BUTLER GROUPS OF INFINITE RANK II...Volume 320, Number 2, August 1990 BUTLER GROUPS OF INFINITE RANK...

transactions of theamerican mathematical societyVolume 320, Number 2, August 1990

BUTLER GROUPS OF INFINITE RANK II

MANFRED DUGAS, PAUL HILL, AND K. M. RANGASWAMY

Abstract. A torsion-free abelian group G is called a Butler group if Bext(G, 7")

= 0 for any torsion group T . We show that every Butler group G of cardi-

nality N, is a ¿?2"8rouP; 'e-' C7 is a union of a smooth ascending chain of

pure subgroups Ga where Ga+X = Ga + Ba , Ba a Butler group of finite rank.

Assuming the validity of the continuum hypothesis (CH), we show that every

Butler group of cardinality not exceeding N^ isa ß2-group. Moreover, we are

able to prove that the derived functor Bext (A , T) = 0 for any torsion group

T and any torsion-free A with \A\ < Kw . This implies that under CH all

balanced subgroups of completely decomposable groups of cardinality < Hm

are 52"ßrouPs-

1. Introduction

All groups in this paper are (torsion-free) abelian groups. Undefined nota-

tions are standard as in [Fu]. A torsion-free abelian group is called a Butler

group if Bext(B, T) = 0 for all torsion groups T. Here Bext is the subfunctor

of Ext of all balanced-exact extensions. It is known [BS] that this definition

coincides with the familiar one if B has finite rank, i.e., a pure subgroup of a

completely decomposable group. The main result in this paper is that under the

continuum hypothesis (CH) each Butler group B (of rank < Nw ) is a B2-group

[A]; i.e., B is the union of a smooth ascending chain of pure subgroups Ba ,

a < k an ordinal and B ,. = B +L for all a < X where L is a Butler groupa+l a a a or

of finite rank. This result partially answers questions raised in [BSS], [A], and

elsewhere. We will, in fact, prove the following:

Theorem [CH], Let G be a torsion-firee group of cardinality < Kw .

(a) The following are equivalent:

(1) Bext(t7, T) = 0 for all torsion groups T.

(2) Bext(C7, T) = 0 for all 1,-cyclic torsion groups T.

(3) G is a B2-group.

(ß) The derived function Bext (G, T) = 0 for any torsion group T. (We

have to use CH only if \G\ > N2.)

Received by the editors October 13, 1988.

1980 Mathematics Subject Classification (1985 Revision). Primary 20K20.The research of the first author was partially supported by NSF Grant DMS 8701074.

The research of the second author was partially supported by NSF Grant DMS 8521770.

©1990 American Mathematical Society0002-9947/90 $1.00+ $.25 per page

643License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

644 MANFRED DUGAS, PAUL HILL, AND K. M. RANGASWAMY

Part (a) of the preceding theorem extends results [A], [AH], [BS], [BSS], [D],

and [DR1]. Part (ß) was shown for \G\ < K, in [AH]. The statement that (3)

implies (1) may be found in [BS]. Under V = L it was shown in [DR2], with

some help from [D], that (2) implies (3) in case |G| < N¡ .

As an immediate consequence of the theorem we obtain the following:

Corollary [CH], If G is a balanced subgroup of a completely decomposable group

and \G\<#W, then G is a B2-group.

The proof of the theorem will show that if the corollary holds in general, we

may drop the cardinality restriction in the theorem.

Albrecht and Hill [AH] showed (without CH) that the preceding corollary is

true if |G| < N, .

In order to prove the above theorem we have to prove numerous auxiliary

results, some of independent interest. These auxiliary results include the follow-

ing: we show that any balanced subgroup of a completely decomposable group

admits a C7(2 °)-family (cf. [FH]) of balanced subgroups. The proof is a mod-

ification of one in [AH]. Using methods different from the proof of this result,

we show that for any torsion-free group G of cardinality Kn , n < co, there is a

filtration G = Uq<n Ga into pure subgroups Ga suchthat Gn+X/Gn has rank

1 and each Ga is separable in G in the sense of Hill; i.e., for all g e G there

is a countable subset S of Ga such that for any h e Ga there is some s = s(h)

in S for which \g + h\ < \g + s\.

The main obstacle we have to overcome is the fact the pure subgroups of

Butler groups (of infinite rank) are in general not again Butler groups. In order

to show that balanced subgroups of Butler groups are Butler again, one needs to

know that Bext (G, T) = 0 for torsion-free groups G. We were able to show

this for all torsion-free groups G of cardinality not exceeding Hw .

If T is a torsion group and G a torsion-free group, any pure subgroup A ç G

gives rise'to a natural map Bext((7, T) L» Bext(^l, T). Under the hypothesis

that A is separable in G, \G/A\ < N( , and each pure subgroup of finite rank of

G is Butler, we are able to show that i* is onto. This enables us to show that

some balanced subgroups of completely decomposable groups are again Butler

groups.

We will prove our theorem by cardinal induction. We slightly modify Hodges'

version [H] of S. Shelah's singular-compactness theorem to fit our situation. The

case of regular cardinals is then treated similarly to [EF, Lemma 9].

2. Preliminaries

Unless otherwise specified, all the groups that we consider here are additively

written torsion-free abelian groups. We generally follow the notation and ter-

minology of [Fu]. A height is a sequence h = (h ), indexed by the set P of

primes p such that « is an ordinal or the symbol oo for each p. If G is an

abelian group and x e G, then \x\ = (hp (x)) is a height, where hp (x) is theLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

BUTLER GROUPS OF INFINITE RANK. II 64S

p-height of x in G. If S is a subset of a torsion-free group G, then (S), (S)t

denote, respectively, the subgroup and the pure subgroup generated by S. If

/ : A —> B is a map and S ç A , then / | S denotes the restriction of / to 5.

If T is a torsion group, p is a prime, and x e T, then T is the p-component

of T and x the p-component of x in T.

A subgroup H of a torsion-free group G is said to be separable [HM] if to

each g g G is a countable subset C of H such that to each h e H, there is

s e C such that \g + h\ <\g + s\.

A torsion-free abelian group G is called a Butler group if Bext(C7, T) = 0

for all torsion groups T. A torsion-free group G is said to be a B2 -group

if G is the union of a smooth chain 0 = Gn c G, c • • • c G C • • • ofU — 1 — — Q —

pure subgroups Ga such that, for each a, Ga+X = Ga + Ba where Ba is a

finite-rank Butler group. In this case we call G = \JGa a ß-filtration. Let

H be a pure subgroup of a torsion-free group G. We say that H satisfies

the torsion extension property (TEP, for short) in G or G has the TEP over

H if, for any Z-cyclic torsion group T, every homomorphism / : H —> T

extends to a homomorphism g : G —> T. In [DR2] it was shown that every

pure subgroup of a finite-rank Butler group G has the TEP in G. A subgroup

S of a torsion-free group G is said to be decent [AH] if to each finite subset X

of G there is a finite-rank Butler group B such that S + B is pure and contains

X . An important result of [DR2] is that in a countable Butler group G, a pure

subgroup H is decent if and only if H has the TEP in G. It is known that a

B2-group G is always a Butler group [BS] and every finite-rank pure subgroup

of G is Butler. It is easy to see that if H is a decent subgroup of G and both

H and G/H are B2-groups, then G is also a B2 -group. In the following Q

denotes the additive group of rational numbers and Z the additive groups of

integers. As usual, CH denotes the continuum hypothesis, i.e., 2N° = N¡ .

3. Pure subgroups of infinite-rank Butler groups

The backbone of the investigation of Butler groups in [DR2] is the theo-

rem that in countable Butler groups the TEP over a pure subgroup implies the

decency of the subgroup. In this section we generalize this important result

to uncountable Butler groups. Using this, we answer in more than one way a

question raised by Arnold [A] and Bican and Salce [BS] for Butler groups of

rankN, , namely, whether countable pure subgroups of such groups are again

Butler. Using a different approach, this was shown in [D].

We begin our investigation with the following useful lemma.

