R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
P AUL H ILL
C HARLES M EGIBBEN
Butler groups cannot be classified by certain invariants
Rendiconti del Seminario Matematico della Università di Padova, tome 90 (1993), p. 67-79
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by Certain Invariants.
PAUL
HILL (1) -
CHARLES MEGIBBEN(2) (*)
SUMMAR,Y - In this paper, we
give
a necessary and sufficient condition for twoBo-
groups to be
quasi-isomorphic.
This new characterization ofquasi-isomor- phism
can beinterpreted
as anequivalence
theorem forquasi-isomorphism,
which extends the
isomorphism equivalence theory
that has heretoforeproved
fruitful in avariety
of differentsettings.
In the present context, this result enables us to prove thatquasi-isomorphism
invariants which have re-cently
been used toclassify
certainstrongly indecomposable
Butler groups do not suffices ingeneral.
Inparticular,
thequasi-isomorphism
invariants thatclassify strongly indecomposable
groups that are imbeddable as corank 1 puresubgroups
of finite rankcompletely decomposable
groups are not ade- quate in the corank 2 case.A Butter groups is a torsion-free abelian group that can be imbedded
as a pure
subgroup
of a finite rankcompletely decomposable
group. An alternative characterization is that Butler groups areprecisely
thosethat are torsion-free
homomorphic images
of finite rankcompletely
de-composable
groups[BUT].
A Butler group G is said to be aBo-group provided,
for eachtype
T,G(T*)
=(G(a): T)
is a puresubgroup.
Inthis paper, we
give
a characterization of when twoBo-groups
arequasi- isomorphic.
Recall that fmite rank torsion-free groups G and G’ arequasi-isomorphic
if andonly
if there existsmonomorphisms ~:
G ~ G’and
~’ :
G’ - G. Moreinsightful
is the observation that G and G’ arequasi-isomorphic
if andonly
if G’ isisomorphic
to asubgroup H
of Gwith
G/H
finite. Our main result will rest on thefollowing theorem,
which can be
thought
of as anequivalence
theorem forquasi-isomor- phism.
THEOREM 1. Let G and G’ be
Bo-groups,
and suppose thatand
are balanced exact sequences where A and A’ are
isomorphic
finiterank
completely decomposable
groups. Then G and G’ arequasi-iso- morphic
if andonly
if thefollowing
condition is satisfied:(3)
There existmonomorphisms
and~’: A’
-~ A such thatB’ 1 t/J(B)
andB/~’ (B’ )
are finite groups.Perhaps surprisingly,
thesufficiency
of condition(3)
isessentially
trivial and does not
depend
on thehypothesis
that the resolutions arebalanced. Indeed
given
amonomorphism §
as in(3),
a routinediagram
chase
yields
ahomomorphism ~:
G -~ G’ such that = Further-more, because
B’/~(B)
isSuite,
each element ofker ~ necessarily
has fi-nite order.
Since, however,
G istorsion-free, if;
is monic.By symmetry,
there is an inducedmonomorphism ~’ :
G’ -G,
and thus(3) implies
thatG and G’ are
quasi-isomorphic. Obviously
then we do notput
Theorem1 forward
primarily
as a method forestablishing
thequasi-isomor- phism
of twogiven
groups. Instead itsgreatest potential
resides inproving
two groups notquasi-isomorphic
that are soclosely
relatedthat other methods fail to
distinguish
them up toquasi-isomorphism.
The demonstration that G and G’
being quasi-isomorphic implies
condi-tion
(3)
isfairly
involved and ourproof
reliesheavily
on anequivalence
theorem established
in [HM3].
For the convenience of the
reader,
beforeproceeding further,
wenow establish the notational conventions that will be in effect
through-
out this paper and recollect certain relevant definitions. All groups considered will be
additively
written abelian groupsand,
with few ex-ceptions,
torsion-free of finite rank.By
aheight
we understand a se-quence s = where P denotes the set of rational
primes
and eachsp
is either anonengative integer
or thesymbol oo.
