Saharon Shelah and Lutz Strungmann- The failure of the uncountable non-commutative Specker...

8/3/2019 Saharon Shelah and Lutz Strungmann- The failure of the uncountable non-commutative Specker Phenomenon

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The failure of the uncountable non-commutative

Specker Phenomenon

By

Saharon Shelah and Lutz Strungmann

Abstract

Higman proved in 1952 that every free group is non-commutativelyslender, this means that, for a free group G and for a homomorphism hfrom the free complete productZ of countably many copies ofZ into G,there exists a finite subset F and a homomorphism h : FZ G suchthat h = hF where F is the natural map from Z into FZ. Due tothe corresponding phenomenon for abelian groups this is called the non-commutative Specker Phenomenon. In the present paper we shall showthat Higmans result fails if one passes from countable to uncountableand with it answer a question posed by K. Eda. In particular, we willsee that, for an uncountable cardinal and for non-trivial groups G

( ), there are 22

homomorphisms from the free complete productof the Gs into the integers.

Introduction

In 1952 Higman [H] proved that every free group G is non-commutatively slen-der where slenderness means that any homomorphism h from the free completeproduct Z of countably many copies of the integers into G depends on finitelymany coordinates only. A similar result was proven by Specker in 1950 [S] forabelian groups. Specker showed that any homomorphism from the productZ of countably many copies of Z into the integers is determined by onlyfinitely many entries. These two phenomenons are called the commutative and

Mathematics Subject Classification (2000): 20E06.

Number 729 in list of publications.Supported by GIF project No. G-0545-173.06/97 of the German-Israeli Foundation forScientific Research & Development

Supported by the Graduiertenkolleg Theoretische und Experimentelle Methoden derReinen Mathematik of Essen University.

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the non-commutative Specker Phenomenon, respectively. Eda [E1] extendedHigmans result by showing that, for any non-commutatively slender group S,for any non-trivial groups G ( I) and for any homomorphism h from thefree -product of the Gs into S, there exist a finite subset F of I and a ho-momorphism h : iFGi S such that h = hF where F is the natural mapfrom iIGi to iFGi (for the definition of -product see 1.2). Motivated bythis result Eda [E1, Question 3.8] asked whether or not the non-commutativeSpecker Phenomenon still holds if one passes from countable to uncountablecardinals replacing Z by the free complete product Z for some uncountablecardinal (see 1.2). Here we shall give a negative answer to Edas question byconstructing, for a given uncountable cardinal and for non-trivial groups G( ), a homomorphism h from the free complete product of the Gs intoZ for which the non-commutative Specker Phenomenon fails. In fact, we will

show that there are 22

of these homomorphisms and so, in particular, we havethat the cardinality of the set of all homomorphisms from G into theadditive group of the ring of integers is the largest one possible. This contraststhe countable case and also the abelian case.

Basics and notations

Let I be an arbitrary set. For groups Gi (i I), the free product is denotedby iIGi (for details on free products see [M]).Given arbitrary subsets X Y of I we put XY : iYGi iXGi tobe the canonical homomorphism. Moreover, we use the notation X I forfinite subsets X of I. Then the set {iXGi : X I} together with thehomomorphisms XY (X Y I) form an inverse system; its inverse limitlim

(iXGi, XY : X Y I) is called the unrestricted free product of the

Gis (see [H]).Eda [E1] introduced an infinite version of free products and defined the freecomplete productiIGi of the groups Gi (for the exact definition see 1.2); itis isomorphic to the subgroup

FI{iFGi lim

(iXGi, XY : X Y I)}

of the unrestricted free product.For the convenience of the reader who is not familiar with the notion of a freecomplete product we recall the definition of words of infinite length and alsothe definition of and some basic facts about iIGi as can be found in [E1].

Definition 1.1 Let Gi (i I) be non-trivial groups such that Gi Gj = {e}for i = j I. The elements of

iIGi are called letters.

A word W is a function W : W iIGi from a linearly ordered set W into

the set of all lettersiIGi such that W

1(Gi) is finite for any i I. If the

domain W of the word W is countable then we say that W is a -word.

