Paul C. Eklof, Matthew Foreman and Saharon Shelah- On Invariants for omega-1 Separable Groups

8/3/2019 Paul C. Eklof, Matthew Foreman and Saharon Shelah- On Invariants for omega-1 Separable Groups

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ON INVARIANTS FOR 1-SEPARABLE GROUPS

PAUL C. EKLOF, MATTHEW FOREMAN, AND SAHARON SHELAH

Abstract. We study the classification of1-separable groups us-ing Ehrenfeucht-Frasse games and prove a strong classification re-sult assuming PFA, and a strong non-structure theorem assuming.

Introduction

An 1-separable (or 1-separable) group is an abelian group suchthat every countable subset is contained in a free direct summand ofthe group. In particular, therefore, an 1-separable group is 1-free,i.e., every countable subgroup is free. The structure of 1-separablegroups of cardinality 1 was investigated in [1] and [8]; most of theresults proved there required set-theoretic assumptions beyond ZFC.(See also [2, Chap. VIII] for an exposition of these results.) Specifi-cally, assuming Martins Axiom (MA) plus CH or the stronger ProperForcing Axiom (PFA), one can prove nice structure and classification

results; these results are not theorems of ZFC since counterexamplesexist assuming CH or prediction principles like . In [1, Remark 3.3]it is asserted that a construction given there under the assumption ofCH (or even 20 < 21) of two non-isomorphic 1-separable groups

is strong evidence for the claim that in a model of CHthere is no possible meaningful classification of all 1-separable groups. It is difficult to see what conceivablescheme of classification could distinguish between [thegroups constructed here].

1991 Mathematics Subject Classification. Primary 03C55, 20K20; Secondary03E35, 03C75.

Key words and phrases. Ehrenfeucht-Frasse games, 1-separable abelian groups,almost free groups.

Thanks to Rutgers University for its support of this research through its fundingof the first and third authors visits to Rutgers.

The second author thanks the NSF, Grant No. DMS-9203726, for partial sup-port.

Partially supported by Basic Research Fund, Israeli Academy of Sciences. Pub.No. 520 .

1


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2 PAUL C. EKLOF, MATTHEW FOREMAN, AND SAHARON SHELAH

But, in fact, the Helsinki school of model theory provides a schemefor distinguishing between such groups. It is our aim here to use the

methodology of the Helsinki school which involves Ehrenfeucht-Frasse games (cf. [9], [11] or [12]) to strengthen the dichotomyreferred to above: that is, to obtain strong classification results assum-ing PFA, and a strong non-structure theorem assuming .

We begin by describing the Ehrenfeucht-Frasse (or EF) games, afterwhich we can state our results more precisely. If is an ordinal and Aand B are any structures, the game EF(A, B) is played between twoplayers and who take turns choosing elements of A B through rounds. Specifically, in each round picks first an element of eitherA or B; and then picks an element of the other structure. Theresult is, at the end, two sequences (a)


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ON INVARIANTS FOR 1-SEPARABLE GROUPS 3

EF(A, B; T) will end after countably many moves. We will say A isT-equivalent to B if A T B. This relation provides a possible way of

distinguishing between the 1-separable groups constructed in [1] underthe assumption of CH (cf. the remark after the quotation above).

By a theorem of Hyttinen [3], the entire class of bounded trees de-termines A up to isomorphism; that is, if A and B are of cardinality1 and A

T B for all bounded trees, then A is isomorphic to B. Thestructure of the class of bounded trees is much more complicated thanthat of the class of well-founded trees (cf. [12]). However, in contrastto the situation for countable structures, there is not always a singletree which suffices to describe A up to isomorphism. Specifically, Hyt-tinen and Tuuri [4] proved (assuming CH) that there is a linear orderA of cardinality

1such that for every bounded tree T there is a linear

order BT of cardinality 1 such that A T BT but A is not isomorphic

to BT. They call this result a non-structure theorem for A. It can betranslated in terms of infinitary languages and says that there is nocomplete description of A in a certain strong language M21 (whichwe shall not define here).

