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The independence of the Whitehead problem from ZFC Dean, Richard J. 1976

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THE INDEPENDENCE OF THE WHITEHEAD PROBLEM FROM ZFC by Richard J. Dean B.Math, University of Waterloo, 1974 A THESIS SUBMITTED IN PARTIAL. FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in THE FACULTY OF GRADUATE STUDIES Department of Mathematics We accept this thesis as conforming to the required' standard THE- UNIVERSITY OF BRITISH COLUMBIA April , 1976 © Richard J. Dean In p resent ing t h i s t he s i s in p a r t i a l f u l f i l m e n t o f the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r agree tha t permiss ion fo r ex tens i ve copying o f t h i s t he s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r ep re sen ta t i ve s . It i s understood that copying or p u b l i c a t i o n o f t h i s t he s i s f o r f i n a n c i a l gain s h a l l not be a l lowed without my w r i t t e n permi s s ion . Department of Mathematics The U n i v e r s i t y of B r i t i s h Columbia 2075 Wesbrook Place Vancouver, Canada V6T 1W5 Date A p r i l , 1976 i i Abstract An abelian group G i s c a l l e d a W-group i f Ext(G.Z) = 0. Whitehead's problem asks which groups are W-groups. Saharon Shelah proved that the answer to Whitehead's problem, for groups of c a r d i n a l i t y w 1 , i s independent of the axioms of Zermelo-Frankel set theory with the axiom of choice. This thesis gives a complete and d e t a i l e d proof, based on Shelah's proof, of t h i s independence r e s u l t . I l l Table of Contents Introduction page 1 Whitehead Groups and t h e i r Structure page 3 (G,Z)-Groups page 19 V = L and W-groups page 30 Martin Axiom and W-groups page 37 The Independence Result page 49 Bibliography page 50 Appendix page 51 i v Acknowledgment I would l i k e to thank Dr. Andrew Adler for h i s time and guidance throughout the past year. I would also l i k e to thank Dr. Stan Page fo r reading t h i s t h e s i s . L a s t l y I would l i k e to thank Ron Aiken f or h i s help i n t r a n s l a t i n g German. •1 Introduction The following r e s u l t was proved by Saharon Shelah i n (11). "The Whitehead problem, for groups of c a r d i n a l i t y IOJ , i s independent of•and consistent with ZFC." In t h i s thesis we present a proof of. t h i s r e s u l t based on Shelah's proof. Certain a l t e r a t i o n s had to be made, as well as a good deal of f i l l i n g i n of d e t a i l s . A d e s c r i p t i o n of some of the a l t e r a t i o n s i s given i n the appendix at the end of the t h e s i s . A Whitehead group, or simply a W-group, i s an Abelian group for which Ext(G,Z) = 0. (Ext(G,Z) = 0 i s a mapping property which w i l l be explained i n d e t a i l i n t h i s thesis.) The Whitehead Problem asks: "Are a l l W-groups of c a r d i n a l i t y ca^  f r e e l y generated." The axioms of ZFC, Zermelo-Frankel set theory with the axiom of choice-, are the axioms on which a l l current mathematics can be b u i l t upon. We w i l l show that within ZFC the Whitehead Problem cannot be solved.. We w i l l do t h i s by showing that within one model of ZFC a l l W-groups are f r e e , and within another model there e x i s t s non free W-groups. Godel exhibited a construction which produced a model of ZFC. His construction i s r e f e r r e d to as V = L and so ZFC + V = L i s consistent. Jensen showed that w i t h i n such a model of ZFC, a combinatorial property c a l l e d 'diamond' holds. We w i l l define and use t h i s property diamond to show that within a modsl of ZFC where V = L holds, a l l W-groups are f r e e l y generated. 2 M a r t i n and Soloway showed that MA (Martin Axiom) + 2 U > ooj-i s c o n s i s t e n t w i t h .ZFC. We w i l l d e f ine and use MA to show the exis t e n c e of non f r e e W-groups in. a model of ZFC i n which MA + 2 W > u)]/ h o l d s . . . And so the s i t u a t i o n i s t h i s . Let X be the f o l l o w i n g statement: " A l l W-groups of c a r d i n a l i t y u>1 are f r e e l y generated." Then: ( i ) ZFC + V = L i m p l i e s X and so ZFC + X i s c o n s i s t e n t . ( i i ) ZFC + MA + 2 W > a)1 i m p l i e s - i X and so ZFC + - i X i s c o n s i s t e n t . Thus X i s c o n s i s t e n t w i t h and independent of ZFC. 3 Whitehead Groups and t h e i r Structure In t h i s section we w i l l give some preliminary facts and d e f i n i t i o n s •.. about Whitehead groups and w.-free groups. We w i l l c l a s s i f y the oi^-free groups of c a r d i n a l i t y 0)^ into three p o s s i b i l i t i e s . In this thesis^ by group we w i l l always mean Abelian group. D e f i n i t i o n (1): G i s c a l l e d a Whitehead group or W-group i f for every epimorphism h:H ->''G such that the. kernel of h i s . isomorphic . to Z (the integers) there e x i s t s a homomorphism g:G ->• H such : that gh:G ->• G i s the i d e n t i t y map on G. D e f i n i t i o n (2): For. a group A. we say A i s the d i r e c t sum of subgroups B and C of A i f : (i) B + C = A where B + C i s the set of a l l sums of the form b + c where b i s in: B and c i s i n C. ( i i ) B C\ C = 0. where' 0, i s the i d e n t i t y element of A. . A A J This i s written as A = B © C. Lemma (1): The group H. as i n the d e f i n i t i o n of a W-group i s a d i r e c t sum of a copy of Z and a copy of G. Proof: Let g and h be as i n the d e f i n i t i o n of' a W-group. Since gh i s the i d e n t i t y map on G, g i s 1-1, so G* = image(g) - G. . By d e f i n i t i o n kernel(h) - Z. We show that H = G * © kernel(h)• C l e a r l y G* and kernel(h) have only 0 i n common, else gh could not be 1-1. Now l e t h* be h r e s t r i c t e d to G*.. If • . 4 x e H , t h e n x = h* ^ ( x ) + (x - , h * ^ h ( x ) ) ' and c l e a r l y h* 1h(x)£ G* and x - h* ^h(x)£ k e r n e l ( h ) . Some P r e l i m i n a r y F a c t s about W-groups ' D e f i n i t i o n ( 3 ) : A group i s c a l l e d f r e e i f i t i s i s o m o r p h i c t o a d i r e c t sum o f c o p i e s o f Z. D e f i n i t i o n (4) : A group i s u ^ - f r e e i f e v e r y c o u n t a b l e subgroup i s f r e e . I f G i s a W-group the n i t i s : ( i ) T o r s i o n f r e e ( i i ) co^-free From now on G w i l l be t a k e n t o be t o r s i o n f r e e and of c a r d i n a l i t y u>\. So we can assume w i t h o u t l o s s o f g e n e r a l i t y t h a t the elements of G a r e a l l t h e o r d i n a l s <. ipi where t o j i s t h e f i r s t u n c o u n t a b l e o r d i n a l : c o j = { a : a < t o } } = t h e s e t of a l l o r d i n a l s l e s s t h a n u ^ . D e f i n i t i o n (5.) : B i s a. pure subgroup of G . i f B fl zG = zB f o r a l l z € Z , where zG = {g£G:g = zx f o r some x€TG}. E q u i v a l e n t ! y B i s pur e i f f o r any z £ Z , b £ B , i f the e q u a t i o n z x = b i s s o l v a b l e i n G the n i t i s s o l v a b l e i n B. (2) page 178 (2) page 178 5 Lemma (2): Let G be a t o r s i o n free group of c a r d i n a l i t y co_'. Then G can be well-ordered as ^ a : a ^ ^ n s u c ^ a w a Y that f o r any l i m i t o r d i n a l 5, = { § a : a < <$} i s a pure subgroup of G. Proof: We define g. by t r a n s f i n i t e induction. The l i m i t ordinals • : "a l e