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.
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 , we say H is a free factor of G provided there exists another subgroup such that is the free product in the natural way (that is, the map induced by the inclusions of H and K is an isomorphism). We call such a writing 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 . 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 , where is of cardinality . A κ-free group G is strongly κ-free provided each subset with 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 . 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. , , ).
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
G is freely indecomposable,
G is strongly κ -free,
the abelianization 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  and ). 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 , and the construction apparently does not have free abelianization; indeed, the proof that their groups are freely indecomposable involves abelianizing. A non-free -free group of cardinality which abelianizes to a free abelian group was produced by Bitton  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 was given by Higman  without any extra set-theoretic assumptions; a strongly -free group of cardinality produced only from ZFC was given by Mekler . 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 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 with (see [14, Theorem 1.1]). Finally, we mention that the construction used in proving Theorem 1.1 allows one, as in , to construct many pairwise non-isomorphic groups of this description using .
2 Some group-theoretic lemmas
The following appears as [7, Lemma 1].
If K is a free factor of G and H is a subgroup of G, then is a free factor of H.
We will use the following construction in our proof of Theorem 1.1.
Suppose that we have a free group with free decomposition with nontrivial and freely generated by . Let be a free group with free decomposition , where is freely generated by . Let y be an element of a free generating set for . Define so that is the identity and .
Property (iv) of the following lemma compares to [1, Lemma 2.9 (3) & (4)].
Let . The map ϕ satisfies the following:
ϕ is a monomorphism;
for all ;
is not a free factor of , but is a free factor for every ;
the equality holds, and the natural induced map is an isomorphism.
Fix a possibly empty set of free generators X for and a possibly empty set Y such that the disjoint union freely generates . Now is the subgroup
and the generators freely generate since they satisfy the Nielsen property (see [11, I.2] and [8, Example 2.8 (iii)]). The set is a free generating set for and maps bijectively under ϕ to the free generating set for . Thus ϕ is a monomorphism, and we have (i). Moreover, since
satisfies the Nielsen property, it is clear that for all , and we have (ii).
For (iii), we notice that the smallest normal subgroup in including , which we denote , is precisely since the free generators listed for are conjugate in to free generators of . Since is a proper subgroup of , this means that cannot be a free factor of . On the other hand, we have generated by
and we claim that is a free generating set for . It is plain that
and since finitely generated free groups are Hopfian, we get that
is a free generating set of the free factor of . Thus, indeed, is a free generating set for , and so
is a free factor of and we have shown (iii).
For condition (iv), certainly the inclusion holds. Moreover, a word w in represents an element of if and only if the sum of the exponents of each element in is 0. By treating each element of as an unreducing letter, a word in
represents an element of if and only if the sum of the exponents of each element of is 0. This is clearly equivalent to having the sum of the exponents of each letter in be 0, so we have . Thus the map is injective. Moreover, since each element of is conjugate in to an element of , the map 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 . We call the nontrivial elements in letters. Each element can be expressed uniquely as a product of letters such that if and only if for all (this is the reduced or normal form of the element). We call the number n the length of g, denoted . 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 as a product of nontrivial elements in , it is easy to determine the normal form of the element by taking a consecutive pair for which are both in or both in and performing the group multiplication in the appropriate group. This either gives the trivial element, in which case we remove the pair from the expression, or it gives a nontrivial element , and we replace with 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 as a word (the normal form) in the letters. For words and in the letters, we will use to represent that and are the same word letter-for-letter when read from left to right.
We say an element of is cyclically reduced if its reduced form is either of length 0 or 1 or begins with a letter in and ends with a letter in . 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 such that ). Each element g of is conjugate to a cyclically reduced element h, and we call the cyclic length ofg.
If is a monomorphism such that
then there does not exist a monomorphism for which .
Suppose on the contrary that such a θ exists. We will treat as a subgroup of since ϕ is a monomorphism and treat and as subgroups of such that and . We have for all , and so . Since and , we see that is cyclically reduced and . Therefore, for each , we know is of cyclic length 2, so , and any cyclic reduction of must be a cyclic shift of . We claim that for all . This immediately gives for all , which is a contradiction.