A similar non-structure theorem for p-groups was proved by Meklerand Oikkonen [10]; their theorem is proved by carrying over to p-groups,by means of a Hahn power construction, the result of Hyttinen andTuuri. Whether the analogous result for 1-free groups is a theoremof ZFC + CH remains open, but when we consider the question for

1-separable groups, we obtain an independence result, which is thesubject of this paper. In the first section we prove (with the help ofthe structural results referred to above) that assuming PFA

ifA andB are 1-separable groups of cardinality1 suchthatA

2+ B, then they are isomorphic (where 2+is the countable ordinal regarded as a linearly ordered tree).

See Theorem 6. Thus a single, simple, tree contains enough informationto classify any 1-separable group in the precise sense that a singlesentence of M21 of tree rank

2 + completely describes A.

In section 2 we show, assuming , that not only does 2

+ nothave the property above, but for any bounded tree T, there are non-isomorphic 1-separable groups A

T and BT of cardinality 1 whichcannot be separated by T, in the sense that AT T BT. (See Theorem7.) The construction in section 2 is strengthened in section 3 to obtaina non-structure theorem (Theorem 8.):

there is an 1-separable group A of cardinality 1 suchthat for every bounded tree T there is an 1-separable


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group BT of cardinality1 which is not isomorphic to Abut is T-equivalent to A.

(Note that A does not depend on T.)We shall make use, at times, of the following simple lemma, where

A denotes the dual of A, i.e. Hom(A,Z).

Lemma 1. Suppose A B and A B C where C/B is 1-free,B/A is countable and (B/A) = 0. If : B C such that [A] A,then [B] B.

Proof. induces a homomorphism: B/A C/A. By the hypotheses,the composition of this map with the canonical surjection: C/A C/B must be zero; that is, [B] B.

1. A structure theorem

An 1-separable group A of cardinality 1 is characterized by theproperty that it has a filtration, that is, a continuous chain {A : .

Ifd divides t(y) mod A+1 for some integer d and linear combinationt(y), then d divides t(y) a where a = +1( t(y)) =

S,+1(

t(y)) for some finite subset S rge().

Proposition 5. Let G and G be 1-separable groups such that G isin standard form. Suppose that they have filtrations {G : 1}and {G : 1} respectively such that the filtration of G attests thatG is in standard form and E = { 1 : G is not a summand ofG} = { 1 : G

is not a summand of G

}. Suppose also that forall limit ordinals , given a ladder on , there is an isomorphism : G+1 G

+1 such that for all n , [G(n)] = G

(n)and

[G(n)+1] = G

(n)+1. Then G and G are filtration-equivalent.

Proof. We can assume that the filtration of G is as in Lemma 4. Weprove by induction on the following:

if < and , 1 \ E and f : G G is a

level-preserving isomorphism, then f extends to a level-preserving isomorphism g : G G.

If = + 1 where / E, then the result follows easily by inductionand the fact that G/G and G

/G

are free. If is a limit ordinal,

choose a ladder on such that (0) > and for all n, (n) / E,


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and extend f successively, by induction, to gn : G(n) G

(n), and

let g = ngn.

The crucial case is when = + 1 where E. Let be as inLemma 4 with (0) > , and let be the corresponding isomorphismgiven by the hypothesis of this Proposition. Let C,n = Kn,n+1. Byinduction, extend f to a level-preserving isomorphism f0 : G(0) G(0) and then extend it to g0 : G(0)+1 G

(0)+1by letting g0

C,0 = C,0. Clearly g0 is level-preserving. By induction extend g0to a level-preserving f1 : G(1) G

(1)and then to g1 : G(1)+1

G(1)+1 by letting g1 C,1 = C,1. Continuing in this way we

obtain level-preserving isomorphisms gn : G(n)+1 G

(n)+1for each

n. Let g = ngn : G G.

