aa r X i v : . [ m a t h . L O ] N ov BASIC SUBGROUPS AND FREENESS,A COUNTEREXAMPLE
ANDREAS BLASS AND SAHARON SHELAH
Abstract.
We construct a non-free but ℵ -separable, torsion-freeabelian group G with a pure free subgroup B such that all sub-groups of G disjoint from B are free and such that G/B is divisible.This answers a question of Irwin and shows that a theorem of Blassand Irwin cannot be strengthened so as to give an exact analog fortorsion-free groups of a result proved for p -groups by Benabdallahand Irwin. Introduction
All groups in this paper are abelian and, except for some motivatingremarks about p -groups in this introduction, all groups are torsion-free.A subgroup B of a group G is basic in G if • B is a direct sum of cyclic groups, • B is a pure subgroup of G , and • G/B is divisible.Of course in the torsion-free case, “a direct sum of cyclic groups” canbe shortened to “free.”Benabdallah and Irwin proved in [1] the following result:
Theorem 1.
Suppose G is a p -group with no elements of infiniteheight. Suppose further that G has a basic subgroup B such that everysubgroup of G disjoint from B is a direct sum of cyclic groups. Then G itself is a direct sum of cyclic groups. “Disjoint” means that the intersection is (0), not ∅ , as the latter isimpossible for subgroups.Later, Irwin asked whether an analogous theorem holds for torsion-free groups. The following partial affirmative answer was given in [2].Note that, unlike p -groups, torsion-free groups need not have basicsubgroups. Theorem 2.
Suppose G is a torsion-free group such that • G has a basic subgroup of infinite rank, and • for every basic subgroup B of G , all subgroups of G disjointfrom B are free.Then G is free. This result is weaker in two ways than the hoped-for analog of The-orem 1. First, not only must there be a basic subgroup, but it musthave infinite rank. (It was shown in [3] that all basic subgroups of atorsion-free group have the same rank.) Second, the assumption thatall subgroups disjoint from B are free is needed not just for one basicsubgroup B but for all of them.The assumption that a basic subgroup has infinite rank is needed.As was pointed out in [2], Fuchs and Loonstra constructed in [5] atorsion-free group of rank 2 such that every subgroup of rank 1 is freeand every torsion-free quotient of rank 1 is divisible. In such a group G , every pure subgroup B of rank 1 is basic, every subgroup disjointfrom B has rank at most 1 and is therefore free, yet G is certainly notfree.It has remained an open question until now whether the second weak-ness of Theorem 2 can be removed. Can “for every basic subgroup” bereplaced with “for some basic subgroup” in the second hypothesis? Inthis paper, we answer this question negatively. Theorem 3.
There exists an ℵ -separable torsion-free group G of size ℵ with a basic subgroup B of rank ℵ such that all subgroups of G disjoint from B are free but G itself is not free. The rest of this paper is devoted to the proof of this theorem. Thegroup G and the subgroup B will be constructed in Section 2 and theclaimed properties will be proved in Section 3.The proof will show a little more than is stated in the theorem.We can arrange for the Gamma invariant Γ( G ) to be any prescribednon-zero element of the Boolean algebra P ( ℵ ) /N S of subsets of ℵ modulo non-stationary subsets. (See [4, Section IV.1] for the definitionand basic properties of Γ.)2. Construction
Our construction is somewhat similar to the construction of ℵ -separable groups in [4, Section VIII.1]. We shall, however, presentour result in detail, not presupposing familiarity with the cited con-struction from [4]. We begin by fixing notations for a set-theoreticingredient and a group-theoretic ingredient of our construction. ASIC SUBGROUPS AND FREENESS, A COUNTEREXAMPLE 3
Notation 4.
Fix a set S of countable limit ordinals such that S isstationary in ℵ . Also fix, for each δ ∈ S , a strictly increasing sequence h η ( δ, n ) : n ∈ ω i with limit δ .The equivalence class of S in P ( ℵ ) /N S will be the Gamma in-variant of the group G that we construct. Since the countable limitordinals form a closed unbounded subset of ℵ , every non-zero elementof P ( ℵ ) /N S is the equivalence class of an S as in Notation 4 and cantherefore occur as Γ( G ) in Theorem 3. Notation 5.