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The class of all words is denoted by W(Gi

: i I) (abbreviated by W) and theclass of all -words is denoted by W(Gi : i I) (abbreviated by W).

Two words U and V are said to be isomorphic (U = V) if there exists an order-isomorphism : U V between the linearly ordered sets U and V such thatU() = V(()) for all U. Identifying isomorphic words it is easily seenthat W is, in fact, a set. Moreover, for words of finite length (i.e. with finitedomain) the above definition obviously coincides with the usual definition ofwords.For a subset X of the set I the restricted word (or subword) WX of W is givenby the function WX : WX

iX Gi with WX = { W : W()

iX

Gi}

and WX() = W() for all WX . Therefore WX W. Using restrictedwords with respect to finite subsets of I we define an equivalence relation onW by saying that two words U and V are equivalent (U V) if UF = VFfor all F I where we may consider UF and VF as elements of the freeproduct iFGi. The equivalence class of a word W is denoted by [W] and thecomposition of two words as well as the inverse of a word are defined naturally.Thus W/ = {[W] : W W} together with the representative-wise definedcomposition form a group.

Definition 1.2 Given groups Gi (i I) the free complete product iIGi isdefined to be the group W(Gi : i I)/ as described above. Moreover, thefree -productiIGi is the group W

(Gi : i I)/ which is a subgroup ofiIGi.If Gi is isomorphic to a fixed group G for all i I then we write IG and IG

instead ofiIGi and iIGi, respectively.

Note, for a finite set I, we obviously have that iIGi and iIGi are isomor-phic to iIGi. In general we have, by [E1, Proposition 1.8], the free completeproduct iIGi is isomorphic to the subgroup

FI{iFGilim

(iXGi, XY :

X Y I)} of the unrestricted free product. Moreover, Eda [E1] proved thateach equivalence class [W] is determined uniquely by a reduced word; a wordW W(Gi : i I) is said to be reduced if W = U XV implies [X] = e for anynon-empty word X, where e is the identity, and it never occurs that the lettersW() and W() belong to the same Gi for neighbouring elements and ofW.

Lemma 1.3 (Eda, [E1]) For any word W W(Gi : i I) there exists areduced word V W(Gi : i I) such that [W] = [V] and V is unique up toisomorphism.

Furthermore, Eda [E1] showed the following lemma where a word W W(Gi :i I) is called quasi-reduced if the reduced word of W can be obtained by

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multiplying neighbouring elements without cancellation.

Lemma 1.4 (Eda, [E1]) For any two reduced words W, V W(Gi : i I)there exist reduced words V1, W1, M W(Gi : i I) such that W = W1M,V = M1V1 and W1V1 is quasi-reduced.

We would like to remark that the free -product IZ is isomorphic to thefundamental group (see [E1]) and the free complete product IZ is isomorphicto the big fundamental group of the Hawaiian earring with I-many circles (see[CC]). Hence free complete products are also of topological interest.

The uncountable Specker Phenomenon

In 1950 E. Specker [S] proved that, for any homomorphism h from the directproduct Z of countably many copies ofZ into the additive group of the ring ofintegers Z, there exist a finite subset F of and a homomorphism h : ZF Zsatisfying h = hF where F : Z ZF is the canonical projection. This resultis called the Specker Phenomenon. It can be easily seen that Speckers resultstill holds if one considers homomorphisms into any free abelian group G insteadof homomorphisms into Z, i.e. free abelian groups are slender. In general, anabelian group G is said to be slender if G satisfies the above property for anyhomomorphisms h : Z G. For generalizations to products of uncountablymany copies of Z within the category of abelian groups we refer to [EM] or[F1].In [E2] Eda introduced a non-commutative version of slenderness and that is

exactly what we shall consider here.

Definition 2.1 A group G is non-commutatively slenderif, for any homomor-phism h : Z G, there exists a natural number n such that h(\{1, ,n}Z) ={e} where e denotes the identity element of G.

