)) C Wh'(1 p[G]).
Y(G)
In particular, we can identify
Y(C) C Im(rG). Since
Y(G)/Im(fC)
is finite, the top row
in (2) now restricts to an exact sequence
1 -+ 7Lp ® Wh'(G)r G Y(G) 4 tors Goker[[p®Kj(IR)
Ki(ip)],
where 0= e o (rG1IY(G)). p
If
is regular, and if
is a product of matrix algebras over
D[G]
fields, then the reduced norm homomorphisms in (1b) are isomorphisms, and so
Coker(t,,)
this case.
is torsion free by Proposition 10.2.
arbitrary 2-group, then
all cyclic H C G;
In principle, information about
p=2
in
p
is
is abelian.
G
contains the image of
Im(TG)
and hence
and
If
G
is an
Y(H) = Yo(H)
for
Im(1'G) ? Y0(G).
it should be possible to use these methods to get Wh'(G)
is a p-group and
G
when
With certain conditions on
torsion in
Im(TG) = Y(G)
In particular, by Theorem 9.1, this always applies if
odd (and regular), or if
prime.
So
Coker[E 0 (Z[fk])* P
Washington [1, Theorem 13.56]).
p
p
an irregular
(see Ullom [1]), the p-power is understood (see also
(ap[fk])*]
But most results which we know of, shown
using Theorem 10.3, seem either to be obtainable by simpler methods (as in Ullom [1]); or to be quite technical. The next theorem gives a precise description of
G
is a nonabelian 2-group.
then of
D[G]
I(fn)
for
Recall (Theorem 9.1) that if
when
exp(G) = 2n,
is isomorphic to a product of matrix algebras over subfields (fn = exp(2ai/2n)),
and over division algebras
D(fk,j)
(C M)
k S n.
Theorem 10.4 10.3.
CG(7L2 ® Wh'(G))
Let
G
be a 2-group, and let
Y(G)
be as in Theorem
Then
k
k rG(22 ®Wh'(G)) = Ker[9G =
ll
i=1
oG'i
: Y(G) -
(Z/2),+ i=1
238
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
where
V1,...,Vk are the distinct irreducible
C[G]-representations which
are quaternionic, and where
OG'i
Y(G) C HO(G;Z2[G]) -4 7/2
is defined by setting
8G'1(g) = I dimm(fr-eigenspace of (g: Vi -> Vi)).
(for
g E G)
r>0
This is based on the exact sequence
Proof
1-
120Wh'(G) G+ Y(G) - tors Coker[22®Ki('R) - 4 Ki(12)] (1)
of Theorem 10.3; where and where
is a maximal order containing
iR C @[G]
6(r(u)) = [u] E K1(N2)'
Write
D[G] = nm-1Ai,
A = Ai = Mr(D),
where
where the
D
some
D
(i. e.,
n
is a field),
then
by Theorem 9.1, and hence
Theorem 1.19.
A.
Fix
are simple.
i,
is a division algebra with center
IRA C A be a maximal 7-order, and let
D = K
g[G],
and set Let
K.
R C K be the ring of integers. K C 1(fn)
NA
(fin = exp(2ai/2n))
is Morita equivalent to
R
If
for
by
So
Coker[220Kj(711A)
I
Ki(111A2)] = Coker[22®R*
)
(R2)*]
is torsion free in this case by Proposition 10.2.
By Theorem 9.1 again, the only other possibility is that (C al)
for some
n > 2
commutative diagram
(so
K = R(fn+(n1)).
D = Q(fn,3)
In this case, consider the
239
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
"
1 - ) 22 0 K1(vA)
) ® Z/2
g2 ® R*
ICA
(2) 1
(g2)*
K1(IA2) where if
X = ONv
is defined by setting By Theorem 2.3,
v(u) < 0.
nr2
if
Av(u) = 1
v(u) > 0;
is an isomorphism, and the top row in
By Proposition 10.2,
(2) is exact.
Av(u) = 0
X
is onto and
Coker(LK)
is torsion
free; and so by (2),
tors Coker[22 0 Ki(MA)
In other words,
-
tors(Coker(cA))
quaternion representation of
IR % A.
Ki(iA2)] =
(7(/2)IVI.
includes one copy of
7l/2
for each
Sequence (1) now takes the form
1 . 22 ®Wh' (G) G , Y(G) e (7V2)k; where k
is the number of quaternion components in
identifying
0
with
[1, Section 3].
9
The details of
IR[G].
as defined above are shown in Oliver & Taylor
o
A second description of
Im(IG),
when
C
is a 2-group, is given in
Oliver & Taylor [1, Propositions 4.4 and 4.5].
The next result is an easy application of Theorem 10.3.
Theorem 10.5
(i)
Fix a regular prime
a: G -* G
For any surjection
Wh'(a)
is surjectiue if
p
:
of
G.
p-groups,
Wh'(G) 3 Wh'(G)
is odd or if
exponent at most 2 otherwise.
p and a p-group
C
is abelian; and
Coker(Wh'(a))
has
240
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
(it) For any
x E Wh'(G),
G
from cyclic subgroups of
if
is a product of elements induced up
x p
is odd or if
G
is abettan;
and
x2
C
is
is a product of such elements otherwise.
Proof
Fix a p-group
and consider the group
G,
C = Coker[j{Wh'(H) : H C G,
H cyclic)
Wh'(G)].
)
By a result of Lam [1, Section 4.2] (see also Theorem 11.2 below),
a finite p-group. Y(G)/Yo(G)
C = 1p O C
So
by Theorem 10.3.
is odd or if
G
By definition,
Y(G)/Yo(G)
2
p
is trivial if otherwise.
To prove (i), it now suffices to consider the case
This proves (ii).
2p ®Wh'(C)
to a subgroup of
isomorphic
is abelian, and has exponent at most
where a and G are both cyclic. onto
is
By Theorem 10.3,
in this case, and so
Zp 0 Wh'(G)
Coker(Wh'(a))
surjects
is finite of order k
prime to Im(Wh'(a))
p.
It thus suffices to show, for any u E (7L[G])*, for some
Assume that
that
up
E
k.
G = Cp,
and
and consider the pullback
G = Cpn-1 ;
square Z[fn]
7L[Cpn] al
(fn = exp(2vi/pn)) l /p[Cpn-1]
Z[Cpn-
This induces a Mayer-Vietoris exact sequence
K1(Z[CP ]) - K1(7L[Cpn-1]) $ K1(7L[fn])
Set
I = Ker[Zfp[Cpn_i] - up],
u E K1(Z[Cpeither (3*(u)
) (Z/Plc n-1]).
the augmentation ideal. or
lies in
(3M(-u)
1+I,
Then for any and this is a
k
group of p-power order. some
k;
In other words,
and we are done.
0
up
E (-1,Ker(0,)) C Im(a*)
for
CHAPTER 10.
The result that cyclic p-groups
THE TORSION FREE PART OF Wh(G)
is onto whenever
Wh'(a)
(for regular
a
241
is a surjection of
is due to Kervaire & Murthy [1].
p)
We now turn to the problem of determining, for a given finite group G,
which elements of
7[G].
K1(7[G])
can be represented by units in
Wh(G)
or
This was studied in detail by Magurn, Oliver and Vaserstein in [1].
The main general results in that paper are summarized in the following theorem.
A simple Q-algebra
Note that
IH.
Q-algebra
with center
K
is called Eichler if there is
v: K '-> C such that either v(K) ¢ 1t,
an embedding ROvK A
A
A
is always Eichler if
or
[A:K] t 4.
and
v(K) C IR
A semisimple
is called Eichler if all of its simple components are Eichler.
Theorem 10.6
Let
be any semisimple Q-algebra, where
A=VxB
B
is the product of all commutative and all non-Etchler simple components in A.
Then for any 7L-order
projection to
21
an element
B,
and only if its image in particular, if
A
in
if
A,
x E K1(21)
B C B
is the image of
under
21
can be represented by a unit if
can be represented by a unit.
K1(B)
is Eichler - i. e., if
B
In
is commutative - then
there is an exact sequence
K1(2I)
2I
Proof
21-*B
SK1(B) -> 1.
See Magurn et al [1, Theorems 6.2 and 6.3].
o
We now list two results containing examples of finite groups where
is or is not generated by units.
Wh'(G)
G
The first theorem
involves 2-groups, and is an application of Theorems 10.4 and 10.5 above.
The second theorem will deal with generalized quaternion groups, and is proven using Theorem 10.6.
Theorem 10.7
(i)
represented by some unit
(ii) 32.
Set
For any 2-group u E (g[C])
G = Q(32) X C2 X C2,
Then Wh'(G)
G
and any
x E Wh'(G),
x2
is
.
where
Q(32)
is quaterntontc of order
contains elements not represented by units in
Z[G].
242
THE TORSION FREE PART OF Wh(G)
CHAPTER 10.
Proof
Point (i) is clear:
cyclic subgroups of
G
definition if
G
is a product of elements induced from
x2
by Theorem 10.5(ii); and
Ki(7L[G]) = (7L[G])*
is abelian.
To prove (ii), fix any element a E Q(32) generate the two factors
t1,t2
by
C2
and let
of order 16, Set
in G = Q(32) X C2 X C2.
x = (1- ti)(1 - t2)(a-a3) E HO(G;7L2[G]). A straightforward application of Theorem 10.4 (in fact, of Theorem 10.3) shows that x E rC(7L2 ®Wh'(G)).
Thus,
if all elements of
represented by units, then there must be a unit Iu) = x G(
64).
(mod
(Magurn et al
But using the relation
[1, Lemma 7.5(b)]),
Wh'(G)
such that
u E (7L[G])*
(g[ ,,J])* = ((7L[f+])*+J)
it can be shown that no such
See Oliver & Taylor [1, Theorem 4.7] for details.
exists.
are
u
a
The following results are similar to those in Theorem 10.7, but for generalized quaternion groups instead of 2-groups. n > 2,
denotes the quaternion group of order
Q(4n)
Theorem 10.8
For any
represented by a unit in
(i)
n
If
n > 2,
7L[Q(4n)].
is a power of
2,
and any
Recall that for any 4n.
x E Wh(Q(4n)),
x2
is
Furthermore:
then all elements of
Wh(Q(4n))
can
then the elements of
Wh(Q(4p))
can
be represented by units.
(ii)
If
p
is an odd prime,
all be represented by units, if and only if the class number
(iii)
For any prime
p ° -1
(mod 8),
Wh(Q(16p))
hp
is odd.
contains elements
not represented by units.
Proof
See Magurn et al [1, Theorems 7.15, 7.16, 7.18, and 7.22].
An obvious question now is whether, for any finite group x E Wh'(G),
x2
is represented by a unit in
7L[G].
G
o
and any
PART III: GENERAL. FINITE GROUPS
One
of
the
procedures
standard
when
working
almost
with
any
K-theoretic functor defined on group rings of finite groups, is to reduce problems involving arbitrary groups to problems involving hyperelementary i. e., groups containing a normal cyclic subgroup of prime power
groups:
For most of the functors dealt with here, one can go even further.
index.
The main idea, when dealing with
reduce computations first to the case where
etc., is to
C11(7[G])(p),
SK1(7Lp[G]),
G
is p-elementary (i. e., a
product of a cyclic group with a p-group); and then from that to the case
where G
is a p-group.
The formal machinery for the reduction to p-elementary groups is set up in Chapter 11.
The actual reductions to p-elementary groups, and then
to p-groups, are carried out in Chapters 12 (for C11(7L[G])).
The inclusion
C11(7L[G]) C SK1(7L[G])
Section 13c to be split in odd torsion.
and 13 (for
SK1(2p[G])) is
then shown in
Finally, in Chapter 14,
some
applications of these results are listed.
Since much of the philosophy behind the reductions in Chapters 12 and
13 is similar,
used in
it seems appropriate to outline it here.
the reduction to p-elementary groups
is
The main tool
induction theory as
This sets up conditions for when
formulated by Dress [2].
.A(G),
.N
being a functor defined on finite groups, can be completely completely computed as the direct or inverse limit of the groups
H C G
lying in some family.
.N(H)
for subgroups
The main general results on this subject are
Theorem 11.1 (Dress' theorem), Theorem 11.8 (a decomposition formula for certain functors defined on 1- or 2p orders), and Theorem 11.9 (conditions for computability with respect to p-elementary subgroups). Using these results,
SK1(7[G])(p)
be p-elementary computable for odd
when p = 2
(Theorems 12.4 and 13.5).
SK1(7[G]) (p)
-
p,
is shown in Chapters 12 and 13 to and 2-IR-elementary comput-able In particular, for odd
liHC SK1(7L[H])
(P)'
p,
244
where
is the set of p-elementary subgroups of
8
and the limits are
G,
taken with respect to inclusions of subgroups and conjugation by elements of
When
G.
the connection between
p = 2,
SK1(Z[G])(2)
and 2-elemen-
tary subgroups is described by a pushout square (Theorem 13.5 again).
The process of reduction from p-elementary groups to p-groups is simpler.
G be a p-elementary group:
Let
Write
is a p-group.
@p[Cn]
ni=1Fi,
=
k SK1(2p [G]) = ®SK1(Ri[ir])
where the
Fi
are fields, and let
Then
be the ring of integers.
R. C F.
where pIn and a
G = Cn x v,
C1 1(Z[G])
and
®C11(ZCd[A]) din
i=1
the first isomorphism is induced by an isomorphism of rings (Theorem 1.10(i)), and the second by an inclusion index prime to
groups
Z[G] C ])dinZCd[A]
(Example 1.2, Theorem 1.4(v), and Corollary 3.10).
p
SK1(Ri[a])
These results then lead to explicit descriptions of
when
p
p
and
G
(Theorems 12.5 and 12.10),
is odd (Theorem 13.9) or
nonabelian
G,
The
have already been described in Theorem 8.6, and the
are studied in Section 13b by comparing them with
C11(ZCd[A])
arbitrary
of orders of
G
C11(Z[ir]).
SK1(2 p[G])
and of
for
Cl1(Z[G])(p)
is abelian (Theorem 13.13).
For
the situation in 2-torsion is as usual incomplete, but
partial descriptions of
C11(Z[G])(2)
in terms of
C11(Z[v])
2-subgroups a can be pulled out of Theorems 13.5 and 13.12.
for
Chapter 11
A QUICK SURVEY OF INDUCTION THEORY
The term "induction theory" refers here to techniques used to get information about
when
.1t
in terms of the groups
.M(G)
is a functor defined on finite groups.
for certain H C G,
.N(H)
Such methods were first
applied to K-theoretic functors by Swan [1], when studying the groups K0(7L[G])
and
G0(7L[G])
systematized by Lam conditions for
finite
G.
whose Frobenius
[1];
Swan's
techniques were
functors gave very general
.4(G) to be generated by induction from subgroups of
lying in some family 5.
for
9,
G
or to be detected by restriction to subgroups in
later, Lam's ideas were developed further by Dress [2], who gave
conditions for when
for subgroups H C G
can be completely computed in terms of
A(G) in
4(H)
9.
The results of Dress are based on the concepts of Mackey functors, and Green rings and modules, whose general definitions and properties are
summarized in Section lla.
The central
theorem, Theorem 11.1, gives
conditions for a Green module to be "computable" with respect to a certain
family of finite groups.
Two examples of Green modules are then given:
functors defined on a certain category of R-orders (when
R
is any
Dedekind domain of characteristic zero) are shown to induce Green modules
over the Green ring
G0(R[-])
(Theorem 11.2), and Mackey functors are
shown to be Green modules over the Burnside ring (Proposition 11.3).
In Section llb, attention is focused on p-local Mackey functors: i. e., Mackey functors which take values in 2P modules.
A decomposition
formula is obtained in Theorem 11.8, using idempotents in the localized Burnside
ring
and
12(c)
this
reduces
the
computation
of
(p);
C11(7L[G])(p),
SK1(7L[G])(p),
group rings over p-groups.
SK1(2 p[G]),
etc. to that of certain twisted
This is the first step toward results in
Chapters 12 and 13, which reduce the computation of least for odd
p)
to the case where G
is a p-group.
SK1(7L[G])(p)
(at
246
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
Induction properties for Mackey functors and Green modules
Ila.
The following definitions are all due to Dress [2, Section 1].
A Mackey functor
(A)
is a bifunctor
from the category
.,N =
of finite groups with monomorphisms to the category of abelian groups, 1'
such that
is contravariant,
is covariant,
A.
-P(G) = #*(G) = -A(G)
for all
and the following conditions are satisfied:
C,
(1)
(ii)
send inner automorphisms to the identity.
.N*
For any isomorphism
a: G -' G',
and any pair
H,K C G,
A --
.N(G)
is equal to the sum, over all double cosets
A(H)
cg
(B)
* .41(c) A-A(g 1KgnH) g 36
denotes conjugation by
A Green ring
ring structure on
reciprocity conditions.
a*(xy) =
IS
'44(C)
.Y*:
for any
the composite
.N(H)
Here,
.N *(a) = A,(a)-1.
The Mackey subgroup property holds for e and
(iii) G,
and
.N*
A(K)
KgH C G, of the composites .N
A(KngHg 1)
*) .I1(K).
g.
is a Mackey functor together with a commutative
for all
C,
and satisfying the Frobenius
More precisely, for any inclusion
for
x,y E '44(G)
x E '&(G),
y E '44(H)
a: H I-- C,
CHAPTER 11.
A QUICK SURVEY OF INDUCTION THEORY
for
(where a*
=
A Green module
(C)
y E 41(G)
a* = 3*(a)) .
`3*(a) ,
together with a
x E (A(H),
247
over a Green ring
44(G)-module structure on
`3
A(G)
is a Mackey functor for all
the same Frobenius relations hold as in (B), but with
G,
A.
such that
y E A(G)
or
instead.
y E AI(H) (D)
For each
Let
T
G,
set
be any class of finite groups closed under subgroups. Then a Mackey functor
T(G) = (H C G: H E T).
called 'e-generated if, for any finite
Al
is
G,
Al
® AI(H)
) .(G)
HE'C(G)
is onto; any
G,
is called *-computable (with respect to induction) if, for
Al
A.
induces an isomorphism
li
A(G) =
AI(H).
HE'9 G)
Here,
the
limit
is
taken with respect
to all maps between subgroups
induced by inclusions, or by conjugation by elements of is
`6-detected, or
finite
G
G.
Similarly,
A
`41-computable with respect to restriction, if for all
the homomorphism
A(G) - Jim
AI(H)
HE'e(G)
(induced by
AI*)
is a monomorphism or isomorphism, respectively.
For convenience of notation, if and
is H -1 C
Indd = Am(i)
H C G
is any pair of finite groups
is the inclusion map, we usually write
: A(H) -) AI(G),
to denote the induced homomorphisms.
ResH = Al*(i)
:
A(G) - A4(H)
248
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
The first theorem can be thought of as the "fundamental theorem" of induction theory for Green modules.
Theorem 11.1
(Dress [2, Propositions 1.1' and 1.2])
Green module over a Green ring groups such that
'&
19,
is '44-generated.
`44
Then
is
.N
Let
be a
.N
be a class of finite
and let
*-computable for both
induction and restriction.
Proof
I
Fix any
:
write
G,
.N(H) --> .N(G)
1
for short, and let
'44 = '44(G)
.N(G) --
and
j
.p(H)
ET
HE
be the induced maps.
Choose elements
for
aH E '44(H),
H E 'B,
such that
IndH(aH) = 1 E 'Q(C).
(1)
HE'
For any x E .N(G), E Im(I)
x = HET
HE'44
by Frobenius reciprocity. H E '8,
i. e., if
To show that
R(x) = 0. I
property is needed.
x = 0
In particular, Thus,
I
double coset representatives for
H\G/K.
restriction, while
Here,
and
R'
composed with conjugation by
Fix any x E Ker(I).
let
gHKi
(1
i S nHK)
be
IHKi
'&).
I'
is one-to-one.
Consider the maps
A(gHKiKgHKi nH)
I HK,
(and similarly for
for all
surjective, the Mackey subgroup
H,K E '44,
RHKI
.M(K)
ResH(x) = 0
is onto and R
is injective and R For each pair
if
I
and
A(H)
R
R
are the induction and restriction maps
gxxi.
Write
x =
[xK, K] E KE'C
denote induction and
.M(K),
KE'C
CHAPTER 11.
where
and
xK E .N(K)
A QUICK SURVEY OF INDUCTION THEORY
ZKEVInd4(xK) = 0.
Then, in
1
249
A(K),
x = I [xK, K] = I [ResG(1).xK, K] K KEC KEG
= I
I [(Re.G o
(by (1))
K]
KEG HEM nHK
[(IHKioRHKi(aH))'xK, K]
KEC HE'C i=1 nHK
[IHKi(RHKi(aH)-RHKi(xK)), K]
KEC HE* i=1 nHK
[IHKi(RHKi(aH)-RHKi(XK)), H]
(by defn. of 1)
HE* KET i=1 nHK H]
E HET KET i=1
[aw (ResH o IndK(xK)), H]
HE'S KEC
I IndK(xK)), H] = 0;
KE
HET
and so
I
is injective.