Lemma 3.1. (i) Let G = \JGa be a B-filtration of pure subgroups Ga . Then G

has the TEP over each G .a

(ii) Suppose G isa B2-group. Then every countable subgroup H is contained

in a countable pure subgroup A which as the TEP in G.

Proof, (i) Let / : Ga —> T be any homomorphism where T is an arbitrary

torsion group. By hypothesis, Ga+X = Ga + Ba ; Ba is pure in G and is a

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finite-rank Butler group. By Theorem 2 of [DR2], / | (Ba n Gf) : Ba n Ga -» T

extends to a homomorphism g : Ba —> T. Then f x : Ga+X —► T given by

fa+\ix+y) = f(x) + g(y) for all x e Ga , ye Bn is a homomorphism extending

/. By transfinite induction, on y > a, we get an extension / : G —► T.

(ii) By [AH], G has an axiom-3 family W of decent and pure subgroups.

Now H c A e i7, where A is countable. The axiom-3 nature of % and the

decency of A enable us to build a ß-filtration of G beginning with A . By (i)

above, G has the TEP over A .

Lemma 3.1 is useful in proving the following generalization of Theorem 7 of

[DR2].

Proposition 3.2. Let A be a pure subgroup of the torsion-free group G suchthat

(i) A has TEP in G.

(ii) A is a B2-group.

(iii) Each finite-rank pure subgroup of G is Butler.

Then A is decent in G.

Proof. We may assume without loss of generality that G/A has finite rank.

Then G = H + A, for some countable pure subgroup H. Since H n A is

countable and A is a 52-group, Lemma 3.1(h) implies that H n A ç B ç A ,

B a countable pure subgroup which has the TEP in A and hence in G. Now

B has the TEP in the countable Butler group (H + B)t and (H + B)t/B has

finite rank. Thus Theorem 7 of [DR2] applies and (H + B)t = Bx + B , where

Bx is a finite-rank Butler group. Then G = H + A = (H + B)t + A = BX+A

and A is decent in G.

Arnold [A] and Bican, Salce, and Stepan [BS], [BSS] raised the question

whether countable pure subgroups of Butler groups are always Butler. Proposi-

tion 3.2 enables us to answer this in the affirmative for Butler groups of rankr^ .

Dugas [D] also answers this for groups of rankN, using another approach.

Theorem 3.3. (i) Suppose G = C/K is a Butler group where C is completely

decomposable and K is a balanced subgroup of C. If K is a B2-group, then

every countable pure subgroup of G is Butler.

(ii) If G is a Butler group of rank hi, , then every countable pure subgroup of

G is also Butler.

Proof, (i) Let H/K be a countable pure subgroup of C/K . Since G is Butler,

K has the TEP in C and hence in H. Also, any finite-rank pure subgroup of

H is Butler. Thus Proposition 3.2 applies and K is decent in H. Since K is

balanced in H, we conclude that H/K is the union of an increasing sequence

of finite-rank Butler groups and hence is itself Butler; cf. [A].

(ii) Albrecht and Hill [AH] have shown that a balanced subgroup of a com-

pletely decomposable group of rankN, is a B2-group. Hence (ii) follows from

(i).As another application of Proposition 3.2, we obtain the following (not very

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BUTLER GROUPS OF INFINITE RANK II 647

Theorem 3.4. An infinite-rank torsion-free group G is a B2-group if and only

if every finite-rank pure subgroup of G is Butler and G = \JGa is a smooth

filtration of pure subgroups Gn where, for each a, Ga has the TEP in G and

G ,JG is countable.a+1 ' a

Proof. We need only prove the sufficiency. We first claim that each Ga is a

52-group. Suppose, by way of contradiction, there is a ß such that Gß is not a

B2-group. We may assume ß is the least such ordinal. Then for each a < ß ,

Ga is a B2-group and has the TEP in Ga+X , Then, by Proposition 3.2, Ga is

decent in G ., . Since, further, G ,JG is countable, the filtration M .. Ga+l ' ' a+\' a v^a</> a

can be refined to a /^-filtration. This would imply that Gß is a #2-group, a

contradiction. Thus each Ga be a B2-group and, by Proposition 3.2, must be

decent in Ga+X, Thus the smooth filtration G = \J Ga refines into a 5-filtration

of G, showing that G is a B2 -group.

Since, for groups of cardinality N,, a smooth filtration of countable groups

is an axiom-3 family, we get an axiom-3 characterization of B2 -groups in terms

of pure subgroups with the TEP.

Corollary 3.5. A torsion-free group G is a B2-group if and only if G satisfies

the TEP over an axiom-3 family of pure subgroups and every finite-rank pure

subgroup of G is Butler.

Before we proceed further we review the following proposition, which is a

slight recast of Lemma 10 of [DR1] and which is also implicit in the works of

Hill and Megibben [HM].

Proposition 3.6. Suppose C is a completely decomposable group of regular car-

dinal k , H is a balanced subgroup C, and G = C/H. Then there are a

closed and unbounded subset E of k and smooth filiations H = [JaeE Ha ,

C = U cf C , and G = M c(r G such that, for each a e E, Hn is balanced

in H, Ca is a direct summand of C, Ga is pure in G, Gn = (Ca + H)/H,

H = HC\C , and \H \ = \C \ = \G \<k.

We obtain from Proposition 3.6 an alternative proof of a stronger form of

Theorem 3.3. In §7 we will prove a stronger version of 3.7.

Theorem 3.7. Let G = C/H be a Butler group of regular cardinality k , where

C is completely decomposable and H is balanced.

(i) If H is a B2-group, then every countable pure subgroup of G is Butler

and G has a smooth filtration of pure subgroups G = \J Ga where each Ga is a

Butler group of cardinality < k .

(ii) If G is a Butler group of rankN,, ¿«c« every countable pure subgroup of

G is Butler.

Proof. By Proposition 3.6, there are a closed and unbounded subset E of k

and smooth nitrations H = M CFH , C = N cF C , and G = NcF G with

the stated properties. Let H = \Jß<k Lß be a ß-filtration of the B2-group H,

so that for all ß < k , Lß+X = Lß + Bß , where Bß is a Butler group of finiteLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


rank. By Lemma 3.1, H has the TEP over each Lß . Now, by Eklof [E],

the two nitrations U L „ and IJ Ha agree on a closed and unbounded subset

É of E. Hence we may assume that the H = \JaeE Ha, C = [ja€E Ca,

and G = \Ja€E Ga satisfy the conclusion of Proposition 3.6 with the additional

property that H has the TEP over each Ha . Since G is Butler, C has the

TEP over H. Then H and hence C will have the TEP over H . Sincea a

Ha is a B2-group, Proposition 3.2 implies that Ha is decent in Ca. Since

Ha is balanced in Ca , we conclude that Ga is a Butler group in which every

finite-rank pure subgroup is Butler.

A natural question is whether the converse of Theorem 3.7 (i) holds. Observe

that if G = \JaeE Ga is a smooth filtration of pure Butler subgroup GQ and

each countable pure subgroup of G is Butler, then Proposition 3.6 provides a

closed and unbounded subset E1 of E and a filtration H = M cír, H where

each H is balanced in H, H = H n C , and C /H = G , so that C hasa 'a a ' a' a a ' a

the TEP over Ha . By Proposition 3.2, Ha is decent in Ca and hence in //.

Thus H = M cF/ H is a smooth filtration where, for each a, H is balanced,

decent, and has the TEP in // and every countable subgroup of H is Butler.

Moreover, H itself is decent in C. Is // a 52-group? When does // fit into

a ß-filtration of C ? These are the questions that we wish to explore next.

Lemma 3.8. Suppose A and B are pure subgroups of G with A ç B. If A and

B/A have the TEP respectively in G and G/A, then B has the TEP in G.

Proof. Let T be any torsion group and / G Hom(B, T). Now / | A : A —> T

extends to a homomorphism h : G —► T. Since h — fi vanishes on A , h— fi' | B

induces hx : B/A -> T, which, by hypothesis, extends to a gx e Hom(G/A, T).