Fix a torsion-free group E of rank 2 such that all rank1 subgroups are free and all torsion-free rank-1 quotients are divisible.Such a group exists by [5, Lemma 2]. Also fix a pure subgroup of E of rank 1 and, since it is free, fix a generator a for it. Since E/ h a i is atorsion-free rank-1 quotient of E , it is divisible and thus isomorphic to Q . Fix an isomorphism ϕ from Q to E/ h a i and fix, for each positiveinteger n , a representative b n ∈ E of ϕ (1 /n !). Since ϕ (cid:18) n ! (cid:19) = ( n + 1) ϕ (cid:18) n + 1)! (cid:19) , there are (unique) integers q n such that b n = ( n + 1) b n +1 + q n a for all n . Fix this notation q n for the rest of the paper. Remark . We shall not need the full strength of the conditions on E .Specifically, we need divisibility only for E/ h a i , not for all the othertorsion-free rank-1 quotients of E . Lemma 7.
The generators a and b n for n ∈ ω and the relations b n =( n + 1) b n +1 + q n a constitute a presentation of E .Proof. Since Q is generated by the elements 1 /n !, E/ h a i is generatedby the images [ b n ] of the elements b n . Therefore E is generated bythese elements together with a .It remains to show that every relation between these generators thatholds in E is a consequence of the specified relations b n = ( n + 1) b n +1 + q n a . Consider an arbitrary relation ca + P n ∈ F d n b n = 0 that holds in E ; here F is a finite subset of ω and c and the d n ’s are integers.The given relations b n = ( n + 1) b n +1 + q n a allow us to eliminateany b n in favor of b n +1 at the cost of changing the coefficient of a . So,at a similar cost, we can replace any b n with a multiple of b m for anydesired m > n . Thus, we can arrange to have only a single b n occurring;that is, the relation under consideration can, via the given relations,be converted to the form c ′ a + d ′ b n = 0. ANDREAS BLASS AND SAHARON SHELAH
Since this relation holds in E , we have d ′ [ b n ] = 0 in E/ h a i . But E/ h a i is torsion-free and [ b n ] = ϕ (1 /n !) is non-zero. So d ′ = 0 andour relation is simply c ′ a = 0. Since h a i is torsion-free, c ′ = 0. Thus,the given relations b n = ( n + 1) b n +1 + q n a have reduced our original ca + P n ∈ F d n b n = 0 to 0 = 0. Equivalently, ca + P n ∈ F d n b n = 0 is aconsequence of the given relations. (cid:3) We are now ready to define the group G and subgroup B requiredin Theorem 3. Definition 8. G is the group generated by symbols x α for all α < ℵ and y δ,n for all δ ∈ S and n ∈ ω , subject to the defining relations, onefor each δ ∈ S and n ∈ ω , y δ,n = ( n + 1) y δ,n +1 + q n x δ + x η ( δ,n ) .B is the subgroup of G generated by all of the x α ’s.We shall sometimes have to discuss formal words in the generatorsof G , i.e., elements of the free group on the x α ’s and y δ,n ’s withoutthe defining relations above. We shall call such formal words expres-sions and we say that an expression denotes its image in G , i.e., itsequivalence class modulo the defining relations. We call two expres-sions equivalent if they denote the same element, i.e., if one can beconverted into the other by applying the defining relations.We shall sometimes refer to the defining relation y δ,n = ( n +1) y δ,n +1 + q n x δ + x η ( δ,n ) as the defining relation for δ and n ; when n varies but δ is fixed, we shall also refer to a defining relation for δ .Given an expression that contains y δ,n for a certain δ and n , we caneliminate this y δ,n in favor of y δ,n +1 by applying the defining relationfor δ and n . In the resulting equivalent expression, the coefficient ofthe newly produced y δ,n +1 will be n + 1 times the original coefficientof y δ,n , and a couple of x terms, namely that original coefficient times q n x δ + x η ( δ,n ) , are introduced as well. We shall refer to this manipulationof expressions as “raising the subscript n of y δ,n to n + 1,” and weshall refer to the introduced x terms as being “spun off” in the raisingprocess.By repeating this process, we can raise the subscript n of y δ,n toany desired m > n . If the original y δ,n had coefficient c , then thenewly produced y δ,m will have coefficient c · m ! /n !. There will also bespun off terms, namely x δ with coefficient c P m − k = n k ! n ! q k , and x η ( δ,k ) withcoefficient c k ! n ! for each k in the range n ≤ k < m . ASIC SUBGROUPS AND FREENESS, A COUNTEREXAMPLE 5 Proofs