Now fix some element
y = (yK)KET
in
4im .11(H).
HE'S
y=
Indd EM
Then, for each K E 0,
if
I(G/M)HI it 0 (mod p)
if
I(G/M) I = 0 (mod p).
H
shows
this
that
there exists
such that
0 (mod p)
1
if
I(G/M)HI
{ 0
if
I(G/W)HI = 0 (mod p).
XH(E) =
In particular, if
XH(E) = 1,
p-hyperelementary, and XH(E) = XC(E)
if
H E
H E 4(C);
then f(C)
and
gHg 1 C M
for some
for some
C C C.
XC(E) = 1
is defined such that
be defined by setting
XH(EC) = 1
0
C C C,
if and only if
so
H
is
Also, by (1),
since by choice of
I(G/M)Cj = j{gM: g 1Cg C M11 = IN(C)I/IMI
We may assume inductively that for each
g;
M:
(mod p).
an idempotent
EC
Then EC
can
H E f(C).
256
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
EC=E-J{EC:CSC, x(E)=1}.
o
Lemma 11.4 will now be applied to split p-local Mackey functors.
Assume that
p
is a fixed prime,
and that
is a p-local Mackey
A
For each
functor which is generated by p-hyperelementary induction. finite
be a set of conjugacy class representatives of
Cy(G)
let
G,
cyclic subgroups C C C EC(G) E O(G)(p)
of order prime to
For each
p.
C E Cy(G),
let
be the idempotent defined in Lemma 11.4, and set
AC(G) = EC(C)-A(G) C 4(G).
If
is p-hyperelementary - if
G
p-group - then for
kin
subgroup of order k
(and set
Proposition 11.5 functor
generated
G = Cn>4a
we write
Ak(C)
4k(C) = 0
Fix a prime
if
C C C
when
.AC(C)
A
®
4(G) =
for any finite
.AC(G)
is the
be any p-local Mackey
Then
induction.
p-hyperelementary computable for induction and restriction.
(i)
is a
k4n).
and let
p,
p-hyperelementary
by
=
ptn and it
where
is
.M
Also:
G.
CECy(G)
(ii)
For any finite
C and each C E Cy(G),
li
AC(G) =
4C(CAa) =
TrEO(C)) Here,
56(-)
A.
elements in
NG(C).
4M)
(or
Assume
applied to inclusions, and to conjugation by
G = Cn>la
is a p-group).
Then for any
automorphism of
4(H);
maps
Ar(C>4 a)
denotes the set of p-subgroups, and the Limits are taken with
respect to
(itt)
Iim
'RE9s(N(C))
is p-hyperelementary (where H = Cm A w C G
and for each
kim
(min),
pjn and
ResH o IndH
w
t s an
the induction and restriction
kIndC
:
- ResH : 4k(G) _-> .k(H)
and
Afk(G)
Afk(H)
are tsomorphisms.
257
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
Furthermore,
Afn(G) = Ker[ ® Res
A4(G) = A4(Cn A a)
:
®A1(Cn/p A a)].
)
pin Proof
Let I denote the class of p-hyperelementary groups, for
(i)
Define
short.
Eo(G) _
Then for any H C G,
E ,(G)EC(G) E O(G)(p).
1
if HEl(G)
0
otherwise.
(1)
XH(Eo(C)) _
In particular, for any
is an idempotent.
Eo(G)
Since
H C *(G).
Also,
ResH(Eo(G)) = 1 E f0(H)
is generated by p-hyperelementary induction,
A
and since IndH(x)
this shows that
for any H E X(G) and any x E A4(H); A4(G) = Eo(G) -A4(C) _
(iii) H = Cm Air
® E (C) -A(G) _
CECy(G) C
G = CnAir
Now assume that
For each
Ki = HflgiHgil = CmApi Then
gi E NG(Ki),
so
i,
where
pi
Alk(H).
Alk(G)
k
Choose double coset representatives i.
is p-hyperelementary, and that
kRes,
IndC
gi E n for all
(2)
Consider the maps
for some min. Alk(H)
® M (G)
CECy(G) C
and
g1,...,gr
for
(3)
H\G/H
such that
write
pi = {xEw : gixgil=x (mod Cm)).
gipig.
are both p-Sylow subgroups of
258
and are therefore conjugate in
Ki,
gi
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
is an inner automorphism of
Ki.
It follows that conjugation by
Ki.
Hence, in (3),
r
= I
kResH o kIndH
r
kIndK o (cg )* o kResK
=
I klnd,
and this is multiplication by
Ek(H)'
E flk(H).
[H/(HflgiHgil)] = i=1
For any
such that XK(Ek(G)) = 1,
KCG
Hence, for such
p-group.
have
just
seen
E flk(G)*
ResH o kIndH
that
and
(mod p).
by Proposition 11.3.
klm)
that
ResH olndH
and
.Nk(G);
is an isomorphism of
itself.
In particular, this shows that for any prime
pin,
' .A(C n/pXn)] = ® Ak(Cn Aa); kin
Ker[Res : 4(Cn >4 v)
kJn/p
and hence that
An(Cn
(ii)
>4 a) = Ker[®Res
Now let
:
A(Cn>4 i)
by
is multiplication by
It follows that
.Nk(H)
flk(G).
multiplication
is
kIndH o kResC
are both isomorphisms between
summing over all
is a
K/Ck
iskinvertible in the p-local ring
E flk(H)*;
kResH
ig 0
= IG/HI
) = I(G/H)KI =
This shows that We
and
K,
I(G/H)K/CkI
XK(
Ck a K
) ®A(Cn/p M7r)] pin
G be arbitrary, again, and define
C
kIndH
and
and (after .N(H)
to
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
259
® EC(G)-O(G)(P)
flo(G) = Eo(G)'G(G)(P)
CECy(G)
Then
is a ring factor of
flo(G)
by (2).
Also, since
and
fl(G)(p),
ResH(Eo(G)) = Eo(H)
.N
for all
0o
Frobenius reciprocity relations show that
is an flo(G)-module
.M(G)
H C G
by (1), the
is a Green ring, and that
is a Green module over flo. If
G
x(Eo(G)) = 0 by (1), and so
is not p-hyperelementary, then
the coefficient of
in
[G/G]
is
Since
zero.
Eo(G) E 0(G) (P)
multiplication by
shows that
[G/H] E R(G)(p)
0o(G) =
is
.4
by definition,
this
is generated by induction from proper
subgroups in this case. Theorem 11.1
IndH o ResH
is
In other
words,
is
Oo
*-generated; and by
*-computable with respect to both induction and
restriction.
In particular, for any C E Cy(G),
MG) G ).14(H)
AC(C) = EC(G)'41(G) H
By definition of
EC(G),
for
H
H im ResH(EC(G))-A(H).
4C,(H)
H E 2I(G),
for some (unique) C' conjugate to Also, if H = Cn>r, where C' C Cn, pin, H D C'
AC,(H) = AC,(C' >4 a)
(4)
if
and is zero otherwise. and it is a p-group, then
C,
by (iii). So (4) now takes the form lim ),NC(H) _
AC(C) _ 25
lim
H
CCH where the limits here are taken with respect to conjugation in The proof for inverse limits is similar.
So far, functor
A.
the summands
NG(C).
13
the results in this section apply to any p-local Mackey When
.N(G) = X(R[G])
.NC(G)
and
.14(G)
for some functor
description in terms of twisted group rings.
as summands of
X on R-orders, then
can sometimes be given a more accessible
(ordinary) group rings
K[G]
Such rings arise naturally when
G
is p-hyper-
260
elementary.
n
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
As usual,
This is explained in the following proposition.
denotes a primitive n-th root of unity.
Fix a prime
Proposition 11.6
K[Cn] _
Write
[lm-1Ki,
where
p-K-elementary if and only if leaves each
K.
invariant.
K
a field
p,
zero, and a p-hyperelementary group
G = C A a
the
of characteristic
(pIn,
are fields.
K.
Then
a
the conjugation action of
G
on
is
K[Cn]
splits as a product
K[G]
In this case,
a p-group).
it
K[G] = K[Cn la] = II Ki[w]t. i=1
where each
is the twisted group ring with twisting map
Ki[lr]t
t
: n --> Cal(Ki/K)
a
induced by the conjugation action of
on
a Dedehind domain with field of fractions integral closure of
then
R,
[lm
R [a]t
i=1 i
Ki. K,
If, Furthermore,
and if
R. C K.
is an R-order in
K[G]
R
is
is the and
m
R[G] = R[Cni4ir] C [] R1[a]t C i=1
Proof
and
it
By Example 1.2, we can write
acts on each K ®Q QCd
a
Then
Gal(Kcd/K)
con
-
H KO QCd; Q
din
via the composite
Aut(Cn) -» Aut(Cd) = Gal(QC d/Q)
is the subgroup of elements in
all field summands of invariant if and only if if and only if
K[Cn] = K ®Q Q[ n]
G = C A a
KOQ
QCd
invariant.
Gal(Qcd/Q) So
a
Im[w -* Aut(Cd)] C Gal(Kcd/K) is p-K-elementary.
which leave
leaves the for all
K. din,
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
K[G] = fl 1Ki[w]t
The splitting
the maximal R-order in
K[Cn]
is immediate.
Also,
261
li=1Ri
is
and so
fi=1Ki;
m R[Cn] C
by Theorem 1.4(v).
n]
II Ri C i=1
o
As an example of how this can be used, consider the functor C11(Z[G])(p).
G = Cn Air,
If
where
the proposition, there is an inclusion orders of index prime to
p.
A1(G) =
is a p-group, then by
Z[G] = Z[Cn A r] C ndInZcd[n]t
of
So by Corollary 3.10,
Cl (Z[Cn AV])
A1(G)
a
and
p4'n
Cll(ZCk[a]t)(P). (P) - ki® In
We thus have two decompositions of divisors of
for each
k.
1(G) - 19 kinAlk(G),
and it is natural to expect that
n;
both indexed on
Alk(G) = Cll(Zck[lr]t)(p)
This is, in fact, the case; but the actual isomorphism is
fairly complicated.
Fix a prime
Lemma 11.7 functor
from
the
bimodule morphtsms to G.
p,
let
category of Z-orders
X
be an additive (covariant)
in semisimple lQ-algebras with
Z(p)-modutes, and write
A1(G) = X(Z[G])
Assume that for any p-hyperelementary group
p-group), the projections
Z[Cn] --» Z(k
for finite
G = Cn k a
v
a
induce an Isomorphism
X(Z[G]) = () X(ZLk[n]t).
(1)
kin
Then there is an isomorphism
R : Aln(G) = Aln(Cn >4T) = X(Z1n[a] t )
which is natural with respect to both induction and restriction in
n,
as
262
CHAPTER 11.
A QUICK SURVEY OF INDUCTION THEORY
well as to the Galots action of
The definition of
Proof
(Z/n) * on ZCn.
as well as the proof that it is an
%3,
The best way to see
isomorphism, are both fairly long and complicated.
what is going on is to first read the proof under the assumption that
n
is a square of a prime, or a product of two distinct primes.
Fix G = C >4r,
where
be a generator. For all min, generated by
and
gn/m
Y.
a is a p-group, and let G E Cn
and
p4n
set
I. e., the subgroup
Gm = Cm>4x C G;
For all
we fix the following
klmin,
homomorphisms:
(i)
Indk
.M(Gk) --> .N(C)
:
and
Resk
.N(C) - .N(Gk)
:
are
the induction and restriction maps Prof
(ii)
:
Cm XT -» Ck X w
appropriate
(iii) of
which is the identity on (any prime
Cm
subgroup of
is induced by the surjection
.N(Cm) --) 14(Ck)
qlm)
(so Prof o Indk = Id
r
Prk
:
is if
(iv) Ik: x(Zck[ir]t) -. X(ZC m[v]t), are the induction and restriction maps for
k > 0,
For each ulmin
let
p(k)
such that
[k,u]
aqr
for
(k,k) = 1)
gn/k E Ck
is the composite Ck = exp(2vi/k)
to
Rk: X(ZCm[ir]t)
X(ZCk[lr]t)
ZCk[x]t C ZCm[A]t
(u,u) = 1,
set
N(Gm)
® X(ZZk[w]t) ulklm
klm
(where
a H
induced by
be the number of distinct prime divisors.
(-1)ll([k,u])-p(k).(Ikk'u]o Prk)
l ' _
and which on the q-Sylow
4(C ) = X(Z[Cm4w]) ) X(ZCk[a]t)
Projk with the map induced by sending
For
v,
denotes the least common multiple).
We claim that
CHAPTER 11.
= limo incl
® 9k(Gm) ulklm
:
263
A QUICK SURVEY OF INDUCTION THEORY
is an isomorphism for all
® X(Zck[AJt)
3 4(Gm)
ulmin
ulklm
such that
the lemma will follow from the case
(u,u) = 1.
In particular,
%3 = %3i .
To simplify notation, we write
X(k) = X(Zck[lr]t)
AU(Cm)
kin);
(any
= ® Ak(Gm)
(any
ulmin).
ulkim
The following naturality relations will be needed in the proof below.
naturality of
with respect to projections
Indm
to
the
X(k)
The is
described by the commutative square
Indn
m
4(Gm)
4(Cn)
=
®Prk ®klnla(m,k).
® X(k) klm
(a(m,k) = kW(k+m))
X(k)
kin
(this is induced by a commutative square of rings). for
seems to hold in general; but if
Resm
(2)
11BPrk
k
No analogous result
q2In
for some prime
q,
then the following square does commute:
Resn
n/q
4(Gn)
3 4(Gq) D/q = Projn/g2 ® ®Prk/q
®Prk X(k) n/q ) ® (D kin
Resn/g2 n/q
A(G
3 A(G
® ®R q
2) $ ® X(k/q). kin
n/q
qJn/k
The commutativity
of
gJn/k
(3)
follows upon comparing bimodules,
vertical maps are isomorphisms by (1). isomorphism (Proposition 11.5(iii)), Rjq Ik/k q
(3)
is an isomorphism whenever
Since
Res /q Indn/q
and
the
is an
(2) and (3) combine to show that q2 In
and
q{(n/k).
In particular,
264
A QUICK SURVEY OF INDUCTION T EORY
CHAPTER 11.
for each such
k,
Ik/q®incl
X(k/q) ®Ker(Rq)
:
(4)
X(k)
Finally, the commutativity of the following square is
is an isomorphism.
immediate from the definition of
Prk:
Pro n
.(C n)
.N( m)
®Prk
(any
=I®Prk
min)
(5)
® X(k) RE214 ® X(k) kin kim
It suffices to prove that Fix
where
u,
= 1.
(u,
is an isomorphism in the case
%3m
If
u = 1,
then
(3u
m = n.
is an isomorphism by
u)
Otherwise, let
(1). r
qlu
be any prime divisor, and define
m,
and
m = n/q,
and
v,
to satisfy
r u=qv, We assume inductively that
Case 1
r n=qm,
q{v,
both are isomorphisms.
and
-
Assume first that
qjm.
1. e., that
r = 1;
g2in,
Consider the following diagram:
v= u/q.
.NV(Gm) -
f2
A(Gm)
4-- ®
X(k)
vlkim
Indn m
vIndm
® Ik(m'k)
Iv1kin .NV(Gn) -
fa
(Ik
® + Iqk) vikim k k
X(k)
.N(Gn)
vikin
® (-lkk+ 1Qk)
1f
4u(G) -
vikim
C
® X(k) ulkin
are inclusion
where
vIndm
maps.
The two small squares commute - the right-hand square by (2) -
is the restriction of
Indm,
and the
fi
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
and the lower rectangle commutes by definition of fi ® vIndm
is an isomorphism
Proposition 11.5(iii));
and
pmof2 = %3m
v
v
(recall
(3
.1&(C ) = Ak(Gn)
265
and for all
Also,
(3.
and the right-hand column is short exact.
Since
are isomorphisms, by assumption, this shows
(3nof3 = Pn
v
by
kim,
v
is also an isomorphism.
that Ru
Case 2 Now assume that
g2In,
and consider the following diagram:
vResn n/g
.NV(G )
.Mv(Gq
n
=f1
f2
g
(6)
vResn/g2
® X(k) n/q ) ®ulkln
.NV(G
® ®R q
- v(Gn/g2) ® ® X(k/q) ulkln
where
®(proj o An)
f1=
and
f2 = vProj,g219 (proj o A1q).
Diagram (6) commutes by (3), and the relations I[k/q,v] o Rk o I[k,v] - R[k,v] k/q Rk/q [k,v]/q k
grIkIn
when
_ (ITV
where f2.
(i. e.,
%Id) o f i
Pv
But now Ru
:
An/q
q.fk).
The maps
fl
and
f2
.NV (Gn)
ulkin
are isomorphisms:
X(k) - ® X(k) , vlkln
are isomorphisms by assumption; and similarly for
is the composite
266
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
.Nu(Gn) = Ker(vResn
(Proposition 11.5(iii))
)
g
fi
) Ker(VResn/g2) ® ® Ker(Rk) n/g k/g ulkln
(by (6))
N1g(Glg) ®u ®InKer(Rg)
=
ru/g g®_ Id
`
® (X(k/g)®Ker(R g)J
)
ulkln
I k q ® incl (by (4))
® X(k);
_
ulkin
0
and is hence an isomorphism.
Proposition 11.5 and Lemma 11.7 now lead to the following theorem, which greatly simplifies the limits involved when applying Theorem 11.1 to calculate
SK1(7L[G]),
hyperelementary subgroups.
K
group, and if g,h E G
G
Recall (Theorem 1.6) that if
terms
in
of
is any finite
is a field of characteristic zero, then two elements
h
are called K-conjugate if
a E Gal(Kcn/K),
etc.,
SK1(2p[G]),
C11(7L[G]),
ga
is conjugate to
for some
Also, for any cyclic a = (g) C G,
where n = IgI.
with
n= IgI = Ia1, we define NG(a) = Na(g) _
Theorem 11.8 fractions
{x CC : yg 1 _ ga,
Fix a prime
p and a Dedehtnd domain R with field of
K of characteristic zero.
the category of
7L (p)-modules.
orders, such that
nB C 21
are
where
X
Let
be an additive functor from
R-orders in semisimple K-algebras with btmodule morphisms
to the category of
tsomorphism
some a E Gal(KCn/K)}.
X(21) -_> X(8).
Assume that any inclusion
for some
n
prime to
Then, for any finite
G,
p,
if
21 C B
induces an g1,...,gk E G
K-conjugacy class representatives for elements of order prime to ni = 1g11,
there are isomorphtsms:
of
p,
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
k X(R[G]) = ® 1
(where
{X(RCn [n]t)
267
,r E 106(gi))}
:
(1)
denotes the set of p-subgroups), and
k {X(RCni[i]t)
X(R[G])
Here,
it E 5(N6(gi))}.
(2)
the limits are taken with respect to inclusion of subgroups, and
conjugation by elements of
For all
RC
n,
denotes the
in K. The first isomorphism is natural with
R
integral closure of
NG(gi).
respect to induction, and the second with respect to restriction maps.
Proof
Write A = X(R[-]),
for convenience.
Fix
G,
and let
be a set of conjugacy class representatives for cyclic subgroups of order prime to
p.
Cy(G)
C C G
By Proposition 11.5,
A(G) =
® AC(C);
(3)
CECy(G)
where for each
C,
n = ICI,
if
A(G) = C
Fix
C E Cy(G),
then
ll A(C>4a). irEJ(N(C)) n
and set
n =
ICI.
By Theorem 11.2,
computable with respect to p-K-elementary subgroups.
any
(4)
44
is
In particular, for
it E I(N(C)),
An(CAir)
(A (CAp): p C nnNG(C)}.
Using this, the limit in (4) takes the form AC(C) - HO(N(C)/N (C);
1
An(C A T)).
(5)
(NG(C))
This time, the limit is taken with respect to inclusion, and conjugation
268
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
by elements in Now write Ri C K.
NG(C).
K ®Q Dn
nr=,Ki,
=
Ki = KCn
where
be the integral closure of
i;
and let
By Proposition 11.6,
R.
R[C>4a] C ]{ R®Z7Lck[a]t C
and
kin
R®g7Lcn[n]t C
for each
r ll Ri[a]t C i=1
So by Lemma 11.7 (applied to the functor 21 I-> X(R®72I)
on 7L-orders),
for each a C 55(N6(C)), Nn(C>an) = X(R®77Qn[a]t ) = X( fl i=1
t (6)
® X(Ri[n]t) = ®X(RCJw]t) i=1
Note that
r,
i=1
the number of field summands of Wn ® K,
the number of equivalence classes of generators of
g^'ga
if
a E Gal(Kcn/K).
conjugation by
permuted in
NG(C),
generators of
C,
X(Ri[a]t)
C under the relation are permuted, under
in the same way that these equivalence classes are
Thus, if
G.
