# Freely indecomposable almost free groups with free abelianization

Samuel M. Corson
From the journal Journal of Group Theory

# Abstract

For certain uncountable cardinals κ, we produce a group of cardinality κ which is freely indecomposable, strongly κ-free, and whose abelianization is free abelian of rank κ. The construction takes place in Gödel’s constructible universe L. This strengthens an earlier result of Eklof and Mekler.

## 1 Introduction

We produce examples of groups which exhibit some properties enjoyed by free groups but which, in other ways, are very far from being free. We recall some definitions before stating the main result. Given a group G and subgroup HG, we say H is a free factor of G provided there exists another subgroup KG such that G=H*K is the free product in the natural way (that is, the map H*KG induced by the inclusions of H and K is an isomorphism). We call such a writing G=H*K a free decomposition of G and say that G is freely indecomposable provided there does not exist a free decomposition of G via two nontrivial free factors.

Given a cardinal κ, we say G is κ-free if each subgroup of G generated by fewer than κ elements is a free group. Historically, a κ-free group of cardinality κ is called almost free [7]. By the theorem of Nielsen and Schreier, every free group is κ-free for every cardinal κ. A subgroup H of a κ-free group G is κ-pure if H is a free factor of any subgroup HX, where XG is of cardinality <κ. A κ-free group G is strongly κ-free provided each subset XG with |X|<κ is included in a κ-pure subgroup of G generated by fewer than κ elements.

Let ZFC denote the Zermelo–Fraenkel axioms of set theory including the axiom of choice, and let V = L denote the assertion that every set is constructible. The theory ZFC + V = L is consistent provided ZFC is consistent [6]. The set-theoretic concepts in the following statement will be reviewed in Section 3, but the reader can, for example, let κ be any uncountable successor cardinal (e.g. 1, 2, ω+1).

## Theorem 1.1 (ZFC + V = L).

Let κ be an uncountable regular cardinal that is not weakly compact. There exists a group G of cardinality κ for which

1. G is freely indecomposable,

2. G is strongly κ -free,

3. the abelianization G/G is free abelian of rank κ.

The hypotheses on the cardinal κ cannot be dropped since a κ-free group of cardinality κ must be free when κ is singular or weakly compact (see respectively [15] and [3]). A group as in the conclusion seems unusual since, on a local level, it is free; on a global level, it is quite unfree and in fact indecomposable, but the abelianization is as decomposable as possible. Theorem 1.1 minus condition (3) was proved in [4], and the construction apparently does not have free abelianization; indeed, the proof that their groups are freely indecomposable involves abelianizing. A non-free 1-free group of cardinality 1 which abelianizes to a free abelian group was produced by Bitton [1] using only ZFC, and Theorem 1.1 can be considered a constructible universe strengthening of his result. The first construction of a non-free almost free group of cardinality 1 was given by Higman [7] without any extra set-theoretic assumptions; a strongly 1-free group of cardinality 1 produced only from ZFC was given by Mekler [13]. The reader can find other results related to almost free (abelian) groups in such works as [5, 12, 2].

We note that it is not possible to produce a group G of cardinality κ whose every subgroup of cardinality κ satisfies conditions (1)–(3) of Theorem 1.1. This is because, for every uncountable locally free group G, there exists a free subgroup HG with |H|=|G| (see [14, Theorem 1.1]). Finally, we mention that the construction used in proving Theorem 1.1 allows one, as in [4], to construct 2κ many pairwise non-isomorphic groups of this description using [16].

## 2 Some group-theoretic lemmas

The following appears as [7, Lemma 1].

## Lemma 2.1.

If K is a free factor of G and H is a subgroup of G, then HK is a free factor of H.

We will use the following construction in our proof of Theorem 1.1.

## Construction 2.2.

Suppose that we have a free group Fa with free decomposition Fa=F0*F1*F2,a with F1 nontrivial and F2,a freely generated by {tn}nω. Let Fb be a free group with free decomposition Fb=F0*F1*F2,b, where F2,b is freely generated by {zn}nω. Let y be an element of a free generating set for F1. Define ϕ:FaFb so that ϕF0*F1 is the identity and ϕ(tn)=yzn+1-1znzn+1.