Twoheights
s and tare
equivaclent, s -~- t, provided (i) sp
=tp
for almostall p
and(ii) ~=00.
if and
only
iftp
= oo. Anequivalence
class ofheights
will be called atype.
As functions to an orderedset, heights
are orderedpointwise
andthis
ordering
ofheights
induces an order on the set of z ifand
only
if there existheights
s e J and t e c such that s ~ t. Since w UU 1 - 1
islinearly ordered,
the ordered sets ofheights
andtypes
are dis- tributive lattices. We shallemploy
thesymbols
A and V for the latticeoperations
in all three of these ordered sets. If x is an element of the torsion-free groupG,
then we associate with it theheight x ~
=where 1 x 1 p
is theordinary p-height
of xcomputed
in G. Welet
typeG (x)
denote thetype
determinedby lx 1.
Ifrank (G)
=1,
thenall nonzero elements of G have the same
type
which we indicate astype (G).
If H is a puresubgroup
of the torsion-free groupG,
then we say that H is balanced in Gprovided
each coset x + H contains an elementy with
y ~ % ~ x
+ h1
for all h E H. In terms oftypes
rather thanheights,
there is anequivalent
formulation of balanceness that is usual-ly
easier toapply; namely,
each coset x + H contains an element y such thattypeg ( y)
=(x
+H ) [AR2,
pp.4-5].
In Theorem 4below,
we
give
another characterization of balancedsubgroups
that involves bothtypes
andheights
but avoidscomputations
in thequotient
groupG/H.
Anepimorphism ~:
G - K is said to be balanced ifker ~
is a bal-anced
subgroup
of G. If H is anarbitrary subgroup
of the torsion-free groupG,
thenH*
will denote the minimal puresubgroup containing
H.Let G be a fixed torsion-free group. With each
height s
we associatethe
fully
invariantsubgroups
andG( s * ) _ - ~G(t): t ; s and t ~- s~.
We shall writeG(s*, p)
for thesubgroup G(s*)
+pG(s).
A nonzero element x E G is said to beprimitive provided
x fi.
G(s*, p)
whenever1 x 1
- sand1 x
If a is atype,
then the ful-ly
invariantsubgroup
E G :typeg (x) > c }
=U G(s)
is a pureSEQ
subgroup; but,
ingeneral, = G(r):r> 7)
need not be pure in G.If G has finite
rank,
then it is a Butler group if andonly
if it satisfiesthe
following
three conditions:(i)
G has finitetypeset {typeG (x):
0 =~ x E
G }; (ii) G(a- *), /G( ~ * )
is finite for eachtype
cr;(’lü)
for eachtype
~,there is a
completely decomposable subgroup G~
such thatG( a)
=GO’ E9
E9
G(~ * )* .
See[AV1]
for aproof
of this fundamental characterization of Butler groups,Clearly
each nonzero element ofG~
hastype a and,
infact,
each such element isprimitive
in G[HM1]. Accordingly,
we referto
G( ~ * )* }
as the set ofprimitive types
of the Butlergroup G. If C
= 1 CQ
where =C~
E9G( o~ * )*
for each cr, then C is eepgcalled a
regulating subgroup
of the Butler group G. If G is atmost com-pletely decomposable
in the sense that it isquasi-isomorphic
to a com-pletely decomposable
group, then eachregulating subgroup
is com-pletely decomposable [AR2, Corollary 3]
andquasi-isomorphic
to G.PROPOSITION 2. Let ~: A -~ G be a balanced
epimorphism
where Ais a finite rank almost
completely decomposable
group and G is aBo-
group. Then A contains a
regulating subgroup
C such that 7r C - G isa balanced
epimorphism
where 7t 1 =7t 1 C.
PROOF. Since G is a
Bo-group,
G= E G~
where =Ga- EB
a-ePG
for eaeh cr e
PG .
Webegin by considering
anarbitrary regulating
sub-group C = of A. Now assume that we have constructed a sub- group B = of A that satisfies the
following
conditions for each.