s s than co]_ are p r e c i s e l y the ordinals of the form co$, where 0 < 0 < co],. Suppose that for every $ < y, and every a. < tog, g^ has been defined, and -G i s a pure subgroup of G. We w i l l show how. to extend.the d e f i n i t i o n so that G i s pure. There are two coy . cases to consider. Case ( i ) : Y i s a l i m i t o r d i n a l . Then of course g^ i s already defined for every a < COY • Since for every 3 < Y» G i s a pure cop , subgroup of G, and G = U :G „ , i t i s t r i v i a l to v e r i f y that . ^ 0<Y w f 3 G i s a pure subgroup of G. COY Case ( i i ) : Y i s a successor. Take a f i x e d well-ordering of G i n order type CO], with the. f i r s t element' i n the ordering ^ 0. Let g be the f i r s t element of G with respect to t h i s f i x e d order which i s not a g^ for any a < CO(Y -1) • Let B be the subgroup of G generated by a n c ^ ^ (where G^ = (j>) . Since G i s torsi o n free B has c a r d i n a l i t y co. So by (1), page 115, B i s contained i n a countable pure subgroup of G, say B*. As zg frG . T OJ(Y~1) for any z i n Z i t i s c l e a r that B*\G .. has c a r d i n a l i t y co, CO(Y-I) and so i t may be enumerated as {g :CO(Y-1) < a < COY). Thus G .= B* a — COY i s a pure subgroup of G. If G i s a (torsion free) group of c a r d i n a l i t y i o l s instead of l a b e l l i n g the elements of G by the ordinals • l e s s than co l 5 i t i s n o t a t i o n a l l y more convenient to assume the elements of G are the ordinals less than . Whenever such notation i s used, i t w i l l be understood that f o r any l i m i t o r d i n a l 6, G^ .= {a:a < <5} -i s a pure subgroup of G. Lemma (2) shows th i s i s . a harmless assumption. C a l l any such naming of G admissible. C l a s s i f i c a t i o n .of a^-Free Groups of C a r d i n a l i t y l O ] . We w i l l i n th i s section c l a s s i f y a^-free groups of c a r d i n a l i t y oo j into three p o s s i b i l i t i e s , c a l l e d unimaginatively P o s s i b i l i t y I, P o s s i b i l i t y I I , and P o s s i b i l i t y I I I . F i r s t we need a remark and then some preliminary group t h e o r e t i c and set theoretic- d e f i n i t i o n s . Remark ( 1 ) : For torsi o n free groups the equation zx = g can have at most one so l u t i o n , f o r zx = g = zy implies zx = zy Implies x = y. So i f zx = g i s solvable i n G, then the unique s o l u t i o n belongs to a l l pure subgroups containing g and thus the i n t e r s e c t i o n of pure subgroups i s again pure. This allows us to make the .following d e f i n i t i o n . . D e f i n i t i o n (6): Let <L,G>^ be the smallest pure subgroup of Gwhich. contains L, L<=LG. Remark (2) : If S i s a pure subgroup of G, then <L,S>jV = <L,G>.V by. Remark (1). We write <L>>V = <L,G>y. i f i t is. clear which group G we are r e f e r r i n g to. D e f i n i t i o n (7): Let S be a subgroup of G, L a f i n i t e subset of G, and a an element of G. We say H(a,L,S,G) holds i f <SUL> 4 = < S > A ® <L>jV but for no b £ < S U L U { a } > A i s . <SLVL'f{a}> J k = <S>A ©<Ll/{b}>.,c • D e f i n i t i o n (8) : A subset C of LOJ Is closed and unbounded i f : ( i ) ' For every non-empty subset S of C sup S £ C L / { t o i } . This says that C i s closed. ( i i ) sup C = a)].. This says that C i s unbounded. D e f i n i t i o n (9) : .A subset A of -u )_ i s stationary i f C f i A ^ <j) for every closed and unbounded subset of t o _ - . • D e f i n i t i o n (10) : Let X be a subset of t o _ . Then X i s co f i n a l i n i o _ i f for a l l a i n to there ex i s t s g i n X such that a <_ 3 . Remark (3) : No countable set. i s c o f i n a l with t o _ . . (3) page 207.. We are now ready to define the three p o s s i b i l i t i e s . D e f i n i t i o n ( l l ) : An toj-free group G of c a r d i n a l i t y •. CO] s a t i s f i e s P o s s i b i l i t y I i f for any admissible naming of G there i s some l i m i t o r d i n a l 6 < to']., and there are elements of G, ct say a n ( a ) ^ o r a l l . a < ai 1, (where n(ct) i s a f i n i t e ordinal) and subsets L^' = ^ a i : ® 4 '^ < n ( a ) } such that: (A) {a^ + G^:a < coi, I j^nta)} i s an independent family i n G/G, . o (B) n ( a a . . ,L ,G.,G) holds f o r a l l a < u, . n(ct) a o 1 Remark (4) : Since 6 < then G^ i s countable and so. we can assume without loss of generality that 6 = co' . Rename G. by o {ct:0 £. ct < co} which can be done as G. i s countable. Now rename' o the rest of G using .the technique of Lemma (2). D e f i n i t i o n (12): An toj-free group G of c a r d i n a l i t y co^  s a t i s f i e s P o s s i b i l i t y II i f G does not s a t i s f y P o s s i b i l i t y I and there i s a stationary subset of co l 5 say A, such that for ct any a i n A there are elements of G,. say a , £ <_ n(a), (where Ct n(a) i s a f i n i t e o r d i n a l ) , and subsets L = {a..:0-<_ i < n(ct) } such a Jc — that: Ct (A) ^ a £ : 0 4 ^ < n(ct)} i s an independent family i n G / G a (B) n(a°'/ \ ,L ,G ,G> holds. n(ct) a a D e f i n i t i o n P o s s i b i l i t y I I I (13): i f i t An co^-free group G of c a r d i n a l i t y coj s a t i s f i e s doesn't s a t i s f y P o s s i b i l i t y . I or P o s s i b i l i t y I I . 9 Lemma (3): The c l a s s i f i c a t i o n of a given group G to the^ three p o s s i b i l i t i e s depends on G only up to.isomorphism. Proof: We must show that under any admissible naming of . G, i t w i l l always s a t i s f y the same p o s s i b i l i t y . There are three cases to consider. Case (i) : Suppose G s a t i s f i e s . P o s s i b i l i t y I. Let h:G -> G*. be an isomorphism. Then •G* can be thought of as a renaming of the elements of G and so by the d e f i n i t i o n of P o s s i b i l i t y I, G* s a t i s f i e s P o s s i b i l i t y I. (Case ( i i ) : Suppose G s a t i s f i e s P o s s i b i l i t y I I . F i r s t we show that i f h:G G*" i s an isomorphism, then the set C defined by C = {6:h|Gr o i s an isomorphism from G^ onto G*} i s a closed.and unbounded subset of ui where hlG. i s the r e s t r i c t i o n of h to G.. G i s closed o o since the union of a chain of isomorphisms i s an isomorphism. Suppose C i s bounded. Choose a < c o _ such that a i s an upper bound for C. For n < c a define a i n d u c t i v e l y as follows: n ag = a a = sup ({h(6):6 <a }'{J (3:h(3) < a . ' } ) n n-1 . n-1 As ag < ui], then < coj for a l l n. That i s i f we assume i n d u c t i v e l y that a , < can , then a i s the sup of a countable set and since n-1 1 n no countable set i s c o f i n a l with coi , then a < co-i . Let a * = sup a . 1 n 1 n<& n Since a < coi for a l l n < co, then a* < coi . Let 3 £ G such that 3 < ct,-c, then. 3 < for some n, and so h(3) < a n +-^ the. d e f i n i t i o n of a t 1 - . S i m i l a r l y i f 3 £ G * such that 3 < OL* then 3 < a for n+1 n some n and so 3 = h(p) where p < a ,. . Thus h G . i s an isomorphism n+1 1 a* ' and so a*€C. Thus a i s not an upper bound f o r C and so C i s unbounded. . 