In this section, we verify the properties of G and B claimed in The-orem 3.3.1. B is free. We show that the generators x α of B are linearly in-dependent, by showing that no nontrivial linear combination of thedefining relations can involve only x ’s without any y ’s. In fact, weshow somewhat more, because it will be useful later. Lemma 9. If x α occurs in a linear combination of defining relations,then so does y δ,n for some δ ≥ α and some n . Furthermore, if y δ,n occurs in a linear combination of defining relations, then so does y δ,m for at least one m = n (and the same δ ).Proof. For the first statement, consider a linear combination of definingrelations in which x α occurs, and consider one of the defining relations,say y δ,n = ( n + 1) y δ,n +1 + q n x δ + x η ( δ,n ) , used in this linear combinationand containing x α . So either α = δ or α = η ( δ, n ). In either case δ ≥ α .Fix this δ and consider all the defining relations for this δ that are usedin the given linear combination. If they are the defining relations for δ and n < · · · < n k , then the y δ,n from the first of these relations is notin any of the others, so it cannot be canceled and therefore occurs inthe linear combination.For the second statement, again suppose that the linear combinationinvolves the defining relations for δ and n < · · · < n k (perhaps alongwith defining relations for other ordinals δ ′ = δ ). As above, the y δ,n from the first of these cannot be canceled. Neither can the y δ,n k +1 fromthe last. So at least these two y δ,n ’s occur in the linear combination. (cid:3) G/B is divisible and torsion-free.
We get a presentation of
G/B from the defining presentation of G by adjoining the relations x α = 0 for all the generators x α of B . The resulting presentationamounts to having generators y δ,n for all δ ∈ S and all n ∈ ω withrelations y δ,n = ( n + 1) y δ,n +1 . For any fixed δ ∈ S , the generators and relations with δ in the sub-scripts are a presentation of Q , with y δ,n corresponding to 1 /n !. With δ varying over S , therefore, we have a presentation of L δ ∈ S Q , a torsion-free, divisible group. Corollary 10. G is a torsion-free group, and B is a basic subgroup.Proof. Since both the subgroup B and the quotient G/B are torsion-free, so is G . B is pure in G because G/B is torsion-free. Since B isfree and G/B is divisible, B is basic. (cid:3) ANDREAS BLASS AND SAHARON SHELAH G is ℵ -free. To prove that G is ℵ -free, i.e., that all its count-able subgroups are free, we use Pontryagin’s criterion [4, Theorem 2.3].We must show that every finite subset of G is included in a finitelygenerated pure subgroup of G .Let F be an arbitrary finite subset of G , and provisionally choose, foreach element of F , an expression denoting it. (“Provisionally” meansthat we shall modify these choices several times during the followingargument. The first modification comes immediately.) Raising sub-scripts on the y ’s, we may assume that, for each δ , there is at mostone m such that y δ,m occurs in the chosen expressions. In fact, withfurther raising if necessary, we may and do assume that it is the same m , which we name m , for all δ . Notice that, although there is stillsome freedom in choosing the expressions (for example, we could raisethe subscript m further), there is no ambiguity as to the set ∆ of δ ’sthat occur as the first subscripts of y ’s in our expressions. Indeed, if δ occurs exactly once in one expression but doesn’t occur in anotherexpression, then, according to the second part of Lemma 9, these twoexpressions cannot be equivalent.Let us say that an ordinal α is used in our (current) provisionalexpressions if either it is in ∆ or x α occurs in one of these expressions.