By Lemma 4, G+1 is generated mod G by a set of elements of theform

t(y) a

d

where a nC,n; hence G+1 is generated mod G

by elements

t((y)) (a)

d.

But then since g(a) = (a) for each such a by construction, we canextend g to g : G+1 G

+1 by sending each y,i in y to (y,i). Since

y is linearly independent over G this is a well-defined homomorphism.

Theorem 6. Suppose A and A are 1-separable groups of cardinality1 and at least one of them is in standard form. If A and A

are2 + -equivalent, then they are filtration-equivalent.

Proof. We can suppose that A is in standard form, and that we havechosen a filtration, {A: 1} which attests to that fact. Moreover,we can assume that if E = { : A is not a direct summand of A},

then (A+1/A) = 0. (Use Steins Lemma [2, Exer. 3, p. 112], andreplace A+1 by a direct summand, if necessary.)

Since A is quotient-equivalent to A by Proposition 3 , we can assumethat there is a filtration {A: 1} of A

such that E = { : A isnot a direct summand of A} and for E, A+1/A

= A+1/A = 0,so in particular (A+1/A

) = 0.

Fix a bijection : (A\ A) (A \ A

) for each < . Let

= { : < < 1}.


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Whenever we talk about moves in a game, we refer to the gameEF2+(A, A

). Given a strictly increasing finite sequence of countable

ordinals 1 < 2 < ... < n, we will say that plays according to and 1, 2,...,n for the first n moves if the k + move of player is kk+1() for k = 0,...,n 1 and (where 0 = 0).

Suppose that is a w.s. for in the game EF2+(A, A). Let C

be the set of all < 1 such that for any integers n > 0 and m 0and any ordinals 1 < 2 < ... < n < , if plays according to and 1, 2,...,n for the first n moves and then plays any elementsof A for the next m moves, then the responses of using are all inA A

.

Then C is a cub: for the proof of unboundedness, note that there areonly countably many possibilities that one has to close under: choiceofn and m, choice of1 < 2 < ... < n, and choice of moves n,n +1, ...n + m 1. (The earlier moves are determined by the kk+1 andby .)

There is a continuous strictly increasing function h : 1 1 whoserange is C. Define h : 1 1 by

h() =

h() + 1 if is a successor and h() Eh() otherwise

Let G = Ah() and G = A

h(). Then for successor , G is a sum-mand of A and for limit G = Ah() =


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groups of cardinality 1 are isomorphic. Thus the result follows fromTheorem 6.

2. A diamond construction: one tree

The result to be proved in this section is the following:

Theorem 7. Assume . For any bounded tree T1 there exist non-isomorphic 1-separable groups G

0 and G1 of cardinality 1 which areT1-equivalent (and filtration-equivalent) and are both in standard form.

Proof. We will present the proof in layers of increasing detail.

(I) Fix a stationary subset E of 1 consisting of limit ordinals andsuch that E is the disjoint union of two uncountable subsets E0 andE1 such that (E1) holds.

Given a bounded tree T (which in practice will be determined by, butnot equal to, T1), we shall identify its nodes with countable ordinalsin such a way that if


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same height, and the partial ordering is defined coordinate-wise. (Asabove, we identify the nodes of T with ordinals.)

Suppose we are able to carry out the construction outlined above forthis T. Then since the G are free, G

is 1-free. Moreover, for / E1


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4. subsets W[] ofG0 for every non-empty finite subset

of which is an antichain in T, satisfying:

(a) for all < , W[] W[];(b) every element of W[] is of the form w,n for

some E0, and n .

The functions f0 will be required to satisfy:

(c) for all , j {0, 1} f0(x0,j) = x

1,j; moreover,

if w,n W+1[] and { : T } = , thenf0(w,n) = v

1,n.

For any finite antichain in T, let W[] =

W[].