10 Now we show that C C\ A, where A i s the stationary set required by the d e f i n i t i o n ' of P o s s i b i l i t y I I , i s a stationary set, and then G* w i l l s a t i s f y P o s s i b i l i t y II using C fl A as the required stationary set. CD A i s stationary because any closed and unbounded set is. stationary. That i s i f ' C^ and : are closed and unbounded sets then H C^ f cj>, for l e t {£^, £J2, ?3 v • • } he an increasing sequence of ordinals such that for n even E € C, and for n odd E 6C„ . n 1 n 2 Then tj) = sup. {£]_., f^,^., .. } = sup {E2> ?4 > ?6> • • • } A N ^ X/J i s i n both C^ and C^ since C^ and C^ are closed. A c t u a l l y C^O C^ i s closed and unbounded. C l e a r l y D C^ i s closed as C^ and. .C^  are closed. If C^H C^ was bounded, by say a < coj., then define a sequence {£J j.,£2> 53> • • • } as before x<rith £j = a to get a c o n t r a d i c t i o n . Thus C /I.C*, for any closed unbounded set C*, i s closed and unbounded. Thus (C 11 A) D C* = A f) (C n C*) ^ <j> . Thus C.-fl A i s stationary and so G* s a t i s f i e s P o s s i b i l i t y I I . Case ( i i i ) : By. d e f i n i t i o n P o s s i b i l i t y III holds i f and'only i f neither P o s s i b i l i t y I nor P o s s i b i l i t y II holds'. The c l a s s i f i c a t i o n into : the three p o s s i b i l i t i e s depends on. G only up to isomorphism. By the d e f i n i t i o n of the three p o s s i b i l i t i e s , an tu^-free group G can s a t i s f y only'one, . and so the three p o s s i b i l i t i e s form a p a r t i t i o n . The following lemma shows that each p o s s i b i l i t y ' i s s a t i s f i e d by a p a r t i c u l a r coi-free. group. 11 Lemma (4): • Each P o s s i b i l i t y i s s a t i s f i e d by some co._-free group. Proof: Again there are three cases to consider. Case ( i ) : We w i l l c o n struct an co_-fr.ee group s a t i s f y i n g P o s s i b i l i t y I. F i r s t we de f i n e a set C of i n c r e a s i n g sequences of n a t u r a l numbers of length co such that the c a r d i n a l i t y of C i s co_, and i f " n and. T are i n C, .n ^  T, then n and T have at most f i n i t e l y many n a t u r a l numbers i n common; that i s n / i x i s f i n i t e . To show that such a set C e x i s t s we give an example. Consider the f o l l o w i n g diagram: The sequences are defined by ta k i n g p o s s i b l e paths. For example: . {1,2,4,8,...} (1,2,4,9 } {1,2,5,10,...} e t c . By the n ^ 1 row of the diagram 2 n sequences or paths are defined, and i n the l i m i t there are 2 W >_ co j sequences. Choose any . co _ sequences. The i n t e r s e c t i o n of any two i s f i n i t e f o r they can agree only up to the p o i n t where t h e i r corresponding paths separate. 12 Let G be generated by: (i) x, for k < 10 m 0 0 k! (i i ) x '= , 1 (—r')x , for m < oo and xeC T k=m m. x(k) Using the notation of the definition of Possibility I and C = {.x(ct):ot < a)].} let:. (i) G^  be the group freely, generated by the x^' s-(i i ) . n(ct) = 0, and so L = <j>. a /. . . \ • ct m . • (ixx) an = x . . tor a < ton. and m fixed. . x(ct) 1 . We must show that G satisfies conditions (A) and (B) in the definition.of Possibility I. That i s : Ct (A) {a n + G.:'ct < OJI } is an independent family in G/G. . o • . . o .(B) n(ao,cj),G ,G) holds for a l l a < ooL . (A) follows from the f i n i t e intersection property of the elements of C. That is i f . z.x1". . +...+ z x™ ' . = g where z,£Z. z. ± 0, 1 i ( a i ) n x(ct ) i i 1 n x. (a.)£C, and g £ G / , then z,x, .+...+ z x , , is a f i n i t e l 6 l x ( a i ) n x ( a ) . . n linear combination of the x, 's that generate G„ . As the x . 's k 6 • x(a_^) are i n f i n i t e linear combinations of the x, 's , then z,x m. . +...+ z x™. . k 1 x(cti) n T(ct n) must be an i n f i n i t e linear combination of the x, 's since for i .^ j x™, . and x™. . agree at only f i n i t e l y many x ' s for x(ct..) C\ x(ct ') x ( a j x (a ) k I J is f i n i t e . This is a contradiction and so (A) holds. Now we show condition (B) holds. As L = <L then a <Gr U L >.=' <G > © <L > . Choose any a n = x™, . . Then 6 ct * . 6 * . a * . J u x(a) / , i s m+1 m m , . , r . . . (m + l)x , s = x / N - x . w . and so by definition of purity x(ct) x(ct) x(a)(m) J xm+\ G<G, U {x™, }>, . Similarly x k . . £<G r U { x m . }>, for a l l x(ct) 6 x(a) * J x(a) 6 x(a) * k ^ m + 1. Using the f i n i t e intersection property for elements 13 of C i t i s c l e a r that no other elements of G w i l l , be thrown i n t o <G.V ( x m . }> . . So i f f o r some x e <G. V • { x m , }> , , <G U {x™, . }> , = 6 x(a) 6 x(a) * ' 6 x(a) *. <G^ >5,. ©• < x > , v ' then we can assume that <x>5,{ = <x> = the group generated by x f o r some x = .2 , z.y. where z.e.Z and each y. i s some i = l i i l J i k k t or x . . . C l e a r l y x w i l l cause only f i n i t e l y many of the x > . k x(a.) . t(a) to be i n <^g>,'- ® <x>A ;and so t h i s i s impossible. Therefore n(x , \ A, G.,G) = n(a 0,L ,G.,G) holds f o r a l l a < ui and so c o n d i t i o n x (cO o :. a o x (B) i s s a t i s f i e d . Now we show that G i s oj_-free. I t i s s u f f i c i e n t to show that f o r any g g' £G , the pure subgroup generated by g.,...,g i n . I n i s f r e e on a f i n i t e number of generators, (4) page 25. Without l o s s of g e n e r a l i t y g ,...,g are independent and so <{g.,,...,g }>, I n 1 n x has rank, n (1) page 116. So l e t b b generate <{g,,.••,g }>, , 1 n 1 n :< that i s < ' * * • ' § n ^ > A = < { h ^ , . . . ,b^}> . We do an i n d u c t i o n on the number of generators. For n = 1 c l e a r l y . <{b^}> i s f r e e s i n c e G i s t o r s i o n f r e e . Assume any pure subgroup on n - 1 generators i s fr e e and l e t <{g. , . . ..g }>, be generated by b. b . I f I n * I n <{b-1,...,b }> i s not f r e e l y generated, by b,,...,b , then f o r I n I n some z.e Z, not a l l zero, .? • z.b. = 0 => z.b. = -z b =^  • i i = l l l i = l l i n n b £ < { b b . ,}>, . Thus " the pure subgroup generated by g,,...,g n 1 n-1 x  r ° I n has rank l e s s than n, a c o n t r a d i c t i o n . So b,,...,b f r e e l y 1 n generates <^S-^ > • • • J § n^ > and by the i n d u c t i o n hypothesis any pure subgroup generated by a f i n i t e subset of G i s f r e e . Thus G i s to . - f r e e . 14 Now let G* be any admissible naming of the elements of G. Choose 6 <'ID, such that x is in G* for a l l n < co . As 1 n 6 {x m, . :a < co i } is uncountable and G„ is countable, we can find x (a) 1 6 uncountably many x™. . 's such that . G • for k < co . let x (a) T (a) ' 6 {x™^^:3 < c o . } be such a set. By letting: ( i i ) n(3) = 0, and so L = <j> 3 /•••\3 m _ ( i n ) a.0 = x T ( g ) f o r 3 < c o . i t follows that G* satisfies Possibility I in exactly the same way as G was shown to satisfy Possibility I. Case ( i i ) : We w i l l construct a group satisfying Possibility II. For this example, the stationary set A required by the definition for Possibility II w i l l be the" set of a l l limit ordinals. First we show that this set A = {6 < co_:6 is a limit ordinal} is stationary. This follows from the observation that any closed and unbounded set C contains a limit ordinal. That is i f . {a.,a2,a3,... } is any countably in f i n i t e subset of C where a. < a.,, , then sup a. = a is a limit 1 1+1 1<GJ 1 ordinal for i f not then a has a predessor a - 1 which would be an upper bound to the sequence. Now for 6 a"limit ordinal, let x _ be a sequence of ordinals o of length co such that sup x „ (n) = 6 where x.