(i) B~
is a direct summand ofC~ . (il) 7t(Ba-)
is a direct summand ofG~ . (iii)
If it should furthermore
happen
that=
GQ
for each e EPG ,
then Theorem 2.2
of [AR2] implies
that n 1= 7t 1 C
is a balancedepimor- phism. Suppose, however,
that there is some r EPG
such that~c (Bz) ~
~
G~.
In this case there is aprimitive element y
of G such that (Bis a direct summand of
G .
Let t= 1 yi 1
E r. Since 7t: A - G is a bal-anced
epimorphism,
there is an x e A such1
= t and7t(x)
= y.By (i),
we have a directdecomposition C-
=Bz
Q3DT
and hence we canwrite x = b + d + z where b E
BT ,
d EDr
and z EA( z * )* . Replacing y by
a nonzero
multiple
ofitself,
we may assume that z EA(z*).
Becauseixl [ = Ibl Idl
/~Izl, each
of the elementsb,
d and z liesin A(t)
andconsequently
z isn A( t ) = A( t * ).
We claim that1 d 1
=1 x 1.
In-deed if this were not the case, then we would have x - b E
A(t * , p)
forsome
prime
pThis,
in turn wouldimply
that y -n ( b )
EE
G(t*, p),
which isreadily
seen to contradict the factthat 7t(BT)
is a direct summand of
G . Having
now confirmed thatd ~ 1 = 1 xl,
wenote that Lemma 2.5
in [HMI] implies
thatA(z)
=B,
Q3(x)*
®DT
Q3Q3
A( z * )*
whereDz
is anycomplement
inDT
of the rank 1 direct summand(d)*.
If we takeBz Cz = B§ Q3 D§ and,
r,B’ = BQ
and
C’c
=Cc,
thenC ’ = É C’c
is a newregulating subgroup
of A. Fur-cEPA
thermore B’ is a direct summand of C’ with conditions
(i), (ü)
and(iü)
still satisfied. A Suite number ofrepetitions
of theforegoing argument
will
yield
anappropriate
C and B with condition(iv)
alsosatisfied;
andthus the
proof
iscomplete.
COROLLARY. If G is a
Bo-group,
then the natural mapis a balanced
projective
cover forG;
thatis, given
a balancedepimor- phism
with Acompletely decomposable,
there exists a bal-anced
epimorphism
0: A - such thatpo
= 7t.PROOF.
By
theproof
ofProposition 2,
A contains a direct summandB = such
that,
for each ~, n mapsBQ isomorphically
ontoG~ .
Thus we may write A = B Q3
xi>* where xi =
0 for each i. Sincen
B
is a balancedepimorphism by [AR2,
Theorem2.2],
wehave,
foreach i e
I, a bi E Bi
such that7r (bi)
=7r (xi)
and1
=In:(xi) 1.
It thenfollows from
[HM1,
Lemma2.5]
that A = B Q3®iEr Di
whereDi
=(xi - bi>*
for all i.Clearly
D =®iEI Di
is contained in ker n andtherefore, identifying
B withG,,
the desired 0 is thatprojection
of A onto B with ker 0 = D.
We need one further technical result before we can finish the
proof
of Theorem 1.
LEMMA 3. Let
(1)
and(2)
be as in the statement of Theorem 1 andassume that the
Bo-groups
G and G’ arequasi-isomorphic.
Then for alltypes J,
PROOF. Let
PG
and theG~.’s
be as in theproof
ofProposition
2 andtake p:
G~ -~
G to be the canonical map inducedby
theidentity
maps of the
G~’s. Then, by
thepreceding corollary,
there is a balancedepimorphism
0: A -+ED,,.PG G (j
suchthat p0
= n. Sincecompletely
de-composable
groups are balancedprojectives [FU,
Theorem86.2],
it fol-lows that there is a direct
decomposition
A = C E9 D where C ==
GQ
and D ç ker = B. Therefore B =Bo
E9 D whereBo
= B f1n C n C.