Property (iv) of the following lemma compares to [1, Lemma 2.9 (3) & (4)].

## Lemma 2.3.

Let F2,a,n=t0,,tn-1. The map ϕ satisfies the following:

1. ϕ is a monomorphism;

2. znϕ(Fa) for all nω;

3. ϕ(Fa) is not a free factor of Fb, but ϕ(F0*F1*F2,a,n) is a free factor for every nω;

4. the equality (ϕ(Fa))=ϕ(Fa)Fb holds, and the natural induced map ϕ¯:Fa/FaFb/Fb is an isomorphism.

## Proof.

Fix a possibly empty set of free generators X for F0 and a possibly empty set Y such that the disjoint union Y{y} freely generates F1. Now ϕ(Fa) is the subgroup

XY{y}{yzn+1-1znzn+1}nω=XY{y}{zn+1-1znzn+1}nω,

and the generators XY{y}{zn+1-1znzn+1}nω freely generate ϕ(Fa) since they satisfy the Nielsen property (see [11, I.2] and [8, Example 2.8 (iii)]). The set XY{y}{y-1tn}nω is a free generating set for Fa and maps bijectively under ϕ to the free generating set XY{y}{zn+1-1znzn+1}nω for ϕ(Fa). Thus ϕ is a monomorphism, and we have (i). Moreover, since

XY{y}{zn+1-1znzn+1}nω

satisfies the Nielsen property, it is clear that znϕ(Fa) for all nω, and we have (ii).

For (iii), we notice that the smallest normal subgroup in Fb including ϕ(Fa), which we denote ϕ(Fa), is precisely Fb since the free generators listed for Fb are conjugate in Fb to free generators of ϕ(Fa). Since ϕ(Fa) is a proper subgroup of Fb, this means that ϕ(Fa) cannot be a free factor of Fb. On the other hand, we have ϕ(F0*F1*F2,a,n) generated by

XY{y}{yz1-1z0z1,,yzn-1zn-1zn},

and we claim that XY{y}{yz1-1z0z1,,yzn-1zn-1zn}{zn,zn+1,} is a free generating set for Fb. It is plain that

{y}{yz1-1z0z1,,yzn-1zn-1zn}{zn}={y}{z0,,zn},

and since finitely generated free groups are Hopfian, we get that

{y}{yz1-1z0z1,,yzn-1zn-1zn}{zn}

is a free generating set of the free factor {y}{z0,,zn} of Fb. Thus, indeed, XY{y}{yz1-1z0z1,,yzn-1zn-1zn}{zn,zn+1,} is a free generating set for Fb, and so

ϕ(F0*F1*F2,a,n)=XY{y}{yz1-1z0z1,,yzn-1zn-1zn}

is a free factor of Fb and we have shown (iii).

For condition (iv), certainly the inclusion (ϕ(Fa))ϕ(Fa)Fb holds. Moreover, a word w in (XY{y}{zn}nω)±1 represents an element of Fb if and only if the sum of the exponents of each element in XY{y}{zn}nω is 0. By treating each element of {zn+1-1znzn+1}nω as an unreducing letter, a word in

(XY{y}{zn+1-1znzn+1}nω)±1

represents an element of (ϕ(Fa)) if and only if the sum of the exponents of each element of XY{y}{zn+1-1znzn+1}nω is 0. This is clearly equivalent to having the sum of the exponents of each letter in XY{y}{zn}nω be 0, so we have (ϕ(Fa))=ϕ(Fa)Fb. Thus the map ϕ¯:Fa/FaFb/Fb is injective. Moreover, since each element of XY{y}{zn}nω is conjugate in Fb to an element of XY{y}{zn+1-1znzn+1}nω, the map ϕ¯ is onto as well. ∎