(n) is the n'th n<co o o ordinal of the sequence x. . Let G be generated by: 6 (i) x for a < co-. a -1 m 0 0 k \ ( i i ) x = .1 (—r)x ' . for m < to , 6 < co n , and 6 6 k=m m. x J(k) . 1 a limit ordinal 15 Using the notation of the definition of Possibility II l e t : (i) x £ G. , X j i G . , for ct < 6 , m < cu , and 6 a limit ordinal ( i i ) n(<5) = 0 , and so L = <J> .o . . . / • • • \ & m . ( m ) ag = x^ , m fixed Since x™£G_ , then {x™ + G.} is an independent family in o o o o G/G and so condition (A) in the definition of Possibility II 6 holds. n(x.,L-,G.,G) .holds using the same arguement as used for o o o ' Possibility I , and so condition (B) is satisfied. The oj^-freeness of G is again similar to Possibility I. Lastly we must show that G doesn't satisfy Possibility I. It is sufficient to show that a given admissible naming of G does not.satisfy i t . Let 'G_ be generated by: o (i) x for a < 5 ct ' (i i ) x^ . . for $ < 6 , m < co , 3 a limit ordinal 3 Now define the x 1s to be increasing sequences for a l l limit ordinals S <•(!>! . Then It(x™,<j>p ,G) holds for any m < w .. The "II" condition cannot hold for any other x™' 's , 3 / <5 , since for 3 < 6 , x™ is 3 - 3 m Ic in Gr , and for 3 > 5 , <G,U {xn}>. = Gr ® <x>, where k is o o p o 3 " the largest element of the increasing sequence x less' than 6 . ' 3 As the x™ 's are the only p o s s i b i l i t i e s for creating the " n " condition, we can conclude that i t is satisfied at only countably many places for each G. . Thus Possibility I cannot hold and so o G satisfies Possibility II." Case ( i i i ) : Let G be the free group on cu_ generators. I t i s s u f f i c i e n t to show that G does not s a t i s f y P o s s i b i l i t y I or II fo r some admissible naming of G. Let' G be.generated by the elements a , 3 < co i . Let G be the group generated by a , 3 cot; 3 3 <_ £, for L not a l i m i t o r d i n a l . That i s G _ = „ffi^ G^  , where — CJcj S<(' G^ i s the subgroup of G generated by the element a. . Let G _ . 3 w5 be the group generated by a , 3 < for 5 a l i m i t o r d i n a l . P 3 3 That i s G = »©_G , where G i s the subgroup of G generated 3 • by the element a . C l e a r l y t h i s i s an admissible naming of G. Claim that n(a,L,G^,G ) does not hold f or any l i m i t o r d i n a l 5, where L i s a f i n i t e subset of G and a i s an element of. G. So suppose for some L and a that II(a,L,Gr,G) holds. If we can o show that only f i n i t e l y many element's are i n the group W where W •= <G. U L U {a}>./<<GpU L> U {a}>, then by the r e s u l t on page 24 CG57G^ i s i n f i n i t e i f "n" holds)' since W ='G /G^ , "11" must f a i l . If L = {a, ,.. . .a } and i f a = a , then each I n o m. a. =.E-,z.a. , where z.CZ, and a. i s the generator of G J . Then the only new elements i n .W w i l l be l i n e a r combinations of the a ' s which make up the a / s . C l e a r l y there are only f i n i t e l y many of. these i n W. Thus the "II" condition f a i l s i n G under t h i s admissible naming, and so P o s s i b i l i t y I or II cannot hold. Also G i s io_-free since any subgroup of a free group i s free, (1) page 74. Thus G must s a t i s f y P o s s i b i l i t y I I I . 17 Lemma (5) : Let G be ooi-free. Then P o s s i b i l i t y I II • i s equivalent to G being the d i r e c t sum of countable groups. Proof: Suppose G i s the d i r e c t sum of countable groups and G i s ct ct o)i-free. Then G = $ VG where each G i s countable. Since G a<co1 ct ct i s to T - f r e e , each G i s free,, so each G i s isomorphic to a countable d i r e c t sum of copies of Z. Thus G i s isomorphic to a d i r e c t sum of U[- copies of Z and so G i s free on to^  generators. By Lemma ( 4 ) case ( i i i ) , G s a t i s f i e s P o s s i b i l i t y III.. Now suppose G s a t i s f i e s P o s s i b i l i t y I I I . F i r s t we show that i f C i s a, closed and unbounded subset of OJ1 , then C* = {6:6 €C and 6 i s a l i m i t ordinal} i s also closed and unbounded. C*. i s closed since C i s closed and the sup of a sequence of l i m i t ordinals i s a l i m i t o r d i n a l . C* i s unbounded since C i s unbounded and the sup of an i n f i n i t e increasing sequence of ordinals of C i s a l i m i t o r d i n a l of C. Thus C* i s closed and unbounded and contains only l i m i t o r d i n a l s . . < Since P o s s i b i l i t y I and P o s s i b i l i t y II f a i l s , we can f i n d a closed unbounded set C .such that i f 6 £ C , there does not. e x i s t a £ G and L, a f i n i t e subset of G, such that n(a,L,G.,G) holds. 6 That i s , i f for every closed and unbounded set such a 6 e x i s t s , then by taking the set of these 6's we get a stationary set which s a t i s f i e s condition (B) of P o s s i b i l i t y I I . By taking L * £ L such that L* i s a maximal independent family i n L, then condition (A) of P o s s i b i l i t y II would be s a t i s f i e d using the L*'s i n place of the L's . Since P o s s i b i l i t y I f a i l s , then P o s s i b i l i t y II would 18 hold for G, a contradiction. Therefore such a C exists. From previous remarks, in this proof we can assume that C contains only limit ordinals. Since sup C = OJ. , then the cardinality of C is coi since no countable set is cofinal with OJ, . So let C = {6 :a < aii } where each 5 is a limit ordinal and a < 3 ^ 6 < 6„ . Now we a a 6 rename G as follows: Rename {3:6 <.' 6 < 6 as (3:toa < 3< w(a +1)}. a — ci+1 — Now we can assume that C = {toa:a < to.-}, and i t is clear that we s t i l l have an admissible naming of G. Now we do an induction to show that G , = G ffi <b, ,'b„, . . . >. coa+oj ua . 1 2 * for some b.'s in G . ^ G . Suppose b,, . . .b have been choosen. i wa+co a>a r r I n Let G n = G ® <b -,...b >., . Now let L = {b,,...b } and let I n " . 1 n a = inf '{6:6 6 G , — G n } . As to a is in C, then n(a,L,G ,G) toa+co tea toa f a i l s and so there must exist b ,.€ <G U L U {a}>, such that n+1 toot , * <G U L V {a}>, = G ® < L U {b ,,}>,= G © <bn,...,b ,b ,,>, . wa x ooa n+1 >c toa 1 n n+1. * Since clearly U G n =. G , we get that. G , = G ® <b,,b0,...>, . n toa toa+to , toa+to toa 1 2 5C Let H = <b, ,b o 5 ...>_. . Then G , =G <B H . Thus wa 1 1 > toa+w coa toa G = G • ffi H and so G is the direct sum of countable groups, (o l_<a<cj1 toa ' 19 (G,Z)-Groups In t h i s s e c t i o n we w i l l define (G,Z)-groups and prove some lemmas about them necessary for the consistency r e s u l t . D e f i n i t i o n (14): A (G,Z)-group i s a group H set G x Z = {(a,b):a£G, b £ Z } such that: (i ) (a,b) + (0,c) = (a,b+c) , ( i i ) The map h:H ->- G defined by h(a,b) = a For a given G/ , H w i l l denote.a (G^,Z)-group, homomorphism w i l l be denoted by h_^  . Lemma (6): Let G^ be a countable subgroup of G^ where G^ i s w i-free and the c a r d i n a l i t y of G„ i s at most W i . Let H be 2 1 a (G^,Z)-group. Then can be extended to a (G^,Z)-group. Proof: F i r s t note that G^  i s f r e e l y generated since i t i s countable and G^ i s wi-free. Thus from the r e s u l t noted before, G . i s a W-group since f r e e l y generated groups are W-groups. The r e s u l t w i l l be proved by t r a n s f i n i t e indection. To si m p l i f y the induction we w i l l , deal with two s p e c i a l cases f i r s t . Let <a + G^ > be the subgroup of G2^ G1 8 e n e r a t e c * by the element a + G , where a i s i n G^. Let G = <{a} U G > be the subgroup of G generated by {{a} U G }. a. j. z. x Case ( i ) : <a + G^ > i s isomorphic to Z. Case ( i i ) : <a + .G > i s c y c l i c of prime order. We w i l l show that i n cases ( i ) and ( i i ) H can be extended to a (G^,Z)-group. with underlying i s a group homomorphism. and the corresponding 20 Proof of case ( i ) : Suppose <a + G^ > i s isomorphic to Z. Then every b £ G has a unique representation as za + c where z e Z and c € G, . a 1 Now define.for b , b„ i n G and k , k i n Z , the following: 1 z • a 1 / (bl,k1) + (b 2,k 2) = (z a + c .k^ + ( z 2 a H-'c ,k 2) d i f ( ( Z l + z 2 ) a + c 3 , k 3 ) where (c^,k^) + ( c 2 , k 2 ) = (c^.k )' i n . I t i s easy to check that t h i s natural extension of H, forms a group. C a l l t h i s group H . 