(In other words, α occurs either as a subscript on an x or as the firstsubscript on a y .) Of course, only finitely many ordinals are used. So,by raising subscripts again from m to a suitable m , we can assumethat, if δ ∈ ∆ and if α < δ was used (before the current raising), then α < η ( δ, m ).We would prefer to omit the phrase “before the current raising,” butthis needs some more work. The problem is that the raising processspins off x ’s whose subscripts may not have been used before but areused after the raising. We analyze this situation, with the intention ofcorrecting it by a further raising of subscripts. The problem is that,in raising the subscript from m to m for y δ,m , we spin off x δ and x η ( δ,k ) for certain k , namely those in the range m ≤ k < m , and thesubscript used here ( δ or η ( δ, k )) may be < δ ′ but ≥ η ( δ ′ , m ) for some δ ′ ∈ ∆.The problem cannot arise from x δ . That is, we will not have η ( δ ′ , m ) ≤ δ < δ ′ . This is because m was chosen so that (among other things),when δ, δ ′ ∈ ∆ and δ < δ ′ , then δ < η ( δ ′ , m ).So the problem can only be that η ( δ ′ , m ) ≤ η ( δ, k ) < δ ′ . Here wecannot have δ = δ ′ because η ( δ, n ) is a strictly increasing function of n and k < m . Nor can we have δ < δ ′ , for then we would have η ( δ, k ) < δ < η ( δ ′ , m ) by our choice of m . So we must have δ ′ < δ . ASIC SUBGROUPS AND FREENESS, A COUNTEREXAMPLE 7
Unfortunately, this situation cannot be excluded, so one further mod-ification of our provisional expressions is needed. We raise the subscriptfrom m to an m so large that, whenever η ( δ, k ) < δ ′ < δ with k < m and δ, δ ′ ∈ ∆, then η ( δ ′ , m ) > η ( δ, k ).This raising from m to m solves the problem under consideration,but one might fear that it introduces a new problem, just like theold one but higher up. That is, the latest raising spins off new x ’s,so some new ordinals get used. Could they be below some δ ′ ∈ ∆but ≥ η ( δ ′ , m )? Fortunately not. To see this, repeat the precedingdiscussion, now with m in place of m , and notice in addition that thenewly spun off x η ( δ,k ) will have m ≤ k < m . As before, the problemcan only be that η ( δ ′ , m ) ≤ η ( δ, k ) < δ ′ with δ ′ < δ . But now thisis impossible, since δ ′ < δ implies δ ′ < η ( δ, m ) ≤ η ( δ, k ), thanks toour choice of m and the monotonicity of η with respect to its secondargument.Rearranging the preceding argument slightly, we obtain the followingadditional information. Lemma 11.
With notation as above, it never happens that δ, δ ′ ∈ ∆ and k < m and η ( δ ′ , m ) ≤ η ( δ, k ) < δ ′ .Proof. Suppose we had δ , δ ′ , and k violating the lemma. We considerseveral cases.If δ = δ ′ then the suppositions η ( δ ′ , m ) ≤ η ( δ, k ) and k < m violatethe monotonicity of η with respect to the second argument.If δ < δ ′ , then η ( δ, k ) < δ < η ( δ ′ , m ) (in fact even with m in placeof m ), contrary to the supposition.If δ ′ < δ and k < m then our choice of m ensures that η ( δ ′ , m ) >η ( δ, k ), contrary to the supposition.Finally, if δ ′ < δ and k ≥ m then δ ′ < η ( δ, m ) ≤ η ( δ, k ), againcontrary to the supposition. (cid:3) What we have achieved by all this raising of subscripts can be sum-marized as follows, where ∆ and “used” refer to the final version of ourexpressions. (Actually, the raising process doesn’t change ∆, but itusually changes what is used.) We have an expression for each elementof F . There is a fixed integer m (previously called m ) such that theonly y ’s occurring in any of these expressions are y δ,m for δ ∈ ∆. If δ ∈ ∆ and α is used and α < δ , then α < η ( δ, m ). Furthermore, by thelemma, if δ, δ ′ ∈ ∆ and k < m and η ( δ, k ) < δ ′ then η ( δ, k ) < η ( δ ′ , m ).These expressions for the members of F will remain fixed from nowon. Thus, the meanings of ∆ and “used” will also remain unchanged.Also, m will no longer change. ANDREAS BLASS AND SAHARON SHELAH