We recall some notions for free products of groups (see [11, Section IV.1]). Suppose that we have a free product L0*L1. We call the nontrivial elements in L0L1letters. Each element gL0*L1 can be expressed uniquely as a product of letters g=h0h1hn-1 such that hiL0 if and only if hi+1L1 for all 0i<n-1 (this is the reduced or normal form of the element). We call the number n the length of g, denoted Len(g). Thus the identity element has length 0 and a nontrivial element has length 1 if and only if it is a letter. Given a writing of an element of L0*L1 as a product of nontrivial elements h0hn-1 in L0L1, it is easy to determine the normal form of the element by taking a consecutive pair hihi+1 for which hi,hi+1 are both in L0 or both in L1 and performing the group multiplication in the appropriate group. This either gives the trivial element, in which case we remove the pair hihi+1 from the expression, or it gives a nontrivial element gi, and we replace hihi+1 with gi in the expression. This process reduces the number of letters in the writing by at least 1 every time, and so the process must eventually terminate, and it terminates at the normal form. We will generally consider an element of L0*L1 as a word (the normal form) in the letters. For words w0 and w1 in the letters, we will use w0w1 to represent that w0 and w1 are the same word letter-for-letter when read from left to right.

We say an element of L0*L1 is cyclically reduced if its reduced form is either of length 0 or 1 or begins with a letter in Lj and ends with a letter in L1-j. A cyclically reduced element is of minimal length in its conjugacy class, and if two cyclically reduced elements are conjugate to each other, then the normal form of one is a cyclic shift of the other (w is a cyclic shift of u if we can write w as a concatenation wv0v1 such that uv1v0). Each element g of L0*L1 is conjugate to a cyclically reduced element h, and we call Len(h) the cyclic length ofg.

## Lemma 2.4.

If ψ:FaL0*L1 is a monomorphism such that

ψ(F1)=ψ(Fa)L0𝑎𝑛𝑑ψ(F0*F2,a)=ψ(Fa)L1,

then there does not exist a monomorphism θ:FbL0*L1 for which θϕ=ψ.

## Proof.

Suppose on the contrary that such a θ exists. We will treat Fa as a subgroup of Fb since ϕ is a monomorphism and treat Fa and Fb as subgroups of L0*L1 such that F1=FaL0 and F0*F2,a=FaL1. We have tn=yzn+1-1znzn+1 for all nω, and so y-1tn=zn+1-1znzn+1. Since y-1L0 and tnL1, we see that y-1tn is cyclically reduced and Len(y-1tn)=2. Therefore, for each nω, we know zn is of cyclic length 2, so Len(zn)2, and any cyclic reduction of zn must be a cyclic shift of y-1tn. We claim that Len(zn)Len(zn+1)+1 for all nω. This immediately gives Len(z0)Len(zn)+n for all nω, which is a contradiction.

It remains to prove that Len(zn)Len(zn+1)+1. To economize on writing subscripts, we will show that Len(z0)Len(z1)+1, and the same proof will work for general n by adding n to the subscripts of t0,z0,t1,z1. Suppose to the contrary that Len(z0)Len(z1). We have z1y-1t0z1-1=z0. Since z1 is nontrivial, it must end with a letter of L0 or a letter of L1.

Case A: z1 ends with a letter of L0. In this case, it must be that z1 ends with y since otherwise we readily see from the reduced form for z0=z1y-1t0z1-1 that Len(z0)=(2Len(z1)+2)-1=2Len(z1)+1>Len(z1) contrary to our assumption that Len(z0)Len(z1). Also, the second-to-last letter of z1 must be t0-1 since otherwise L(z0)=(2Len(z1)+2)-3=2Len(z1)-1>Len(z1) since Len(z1)2. Thus we may write z1 as a reduced word z1=w(t0-1y)k with k1 and maximal. Notice that w is nonempty, for otherwise z1=(t0-1y)k and z0=y-1t0, and therefore z0 and z1 commute instead of generating a free subgroup of rank 2. The word w must end with a letter from L0 since w(t0-1y)k is reduced. Moreover, Len(w)2 since otherwise w=gL0 and z1 is conjugate to (t0-1y)k-1t0-1yg, and cyclically reducing this word cannot produce a cyclic shift of y-1t1. Then the second-to-last letter of w is an element from L1. If the last letter of w is y, then the second-to-last letter of w is not t0-1 by maximality of k, and we get

Len(z0)=(2Len(z1)+2)-4k-2-1=2Len(z1)-1-4k.