1 a Then (b,k) + (0,m) = (za + c,k) + (0,m) .= (za + c,k + m) ' = (b,k + m) , since i n (c,k) + (0,m) = (c,k + m). Also the mapping h :H -> G defined by h (b,k) = b i s c l e a r l y a homomorphism, and so H i s a (G ,Z)-group. a a Proof of case ( i i ) : Suppose <a + G^ > i s c y c l i c of prime order p. Since h^ '.H^ .-*- G^ has kernel isomorphic to Z and G^ i s a W-group, then there e x i s t s g :G -> H such that 'h g = 1 . Let g (c) = (c,m(c)) 1 i 1 1 1 Gi I for c £ G ^ . Every b £ G^ has a unique, representation as za + c where 0 <_ z < p , c £ G ' . Now for b, , b„ i n G and k, , k- i n Z ~~• 1 1 2 a 1 . 2 define: (bl,k1) + (b 2,k 2) def ( z x a + c ^ k ^ + ( z 2 a +c ,k 2) ^ ' ( ( z ^ + z 2>a + c^ + c2'^i + k2 ~ m ^ c p - m(c 2) + m(c^ + c 2) . + f ( z l + z 2 ) ) wiere 0 <.-z •, z < p , and f (n) = 0 when n < p and f (h) ,=. M€Z otherwise; where M i s an a r b i t r a r y constant which once chosen remains the same for a l l such defined sums. We w i l l show that t h i s set, c a l l i t H , forms a (G ,Z)-group under t h i s defined operation. To a a show H i s a group, the only non t r i v i a l thing to show the existence of inverses. Let (b,k) = (za + c,k) be i n H and - c be the a inverse of c i n • G . ; Then: (za + c,k) + ((p - z)a - c, -k - M) = ((z + (p - z))a + c - c,k - k - M - m(c) - m(c) + m(c - c) + f ( z + (p - z))) = (pa, -M + f(p)) = (0, -M + M) = (0,0) This the inverse of (b,k) i s ((p - z)a - c, -k - M) , and so H i s 3-a group. Now l e t (b,k) = (za + c,k) and (0,t) be i n H,. Then: a ' (b,k) + (0,t) = (b,k + t - m(c) - m(0) + m(c) + f(z>) = (b,k + t) , as z < p . Also the mapping h :H G, , defined by h (b,k) = b i s a homomorphism, a a 1 a so H i s a (G ,Z)-group. a a Since G^ i s .coj-free and G^ i s countable, then G^ i s countable and so i t i s f r e e l y generated. Thus G i s a W-group. a Now we do the-induction. F i r s t we f i n d a sequence of elements, of G^ , say A = {a^:6 < a,a an ordinal}, such that G^U A generates G„., and such that i f J . = <G, U {a : p < 6}> for a l l 6 < a, then 2 o 1 p <a^ + J^> i s i n f i n i t e c y c l i c or c y c l i c of prime order. The sequence A i s defined as follows: 22 Assume a„ has been defined for a l l g < 6. Let b = inf{a:a£ G.\ J.} g 2 6 If <b + J > i s i n f i n i t e c y c l i c or c y c l i c of:prime order, l e t a. = b. o , • . • • o If not, then <b + J > i s c y c l i c of non prime order, say of order np o where p i s prime. Then l e t a. = nb. So ' <a '+ J > has prime order p. 6 6 6 It i s clear that card (A) < oj; , since G^V A generates G^ and cardCG^) <_ o)_ . Let = , a (J^,Z)-group. We w i l l define to be a (J,Z)-group for a l l g <_ a. 3 • • . ~ (a) .3 i s not a l i m i t o r d i n a l : Since 3 i s not a l i m i t o r d i n a l , i t has a predessor. So we can suppose a (J 1,Z)-group, K , has been g-1 ii- L defined. Then by construction of the sequence A, <a . + J > i s .3-1 3-1 i n f i n i t e c y c l i c , or i s c y c l i c of prime order. I f i t i s i n f i n i t e c y c l i c then case ( i ) can be applied d i r e c t l y to K to show i t g-1 can be extended to a (J^,Z)-group K . If i t i s c y c l i c of prime order, then as ^ s countable, i t i s f r e e l y generated and so i t i s a W-group. Then there e x i s t s g :J K such that p-1 3-1 g-1 h g _ l § 3 _ l = * j where as usual h ^ i K ^ r> J ^ _ 1 and h(a,b) .= a. g-1 Thus we can apply case ( i i ) using g as the required map, and so •p-1 extend K to a (J,Z)-group K . (b) 3 i s a l i m i t o r d i n a l : Define K. •==IVr>Kr . I t i s easy to check • g 6<g 6 that K i s a (J,Z)-group. 3 R So i n d u c t i v e l y we can define a (J^,Z)-group. C a l l i t H . As. the set Gj f A generates G^ , then = G^ , and so i s a (Gn,Z)-group and the lemma i s proved. 23 Lemma(7): Let be a (G^,Z)-group. Let h^ and g^ be homomorphisms, n i : H ^ G-[ a n < ^ h (a,b) = a, g^:G H-^ ' , such that h g = 1 1 1 G-L . Let G^ be toi-free and card(G2) 4 to1 . Suppose n(a,A,G^,G2,) holds. Then can be extended to a (G2>Z)-group such that for no homomorphism g N:G -> H„ does :h g = 1 where g-2 2 Z *.Z Z z extends g^ and as usual h^Ca.b) = a. Proof: Let: (i) ' A = {a, ,...,a } 1 m ( i i ) G 3 = <G 1 U A>?v = <G 1> A 0 <A>5V ( i l l ) G. = <G„ U {a}> 4 3 (iv) G 5 = <G U A U {a}>i< mo v Let H. be a (G.,Z)-group. Consider the homorphisms g:G. -> H, 4 4 4 4 that extend g and such that h g = 1 . Any such g i s uniquely 1 4 determined by where g maps a.,...,a and a. That i s i f b€ G. , ' 1 m . 4 then b = c + za where c £ G^ and z £ Z . So b = d + x + za where d £ < G^ >,' { a n c ^ x £ < ^ > * ' x e <^- >* implies nx i s a l i n e a r combination of the a^'s for i = ,l,...,m. Thus the g(a_^)'s determine g(nx) = ng(x) , and so they determine g(x) since there i s a unique so l u t i o n to ng(x) = y. As d'£<G^> A , g(d) i s already determined by g^ . Let a n = a. As h g •= 1 and h. (b,z) = b, then g(a.)£{(a.,z):z£Z} U 4 G^ 4 - l l for i = 0,...,m. So each g(a^) can be defined i n only countably many ways and since there are only f i n i t e l y many a-^' s » there can be. only countably many such . g's.. C a l l them {g n:n < io) = R. Now we w i l l show that Gr/G, must be i n f i n i t e . Then we w i l l 5 4 ' make some observations about the structure of G^/G^ a n < ^ c l a s s i f y i t into two p o s s i b i l i t i e s . 24 G5'/G4 i s i n f i n i t e : Since G 5 = <G U A U {a}>^ $ < G 1 > . V ® < A u {a}>A-. , then there e x i s t s x£ <G^ A V {a}>5V , ;x^<G >A © <A U {a}>^ such that nx = g + c + ka where for some n ^ 1, g £ < G ^ > A , c 6 <A>4 , and k ^  0. Let n be the smallest p o s i t i v e integer for .which there i s such a g, k, and c. Then the greatest common d i v i s o r of n and k i s 1, for i f not then say m divides n and k. Then • m((—)x - (—)a) = g + e, and since <G. f A>. = <G > u ® <A>J , then m m 1 * 1 * . * ' there e x i s t s g ,6 <G >. and c,€ <A> . such that (—)'x - (—)a = g, + c • , • 1 . 