Let M be the set of • all the x ’s and y ’s occurring in the (final) expressions for ele-ments of F , • x δ for all δ ∈ ∆, and • x η ( δ,k ) for all δ ∈ ∆ and all k < m .Clearly, M is a finite subset of G and the subgroup h M i that it gen-erates includes F . To finish verifying Pontryagin’s criterion, we mustshow that h M i is pure in G .We point out for future reference that the only y ’s in M are y δ,m forthe one fixed m and for δ ∈ ∆.Suppose, toward a contradiction, that h M i is not pure, so thereexist an integer r ≥ g ∈ G such that rg ∈ h M i but g / ∈ h M i . Choose an expression ˆ g for g in which (by raising subscriptsif necessary) no two y ’s occur with the same first subscript δ . In fact,arrange (by further raising) that the second subscript on all y ’s in ˆ g isthe same n , independent of δ . Also choose an expression ˆ d for rg whereˆ d is a linear combination of elements of M . We may suppose that ˆ d isminimal in the sense that the number of elements of M occurring in ˆ d is as small as possible, for any r , g , and ˆ d as above.Consider any y δ,n that occurs in ˆ g . According to Lemma 9, we musthave δ ∈ ∆, because the difference r ˆ g − ˆ d is a linear combination ofdefining relations.If n ≤ m , then we can raise the subscript n to m in ˆ g , obtaininga new expression ˆ g ′ for the same element g . Since r ˆ g ′ − ˆ d is a linearcombination of defining relations and since it no longer contains y δ,k forany k = m (and the same δ ), we can apply Lemma 9 again to concludethat y δ,m has the same coefficient in r ˆ g ′ and in ˆ d . So, if we delete theterms involving y δ,m from both ˆ g ′ and ˆ d , we get another counterexampleto purity with fewer elements of M occurring in ˆ d . This contradictsthe minimality of ˆ d .We therefore have n > m . Now consider what happens in ˆ d if we raisethe subscripts of all the y δ,m terms to n . Call the resulting expressionˆ d ′ . (Note that ˆ d ′ will no longer be a combination of the generators listedfor M .) The same argument as in the preceding paragraph shows thateach y δ,n has the same coefficient in r ˆ g and ˆ d ′ . Therefore, if we removeall the y terms from both ˆ g and ˆ d ′ , obtaining ˆ g − and ˆ d − , then r ˆ g − andˆ d − denote the same element in G . But we saw earlier that the x ’s arelinearly independent in G , so r ˆ g − and ˆ d − must be the same expression.In particular, all the coefficients in ˆ d − must be divisible by r . Theseare the same as the coefficients of the x terms in ˆ d ′ . ASIC SUBGROUPS AND FREENESS, A COUNTEREXAMPLE 9
Let δ be the largest ordinal such that y δ,m occurred in ˆ d . Let c bethe coefficient of y δ,m in ˆ d .When we raised the subscript of y δ,m from m to n in going from ˆ d toˆ d ′ , the first step spun off (a multiple of x δ and) cx η ( δ,m ) . The subscript η ( δ, m ) here is larger than all the other elements δ ′ ∈ ∆ that occur assubscripts of y ’s in ˆ d , because of our choice of δ as largest and our choiceof m . It is also, by choice of m , not among the α ’s for which x α ∈ M .As a result, no other occurrences of x η ( δ,m ) were present in ˆ d or arose inthe raising process leading to ˆ d ′ . (Raising for smaller δ ′ spun off only x ’s whose subscripts are ordinals smaller than δ ′ < η ( δ, m ), and latersteps in the raising for δ spun off only x ’s with subscripts > η ( δ, m ).)This means that the coefficient of x η ( δ,m ) in ˆ d ′ is c . Since we alreadyshowed that all coefficients of x ’s in ˆ d ′ are divisible by r , we concludethat r divides c .Now we can delete the term cy δ,m from ˆ d and subtract cr y δ,m from g to get a violation of purity with fewer terms in its ˆ d . That contradictsour choice of ˆ d as minimal, and this contradiction completes the proofthat h M i is pure in G . By Pontryagin’s criterion, G is ℵ -free.3.4. G is ℵ -separable. A group is κ -separable if every subset of size < κ is included in a free direct summand of size < κ (see [4, Sec-tion 4.2]). So we must prove in this subsection that every countablesubset of G is included in a countable free direct summand of G . Webegin by defining the natural filtration of G . Definition 12.