But, by
Theorem 4.2 in[HM2], Bo
f1C( a) = Bo
nfor all
types
c, andconsequently BnA(,7)IB
== for all
types
a.Similarly,
we have a directdecomposition
where C’ ==
G;
and B’nA(,7
== D ’ ( ~)/D ’ ( ~ * )
for alltypes
a.Since, however,
G and G’ arequasi-iso- morphic,
C = C’ . Therefore A = A’implies
D = D’ .Clearly (3’)
fol-lows from the
foregoing
observations.We are now in
position
tocomplete
theproof
of Theorem 1.Assuming
that G and G’ arequasi-isomorphic,
fix amonomorphism
~’:
G’ - G and letG1
=~’ (G’ )
whereGIGI
is finite.By symmetry,
it suffices to construct a
monomorphism ~:A/ 2013>A
such thatB/§’(B’)
is finite. Since B is balanced in and the latter is almost
completely decomposable
becauseA/A
*= G/G1,
anapplication
of
Proposition
2yields
a balanced exact sequencewhere A.
By Proposition
1.7 in[HM2],
B’ andBI
areweakly
*-pure
subgroups (see
Definition 1.3 in[HM2])
of A’ andA1, respect- ively.
As B’ andBi
are balancedsubgroups,
theisomorphism
G ’ =G1
can be construed as an
isomorphism
thatrespects heights
in the sense thatfor all
heigths
s.Finally
anapplication
of Lemma 3 shows that all thehypotheses
of Theorem 1.5 in[HM2] (see [HM3]
for aproof)
are satis-fied and hence that theorem
implies
the existence of anisomorphism
~’ : A’ -~ A1
with~’(B’)=J5i. Noting
thatB/B1 = (ker n)/(A1
nn ker 7r)
=(ker 7r
+ is finite since A *lAI is,
we can viewy’
as thedesired
monomorphism
from A’ to A withB/~’ (B’ )
finite,.Given a torsion-free group
G,
there is a standard method for con-structing
a balancedepimorphism
7r: A - G with Acompletely
decom-posable.
When G is a Butler group, this can be done in aquite explicit
manner so as to insure that A also has finite rank
(see
Theorem 1.2 in[AVI]).
On the otherhand,
theproblem
ofdetermining
when a par- ticular puresubgroup
B of acompletely decomposable
group A is bal- anced is less often dealt with in the literature. Thedifficulty
in the lat- ter situation is that we are notgiven, a piori,
anyspecific
information about the structure of thequotient
group G =A/B.
To be sure,heights
in
A/B
arecomputable
via the formulaa
+B 1 =
but it is the
unwieldy
nature of this formula which is the heart of thedifficulty
indetermining typeA/B (a
+B).
There is nevertheless a veryelementary
criterion for balanceness that can often beapplied quite
ef-ficiently
when one isgiven adequate
information about thegenerators
of B. We say that the pure
subgroup
B isprebalanced
in the torsion- free group Aprovided
for each a E A there exists a Suite collection of elementsbl, b2, ... bn
in B such thatsup ( [ a
+ k =1, 2,
... ,n ~ _
= sup{|a+b|:bEB}.
THEOREM 4. A pure
subgroup
B of the torsion-free group A is abalanced
subgroup
of A if andonly
if thefollowing
two conditions aresatisfied.
(i)
B isprebalanced
in A.(ii)
For eacha E A B B,
the coset a + B contains an element of maximumtype.
PROOF.
Clearly
if B is balanced inA,
then conditions(i)
and(ii)
are satisfied since there is asingle bo E B
witha + bo 1
=sup { a
+b ~ :
b Ee
B ~ _ ~ a
+B
and hencetypea (a
+bo )
=typeA~B ( a
+B ) ~ typeA (a
+b )
for all b E B.
Conversely,
assume that conditions(i)
and(ü)
are satis-fied,
and consider anarbitrary a E A B B.
Selectbl, b2 ,
... ,bn
in B suchbk 1:
k =l, 2,
... ,n ~
and observe that with- out loss ofgenerality
we may assume that r =typeA ( a
+b)
for all b E B.Consequently,
each of thesets,
for the variousk’s,
is finite and = 00
whenever + bk ~ P - ~ . Obviously
it suffices to show thattypeA/B (a
+B)
= r. Sincea
+lai,
, this amounts toverifying
thefollowing
twofacts;
(iii)
Ifp is a prime
such thata
+B 1 p
= 00, thenfinite.