It remains to prove that . To economize on writing subscripts, we will show that , and the same proof will work for general n by adding n to the subscripts of . Suppose to the contrary that . We have . Since is nontrivial, it must end with a letter of or a letter of .
Case A: ends with a letter of . In this case, it must be that ends with y since otherwise we readily see from the reduced form for that contrary to our assumption that . Also, the second-to-last letter of must be since otherwise since . Thus we may write as a reduced word with and maximal. Notice that w is nonempty, for otherwise and , and therefore and commute instead of generating a free subgroup of rank 2. The word w must end with a letter from since is reduced. Moreover, since otherwise and is conjugate to , and cyclically reducing this word cannot produce a cyclic shift of . Then the second-to-last letter of w is an element from . If the last letter of w is y, then the second-to-last letter of w is not by maximality of k, and we get
If, on the other hand, the last letter of w is , then we see that
so, in either case, we know . Since we are assuming , we have .
Write with maximal. Certainly, u is nontrivial since otherwise would commute with . We also claim that u must begin and end with a letter from . It is clear, since is written in reduced form, that u must end with a letter from since . Also, if , we have u beginning with a letter from for a similar reason. But if , then we have , and were it the case that u started with a letter in , then this writing would already be cyclically reduced and would have cyclic length greater than 2 since u is nontrivial. Thus, in all cases, u must begin and end with a letter from .
If , then . If , then, since , we get . Then is conjugate to an element of length 1, contradicting the fact that is of cyclic length 2. Thus .
If , then, by maximality of m, we conjugate to a word
where u might or might not start with , but if it starts with , the second letter of u would not be . Thus, whether or not u starts with , we know that has cyclic length at least 3 despite being conjugate to , a contradiction. Thus we know precisely that , and so is conjugate to . If u does start with , say for some word v, then we get conjugate to . Since the cyclic length of is 2, we know that v is nonempty, must start with a letter from and that letter must not be by the maximality of m. Then with , and begins and ends with a letter in . Thus is conjugate to which is a cyclically reduced word. Since has cyclic length 2 and cyclic reduction , we obtain that and . But now
so that is expressed as a product of elements in , contradicting Lemma 2.3 part (ii). Therefore, u must start with some , say . Now we conjugate to which is cyclically reduced (by considering as a single letter in ). But cannot possibly be a cyclic shift of , regardless of whether is empty or not. This finishes the proof of Case A.
Case B: ends with a letter of . 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 must end with , and the second-to-last letter must be . Write with maximal. As before, w is nonempty. Again, we have from which . Writing
with maximal, we again see that u is nonempty and that . Also, u must begin and end with letters from . Therefore, is conjugate to . If for some v, then we get conjugate to , and as m was maximal, the word v starts with an element of which is not y. Then with and nontrivial since u, and therefore v, must end in a letter from . Conjugating to the cyclically reduced word , we must have that this word is a cyclic shift of . Then , but cannot be since g was nontrivial. Therefore, it must be that for some . Conjugating to the cyclically reduced word , it must be that this word is a cyclic shift of . Then v is empty and . Therefore,
and we have , 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.
(see ) 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 which is an upper bound on E. The cofinality of an ordinal α is the least cardinal κ for which there exists an unbounded 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]).
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.
Theorem 3.3 (ZFC + V = L).
If κ is an uncountable regular cardinal and is stationary in κ, there exists a sequence such that and, for any , the set is stationary in κ.
From a sequence given by Theorem 3.3, one obtains a sequence of ordered pairs for which and, given any subsets , the set is stationary in κ. To see this, we give the product the lexicographic order. There is an order isomorphism under which for every limit ordinal . Let be given by for each , so has disjoint image from . Notice that, for each limit ordinal and , we have . Letting be the intersection of E with the set of limit ordinals below κ, we have that is stationary as the intersection of a stationary with a club. For , define and . Given , we let and notice that
is stationary in κ. Letting for 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 for which each element of E is a limit ordinal of cofinality ω and, for each limit ordinal , the set 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 , and for limit ordinal . The set is obviously a club set in κ. We will define a group structure on each so that will be a subgroup of whenever , and thus the group structure on will be well defined.