Let g2 be the natural map G —► G/A followed by gx . Then g = h—g2 satisfies,

for all b e B , g(b) = h(b) - g2(b) = fi(b).

Proposition 3.9. Let G be a torsion-free group such that any finite-rank pure

subgroup is Butler. Suppose A is a balanced subgroup or A isa B2-group. Then

A fits into a B-filtration of G ; i.e., G = \Ja<Á Ga with GQ = A, Ga+X = Ga+Ba ,

Ba finite-rank Butler ifand only if A has the TEP in G and G/A isa B2-group.

Proof. Clearly, if A fits into a ß-filtration with A = G0, Ga+X = Ga + Ba,

Ba finite-rank Butler, and G = lJQ</i Ga , then \ja<x(Ga/A) is a ß-filtration of

G/A which is therefore a B2-group. By Lemma 3.1, A will have the TEP in

G.Conversely, Let G/A = \Ja<fGa/A) be a .S2-filtration so that, for each a,

G ,,/A = G /A + B ¡A where B ¡A is finite-rank Butler. If A is balanced

then B = A + B' , where B' is finite-rank Butler, and so G ,, = G + B1 ,a a ' f» cc+1 a a

proving that IJ G with GQ = A is a Z?2-filtration of G from GQ onwards.

If, on the other hand, A is a B2-group, then Proposition 3.2 applies so that A

is decent in G. Thus, as before, Ba = A + B'a, B'n finite-rank Butler. Then

IJ Ga with Gf = A is a 52-filtration of G and this second-case G itself is a

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Corollary 3.10. Let B = \Ja<x Ba be a smooth filtration with B0 = 0 and, for

each a, Ba is pure and has the TEP in B. If Ba+X/Ba isa B2-group for all

a, then B itself is a B2-group.

The following proposition provides a partial answer to the question raised

earlier.

Proposition 3.11. Suppose 0 —* H —► C —> G —* 0 is a balanced exact sequence

of torsion-free groups of cardinality K^, n < co. If C and G are B2-groups,

then so is H.

Proof. Identify H with a subgroup of C and take G = C/H. We will do

an induction on «, the result being true for « = 0. Suppose G = \Ja<0) G ,n

C = \Ja<w Ca are ß-filtrations of G and C, respectively, so that for each a,

Ga and Ca are decent and have the TEP respectively in G and C ; for a < ß ,

Gß/Ga and Cß/Ca are B2-groups, Ga and Ca are B2-groups themselves,

and |CJ = \Ga\ < Nn . The proof of Proposition 3.6 shows that there are

closed and unbounded subset E of con and smooth nitrations H = \JaeE Ha ,

C = \JaéE Ca G = \JaeE Ga such that Ha = Hr\Ca, Ha balanced in Ca,

Ga = (H+Cf/H = CJHa . Since Ga is a ß2-group (and hence Butler) and Ha

is balanced in C , H has the TEP in C and hence in C. By induction, H

is a B2-group for each a. Now Ha+X/Ha is a balanced subgroup of Ca+X/Ha .

Moreover, Ca+X/Ha is a decent extension of the B2-group CJHa by the B2-

group Ca+X/Ca and hence is itself a B2-group. Since (CQ+x/Ha)/(Ha+x/Ha) =

Ca+X/Ha+X = Ga+X is a B2-group, cardinal induction implies that Ha+X/Ha is

a B2 -group.

4. Properties of Bext(-, T) and «-maps

It is well known [Hu] that if 0 -* A -^ B —* C —> 0 is balanced exact,

then for any torsion T, 0 -► Hom(C, T) — Hom(5, T) -> Hom(^, T) -^p* a*

Bext(C, T) —► Bext(i?, T) —> Bext(^ , T) is exact. We want to investigate what

can be said without the balancedness of 0 -» ^ —►/?—>C—>0.

The concept of an «-map is introduced and its relationship to separable

subgroups and, in particular, balanced subgroups is investigated. The notion of

an «-map appears to have the potential to play a dominant role in the future

investigations of infinite-rank Butler groups.

Lemma 4.1. Consider the following row exact commutative diagram:

0 -► F — M' —2—» G -» 0

0 — T - M —2—» G/A -» 0

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where the right square is a pullback, T any group and G/A torsion-free. If the

bottom row is balanced exact, then the top row is balanced exact.

Proof. Let R be a torsion-free group of rank 1 and cp : R —> G a homomor-

phism. Since the bottom row is balanced, there is a map \p : ß(cp(R)) —> M

with ßcp = a\p. Thus there exists a map y : R —> M' with r\y = <p since M1

is a pullback. This shows that the top row is balanced.

The following is an immediate consequence:

Corollary 4.2. The map ß: G -* G/A, G/A torsion-free, induces a map

ß*: Bext(G/A, T) -> Bext(G, T)

for any torsion group T.

We will need the following later:

Corollary 4.3. // G is a Butler group and has the TEP over a pure subgroup A,

then G/A is a Butler group.

Proof. Since A has the TEP over G, the following is exact:

0^Ext(G/A, T)CexI(G, T).

Since G/A is torsion-free we may apply Corollary 4.2 and obtain

0 - Bext(G/A, T) C Bext(G, T)

is exact. Thus Bext(G/A, T) = 0 and G/A is a Butler group. D

If A ç G is as above, we want to investigate when

Bext(G, T) £ Bext(A , T) -> 0

is exact for some 0 —► A -2+ G. This turns out to be quite involved.

First we need some notation. Let F be a reduced torsion group and A

reduced torsion-free. We will fix a mixed group M that fits into 0 —► T —» M A

A —> 0. For any group G, let G be the cotorsion hull of G ; see [Fl]. Then

G is a pure subgroup of G and G/G is divisible. We obtain a commutative

diagram with exact rows:

0^ f^M^*A-+0,p

u u u

o^ r->M^ A ->0.

n is the map induced by n on the cotorsion hull and p is a splitting map, i.e.,

ñp = id-. We will use this to obtain a description of M as a subgroup of M :

There is a map cp : A -> f such that M = T + (p(b) + cp(b) \ b e A). Here

tp : A —> f is a map such that tp* : A —> f/T defined by cp*(x) = cp(x) + T is a

homomorphism. If we pick another splitting map p : A —> M , then again M =License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


T + (p'(b) + cp'(b) \b e A) and an easy computation shows cp(b) = cp'(b) mod T

for all b G A . We will call such maps cp and cp' equivalent since they yield the

same mixed group M in the way described above. We find the maps cp : A —> T

to be of some help since they allow us to "code" most of the structure of M

into a single map. In what follows we will omit the splitting map p ; i.e., we

will set p = id-. Since we are interested in the elements of Bext(^, T) we will

investigate what maps p: A —> T belong to mixed groups M with T balanced

in M.

Definition. Let A be a torsion-free group. A map cp : A —> T is called an

«-map if

(j) The composite a h^ cp(a) + T is a homomorphism from A into T/T.

(ij) If W\p = °°, then tp(a)p eTp+p°°Tp.

(iij) For almost all primes p , \a\ < |^(a)| for any a e A .

We call two «-maps cp, \p : A -* T equivalent if for all a e A, cp(a) =

\p(a) mod T. Note that each «-map is equivalent to an «-map y/ and | t/(<z)| >

|a| for all a e A and all primes p .

The next proposition relates «-maps from A into T to elements in

Bext(A, T). There we will use the following observation: If X, Y are pure

subgroups of some group G containing its torsion part T with X n Y = T and

Y/T divisible, then X + Y is pure in G. We omit the straightforward proof.

Proposition 4.4. Let T be a reduced torsion group and B reduced torsion-free.

(j)//£:0->r->M-+rj->0 is exact, then E is balanced exact if

and only if there is an h-map tp : B —> T such that as a subgroup of M =

Ext(Q/Z, M)^f@B, M = T + ({b + tp(b) \ b G B}).(Ü) If the short exact sequence E is balanced exact, then E splits iff the h-

map cp is equivalent to a homomorphism cp : B —► T, i.e., cp(b) + T = cp(b) + T

for all be B.