’ " "
Since 1 a
+= 1 a
+ +b2 P
V ...(iii) is
a con-sequence of our observation that
a
+ = 00implies 1 a 1 P
= 00 and(iv)
holds because P =Pl
UP2
U ... UPn .
In
applying
Theorem4,
it ishelpful
to have a more concrete formu-lation of condition
(i).
This issupplied by
thefollowing essentially
triv-ial observation.
PROPOSITION 5. Let B be a pure
subgroup
of the torsion-free group A and suppose that to each aE ABB
therecorresponds
a fmitesubset
Q
of P that satisfies thefollowing
two conditions:(a)
If thena +
p for all b E B.("B) If p e Q,
thensup ( [ a
+b ~ P :
+b’ 1 p
for some par-ticular b’ e B.
Then B is a
prebalanced subgroup
of A.PROOF. Let
..., pn} = Q
and choose fork = 1, 2, ..., n
anelement bk
E B such thata
+bk 1 Pk =
supb 1 Pk :
b EB }.
Then tak-ing bo = 0,
we see that..., n} = sup { ~ a + + b 1 beB}.
Using
Theorem1,
we can show that there existstrongly
indecom-posable Bo-groups
G and G’ which are notquasi-isomorphic
inspite
ofhaving
in common all thequasi-isomorphism
invariants that havefig-
ured in recent classifications of
special
classes ofstrongly indecompos-
able Butler groups. The
tecnique
used to construct balancedsubgroups
in the
following
existence theorem isquite general
and lends itself tomany variations.
(Previously unexplained
notation andterminology appearing
in the statement of our finaltheorem,
as well as its relevance theliterature,
will be discussed in theensuing proof.)
THEOREM 6. There exist two
strongly indecomposable
Butlergroups G and G’ that are not
quasi-isomorphic,
but whichsatisfy
thefollowing
conditions:(1)
rank(G)
= rank(G’).
(2)
Thetypesets
of G and G’ areequal.
(3)
Thecotypesets
of G and G’ areequal.
(4)
G and G’ haveisomorphic endomorphism rings.
(5)
G and G’ have the same Richmantype.
(6)
rank G(M)
= rankG’ (M)
for all subsets M of thetypesets
of Gand G’.
(7)
rank G[M]
= rank G’[M]
for all subsets M of thecotypesets
of G and G’.
(8) o-)
=rG , ( ~, ~)
for alltypes
r and ~.Select four rank 1 groups
Ai, A2, A3, A4
ofincomparable idempotent type r1,
1’2, 73e ’74,respectively,
such that zi/B 1’j
= 1’0 =type (Z)
when-ever i =
j.
Then fix elements ai EAi
for i =1, 2, 3, 4
such thateach |ai| 1
involves
only
O’s and 00 ’s. We also assume, for reasons that willshortly
be evident that
a3 ~ p
= 00 for p =2,
3 and 5. In order to construct ourfirst group
G,
we let B denote the puresubgroup
of A0153
A4 generated by b1
= a, + a2 +a3 and b2 = a2
+3a3
+ a4 . We shallrepresent
anarbitrary
element a E A in the form a = si ai + + + S3 a3 + S4 a4 where thesi’s
areappropriate
rational numbers. Indeed wemay
write,
for each i =1, 2, 3, 4, si
= wheregcd (nü mi)
= 1 andmi is a
positive integer
withprime
factors fromlai 1 p = }. By considering
theequation
na =t1 b1
+t2 b2 , a simple computa-
tion establishes the
following:
(0)
a E B if andonly
ifs2 - sl - s4 = 0
ands3 - s1 - 3s4 = 0.