The factors
is equal to
there are
m
K-conjugacy classes (in
G)
of
then (5) and (6) combine to give an isomorphism
AC(G) = 9 14 {X(Rcn[n]t)
,r E 51(N6(C))),
(7)
where again the limit is taken with respect to inclusion, and conjugation
by elements of
NG(C).
Formula (1) now follows upon combining (3) and
(7); and formula (2) (for restriction) is shown in a similar fashion.
Induction properties with respect to p-elementary groups - i.
subgroups of the form n x a
where
p4'n
and
play an important role in Chapters 12 and 13.
a
0
e.,
is a p-group - will
The next theorem gives a
269
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
simple criterion, in terms of twisted group rings, for checking them.
Theorem 11.9 of fractions
K.
Fix a prime Let
p,
R with field
and a Dedekind domain
X be an additive functor on R-orders with btmodule
morphtsms satisfying the hypotheses of Theorem 11.8.
For any
be a primitive n-th root of unity, and let
denote the integral
closure of
(i)
R
RCn
let
n
in K. Then
X(R[G])
generated by
is
t: v -i Gal(Kcn/K)
ind
:
with
induction from
(computable for) n
p-elementary subgroups if and only if for any v, and any
n,
with
any p-group
the induction map
p = Ker(t),
HO(n/p; X(Rcn[P]))
pin,
' X(RC n[w]t)
is surjecttue (bijecttue).
is detected by
X(R[G])
(ii)
p-elementary subgroups if and only if for any
t: v - Gal(Kcn/K)
v, and any
res
with
for)
(computable
n
with
restriction
pin,
to
any p-group
the restriction map
p = Ker(t),
: X(Rcn[w]t) - H0(ir/p; X(Rcn[P]))
is injectiue (bijectiue).
We prove here point (i) for computability; the other claims
Proof
are shown similarly.
Write
4(G) = X(R[G]).
computable by Theorem 11.2; and Proposition 11.5(iii).
So A
A
Then
Ak(C mr) = .Nk(Ck >4v)
9
li
by
(1)
An (H)
where
is the set of p-elementary subgroups of
H 9 Cn >4 v,
kin
is p-elementary computable if and only if
for a n y p-K-elementary group of the form G = n A v;
For
if
n
A (Cn41r) 255
p-group, and
is p-K-elementary
.Nn(H) = 0
unless
nhIHi;
pin,
v
is a
G.
i. e., unless n C H.
270
A QUICK SURVEY OF INDUCTION THEORY
CHAPTER 11.
Hence, if
then
p = Ker[t: A --1 Gal(KCn/K)],
25
HEM
An(H)
a1C
An( n x a)
where the limit is taken with respect to inclusion and conjugation in
G.
In other words,
O H 11
An (H)
a5 liM An(n x a) H H0(ir/P; .Nn (C x p)); _P
and the result follows from the isomorphisms
.Un(CnM r) H X(ROZZrn[w]t) H X(Rrn[ir]t)N
.Nn(Cnxp) H X(R®azcn[P]) H X(RCn[p])N
(where
of Lemma 11.7.
N = v(n)/[Kcn:K])
0
Finally, we list some specific applications of Theorem 11.8, which
will be used in later chapters. SK[p]
will be needed.
and p
If
21
For technical reasons, a new functor
is any Z-order in a semisimple Q-algebra,
is a prime, set
SKiP](21) = Ker[SK1(21)
1
II SKI(uq)J(P) q;eP
In particular, there is a short exact sequence
1 -> C11(21)(P) By Theorem 3.14, primes
q X p;
and so
) SKiP](21) - ) SK1(i )(P) -' 1. p
for any finite
G,
SKI(2q[G])(p) = 1
for all
in this case.
This is,
SKlp](Z[G]) = SK1(7L[G])(p)
however, not always the case for twisted group rings.
is any odd prime, and C2 C Gal(QCq/Q),
Theorems 2.5 and 2.10 that
For example, if
q
then it is not hard to show using
SKI(2gcq[C2]t) H Il(q-1).
So in this case,
CHAPTER 11.
SKi2](Zcq[C2]t)
Fix a prime
denotes the set of
5$(H)
p
and a finite group
p-subgroups.
of conjugacy classes of cyclic subgroups of
n1 = Ia, I
271
SK1(ZCQ[C2]t)(2).
Theorem 11.10 H C G,
A QUICK SURVEY OF INDUCTION THEORY
G
Let
G.
For any
a1,.... ak
be a set p.
Set
and Cp(Q OZ -)
are
of order prime to
and N1 =NG(a1). Then k
(1)
C11(ZCG])(p)
® E-(IJ,cl,(ZCn`[w]t(p)'
(2)
SK1(a[G])(P)
®
k
SK1p](Zcni[w]t),
and
Eurn
k (3)
Cp(Q[C])
liter
cp(C
[w]t).
i=1
Proof
By Proposition 1.18,
Cl1(-)(p),
SKip],
all functors on the category of Z-orders with bimodule morphisms. elements
g,h E C
subgroups (by Theorem 1.5(1)).
is trivial for by Corollary 3.10. Theorem 11.8.
a
Also,
are Q-conjugate if and only if they generate conjugate
So
The condition that
Cp(Q®Z-);
X(21) = X(B)
whenever
and holds for C11(-)(p) and
SKEP]
the above decomposition formulas follow from
Chanter 12 THE P-ADIC QUOTIENT OF SK1(Z[G]):
FINITE GROUPS
The induction techniques of Chapter 11 will first be applied to describe
SK1( p[G]) = SK1(Z[G])/C11(Z[C]),
for finite groups
Kj(7Lp[G])(p),
functors are shown subgroups.
as well as
K1(2 [G])(p)
and
In particular, all three of these
G.
to be computable
induction from p-elementary
for
A detection theorem would be still more useful; but in Example
G
12.6 a group
is constructed for which SK1(2p[G])
is not detected by
restriction to p-elementary subgroups.
These results lead to two sets of formulas for torspKj( p[G]).
and
SK1(2 p[G])
The formulas in Theorem 12.5 are based on the direct sum
decompositions of Theorem 11.8, and involve only the functors Hb and (_)ab.
They are the easiest to use when describing either
torspKj(2 p[G])
as abstract groups.
show, for example,
that
SK1(2p[G])
or
As applications of these formulas, we
SK1(2 p[tj) = 1
contains a normal
Sp(G)
if
abelian subgroup with cyclic quotient (Proposition 12.7), or if
G
is any
symmetric or alternating group (Example 12.8).
In Theorems 12.9 and 12.10, alternative descriptions of the groups Ki(7Lp[G]),
and
SK1(7Lp[G])
homology groups of the form The formula for determine
SK1(7Lp[G]),
whether
torspKj(2 p[G])
a given
are derived,
torspKi(7Lp[G])
Hn(G;2p(Gr)),
where
in terms of
Cr = {g E G: p.lgl}.
for example, can be applied directly to element
vanishes.
The
new
formula
for
is derived from two exact sequences which describe the
kernel and cokernel of
r
for arbitrary finite
G : Ki(2p[G]) -4 HO(G; p[G])
G,
and which generalize the exact sequences of
Theorems 6.6 and 6.7.
As was seen in Chapter 11, results on p-elementary induction are
CHAPTER 12.
technical
two
Each is
lemmas.
273
This is the subject of the
obtained by studying twisted group rings. first
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
in two parts:
stated
part
(i)
contains (most of) what will be needed in this chapter, while part (ii) in each lemma will be needed in Chapter 13 (in the proof of Lemma 13.1).
For convenience, for any finite extension
F
will be used to denote the group of all automorphisms of
whether or not the extension is Galois. a C Gal(Ff%),
R C F
Fix a prime
is unramified.
-
for any
In this situation,
let
be any finite extension of
F
t: a --> Gal(F//p)
Let
be any
denote the induced twisted group ring; and set
R[p] C R[a]t
induces a surjection
: K1(R[p]) -v K1(R[r]t)
For any a-inuartant radical ideal
(ii)
P
Then the following hold.
ind
for all
fixing
is a p-group, and such that the extension F/Fv
it
R[a]t
Let
The inclusion
(i)
p,
be the ring of integers.
homomorphism such that
p = Ker(t).
F
Gal(F/&)
Fv will denote the fixed field.
Lemma 12.1 and Let
Qp,
of
g E a),
set
I
IgEy
C R[a]t.
I C R[p]
(i. e.,
gIg 1 = I
Then
indI : K1(R[p],I) -v K1(R[a]t,I)
is onto.
Proof
and
q C S
Set
S = Ra,
the ring of integers in
be the maximal ideals.
Then
FT,
p = qR,
p C R
and let since
F/F7
is
unramified, and
a/p = Gal(F/FT) = Gal((R/p)/(S/q)).
We first prove point (ii). generates
p
(R/p)*.
in a, and set
Choose some
Fix coset representatives
r E R 1 =
(1)
whose image
F E R/p
90'91'92 " " 'gm-1
for
274
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
si = r l.t(gi)(r) - 1 E R
for each
i Z 1.
si E R*
Here,
since
t(g1)(F) # F
Now fix any ir-invariant radical ideal [u] E K1(R[a]t,I)
(u E 1+1).
in
I C R[p],
R/p by (1). and any element
We want to construct a convergent sequence
such that for each k Z 1,
U= u1,u2,u3,...,
uk E 1+ I+
I
and
[u] = [uk] E K1(R[w]t,I).
To do this, assume that
m-1
(x0EI, xiEIk for
xigi i-0
uk = 1 +
Then
has been constructed. m-i U.K a
1SkSm-1).
m-1
fi (l+ xigi) _ (1+x0) II (1 + xisil(r l't(gi)(r)-1)g1) 1=0
i=1
1
(1 + r-I(xisilgi)r - (xisiIgi))
11
_ (1 + i=1
m-1
(1 + x0)
[r 1,1+ xisilgi].
(mod
Ik+l)
i=1
Thus, Ik).
for some
[uk] = [uk+l]
Hence, since
and since
Ik
(1+I)flE(R[lr]t,T)
-> 0 as
k --> a
indI
To prove (i), let R[p]
with
uk a uk+1
(mod
is p-adically closed (Theorem 2.9),
is radical),
(I
Im[indI: K1(R[p],I) - Ki(R[n]t.I)]
[u] = [lliimm uk] E
It follows that
uk+l E 1+ 1 + !k +1
is surjective.
J = (jrigi
(see Example 1.12),
:
jri E p)
be the Jacobson radical of
and consider the following commutative diagram:
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SKI(Z[G]):
275
Kl(R[P],J) - K1(R[P]) _+ K1(R[P]/J) -' 1 lid,/p
find
(2)
IindJ
Kl(R[v]t,J) -'' Kl(R[ir]t) The rows in (2) are exact, so
where
is onto if
ind
and
R[p]/J = R/p
- KI(R[v]t/1) - 1. ind./p
is.
Also,
R[ir]ta = R/p[./p]t = Mm(S/q),
The composite
m = Iv/pI = [R:S].
hidA/o K1(R/p[ir/p]t) = K1(S/q)
KI(R/p)
is the norm map for an inclusion of finite fields, and hence is onto.
0
The next lemma will be used to get control over the kernels of the induction maps studied in Lemma 12.1.
Lemma 12.2 12.1.
Let
p g Y,
R C F,
be as in Lemma
t: a -' Aut(F)
and
Then the following hold.
with the
K1(R[p]),
(i)
aip-actton induced by
it/p --> Aut(R) x Out(p),
is cohomologtcatly trivial.
(it)
I2 2 ...
of
then there is a sequence
SK1(R[p]) = 1,
If
u-inuartant ideals, where
n
R[p] 2 J = I1
2
is the Jacobson radical, such
J
CO
that
flk=lIk = 0,
and such that for all
and
Ik+l 2 IkJ + JIk
Proof
with
Set
E = Fw.
TrF/E(r) = 1
Since
(see
F/E
k:
(1)
ir(A/P;KI(R[P]/Ik)) = 1.
is unramified, there exists
Proposition
1.8(i1i)).
If
M
r E R is
any
276
CHAPTER 12.
THE P-ADIC QUOTIENT OF SK1(7L[G]):
R[tr/p]t-module, then for all
FINITE GROUPS
x E :d
I g(r-(g lx)) = gEtr/p
(t(g)(r)).x = x.
I gEtr/p
In other words, the identity is a norm in
End(M);
and so
M
is cohomo-
logically trivial (see Cartan & Eilenberg [1, Proposition XII.2.4]).
In particular, if
is abelian, then this applies to any power of
p
the Jacobson radical J C R[p]; and so H*(n/P;(l+Jk)/(1+Jk+1)) = Ir(w/p;Jk/Jk+l) = 0
acts effectively on the finite field
for all
Also,
and
is easily seen to be it/p-cohomologically trivial.
(R[tr]/J)*
trip
R[p]/J; Thus,
K1(R[P]/Jk) - (R[P])*/(1+Jk)
is cohomologically trivial for all (R[p])*
K1(R[p]) _
is cohomologically trivial.
Now assume that
p
of central commutators in
Let
is nonabelian. p
of order
p,
a: p -I p/a be the projection, and set and
a a v;
and in the limit
k,
a X 1
lemma holds for
by Lemma 6.5.
{z1,...,zk} C p
be the set
set a = (z1,...,zk) a p,
let
Then
I. = Ker[R[p] -+ R[p/o]].
We may assume inductively that the
R[p/o].
Define
k 9 = SI C R[p]
:
ll
I v-invariant; I = p2IO+ 1 (1-zi)IQ, some IQ C R[p]I: B=1
l
a family of ideals in
R[p].
For all
I E .1,
the group
Ki(R[P],I) = Im[K1(R[P],I) - K1(F[P])] C Ker[Kj(R[P]) -* Ki(R/p2[P/a])]
is torsion free: injects into
tors(Ki(R[p])) = tors(R*)x pab by Theorem 7.3, and this
Ki(R/p2[p/a]).
Hence, by Theorem 2.8 and Proposition 6.4,
THE P-ADIC QUOTIENT OF SK1(R[G]):
CHAPTER 12.
log, : K1(R[P],I) -1 HO(P;I) = Im[HO(P,I)
is an isomorphism.
In particular,
trivial for
since
I E 9,
K1(R[p],I)
HO(p;I)
is an
FINITE GROUPS
277
- HO(P;R[P])]
it/p-cohomologically
is
R[lr/p]t-module.
By Proposition 8.1, the map
SK1(Ra)
surjective.
is
Ki(R[p/a])
:
SK1(R[P])
Ki(R[p/a])
Hence,
Ki(R[p],Ia)
and
0 SK1(R[p/v])
(3)
=
Both
Ki(R[p])/Ki(R[p],Ia).
the first by
are cohomologically trivial:
the induction hypothesis and the second since
Ia E J.
So
by (3).
So we may assume
Ki(R[p])
is
cohomologically trivial. If
SK1(R[p]) = 1,
then
SK1(R[p/a]) = 1
J(R[p/a]) = Ii 7 I2 2 ...
inductively that there are ideals satisfy (1).
=
jk-M.
Im
such that
Im C p2R[p/a],
In particular,
1 S k S m.
for Ik
Fix m
for
k >m,
then
and set
Ik = (Ra)-1(Ik)
So if we set
Ia C 1m g Ia + p2R[p].
for all such
Ik E 9
k,
which
and
Kl(R[P]/Ik) = K1(R[P])IKi(R[P],1k)
is cohomologically trivial. for
But
K1(R[p]/Ik)
k S m by assumption; and hence the
is cohomologically trivial
satisfy conditions (1).
Ik
This will now be applied to describe the functors
SK1,
K1,
o
and Ki
on twisted group rings.
Theorem 12.3
Qp,
Fix a prime
any homomorphism such that F/Fr
let
p,
F
be any Finite extension of
and let R C F be the ring of integers. Let t: it -> Gal(F/Qp) is unramifted.
twisted
group
isomorphisms
ring.
Set
a
is a p-group, and such that the extension
P = Ker(t),
Then
be
the
and let
inclusion
R[a]t R[p]
denote the induced C R[lr]t
induces
278
CHAPTER 12.
THE P-ADIC QUOTIENT OF SK1(Z[G]):
(1)
indSK : H0(v/p;SKl(R[P]))
(2)
indK : HO(w/p;Kl(R[P]))
(3)
indK,
(4)
trfK,
Proof
'
FINITE GROUPS
SK1(R[n]t)
) K1(R[n]t)
: HO(w/P;Ki(R[P]))
) Ki(R[ir]t)
H0(A/P;Ki(R[P]))
Ki(R[a]t)
Using Lemma 8.3(ii), choose an extension
1 -4a--->a- -(X-+ 7r-4 1 a(p) and
of p-groups, where H2(ao) = 0.
ao = aIp,
In particular, by Lemma 8.9,
a C Z(p)
and
as bimodules
(see
such that
SK1(R[p]) = 1.
The composites
KI(R[P])
Kj(R[p])
and, K1(R[n]t)
ind, Ki(R[w]t)
are induced by tensoring with
R[a]t
t K1(R[P]) t Kj(R[p]) or R[ir]t
Proposition 1.18), and are hence the norm homomorphisms n/p-actions. (= Ki(R[p]))
So
Ker(ind) C Ker(N,R/P)
and Kj(R[p])
in both cases.
NT/p
for the
Since
Kj(R[p])
are cohomologically trivial by Lemma 12.2,
Ker(N,a/p) _ (g(x).x I : gEir/p, xEKi(R[P])) Ker[ind: Kj(R[p]) --1 Ki(R[a]t)] and similarly
for
K1(R[p]).
HO(ir/p;Ki(R[p])) = Im(N,/P).
and it now follows that
Also,
since
HO(a/p;Ki(R[p]))
=
1,
The induction maps are onto by Lemma 12.1,
CHAPTER 12.
THE P-ADIC QUOTIENT OF SK1(7L[G]):
indK : H0(ir/p;K1(R[P]))
279
FINITE GROUPS
' K1(R[5]t), (5)
indK,
: HO(a/p;Ki(R[p]))
trfK,
: Kj(R[n]t)
and
Ki(R[,r]t),
10(A/p;Ki(R[p]))
are isomorphisms.
Now set
I = Ker[R[p] - R[p]]
and
and consider the
I =
following diagrams with exact rows:
K1(R[p],I) - HO(n/p;K1(R[P])) - HO(n/p;Kl(R[p])) '-' 1 =lindK
IindI
- K1(R[a]t) ---i K1(R[-]t) - 1
K1(R[a]t,I)
1 -4 H0('a/p;SKl(R[p])) - H0(n/p;Kl(R[p])) - HO(ir/p;Ki(R[p])) - 1 find
a- indK,
Ii1dK
1 ) SK1(R[a]t) ------------ ) Kl(R[w]t) ) Ki(R[a]t) -+ 1.
Then
indK
and
are isomorphisms by (5),
indK,
indK and
12.1, and hence
is onto by Lemma
indl
indSK are also isomorphisms.
a
Theorem 12.3 applies in particular to twisted group rings of the form occurring in Theorem 11.8: Theorem 1.10(i).
finite
K1(R[G])(p),
is unramified if
p4'n
by
So Theorem 11.9 now implies as an immediate corollary:
Theorem 12.4
in any
Fcn/F
Rfn[a]t
If
p
is any prime, and if
extension
and
of
Kj(R[G]) (p)
Qp,
then
are alt
induction from p-elementary subgroups, and
R the
is the ring of integers functors
SK1(R[G]),
computable with respect Kj(R[G]) (p)
with respect to restriction to p-elementary subgroups.
to
is computable
O
As another application of Theorem 12.3, the decomposition formulas of
280
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
Theorem 11.8, when applied to
SK1(R[G])
and
tors Kj(R[G])(p),
take the
following form:
Theorem 12.5
Fix a prime
R C F
Qp,
and let
let
g1,...,gk
Zi = CG(gi).
(t)
For any finite group
G,
G
and set
p,
xgi x 1 =g,, asome a E Gal (Kcn; /K) }
N. = NF (g) G i = {x E G: and
be any finite extension of
F-conjugacy class representatives for elements in
be
of order prime to
F
let
p,
be the ring of integers.
(ni = Igi 1)
Then
k SK1(R[G]) = ®HO(N1/Zi; H2(Zi)/H2b(ZI))(P);
and
i=1
k (it) tors(Kj(R[G]))(P)
[(uF)P]k 0 ®HO(Ni/Zi; b)(PY
Proof
and let
Set
p-subgroups.
ni = IgiI,
#(N1)
and
§$(Zi)
Then by Theorem 11.8,
k SK1(R[C])
1
SK1(Rrn.[R]t)
k
® li HO (w/(nnzi);SKI(RCn, [,rnZi1)) i=1 rE ; ) k
® HO(Ni/Zi; i=1
li
SK1(Rcn,.[P]))
pETTR;)
k HO(Ni/Zi; i=1
,,aa
li
pd;)
H2(P)/n2b(P))
be the sets of
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[C]):
281
k /
((
l
i=1 H0(Ni/Zi; H2(Zi)' Lb(Zi))(P).