If, on the other hand, the last letter of w is gL0{1,y}, then we see that

Len(z0)=(2Len(z1)+2)-4k-1=2Len(z1)-4k+1,

so, in either case, we know Len(z0)2Len(z1)-1-4k. Since we are assuming Len(z1)Len(z0), we have Len(z1)4k+1.

Write z1=(y-1t0)mu(t0-1y)k with m0 maximal. Certainly, u is nontrivial since otherwise z1 would commute with z0. We also claim that u must begin and end with a letter from L0. It is clear, since z1 is written in reduced form, that u must end with a letter from L0 since t0-1L1. Also, if m>0, we have u beginning with a letter from L0 for a similar reason. But if m=0, then we have z1=u(t0-1y)k, and were it the case that u started with a letter in L1, then this writing would already be cyclically reduced and z1 would have cyclic length greater than 2 since u is nontrivial. Thus, in all cases, u must begin and end with a letter from L0.

If m>k, then Len(z1)=2m+2k+Len(u)>4k+1. If k=m, then, since 4k+1Len(z1)=2m+2k+Len(u)=4k+Len(u), we get Len(u)=1. Then z1 is conjugate to an element of length 1, contradicting the fact that z1 is of cyclic length 2. Thus k>m.

If k>m+1, then, by maximality of m, we conjugate z1 to a word

u(t0-1y)k-m-1t0-1y,

where u might or might not start with y-1, but if it starts with y-1, the second letter of u would not be t0. Thus, whether or not u starts with y-1, we know that u(t0-1y)k-m-1t0-1y has cyclic length at least 3 despite being conjugate to z1, a contradiction. Thus we know precisely that k=m+1, and so z1 is conjugate to ut0-1y. If u does start with y-1, say uy-1v for some word v, then we get z1 conjugate to vt0-1. Since the cyclic length of z1 is 2, we know that v is nonempty, must start with a letter from L1 and that letter must not be t0 by the maximality of m. Then vtv0 with tL1{1,t0-1}, and v0 begins and ends with a letter in L0. Thus z1 is conjugate to v0(t0-1t) which is a cyclically reduced word. Since z1 has cyclic length 2 and cyclic reduction y-1t1, we obtain that v0=y-1 and t0-1t=t1. But now

z1=(y-1t0)k-1u(t0-1y)k=(y-1t0)k-1y-1(t0t1)y-1(t0-1y)k

so that z1 is expressed as a product of elements in Fa, contradicting Lemma 2.3 part (ii). Therefore, u must start with some gL0{1,y-1}, say ygv0. Now we conjugate ut0-1ygv0t0-1y to v0t0-1(yg) which is cyclically reduced (by considering (yg) as a single letter in L0). But v0t0-1(yg) cannot possibly be a cyclic shift of y-1t1, regardless of whether v0 is empty or not. This finishes the proof of Case A.

Case B: z1 ends with a letter of L1. The reasoning in this case follows that in the other case more or less, and we give the sketch. Arguing as before, we see that z1 must end with t0, and the second-to-last letter must be y-1. Write z1w(y-1t0)k with k1 maximal. As before, w is nonempty. Again, we have Len(z0)2Len(z1)-1-4k from which Len(z1)4k+1. Writing

z1(t0-1y)mu(y-1t0)k

with m0 maximal, we again see that u is nonempty and that k=m+1. Also, u must begin and end with letters from L1. Therefore, z1 is conjugate to uy-1t0. If ut0-1v for some v, then we get z1 conjugate to vy-1, and as m was maximal, the word v starts with an element of L0 which is not y. Then vgv0 with gL0{1,y} and v0 nontrivial since u, and therefore v, must end in a letter from L1. Conjugating vy-1 to the cyclically reduced word v0(y-1g), we must have that this word is a cyclic shift of y-1t1. Then v0=t1, but y-1g cannot be y-1 since g was nontrivial. Therefore, it must be that utv for some tL1{1,t0}. Conjugating uy-1t0=tvy-1t0 to the cyclically reduced word vy-1(t0t), it must be that this word is a cyclic shift of y-1t1. Then v is empty and t0t=t1. Therefore,

z1=(t0-1y)k-1u(y-1t0)k=(t0-1y)k-1(t0-1t1)(y-1t0)k,

and we have z1Fa, which contradicts Lemma 2.3 part (ii). This completes the proof of Case B and of the lemma. ∎