Furthermore,
since the setsPl, P2, P3, P4
arepairwise disjoint,
ele-mentary
number theoreticalarguments
show that each of sl, S2, S3 and S4 is aninteger provided
the twoequations
in condition(0)
are satisfied.Thus when a e
B,
it iseasily
seen that a = sibl
+ S4b2 ,
and therefore B 0153(b2~-
The next four observations are almost
equally routine,
butrequire
our
assumption
that 2 and 3 are inP3 . Assuming
that a fi.B,
then thefollowing
hold:For
example,
if a +ti bl
+t2 b2
is an element ofAi,
then theequation
S3 - S2 -
2 s4
= 0 follows from the fact that S2 +t1
+t2
= S3 +tl
+3 t2
== S4 +
t2
= 0.Conversely
suppose that S3 - S2 -2s4
= 0. Then anothersimple
number theoreticalargument exploiting
the fact that2 E P4
al-lows us to conclude that S2, s3 and S4 are
integers. Taking t1
= s4 - S2 andt2
= - s4 , we see that a +t1 bl
+t2 b2
is inAi .
Since s3 +28i - 3s2
=-(s3-s1-3s4)-3(s2-sl-s4)
ands3-s2-2s4=(s3-sl-3s4)- - ( s2 -
sl -s4 ),
itquickly
follows from(0)-(4)
that if a EABB,
then thereis at most a
single
i =1, 2, 3, 4
such that(a
+B)
f1Ai ~
0.Consequently,
condition
(ü)
of Theorem 4 is satisfied.To
complete
theproof
that B is balanced inA,
we needonly verify
that B is
prebalanced.
We shallaccomplish
thisby showing
that condi- tions(a)
and([3)
ofProposition
5 are satisfied. Towards thisend,
wenote that we have a
partition ~Po, Pl, P2, P3, P4 ~
of P whereand,
whenever s is a rational number and p is aprime,
we = kto indicate that s =
pk(mlm)
where neither n nor m is divisibleby
p.We shall
require
thefollowing
crucial facts:We shall
verify (0’)
and(1’).
Write thenand
similarly
On the other
hand, if p
EP1,
thenHence,
in this case,1 a
+b ~ p ~ k implies
We can now show each a E
A B B
satisfies the conditions ofProposi-
tion 5. First consider the situation where
( a + B ) n Ai = ~
fori =
1, 2, 3, 4.
In thisinstance, (0)-(4) imply
that all four of the rationals~1 = 83 - ~2 " 2~4, ~2 = ~3 " ~1 " 3~4, ~3 = ~2 - ~l - ~4,
andr4=s3+
+
2 sl - 3 s2
are nonzero, and furthermore1 a 1 p
is finite for eachprime
psince 1:
1:j = 1:0 wheneveri ~ j . Then, by (0’)-(4’),
theset Q
of thoseprimes
p for which there is some b E B witha
+b ~ ~
>a ~ p
must befinite,
since any such p must have theproperty
that forsome i =
1, 2,
3 or 4.(Note
that s ~ 0implies 1 si p
= 0 for all butfinitely
manyprimes p.)
But evenif peQ,
is attained for some b’ E B because this supremum cannot exceed the maximum of
Ir1lp,
Theargument
where(a
+B)
nAi ~ ~
for some i isonly slightly
different. Fordefiniteness,
consider the case where(a
+B)
f1 0. Without loss ofgenerality
we may assume that a = SI
0,
so that S2 = S3 = S4 = 0. In thiscase, we and which is
certainly
finite since 0. When p E
Q,
and hencethe supremum is attained for some b’ E B. On the other
hand,
ifthen either or else for all Thus
taking
G =A/B,
we see that the canonical short exact sequenceis balanced exact
and, by Proposition
1.7 in[HM2] (or
from moreelementary considerations),
G is aBo-group.
To construct the
companion
group G’= A’ /B’,
we take A’ = A ==
Ai Q3 A2 EB A3
Q3A4
and let B’ be the puresubgroup generated by b1
== b, = a, + a,2 + a3
andb2=a2+5a3+a4.