Proof, (j) Let E be balanced exact. The mixed group M is a pure subgroup

of its cotorsion hull M = T® B. Since T/T is divisible and M/T reduced,

MnT = T. If n : M —> B is the natural projection along T, then n(M) = B .

The remark above gives that M + f is pure in M and hence B = (M + f)/T

is pure in M/f = B. Since T is balanced in M, to each b e B there exists

mbe M suchthat \mf = \b\ and mb + T = b . We may write mb = n (mf)+b,

where ri' : M —► f is the projection. Clearly, \n'(mf)\ > \mb\ = \b\. Then the

map tp : B —> f defined by cp(b) = n'(mb) satisfies \cp(b)\ > \b\ and theinduced map Ip : b >-+ cp(b) + T is a homomorphism. Thus cp is an «-map and

M =T+(mb:beB) = T + ({cp(b) + b : b e B}).Conversely, suppose 0—* T —> M ^ B —>0 is exact and there is an «-

map cp : B — f with M = T + (cp(b) + b \ b e B) ç M. Then M is

a mixed group with M/T s B. If 0 ¿ b e B, then \cp(b)\p > \b\p for

almost all p. If \cp(b)\p < \b\p , then \b\p < oo because tp is an «-map. SinceLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


<p(b)p G f is divisible mod F and there are only finitely many such primes

with \cp(b)\p < \b\ , we find t e T with \t + cp(b)\ > \b\ for all primes p.

Thus cp(b) + t is proper in cp(b) + T and T is balanced in M. The proof of

(jj) is straightforward.

The following theorem is crucial for this paper. It will enable us to discover

many subgroups of Butler groups that are again Butler groups.

Theorem 4.5. Let A be a pure and separable subgroup of the torsion-free group

B with \B/A\ < N[. Let cp : A —> T be an h-map and assume that each pure

subgroup of finite rank of B is Butler. Then there exists an h-map \p : B —> f

with ip | A = tp.

Proof. We may assume that B/A is countable. For each coset b + A e B/A

there exists a countable subset C = C(b) ç A such that for each a e A there

exists an element a0 e C with |6 + a| < \b + a0\. Let B0 be the pure and

countable (!) subgroup generated by a representing system of the cosets in B/A

together with the countably many elements in C = C(b), b running over the

representing system of B/A . Let A0 = Ar\B0. Then AQ is pure in A and our

hypotheses imply that A0 and BQ are (countable) Butler groups. Moreover,

B = A + Bf . Since AQ is a Butler group we may apply 4.4(jj) and assume that

tp | Af e Hom(A0, T). Since T is pure-injective there is a y/ e Hom(ß0, T)

with \p | Af = tp | Af . For each pair (b0, a) G BQxA we fix a0 = aQ(b0, a) e A0

with |è0 + a| < |60 + a0|. Note that such an a0 e A0 exists by the choice of A0 .

Now define p(bQ + a) = ip(b0 + af + <p(a - af . Then p : B —► f is a map.

Suppose b = bf + ae Bf. Then a e A0 and we have p(b0 + a) = ip(b0 + af) +

cp(a-af) = ip(bf + a) since <p is a homomorphism with ip | AQ = tp | A0 . Thus

p | Bf = ip . If b0 + a e A , then b0 e B0 n A = AQ and

p(b0 + a) = \p(bf + af + <p(a - af,

= cp(bf + a0) +tp(a-a0) = tp(bf + a) modF.

Thus p | A is equivalent to cp and for M = T + {(cp(a), a) \ a e A}, M' =

T + {(p(b), b) | b e B} we have that the diagram

0— T^M — A^O

Il n rV

o^t^m'^b^o

is commutative. In order to finish the proof we have only to show that p is

an «-map. (After that we may change p into an equivalent «-map p with

p | A = cp .) Let b e B , b = b0 + a, b0€ B0, a e A. Then \b\ = \bQ + a\ <

\b0 + a0\ where aQ e AQ . We obtain

\p(bf + a)\p> min{|^(è0 + af\p, \tp(a - af\p} > min{\b0 + a0\p , \a - af\p

for almost all primes p . If q is a prime with \b0 + a\ = oo , then \b0 + a0\q =

oo = \a - Of\q and tp(a - a0)q, y/(b0 + afq eTq + q°°Tq. Thus p(b0 + a)q e

T + q°°f . This completes the proof.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Corollary 4.6. Let B be a Butler group such that for any countable subset X

of B there is a filtration B = \Ja<xBa such that \Ba+x/BJ < N,, X c Bf,

\Bf\ < N,, and each Ba is a separable subgroup of Ba+X . Then each Ba is a

Butler group.

Proof. Let A be a pure subgroup of finite rank of B . Then we have a filtration

B = [ja<^Ba such that each Ba is separable in Ba+X and BQ = A. By [D,

Theorem 2] and [B] (see also [A]) we conclude that A is a Butler group. Thus

any pure countable subgroup of B is Butler. Let T be torsion and let 0 —►

T -» M —> B -^ 0 be balanced exact. By Theorem 4.5, there is a commutative

diagram0 -► T --► M -► B -► 0

0 -► T -► M J.. -> B ..a+l a+l

where Ba —> Ba+X is the embedding and the bottom row is balanced exact. A

transfinite induction provides a commutative diagram

0 -► T --► M -► B -► 0

I IM -► B

where the bottom row is balanced exact. Since B is Butler, M = T © B1 for

some subgroup B1 isomorphic to B . Since T ç Ma we obtain

M =MnM =(T + B')nM =T®(B'nM).

Thus Ma splits and Bn is a Butler group. G

Example. A group B with pure subgroup A, rk(B/A) = 1 and there is an

«-map cp : A —» T = Y[ fp which does not lift to B .

Construction. Let AF be a maximal family of almost disjoint maps / : co —>

co, i.e., fi+geAW = {i\ fi(i) = g(i)} finite.Generators, x , xn (n < co), xf (fie AF).

Define B = (x, xn , xf | n < co, fi e AF).

Let n be an infinite set of primes,

n = {P,j \i, j <co}.

Relations, (x + xf/pn¡ (i < co) and (x + xf)/pnf(n) (n < co), i.e.,

I x + x x + xf \B = (x,xn,xf,-^ ,-J- | « < co, i < co, f eF ) .

\ " f p».i Pn,m] I

Let A = (xn, x{ | « < co, f G AF)t. Then

A = lxn, xf,X" ~ Xf Xnr ~ Xg | « < co,g, fie F, /(«) = g(n)\ .\ Pn,f(n) Pn.f(n) /



Pick elements s; , Z- G T with \si} - t¡.\

Define

and

<PiXn\

<Pixf)

su if i = n,

0 if i ¿ «

ln,f(n) K(i,j) = (n,f(n)),

0 elsewhere.

cp induces a homomorphism cp : A1 = (xn, xf \ n < co, f e AF) —► T/T.

Since T/T is divisible, there is cp" : A —> T/T extending cp'. If a e A,

a g {xn, Xj- | « < co, f G F}, define tp(a) to be any element of the coset

cp"(a). Since A is homogeneous of type 0, tp is an «-map, cp : A —► T.

Now suppose cp lifts to an «-map ip : B -* f. Then |i//(x) + f3(*„)L >

0 for all i > mfn), i.e., V(*)Pn , = 5„, mod^„,^ , for ' > ™tyin); on

the other hand, define g : co —> co via #(/') = m0(i) => 3f e F such that

W = {i < co | /(/) = ^(/')} is infinite. We have \ys(x) + cp(xf)\ > 1 for

n > kf(f) => ip(x) = t f(n] modpn f, AT . Since co is infinite, there is

i eW, i > kf(f), and for « > /c0(/) we have

^X\,,-Sni m0dPniTPn¡ and

^^„.«-^ modpmfP„y

This implies 5n( = tni modpniT , which is a contradiction to the choice of

sm and tHf

5. Separable subgroups of torsion-free groups

Our goal in this section is to prove the following

Theorem 5.1. Assume CH holds. If G is a torsion-free group of cardinality Nn ,

n < co, then there is a smooth chain of separable pure subgroups Ga, a < Nn ,

such that G = IJ ^m G and G ,./G has rank 1.