Eda proved that non-commutatively slender groups are torsion-free and thatnon-commuative slenderness for abelian groups is the same as the ordinary(commutative) slenderness (see [E1, Theorem 3.3. and Corollary 3.4.]). More-over, he proved that non-commutatively slender groups have the following niceproperty:

Proposition 2.2 (Eda, [E1]) LetGi (i I) be non-trivial groups, let S be a

non-commutatively slender group, and let h : iIGi S be a homomorphism.Then there exist a finite subset F of I and a homomorphism h : iFGi Ssuch that h = hF where F is the canonical map fromiIGi to iFGi.

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Another interesting result is that the restricted direct product and the freeproduct of non-commutatively slender groups Sj (j J) are non-commutativelyslender (see [E1, Theorem 3.6.]). However, the first fundamental result on theclass of non-commutatively slender groups was already obtained by Higman [H]in 1952; it is stated below.

Theorem 2.3 (Higman, [H]) Every free group is non-commutatively slender.

Contrary to Higmans result above we will show that the non-commutativeSpecker Phenomenon fails if one replaces the product of countably many groupsby products of uncountably many. To be more precise: we actually show that,

for an uncountable cardinal , there are 22

homomorphisms from the freecomplete product G of non-trivial groups G ( ) into the additive

group of the ring of integers.To make the proof more transparent we first construct one homomorphismfor which the Specker Phenomenon fails and then modify the construction toobtain our main result.

Theorem 2.4 Let be any uncountable cardinal and G ( ) non-trivialgroups. Then there exists a homomorphism : G Z for which theSpecker Phenomenon fails.

Proof. Let G ( ) be a collection of non-trivial groups with identityelements e and choose elements g = e ofG for each . For any regularuncountable cardinal we define the word M G as follows:

M : (,


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Dealing with Occ+ two convex subsets J1

, J2

of (X, 0. Since [n, ) for all n we can find jn Jn such that

X(jn) = M,n() = M,()

for all n . But all Jn are pairwise disjoint and therefore X1(G) is infinitewhich contradicts the definition of a word (see 1.1). Thus Occ+ (X)/ andalso Occ (X)/ are finite sets.We now define : G Z as follows:

W |Occ+ (X)/ | |Occ (X)/ |

where X is the reduced word corresponding to W. Note that is well de-fined by Lemma 1.3. Moreover, it follows immediately from the definition that(X

1) = (X) and also the Specker Phenomenon obviously fails for .Note that in general the sets Occ+ (X) and Occ

(X) are not of the same size,

e.g. (M,) = 1. It remains to show, however, that is a homomorphism.Therefore let X and Y be reduced words. By Lemma 1.4 there exist reduced

words X1, Y1 and M such that X = X1M and Y = M1Y1 and X1Y1 is quasi-reduced. Now it is easy to check that (XY) = (X1Y1) by definition andthe fact that XY = X1MM

1Y1. Hence

(XY) = (X1Y1) = (X1) + (Y1) = (X) + (Y),

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as X1

Y1

is quasi-reduced and thus the reduced word ofX1

Y1

is obtained withoutcancellation. 2

We would like to remark that the uncountability of in Theorem 2.4 is essentialfor the definition of the homomorphism and can not be omitted because ofHigmans theorem. Modifying the proof of Theorem 2.4 we obtain:

Theorem 2.5 Let be any uncountable cardinal andG ( ) be non-trivial

groups. Then there are 22

homomorphisms from the free complete product ofthe Gs into the additive group of the ring of integers. In fact there is anepimorphism from G onto the free abelian group of 2

copies of theintegers.

Proof. Let be uncountable and {G : } be given as stated. Wechoose the following family of reduced words M for 2. First we choosenon-trivial elements e = g G for . Let {I : 1} be a family ofpairwise disjoint subsets of each of which has cardinality . It is well-known(see e.g. [EK]) that for every 1 we can find a family {I, I : 2} ofsubsets ofI such that any finite Boolean combination of them is of cardinality and moreover there is I that belongs to each I, for every 2. Forevery 1 and for every 2 we choose a word M, such that its domainM, equals I, and M,() = g for I,.Then the composition M =

1

M, is a well-defined reduced word in W(G :

) for every 2. Before defining the claimed homomorphism let us first

state the crucial condition satisfied by the Ms ( 2):

(i) the domain M of M is well-ordered of order type 1 (the ordinalproduct) for each 2 and therefore has uncountable cofinality.