For any countable ordinal ν , let G ν be the subgroupof G generated by the elements x α for α < ν and the elements y δ,n for δ ∈ S ∩ ν and n ∈ ω . (In writing S ∩ ν , we use the usual identificationof an ordinal with the set of all smaller ordinals.)Clearly, G λ = S ν<λ G ν for limit ordinals λ , the sequence h G ν : ν < ℵ i is increasing, and it covers G , so we have a filtration. Because G is ℵ -free and each G ν is countable, each G ν is free. Furthermore, everycountable subset of G is included in some G ν . So to complete the proofthat G is ℵ -separable, we need only show that there are arbitrarilylarge ν < ℵ such that G ν is a direct summand of G . In fact, we shallshow that G ν is a direct summand whenever ν / ∈ S . Recall that thestationary S in Notation 4 was chosen to consist of limit ordinals, so,in particular, G ν will be a direct summand for all successor ν .Fix an arbitrary ν / ∈ S . We shall show that G ν is a direct summandof G by explicitly defining a projection homomorphism p : G → G ν that is the identity on G ν . For this purpose, it suffices to define p on the generators x α and y δ,n of G and to show that the defining relationsof G are preserved.Of course, we define p ( x α ) = x α for all α < ν and p ( y δ,n ) = y δ,n forall δ ∈ S ∩ ν and all n ∈ ω , so that p is the identity on G ν . For α ≥ ν ,we set p ( x α ) = 0. Finally, for δ ∈ S − ν and n ∈ ω , we set p ( y δ,n ) = X k ≥ n k ! n ! p ( x η ( δ,k ) ) . Although the sum appears to be over infinitely many k ’s, only finitelymany of them give non-zero terms in the sum. Indeed, since ν / ∈ S and δ ∈ S − ν , we have ν < δ ; therefore, for all sufficiently large k ∈ ω , wehave ν < η ( δ, k ) and so p ( x η ( δ,k ) ) = 0.It remains to check that p respects the defining relations of G , i.e.,that, for all δ ∈ S and all n ∈ ω , p ( y δ,n ) = ( n + 1) p ( y δ,n +1 ) + q n p ( x δ ) + p ( x η ( δ,n ) ) . If δ < ν this is trivial, since all four applications of p do nothing. δ = ν is impossible as δ ∈ S and ν / ∈ S . So we assume from now on that δ > ν . In this case, the term q n p ( x δ ) vanishes and what we must checkis, in view of the definition of p , X k ≥ n k ! n ! p ( x η ( δ,k ) ) = ( n + 1) X k ≥ n +1 k !( n + 1)! p ( x η ( δ,k ) ) + p ( x η ( δ,k ) ) . But this equation is obvious, and so the proof is complete.3.5. G is not free. Using the filtration from the preceding subsection,we can easily show that G is not free because its Gamma invariant,Γ( G ), is at least (the equivalence class in P ( ℵ ) /N S of) S . (See [4,Section IV.1] for Gamma invariants and their connection with freeness.)Indeed, for any δ ∈ S , the quotient group G δ +1 /G δ is generated by x δ and the y δ,n for n ∈ ω , subject to the relations y δ,n = ( n + 1) y δ,n +1 + q n x δ , because the remaining term in the defining relation for G , namely x η ( δ,n ) , is zero in the quotient. But this presentation of G δ +1 /G δ is,except for the names of the generators, identical with the presenta-tion of E in Lemma 7. Since E isn’t free, G/G δ isn’t ℵ -free, and so δ ∈ Γ( G ).Although the preceding completes the verification that G isn’t free,we point out that Γ( G ) is exactly (the equivalence class of) S . Indeed,we showed in the preceding subsection that, when ν / ∈ S , then G ν isa direct summand of G . Thus, the quotient G/G ν is isomorphic to asubgroup of G and is therefore ℵ -free. ASIC SUBGROUPS AND FREENESS, A COUNTEREXAMPLE 11