We will prove this theorem after some preparation. First we need a

Definition. Let H be a pure subgroup of the torsion-free group G. Let g e G.

For each countable subset C of H, we define a map gc from C into the set

of all height sequence by gc (x) = \g + x\ for all x eC. We H hyperbalanced

if for each g e G and for each countable subset C of H there is some « G H

with g* = h*c .

Lemma 5.2. Let 0 / H be a subgroup of G. Then there is a hyperbalanced

subgroup L of G with H ç L and \L\ <\H\ °.

Proof. By induction on a < cox we define a smooth chain of subgroups Ha with

Hf = H and Hx = \}a<xHa if A < w, is a limit ordinal. Moreover, we willLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


have \Ha\ < \H\ °. Suppose Ha is already defined. Then S = {gc \ g eG, C

a countable subset of HJ has cardinality \Haf» .(2*0f0 < (|//|N°)2*° = \H\*°.

Thus there is a subset IF of G with \W\ < |//|N° and S = {g* \ g e W, C a

countable subset of H} . Now define H x to be the pure subgroup generated

by Ha and S. Note that \Ha+x¡ < \H\*°. We will show that L = \Ja<w^ Ha

is hyperbalanced in G: Let C be a countable subset of L. Since cf(cox) =

co, > co there is a < co, such that C c H . If g e G then the construction1 1 — a °

of H .. shows the existence of an « G // ,, CL with ?„ = «„ . Thus L isa+l a+l — °c c

hyperbalanced and |L| < cox ■ \H\*° = \H\H°. o

The argument above also shows

Observation 5.3. If {Ha, a < X} is a smooth chain of subgroups of G, cf(X) >

co, and each Ha+X is hyperbalanced in G, then \Ja Ha is hyperbalanced in G.

The following is crucial:

Lemma 5.4. Any hyperbalanced subgroup H of G is balanced in G.

Proof. Assumed not. Then there is a coset 0 ^ g + H without proper element.

By induction on a < cox we will construct elements gae g+H with \ga\ < \gß\

for a < ß . Let g0 be any element in g + H and suppose that g has been

defined for all y < a .

Note that C = {gy - g \ y < a} is countable. Since H is hyperbalanced,

there is some hx e H with g* = (hxfc, i.e., \g + (gy - g)\ = |«, + (gy - g)\

for all y < a. This implies \g - «J > \g \ for all y < a. Since there is no

proper element in g + H, g - hx is not proper and there is some h2 e H

with \g + «2| f. \g - hx\. Set C' = C U {-«,, «2} and use again that H is

hyperbalanced and we find «3 e H with \g + «3| > \g + hf\, \g - «,|, and

\gy\ for all y < a, Now suppose \g + hf = \g \ for some y < a. Since

\g + h3\ > \g - hx\ > \gy\ we obtain

\g + h2\ <\g + «3| = \gy\ <\g-hx\£\g + h2\,

a contradiction. This show |g+«3| > \gy\ for all y < a. We may set ga = g+h^

and the induction works. This also completes the proof of the lemma, since

there is no uncountable ascending chain of height sequences. D

Let G be a torsion-free group. By induction on ordinals a < cox (= Kj) we

define ^"-subgroups of G : The B -subgroups of G are precisely the balanced

subgroups. If X < cox is a limit ordinal, then the ZT-subgroups are all the

subgroups of G that are ./3a-subgroups for some a < k. The Ba+1 -subgroups

are the unions of countable chains of ^"-subgroups. We call a subgroup A of

G a 5°°-subgroup if A is a ^"-subgroup for some a < cox .

Lemma 5.5. (i) The class of B°°-subgroups is closed with respect to unions of

countable chains.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


(ii) If A is a B°°-subgroup of B and B is balanced in C, then A is a

B°°-subgroup of C.

Proof, (i) follows since cf(cox) > co. To prove (ii) we use induction. If A is

a B -subgroup of B, i.e., A is balanced in B , then A is balanced in C since

"balanced" is transitive. We may assume that A is a Ba+ ' -subgroup of B and

(ii) is true for ^"-subgroups. Now A = \Jn<ü)An, An, ßQ-subgroups of B.

Thus An are 5°°-subgroups of C and A is a 2?°°-subgroup of C.

Theorem 5.6. Assume CH holds. Let G' be a torsion-free group such that G,

Hm , G + Hm (m < co) are balanced subgroups of G1. Let \G\ = Nn, 2 <

« < co. Then there is an K -filtration G = M ^« G of G, \G I = N , , and

Ga(Ga + HJ is a B°°-subgroup of G(G + HJ for all a and \Ga+x/GJ < N,.If X is a countable subset of G, then we may have that X ç G0 .

Proof. By induction on «. We may assume H0 = 0. Suppose |G| = N2.

Lemma 5.2, Observation 5.3, and a standard back-and-forth argument provide

a filtration G = (Ja<!<2 Ga where |GJ = N, , X ç G0, and (Ga + HJ/Hm is

hyperbalanced in (G + Hm)/Hm for all m < co whenever cfi(a) ^ co. Here

we had to use CH to obtain Wf = Kn in order to avoid possible jumps of our

cardinals in 5.2. Since hyperbalanced implies balanced (Lemma 5.4), we have

for cf(a) t¿ co : Ga+Hm is balanced in G+Hm . Thus for all a < K2, Ga (resp.

Ga+Hm) isa B'-subgroup of G (resp. G+Hm ). Since |GJ = N, , |GQ+1/GJ <N[ and the theorem holds for n = 2. Now suppose the theorem holds for « .

Let |G| = Nn+1 . As before, we get an Kn+1-filtration G = \Ja<H Ga where

(Ga + Hm)/Hm are hyperbalanced in (G + Hm)/Hm whenever cfi(a) / co.

Thus Ga + Hm is balanced in G + Hm whenever cjf(a) ± co. We now show

that we can refine this filtration to meet the requirements of our theorem. We

distinguish two cases.

(I) cfi(a) f co. Here we apply the induction hypothesis where Gn+1 plays

the role of G and Ga , Ga+Hm play the role of Hm . Thus we find Ar, r <Wn,

and Ar isa Z?°°-subgroup of Gn+x + (Ga + Hm) and Ar + Gn isa ß°°-subgroup

of Ga+X + Ga . Now set Gar = Ga + Ar. Then Ga r is a B°°-subgroup of Ga+1_

and Hm + Gnr is a 5°°-subgroup of Ga+X + Hm . Since Ga+X , Ga+X +Hm are

balanced in G', Hm + Ga r (resp. Ga r ) is a ß°°-subgroup of G1 by Lemma

5.5. Since \Ar+x/Ar\ < N, we conclude that also |GQ r+l/Ga r\ < Xx.

(II) cfi(a) = co. Here Ga = \Jn<(l) Ga and the an are in case I. We apply

the induction hypothesis where Gt+1 plays the role of G and Ga , Hm + Gn

play the role of the Hm , (m, n) e co x co. Again, we find an Nn-filtration

Ga+l - Ur<K¿r where \Ar+x/Ar\ < N, , Ar + Hm + Gn^ is a 5°°-subgroup

of Gn+X + Hm , and Ar + Gn is a ß°°-subgroup of Ga+X . This implies that

ön<JAr + GaJ + Hm isa 5^-subgroupof Ga+X+Hm and {Jn<wAr + Ga>¡ isa

B°°-subgroup of Ga+X . Thus we may set Ga r = [jn<wiAr + Gn ) and because of

Lemma 5.5, Ga r isa 5°°-subgroupof G' and Hm + Gn r isa B°°-subgroup ofLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


G'. Again, since \Ar+x/Ar\ < Nj, we obtain |GQ r+x/Ga r| < N,. This finishes

the proof of the theorem. D

Corollary 5.7. Assume CH holds. Let G be a torsion-free group, \G\ < Nw,

and X a countable subset of G. Then G has a filtration G = (] „„ G„ with

Ga+X/Ga torsion-free of rank 1, Ga separable in G, and X ç G0, where G0 is

countable.