1 * 1 * m m' 61 1 ' and so (—)x = g + c, + (—)a, contradicting the minimality of n. m l l m • Thus there e x i s t s integers w and z, such that nw + kz = I. Now consider wa + zx: nx = g + c + ka , so wg + wc = wnx - wka , and so wg + wc + k(wa + zx) = wnx -wka + kwa +kzx = wnx 4- kzx ' = (wn + kz)x = . x Similarly:' -zg - zc + n(wa + zx) = -znx + zka + nwa + nxz = (zk + nw)a = a So i f x was the only new element i n <G^ C A L/ {a}>.,. , then <G\^U A U (a}> A = <G1>,,{ © <A {wa +zx}> A , a contra d i c t i o n to II (a,A,G^ ,0 s) , since wa + zx i s i n <G^U A U {a}>A • So there must be a y such.that my = g + c + sa + tx, or my = g^ + c^ + u(wa + zx) for some u Z. Using the same method we can fi n d an element b i n <G U A U {a}>^ such that y and wa + zx are i n ® <A (/. {b}>^ . Since t h i s process can be repeated for any f i n i t e number of such elements, i t follows that there must be i n f i n i t e l y many of them else we get a con t r a d i c t i o n to the II condition. Thus .Gr/G. i s a • b 4 countably i n f i n i t e t o r s i o n group. D e f i n i t i o n (15): A group G i s d i v i s i b l e i f for every x i n G and every integer n, there e x i s t s elements i n G that s a t i s f y the equation ny = x. From Kaplansky (5), we have the following two r e s u l t s : (a) Any abelian group G has a unique largest d i v i s i b l e subgroup M, and G = M ® N where N has no d i v i s i b l e subgroups. (5) page 9. (b) Any d i v i s i b l e group i s a d i r e c t sum of groups, each isomorphic o  to the .additive group of r a t i o n a l s Q, or to Z(p ), the group of a l l p ^ roots of unity f o r various primes p. (5) page 10. As G^/G^ i s a to r s i o n group i t cannot have a subgroup isomorphic to Q. So i f G^/G^ has a non t r i v i a l d i v i s i b l e subgroup, then by Kaplansky s two r e s u l t s , G^/G^ contains a copy of Z(p ) for some CO prime p and the copy of Z(p ) i s a d i r e c t summand of the group. So suppose G^/G^ has no non t r i v i a l d i v i s i b l e subgroups. D e f i n i t i o n (16): A group G i s reduced i f i t has no non t r i v i a l d i v i s i b l e subgroups. From Kaplansky we have the following result". (c) I f G i s a reduced group which i s not torsion free, then G has. a f i n i t e c y c l i c summand. (5) page 21. Since G^/G^ i s a reduced t o r s i o n group, then by (c) G has a f i n i t e c y c l i c summand. Now apply (c) to the other summand. Repeated a p p l i c a t i o n of (c) to the i n f i n i t e remaining summand of G^ /G^ . shows that G^/G^ contains an i n f i n i t e d i r e c t sum of f i n i t e c y c l i c groups. From each of these choose an element of prime order. Thus we can assume that there e x i s t s i n f i n i t e l y many d i s t i n c t elements, say a^ for n < co, such that p a € G, where each p i s prime. . n n 4 n Now we can say that one of the following p o s s i b i l i t i e s occurs i n G./G.: D 4 (I) Gj_/G^ contains i n f i n i t e l y many elements of prime order. CO or (II) 'G /G^ contains a copy of Z(p ) for some p. l e t (I) hold i n Gr/G. . Let a , n < co, be the elements of 5 4 n . prime order. That i s the element a + G, has order p i n Gn/G, . n 4 n 5 4 ' * . Let G be generated by G. U { a , , a We w i l l i n d u c t i v e l y n ^ 4 O n - 1 ft ft ' * define a (G ,Z)-group H , using Lemma (6) so that H ' n+1 n+1. n+1 * • extends . F i r s t use Lemma (6) to extend to a (G^,Z)-group ft which we w i l l c a l l . C l e a r l y t h i s can be done as G^ and G^ meet a l l the conditions of Lemma (6). That i s G^ i s a countable subgroup of G^ , and' G^ i s c o j-free as i t i s a subgroup of the c o i-free group G^ . Also i s given to be a (G^,Z)-group. Thus • * >v * .ft H„ e x i s t s . Assume in d u c t i v e l y H i s defined. G and G ,, 0 n n n+1 ft s a t i s f y the conditions of Lemma. (6) using as the required ft ft ft (G n,Z)-group. So by Lemma (6) can be extended to a (G^ +^,Z)~ ft . f t group, H . As a. + G. has order p , a + G has order p • n+1 n 4 n' n n n ft - ft Let M be the constant used i n Lemma (6) to extend H to H ,, . n n n+1 ft. ft Inductively we define H . Again apply Lemma'(6) to extend H CO CO to a (G,_,Z)-group, say . H,. , which extends a l l the H n ' s -27 If g :G,. H •' i s a homomprphism extending g^ such that h.gr = l r then for some n, g extends g n ; that i s g| T/ c = g n 5 -> 0 where g n £ R as defined e a r l i e r i n the proof.. Now we w i l l show i t . * ' ft . •'< using g 5 l H * = § n as the required map i n extending to H + _ > n (see Lemma (6), case ( i i ) ), that constants M can be chosen •n i ft such that g r - L * has no extension to G for each n < to. This 5 ii n+1 n • * n w i l l show that g^ and thus g has no extension to G^ and so such a gj. does not e x i s t . As a + G. i s of order p , then l e t p a = b 6 G, . Let n 4 n n n n 4 n ft g (b ) = (b ,k ) and g (a ) = (a ,c ). Since b i s i n G, and n n n n n n n n 4 ft xi n g extends g we have g (b ) = g (b ) and so: n n n n g n(b ) = (b ,k ) by d e f i n i t i o n • n ' n n = g X(b ) as b € G. n n . n 4 ft = g (p a ) as p a -= b n n n .n n n ft ft = P g (a ) as g i s a homomorphism n n n n 1 = p (a ,c ) by d e f i n i t i o n n n n = (b ,p c + M ) by d e f i n i t i o n of "+" i n H" as n n n n n+1 as defined i n Lemma (6), case ( i i ) . So i n H ,,: n+1 p (a c ) = (a ,c ) +.....+(a ,c ) (p times) n n n n n. n n n = (a + a ,c + c + f ( l + 1)) + (a ,c ) + + (a ,c ) n n n n n n n n = (2a ,2c ) + (a ,c ) + + (a ,c ) n n n n n n = ( ( P n - . D a n , ( P n - D O + (an,cn) = ( V n ' P n C n + f ( P n } 0 . = (b ,p c + M ) n n n n 28 R e c a l l that the constant M as chosen i n Lemma (6) case ( i i ) n was a r b i t r a r y . By the c a l c u l a t i o n on the previous page we have that k = p c + M and so k E M (mod p ). By choosing M = k + 1 , n n n n n n n n n t h i s i s impossible,- and so g cannot be extended to G and so n n+1 n • & n g cannot be extended to G ,, and so g cannot be extended to Gn n+1 - 5 Now suppose (II) holds. That i s G^/G^ contains a copy of 0 0 / . C O Z(p ) for some p. Then from the structure of Z(p ) there are elements, say a for n < OJ, such that: n (a) p a Q - b Q € G 4 (b) pa - a = b 6 G '• n n-1 n 4 That i s an i s a p t b root of unity and a i s the ( p n ) ^ root of 0 n * unity such that (pa - a ,) = 0(mod G,). Again l e t G be generated n n - i 4 n • * * by G. U {a ,...,a }. and l e t H be a (G ,Z)-group constructed 4 0 n-1 n n . in d u c t i v e l y as before using the constants M . We w i l l again show n that by proper choice of the M n ' s » that g n £ R has no extension to * * Gn + l a n d thus no extension to G . As before l e t g (a_) = (a ,c ) n n and g n ( b n ) ="g*(b n) = (b n,k n)-- Then-: = -= pg^(a 0) , =: p ( a o ' c o ) = ( b o> p c o + V And so k = pc Q-+ M Q or k Q = MQ(mod p). Also: g n(b ) = (b ,k ) • by d e f i n i t i o n n n n = g\bn) ^ b n £ G 4 29 = V p a n " Vl> b y ( b ) ft ft ft = P8 ( a ) ~ S (a -j) as g i s a homomorphism n n n n n-1 = p(a ,c ) - (a ,c ) by d e f i n i t i o n n n n-1 n-1 . = (b + a ,pc. .+ M ) - (a . ,c .) n n-1. n n n-1' n-1 "+" i n H n+1 = (b.,pc + M . - c ) "+" i n using fact n n n. n-1 n+1 ft that a , H n-1 n And so k = pc + M - c , or k + c , = M (mod p). n n n n-1 n n-1 n Thus we have: (1) k^ = M^mod p) (2) k + c = M (mod p) n n-1 n Keeping i n mind the M^'s were chosen a r b i t r a r i l y we can do the following. For (1) choose M,. = kn + 1 and for (2) choose M = k + c , + D O n n n-1 ft n C l e a r l y In both cases no such k's e x i s t . Thus g^ , and so g , cannot be extended to . F i n a l l y use Lemma (6) to extend H,. to a (G^,Z)-group, say Let g„ be any. homomorphism g~:G ->- H such that h g — 1' and I. , 2 2 2 2 2 G2 g 2 extends g^ . Then g 2 extends some g n 6 R. We have just shown n * * ft that g cannot be extended to g ,, such that h •, ,e = 1 * n+1 n+1. n+1 G n+1 * n -As extends ^ n + ^ l t : follows that g cannot be extended to §2 ' Thus g 2 does not exist . . Therefore H 2 s a t i s f i e s the requirements of the lemma. 