Subgroups of G disjoint from B are free. Suppose, towarda contradiction, that H is a non-free subgroup of G disjoint from B .So Γ( H ) = 0. The Gamma invariant here can be computed using anyfiltration of H ; we choose the one induced by the filtration of G alreadyintroduced. So we set H ν = G ν ∩ H and conclude that the set A = { ν < ℵ : H/H ν is not ℵ -free } = { ν < ℵ : For some µ > ν, H µ /H ν is not free } must be stationary.Thanks to our choice of the filtration h H ν i , we have, for all ν < µ < ℵ , H µ H ν = H µ H µ ∩ G ν ∼ = H µ + G ν G ν ⊆ G µ G ν , the isomorphism being induced by the inclusion map of H µ into H µ + G ν . We already saw that, when ν / ∈ S , the groups G µ /G ν are free;therefore, so are the groups H µ /H ν . Thus, A ⊆ S .Temporarily fix some ν ∈ A . For any µ > ν , we have an exactsequence 0 → H ν +1 H ν → H µ H ν → H µ H ν +1 → . Since ν ∈ A , the middle group here is not free for certain µ . The groupon the right, H µ /H ν +1 , on the other hand, is free because ν + 1 / ∈ S .(Recall that S consists of limit ordinals.) So the exact sequence splitsand therefore the group on the left, H ν +1 /H ν , is not free.Since ν ∈ S , we know, from a calculation in the preceding subsection,that G ν +1 /G ν is isomorphic to E , and we saw above that H ν +1 /H ν isisomorphic to a subgroup of this (via the map induced by the inclu-sion of H ν +1 into G ν +1 ). Since all rank-1 subgroups of E are free but H ν +1 /H ν is not free, H ν +1 /H ν must have the same rank 2 as the wholegroup G ν +1 /G ν . So the purification of H ν +1 /H ν in G ν +1 /G ν is all of G ν +1 /G ν .In particular, this purification must contain the coset of the element x ν ∈ G ν +1 . That is, there must exist an integer n = 0 and an element g ∈ G ν such that nx ν − g ∈ H ν +1 .Now un-fix ν . Of course the n and g obtained above can depend on ν , so we write them from now on with subscripts ν . Thus we have, forall ν ∈ A , some n ν ∈ Z − { } and some g ν ∈ G ν such that n ν x ν − g ν ∈ H ν +1 . Because A is stationary and all values of n ν lie in a countable set,there is a stationary A ′ ⊆ A such that n ν has the same value n forall ν ∈ A ′ . Furthermore, by Fodor’s theorem, there is a stationary set A ′′ ⊆ A ′ such that g ν has the same value g for all ν ∈ A ′′ . (Inmore detail: For each ν ∈ A ′ ⊆ S , we know that ν is a limit ordinal,so G ν = S α<λ G α . Thus, g ν ∈ G r ( ν ) for some r ( ν ) < ν . This r isa regressive function on A ′ , so by Fodor’s theorem it is constant, saywith value ρ , on a stationary subset. For ν in this stationary set, g ν hasvalues in the countable set G ρ and is therefore constant on a smallerstationary subset A ′′ .)Consider any two distinct elements ν and ξ of A ′′ . Since n ν = n ξ = n and g ν = g ξ = g , we have that H contains both nx ν − g and nx ξ − g .So it contains their difference n ( x ν − x ξ ). Since n = 0 and ν = ξ ,this contradicts the assumption that H is disjoint from the subgroup B generated by all the x α ’s. References [1] K. Benabdallah and J. M. Irwin, “An application of B -high subgroups ofabelian p -groups,” J. Algebra
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Mathematics Department, University of Michigan, Ann Arbor, MI48109–1043, U.S.A.
E-mail address : [email protected] Einstein Institute of Mathematics, Edmond J. Safra Campus, TheHebrew University of Jerusalem, Jerusalem 91904, Israel and Math-ematics Department, Rutgers University, New Brunswick, NJ 08854,U.S.A.
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