Proof. First, let |G| = Nn < Nw. If |G| < N¡, there is nothing to show since

each countable subgroup is separable. If |G| = Nn , we have only to use Theorem

5.6 and observe that B°°-subgroups are separable and countable extensions of

separable subgroups are again separable. If |G| = «B, we may use Lemma 5.2,

Observation 5.3, and Lemma 5.4 to write G = \Jn<0JGn , Gn balanced in G,

|GJ = N„ . We apply Theorem 5.6 to Gn+X/Gn and find G„ C Anr c Gn+X

with An r/Gn a fi^-subgroup of Gn+X/Gn . We want to show that this implies

that An r is a 5°°-subgroup of Gn+X .

If An r/Gn is a B -subgroup of Gn+X/Gn , then An r/Gn is simply balanced

in Gn+X/Gn , and since Gn is balanced in Gn+X, we have that An r is balanced.

Now assume that Anr/Gn = \Ji<l0iAi/Gn) is a 5a+1-subgroup of Gn+X/Gn.

Our induction implies that Ai is a Z?00-subgroup of Gn+X ; i.e., An r = \Ji<(aAi

is a Z?°°-subgroup of Gn+X . Thus G = \Jn rAn f is a filtration into B°°-

subgroups of G with \An r+x/An r\ < N1 . This filtration may be refined to

prove the theorem.

Our methods break down if |G| = Nw+1 , since one cannot show in general

that there are many or even one balanced subgroup of cardinality Nœ of G

when |G| = Nw+1 , as the following example shows. Again, we assume Nn° = Kn

for all « < co, i.e., CH, to obtain N^0 = Nw+1 . Let F be a free group of rank

Kw and G the cotorsion-hull of F. Then |G| = |F|N° = Kw+1 . Now suppose

A ç G is balanced, \A\ = Ww . Since A is pure in G, A , the cotorsion-hull of

A , is contained in G ; i.e., A/A c G/A . Now A/A is divisible, which implies

A = A since A is balanced. Hence A is torsion-free cotorsion of size Nw . In

our setting, this is not possible for cardinality reasons.

In conclusion, there are torsion-free groups of cardinality NM+1 without any

balanced subgroups of cardinality Hw (at least under CH). In our example

above, A is still separable in G since A is a summand of G and A/A is

divisible. At this time, we do not know of an example of such a G without

large separable subgroups. An immediate consequence of Corollary 5.7 and

[AH, 6.3] is the noteworthy

Theorem 5.8. Assume CH holds. If G is a torsion-free group of cardinality <NW,

then Bext (G, T) = 0 for any torsion group T.

Remark. For |G| = N, this is contained in [AH, Theorem 6.3]. Using the long

Hom-Bext sequence established in [Hu], we obtain the following corollary.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Corollary 5.9. Assume CH holds. Then any balanced subgroup of a completely

decomposable group of cardinality < Nw is a Butler group.

6. Singular compactness

Here we state a version of Shelah's singular compactness analog to the version

in [H] tailored to our purposes.

Definition 6.1. Let X be a cardinal and B a group. A family AF of subgroups

of B is called a A-family if

(a) OeAF.

(b) If AeF, then \A\<X.(c) If p < X and Aa , a < p, is an ascending chain of subgroups in F, then

\Ja<fl AaeAF if for some k , \Aa\ < k < X for all a < p, \<p regular.

(d) If A e AF, and X ç B with \X\ < X, then there is C G AF with

|C| < \A\\X\ and AuX CC.

We adapt a proof in [H] to prove the next theorem. We give only an outline

to indicate where there are changes.

Theorem 6.2 (Hodges/Shelah). Let X < Nœ be a singular cardinal, B an abelian

group of cardinality X, and AF a X-family of subgroups of B. Let F* be the

family of unions of countable chains of AF.

(j) For each A G AF* there is an axiom 3-fiamily AF(A) of subgroups of A ;

cf [AH],(ij) // An c An+X G AF for all n < co and if An G F(An+x) for each «,

/«<?« there are axiom ^-families F(Af) of An and AF(Af) ç AF(A) where

A = U„<w An • Provided \A\<X.(iij) If k is a regular cardinal < X and Ci G AF, i < k , is a smooth chain

with |C(| < k for all i < k , we set C = \Jj<K C; and define AF(C) \ C¡ = {X \X e AF(C), X ç Cf . If C,. G F(C), then AF(C) | C, is an axiom 3-subfamilyof AF(CA). Note that C eAF since |C;| < k < X and k regular.

Then there is a smooth chain A', i < cf(X), with A' eAF*, A' e AF(Al+l),

and B = (Ji<cfwAl.

Proof. (We adopt the proof in [H] up to some minor changes.) Let k be a

regular cardinal < X. We define the /c-Shelah game on B : Player I picks

subgroups Bi of cardinality < k for i even, / = 0, 2, 4, ... , and player II

picks B{ e AF for i odd such that Bi ç Bi+X for all i < co. Player II wins

if Bj e ^(Bl+X) for all odd i. Observe that (Ji<lüB¡ e AF* since B2 eAF

and \Bt\ < k < X. Lemma 1.2 in [H] still holds: the /c-Shelah game on B is

determinate; i.e., if player I has no winning strategy, then II has one! Thus, we

have only to show that player I has no winning strategy. By way of contradiction,

let us assume I has a winning strategy s. We have to show that II can defeat

s . Let's play! Player I moves first and picks B0 . Now player II (he's got all

the time in the world) gets working. By transfinite induction he constructs a



smooth chain C¡ (i < k) of subgroups of B with C¡ e AF*, Ci e AF if

(/) t¿ co, and |C,-| < k for all / < k such that: For each j > k and each finite

set of ordinals jx < ■■■ < jn < j each subgroup showing up in an initial piece

(of length 2« ) of the /c-Shelah game where I plays his strategy 5 and starts

with Bn and II plays C, , ... , C, is contained in C,. This can be done sinceU h ¡n '

j < k , k regular and (d). Because (c), C = \Ji<K Cte AF . Since there is the

axiom 3-family AF(C) on C, we may assume—passing to a on ô ç k—that

C,. G AF(C) for ail i < k . Thus C, G AF(C) \ Cl ç P'C) \ Ci+X ç AFC(Cl+x),

i.e., CteAF(Cl+x).Now II easily defeats I by choosing Bi+X tobe Ca(/) where a(i) = l+inf{a <

k \ Bq, Bx, ... , B¡ ç Ca}. Now let (k¡) , i < cf(X), be a strictly increasing

sequence of regular cardinals, sup{K¿ | i < cfi(X)} = X and k0 > cf(X). Since

AF is a A-family on B, we may write B = U,-<cm) G' > \G'\ = Ki> C' e

AF. We want to extend C' to A', i < cf(X), smooth chain, A' e AF*,

and A' e AF(A,+i) for all i < cf(X). Each A' will be the union of a chain

C = Cf CB'fCAfC C\ C B\ ç A\, ç • • • with

(i) If i - 1 exists, then C[, B'0, C\ ç B\, ... is a play of the K(+-Shelah

game on B where II uses his winning strategy picking the B'k 's (I picks the

C¿'s). If i is a limit or 0, set B'k = C'k for all k < co. We make player I

(each upper index stands for a separable game) pick the C'k 's. (As in [H], the

Aln will be defined and used later to ensure that the chain {A1} is smooth.) We

also make I pick the C'k 's such that:

(ii) Whenever AJn is defined for each j < cfi(X), then C'n+X 2 U/<,-^ and

for k = 0,l,2,...,n C'n+xnB'k+' e AF(Blk+x ) where <F(b£1) C AF([)k<w B'k)

is the family in hypothesis (ij). This can be done by using a standard back-and-

forth argument using the fact that AF(B'k+x ) is closed with respect to (countable)

unions.