Here, the last step follows since
and
H2(-)(p)
computable for induction from p-subgroups by Theorem 11.1: modules over the functor
The formula for
both are
I-I2b(-)(P)
they are Green
HO(-;Z (p)).
torspKi(R[G])
is derived in a similar fashion, but
using inverse limits (and Theorem 7.3).
In contrast to the results for induction, the following example shows that
SK1(2 p[G])
Example 12.6
SK1(7Lp[p]) # 1.
is not in general detected by p-elementary restriction.
Fix a prime
p,
n=pp-1,
Set
product induced by the action of G = p x H.
Then
SK1( P[G])
p-elementary subgroups of
Proof
and let
let
be any p-group such that
p
H = Cn>4 Cp
Cp - Gal(4p(cn)/4) is
detected
not
be the semtdirect on by
(cn),
and set
restriction
to
G.
By Theorem 11.9, it suffices to show that the transfer map
trf : SK1(2 pcn[pxCp]t) - SK1(2 pcn[p]) is not injective.
H2(p)/ 2b(p)
Since the conjugation action of
Cp
on
SKl(Zpcn[P])
-
is trivial, the inclusion induces an isomorphism
SK1(2 pCn[P X Cp]t) = SK1(aprn[P]) $ 1
by Theorem 12.3.
The composite
SK1(7Lpcn[P])
1
8K (2prn[P x Cp]t)
is the norm homomorphism for the
try
C-action on
osition 1.18); is hence multiplication by
p,
SK1(7LpCn[P])
SK1(2 pcn[P])
(use Prop-
and not injective.
The next proposition gives some very general conditions for showing
282
CHAPTER 12.
that
SK1( p[G]) = 1.
SL(2,q)
and
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
Note, for example, that it applies to the groups for any prime power
PSL(2,q)
Proposition 12.7
Let
group.
Then
a g; G.
In particular,
p
SK1(2 p[G]) = 1
q.
be any prime, if
and let
SKI(2 [w]) = 1 p
SKI(Zp[G]) = 1
G
be any finite
for all p-subgroups Sp(G)
if the p-Sylow subgroup
has a normal abeltan subgroup with cyclic quotient.
Proof
all p-subgroups W.
SK1(2 [G]) = 1 p
By Theorem 12.5(1),
and this holds if
a C G;
If a p-group
H2(a)
if
(
SKI(Zp[w]) = 1
) = 1 for
for all such
contains a normal abelian subgroup with cyclic
a
quotient, then SKI(Zp[T]) = 1
by Corollary 7.2.
0
As a second, more specialized example, we now consider the symmetric
Note that Proposition 12.7 cannot be applied in
and alternating groups.
this case, since any p-group is a subgroup of some
Example 12.8
For any n 2 1
and any prime
Sn.
p,
SK1(2 p[Sn]) = SK1(2 p[An1) = 1.
Proof CS.(g)
g E Sn
For any
of order prime to
p,
the centralizer
is a product of wreath products:
CS.(g) = Cm12Sn1 x ... X Cmk2Snk;
where for each
and hence
i,
SK1(2 p[Sn]) = SK1( p[An]) = 1,
if
Abschnitt 1]. the maps
H2(Sn)
and
H2(An)
So by Theorem 12.5(1),
H2(G)/b(G) = 0
product of symmetric groups, or is of index
The groups
pFmi.
2
whenever
G
is a
in such a product.
have been computed by Schur in [1,
It follows from the description there that for a n y
n Z 4,
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
H2(C2 x C2) -i H2(A4) --' H2(An) (2) (induced by inclusion) are all isomorphisms.
n = 6
2-group unless
and A6 and A7
or
o H2(Sn)
both have order
H2(A7)
have abelian 3-Sylow subgroups.
is a
H2(An)
Furthermore,
and
H2(A6)
7,
283
So for all
6,
n,
H2(An)/b(An) n-' H2(Sn)/b(Sn) = 0.
By Proposition 8.12,
the functor Hb is multiplicative with
respect to direct products of groups.
G
is a product of
H2(G)/H2b(G) = 0
Thus,
symmetric or alternating groups.
whenever
C
If
is a
semidirect product
G = (Anl x ... x Ank
then since
H1(Anl x ... x Ank)
H2(
(
for such
) = 0
x ... X Snk ;
is generated by
k-1
and
C,
nl
H2(G)
has odd order,
H2(AnIx...xAnk) Thus,
S
(Y
H2((C2)
)
Wh(Sn)= 1
Example 12.8 was the last step when showing that n.
We have already seen that
Wh'(Sn)
computation of
for all
is finite (Theorem 2.6) and
Cll(Z[Sn])=1
torsion free (Theorem 7.4); and that
a
and this finishes the proof.
(Theorem 5.4).
The
will be carried out in Theorem
SK1(Z[An]) = C11(Z[An])
14.6.
To end the chapter, we now want to give some alternative, and more direct, descriptions, of for arbitrary finite
G.
SK1(p[G]), For any
G,
torspKi(Zp[G]),
and any fixed prime
denote the set of p-regular elements in prime to
p.
For any
g E C,
G:
i.
g,.,g, E (g)).
Ki(Zp[G])
p,
C,
will
e., elements of order
g,.,g, E C will denote the unique elements
such that g,. E G,, g, has p-power order, g = g,.g, , (note that
and
For any
R,
H1(G;R(G,.))
and
[g,. , g, ] = i
denotes the homology
284
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(7L[G]):
C
group induced by the conjugation action of convenient to represent elements of
on
H1(G;R(Gr))
02
H1(G;R(Gr)) = H1(Z[G]®7 Z[G]®z R(Gr)
It will be
R(Gr).
via the bar resolution:
- z[G]®7 R(Gr) i R(Gr)),
where 01(g®x) = x-gxg 1 and 02(g®h®x) = h®x-gh®x+g®hxh 1. When
R
then
0 denotes the automorphism of
is the ring of integers in a finite unramified extension of
Hn(G;R(Gr))o = Hn(C;R(Gr))/(1-4);
Theorem 12.9 a finite group
Fix a prime
G.
Let
Hn(G;R(Gr))40 = Ker(1-4) C Hn(G;R(Gr)).
an unramified extension
p,
R C F
be the ring of integers.
and
'RG : HO(G;R[G]) -' H1(G;R(Gr))
by setting, for
As usual, we write
on coefficients.
41(jrigl) = 1*(rl)g!
induced by the map
Hn(G;R(Gr))
F
Qp,
and
Define
ORG : HO(G;R[G]) -i HO(G;R/2(G,))
r; E R and g; E G:
w(Gr.gi) = Yg: ® r; (g: ),
O(Zrig:)
and
=
(F, E R/2).
Zri (g. )r
Then
(t)
There are unique homomorphtsms
URG : K1(R[C]) -- i H1(G;R(Gr))
and
which are natural with respect characterized as follows.
u
1
= I js j h; ,
where
u([u]) = I g;
For any
9RG:Ki(R[G]) -' HO(G;R/2(Gr)),
to group homomorphtsms, and which are u E CL(R[G]),
write
r; , s; E Mn(R) and gi,hj E G. E H1(G;R(Gr)).
u = Jir;g;
and
Then
(Tr: Mn(R) --> R)
i,j
If
p=2,
then for any commuting pair of subgroups
H,irc G,
where
IHI
CHAPTER 12.
THE P-ADIC QUOTIENT OF SK1(Z[G]):
FINITE GROUPS
is odd and it is a 2-group, and any x E J(R[H x v]), image of x under the composite
,r,,.
J(R[H x v])
J(R[H]) = 2R[H]
(it)
The sequence
1 -
K1(R[G])(P)
T u $8
1(1
A
R[H] c R(G,.)
285
is the
8(1 +x)
*+ HO(G; R/2(G,.)) .
HO(G;R[G]) 0 H1(G;R(Gr)) 0 HG(G;R/2(Gr))
,-1
0
0
$-1
H1(G;R(Gr)) $ HG(G;R/2(Gr))
0
is exact.
(iii)
There is an exact sequence
Kj(R[G])(P)
0 --' H1(G;R(Gr))40 $ HO(G;R/2(Gr))
r ' HO(G;R[G]) -' H1(G;R(Gr))4, 0 HO(G;R/2(Gr))4 - 0.
In particular,
H1(G;R(Gr))40
torspKi(R[G]) =
S HO(G;R/2(Gr))41.
Proof Using the relation gh®x = h0x+g®hxh 1, x E R(Gr),
one easily checks that the map
defined in (i) is a homomorphism. GL(R[G]) ab.
and
If
G
Hence, this factors through K1(R[G]) =
is p-elementary, then
so that
SK1(R[Gab]) = 1,
for g,h E G and
uRG: GL(R[G]) -i H1(G;R(Gr))
H1(G;R(Gr)) = H1(Gab;R((Gab)r))
SK1(R[G]) C Ker(uRG).
Since
is generated by p-elementary induction, this shows that through Ki(R[G]) = K1(R[G])/SK1(R[G])
To see that G= H x v where
and so
9RC IHI
for arbitrary finite
is well defined when
is odd and
it
uRC
p=2,
is a 2-group.
Then
SK1(R[G]) factors
G.
assume first that J(R[H]) = 2R[H],
286
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
K1(R/4[H])(2) = K1(R/4[H],2) = HO(H;2R[H]/4R[H]) Q--' HO(G;R/2(Gr))
by Theorem 1.15.
This shows that
in particular when
G
2-elementary computable,
G
6
Since
Kj(R[-]) (2)
G.
G = Cn x a where
pjn and
is isomorphic to a product of rings
R[G]
is
now automatically extends to a homomorphism
is p-elementary - if
p-group - then
is well defined in this case; and
is 2-elementary.
defined for arbitrary finite
If
0RG
various unramified extensions
R./R.
is a
w
R1[x]
for
So in this case, sequence (ii) is
exact by Theorem 6.7, and sequence (iii) by Theorems 6.6 and 7.3.
All terms in sequence (ii) are computable with respect to induction from p-elementary subgroups.
Hence, since the direct limits used here are
right exact, (ii) is exact except possibly at
K1(R[G])(p).
But then (ii)
is exact if and only if (iii) is, if and only if
IKer(r)I = IH1(G;R(Gr))"I'IHO(G;R/2(Gr))"I
Ker(r) = Ker(log) = torspK'(R[G])
Also,
(1)
(Theorem 2.9), and so (1)
follows from a straightforward computation based on Theorem 12.5(11). details, see Oliver [8, Theorem 1.7 and Corollary 1.8].
The above definition of
n
was suggested by Dennis' trace map from
v
K-theory to Hochschild homology (see Igusa [1]). find a correspondingly satisfactory definition for
We have been unable to A.
We saw in Theorem 6.8 that a restriction map on defined, which makes
For
HO(G;R[G])
can be
rRG natural with respect to transfer homomorphisms.
Unfortunately, there is no way to define restriction maps on the other terms in sequence (ii) above, respect to the transfer. (r,v,49)
to make the whole sequence natural with
If there were, the proof of the injectivity of
would be simpler, since inverse limits are left exact.
We now end the chapter with a second description of
any finite G and any unramified
R,
set
H2(G;R(Gr))4, = H2(G;R(Gr))/(0-1);
SK1(2 [G]). p
For
CHAPTER 12.
where
is induced by the automorphism
287
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[G]):
$(rg) = .p(r)gp
R(Gr).
of
In
analogy with the p-group case, we define
H2b(G;R(Gr))t = Im[
®
I
H2(G;R(Gr))$]
H2(H;R(Hr))
HCG
H abeltan = ((g^h)®rk E H2(G;R(Gr) )0
:
The following formula for
SK1(R[G])
g, h E G, k E G,. , r E R,
(g,h,k) abel tan).
is easily seen to be abstractly
the same as that in Theorem 12.5(1), but it allows a more direct procedure
for determining whether or not a given element in
This procedure is analogous to Proposition 8.4.
SK1(R[G])
that in the p-group case described in
Note, however, that in this case, once
has been lifted to
TG(a).
Knowing
not in general suffice to determine whether or not SK1(R[G]) - no matter how large G
Fix a prime
G,
TG(a)
u
G,
it
is
alone does
vanishes in
is.
and let
p,
in any finite unramifted extension
u E SK1(R[G])
for some appropriate
a E K1(R[G])
necessary to evaluate both G(a) and
Theorem 12.10
vanishes.
F
of
R
be the ring of integers
Then, for any finite group
there is an isomorphism
0G : SK1(R[G])
H2(G;R(Gr)) V
which is described as follows. Let extension of finite groups such that Im[H2(6;R((;r))
Consider the homomomorphtsms
1 --> K
-2b(G;R(Gr))O;
a G -> 1 be any
) H2(C;R(Gr))] C H2b(G;R(Gr))
(1)
288
CHAPTER 12.
FINITE GROUPS
THE P-ADIC QUOTIENT OF SK1(Z[C]):
Ker[HO(G;R[G]) -t HO(C;R[G])]
Kab
6
®ZG R(C )
' 2(C;R(Gr) )6)
ab
6a(Ha2b(G,R(Gr))+ (l-+)H2(G;R(Gr)))
qb(G;R(Gr)),
(6)-1)o c'1
(2)
Ker[H1(G;R(Gr)) -t H1(G;R(Gr))]
a (r(z-1)g) =
Here,
for any zEK,
to
and
c
are induced b
Ha
H2(G;R(Gr))
and gEG; and
rER, nce
the five term exact se
H2(G;R(Gr)) S b®ZGR(Cr) (3)
C Hl(G;R(C,)) of Theorem 8.2. homomorphism
of Theorem 12.9(1).
and any ttfttng to
(Sab)-1(@a
Proof S ab
rG(u)+
Then, for any
[u] E SK1(R[G]),
G(u))) E H2(G;R(Gr)) V
By (1) and (3),
is a monomor Phism
Sab
and
To see that
Ia = Ker[R[G] -' R[G]]
for convenience.
The map
sending
to
(z-l)g
be induced by the
[u] E K1(R[G]),
eG([u]) _
a
H1(C;R(Gr))
DG: K1(R[G]) -+ HI(G;R(Gr))
Let
vRG
HI(a)
and
Ia
z O a(gr),
fa
(6)-1)oc-I
-2b(G;R(Gr)),t.
are well defined, and
is well defined
first set
HO(G;Ia) = Ker[HO(G;R[G]) -+ HO(G;R[C])]
HI(K;R(Gr)) = Kab ®R(Gr),
defined by
is easily seen to be well defined; and
induces a homomorphism
wa : HO(G;Ia) '-' HO(G;HI(K;R(Gr))) =
Consider the following commutative diagram:
Kab
OZG R(Gr)
CHAPTER 12.
THE P-ADIC QUOTIENT OF SK1(7[G]):
289
HO(G;Ia) --' HO(G;R[ ]) -0 HO(G;R[G])
H1(G;R[G])
IWa
Iw
I)'
H2(G;R(Gr)) -fwhere
FINITE GROUPS
b®ZGR(G,)
`-' HI(G;R(Gr))
G is as in Theorem 12.9; and where
X(g®rh) = (a(g)^h) O
E
r E
(4) are exact, and
E C
such that
HO(G;Ia) = Coker(aa);
The rows in
[a(g),h] = 1. so
wa
factors through a
homomorphism 3a as in diagram (2). For any
u E K1(Ra)-1(SK1(R[G])),
G(u) E Ker(H1(a)) = Im(c).
By
the exact sequence in Theorem 12.9(11),
is TG(u)+
G(u))
R(G )
Kab®
E Ker
7LG
I 6a( H2
So to see that
r
Hl(G;R(G,.))
l
(G;R(Gr)) + (1-$)H2(G;R(Gr))))
©G
= Im(bab
is uniquely defined - with respect to a given
a,
at
least - it remains only to check that
a TG(u)+ (4-1)(c-luG(a)) = 0
To prove this, set
for any u E Kl(R[G],Ia).
G = {(g,h) E G: a(g)= a(h)},
is a pullback square. Set I = Ker(R[P2]) C R[G]. surjective (split by the diagonal map), 6P2 = 1, homomorphism
(5)
so that
Since and
132
/31
is split induces a
290
THE P-ADIC QUOTIENT OF SKI(Z[G]):
CHAPTER 12.
FINITE GROUPS
Ker[H1(G;R(Gr)) - H1(G;R(Gr))] = Kab ®Z R(Gr)
13,
Kab®ZGR(Gr) ).
ba(` ab(G,R(Gr)) + (1-$)H2(G;R(Gr)) Then any u E K1(R[G],Ia)
6 E K1(R[G],I);
lifts to
($-1)(L-luG(u))
@a rG(u)+
($-1)(L-IuG(u))) = R1*(0)
=
by Theorem 12.9(ii); and this proves (5). We have now shown that there is a well defined epimorphism
OG : SK1(R[C]) - H2(G;R(Cr))
such that
0G([u]) = [Sab(rG(u))]
lifting to
[u] E K1(R[G]).
surjection
a'
onto
G,
for any
-(G;R(Gr))$
This is independent of the maps
0G
given a second
a:
defined using
a
and
each be compared to the map defined using their pullback. existence of
a
a'
Also,
can the
satisfying (1) follows from Lemma 8.3.
To show that
OG
is an isomorphism, it remains to show that the two
groups are abstractly isomorphic. SK1(R[G])
and any
[u] E SK1(R[G])
But this follows from the formula for
in Theorem 12.5; the formula
H2(G;R(Gr))
m ®H2(CG(gi)) 0 R(gi) i=1
(when
g1,...,gm
description of
are conjugacy class representatives for
NG(gl)
H2(G;R(Gr))
both p-elementary computable,
0G
8.6.
o
and the
in Oliver [8, Lemma 1.5].
Alternatively, since
that
Cr);
is an isomorphism.
V -2b(G;R(Gr))$
and
SK1(R[C])
it suffices to show for p-elementary
are
C
And this is an easy consequence of Theorem
Chapter 13
C11(Z[G]) FOR FINITE GROUPS
The goal now
is
and
C11(Z[G])(p)
to
far as possible computations
reduce as
SK1(Z[G])(p),
first
p-elementary, and then to the p-group case.
computable if
and
C11(Z[G])(p)
is odd; and that
p
G
of is
The reduction to p-elementary
The main result in that section,
groups is dealt with in Section 13a. Theorem 13.5, says that
the case where
to
SK1(Z[G])(p)
SK1(Z[G])(2)
are p-elementary
can be described in
terms of 2-elementary subgroups via a certain pushout square.
Section 13b deals with the reduction from p-elementary groups to p-groups. of
In particular, explicit formulas for
odd) and 13.13 the
special
(G
abelian).
problems
C11(Z[a]) - when
a
(p
Theorems 13.10 and 13.11 deal with some of
which arise
when
is a 2-group and
algebraic number field in which
are given in Theorems 13.9
a g G,
for p-subgroups
C11(Z[ir])
in terms
C11(Z[G])(p),
2
R
comparing
with
C11(R[ir])
is the ring of integers in an
is unramified.
In Section 13c, the extension
1 -> cll(Z[G])
' SK1(Z[G])
®SK1( p[G]) - 1 p
is shown to be naturally split in odd torsion.
An example is
constructed (Example 13.16) of a 2-elementary group
C
for which
no splitting which is natural with respect to automorphisms of
13a.
2
then
has
G.
Reduction to p-elementary groups
As seen in Theorem 11.9, reducing calculations to p-elementary groups
involves "untwisting" twisted group rings. for
C11(R[ir]t)
and
SK1(R[ir]t)
localization sequence for
SK1(-)
Before results of this type
can be proven, the other terms in the must be studied.
The main technical
292
CHAPTER 13.
C11(7L[G]) FOR FINITE GROUPS
results for doing this are in Lemma 13.1 and Proposition 13.3.
Fix a prime
Lemma 13.1 extension of a,
and let
and let
and Let
t:
R[A]t
p,
R C F
it --> Cal(F/&)
let
be any finite unramifted
F
be the ring of integers. be any homomorphtsm.
denote the induced twisted group ring.
Fix a p-group p = Ker(t),
Set
Then the following
hold.
For any radical ideal
(t) g E n,
set
I C R[p]
E R[a]t.
I =
Then
such that the
gIg 1 = I
inclusion
for all
R[p] C R[lr]t
induces an isomorphism
K1(R[a]t,I).
indI : H0(n/p;K1(R[p],I))
(ii)
SK1(R[p]) = 1,
If
then the inclusion
induces
R[p] C R[w] t
an epimorphism
indK2 : K2(R[p]) -" K?(R[ir]t).
Proof
Let
For any pair
J C R[p]
denote the Jacobson radical, and set
Io C I C R[p]
aI/Io
of
a-invariant ideals of finite index, let
: H0(a/p;Kl(R[p]/Io,I/Io)) -* K1(R[ir]t/Io,I/Io)
13I/I0
. K2(R[p]/Io,I/Io) - K2(R[a]t/IIo,I/Io)
be the homomorphisms induced by the inclusion will be proven in three steps. for
K.(R[p]/Io,I/Io),
Step 1
isomorphism, and that I = 1 0 I
R[p] C R[a]t.