Now we will outline how we do the construction so that G0 and

G1 are not isomorphic. Before we start, we choose a function withdomain E0 which maps onto the set of all -sequences n : n offinite subsets of T such that

n n is an antichain; we also require

that if () = n : n , then each n .

Suppose now that we have defined G for . If = E0,then G+1 will be defined as indicated above and is such that (as wewill prove)

(II.1) for all e {1, 1}, there is no isomorphism ofG0+1 with

n Zv

1,nC for any C, which for all n

takes w,n to ev1,n.

Moreover w,n will be put into W+1[n]. (This is the only way thatan element becomes a member of a W[].)

If = E1 and < , we introduce the notation A, = {t : t is


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but G1/G1+1 is 1-free by construction. If for any such (II.2) holds,then H G0+1 would extend h = H G

0, contradicting (II.3).

Since (II.2) fails, for all S and all < there exists e {1, 1}and a finite subset of A, such that H(w,n) = ev

1,n for all w,n

W[]. Now there is a cub C such that for all C, all e {1, 1},all < , and all finite subsets of A, , if H(w,n) = ev1,n for somew,n W[], then H(w,n) = ev1,n for some w,n W[]. Thus forall C S and all < , there exists e {1, 1} and a finite subset of A, such that H(w,n) = ev

1,n for all w,n W[]. Since C S

is uncountable, it follows easily that there exists e {1, 1}, and anuncountable set { : < 1} of pairwise disjoint finite antichains suchthat H(w,n) = ev

1,n for all w,n W[] for all < 1. Since T has

no uncountable branches, by a standard argument (see, for example, [5,Lemma 24.2, p. 245]), there is a countably infinite subset {n : n }of 1 such that

{n : n } is an antichain. There exists E0

such that () = n : n . Now H G0+1 is such that for all

n , H(w,n) = ev1,n since w,n W+1[n]; this contradicts (II.1),since

n Zv

1,n is a direct summand of G

1+1, and hence of G

1 (by2).

(III) The next step is to describe in detail the recursive constructionof the data satisfying the properties 1, 2, 3 and 4, as well as (II.1) and(II.3). So assume that we have defined G, and W[] for < and

f for + 1 < .

There are several cases to consider.

Case 1: is a limit ordinal. We let G =


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Case 2: = + 1 for some / E. As described above, define

G

= G

Zx,0

Zx,1

.

Let W[] = W[] for every (= if is not a subset of ).Assuming (III.1), we have f0 as desired.

Case 3: = + 1, where E0. In this case, as stated before,

G = G n

Zu,n Zv,n

and recall that w,n is defined to be 2u0,n+1 u

0,n. Say () =

n : n . Define

W[] =

W[] {w,n} if =

nW[] otherwise.

Assuming (III.1) (with = ), we can define f0 . Now let us see why(II.1) holds. Suppose to the contrary that there is an isomorphism H :G0 G1 contradicting (II.1). Now

n Zv

1,n is a direct summand of

G1 and hence (by 2) a direct summand ofG1. Thus H1[

n Zv

1,n] is

a direct summand ofG0. But by assumption on H, H1[

n Zv1,n] =

n Zw,n and the latter is not a direct summand of G0 because the

coset ofu0,0 is a non-zero element ofG0/

n Zw,n which is divisibleby all power of 2 by definition of the w,n.

Case 4: = + 1, where E1. If (II.2) fails, let G+1 = G

.

Otherwise, let be as in (II.2). We introduce some ad hoc notation.For any finite subset ofA, , let f be the function whose domain isthe subgroup generated by {x0,j : / E, < , j {0, 1}} W[]such that f(x

0,j) = x

1,j and f(w,n) = v

1,n. Notice that for all u

dom(f) and all , if T and u dom(f0 ), then f(u) = f

0 (u)

by 4(c). Let ,n be as before (finite subsets forming a chain whoseunion is A,); for short, let n =

,n . We claim that:

(III.2) given m, m Z \ {0}, n , y G1 , for suffi-

ciently large < there exists k0

dom(fn) G0+2

such that k0 is pure-independent mod G0+1 and is such

that mh(k0) = mfn(k0) + y. Moreover, fn(k

0) ispure-independent mod G1+1.