Now we repeat the construction given in Theorem 2.4 replacing by M andfor a reduced word X and 2 we define

Occ+ (X) = {J (X,


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we can see that the sets Occ+

(X)/

and Occ

(X)/

are finite for anyreduced word X and 2. Moreover, the maps : G Z definedby

V |Occ+ (X)/ | |Occ (X)/ |

where X is the reduced word corresponding to V, are well defined homomor-phisms since I, for every 2.

To obtain 22

homomorphisms we will show that there is a surjection onto thefree abelian group of 2 copies of the integers. Define

: G

2

Z

via

(V)() = (X)for a word V where X is the reduced word corresponding to V. As all thes ( 2) are homomorphisms, so is and we claim that is actually ahomomorphism from the free complete product of the Gs onto the direct sumof 2 copies of the integers

2

Z. First assume that this mapping is not into,

then there is a reduced word X and a sequence of pairwise distinct ordinals n(n ) such that n(X) = 0 for all n . Thus for each n there is aconvex subset Jn X such that w.l.o.g.

X Jn= Mn,n

for some n Mn,n Mn (n 1). But now, if 1 such that n < for all n , then the element g (since belongs to all sets I,n) appears

infinitely many times in X, i.e. X1(G) is infinite - a contradiction. Thus theimage of is contained in the direct sum of 2 copies of the integers.On the other hand, by the choice of the sets I, we certainly have that

Occ+ (M) = 0 = Occ (M)

for distinct , 2. Moreover,

Occ+ (M) = 1 and Occ (M) = 0

for any 2. Thus we obtain

(M) = (0, , 0, 1, 0, )

2

Z

and therefore is obviously surjective. Since there are 22 homomorphismsfrom the direct sum of copies ofZ to Z itself we are done. 2

We have a short remark.

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Remark 2.6 Note that the above proof gives us that

(i) the free complete product G = G contains a free subgroup H (thegroup generated by the words M) and;

(ii) there is a projection onto H.

The following theorem gives us a complete description of all interesting ho-momorphisms from G = G to the integers for uncountable and groupsG ( ). By interesting we mean interesting with respect to the SpeckerPhenomenon, i.e. if W is a finite subset of , then all homomorphisms fromthe subproduct GW = WG to the integers extend naturally to a homo-morphism from G to Z but these homomorphisms are not of particular interestfor us and well-understood, hence we will restrict ourselves to homomorphisms

from G to Z which are already zero on every finite subproduct of G.First note that the definition of M in the proof of Theorem 2.5 did notreally depend on the particular word M but only on the fact that M haduncountable cofinality. It is immediate to see that for any word M such thatthe domain M has uncountable cofinality we can define such a homomorphismM : G Z. Hence we define the following set:

IG = {M G : cf(M) 1}

and letG : G

MIG

ZM

be defined by G(V) = (M(X) : M IG) where X is the reduced wordcorresponding to V. Now G is well-defined and we have the following theorem.

Theorem 2.7 Let G = G for some uncountable cardinal and groupsG ( ). Then any homomorphism : G Z which is zero on every finitesubproduct of G factors through G.

Proof. For simplicity we assume that all words are already in reducedform. First we will show that the kernel of G is exactly Ker(G) = {M G : M contains no monotonic sequence of length 1}, where a monotonic se-quence of length 1 is just a subset ofM which is isomorhic to 1 or its inverse.Clearly we have that Ker(G) is contained in {M G : M contains no mono-tonic sequence of length 1}. Conversely, if M is a reduced word that contains

no monotonic sequence of length 1 and G(M) = 0, then there exists a re-duced word N IG, i.e. N has uncountable cofinality, such that N(M) = 0.But then Occ+N(M) or Occ

N(M) is non-trivial and hence M must contain a

monotonic sequence of length 1 - a contradiction. It is now enough to provethat any homomorphism f : G Z which is zero on any finite subproduct of

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G acts trivial on Ker(G

). For this assume that M is a reduced word whichcontains no monotonic sequence of length 1 such that f(M) = 0 for somehomomorphism f : G Z. We distinguish between three cases:

Case a: There exist subwords Nn of M (n ) such that

(i) Nn is a convex subset of M;

(ii) the Nns (n ) are pairwise (almost) disjoint;

(iii) f(Nn) = 0 (without loss of generality f(Nn) > 0).