To simplify notation, we write
The lemma K1(I/Io)
etc.
Assume first that
finite index such that
Write
if = yr/p.
Io C I C R[p]
IJ + JI C Io.
are
a-invariant ideals of
We want to show that
PI/Io
is surjective.
and
Yo = Io ® To,
where
I =
al/1o
is an
and
CHAPTER 13.
To = jgCV,PI0-g.
293
C11(Z[C]) FOR FINITE GROUPS
By Theorem 1.15,
K1(I/Io) = (I/Io)/[R[p]/Io,I/Io] = HO(p;I'/I)
and
(since
generates
R[A]t
as an additive group)
K1(I%Io) = (I/IIo)/[R[a]t/IIo.I/IIo] = HO(A;I/Io) $ HO(R'a;I/ o).
In particular, this shows that
aI/Io
is a monomorphism.
aI/Io
But
is
surjective by Lemma 12.1(11), and is hence an isomorphism.
By Example 1.12,
J C R[a]t
Jacobson radical
and
has
Hence, by Theorem 3.3,
R[a]t.
{g,v}
g E p
for
that
and
g E amp.
t(g)(r) % r
s= Then, since
is generated by symbols
To show that
{u,l+gg} = 1
RI/Io
whenever
{1+p,v}
is onto, it
u E (R[p])E
As in the proof of Lemma 12.1, choose (Gal(F/&) = Gal((R/pR)/Fp));
(mod pR)
and the
the same generators as an ideal in
K2(I/Io)
and v E 1+I/Io.
will thus suffice to show that I/Io,
(as an R[p]-ideal);
J = (p,g-1: g E p)
r E R
such
and set
r-l. grg 1- 1 E R*.
r-l. t(g)(r) - 1=
rur-1 = u,
{u, l+gg} = {u , I+ s 1(r 1.grg 1- 1)gg}
= {u, l+r
In order to prove (i), we first show that aI/Io
Step 2
isomorphism for any pair such that
Io C I,
l+s 1fg}-I = 1.
1(s
[I:Io] < -.
set
Io C I C R[p]
ir-invariant radical ideals
This will be done by induction on
I1 = Io + IJ + JI
following diagram:
of
(so
is an
II/Iol.
Fix
Io C I1 C I), and consider the
294
CHAPTER 13.
Cl1(Z[G]) FOR FINITE GROUPS
K2(I/I1) --1 HO(?r;Kl(I1/Io)) -+ HO(W;K1(I/Io)) -4 HO(a;Kl(I/I1)) -4 1 1aj1/Io
-IocI/I1
la1,10
(1)
1131/li
K2(I/Ti) ) Kl(I1%Io) -- K1(ILI0) The top row is exact, except possibly at is an isomorphism and
aI/I1
/3I/I1
isomorphism by the induction hypothesis,
Kl(I/I1) -' 1.
By Step 1,
HO(a;Kl(I1/Io)).
is onto.
and so
a
I/Io
is an
aI1/Io
Also,
is an isomorphism
by diagram (1).
Now, for any ir-invariant radical ideal K1(R[p],I) = 4Lm Kl(I/Io)
and
I C R[p],
K1(R[n]t,I) = Jim Kl(I/IIo)
Io C I
by Theorem 2.10(1 i i) , where the limits are taken over all finite index.
K1(I/Io)
aI =
Lm
Also,
are aI/Io
HO(A;-)
finite.
of
commutes with the inverse limits, since the
Since
the
aI/I0
are
all
isomorphisms,
is also an isomorphism.
Step 3 Now assume that
SK1(R[p]) = 1.
By Lemma 12.2(11), there is
a sequence
R[p]2J=I1;? I2;? ... of a-invariant ideals, such that and such that
flk-lIk = 0,
all
k.
We claim that
when k = 1 Fix
since
k
0,
JIk-l+Ik-lJ C Ik
Kl(R[p]/Ik)
/3R[p]/Ik
K2(R[p]/J) = 1
is
k,
such that
if-cohomologically trivial for
is surjective for all
k;
this is clear
(Theorem 1.16).
and assume inductively that
Consider the following diagram:
for all
13R[p]/Ik-1
is onto.
CHAPTER 13.
295
Cl1(7L[G]) FOR FINITE GROUPS
K2(Ik-l/Ik) - K2(RLP]/Ik) -' Ho(W;K2(RLP]/Ik-1)) - HO(v'Kl(Ik-I/Ik))
19Ik-1/Ik 1
KI(R[v]t/ik-1) -
K2(Ik-l/-Ik) - K2(R[v]t/k)
(2)
- JaIk-1/1'
j13R[P]/I1c_1
RRLP]IIk
Kl(lk-l/-Ik)
Since the last two terms in the exact sequence
K2(R[p]/Ik-1)
-' Kl(Ik-l/Ik) -
K1(R[P]/Ik) -.' Kl(RLP]/Ik-1) -' 1
are v-cohomologically trivial, by assumption, the top row in (2) is exact at
and
HO(v;K2(R[p]/Ik_l)).
onto, and so the same holds for
Under certain circumstances,
is
ind
is onto.
im 13RLP]/Ik
K2 =
O
twisted group rings actually become
This is the idea behind the next lemma.
Lemma 13.2
Let
R = Ljn
i =1
subgroup such that
gRg
conjugation action of generates
13R[P]/Ik-1
"R[P]/Ik.
In particular, in the limit,
matrix rings.
is an isomorphism
aIk-1/Ik
We have assumed inductively that
Is onto.
pIk-1/Ik
Also, by Step 1,
1
Ri C S
= R
for all
permutes the
it
g E it,
Ri
and such that this Assume that
transitively.
as a right R-module, and that
S
v C e be a
be rings, and let
for any
gRI = R1
it
g E it
21
such that
which sends
such that
gR1g 1 = RI. -
R
to the diagonal.
giR1gi1
r = (r1,...,rn) E R
a: S
Then there is an tsomorphism
= Ri,
then
g1,...,gn E it
More precisely, if
a
1 Mn(R1)
are
can be defined such that for any
(r1 E Ri),
a(r) = a(rl,...,rn) = diag(g1 rlgl,...,gnlrngn).
Proof
Fix elements
e1,...,en C R,
such that
g1,...,gn C gIR1gi1 = RI
particular, if v' = {gEi: gRlg 1=R1},
it,
and
and central Ri = Rei
then the
gl
idempotents
for each
I.
In
are left coset
296
C11(71[G]) FOR FINITE CROUPS
Ch AFTER 13.
representatives for
n'
a basis for
as a right
Se
in
ir.
We first claim that
{glel,...,gnel}
is
The elements generate by
R1-module.
1
assumption
(gRl = R1
g E a').
for
To see that the
i
independent, note for any
eigjrj =
r1,...,rn E R1
such that
gjrj
if
i = j
0
if
i 0 j
)'gj =
so that for all
In particular,
i
J1g1r1 = 0,
(gjrjgj
girl = eigiri =
i,
are linearly
g e
that
E gjRlgj
Rj),
0.
if we consider
as an
Se
(S,R1)-bimodule,
this
1
induces a homomorphism
a : S ) EndR,(Sel) = Mn(Rl).
Furthermore, n n S = ® Sei = ® S.gielgi i=1
and for all
i=1
i,
j,
agjRlgilgkRd
and
=
As
is
SKI(R[lr]t),
with
a
SK1(R[p]).
gjRlgi
;
if k = i
gR J 1
-1
k # i.
if
is an isomorphism.
R[a]t
n
i=1 j=1
k,
suggested by Theorem where
n
-1
i=1
0
This shows that
n
= ®
-1
(elgi gkel = gi eiekgk = 0)
The formula for
11.9,
the
goal
ajR
now
is
is a global twisted group ring and
is clear.
to
compare
p = Ker(t),
The next proposition does this for the other terms in
the localization sequence of Theorem 3.15. Recall the definition of Steinberg symbols in Section 3a:
{u,v} = [¢ 1(diag(u,u 1,1))
for any (not necessarily commuting)
,
$ l(diag(v,l,v 1)), E St(R)
u,v E R*.
Here,
: St(R) ---> E(R)
denotes the standard projection. by
this
defining,
u,v E GLn(R),
for
p
It will be convenient here to extend
arbitrary
and
n >1
arbitrary
matrices
{u,v} E St(Mn(R)) = St(R).
Proposition 13.3 in which
297
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
Fix a prime
p,
a p-group
is unramifted, and a homomorphism
R C K
be the ring of integers, set
R[a]t
be the induced twisted group rings.
t: a -4 Gal(K/f).
p = Ker(t),
i HO(a/p;Cp(K[P])) -'' Cp(K[ir]t),
K
a number field
n,
and let
K[w]
t
Let
and
Let
K2(Rp[P]) -' K2(Rp[lr]t),
i
iSKp : HO(n/p;SK1(Rp[P])) -i SK1(Rp[.r]t)
be the homomorphisms induced by the inclusion
p = 2
is surjectiue, and is an isomorphism if
iCp
(t)
and
(ii)
R[p] C R[a]t.
p
Then
is odd, or if
has no real embedding;
K'r
iSKp
is an isomorphism; and
there is an isomorphism
(iii)
In (iii), for any a E SKl(Rp[p]),
f = 1k=1gi®a1 E H1(1r/P;SK1(Rp[P])),
and
1,
a(f)
where
g.Eir,
is defined as follows.
i
Fix matrices
ui E GL(Rp[p])
which represent the
[gl,ul][g2,u2]...[gk,uk]
for some
X E St(Rp[p]).
ai,
and write
= (X) E E(Rp[P])
Then
k a(f) =
ai) _
1 E K2(Rp[a]t).
298
Clt(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
Point (i) will be proven in Step 1.
Proof
(iii) when
are then shown in Step 2; and the general
SK1(Rp[a]) = 1,
n
product of
p.
Set n = In/pI
Step 1
P
where the products are taken over all prime ideals in
Rp - IPIPII,
R which divide
9P = n PIP
Recall (Theorem 1.7) that
case of (iii) is shown in Step 3.
and
Point (ii), and point
and
copies of
K,
Ko = Kim,
for short.
KKK[p] = (K[p] )n;
permuted transitively under the conjugation action of
Then KOKo K
is a
and the factors are v.
By Lemma 13.2,
KOKOK[7r]t - Mn(K[P]), and the composite
In the following commutative diagram:
transfer map.
trf» p(K[P]) = Cp(K OK. K[,r]t)
CP(K ®Ko K[P])
Itrf
trf I
iC0
p(K[P])
4 CC(K[n]t).
the transfer maps are all onto by Lemma 4.17, and so If
p
is the
CP(K OKo K[p]) --> Cp(K ®Ko K[ir]t) = CP(K[p])
p = 2
is odd, or if
and
Ko = K'r
iCP
is onto.
has no real embedding,
then by Lemma 4.17 again,
CC(K[a]t) - HO(Cal(K/Ko);CC(K ®Ko K[v]t)) - HO(Gal(K/Ko);CC(K[P]))
To see this when
p = 2,
note that Conjecture 4.14 holds for
Theorems 1.10(ii) and 4.13(11):
since
K2[n]t - [p12Kp[A]t,
factor is a summand of some 2-adic group ring of the form the description of the isomorphism the action of n/p.
So
iCP
Gal(K/Ko)
on
K ®Ko K[ir]t - Mn(K[p])
CP(K[p])
K[r]
t
by
and each
F[CC)4n].
By
in Lemma 13.2,
is just the conjugation action of
is an isomorphism in this case.
CHAPTER 13.
Step 2 orbit of
Fix some prime
p
pip
in
and let p=pl,...,pn be the
R,
under the conjugation action of
stabilizer of
w.
.
n
f SKi(Rq[a]t).
is an isomorphism by Theorem 12.3; and
Lemma 13.2: C Aq[ir]t
since
q = pl...p .
Consider the following homomorphisms:
H0(ir/P;SK1(Rq[P])) -f-L-- HO('R/p;i®SKl( pa[Ai]t
fl
be the
ai C w
Let
pi - wi = {g Eir: gipigi1 =pi} - and set
Then Rq = [[i=1Rp
Here,
299
C1i(Z[G]) FOR FINITE GROUPS
Bq[a]t = Mn(Rp[ai]t),
is an isomorphism
f2
li=1Rpi[,ri]t
and the inclusion
is the inclusion of the diagonal.
If
SK1(Rp[p]) = 1,
then by
a similar argument,
®K2(Rpi[nl]t)
K2(Rq[P]) = ®K2(Rp;[P])
R, this shows that
in
surjective if
K2(Rq[a]t)
After summing over all ir-orbits of primes
are surjections by Lemma 13.1.
pip
-"
is an isomorphism, and that
iSKp
iK2
is
SK1(Rp[p]) = 1.
Step 3 Now assume that
SK1(Rp[p]) # 1.
Using Lemma 8.3(11), choose
an extension
a ) i t
such that
a C Z(p)
a
and
1
where
H2(ao) = 0.
Then
p = a 1(P).
SK1(Rp[p]) = 1
ao = aIP,
by Lemma 8.9.
Consider the following commutative diagram:
K2(Rp[P]) - K2(Rp[P]) -1L H0(a/P;K1(Rp[P],I)) -L-> HO(ir/P;Kl(Rp[P])) 113
1i2
= 14
K2(Rp[i]t) ---' K2(Rp[n]t) a-> K1(Rp[ir]t,I) Here,
I = Ker[Rp[p] - Rp[p]]
= is
-+ K1(Rp[ir]t).
and 1 = Ker[Rp[ir]t - Ap[a]t].
Then
i3
300
CHAPTER 13.
C11(7L[G]) FOR FINITE GROUPS
is onto by Step 2 (SK1(Rp[p]) = 1), 13.1(1), and
is an isomorphism by Theorem 12.3.
i5
and
Ki(Rp[p])
is an isomorphism by Lemma
14
K1(Rp[p])
(= Ki(Rp[p]))
are
Furthermore,
w/p-cohomologically trivial
by Lemma 12.2(1); and so a diagram chase gives isomorphisms
Coker(i2) = Ker(T)/Im(/i)
H1(w/p;Kl(Rp[p])) = H1(ir/p;SKl(Rp[p]))
To check the formula for to
ai E GL(Rp[p]),
gi E a,
a(f) = a(ggi ®ai),
and X E St(Rp[p]).
[Sl,al]...[~gk,]'$(X)-1
and as an element of
K1(Rp[flt,I)
Now recall the functor that for any g-order
21,
1 --> Cl1(a)(P)
group and
R
SKip]
-
lifts to
g
this pulls back to
of Theorem 11.10.
o
This was defined so
there is a short exact sequence
SKip](2I) -- SKi(9p)(P) -- 1.
Fix a prime
R C K
p-group, fix a homomorphism
whenever
p
G
K
and a number field
be the ring of integers. t:
rr -+ Gal(K/@),
be the induced twisted group ring.
is a finite
and set
where Let
p
a
p = Ker(t).
is
be a Let
Then
Cl1(R[ir]t)(p)
is surjectiue, and
iSK :HO(lr/p;SKI(R[p])) - SKip](R[a]t)
is surjectiue, and
iCl :HO(7r/p;Cll(R[p]))
is an isomorphism if
(ii)
X
is the ring of integers in a number field.
Theorem 13.4
(i)
and
u1,
E K1(Rp[P],I);
SK[p](R[G]) = SK1(R[G])(p)
unramifted, and let
R[rr]t
Then
1 E K2(Rp[a]t).
a(f) =
By Theorem 3.14,
gi,
lift
is an isomorphism if
p
p
is odd; and
is odd or if 0 has no real embedding.
CHAPTER 13.
Here,
iCl
and
are induced by the inclusion
iSK
301
C11(7L[G]) FOR FINITE GROUPS
R[p] C R[w]t.
In
general, the following square is a pullback square:
HO(w/P;CP(K[P])) -!-- HO(ir/p;SKl(R[P]))
IiSK
I
82
(1)
SK4P](R[1r]t).
P(K[w]t) Proof
First consider the following two commutative diagrams, whose
rows are exact by Theorems 3.9 and 3.15:
K2(R [P])
) HO(ir/P;CP(K[P]))
HO(a/P;C11(R[P])) -' 1
IiK2
(2b)
(2)
jici
Iicp 82
C11(R[a]t)(P) - 1
K2(RP[w]t) -- CP(K[ir]t)
a'
H1(w/P;SK1(RP[P])) HO(v/p;C11(R[P])) iCl
1 -, C11(R[a]t)(P)
Here,
-L (3a)
-
by Proposition
isomorphism.
HO(,r/p;SK1(R[P])) -+ HO(ir/P;SKI(RP[P])) _"' 1 ji
SKIP](R[w]t
13.3,
It follows that
1CP iCl
SKl(RP[A]t)(P)
is
and
surjective and iSK
iSKp
1.
is an
are surjective.
From the exactness of the rows in (2) and (3), we see that (3a) is a pushout square, and that (2b) is a pushout if the
"obstruction"
Hl(a/p;SK1(RP[p])) Hl(w/p;SKl(RP[p]))
to
(2b)
being a pushout
(Proposition
iK2
is surjective.
square
13.3(iii)).
on
is Coker(i) the
also occurs in (3), where it generates
we somehow can identify these two occurrences of
Thus,
other
hand,
Ker(f).
So if
Hl(ir/p;SKI(AP[p])),
302
CHAPTER 13.
C11(7L[G]) FOR FINITE GROUPS
then (2b) and (3a) will combine to show that (1) is a pushout square. To make this precise, consider the following square:
H1(ir/P;SK1(Rp[P])) 8 + HO(wr/p;Cll(R[P])) proj
a l-=
(4)
Coker(iK2) -) HO(ir/p;Cll(R[p]))/1(Ker(iCP )). Here
a
is the isomorphism of Proposition 13.3(111), and
by diagram (2).
Assume for the moment that (4) commutes.
$3
is induced
Then
Im(proj o 8') = Im((3) = Ker(iC1)/01(Ker(i,)).
It follows that Ker(iC1) = a1(Ker(iCP))+Im(8') = c71(Ker(iCP ))+Ker(f);
and hence from (3) that Ker(iSK) = f(Ker(iCi)) = foc71(Ker(iCP)) = 81(Ker(iCP)). Coker(81) = Coker(02)
Since
this shows that (1) is a pushout
by (3),
square.
If iCP
p
is odd, or if p = 2
and
has no real embedding, then
Kx
is an isomorphism by Proposition 13.3(1), and hence
isomorphism.
If
involution fixes
p
is odd,
iCl
8'
=
1
is also an
in (3) - the standard
(Theorem 5.12) and negates
C11(R[p])
(Theorem 8.6) - and so
then
iSK
SK1(Rp[p])
is an isomorphism.
It remains to prove that (4) commutes.
Fix
k =
E H1(ir/P;SK1(Rp[P]))
and represent each SK1(Rp[p])
a1
by some
by Theorem 3.9).
(so
fl(gi(ai)'a11)= 1E SKl(Rp[p]));
u1 E GL(R[p])
Write
(SK1(R[p])
surjects onto
[gl,ul]...[gk,uk]
303
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
(X E St(Rp[p])).
= $(X) E E(Rp[p])
By Proposition 13.3(111),
{g1,u1}...{gk,uk}.X l E
a(f) =
(Rp[A]t),
SK1(R[p][p]) = 1
Also,
lifted to
by Theorems 4.15 and 3.14, so the
Then a(f)
x1 E St(R[p][p]).
n = gl(xl).x1-
u1
can be
lifts to
1.g2(x2).x21 ...gk(xk).xkl'X
1 E 1(KpLPJ);
and the elements gl( xl).xil.g2(x2).x21...gk(xk).1 E St(R[p]CPJ),
are both liftings of
X E St(Rp[p])
From the description of
n[gi,u1] E CL(R[p]).
81
in Theorem 3.12, it now follows that
k Roa(f) =
l(n) =
[gl,ul]...[gk,uk]
=
A i=1
0
= 8'(f) E HO(n/p;Cll(R[p]))
Diagram (1) above need not be a pushout square if by
C11
p = 2).
(when
is replaced
SKI
This is the basis of Example 13.16 in Section
13c.
Recall that a 2-hyperelementary group Cn >iir 2-lt-elementary if
Im[a
con
(21n,
Aut( n) = (Z/n)*] C {J1}.
w
a 2-group) is
Theorems 13.4,
11.9, and 11.10 now combine to show:
Theorem 13.5
For any ftntte group
G,
C11(Z[G])
are generated by tnductton from elementary subgroups of prime
p,
C11(Z[C])(p)
and
SK1(Z[G])(p)
and C.
SK1(Z[G])
For any odd
are p-elementary computable;
304
Cl1(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
while
for
SK1(7[G])(2)
p = 2,
Also, if
2-IR-etementa.ry computable.
subgroups of
2-elementary
is
and
generated
denotes the set of 2-elementary
9
then the following is a pushout square:
C,
C(Q[H]) aa li 2I
l
HE
SK1(Z[H]) (2)
HE (1)
' SK1(7[G])(2).