The following properties will hold on the groups for all :
is free of infinite rank;
is a proper subgroup of ;
is a free factor of if and only if ;
We let be free of countably infinite rank. Suppose that we have defined the group structure on for all such that the above conditions hold. If and , then we give the group structure obtained by any bijection such that is the identity map. Such a bijection exists since . Conditions (i), (ii) and (iv) obviously hold. Notice also that, for , we have a free factor of if and only if is a free factor of (this uses Lemma 2.1), and so (iii) also holds.
If and , then we consider two subcases. Firstly, suppose that there exists a strictly increasing sequence for which
with both free factors nontrivial for all .
From this, it follows immediately that . Since is a strictly increasing sequence of sets, we know for each that it is either the case that
By choosing a subsequence, we can assume without loss of generality that
(else we exchange the roles of and in the construction that follows). Since each , we know by our induction that is a free factor of and also of . Then, by Lemma 2.1, we know that is a free factor of and of . Inductively select a free basis Q of such that is a free generating set for for each . For each , select , and let . Let be a free generating set for . By how the set was defined, we know . Hence we let be given a group structure using a bijection such that is the isomorphism noted above and the inclusion map satisfies , 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 is not a free factor of by Lemma 2.3 (iii). Notice that, for each , we have that is a free factor of by Lemma 2.3 (iii). For each , we know is a free factor of
and for each , there exists for which . It follows by Lemma 2.1 and by our induction that, for , we have a free factor of if and only if , and condition (iii) holds.
On the other hand, suppose and and no such increasing sequence exists. Since is of cofinality ω and is not stationary in δ, we may select a strictly increasing sequence such that . Then . As each is a free factor of and the are strictly increasing, we may select by induction a free generating set Q for such that is a free generating set for . Pick and , and letting and , we can let be a group freely generated by such that the inclusion map is the map ϕ from Construction 2.2. The check that the induction conditions still hold is as in the other subcase.
When is a limit ordinal, the binary operation on is defined by that on the for . By how was chosen, we know is not stationary in β, and so we select a club set for which . By induction, we know that, for with , we have that is a proper free factor of , and since C is closed, we have for any 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 for which is a free generating set for for each . Then (i) and (ii) hold. For , it is clear by Lemma 2.1 that is a free factor of if and only if is a free factor of for some with , and since is a free factor of for each , condition (iii) holds. Condition (iv) follows by induction since . 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 with nontrivial subgroups of κ. Letting be such that and are both nontrivial and
it is straightforward to verify that C is club in κ. Since the set is stationary in κ, we know is stationary and therefore unbounded in κ. Then the closure is club in κ, and so is . Then there exists with and and . Then , and so . As , we know that δ has cofinality ω in κ. Certainly, , and so there exists a strictly increasing sequence with such that . Then
By our construction, we know that includes into in such a way that is not a subgroup of (using Lemma 2.4 and letting or depending on whether we selected or, respectively, in the construction), and this is a contradiction. Thus κ is freely indecomposable, and we have part (1).
For part (2), we let with . By the regularity of κ, select large enough that . Notice that . Any subgroup H of κ with and satisfies for some by the regularity of κ. Since is a free factor of by our construction, we have by Lemma 2.1 that is a free factor of H. Thus is κ-pure, and the group κ is strongly κ-free, and we have verified (2).
For part (3), we notice that, since
the equality holds for all . In particular, the homomorphism induced by the inclusion map is injective, and so the abelianization of κ is the increasing union of free abelian subgroups . In our construction, when and , the map induced by inclusion is an isomorphism by Lemma 2.3 (iv). When and , we have as a proper direct summand of . As well, we have for limit . 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 with . We have verified (3) and finished the proof of the theorem. ∎
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.
The author thanks the anonymous referee for their careful reading and suggested improvements which enhanced the presentation of the paper.
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