Hence the composition N of the words Nn (n ) is a well-defined word inG and applying Theorem 2.3 together with [E1, Proposition 1.9] leads to a

contradiction.

Case b: There is an initial segment M

of M such that

(i) f(M) = 0, where M = M M ;

(ii) for every proper initial segment N of M

we have f(N) = 0, whereN = M N;

(iii) M

has no largest element or for every t M

there exists a convex

subset Nt {m M

: m t} such that f(Nt) = 0, where Nt = M Nt .

Then the cofinality of M

has to be 0 by the assumptions and we choose anincreasing, unbounded sequence {t

n: n } in M

and put

Nn = {m M

: m tn} or Nn = Ntn (hence f(Nn) = 0).

In both cases we easily obtain a contradiction. Similarly we can use the samearguments for the inverse of M to get a contradiction.

Case c: Neither case a nor case b is satisfied. Then it is easy to see that the set

J = {t M : f(Mt) = 0}

is finite, where Mt = M {t} (e.g. use Ramseys Theorem). So without loss ofgenerality we may assume that J is the empty set. We let

I = {A M : A convex and B A convex f(M B) = 0}.

Then I contains all singletons t M, the empty set and it is downwards closed.Moreover, if A and B are elements of I and A B is convex, then A B I.Finally every initial segment of M with no largest element has an end segment

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in I and hence M I - a contradiction. Similarly we obtain a contradictionusing the inverse M1 instead of M. This finishes the proof. 2

For completeness let us state the following remark.

Remark 2.8 If h : G Z is any homomorphism, then an application ofTheorem 2.3 shows that the set { : h(G) = {0}} = F is finite. Hence,regarding FG as a subgroup ofG we let h0 = (h FG)F. Then h h0satisfies the assumptions of Theorem 2.7 and thus factors through G.

Acknowledgement: The authors would like to thank the referee for his help-ful comments and suggestions and in particular for his patience.

References

[CC] J. W. Cannon and G. Conner, The big fundamental group, big Hawai-ian earrings, and the big free groups, Topology Appl. 106 (2000), 273291.

[CE] G. Conner and K. Eda, Free subgroups of complete free products,preprint (2000).

[E1] K. Eda, Free -products and noncommutatively slender groups, J. ofAlgebra 148 (1992), 243263.

[E2] K. Eda, The non-commutative Specker Phenomenon, J. of Algebra 204

(1998), 95107.[EK] R. Engelking and M. Karlowicz, Some theorems of set theory and

their topological consequences, Fund. Math. 57 (1965), 275285.

[EM] P.C. Eklof and A. Mekler, Almost Free Modules, Set-TheoreticMethods, North-Holland Amsterdam - New York (1990).

[F1] L. Fuchs, Infinite Abelian Groups - Volume I and II, Academic Press,New York - London (1970 and 1973).

[F2] L. Fuchs, Abelian Groups, Hungarian Academy of Science, Budapest(1958).

[H] G. Higman, Unrestricted free products and varieties of topologicalgroups, J. London Math. Soc. 27 (1952), 7381.

[M] W.S. Massey, Algebraic Topology: An Introduction, Graduate Texts inMathematics, Springer Verlag New York - Heidelberg - Berlin (1967).

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[Sh] S. Shelah, Classification Theory, Studies in Logic and The Foundationsof Mathematics, NorthHolland Amsterdam - New York - Oxford - Tokyo92 (1990).

[S] E. Specker, Additive Gruppen von Folgen ganzer Zahlen, Portugal. Math.9 (1950), 131140.

Saharon ShelahDepartment of MathematicsHebrew UniversityJerusalem, Israeland Rutgers University

Newbrunswick, NJ U.S.A.e-Mail: [email protected]

Lutz StrungmannFachbereich 6, MathematikUniversity of Essen45117 EssenGermanye-Mail: [email protected]

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