Supposing this is true we will prove it in part (IV) let us defineG+1. Fix a ladder on . Also, enumerate in an -sequence all triplesr,d,v where r , d Z \ {0}, and g G1 so that the nth tripler,d,g satisfies n > r. By (III.2) we can inductively define primes


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pn, ordinals n (n), and elements k0,n dom(fn) G0n+2

pure-

independent over G0n+1 such that (if the nth triple is r,d,g), pn does

not divide mh(k0,n) mfn(k0,n) y where

m =n1

i=0 pim = d

n1i=r pi

y =n

j=0(j1

i=0 pi)h(k0,j) + g d

nj=r(j1

i=r pi)fj (k0,j).

(Note that since G1 is free, every non-zero element is divisible by onlyfinitely many primes, so we can take pn to be any sufficiently largeprime.) Then we let G0+1 be generated by G

0 {z

0,n : n } modulo

the relationspnz

0,n+1 = z

0,n + k

0,n

and G1+1 is defined similarly, except that we impose the relations

pnz1,n+1 = z

1,n + fn(k

0,n).

We need to show that h does not extend to a homomorphism: G0+1 G1+1. If it does, then h(z

0,0) = dz

1,r + g for some r , d Z \ {0},

and g G1. Let n be such that r,d,g is the nth triple in the list.Now, in G0+1 we have

(ni=0

pi)z0,n+1 = z

0,0 +

nj=0

(

j1i=0

pi)k0,j

so, applying h, we conclude that pn divides

dz1,r + g +n

j=0

(

j1i=0

pi)h(k0,j).

On the other hand, in G1+1 we have pn divides

dz1,r + d

nj=r

(

j1i=r

pi)fj(k0,j)

so, subtracting, we obtain a contradiction since pn divides mh(k0,n)

mf0n

(k0,n

) y, where m, m, and y are as above.We let W+1[] = W[] for any subset of (and = if ).

By (III.1) we can define f0 .

(IV) In this layer we will prove (III.1) and (III.2).First let us prove (III.2) since for the purposes of proving (III.1) we

will need more information about the nature of the elements k0,n. Fix


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m, m, n , y , as in (III.2); there are several cases. In the first two caseswe can use any < .

Case (i): y = 0. If neither x0+1,0 nor x0+1,1 will serve for k0, thenx0+1,0 x

0+1,1 will.

Case (ii): y = 0, m = m. Let k0 = x0+1,0. Then by construction,k0 generates a cyclic summand of G0 ; hence fn(k

0) and h(k0) bothgenerate cyclic summands of G1. Hence mh(k

0) = mf0n(k0).

Case (iii): y = 0, m = m. Pick sufficiently large so that thereexists w,j G0+1 W[n] such that fn(w,j) = h(w,j). If x

0+1,0

will not serve for k0 (i.e., h(x0+1,0) = x1+1,0), then we can take k

0 to

be x0+1,0 + w,j.

Case (iv): y = 0, m = m. Similarly k0 can be taken to be of theform x0+1,0 or x

0+1,0 w,j where fn(w,j) = h(w,j).

Now if we examine the construction in Case 4 of (III) and the proofabove we see that

(IV.1) each k0,n can be (and will be) taken to be of the

form x0n,jn ,n where ,n is 0, x0,j or w,j for some

, j.

We will say that w,j is a part of k0,n in case ,n is w,j.

Before beginning the proof of (III.1), let us observe the followingfacts:

(IV.2) Given E0 and N , there is an isomor-phism g :

n Zu

0,n Zv

0,n

n Zu

1,n Zv

1,n

such that g(w,n) = v1,n for n N and g

(u0,n) = u1,n

for n N + 1.