C2(l[G])
Proof
and
For odd
this is an immediate consequence of Theorems 13.4
p,
As for 2-torsion, square (1) is a pushout square by Theorem
11.9.
13.4 and the decomposition formula for
SK[2]
Note
of Theorem 11.10.
that direct limits are right exact, so a direct limit of pushout squares is again a pushout square.
Recall the formula
C(QCG]) =
of Lemma 5.9: functor
where
RCS(-)(2)
Tensoring by
7/n
["'(G) 0 7l/n] u
(Un)
RC/R(G) = RC(G)/R.(G),
21n,
and
2-IR-elementary computable by Theorem
is
2-IR-elementary computable.
Square (1) remains a pushout if the limits are SK1(8[G])(2)
is also comput-
able with respect to induction from 2-IR-elementary subgroups of
Square (1) above need not be a pushout square if C11(7L[G])(2);
computable.
is thus
C2(l[G]) = C(D[G])(2)
taken over 2--R-elementary subgroups; and so
2-IR-elementary
11.2.
and taking coinvariants are both right exact functors,
so they commute with direct limits; and
replaced by
The
exp(G)ln.
and
C11(7L[G])(2)
Counterexamples
to
G.
SK1(7L[G])(2) is
both
not
0
is
in general
of
these
are
constructed in Example 13.16 below.
What would be more useful,
of
course,
would be a result
and
subgroups.
Unfortunately, just as was the case for
SK1(7L[G])
that
were detected by restriction to elementary
Cl1(7L[G])
SK1(2 p[G])
(Example
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
C11(Z[G])(p)
12.6),
is not in general p-elementary detected.
Reduction to p-groups
13b.
The goal now is to compare
is a p-group and R
K
305
p
in which
Cl1(Zc3[a])
p
if
When
p
whenever
w
is odd (Theorem 13.8); and that when p = 2,
(Theorem 13.10).
three groups (when
Cll(Z[a]),
The main results in this section are that
is isomorphic to one of the groups
Cll(R[a])
13.12.
to
is the ring of integers in any algebraic number field
is unramified.
C11(R[a]) = C1l(Z[a])
or
C11(R[a])
n
Cll(Z[ir]),
C11(ZC7[ir]),
The differences between these last
is a 2-group) are examined in Theorems 13.11 and
is odd or
G
is abelian, these results then allow a
complete reduction of the computation of
C11(Z[G])(p)
to the p-group
case.
The main problem here is to get control over the relationship between K2(Rp[a])
and
K2(7Lp[a])
in the above situation. In fact, these two
groups can be compared using Proposition 13.3(iii) from the last section. But first, some new homomorphisms, which connect
K2(7Lp[n])
with
H2(A),
must be defined.
For any group
n,
X7 :
H2(w) - K2(Z[v])/{-l,a}
One way to define
will denote the homomorphism constructed by Loday [1]. Aw
is to fix any extension 1 --> R -+ F a n ---+ 1,
free group on elements
a2,...,an;
where
F
is the
and let
A: F - St(Z[a]) be
the
homomorphism defined by
particular,
and
alai)-1
(A(ai)) E E(Z[ir])
setting
A(ai)
=
h11(a(ai)).
is a diagonal matrix with entries
in the first and i-th positions (and
1
elsewhere).
In
a(ai)
Then
306
for any a C R, and so Also,
Clj(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
$(A(a)) = diag(1,a(a2)j2,...,a(an)j")
A([R,F]) C ({g,g) ={-1,g}: g C C) _ (-1,r) and so
A(R n [F,F]) C Ker(#) = K2(Z[u]),
A : H2(A) - (Rn [F,F])/[R,F] Note that
When
a(g^h) = {g,h} tr
Aa : H2(r) -'' K2(Z[ir])/{-l,n} w
g,h E ir.
we let
p,
(Z
RV
denote the
K2( p
is constructed in the next lemma.
can be thought of as a
9,w: K2(Zp[v]) -> H2(ir)
by Theorem 3.1(i,iv). induces a homomorphism
for any commuting pair
composite
ji C Z;
' K2(Z[A])/{-l,ir}.
is a p-group for some prime
A splitting map for
A
for some
K2
This map
version of the
homomorphism
v
of Theorem 6.7: Jrigi E (2 p[a])*.
:
K1(Zp[ir]) ) nab
defined by setting
(i')1/lr'
u (jrigi) =
One can also define
9,m
for any unit
using Dennis' trace map from
K-theory to Hochschild homology (see Igusa [1]); but for the purposes here the following (albeit indirect) construction is the easiest to use.
Lemma 13.6
Fix a prime
p and a p-group
a.
Then there is a unique
homomorphism
0 = 0
such that
: K2(2p[n]) --) H2(A);
for any central extension
p-groups, the following diagram commutes:
1 -- 4 a - 4 it
n -9 1
of
CHAPTER 13.
K2(p[n]) 'I
K1(2 p[a],I,)
1 K1(
p[A])
ua
9A
Vi;
H2(ir) 0 a a
Ia = Ker[ p[A] -+ Zp[a]],
Here,
ri E p, g1 E W,
and
307
C11(Z[G]) FOR FINITE GROUPS
u;R
ab
(1)
.
is the map defined above, and for
zi E a,
ua(1 + 2ri(zi-1)gi) = Hzi'.
In addition, the following two relations hold for
factors through K2(2pand the composite
9A
(t)
07:
8
H2(ir) "
H2(w)
is the identity.
For any
(it)
u E (Zp[p])*,
g E n,
any
p g; a
be as above, and let ua
the
middle
this shows that
be the augmentation ideal.
0
Ia I /I Id = HO(i; Ia/I Ia) a > a isomorphism
ua
for
r1E
follows p,
from Theorem
ziE a,
is well defined.
can be defined uniquely to make
injective.
Ia
factors as a composite
wa(Zri(zi-1)gi) = flzi'
9,r
and any
1 - a - 7r a> a -> 1, let
I = Ker[Zp[ir] --» Zp]
K1(ap[a] Ia) proj where
[g,p] = 1,
0ir({g,u}) = g"up(u).
Proof For any central extension
Then
such that
1.15,
and giE G.
(2)
and where
In particular,
Since the rows in (1) are exact, (1)
commute whenever
There exist central extensions with
ba
ba
is
injective by Lemma
308
CHAPTER 13.
C11(7L[G]) FOR FINITE GROUPS
8.3(i); and two such central extensions are seen to induce the same by comparing them with their pullback over
07
n.
It remains to prove the last two points.
Let
(i)
injective.
n a a
be such that
Fix x E H2(a),
sa: H2(a) -i a = Ker(a)
is
and write
sa(x) = [gl,hl]...[gk,h ] E afl[ir,ir].
Ir
By the above definition of
(and Theorem 3.1(iv)):
7 (x) a {a(gl),a(h1)}...{a(gk),a(hk)) E K2(2p['r])/{-l,f,r}.
Then OR
ua([g1,h1]...[gk,hk])
r (x)) = ua aoX7(x) =
([gi,hi] E
K1(2 pp],Ia)
[gl,hl]...[gk,hk] = sa(x) E a;
=
and so 0 $ (x) = x.
(ii)
Now let
g
u = Irihi E (2 p[p])*. choose liftings
[g,u] = 1 + (gug
p
be such that
1 -- a -+ a al n
Let
g,hi E it
a({g,u}) _ [g,u]
I
and
of
g
and
hi;
[g,p] = 1, and fix --> 1 be as before,
and set
u = Jrihi.
in diagram (1); and
1
- u) u 1 = 1 +
(jri)-1(gg
again denotes the augmentation ideal).
- u)
So
(mod
Then
CHAPTER 13.
(6«)-1(U«(Cg,u])) =
e({g>u}) =
309
C11(7L[G]) FOR FINITE GROUPS (s«)-l(ua(l +
(yr.)-1(gug
- u)))
1 l
1
(by (2))
_
(ba)-1(ua(1 +
(sa)-l(fl[8,hil]ri)1/2r;
As was hinted above,
=
([I(g^hi)ri)l/lri
0
= g^uP(u).
is needed here mainly as a tool
0
for
describing the cokernel of certain transfer homomorphisms in K.
Lemma 13.7
Fix a prime
p,
and let
any algebraic number field in which pIp
in
R,
p
R
be the ring of integers in
is unramified.
For each prime
set
kp = ordp([R/p:7L/P]) = ma"{i : PiI[R/p:Z/P] = [Kp: p]}; and set
Then for any p-group
k = minplp(kp).
Proof
6k
K2(2p[n]) k ' vPk ® (H2(A)/x2b(n)) -4 0
K2(RP[-])
is the reduction of
Rp = [PIPRp
Since
trf YZ p171)
is exact for each
the sequence
e11
trf
K
is exact, where
n,
Since
p.
cyclic Galois extensions of
0V.
(Theorem 1.7), it suffices to show that
0.1
PO (H
KvA
2(')//"2
is unramified, this involves only C Kp,
If
p1[Kp:F],
and
S C F
9;F is the ring of integers, then
trf
is surjective,
since
:
K2(RP[n]) -b K2(S[ir])
the composite
trf o incl
is multiplication by
310
CHAPTER 13.
on the pro-p-group
[Kp:F]
the exactness of (1) when
In this case, write ring
Rp[,rx C]t.
C11(Z[G]) FOR FINITE GROUPS
[Ikp:Q] = pk
for some
k.
C = Cal(R /& ), and consider the twisted group
Then Rp[G]t
40.14]), and so
In particular, it suffices to prove
K2(S[,r]).
is a maximal order (see Reiner [1, Theorem
Rp[G]t = Mpk(Zp)
by Theorem 1.9.
The transfer thus
factors as a composite
trf
:
is K2(Rp[axC]t)
K2(Rp[,r])
= I (Mpk(Zp['r]))
(
pCA])
Proposition 13.3(iii) now applies to show that
Coker[trf: K2(Rp[,r])
-
2(zp[,r])] = H1(G;SK1(Rp[ir]))
GO (H2(w)/b(,r))
(Theorem 8.6)
Z/pk®(H2(v)/H (v)) The exactness of (1) will now follow, once we have shown that the composite
0°
trf K2(ZpC'r]) ;i'
12(Rp[w]) vanishes.
extension
To see this,
assume that
1 -> (z) - a a a -i 1
is surjective and
a commutator in
Ker(ba) 7 H2b(a) W.
upk ®(H2(w)/H (w) )
H2(a)/H2b(,r) A 0,
such that
ba: H2(w) --» (z) = Cp
We can thus assume, by induction on
holds for
z
is not
SK1(a): SK1(2 p[a]) -4 SK1(Zp[a])
is injective by Theorem 7.1, and its image has index 8.1.
Then
(use Lemma 8.3(1)).
The induced map
and fix an
p
ISK1(2 p[,r])I,
by Proposition that the result
W.
Consider the following commutative diagram with exact rows:
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
311
aR ) Kl(Rp[A],(1-z)) - K1(Rp[3])
K2( p[TM]) R "' K2(Rp[a])
It3
It2
Iti
i
where the
since It
K1(Rp[n],(i-z))
follows
and so this subgroup generates
K2(Rp[a])
that
RV (H2(0) C K2(2p[,r]);
is torsion free
(HO(W;(1-z)Bp[n])
(z)
is
is not a commutator);
z
By Proposition 6.4, the only
are transfer homomorphisms.
t;
torsion in
a
11
K2(2 p[a]) -!Z- Kl(2 p[ai,(1-z));
K2(2 p[ir])
is
generated by
Im(OR).
iR(K2(Rp[ir]))
and
and hence that
Im(t2) = (iZ(Im(t1)), K2(2p[lr])pk).
Clearly,
k
K(2p[1r])p C Ker(87); and i1(Im(t1)) C iZ(Ker(O7'))
by the induction assumption. Lemma C1l(Z[ir]),
13.7
will
when
a
Theorem 13.8
K
in which p
O
now be applied
is a p-group and
this is easiest when p
Ker(O;)
p
to
compare
Cll(R[a])
is unramified in
R.
with
As usual,
is odd.
Fix an odd prime
is unramtfied.
Let
p,
a p-group
a,
and a number Field
R C K be the ring of integers.
Then
the transfer homomorphism
trf
:
C11(R[ir]) -- Cll(Z[u])
is an tsomorph.tsm.
Proof
Consider the following commutative diagram of localization
sequences (see Theorem 3.15):
312
CHAPTER 13.
C11(7L[G]) FOR FINITE GROUPS
1 C11(R[A]) -' 1
c(A [w]) - Cp(K[rr]) trfK2 p[ir])
K2(2
with center
F = L(fn)
F.
p
and since
9.1;
- p(Q[Tr])
(En = exp(2rri/pn))
n
for some
F.
So
of
l[a]
by Theorem is a field
K ®. F = Z(K ®1 A)
K,
with the same p-th power roots of unity as trfCp
A
For any simple summand
is unramified in
by Theorem 4.13; and
(1)
) C11(7L[r]) -9 I.
is onto by Lemma 4.17.
trfCp
Here,
trfCl
ltrfCP
Cp(K[rr]) = Cp(I[ir])
is an isomorphism.
It follows from diagram (1)
trfCl
that
is onto, and also (using
Lemma 13.7) that there is a surjection
Z/pk ® (H2(
(a)) = Coker(trfK2) f s Ker(trfC1).
The standard involution is the identity on and is
(-1)
in Lemma 13.6).
Hence
by Theorem 5.12;
by Lemma 13.7 (and the description of
Coker(trfK2)
on
C11(R[w])
f =1, and
a
is injective.
trfCl
0
This can now be combined with Theorems 11.10 and 13.4, to give the following explicit description of
Theorem 13.9 a1,...,ak C G
55(Zi)
Fix a finite group
C
in terms of p-groups.
and an odd prime
p,
and let
be conjugacy class representatives for the cyclic subgroups
of order prime to let
C11(7L[G])(p)
p.
For each
i,
set
Ni = NG(ai),
Z.
= CG(ai),
and
Then
be the set of p-subgroups.
k C11(7L[G])(P)
k HO(Ni/Zi;
i=1 The
formula
C11(7L[G])(p)
in
Theorem
o
Cl1(7L[ir])).
lei
rrE°P(Z; )
13.9
gives
a
quick
way
of
computing
as an abstract group, but it is clearly not as useful if one
wants to detect a given element.
The best thing would be. to find a
CHAPTER 13.
generalization of
313
C11(7L[G]) FOR FINITE GROUPS
the formula
for p-groups in
C11(7L[G]) = Coker(y(,)
The main problem in doing this is to find a satisfactory
Theorem 9.5. definition of
PC
The closest we have come is to
in the general case.
show that for any finite
C,
Cl1(7[G])[2] = Coker(H1(G;7L[G]) G ) [R,(G) ® Z/n]
*)[2] (8/n)
for any
n
such that
commuting pair irreducible subspace
g,h E C
as follows.
C[G]-representations.
fixing
and
h,
exp(2ai/d)-eigenspaces for
*G(g ®h) =
where
exp(C)In,
For each Vi
g: Vi -i
Vi,
be the distinct
V1,...,Vm
Let
let
is defined for any
PG(g0h)
i,
let
C Vi
be
for all
din.
Vi g Vi
E [RC,(G) ® 7L/n]
m
of
the
Then
i=1 Under the isomorphism
sum
the
be the
*.
(7L/n)
* = C(®[G])
[RC,,,(G) 0 7L/n]
of Lemma 5.9, this
(Z/n) is easily seen to be equivalent to the definition of
9.2 when
G
is a p-group.
Also,
*G
in Definition
is natural with respect to
yG
inclusions of groups; and so the isomorphism
C11(7L[C])[2] = Coker(+G)[2]
follows from Theorems 9.5, 13.5, and 13.8.
In contrast to Theorem 13.8, when
rr
is a 2-group, it turns out that
there can be up to three different values for
(in which 2 is unramified).
C11(R[a])
for varying
R
These are described more precisely in the
following theorem.
Theorem 13.10 unramified; and let
Fix a 2-group R C K
w : K2 21w]) - C2(ID[v]) Let
a
and a number field
be the ring of integers.
and
K
where 2 is
Consider the maps
K2 (2[A]) -'
314
CHAPTER 13.
C11(7L[C]) FOR FINITE GROUPS
0"
Ga
0 DJ2 ®(H2(w) /,,ab(A) )
be the homomorphism induced by
(i)
C11(R[w]) = C2(D[ir])/Im(p)
(ii)
C11(R[w])
is purely imaginary and
R;
and
(iii) C11(R[w]) = K2(k[w])/(Ker(e"))
if
F C @(fn)
p12
in
is even for all primes
[R/p:ff2]
Proof
p12
in
A
of
For any simple summand n
some
for
particular, since
2
prime
p12
K and
in
K.
is unramified in
Furthermore,
F both do.
K
K % F
is purely imaginary
R.
with center
(Q[w]
K % F = Z(K ®Q A)
K,
F2,
F,
F,
by Theorem 9.1.
(fn = exp(2wi/2n))
with the same 2-power roots of unity as
if
K has a real embedding;
if
K
is odd for some prime
[R/p:F2]
and
Then
9
is a field
KP % F
and
In
for any
has a real embedding if and only
It follows that
C2(K[w]) - C2(Q[w])
if
K has a real embedding;
and (1)
K
C2(K[w]) = K2(Kp[w])(2) =
is purely imaginary.
In the first case, the isomorphism is induced by the transfer map (which is onto by Lemma 4.17).
In the second case, we define an isomorphism
a
to be the composite
a : C2(K[w]) = Coker[K2(K[w]) - ® K2(KP[w])](2) P
0
12(Kp[w])(2)
=ltrf
for any prime of
p,
p12
in
K.
To see that
a
note that for any simple summand A
is independent of the choice of
@[w],
either
315
Cl1(7[G]) FOR FINITE GROUPS
CHAPTER 13.
C2(K %A) = L/2 = K2(A2)
C2(K % A) = K2(A2));
(and so there is only one possible isomorphism
else aIC2(K % A)
or
is the composite
t
C2(K % A)
C2(A)
K2(A2).
r°
Now consider the following diagram:
H2(n)
7
proj
(2)
s K2(R2[A])
where map.
) 2(g2(ir])/{-1,±r}
a/2k ®(H2(A)/l b(n))
k = max{i: 211[R/p:W2], all p12 in R},
and
tI
0,
is the transfer
By Lemma 13.7, the row in (2) is exact, and the triangle commutes.
Furthermore,
5v
factors through
K2(7L[a]),
K2(7L[n]) --) vanishes by construction
j SK1(7L[n],n)(2) = C2(D[w])
(K2(2p[1r])(2) = 1
V2(221r]) = (Im(tl) Similarly, since
Ker(V)
,
for odd
Im(am)) _ (Im(tl)
,
p).
It follows that
Ker(4p)).
C (Im(tl),
has a real embedding, there is a commutative diagram
K2(R2[a]) - C2(K[w])
Itt
(3)
differ by exponent 2 (Theorem 9.1),
and
Im(tl) C Ker(O") = Ker(01) C (Ker(9k), K
and the composite
It2
C11(R[n]) -> 1
It3
C2(D[w]) --> Cli(Z['m]) - 1,
316
CI1(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
where the
are transfer maps, and
ti
Im(p)= If
by (3), and so
Then
is an isomorphism by (1).
t2
is an isomorphism.
t3
K has no real embedding, then we use the diagram
K2(R2[ir])
P Rat
I Cll(R[a]) -' 1
C2(K[ir])
c(82[r]) -' A[A]) where the top row is exact, and the square commutes by definition of If
[R/p:ff 2]
is odd for some
onto by (2), and so
p12
in R
C11(R[a]) = Coker(4p).
(i. e., if k= 0), then And if
C11(R[w]) = KA[w]/mPo t1) =
In particular, for any 2-group unramified in C11(gc7[ir])
R),
C11(R[a])
and any
is isomorphic to
(case (ii)), or
C11(7Lc3[7])
R
is
then by (4),
k > 0,
(Ker(A")).
it
a.
t1
o
(such that 2 is
CI1(Z[ir])
(in case (i)),
(case (iii)).
Theorem 13.10
gives algorithms for computing these groups, and the "unknown quantity" in
all of them is for describing
This is why one can hope that any procedure will also extend to the other two cases.
Cl1(7L[a])
C11(ZC7[A]) = Cl1(7C3[ir])
that
n
K2 (22[a]).
if
Note
H2(v) = Hb(a) - in particular, if
is abelian. We now want to carry these results farther, and get lower bounds, at
least,
for the differences between these groups
consider the case where
Theorem 13.11
Let
is abelian (so
it
a
C11(R[a]).
We first
C11(R[a]) = SK1(R[a])).
be an abelian 2-group, and set
Then
2' SK1(ZC7[n]) 2' SK1(7[ir]) ® SK1(Z 3[v/n2])
SK1(Z[n]) ®
(8J2)2'`-1-k-(a).
k = rk(ir).
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
Here,
If n =
ir2 = {g2: g E a} .