30 V = L and W-groups In t h i s section we w i l l show that under the assumption V = L, groups s a t i s f y i n g P o s s i b i l i t y I or P o s s i b i l i t y II are. not W-groups. Lemma ( 8 ) : If G s a t i s f i e s P o s s i b i l i t y I or II then G can 6 be named so that for any l i m i t o r d i n a l 6, there e x i s t s an element a ' and a f i n i t e subset L. such that II(a ,L r, G.,G_, ) holds. o 0 0 0+0) Proof: There are two cases to prove. Case ( i ) : Let G s a t i s f y P o s s i b i l i t y I. Thus G i s named such that for any l i m i t o r d i n a l 6 < ui'-i , G i s a pure subgroup and for o some l i m i t • o r d i n a l g, G i s the p a r t i c u l a r pure subgroup required p by conditions (A) and (B) i n the d e f i n i t i o n of P o s s i b i l i t y I.. That i s : (A) {a + G :ct < a} , I 4 n(ct)} i s an independent family i n G/G . . Jo p p (B) (a , . ,L',GJG) holds for a l l a < wi where L = {a„:£ < n(ct)}. Let $ = {6:6 i s a l i m i t o r d i n a l , 6 < , and G-- does not s a t i s f y conditions (A) and (B)}. That i s there do not exist s u i t a b l e a£' s> a < cu 1 , such that G^ could replace G^ i n the above. We claim that $ i s bounded i n oon . If. $ i s unbounded then card($)= 10j , say $ = {6 :a < aii } where 6 < 6 i f a-i < a? . Rename G so ct 1 • cti ' 012 • that. {6:6 <e'<6 ,.} becomes {9:CJ < 0 < oj(ct+l)}. This renames G err : ct+1 a — so that- under the new ordering, i f 6 i s a l i m i t o r d i n a l , then G R 0 cannot s a t i s f y conditions (A) and ( B ) . As G must s a t i s f y P o s s i b i l i t y I under any naming that i s admissible, by the d e f i n i t i o n of P o s s i b i l i t y I, t h i s i s a c o n t r a d i c t i o n . Thus $ i s bounded, by 3i say p < oi. where p is a limit ordinal. Now rename G so that G . becomes G and {8:p+toa <_ 9 < p+co(a+l)} . becomes {8:co+coa <_ 6 < co+co(a+l)}. p. co — Under this admissable naming i f 6 is a limit ordinal then G^  w i l l satisfy conditions (A) and (B) for some a ' s, a < oy_ . Now consider G . As G satisfies conditions (A) and (B), CO CO then there exists a and L such that II(a,L,G ,G) holds. Choose a co limit ordinal 6 < coi so that <G U L U {a}> £ Gr . Thus II(a,L,G ,G) x (0 * 0 CO holds. Now rename G so that {9:<o <^  9 < 6} becomes {9 :co <_ 6 < co+co} and {9:6+coa. <= 9 < 6+co(a+l)} becomes {9:co+co+coa 4 9 < co+co+co (a+1) } for a < coi . Under this naming n(a,L,G ,G , ) holds. Now we do the x CO CO+CO induction step. Suppose G has been named so that for a l l 6 < g 6 6 there exists a and L. such that II(a ,L„,G .„,G „, ) holds.: Since o 6 coo coo+co G satisfies conditions (A) and (B), then there exists a and L cop such that II(a,L,G n,G) holds. As before choose a limit ordinal p < coi cop such that <G L {a}>. G and so (a.L.G „,G ) holds. Rename cog * p cog p G so that G remains unchanged, {9: cog ^ 9 < p} becomes {9:cog <_ 8 < cog+co}, cog ' , — and {9:p+coa <; 9 < p+to(a+l)} becomes {8 : to g+co+coa ^ 8 < cog+co+co (a+1) }. Thus n(a,L,G „,G n, ) holds. cog cog+co 6 Thus we can assume G can be named so that II (a ,Lr)G.,G ) holds o 6 o+co 6 for any limit ordinal 6 < coi and suitable a 's and L.'s. o • Case ( i i ) : Let G satisfy Possibility II. The proof is almost the same. Let A be the required stationary set in the definition of Possibility II. : Let A = {6 : a < coi }, and of course the 6 's are a 1 a limit ordinals with 6 < 6 i f a i < a ? . From condition ( B ) in a \ a 2 the definition of Possibility II, there exists a and L such that •n(a-,L,G. ,G) holds. Choose 6„ so that <G„ u L U {a}>, £ G t . 6 0 6 60. » 6g Rename G so that G._ becomes G , and {0:6n < 9 < 6„} becomes. 6 0 • co u — B {0:to 4 6<(o+u)}_, and. {0:6' +uia < 0 < 6^+co(a+l)} becomes {0:to+co+toa < 6 < oo+co+oo (a+l) } for a l l a < oil . Then II(a,L,G ,G ) holds. .The induction step i s CO co+co the same as. case (i) except.that the l i m i t ordinals w i l l always be chosen from the 6 Vs .of A. a D e f i n i t i o n (17) : An i n f i n i t e c a r d i n a l K i s regular i f no set of c a r d i n a l i t y l e s s than K i s c o f i n a l i n K. As. noted i n Remark (3), no countable set i s c o f i n a l i n coi , and so co^  i s regular. D e f i n i t i o n (18): Let K be an i n f i n i t e c a r d i n a l number and l e t A be a subset of K. IE there i s a sequence , a A, such that S i s a subset of a, and for each subset X of K the set a {a:X fi a = S } i s stationary i n K, then we say <X 7(A) holds. ct - K From Jensen (6) page 293> we get the following r e s u l t : "Assume V = L and l e t K be a regular i n f i n i t e cardinal.- Then O t r(A) holds for every stationary subset A of K." K So t h i s r e s u l t holds when K i s the regular c a r d i n a l co1 . Now consider a group G of c a r d i n a l i t y co1 . Let A be a stationary set .of • l i m i t o r d i n a l s . As shown before the r e s t r i c t i o n of any stationary set to i t s l i m i t ordinals i s again stationaty, so there are many such A's. Consider the set G x Z = {(a,z):a £ G,z€Z}. 33 Name the elements of G x Z as follows: .(a) Name = {(a,z):a < co} as (3 :3 < co} which i s e a s i l y done as i s countable. (b) Suppose {(a,z):a < co6} has been named as (3 :3 < co6}, then, name the elements of J r , , = {(a,z):to6 <_ a < co(5+l)} o+l — as {3:w6 < 3 < co(6+1) } which i s e a s i l y done as J r i l is. — o+l countable. So by t h i s inductive naming process the elements of G x Z are the ordinals {a:a < coi } and each G • x Z has been named as the ordinals o {a:a < 6} for a l l l i m i t ordinals less than co_ . Let H be such a naming of the elements of G x Z, and so the set A i s stationary, i n H = co. = {ot:a < co.}. Assume V = L and apply Jensen's r e s u l t to H and A. Thus 0 T 1(A) holds. Let g:G G x Z be a function such rl that g(a) = (a,z^) where the z a ' s are i n Z. Then g can be viewed as a set of ordered p a i r s , say L = {(a,z ) :a < COT and z eZ}, a a and so g can be viewed as a.subset of H, say Y, where 66 Y i f and only i f 6 i s the name i n H of some (a,z ) i n L. By Jensen's a r e s u l t there e x i s t S , for a£ A, such that S S a and for any a a • X £. H, the set {a:X fi or- S_^ } i s stationary. In p a r t i c u l a r , ft A = {a:Y ft a 1 S stationary. Since a i s a l i m i t o r d i n a l and : a = ( 3 : 3 < a} = {(6,z) : 6 < a , z € Z} = { (6,z) : 6 € G , z €Z} = G x Z a a and Y = { 3 £ H : 3 = (6,z r) for &€G }, then for a e A\ Y 0 a = S = o a a ( 3 6 H: 3 = (6,z.) f o r i e G } . Thus Y fi a = S = g L can be viewed as o a a 1G a '* a function S :G ->• G x Z. Let S = g f o r a l l a £ A . As g was . a a. ct a a a r b i t r a r y , then for aiy function g:G ->- G. x Z where g(a). has shape 34 (ct,z),the set '{<5 < u>i:g = g '= S } i s stationary and since g i s a Lr 0 0 6 ft* function so i s g.:G. x Z a function. So l e t A = { 6 € A : S . = g„ ;0 o o o o i s a function.from G r into G. x Z}. Thus we can make the following 6 6' statement: . • (J) " I f V = L, there are functions g :Gr -> G r x Z, 