Indeed, we can define g(u0,n) = 2g(u0,n+1) v

1,n for n N (and the

other values appropriately).

(IV.3) Given an isomorphism g : G0

G1

where E1,we can extend g to an isomorphism g : G0+1 G1+1

provided that (using the notation of Case 4) g(k0,n) =

fn(k0,n) for almost all n .

Indeed, ifg(k0,n) = fn(k0,n) for all n N, we can define g

(z0,n) = z1,n

for n N and g(z0,n) = png(z0,n+1)g(k

0,n) for n < N by downward

induction. We will apply (IV.3) to the situation of (III.1), with g =gB, = , + 1 = ; if we are in Case 4, then the hypothesis on g in


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(IV.3) will hold if there exists t B such that t (where is as inCase 4).

We return to the notation of (III.1). Let = sup{t + 1 : t B};then domgB = G

0. Assume first that = . In case G

+1/G

is free

there is no problem extending gB; in the other case = E1 and bythe remarks above we can extend gB since there exists t B such thatt (since sup B = ).

We are left with the case when < . We will first define an exten-sion of gB to a partial isomorphism gB whose domain is

dom(gB) +

{x0,j : , j = 0, 1}{u0,n : E0 + 1, n }{v0,n : E0 + 1, n }

Notice that every k0,n for , n belongs to the domain of gB.We let gB(x

0,j) = x

1,j for all , j. By enumerating in an -sequence the

set (E0 E1) ( + 1) we can define by recursion the values gB(u0,n)

and gB(v0,n) so that:

gB(w,n) = v1,n whenever w,n W+1[] for some withB = ;

for all E0 with , for almost all n , gB(u0,n) =u1,n; and

for all E1 with , for almost all n , if (for some, m) w,m is a part ofk

0

,n

, then gB(w,m) = v1

,m

.

The first condition is required by 4(c). In view of (IV.2), there isno conflict between the first two conditions because for any E0,

n n is an antichain, so there is at most one n such that

nB = .

To be sure that the third condition can indeed be satisfied, we needto consider the case that for some E1, there are infinitely manyn such that there exists wn,mn which is a part of k

0,n and belongs to

the domain of gB . Say this is the case for n belonging to the (infinite)set Y (for a fixed ). Then for each n Y tn B such thattn n. Suppose that the construction of G+1 uses A, = n

,n .

Selecting one n Y, we see that since ,n n , n > and

hence tn A, . Therefore there exists M such that for all n M,tn

,n . But then, for n Y with n M, tn n

,n , so

tn tn and thus tn T tn. By the construction in Case 4 and by4(c), gB(wn,mn) = v

1n,mn

for n Y, n M. Moreover, there is noconflict between the last two conditions because, by construction, if E1 and E0, then w,m W[

,n ] if and only if

,n =

m, but

the elements of {m : m } are disjoint and the ,n form a chain

under .


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It remains to extend gB to f0 by defining f

0(z

0,n) for , n

. This is possible by observation (IV.3) because of the construction

of gB.

(V) We will define the projections , by induction on and thenverify the conditions to be in standard form (see section 1 or [2, Def.1.9(ii), p. 257]). We refer to the cases of the construction in part (III).In Case 1, we take unions. In Case 2, for < +1 we let ,+1 be the

extension of, which sends each x,j to 0 . (Here,

, is the identity.)

In Case 3, for we let ,+1 be the extension of , which sends

each u,n and each v,n to 0.

Finally, for Case 4, we use the notation of that case. We define

0,+1(z0,n) =

mj=n dn,jk0,j where m is maximal such that m+2

and dn,j =j1

i=n pi (and dn,0 = 1 ). (Compare [2, pp. 249f].) Thedefinition of 1,+1 is similar, replacing k

0,j by fj (k

0,j). Let Y

=

{z,n : n }. Then we can easily verify the conditions of [2, Def.1.9(ii), p. 257] using the information in the proof of (III.2) about theform ofk0,j.