Theproof
thatB’ =(&~}
(Bis balanced in A’ is
substantially
the same as thatgiven
abovefor B in
A, except
in this case a’ - s, al + S2 a2 + 83 a3 + s4 a4 lies in B’ if andonly
if S2 - Sl - S4 = 0 and s3 - s, -5s4
= 0.Obviously
B =B’,
andsince both 3 and 5 are in
P3, b2
andb2
even have the same associatedheight
vector in the senseof [HM2]. Nevertheless,
Theorem 1implies
that the
Bo-groups
G and G’ fail to bequasi-isomorphic.
Infact,
therecannot exist any
monomorphism ~: A ----> A’ mapping
B into B’. Indeed since theAi’s
haveincomparable types,
if~:A2013~A~
is a monomor-phism,
then there must exist nonzero rational numberskl, 1~2, k3
and1~4
such = for i =
1, 2, 3, 4.
But then the above conditionsfor r./J(b1)
=k1 al
+k2 a2
+1~3
a3 to lie in B’yield k1
=k2
=k3 ;
while therequirement
=k2 a2
+3k3 a3
+k4 a4
be an element of B’ im-plies k4
=1~2
and51~4
=31~3 .
Since theki’s
are nonzero, theseequations
contradict the fact the 5 ~ 3.
Inspite
of the fact that G and G’ are notquasi-isomorphic,
the twogroups are
quite
alike in structure.Clearly rank (G )
=rank (G’ )
=2,
and the groups have the sametypeset {r0, r1,
72, ’r3,’t’ 4}. By
Theorem3.2 in
[AR 1 ],
it follows that both G and G’ arestrongly indecomposable
with
endomorphisms rings
andquasi-endomorphism rings isomorphic, respectively,
to Z andQ. Obviously _ ~ ai
+B)*
for i =1, 2, 3, 4
and
G(zo)
= G. Since thecorresponding
observations hold forG’, rank G(c)
= rank G ’(,r)
for alltypes
r. Recall that thetype p.
is said tobe a
cotype
of Gprovided
there exists a puresubgroup
H such thatand
By
Remark(4)
on page 108of [AVl ]
and the fact that fori = 1, 2, 3, 4, type (G/G(cj))
x fi. and hence each of the fourtypes
are
cotypes
of G.Similarly,
theseli j’s
arecotypes
ofG’,
and since the set ofcotypes
of a Butler group is closed under supremums, g o = z 1 V V T2 V T 4 is also acotype
of both G and G’ . For anarbitrary
torsion-free group G and
type g,
thefully
invariant puresubgroup G[g]
is de-fined as the intersection of the kernels of all
homomorphisms
of G into arank 1 group of
type
and when G is a Butler groups, it is known that~G(~): ~ [AV1, Proposition 1.9]. Consequently,
for ourparticular
rank 2 groupG,
= 0 and =G( T i)
for i =1, 2, 3, 4;
and
similarly
for G’.Notice,
because rank(G)
=2,
thatand likewise for each ri
V Tj
with i ~j.
Thus none of the ri VT’s
can becotypes
ofG;
nor for that matter can any of the Since thecotypes
ofany Butler group lie in the lattice
generated by
itstypeset [AV1, Corollary 1.5],
it follows ~U 1, P. 2, P. 3,P. 4}
is the set ofcotypes
of G and also of G’.Consequently,
we have rankG[ ~u 1
= rank G’[,u]
for alltypes 03BC (see
Remark(1)
on page 107of [A VI]).
The groups G and G’also have the same Richman
type (see
page 12of [AR1]).
Indeed F ==
( al
+ +B)
is a freesubgroup
ofG and,
from the definition ofB,
it follows that~c -1 (F ) _ (ai,
a2, a3,a4).
ThereforeG/F
isisomorphic
4
to the divisible torsion group D = 1=1
Similarly, G’/F’
= Dwhere F’ -
+ B’>
0153a2 + B’>,
and so G and G’ are bothquotient-
divisible
[BP11.
Recently, strongly indecomposable
Butler groups G that can be re-alized
by epimorphisms
7r: A -+G,
where A is a finite rankcompletely decomposable
group andrank (kern) = 1,
have beenclassified [AV2]
up to
quasi-isomorphism by
the invariantsrank G(M)
whereG(M) =
= (G(,r):
r EM)
and M is anarbitrary
subset of thetypeset
of G[AV4].