## 3 Proof of the main theorem

The groups that we produce for Theorem 1.1 will follow the induction used in [4, Theorem 2.2], using Construction 2.2 at the key stages. We will use combinatorial principles which follow from ZFC + V = L to rule out any possible free decomposition while ensuring strong κ-freeness and free abelianization. We first review some concepts from set theory.

## Definitions 3.1.

(see [9]) Recall that a cardinal number is naturally considered as an ordinal number which cannot be injected into a proper initial subinterval of itself. A subset E of an ordinal α is bounded in α if there exists β<α which is an upper bound on E. The cofinality of an ordinal α is the least cardinal κ for which there exists an unbounded Eα of cardinality κ. An infinite cardinal κ is regular if the cofinality of κ is κ. A subset C of ordinal α is club if it is unbounded in α and closed under the order topology in α. The intersection of two club sets in an uncountable regular cardinal is again a club set. A subset E of ordinal α is stationary if it has nonempty intersection with every club subset of α. The intersection of a club set and a stationary set in a regular cardinal is again stationary. We mention that weakly compact cardinals are inaccessible, and it is therefore consistent to assume that a universe of set theory does not contain any (see [9, Chapters 9, 17]).

The following is a theorem of Solovay (see [16] or [9, Theorem 8.10]).

## Theorem 3.2.

If κ is an uncountable regular cardinal, then each stationary subset of κ can be decomposed as the disjoint union of κ many stationary subsets of κ.

We quote Jensen’s κ(E) principle ([10, Lemma 6.5] or [9, Assertion 27.16]), remark an easy consequence (see [4, page 97]) and quote one more result of Jensen (see [10, Theorem 5.1], [4, Theorem 1.3]).

## Theorem 3.3 (ZFC + V = L).

If κ is an uncountable regular cardinal and Eκ is stationary in κ, there exists a sequence {Sα}αE such that Sαα and, for any Jκ, the set {αEJα=Sα} is stationary in κ.

## Remark 3.4.

From a sequence given by Theorem 3.3, one obtains a sequence of ordered pairs {(Tα0,Tα1)}αE for which Tα0,Tα1α and, given any subsets A,Bκ, the set {αEAα=Tα0andBα=Tα1} is stationary in κ. To see this, we give the product {0,1}×κ the lexicographic order. There is an order isomorphism f:{0,1}×κκ under which f(0,α)=α for every limit ordinal α<κ. Let fi:κκ be given by fi(α)=f(i,α) for each i{0,1}, so f0 has disjoint image from f1. Notice that, for each limit ordinal α<κ and Xκ, we have fi(Xα)=fi(X)α. Letting E0 be the intersection of E with the set of limit ordinals below κ, we have that E0 is stationary as the intersection of a stationary with a club. For αE0, define Tα0=f0-1(Sα) and Tα1=f1-1(Sα). Given A,Bκ, we let J=f0(A)f1(B) and notice that

{αE0Aα=Tα0 and Bα=Tα1}={αE0f0(Aα)f1(Bα)=Sα}={αE0(f0(A)α)(f1(B)α)=Sα}={αE0Jα=Sα}

is stationary in κ. Letting Tα0==Tα1 for αEE0 gives the desired sequence of ordered pairs.

## Theorem 3.5 (ZFC+ V = L).

If κ is an uncountable regular cardinal which is not weakly compact, then there is a stationary subset Eκ for which each element of E is a limit ordinal of cofinality ω and, for each limit ordinal α<κ, the set Eα is not stationary in α.

## Proof of Theorem 1.1.

Let κ be as in the hypotheses of Theorem 1.1. We inductively define a group structure on κ, and this will serve as our group G. Let γ0=ω, γα+1=γα+γα and γβ=α<βγα for limit ordinal β<κ. The set {γα}α<κ is obviously a club set in κ. We will define a group structure on each γα so that γα will be a subgroup of γβ whenever α<β, and thus the group structure on κ=α<κγα will be well defined.