Now define A1 = U„<(X = Un<(0C[ = (J„<C[X • We have A' = A' n

A>+[ = A'ni\Jk<0)B'k+l) = ok<JA'nB'k+i) = {k<wiii}k<n<wC'n)^B'k+i) =

Vk<jnk<n<JC'nnB'k+l)). Since C'H n B'k+X G AF(Blk+l) for all /c < », we

conclude C'n n B'k+l e ^(A,+l), A,+ l = \Jk<wK+l > and since ^iA'+l) is

closed with respect to countable unions of subgroups of size k; , we conclude

that A1 e AF(Al+l). Now we may complete the proof as in [H, p. 210] by

defining the A'n 's to ensure that {A' \ i < cf X} is smooth.

We are interested in

Corollary 6.3. Let B be a Butler group of singular cardinality X and assume

that B admits to a X-family AF of B2-subgroups. Then B is a B2-group.

Proof. For A e AF we let AF(A) be an axiom 3-family of decent subgroups as

introduced in [AH]. We have to show that 6.2 applies: (j) takes care of itself. To

verify (ij), let An c An+[ , « < co, be a chain in AF and An e AF(An+x). ThusLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


An is decent in (An+X) and—better—shows up in a B2-filtration of An+X .

Thus A = \Ja<pHa = \Jn<wAn, Ha+X = Ha + Ga, Ga pure of finite rank

and there is a sequence an < an+x < p with p = sup{an \ n < co} and

Ha = An. Use this B2-filtration of A and the An 's to define "closed" subsets

of p and an as in [AH]. This gives rise to an axiom 3-family AF(A) ç AF(A),

AF(A) | An = F'(An+x) c AF(An+x), and since S ç an is "closed" iff 5 is closed

in p, AF(Af) C AF(A) C AF(A). This shows that (ij) holds. Let the C; 's in

(iij) play the role of the An and (iij) follows.

7. The structure of Butler groups

We will give a cardinal induction to show that under CH every Butler group

of cardinality < Nw is a B2-group. First we want to cover the "regular case."

Theorem 7.1. Let G be a Bulter group of regular cardinality k suchthat G has

a filtration G = UQ<K Ga and each Ga is a Butler group with | GJ < k . Then

there is a cub C ç k and for any a e C, we have that Ga has the TEP in each

Gß, ß>a.

Proof. Let E = {a e k\ There is a a < ß < k such that Ga does not have TEP

in Gß}. By way of contradiction we assume that E is stationary in k. We may

assume that if a e E, then G does not have the TEP in G ,. . Thus there exista a+1

a ( X-cyclic) torsion group Ta of cardinality < | GQ+, | and a homomorphism

tp : G —► T that does not lift to G .. . Now define S = 0„^ T„ . (Here~a a a a+l a VJ^p<a p ^

we borrow an idea from [EF].) By transfinite induction we construct a direct

system {Mfa < k); Ylß} ([Fl, p. 53 ff]) as follows: For any a < ß < k we

have

(1) Ma is a mixed group with torsion subgroup S and Ma/Sa = Ga .

(2) 0 ->• S ->■ M -» G ^ 0 is balanced exact.

(3) All nf are embeddings and nf | Sa = id.

(4) If A is a limit ordinal, then M, = lim M and TÏ1 : Mn —> M, , a < X,

are the natural maps induced by Yla , a < ß < X.

(5) For all a < ß we have UßfMa) + Sß = ïlpa{Ma) ©Sßa where Sßa =

®a<r<ß 'r •

Note that (5) implies that for any limit ordinal X and a < X, Yfa(Ma) + Sx =

YlÀa(Ma)®SÀ a and therefore SÁ is balanced in Mx since n^(SJ is balanced

in Ylka(Mf). (Keep in mind that all n^ are embedding.)

(6) If Ma is already defined and a e E, then Ma = Sa © Ba and Ba =

Ga. Let pa : Ga —* Ba be an isomorphism. This isomorphism extends to

an isomorphism pa+x : Ga+X -» Ba+X where Ba+X D Bn . We set Ma+X =

5a+i©5a+i and define na+' (5, ¿) = (s+tpfp^ l(b)),b) for all s G Sa, beBn.

Clearly, n^+1 : Mn —> Ma+X is an embedding.

If a $ E , we define na+1 as above with cpit = 0 . We have to show that (5)



is satisfied. Let ß < a. Then

so+i +iCH) = n:+1(n;(M,)) + Sa+X ç Ma + Sa+X

= K + sa+l,0 = iMa + sa)®sa+Xa.

By induction hypothesis, Yfß(Mß)+Sa = Ylaß(Mß)+Saß . Thus Sa+x+Yfß+i (Mß)

= naß(Mß)®SaJ}®Sa+Xa = Ylaß(Mß)®Sa+xß. If'A is a limit and ß < A,

then

liß(Mß) + Sx = Ulß(Mß)+SkJi= [J (UXß(Mß)®Saß)ß<a<X

= Uß(Mß)® [J sOi0 = itß(Mß)®sXiß.ß<a<X

Let M, = lim M . Then M = Il Tlk{M ) and n*(M )+Sr = Y1K(M )©

SK a. Therefore SK = ®Q<iC Ta is balanced in AfK with MJSK = G. Since

G is a Butler group, there is a splitting map \p : G —► A/K . Since «: is regular

and |MQ| < k for all a, there is a cub C ç k such that y(Ga) ç Ylxa(Ma) for

all a G C. Since E is stationary, there is some a e E n C . Let ß e C with

ß > a . Then ty | G„ : G„ —» n»(Af«) is a splitting map. We may apply (n^)~

and we may assume that \p : Gß —► Af„ and \p \ Ga —» IT^AfJ is a splitting

map. We have t//(g) = («(^), Pß(g)) for all g e Gß and « = «0 © «f : Gß -*

sa+i ®sß,a+i> hi\Ga = °- l{ 8 €Ga> theo his) € 5a. We may subtract

«. from 1/ and thus are able to assure «. = 0. Then ip : G ^, —> Af ,. is a1 ~ I " a+l a+l

splitting map and

naa+x(Ma) = {s + tpa(p-J(b)),b) \beBa,seSjcSa

®Ta®{(h(p-x(b)),b)\beBa}

and flaa+x(Ma) = Sa © {h(p-x(b),b) | ô G 5a+1}. Thus cpa(p~x(b)), b) =

h(p~x(b)) modSQ for all b e Ba. Let n : 5Q+1 —> Ta be the projection. Then

Ylotpa = tpa and cpa(p~x(b)) = (Yloh)(p-ax(b)) for all b e Ba. This implies

<pa = (Yloh) \ Ga and ç?Q lifts to Ga+X, a contradiction. This shows that E is

not stationary and therefore we may assume that G = UQ<K Ga ls a filtration

where each G has TEP in G ,. .a a+l

Next we will modify Lemma 3.1 in [AH]. We refer to [AH] for a definition

of Hill's compatibility relation ||.

Lemma 7.2. Let A, B, and H be pure subgroups of the torsion-free group G.

(a) If A is a balanced in G and A\\H, then A n H, is balanced in H.

(b) If A\\H and H + A\\B then H\\A + B.

Proof. To prove (a), let heH-(HnA) = H-A. For any a e H n A ç A

we have \h + a\ <\h + a0\ where a0 = afh) e A since A is balanced in G.

There is some element a, G A n H such that \h + a\ < \h + af < \h + ax\

for all a e A n H. Thus A n H is balanced in H. To show (b), fix someLicense or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


h e H, a e A, b e B. Then \h + a + b\ = \(h + a) + b\ < \h + a + b0\where b0 e (H + A) n B , i.e., b0 = «0 + a0 e B and h0 e H, a0 e A . Thus

\h + a + b\ < \h + a + hQ + a0\ < \(h + h0) + (a + a0)\ < \h + «0 + x\ for some

x G An H since H\\A. This implies \h + a + b\ < \h + (bQ -af) +x\ and since

üf, x e A, bf e B we have y = bQ - a0 + x = «0 + x e (A + B) n H. This

shows H || A + B .