Dually, strongly indecomposable
Butler groups G that can be imbed- ded as a corank 1 puresubgroup
of a finite rankcompletely decompos-
able group have been classified up to
quasi-isomorphism by
the invari- ants rankG[M]
whereG[M ]
= and M is anarbitrary
set of
cotypes
of G. Another class of corank 1 Butler groups, the so-called
CT-groups of [AV3],
have been classified up toquasi-isomor- phism by
the invariantsrG(T, a)
= +G[r])/G[(y]).
From theanalysis given
in thepreceding paragraph
for ourparticular
rank 2groups G and
G’,
it is immediate thatrG (T, cr)
= rG,(T, ~)
for alltypes
rand ~, and that
rankG(M)
=rankG’(M)
andrank G[M]
=rank G’ [ M]
for every set of
types
M.Theorem 6
suggests,
as do earlier observationsby
Arnold and Vin-sonhaler,
that any further progress in the classification ofstrongly
in-decomposable
Butler groups willrequire
the introduction of new in- variants. Indeedby
theexample
constructed above and theduality theory of [AV4],
invariants defined in terms of theG(cl’s
andG[gl’s
will not suffice to
classify
thosestrongly indecomposable
groups G thatare imbeddable as corank 2 pure
subgroups
of finite rankcompletely decomposable
groups. There are, of course, earlier clasifications up toquasi-isomorphism
ofquotient-divisible
groups[BP1]
and ofgeneral
rank 2 groups
[BP2].
But these older invariants haveproved
rather in-tractable and have
consequently
fallen out of favor.REFERENCES
[AR1] D. ARNOLD, Finite Rank Torsion Free Abelian
Groups
andRings,
Lec-ture Notes in Mathematics, 931,
Springer-Verlag,
New York (1982).[AR2] D. ARNOLD, Pure
Subgroups of
Finite RankCompletely Decomposable Groups,
Lecture Notes in Mathematics, 874,Springer-Verlag,
NewYork (1981), pp. 1-31.
[AV1] D. ARNOLD - C. VINSONHALER, Pure
Subgroups of
Finite Rank Com-pletely Decomposable Groups
II, Lecture Notes in Mathematics, 1006,Springer-Verlag,
New York (1983), pp. 97-143.[AV2] D. ARNOLD - C. VINSONHALER, Invariants
for
a classof torsion-free
abelian groups, Proc. Amer. Math. Soc., 105 (1989), pp. 293-300.
[AV3] D. ARNOLD - C. VINSONHALER,
Quasi-isomorphism
invariantsfor
aclass
of torsion-free
abelian groups, Houston J. Math., 15 (1989), pp.327-340.
[AV4] D. ARNOLD - C. VINSONHALER,
Duality
andinvariants for
Butlergroups, Pacific J. Math., 148 (1991), pp. 1-10.
[BP1] R. A. BEAUMONT - R. PIERCE, Torsion
free rings,
Illinois J. Math., 5 (1961), pp. 61-98.[BP2] R. A. BEAUMONT - R. PIERCE, Torsion
free
groupsof
rank two, Mem.Amer. Math. Soc., 38 (1961).
[BUT] M. C. R. BUTLER, A class
of torsion-free
abelian groupsof finite
rank,Proc. Lond. Math. Soc., 15 (1965), pp. 680-698.
[FU] L. FUCHS,
Infinite
AbelianGroups,
Vol. II, Academic Press, New York (1973).[HM1] P. HILL - C. MEGIBBEN, Torsion
free
groups, Trans. Amer. Math. Soc.,295 (1986), pp. 735-751.
[HM2] P. HILL - C. MEGIBBEN, The
classification of
certain Butler groups, J.Algebra,
to appear.[HM3] P. HILL - C. MEGIBBEN,
Equivalence
theoremsfor torsion-free
groups, to appear.Manoscritto pervenuto in redazione il 16
gennaio
1991 e, in forma revi-sionata il 15