(so
(C2)k
a2 = 1),
317
then the following
triangle commutes: 0
('ZLA])
\Jv
H2(ir)
(1)
K2(&2[n])/{-l,kir} =
where
V(g-h) = {g,h}
g,h E n
For convenience, set
Proof projection.
and, is tnjecttue.
and
i Z 2,
more details.
A
C(A) = (gi) = K2(A2).
A = Q
either
C(A) = 1;
and is
A = Q(fi)
or
for
See the table in Theorem 9.1 for
In particular, this shows that
f1®f2 :
C(Q[w]) $ K(%[n])
is an isomorphism; where by
Q[w],
of
in which case
(Q[7r],
and let a: A -4 R be the
n = 11/'R2,
For each simple summand
a simple summand of some
for
(Z/2)2 " -1-k
f1
is the usual projection and
f2
is induced
a.
Let 1.12).
J = (2, 1-g
:
gEir) C 12[a]
be the Jacobson radical (Example
From the relation
(g-l) + (h-1) = (gh-1) - (g-1)(h-1) ° (gh-1)
we get that
(mod
J2)
(a2[,r]) = 1 + J = (±g, u: g E ir, u (: 1 + J2).
(for
g,h E a)
So by Corollary
3.4, K2(7L2[a]) = ({±g,±h}, {g,u}
: g,hEa, uEl+J2).
For any g,hEi, {±g,±h} E Im[K2(Z[ir]) - K2(22[ir])] 9 Ker(flopv)
(2)
318
CHAPTER 13.
by definition of
C(R[w]).
uEl+J2,
maps to
42(1]
(=
{g,u}
(42)2k
C11(Z[G]) FOR FINITE GROUPS
Also,
{w,l+J2} C Ker(f24,d,
{il,l+4Z2} = 1
So by Theorem 13.10,
).
at each simple summand 2 of (fl,f2)
induces an isomorphism
Coker[4,R: K2(Z2Cw])
1(ZS3Cw]) =
since for any
-'
KZ(kCw]),
i(ZCi]) ® SK1(Zc3Cn])
Now assume that
clearly commutes on symbols
we have seen that 13.6(11).
gyp({g,u}) = 1;
u C 1+ J2
For any
{ig,ih}.
This
and consider triangle (1).
w = it = (C2)k,
and
and any
9w({g,u}) = g^uw(u) = 0
g E w,
by Lemma
This shows that (1) commutes; and hence by Theorem 13.10 that
SK1(ZC3[w]) = SK1(ZC7[w]) =
Coker(V).
k
Since
K2(k[w]) = (Z/2)2
2[w] =
remaining claims - the injectivity of
by Theorem 4.4.
V and the ranks of
So the
Coker(V)
and
will all follow, once we have shown, for any basis {g1,...,gk}
for
3=
w,
that the set
, {-l,gi} , {g1,gi} E KcA[w]) :
is linearly independent in
1
i < j Sk}
K2(&2[w]).
To see this, define for each
s E F a character
Xs: w -' fill
follows:
s =
Xs(ge) = 1
s = {-1,g1}:
Xs(gi) = -1,
(all
e)
Xs(ge) = 1
(all
a #i)
as
CHAPTER 13.
s = {gi,gj}:
Then, if
by
xs(gi) = Xs(gj) = -1,
[n]) -- K2(@2)= {tl}
s: K2(
we see that
XS,
ordering)
x(s)
KS(s) _ -1
= 1
for all
linearly independent in
For nonabelian
t
ks(ge) = 1
C11(ZC3[ir]).
a Ai,j)
denotes the homomorphism induced s,
while (under an obvious
< s
V.
This shows that
in
P
the best we can do in general is to give lower
Recall that for any 2-group
C11(Z[a]),
C11(7Lc7[ir]),
the Frattint subgroup
ir,
is the subgroup generated by commutators and squares in i. e., the subgroup such that
Theorem 13.12
a be any 2-group, and set k = rk(a/Fr(ir)). Set
Let
S S R S (2).
a;
a/Fr(a) = Z/20 nab.
R = rk(Im[H2(a) --> H2(n/Fr(a) )] ); so that
is
13
bounds for the "differences" between the groups and
(all
for each
K2(@2[a]).
n,
319
C11(7L[G]) FOR FINITE GROUPS
S = rk(Im[112b(a) -> H2(ir/Fr(ir) )] ),
Then there are surjecttons
C11(Zc7[n])
(Z/2)2k-1-k-R
11 Cll(Z[v]) ®
and
C11(Zc3[v]) = C11(ZC21[A]) --» C11(ZS7[n]) ®
In particular,
Proof projection.
C11(Zc3[7]) $ Cll(ZC7[a])
Set
a = it/Fr(a) = (C2) k,
if
(Z/2)R-S.
R > S.
and let
a: n -+ fr
be the
Let
sA : K2ca2['r]) --p H2(w), be the homomorphisms of Lemma 13.6.
s : K2(Z2[A]) - 112(i) Consider the following commutative
C11($[G]) FOR FINITE GROUPS
CHAPTER 13.
Id
H2(n)
'r
HZ(a)
/01K a
1
K2(2
K2(22Cn])/{-l,fa}
C2(D[w]) u
A»
H2(r)
fz
f1
Note in
and is injective by Theorem 13.11.
V(g-h) = {g,h},
Here,
2[R])/{-1,±ii}
(1)
particular the following three points: k
(Z/2)2 -1-k'
K2(k[W
(a)
and
VoO_ = p_,
by Theorem
13.11.
Al 7 = Id
(b)
(c) pA = 1
by Lemma 13.6(1).
7
since
factors
through
K2(Z[w]),
and the
composite
K2(7L[n])
) K2(g2[w])
) 4im SK1(71[a],n)(2) = C2(Q[w])
'P
n
vanishes by construction
(K2(2 p[a])(2) = 1
for odd
p).
Now consider the homomorphism
(f1,f2): KZ(k[ir])/{-1,fr}
Write of
@[A]
4.13),
[I
D[a] = A x Q[R],
where
not isomorphic to (f1,f2)
1 c2(QCrJ) ®
Q.
A
2
is the product of all simple summands
Then, since
C2(Q[A]) = C2(A)
(Theorem
factors through a product of epimorphisms
(A2)/{-1,±r} -» C2(A)] x
[K2(k[W])/{-l,fir} - K2c(E [R])/{-l,fi}].
CHAPTER 13.
This shows that
C11(Z[G]) FOR FINITE GROUPS
321
is onto; and hence (using Theorem 13.10(11))
(f1,f2)
that there are surjections
» Coker((f1,f2)4)
C11(ZE7[ir]) = Coker(4p)
S Coker(f24)
s
H a
= Coker(.p) S Coker[H2(a)
H2(n)
(Z/2)2'`-1-k-R
Cll(Z[v]) S
To compare
(by (d))
C11(Zc3[a])
with
(by (a))
C11(7Lc7[u]),
set
D = Ker[A" : H2(a) -i Z/20 (H2(1r)/b(1r) ), ,
so that
Cl1(Zr3[a] ) = K2(Q2[n])/v(O- (D))
any splitting
(K2(k[a])
by Theorem 13.10(111). Im(f2 oip)
$3:
inclusion
Then there is a surjection
has exponent 2).
Cl (Zr[n]) = K(@2[A]) 1 3
the
of
Fix
(A
-1
n
(D))
(proj pof2)
» Coker(p) ®
_
F2oc(0' 7 (D) ) Im[H2(a) -i H2(a)]
Im(VoH2(a)) 19
VoH2(a)(D)
(Z/2)R-S
Cl1(ZZ7[n]) S
Im[H2b(A)
0
The groups constructed in Example 8.11 (when property that
R > S
C11(Zc7[ir])
for
have the
C11(ZC3[wr]) $
This difference
for
v.
construction in Example 13.16 below of a group G
When G
p = 2)
in the above theorem, and hence that
such
C11(Z[G]) C SK1(Z[G])
-* H2(r)]
is
the basis
the
for which the inclusion
has no natural splitting.
is a finite abelian group, Theorems 13.8 and 13.11 yield as
322
CHAPTER 13.
Cl1(Z[G]) FOR FINITE GROUPS
a corollary the following formula for
SK1(Z[G])(p)
Note that this reduces the computation of
SK1(Z[C])
(= C11(Z[G])(p)).
(for abelian G)
to
the p-group case - which is handled by Theorems 9.5 and 9.6.
Theorem 13.13
Fix an abeltan group
G
and a prime
Write
piIGi.
where v is a p-group and p4IHI. Set k = rk(ir), and let n
G = Hxw,
denote the number of simple summands of
Then
Q[H].
n
® SKJR[Ir )
if
p
if
p = 2.
is odd
sKl(ZCGJ)(P)
® SK1(Z[v)) Proof
0
@[H] = fi=1K
Identify
be the ring of integers.
R.
Ki
in
K[H],
and
[1[a]: Z[G]]
where Then
is prime to
SK (Z[G])()
the
Ki
are fields.
Let
V = fl 1Ri is the maximal order
p by Theorem 1.4(v).
Hence
n ® SK1(R1CAJ) i=1
1
by Corollary 3.10, and the result follows from the formula in Theorem 13.11
13c.
or Theorem 13.8
(p = 2)
(p
a
odd).
Splitting the inclusion C11(Z[G]) C SK1(Z[G])
So far, C11(Z[G])
all results about
and
SK1(Zp[C.])
deal with its components
SK1(Z[G])
separately.
It is also natural to consider
the extension
1 -- Cll(Z[G])
SK1(ZCGJ) E ® sK1($p[GJ) - 1; P
and in particular to try to determine when it is split.
The key to doing
this, in odd torsion at least, is the standard involution.
Theorem 13.14 For any finite group C,
Fix
is multiplication by
C11(7[C])
This will be shown for
Proof time.
whose restriction to
G,
there is a homomorphism
SK1(Z[G]) ) C11(Z[G]),
sG :
natural in
323
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
one prime
SK1(7L[G])(p),
p
2.
at a
and consider the short exact sequence
p,
e
1 --> C11(7[G])(p)
SK1(7L[G])(p)
SK1(2p[G]) -i 1
G
)
of Theorem 3.15.
Step 1
G
Assume first that
is a p-group and
pin.
antiinvolution T
on @[G]
is p-elementary:
T*
where
v
Instead of the usual involution, we consider the defined by:
(ai E (D,
T(Zairigi) = lair igi
We claim that
G = Cn x w,
r i E C 1,
acts via the identity on
gi E n).
Cl1(7L[G])(p),
and via
negation on SK1(2 p[G]).
Step lA r E Cn
and
Let g E Y.
involution on C11(7[G])(p)
a E Aut(G) Then
@[G].
@[a]
for
with the usual T* = a*
on
Cp(D[G]).
By construction, center of
is the composite of
a(rg) = r lg
In particular, by Theorem 5.12,
@[G].
and
T
be the automorphism:
Q[a]
fixes all p-th power roots of unity in the
So by Theorem 4.13,
a*
is the identity on
Cp(Q[G]),
and hence (by the localization sequence of Theorem 3.15) on CI1(7[G])(p). It follows that
Step lB Fi
T. = id
on
Write & [Cn]
CI1(7L[G])(p).
= nk=1FiI
are unramified field extensions of
and
Zp[Cn and
]
= ni=1Ri,
Ri C Fi
where the
is the ring of
324
C11(Z[G]) FOR FINITE GROUPS
CHAPTER 13.
Then this induces decompositions
integers.
k Zp[G] =
By construction,
k
T
SK1(2p[G]) =
and
II Ri[a] i=1
II SK1(Ri[TM]) i=1
leaves each of these summands invariant, and acts on
each one via the identity on coefficients and by inverting elements of
T*
By Theorem 8.6,
acts on each
T.
and hence on
SK1(1 p[G]),
SK1(2p[G]) - SK1(Z[G])(p)
as follows:
SK1(R1[ir]),
by negation.
Stev 1C
Now define
given x E SK1(2 p[G]),
sG:
lift
x
to x E SKI(Z[GI)
and set (p),
SG(x) =
This is independent of the choice of lifting by Step lA, and its composite
with the projection to squaring) by Step 1B.
is multiplication by 2
SK1(2p[G])
By construction,
sG
(1.
e.,
is natural with respect to
homomorphisms between p-elementary groups.
Step 2 Now let
G
be an arbitrary finite group, and let
set of p-elementary subgroups of
G.
8
be the
By Theorem 12.4,
8K1(p[C]) = l
sKl(Zp[H]),
HE
where the
limit
is
taken with respect
to
inclusion and conjugation.
Hence, by Step 1, there is a well defined homomorphism
sG
sH
: SK1(Z [C]) - H
SK1(Z[H])(P)
_Ind
SK l(Z[G])(P)'
H
where s
G
sG
is natural and
2G o sG
: SK,(Z[G])(p) --> Cl1(Z[G])(p)
is multiplication by
can be defined by setting:
SG(x) = x2-(sC 0 QG(x))-l.
13
2.
So
CHAPTER 13.
325
C11(7L[G]) FOR FINITE GROUPS
An immediate corollary to Theorem 13.14 is:
Theorem 13.15
For any finite group
p-power torsion in
G
and any odd prime
the
splits naturally as a direct sum
SKI(Z[G])
SKI(Z[G])(P) = C1I(Z[G])(P) ® SKI(ap[G]).
The problem remains to describe the extension in 2-torsion, in general.
p,
a
C11(Z[G]) C SKI(Z[G])
It seems likely that examples exist of 2-groups
where the inclusion C11(Z[G]) C SKI(Z[G])
has no splitting at all.
This
problem is closely related to Conjecture 9.7 above, and the discussion following the conjecture. Cl1(Z[G]) C SKI(Z[G]) of
H2b(G) C
In particular, the splitting of the inclusion
seems likely to be closely related to the splitting
H2(G)-
The following example shows, at least, that the inclusion C SKI(Z[G])
need have no natural splitting in 2-torsion:
C11(Z[G])
more precisely,
no splitting which commutes with the action of the automorphism group At the same time,
Aut(G).
SK1(-)
is replaced by C11(-).
Example 13.16 Let
Im[H2(a)
Set
it illustrates how Theorem 13.5 can fail if
-- H2(ir/Fr(ir))]
(it)
Im[H2b(a)
and Go = C7x C3 x 7 r a G
G = C7 x Sa x 7r,
(t)
a be any 2-group with the property that
C11(Z[C])(2)
-i H2(a/Fr(,r))]. (S3 = C3 A C2) .
Then
is not 2-IR-elementary computable;
the square
lim
C2(Q[H])
» li CII(Z[H]) (2)
H (1) 1
C2(Q[G])
1
% ClI(Z[G])(2)
326
CHAPTER 13.
is not a pushout square, subgroups of
C11(7L[G]) FOR FINITE GROUPS
where
9
denotes
the set of 2-elementary
and
G;
the extension
(iii)
1 -i C11(7[Go]) --' SK1(7L[Go]) - SK1(12[Go]) - 1
(2)
has no splitting which is natural with respect to automorphisms of
a = Gal((21/QC7) = C21
Set
Proof
induced twisted group ring.
acts trivially on
and let 7S21[7r xa]t be the
Then a acts trivially on
C2(D21[a])
by
Go.
Theorem 4.13).
Cll(Zc21[ir])
(it
Furthermore, there is
an inclusion t
ZC21[lr x a]
C M2(ZC7[n])
of odd index (see Reiner [1, Theorem 40.14]); and so by Corollary 3.10 and Theorem 13.12:
cll(7LC21[n x a]t) = C11(ZZ7[w])
Note that
C = C21 >4 (w x a).
shows that
C11(7[G])(2)
2-It-elementary
subgroups.
Cl1(7c21[w]) = HO(a;Cll(7S21[A]))
(3)
Just as in the proof of Theorem 11.9, this
is not computable with respect to induction from Also,
since
C2(7L[G])
is
2--R-elementary
computable, this shows that square (1) above is not a pushout square. Now, by Theorem 13.4,
SKi2](z521[n x a]t)
HO(a;SKl(ZC21[w]));
a n d similarly (by Proposition 13.3(11)) for
X C I
2a]t).
Together
with (3) above, this shows that the sequence
1 - H0(a;C11(7c21[n])) -' H0(a;SK1(7C21[7T])) -> H0(a;SKI(a2C21[A]))
is not exact.
This implies in turn that the exact sequence
1
CHAPTER 13.
327
C11(7L[G]) FOR FINITE GROUPS
1 - ) Cl1(gc21[r]) -) +SK1(7C21[ir]) --- sKl(22c21[n]) ' 1
has no splitting which commutes with the action of
direct summand of
22[Go]);
But (4) is a
a.
direct summand of sequence (2) above by Corollary 3.10
(4)
is a
(12c21[7]
and so (2) has no natural splitting.
o
More concretely, consider the group
a = (a,b,c,d I a2 =b 2= c2 =d 2= Then
nab
a: 4 --a
= (C2)4,
,gab
(generated by
n
and
[ir,ir] = Z(a) = (C2)5
is the projection, a-b+c^d);
[a,b][c,d] = 1)
then
H2(a)
(see Example 8.11).
has image of rank one
while its restriction to
satisfies the hypotheses of Example 13.16.
If
H2b(a)
is zero.
So
Chanter 14 EXAMPLES
We now list some examples of calculations of
These
SK1(Z[G]).
illustrate a variety of techniques, and apply many of the results from earlier chapters.
We have already seen, in Theorem 5.4, that
C11(Z[G]) = I
is a product of matrix algebras over
R.
to some conditions which imply that
SK1(Z[G]) = 1
if
R[C]
The first theorem extends this or
Wh(G) = 1.
It
shows, for example, not only that the Whitehead group of any symmetric group vanishes, but also that
Wh(C)
vanishes whenever
of symmetric groups, or a product of wreath products
Theorem 14.1
Define classes
9, Q, 9)
G
Sm2Sn,
is a product of matrix algebras over
I+;
Q = {G : @[G]
is a product of matrix algebras over
a}
(
C(g)) = 0,
etc.
of finite groups by setting:
9 = {G : IR[G]
2) = {G : H,(CG(g))
is a product
g E
all
C 91;
and
4.
Then
(i)
Wh(G) = 1
for any G E Ql 1),
and SK1(Z[G]) = C11(Z[G])
(ii)
for any G E kfl),
SK1(Z[G]) = 1
for any G E 0;
all symmetric groups lie in
Q11%,
and all dihedral and
symmetric groups Lie in in!); and (tit)
all three of the classes
91,
Q,
and
)
are closed under
products, and under wreath products with any symmetric group
Sn.
Ci{APTER 14.
Proof
for any G E Q,
By Theorem 12.5(1),
G E W.
then
If
SK1(Z[C])/C11(Z[G]) = 0pSK1(!p[G]) = 1
Wh'(G) = Wh(G)/SK1(Z[G])
has rank zero (Theorem 2.5); and so
For convenience,
W
Proposition 8.12,
products
(note
write
Wh(G) = SK1(Z[G])
and any
GET
in this case.
for any
7I(G) = H2(G)/I1ab(G)
By
G.
is closed under taking
When checking
CC H(g,h) = CG(g) X CH(h)).
for a n y
If
is torsion free (Theorem 7.4) and
is multiplicative; and so 9
that
((CG2S (g)) = 0
by Theorem 5.4.
C11(Z[G]) = 1
then
G E 51,
329
EXAMPLES
that
we are quickly reduced
g E GZSn,
to the following two cases:
(a)
$=(g1,...,gn)EGnCC Sn:
is a product of
then
wreath products (by symmetric groups) over the centralizers
(b)
a=(12...n)ES
where
and
(g,CG,n($)).
CG(gi).
then
=W(C(;($l...gn)) =0.
W(CG.n($))
Using Proposition 8.12 again, we see that 2Y(CS) =0 if X(G)=O (any p-Sylow subgroup of G2 p2...ZCp). Clearly,
GZSn
is contained in a product of wreath products
Together, these relations show that
1
Q
and
are closed under products.
product of matrix rings over
and so
Sn E Q C lt.
for all
Finally,
n,
Also,
if
CE).
Q[Sn]
is a
(see James & Kerber [1, Theorem 2.1.12]);
0
Using this, it is an easy exercise in manipulating
twisted group rings to check that @[GZSn]
GZSn ES
Q
(or
is a splitting field for
IR)
if it is a splitting field for
for each
n,
D(2n) E W,
since
@[G]. X(G) = 0
IR[D(2n)]
is easily seen to be a product of matrix algebras over
The condition that
0[G]
to this is the central extension
R.
be a product of matrix algebras over
does not by itself guarantee that
Wh(G) = 1.
G
whenever
contains an abelian subgroup of prime index (see Proposition 12.7).
And 0
Q
The simplest counterexample
330
CHAPTER 14.
1
-
EXAMPLES
(C2)4-4 G- (C2)4--1 1;
defined by the relations:
[b,ac][c,d]
G = (a,b,c,d : a2 = b2 = c2 = d2 = 1
= [a,cd][b,d]).
`
A straightforward check shows that for
and so
r = 1,2,4;
D[G]
is a product of copies of
Wh(G) = SK1(Z2[G])
by the same arguments as in
But using Lemur 8.9, applied with
the above proof.
one can show that
Mr (Q)
(C2)4,
G = G/[G,G]
SK1(22[G]) n' - (Z/2)- (Z/2).