This completes the proof of Theorem 7.

3. A non-structure theorem

Our goal is to generalize the construction in the previous section toprove:

Theorem 8. Assume . There exists an1-separable group G0 and foreach bounded tree T1 an 1-separable group G

T1 which is T1-equivalentto G0 but not isomorphic to G0. Moreover, all the groups are of cardi-nality 1 and in standard form.

Proof. We assume familiarity with the previous proof and outline themodifications, in layers of increasing detail.

(VI) Fix a stationary subset E of 1 consisting of limit ordinals

(> 0) and such that E is the disjoint union of two subsets E0 and E1such that cardinality (E0) and (E1) hold. ((E0) is not essential,but convenient.)

We need only consider bounded trees T on 1 such that if


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By induction on {0} E we will define the following data:

(1) continuous chains {G : + 1} of countable free groupssuch that for all < + 1, G/G is free if / E1, and if E1, then G

+1/G

has rank at most 1.

(2) projections , : G G

for + 1 and / E1

such that: for < , , ,; and for < ,

, , =

, ;

(3) for each E and each an isomorphism f : G0+1

G+1 satisfying:if 1


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(a) for all < , W[] W[];

(b) every element of W[] is of the form w,n for

some n and some E0 such that T = T.

The functions f will be required to satisfy (as before):

(c) for all , j {0, 1} f(x0,j) = x

1,j; moreover,

if w,n W+1[] and { : } = , thenf(w,n) = v

1,n.

Moreover, in order to carry out the inductive construction we will alsorequire the following for all E, :

(d) if E0 with + 1 and T = T, thenf(u

0,n) = u

1,n for all n ;

(e) for all pairs 1, 2 with sup{t : t


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pnz0,n+1 = z

0,n + k

0,n (which keep h from extending) and G

+1 is gen-

erated by G {z,n : n } subject to relations pnz

,n+1 = z

,n + k

,n

(where k,n = fn(k0,n)).

For the purposes of later stages of the construction we also define,for any 1 > such that 1 E and T1 = T, elements k

1,n G

1 .

We know that k0,n has the form x0n,jn

,n where ,n is either 0, x0,j,

or w,j for some , j (cf. (IV.1)). In case ,n is 0, let k1,n = x

1n,jn

;

in case ,n = x0,j , let k

1,n = x

1n,jn

x1,j. Finally, if ,n = w,j, let

k1,n = x1n,jn

,n where

,n = w1,j if T1 = Tv1,j if T1 = T

and w1,j = 2u1,j+1 u

1,j. We will be able to show (in the next section)

the following:

(VII.1) for any branch B in T1 with = sup{t + 1 :t B}, gB = {f1 : B} is such that for almost alln, gB(k

0,n) = k

1,n.

(This is evidence of what, in view of (IV.3), will enable us to extendfunctions.)

Case 2: < . We need to define G for + 1 + 1 byinduction on . If we have defined G for < , and does notbelong to E

1, we follow the prescription in (**) or (****). If E

1,

then T = T (by definition of). By induction G0+1 is constructedas in Case 1 and we have k,n as there (with playing the role of

1 and playing the role of ). In particular, G0+1 is generated by

G0 {z0,n : n } subject to relations pnz

0,n+1 = z

0,n + k

0,n. We

define G+1 to be generated by G {z

,n : n } subject to relations

pnz,n+1 = z

,n + k

,n. Finally, we define G

+1 as in Case 1.

The definition of the W[] will be as in (III); specifically, W+1[] =

W[] unless = E0, T = T and = n for some n, in

which case W

+1[] = W

[] {w,n}.

(VIII) We have defined the groups and the sets W[]; the last stepis to show that the partial isomorphisms f can be defined satisfyingthe conditions in 4.