Select stationary E0κ satisfying the conclusion of Theorem 3.5 and such that κE0 is stationary in κ (using Theorem 3.2). It is easy to check that E={γα}αE0 is also stationary in κ. Let {(Tγα0,Tγα1)}γαE be a sequence as in Remark 3.4.

The following properties will hold on the groups for all α<β<κ:

1. γβ is free of infinite rank;

2. γα is a proper subgroup of γβ;

3. γα is a free factor of γβ if and only if γαE;

4. γα=γβγα.

We let γ0 be free of countably infinite rank. Suppose that we have defined the group structure on γα for all α<β<κ such that the above conditions hold. If β=δ+1 and γδE, then we give γβ the group structure obtained by any bijection f:γβγδ* such that fγα is the identity map. Such a bijection exists since |γβγδ|=|γδ|. Conditions (i), (ii) and (iv) obviously hold. Notice also that, for α<β, we have γα a free factor of γβ if and only if γα is a free factor of γδ (this uses Lemma 2.1), and so (iii) also holds.

If β=δ+1 and γδE, then we consider two subcases. Firstly, suppose that there exists a strictly increasing sequence {αn}nω for which

1. γαnE,

2. nωαn=δ,

3. γαn=(γαnTγδ0)*(γαnTγδ1) with both free factors nontrivial for all nω.

From this, it follows immediately that γδ=(γδTγδ0)*(γδTγδ1)=Tγδ0*Tγδ1. Since {γαn}nω is a strictly increasing sequence of sets, we know for each nω that it is either the case that

γαn+1Tγδ0γαnTγδ0or thatγαn+1Tγδ1γαnTγδ1.

By choosing a subsequence, we can assume without loss of generality that

γαn+1Tγδ1γαnTγδ1for allnω

(else we exchange the roles of Tγδ0 and Tγδ1 in the construction that follows). Since each γαnE, we know by our induction that γαn is a free factor of γδ and also of γαn+1. Then, by Lemma 2.1, we know that γαnTγδ1 is a free factor of γδTγδ1 and of γαn+1Tγδ1. Inductively select a free basis Q of γδTγδ1 such that QγαnTγδ1 is a free generating set for γαnTγδ1 for each nω. For each nω, select tnQγαn+1Tγδ1γαn, and let X=Q{tn}nω. Let {y}Y be a free generating set for γδTγδ0. By how the set {γα}α<κ was defined, we know |γβ|=|γβγδ|=|γδ|. Hence we let γβ be given a group structure using a bijection f:γβF(X)*F({y}Y)*F({zn}nω) such that fγδ is the isomorphism γδF(X)*F({y}Y)*F({tn}nω) noted above and the inclusion map ι:γδγβ satisfies ι=f-1ϕfγδ, where ϕ is the map from Construction 2.2.

Certainly, conditions (i), (ii) hold, and condition (iv) holds by Lemma 2.3 (iv). Also, we know γδ is not a free factor of γβ by Lemma 2.3 (iii). Notice that, for each nω, we have that X{t0,,tn-1}{y}Y is a free factor of γβ by Lemma 2.3 (iii). For each nω, we know γαn is a free factor of

X{t0,,tn-1}{y}Y,

and for each α<δ, there exists nω for which α<αn. It follows by Lemma 2.1 and by our induction that, for α<δ, we have γα a free factor of γβ if and only if γαE, and condition (iii) holds.

On the other hand, suppose β=δ+1 and γδE and no such increasing sequence {αn}nω exists. Since δE0 is of cofinality ω and E0δ is not stationary in δ, we may select a strictly increasing sequence αnE0 such that nωαn=δ. Then γαnE. As each γαn is a free factor of γδ and the αn are strictly increasing, we may select by induction a free generating set Q for γδ such that Qγαn is a free generating set for γαn. Pick yγα0Q and tnγαn+1Qγαn, and letting X= and Y=Q({tn}nω{y}), we can let γβ be a group freely generated by X{zn}nω{y}Y such that the inclusion map ι:γδγβ is the map ϕ from Construction 2.2. The check that the induction conditions still hold is as in the other subcase.