The next theorem gives necessary and sufficient conditions for when
in the case of an abelian group
SKI(Z[G]) = 1
also gives some conditions for when does not vanish for nonabelian G,
SKI(Z[G])
C.
Note that while it
or
C11(Z[G])
does or
a comparison of Theorems 14.1 and 14.2
indicates that a complete answer to this question is quite unlikely.
Theorem 14.2
Fix a finite group
(i) If each Sy l ow subgroup of (any n 0), then SKI(Z[G])(p) = 1. (ii)
C
If
G =
Gab = (C2)k
for some
(a) n;
If
C
or
(b)
Proof
has the form
p,
for some
Cpn
and if n,
or
Cp x Cpn
C11(Z[G]) = 1, or
p = 2
and
k.
is abelian, then
C = (C2) k
(i)
Cp X Cpn
Cpn
each Sylow subgroup of or
C
is a p-group for some prime
then either
(iii)
G.
G
has the form
for some
By Theorem 5.3,
from p-elementary subgroups.
SKI(Z[G]) = 1
C n p
if and only if either or
Cp x Cpn
for some
k.
SKI(Z[G])(p)
is generated by induction
Hence, it suffices to show for all
n >1
CHAPTER 14.
SK1(Z[Cn]) = SK1(Z[Cp x n]) = 1.
that
n
EXAMPLES
is a power of
(ii)
331
This follows from Example 9.8 when
and from Theorem 13.13 in general.
p;
For nonabelian
this was shown in Example 9.9.
G,
abelian case, recall first that a surjection G - G'
Cl1(Z[G]) -» C11(Z[C'])
induces a surjection C11(Z[G]) 9.8(ii)),
is nonvanishing if onto
C4 x C2 x C2
G
of finite groups
surjects onto
(Cp)3
p = 2
(if
SK1(Z[G])(p) = 1
G
and
Cpn,
and
Cp X Cpn,
if and only if
is not a 2-group)
or
Note that the exact exponent of G,
is odd
(C2)k.
G
and any
SK1(Z[Sp(G)]) = 1
rk(S2(G)) S 2.
Sp(G) - Cpn
this holds if and only if
p
if
(iii) By Theorem 13.13, for any finite abelian group phIGI,
(Example
Cpl x Cpl
The only abelian p-groups which do not
surject onto one of these groups are
prime
So
(Corollary 3.10).
(Example 5.1), or onto
(Alperin et al [3, Theorem 2.4]).
In the
and
By (i) and (ii),
or G = (C2)n.
Cp x Cpn,
0
for arbitrary abelian
SK1(Z[C]),
is computed in Alperin et al [3, Theorem 4.8] (see Example 5 in the
introduction).
We next give a direct application of the results about twisted group
in Chapter
rings
SK1(Z[G])
when
We want
13.
the 2-power
to describe
torsion
is dihedral, quaternionic, or semidihedral.
S2(G)
in
Note
that this includes all groups with periodic cohomology, in particular, all
groups which can act freely on spheres - and that was the original motivation for studying this class.
The following lemma deals with the
twisted group rings which arise.
Lemma 14.3 field
K
R
Let
in which
be the ring of integers in an algebraic number 2
is unramifted.
quaternionic, or semtdthedrat 2-group. homomorphism, set
p = Ker(t),
and let
Let K[ir] t
Let
x
be any dihedral,
t: a -4 Gal(K/@) and
R[,r]t
be the induced
Then
twisted group rings.
_
is nonabeltan and Kw ¢ ll
Z/2
if
1
otherwise.
p
Cl1(R[w]t)(2) 1
be any
332
EXAMPLES
CHAPTER 14.
Assume first that
Proof
by Example 5.8.
C11(R[ir]) = 1
jC11(R[a]))
13.12
2
If
K
K
If
has a real embedding, then
has no real embedding,
by Example 5.8 again; while
1C11(R[v])I
then
by Theorem
2
maps trivially to H2(v b) = 7/2).
(H2(7)
Now assume that Ker(t).
t = 1.
t: it --> Gal(K/@)
By Theorem 13.4,
Cll(R[p])
is nontrivial, and set
surjects onto
p =
and
C11(R[ir]t),
C11(R[ir]t) = HC(ir/p;C11(R[p])) = ?/2
if
KA
where
So it remains only to consider the case
has no real embedding.
is nonabelian, where Kw C IR, but where K !Z R.
p
Assume this, and consider the pushout square of Theorem 13.4: a
HD(A/p;C2(K[p])) mss HD(v/p;Cll(R[p])) = Z/2 jiC2
C2(K[w]t)
We saw, when computing
» Cl1(R[ir]t)(2).
C11(R[p])
in Example 5.8,
that
0P
can be
identified with the composite
4
C2(K[p]) -s C2(K[pab]) = ®C2(K)
(note that order
8:
pab = C2 X C2). since
[w:p] = 2,
nonabelian group of K[pab] C
Also,
order
8
s_m4
it/[p,p] = D(8), 17abI
= 4,
and
which contains
C2(K) = 8/2;
the dihedral group of D(8)
is
C2 X C2-
the only
The pair
now splits as a product of inclusions
(K x K) x (K x K) C (M2(K)) X (M2(KA) X M2(KA)) .
Since C2(k) = 1 and hence that
(Kw C IR),
this shows that
Cl1(R[ir]t)(2) = 1.
o
Ker(iC2) ¢ Ker(ap)
in (2);
CHAPTER 14.
333
EXAMPLES
Applying this to integral group rings is now straightforward.
Example 14.4
be a finite group such that the 2-Sylow
G
Let
G are dihedral, quaternionic, or semtdthedrat.
subgroups of
Then
1(ZCG])(2) = C11(ZCG])(2) = (7/2)k,
is the number of conjugacy classes of cyclic subgroups a g G
k
where
such that
and
(a)
Jul
is odd,
(b)
with gxg 1 =
there is no g E NG(a)
(c)
Proof
Note first that
has nonabelian 2-Sytow subgroup,
CG(a)
SK1(12[G]) = 1
X-1
ni = la iI
and
Ni = NG(a,),
al,...,ak
if
conjugacy class representatives of cyclic subgroups of and if
x E a.
by Proposition 12.7, so that
By Theorem 11.10,
SK1(Z[G])(2) = C11(Z[G])(2).
for all
G
are
of odd order,
then
k C11(ZCG])(2) =
C11(Zrn.[A]t(2),
llii
i=®1
wEA(N
)
denotes the set of 2-subgroups.
where
immediate consequence of Lemma 14.3.
The result is now an
0
We now finish by giving examples of two more specialized families of groups for which
is nonvanishing in general, but still can be
SK1(Z[G])
computed.
Theorem 14.5
(i)
For any prime power
SK1(Z[PSL(2,q)]) = 1/3
p = 3 and k
is odd,
k
5;
and
q = pk,
SK1(Z[SL(2,q)]) = Z/3 x 113,
and
(ii) SK1(Z[PSL(2,q)]) = SK1(Z[SL(2,q)]) = 1
Proof
Write G = PSL(2,q)
if
and G = SL(2,q),
otherwise.
for short.
By Huppert
[1, Theorem II.8.27], the only noncyclic elementary subgroups of
G
and
334
G
EXAMPLES
CHAPTER 14.
are dihedral and quaternionic 2-groups, elementary abelian p-groups,
and (in
particular,
C2
products of
a)
with elementary abelian p-groups.
SK1(Z[G]) = C11(Z[G]),
and similarly for
In
by Proposition
a,
Furthermore, by Theorem 14.2 and Example 14.4, this list shows that
12.7.
C11(Z[H]) = 1
H
for all elementary subgroups
possibly for p-elementary subgroups when p generated
by
C11(Z[G])
are p-groups, and vanish if
elementary
Assume now that
induction
p
is odd and
the decomposition formulas for
in
p = 2
a,
except
C11(Z[G])
is
and
Cl1(Z[G])
5.3), or
or
Since
is odd.
(Theorem
G
k = 1.
Most of the terms vanish in
k > 1.
C11(Z[G])(p)
and
Cl1(Z[G])(p)
of
Theorem 13.9; leaving only
C1 (Z[G]) = 1
C1 (Z[p])
li
1
PWG)
is the set of p-subgroups), and
(where
Cl1(Z[G]) -
Cl1(Z[p]) x
11i
P
Cl (Z[p]).
P576)
76)
Since these limits are all isomorphic, it remains only to show that
-
Z/3
if
1
otherwise.
PE(G)
1
Furthermore, the p-Sylow subgroups of two p-Sylow subgroups of of
SL(2,q)
G
C
k odd
p( li
G)
F*2
q
are isomorphic to
Fq,
and any
intersect trivially (any nontrivial element
of p-power order fixes some unique 1-dimensional subspace of
Hence, for any p-Sylow subgroup
(ffq)z).
Here,
k Z 5,
p = 3,
C11(Z[p])
li
P C G,
q Cl1(Z[p]) - HO(N(P)/P; C11(Z[P])) - HO(F 2; Cli(Z[ff])).
denotes the group of squares in
Now, since e2 has order prime to q
p,
FQ.
EXAMPLES
CHAPTER 14.
335
*2 (elements fixed by FQ2)
HC(F 2; C11(Z[Fq])) = Cil(Z[Fq])F
*
*2
Coker[(Fq 0 Z[Fq])F° -'
Also, any element
a E Fp
leaves no fixed elements Fp fl FQ2 # 1;
Cp(@[Fq])
acts on if
a 0 1.
and this is the case if
via
So
(x N x ),
and this
HO(FQ2;Cl1(Z[Fq])) = 1
if
p25, or if p=3 and k
is
even.
Now assume that
p = 3,
the nontrivial summands of
k 2 3,
@[Fq]
and k
is odd.
Then
ffQ2
permutes
simply and transitively; so that
*2
p(D[Fq])F° = Cp(D3) = Z/3.
Furthermore,
by
Alperin
et
al
[3,
Proposition
2.5],
there
an
is
isomorphism
Im1F 0 Z[F ] ' Cp(D[Fq] )] = S (Fq) (the p-th symmetric power) which is natural with respect to automorphisms of
This now shows that
ffq.
HO(F 2; C11(Z[Fq])) = (Z/3)r,
where
.2 r = 1 - rkffp(Ss(wq) ° ).
If we regard V = Ifq as an F [FQ2]-module, tensor up by the splitting field that
Fq,
Sp(Fq)
and then look at eigenvalues in the symmetric product, we see has a component fixed by
ffQ2
if and only if
The last example is given by the alternating groups. same phenomenon: prime
3.
k = 3.
13
These show the
the only torsion in their Whitehead groups is at the
336
Theorem 14.6
n
EXAMPLES
CHAPTER 14.
Fix
n > 1,
and let
An
be the alternating group on
Then
letters.
r
Z/3
if
n = : 3m'
ml > ... > mr 2 0,
27,
i-1
1(mi) odd
SK1(Z[An]) otherwise.
1
For more
We sketch here the main points in the proof.
Proof
details, see Oliver [3, Theorem 5.6].
SK1(Zp[An]) =
SK1(Z[An]) = C11(Z[An]):
(1)
for all
1
p
by
Example 12.8.
Since
(2)
over
Q
and Q[Sn]
[Sn:An] = 2,
is a product of matrix algebras
(see James & Kerber [1, Theorem 2.1.12]),
matrix algebras over fields of degree at most 5,
then p(Q[An]) = Cll(Z[An])(p) =
If
(3)
subgroups of
a1,...,ak An,
2
Q[An] over
Q.
is a product of
Hence, if
p >
1.
are conjugacy class representatives for cyclic
and if
mi = jail,
k
Z(Q[An])
i=1
then
N(ai (Qcm: )
To see this, note that both sides are products of fields of degree at most 2
over
Q.
Hence, it suffices to show that both sides have the same
number of simple summands after tensoring by any quadratic extension of K,
This follows from the Witt-Berman theorem (Theorem 1.6):
Q.
the
number
of
irreducible
K[An]-modules
equals
the
K
for each number
of
K-conjugacy classes in An.
(4)
By Theorem 4.13,
C2(Q[An])
purely imaginary field summands of
has rank equal to the number of
Q[An];
and by (3) this is equal to
CHAPTER 14.
EXAMPLES
337
the number of conjugacy classes of cyclic subgroups g
that
is not conjugate to
cycles of lengths k1 > ... > ks, i,
and
Z1(ki-1)/2
li
a)
irE
ClI(ZCk[ir]t)(2) = 1
C2(QCk[w]t) =
vE (Q{n)
Na
By (3) again,
On the other hand,
1/2
C2(Q[An1)
is centralized by
a
where
such that
An
under
C11(g[An])(2) = 1.
C3(Q[An]) = (7/3)s,
conjugacy classes of cyclic C3 C a
(k = lal)
These terms thus account for all of
it R.
has
CA. (a)
C2((lk)w) =
1
the decomposition of Theorem 11.8; and so
(5)
is odd for all
ki
and so by Theorem 13.4:
a,
denotes here the set of 2-subgroups).
(55(Na)
since
such that
is a product of disjoint
Dc,=n,
In particular, the centralizer
is odd.
odd order for each such
Each such g
-l.
g
a = (g) C n such
is the number of
s
311a1,
and such that
An easy check then shows that
N(a).
s
Z/3
if
n = 13m', i
m1 > m2 > ... > ms > 0,
odd 1
1
C3(Q[AnI) = otherwise.
1
Assume that
are as above.
C3(D[An]) = 7/3: P C An
Let
P = P1 x ... X Ps,
where
pal
i.
P0,
For example, if
where the
A(3m').
is a 3-Sylow subgroup of
[xi,gj] = 1 then
if
of
Pab and
1 0 j,
Pab
x1,...,xm
and
= (C3)3
mi
Then
Also,
(C2)m.
NST(P)/P =
and
g1,...,gm
n = 12,
n = Ii=13m',
be the "standard" 3-Sylow subgroup.
= (C3)m
In fact, there are bases
such that in
P i
write
of
xigixil = giI
is generated by
N5 (P)/P for all
338
CHAPTER 14.
gl = (1 2 3),
while
g2 = (1 4 7)(2 5 8)(3 6 9),
NSn(P)/P = (C2)3
g3 = (10 11 12);
is generated by
xl = (1 2)(4 5)(7 8),
For any
EXAMPLES
x2 = (1 4)(2 5)(3 6),
el,.... em E Z/3,
let
x3 = (10 11).
denote the irreducible
V(el,...,em) e
Q[Pab]-module with character there is an element of
X(gi) = (c3) i'
NAn(P)/P
negates the corresponding
Z/3
If
ei = 0,
any
then
which negates the character, and hence summand in
The remaining irreducible representations of
m C3(®[Pab]) _ (Z/3)(3 -1)1 Pab
are permuted simply and transitively by NA.(P)/P;
ei = it
for all
i)
P'f1P
is
gi.
It
and hence
HO(N(P)/P; C3(@[Pab])) = Z/3.
If
P' A P
is any other 3-Sylow subgroup in
then
An,
contained in the subgroup generated by some proper subset of the follows that the induced map
C3(@[P'f1P]) ) HO(N(P)/P;
is trivial.
C3(Q[Pab]))
Hence, there is a natural epimorphism
li C (Q[P]) pc(n)3
--"
0A
HO (NA. (P)/P; C3 (R[ib])) = Z/3.
But by Theorem 11.8, this limit is a direct summand of
C3(D[An]) = Z/3.
So with the help of Theorem 9.5 we now get
C11(Z[An]) =
li
C11(D[P]) = Coker[K2(Z3[P]) -i
Coker[H1(P;Z[P])
li
P ,, )
C3(Q[P])]
'' HO(N(P)/P; C3(D[Pab])) = Z/3].
EXAMPLES
CHAPTER 14.
339
The calculation now splits into the following cases:
n = 3.4:
P = C3,
n = 12.13:
C11(Z[A1])(3) = 1
so
Z(g3 0 glg2g3)
Hence,
n > 28:
so
Im(Z) = y(P 0 1) = 0,
In this case,
and
m = Iml > 5.
(where
C11(Z[An]) = 1.
The image of any abelian subgroup of
n = 27.28: Pab.
HO(N(P)/P; C3(a[Pab]))
generates
are the elements defined above);
gi
by Theorem 14.2(1).
P
is cyclic in
C11(Z[An]) = Z/3-
By Alperin et al [3, Proposition
2.5],
Im[K2(2 31 Pab]) _ C3(@[Pab])] = S3(pab),
where since of
S3(Pab) = S3(F3m) m > 5,
N(P)/P.
denotes the symmetric product.
S3(Pab)N(P)/P
S3(F3m)N(P)/P
Furthermore,
= 0 by the above description
There are thus surjections
Z/3 = C3(D[An])
C11(Z[An]) - HO(N(P)/P; C11(Z[Pab]))
Coker[S3(Pab)N(P)/P -' C3(Q[pab])N(P)/P] = Z/3;
and so
C11(Z[An]) = Z/3
in this case.
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INDEX
Artin cokernel 143 Artin-Hasse symbol formulas 100, 210 bimodule-induced homomorphisms 37, 135, 251 Brauer's induction theorem 139 Brauer's splitting theorem 25 Brauer-Witt theorem 31 Burnside ring 252 idemponents in 254-255 C(A) 73 CD(A) 87 C11(7L[G]) 6, 73 standard involution on 151
complex representation ring
9,
134ff
congruence subgroup problem
3,
115ff, 118 cyclic homology 169-171 cyclotomic fields 23, 25, 31 Dennis-Stein symbols 68 Dress' theorem on CC-computable functors 248 Eichler condition 241 elementary groups 139 five term homology exact sequence 187
Frobenius automorphism 154 Green rings and modules 246-247 Hasse's structure theorem for p-adic division algebras 30 Hasse-Schilling-Maass norm theorem 43-44 Higman's theorem 6, 173 Hopf's formula for H2(G) 186-187 hyperelementary groups 250 index of a simple algebra 24, 31 induction theory 245ff integral p-adic logarithms 153ff, 160, 229ff, 284-285 for K2 169-171
for the center of a group ring 168
relative version 166 involutions on rings 148 Jacobson radical 32 for p-group rings 35 K-conjugate elements 26 K-elementary groups 250
K1(R,I) 33 K2(R) 33, 64 70ff of finite rings 35, 68-69 of local fields 94 division algebras p-adic of 100-101, 110-111 68-70, symbol generators for 212 localization sequences 36, 71-72, 76, 87, 88 boundary map in 81, 84 for maximal orders 36, 71-72 Mackey functors 246 p-local 254ff splittings for 256 Matsumoto's theorem 91 24, 36, 61, 71, maximal orders 124 division subfields of maximal algebras 24 Moore's reciprocity law 116-117 Moore's theorem on K2 of local fields 94 Morita equivalence 39 norm residue symbols 92ff, 128 norms and traces for extensions of 0P 29 orders 21, 24 p-adic completion 27 p-adic division algebras continuous Kc2
Hasse's structure theorem 30 maximal order in 30 K1 of 43 K2 of 100-101, 110-111 p-adic extensions totally ramified 28, 197 unramified 28, 29, 154, 197 p-adic logarithms 50ff, 57 p-elementary groups 139 p-elementary computable functors 269, 279, 303-304 p-hyperelementary groups 250 250, 251, p-K-elementary groups 260 Quillen's localization sequences 36, 71-72, 88 reciprocity map 29 reduced norm 5, 40ff kernel of 43, 48
Riehm's theorem 58-59 s-cobordism theorem 2 semilocal rings 34 semisimple algebras 21 simple homotopy equivalence 2 six term homology exact sequence
standard rings
on C11(7L[G])
for
group
151
196 on SK1(1,[G]) 148ff, 182 on Wh(G) Steinberg symbols 64ff generators for K2(R) 68-70, 212 in fields 91 relations among 65 symbols on fields 91 tame symbol 99-100, 132 transfer homomorphisms 38, 98-99, 125, 140, 165, 197, 198, 311 on Steinberg symbols 65 259ff, 266, twisted group rings 269, 277-278, 297, 300-301, quaternion, dihedral, of and semidihedral groups 301 units in Whitehead groups 241-242
187
SK1(Z[G])
construction and detection
involution
of
elements 12-13, 18, 127ff, 203 exponent of (for abelian C) 15 for abelian p-groups 215, 217,
for abelian groups 322
for alternating groups 336 for cyclic groups 142 for dihedral, quaternion, and semidihedral, and related groups 144, 228, 333 for symmetric groups 328 for wreath products 203, 328 p-elementary computable 303-304 SK1(3,I) 73 Skolem-Noether theorem 22 splitting fields 24
Vaserstein's relation 34-35 Wall's theorems 86, 180 Wedderburn decomposition 22 for cyclic groups 23, 31 for p-groups 205-206 for p-adic group rings 31 Whitehead groups, rank of 49 Whitehead torsion 2 Witt-Berman theorem on numbers of representations irreducible 26
349