First let us verify (VII.1). Let and 1 be as in Case 1 of (VII) andsuppose B is a branch in T1 with = sup{t + 1 : t B}. Then gBis an isomorphism : G0 G

1 and we want to show that gB(k

0,n) = k

1,n


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for almost all n. Recall that k0,n has the form x0n,jn

,n where ,nis either 0, x0,j, or w,j for some , j; the only case we need to worry

about is when ,n = w,j.Let = sup{ < : T = T1 }; so < and G

1 = G

for

. Suppose that G0+1 and G+1 are defined using A

, =

n

,n

as in Case 1 of (VII) and Case 4 of (III). We consider several cases.First, suppose that there exists t A, with t . Then for almost

all n, t ,n and thus if w,j W [

,n ] then > t ; hence

T1 = T and it follows from 4(d) that gB(k0,n) = k

1,n. If this

case does not hold then A, so A, = A

, is an antichain in

T1 = T . If there exists t B with t < , then thereexists t B with t A, and hence t

,n for almost all n; it follows

easily that for almost all n gB(k0,n) = k1,n (considering separately thecases when and > ). In the remaining case, if = inf{t B : t }, then so we have sup{t : t . This completes the proof of (VII.1).

Now we need to verify the analog of (III.1). Letting and beas in (VII), we need to define f for . Fix and letB = {t < : t


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(e) for all pairs 1, 2 with sup B 1 < 2 ,it is the case for almost all n that for all w,m

W+1[1,2n ] we have gB(w,m) = v1,m.

Now gB(k0,n) is defined for all E1 with . We need to

define f(z,n) for all such sup B. In view of (IV.3), we can do

this provided that gB(k0,n) = k

,n for almost all n . We consider

separately the cases: T = T; and T = T. The first case is asin (IV); the last is as in the proof of (VII.1) (with playing the role of1, playing the role of and using (d

) and (e)).This completes the proof of Theorem 8.

References[1] P. Eklof, The structure of 1-separable groups, Trans. Amer. Math. Soc. 279

(1983), 497523.[2] P. C. Eklof and A. H. Mekler, Almost Free Modules, North-Holland (1990).[3] T. Hyttinen, Model theory for infinite quantifier languages, Fund. Math. 134

(1990), 125142.[4] T. Hyttinen and H. Tuuri, Constructing strongly equivalent nonisomorphic

models for unstable theories, Ann. Pure Appl. Logic 52 (1991), 203248.[5] T. Jech, Set Theory, Academic Press (1978).[6] M. Karttunen, Model theory for infinitely deep languages, Ann. Acad. Sci.

Fennicae Series A I: Math. Diss. 50 (1984), 897-908.[7] A. H. Mekler, Proper forcing and abelian groups, in Abelian Group Theory,

Lecture Notes in Mathematics No. 1006, Springer-Verlag (1983), 285303.[8] A. H. Mekler, The structure of groups that are almost the direct sum of count-

able abelian groups, Trans. Amer. Math. Soc. 303 (1987), 145160.[9] A. Mekler, S. Shelah and J. Vaananen, The Enrenfeucht-Frasse game of length

1, Trans. Amer. Math. Soc. 339 (1993), 567580.[10] A. Mekler and J. Oikkonen, Abelianp-groups with no invariants, J. Pure Appl.

Algebra 87 (1993), 5159.[11] J. Oikkonen, Enrenfeucht-Frasse-games and nonstructure theorems, preprint.[12] J. Vaananen, Games and trees in infinitary logic: a survey, to appear in Quan-

tifiers (ed. by M. Krynicki, M. Mostowski and L. Szczerba), Kluwer Acad.Publ.

University of California, Irvine

University of California, Irvine

Hebrew University and Rutgers University

Documents

Paul C. Eklof, Matthew Foreman and Saharon Shelah- On Invariants for omega-1 Separable Groups