When β<κ is a limit ordinal, the binary operation on γβ=α<βγα is defined by that on the γα for α<β. By how E0 was chosen, we know E0β is not stationary in β, and so we select a club set Cβ for which CE0=. By induction, we know that, for α,δC with α<δ, we have that γα is a proper free factor of γδ, and since C is closed, we have γδ=α<δ,αCγα for any δC which is a limit under the ordering of κ restricted to C. Then, by induction on C, we can select a free generating set Q for γβ=αCγα for which Qγα is a free generating set for γα for each αC. Then (i) and (ii) hold. For α<β, it is clear by Lemma 2.1 that γα is a free factor of γβ if and only if γα is a free factor of γδ for some δC with δ>α, and since γδ is a free factor of γβ for each δC, condition (iii) holds. Condition (iv) follows by induction since γβ=α<βγα. This completes the construction of the group structure on κ.

We verify that conditions (1)–(3) of the statement of Theorem 1.1 hold. Imagine for contradiction that κ=A*B with A,B nontrivial subgroups of κ. Letting α be such that γαA and γαB are both nontrivial and

C={αα<κγα=(γαA)*(γαB)},

it is straightforward to verify that C is club in κ. Since the set κE0 is stationary in κ, we know D=CE0 is stationary and therefore unbounded in κ. Then the closure D¯ is club in κ, and so is {γα}αD¯. Then there exists γδE with δD¯ and Aγδ=Tγδ0 and Bγδ=Tγδ1. Then δE0, and so δE0D¯. As δE0, we know that δ has cofinality ω in κ. Certainly, δD=CE0, and so there exists a strictly increasing sequence {αn}nω with αnCE0 such that nωαn=δ. Then

γαn=(γαnA)*(γαnB)=(γαnTγδ0)*(γαnTγδ1).

By our construction, we know that γδ includes into γδ+1 in such a way that γδ+1 is not a subgroup of A*B (using Lemma 2.4 and letting (A,B)=(L0,L1) or (A,B)=(L1,L0) depending on whether we selected {tn}nωTγδ1 or, respectively, {tn}nωTγδ0 in the construction), and this is a contradiction. Thus κ is freely indecomposable, and we have part (1).

For part (2), we let Xκ with |X|<κ. By the regularity of κ, select α<κ large enough that γαX. Notice that γα+1E. Any subgroup H of κ with |H|<κ and Hγα+1 satisfies Hγβ for some β>α by the regularity of κ. Since γα+1 is a free factor of γβ by our construction, we have by Lemma 2.1 that γα+1 is a free factor of H. Thus γα+1 is κ-pure, and the group κ is strongly κ-free, and we have verified (2).

For part (3), we notice that, since

γα=γαγβfor eachα<β,and sinceκ=β<κγβ,

the equality γα=γακ holds for all α<κ. In particular, the homomorphism induced by the inclusion map γα/γακ/κ is injective, and so the abelianization of κ is the increasing union of free abelian subgroups γα/γα. In our construction, when β=δ+1 and γδE, the map induced by inclusion γδ/γδγβ/γβ is an isomorphism by Lemma 2.3 (iv). When β=δ+1 and γδE, we have γδ/γδ as a proper direct summand of γβ/γβ. As well, we have γβ/γβ=α<βγα/γα for limit β<κ. Thus we may inductively select a free abelian basis for the abelianization of κ, and the free basis will be of cardinality κ since there are κ many β for which β=δ+1 with γδE. We have verified (3) and finished the proof of the theorem. ∎

Communicated by Alexander Yu. Olshanskii

Funding source: European Research Council

Award Identifier / Grant number: PCG-336983

Funding source: Ministerio de Economía y Competitividad

Award Identifier / Grant number: SEV-20150554

Funding statement: This work was supported by ERC grant PCG-336983 and by the Severo Ochoa Programme for Centres of Excellence in R&D SEV-20150554.

# Acknowledgements

The author thanks the anonymous referee for their careful reading and suggested improvements which enhanced the presentation of the paper.

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