Definition 7.3 [FH]. Let G be a group, A a cardinal, and F a family of

subgroups of G. We call AF a G(A)-family if

(a) OeAF.

(b) If {FfieI}CAF isa chain then U,€/ F¡ G AF .

(c) If F eAF and A ç G, \A\<X, then there is F' e & with A + F/F C

F /F and \F'/F\ <X.

Definition 7.4. Let C = ®(€/ R, be a completely decomposable group. For any

subset J ç I we set C(J) = 0/ey R, ■ If H is a subgroup of C, we call

a subset J ç I an //-special set if C(J) \\ H. Moreover, we define H(J) =

HnC(J).

Theorem 7.5. Let H be a pure subgroup of the completely decomposable group

C = ©,e//v(. F«c« H admits a G(2 °)-family AFH of balanced subgroups

induced by a G(2*°)-family AFc of direct summands of C, i.e., AFH = {X n H \

Xe.Fc}.

Proof. Let AFH be the family of //-special subsets of / and AFH = {C(X) \ X e

AFH} . We want to show that AF* is a G(2N°) family of C . We have only to

show 7.3(c). So let C(J) eAF* and A ç C, |^| < 2^°. By induction on « < co

we define subsets Zn C /, \Zf < 2 ° : Let ZQ be minimal with A C C(Zf).

Suppose Zn is already defined. For g e C(Zn) we find a set {ha | a < 2 °) ç

H + C(J) such that for any y e H + C(J) we have \y + g\ < \ha + g\ for

some a < 2 ° . (Note that there are only 2 ° many height sequences.) Now

let Zn+X be minimal with C(Zf u {hag \ g e C(Zn),a < 2«°} ç C(Zn+x).

Note that |Z„+1| < 2*°. Let Z = \Jn<(1)Zn. Then H + C(J)\\C(Z) and

H\\C(J) since J e FH . By 7.2(b), H \\ C(J) + C(Z) = C(J U Z) and by

7.2(a), H(J*) = //n C(J*), J* = JUZ is a balanced subgroup of H. Now

we are done since \J - J\ < \Z\ < 2 °.

Corollary 7.6 (CH). Let H be a balanced subgroup of a completely decompos-

able group such that H is a Butler group. Then H admits a G(Kx)-fiamily ofi

Butler groups.

Proof. Observe that a countable extension of a balanced subgroup of H is

separable in H and Corollary 4.6.

We are now ready to prove our main theorem.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


Main Theorem (7.7). Assume CH holds. Then any Butler group of cardinality

< Nw is a B2-group.

Proof. By cardinal induction. Let G be a Butler group of cardinality k and

assume that each Butler group of cardinality < k < N^ is a B2-group. Note

that k = N1 has this property. Let 0—►//—► C —► G —> 0 be a balanced

resolution of G ; i.e., C is completely decomposable and H is balanced in C.

We may assume that \H\ = \C\ = k . By Theorem 7.5 there is a G(N^-family

AFc of summands of C such that AFH = {H n X | X e AFf} is a G^^-family

of balanced subgroups of H. Therefore we may—for any S e FH—write

H = U H with H e AF„ and \H ,.IH I < K, and //n = S. Since*—'a<K a a H ' a+l' a1 — 1 U

|G| < N , Theorem 5.8 implies that Bext (G, T) = 0. Since H is balanced

in C, the long Hom-Bext sequence implies that Bext(H, T) = 0 ; i.e., H is a

Butler group. By Corollary 4.6 we obtain that each Ha is also a Butler group;

i.e., any S e AFH is a Butler group and by induction hypothesis, H has a

G(K,)-family AFH of B2-groups.

If k < Nw is singular, Corollary 6.3 implies that H is a /?2-group. If k

is regular we may apply Theorem 7.1 and there is a filtration H = \Ja<KHa

with H e AF„, \H I < k , and each H has TEP in H ,. Since H ,. is aa H ' ! a1 ' a a+l a+I

pure subgroup of C, each pure finite-rank subgroup of Ha+X is a Butler group

of finite rank. Moreover, H , ///Q is a Butler group of cardinality < k by

Corollary 4.3. Thus we may apply Proposition 3.9 and continue any 5-filtration

of H to one of H ,. . This and an obvious induction over a < k show thata a+l

H is a ß2-group. By Corollary 3.5 we may assume that each A e AFH has the

TEP in H. Recall A = HnX, X e Fc. If X e AFc such that XC\H has TEP in7/, then Xn// has TEP in C since // does. Therefore X/XnH = X + H/H

is a Butler group and we may assume that G admits a G(N1)-family ^ of

Butler groups, namely AFG = {X + H/H | X e AFC, \X\ < k} . By induction

hypothesis, all groups in AFQ are B2-groups. If k is singular, G is a B2-group

by Corollary 6.3. Thus we may assume that k is regular. Again, Theorem 7.1

and properties of AFG show the existence of a filtration G = \Ja<K Ga into B2-

groups Ga, \Ga\ < k and each Ga has the TEP in Ga+X . By Theorem 3.7(i),

each pure finite-rank subgroup of G is Butler since H is a B2-group. In the

same way as above for H, we utilize 4.3 and 3.7 to show that G is a B2-group.

References

[A] D. Arnold, Notes on Butler groups and balanced extensions, Boll. Un. Math. Ital. A (6) S

(1986), 175-184.

[AH] U. Albrecht and P. Hill, Butler groups of infinite rank and axiom 3, Czechoslovak Math. J.

37(112) (1987), 293-309.

[B] L. Bican, Splitting in mixed groups, Czechoslovak Math. J. 28 (1978), 356-364.

[BS] L. Bican and L. Salce, Infinite rank Butler groups, Proc. of Abelian Group Theory Con-

ference, Honolulu, Lecture Notes in Math., vol. 1006, Springer-Verlag, Berlin, 1983, pp.

171-189.

[BSS] L. Bican, L. Salce and J. Stepan, A characterization of countable Butler groups. Rend. Sem.

Mat. Univ. Padova 74 (1985), 51-58.License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use


[DR1] _, On torsion-free abelian k-groups, Proc. Amer. Math. Soc. 99 (1987), 403-408.

[DR2] _, Infinite rank Butler groups, Trans. Amer. Math. Soc. (to appear).

[D] M. Dugas, On some subgroups of infinite rank Butler groups, Rend. Sem. Mat. Univ. Padova

79(1988), 153-161.

[E] P. Eklof, Set theoretic methods in homological algebra and abelian group, Presses Univ.

Montreal, Montreal, 1980.

[EF] P. Eklof and L. Fuchs, Baer modules over valuation domains, Ann. Mat. Pura Appl. 150

(1988), 363-373.[Fu] L. Fuchs, Infinite abelian groups, vols. I and II, Academic Press, New York, 1977 and 1973.

[FH] L. Fuchs and P. Hill, The balanced-projective dimension of abelian p-groups, Trans. Amer.

Math. Soc. 293(1986), 99-112.

[HM] P. Hill andC. Megibben, Torsion-free groups, Trans. Amer. Math. Soc. 295 (1986), 735-751.

[H] W. Hodges, In singular cardinality, locally free algebras are free, Algebra Universalis 12

(1981), 205-220.

[Hu] R. Hunter, Balanced subgroups of abelian groups, Trans. Amer. Math. Soc. 215 (1976),

81-98.

[S] S. Shelah, A compactness theorem for singular cardinals, free algebras, Whilehead problem

and transversals, Israel J. Math. 21 (1975), 319-349.

Department of Mathematics, Baylor University, Waco, Texas 76793

Department of Mathematics, Auburn University, Auburn, Alabama 36849

Department of Mathematics, University of Colorado, Colorado Springs, Colorado

80933-7150


BUTLER GROUPS OF INFINITE RANK II...Volume 320, Number 2, August 1990 BUTLER GROUPS OF INFINITE RANK...

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