6 £ A'°" £ A, 6 6 such that f o r any function g:G -> G x Z where g(a) = (a,z), the set {5 < toj_:g | = g } i s stationary. G 6 6 Theorem. (1) : Assume V = L.. Then i f G s a t i s f i e s P o s s i b i l i t y I or P o s s i b i l i t y I I , then G i s not a W-grbup. Proof: Suppose G s a t i s f i e s P o s s i b i l i t y I or I I . • By Lemma (8) G can be named so that for any l i m i t o r d i n a l ' 6 < coj , there e x i s t s 6 6 a and L. such that II(a ,L.,G.,G.., ) holds. Let A. be the 6 6 6 6+0) stationary set cons i s t i n g of a l l l i m i t o r d i n a l s . Thus by Jensen, since V = L, we can assume (J) as above. Let . H. = G ' x Z, and 6 6 ft ft so f o r 5 ^ A , g i s a function from G. into H r and g(a) = (a,z) o o 6 • * * • ' • Let K be the set of these functions, K = { g . : 5 £ A }. o We w i l l now construct a (G,Z)-group, H , such that there does u l . . not e x i s t a map g:G -> H .. such that hg = 1„ and as usual h(a,z) = a. oil G ' If we can construct such a H , then G cannot be a W-group. We do the construction by t r a n s f i n i t e induction. Define a (G ,Z)-group, 0) H , a r b i t r a r i l y . Suppose we have defined a (G ,Z)-group, H oi • o>a • oia for a l l a < 5, such that H extends H „ for a l l 3 < a. oia 0)3 35 Define H' as follows: coo Case ( i ) : Suppose 6 i s not a l i m i t o r d i n a l , so we can assume H , „ 1 N i s well defined. As before l e t H = G x Z and h:H -> G co(6-i) with h(a,z) = a. (a) If c o ( 6 - l ) ^ A or i f .to (6-1) 6 A and the corresponding function g / r 1 N i n K i s not a homomorphism such that co (6-1) hg , . = 1 , then extend H to H co(6-l; G 1 N co(6-l) co6 co (6-1; a r b i t r a r i l y using Lemma (6).. •k-k (b) If co(6-l)£A and the corresponding function , g , . i n K i s a homomorphism such that hg . „ ,. = 1_ co(6 1) w ( 6 - 1 }  t h e n . ' g c o ( 6 - l ) ' a " ' Lco(6-l) ' G<o(6-l) ' a n d Gco6 s a t i s f y the conditions of Lemma (7). That i s : (i) G . „ 1 X i s a countable subgroup of the cor-free co(6-l) group" 6 u f i , and i s a ( G ^ ^ ,Z)-group. ( i i ) i K a ^ 6 " 1 - * ,L 1 N,G ..,G J holds by Lemma. (8) to(6-l) co(6-l) co6 as stated i n the beginning of the proof.. ( i i i ) g s :G, ..,.->- H . i s a homomorphism co (6-1) to(6-l)- co(6-l) such that hg ••. . = 1 • t 0 ( < 5 _ 1 ) G(o(6-l) So we can apply Lemma (7) to extend- H .. '. to H „ co(6-l) co6 so that §u(,5__) cannot be extended to a homomorphism gco6:Gco6 * \& S U C h t h a t H^6 = XG , • co6 Case ( i i ) : Suppose 6 i s a l i m i t o r d i n a l . Define H . = I J H , : — coo , toa ' a<6 and as before H .. i s a (G .,Z)-group extending H for a l l co6 co6 • coa a < 6. Let H be the (G,Z)-group constructed by this induction. Suppose g:G -> H i s a homomorphism such that hg = 1_ . Then ui\ G . • • . . ft A = {6:g - g.} i s stationary by ( J ) . In p a r t i c u l a r A i s u . ' 0 6 ft . . . non empty, say 6 i s i n . A . Thus by the construction of H , 0)1 g„ cannot be extended to a homomorphism g„, :G. ->• H., ; such 0 0+OJ , 0+0) 0+0) that hg., = 1^ . Thus g. cannot be extended to g such that o+o) G., o 6+0) hg = 1^ , , and so g i s not an extension of g. , contradiction. Thus no such g ex i s t s and so G i s not a W-group. 37 Martin ; Axiom and W-groups In this section we w i l l show that under the assumption of the Martin Axiom and 2 W > oi] ,. any group satisfying Possibility II is a W-group. Definition (19) : .Let P be a poset (partially ordered set), and let a,b£P. We say a and b are contradictory i f they. have, no common upper bound in the poset P. Definition (20): Let P be a poset and let D be a subset of P. We say that D is a dense subset of P' i f for any a in P there-is a b in D such that a < b. Definition (21) : Let A be a cardinal number.. Let MA be . A the following assertion: "Let P be any poset of cardinality A. Suppose in P there is no subset of OJ^  pairwise contradictory elements. Also suppose {D :ct < A} are dense subsets of P. Then there exists a subset a B of P such that BHD ^ <j>. for a l l a < A, and such that any. two.members of B have a common upper bound in B." Such a set B is called a generic subset of P (with respect to the D 's). MA (Martin Axiom) says that MA, holds for any A < 2 U. 38 Theorem (2) : Assume the Martin Axiom and 2 W >. t6_ . If G has c a r d i n a l i t y u)_ , i s oj.-free, and does not s a t i s f y P o s s i b i l i t y I. then G i s a W-group. Proof: Suppose- G s a t i s f i e s P o s s i b i l i t y I I I . By Lemma (5) G i s the d i r e c t sum of countable groups. As . G i s u^-free then each summand free, and so G i s free. Thus G i s a W-group. So we can assume G s a t i s f i e s P o s s i b i l i t y I I . Let H be a group whose set of elements i s G x Z, and l e t h:H -> G be defined by. h(a,b.) = a. Now we define a poset P. The elements of P are homomorphisms g from f i n i t e l y generated pure subgroups. •I of. G into H such that hg = 1 . If g^ and &2 belong to P, write, g^ <. g^ i f g^ extends- g^ . We w i l l now show that the c a r d i n a l i t y of P i s u)_ . F i r s t we compute the number of f i n i t e l y generated pure subgroups of. . G. As each subgroup i s countable and G i s t h e i r union, there must be at le a s t OJ_ . d i f f e r e n t f i n i t e l y generated pure subgroups. There cannot be more than co_ f i n i t e l y generated pure subgroups of G since there are only co_ f i n i t e subsets of G. Now each of these pure subgroups i s f r e e l y generated by a f i n i t e set as G i s oj_-free. For a given pure subgroup I, with generators b , ... ,b , the homomorphisms of I into H are uniquely determined by the images of the generators. If g:I ->- H i s i n P, then for each generator b_^  ,g(b )£{ (b ,z) : z £ Z) . Thus there are only countably many choices f o r the image of each b_^  , and since there are only f i n i t e l y many b ''s,-' there are only countably many d i s t i n c t mappings of the b_/s. Thus there are only countably many g's for each I. Since the number of f i n i t e l y generated pure subgroups i s uii. , then the c a r d i n a l i t y of P i s ooi . We now define subsets D of P for a < •u>.y as follows: = {g£.P:a i s i n the domain of g}. ' • . We now show: (a) Each D , a < u>i , i s dense i n P. a , (b) There do not ex i s t OJn pairwise contradictory elements of P Proof of (a) : Let a< ui^ and l e t g be i n P where Then I i s pure and f r e e l y generated by say a^,...,a . We must show that there e x i s t s g i n such that g' <_ g. If a"6Dom g^ (domain of g^) , then g^eD^ and g T 4 g T , so we can l e t g = g. So suppose now a<^Dom g^ and consider I = <a^,. . ..,an>a>A . If I i s a free group generated by a^,...,a^,a^ +^ then define g as follows: (i) Let g(a ±) = g (a±) for i = l , . . . , n . ( i i ) Let g ( a n + 1 ) = (a ^z.) for any z Z. As the a £ » 1 = l>--«»n+l, generate the free group I , then this , it mapping of the generators can be extended to a homomorphism g:I -> H. is C l e a r l y g extends g^ . and a £ Dom g = I •. Thus g £ D ^ and ^ g. •k It remains to show that I i s f r e e l y generated'by a,,...,a ,a 1 n n+1 for some a n + ^ • Since a ^ I , - then a i s independent of {a^,....,a } i< since I i s pure. So I contains at le a s t n+1 independent elements * • ' If x £ I "^1, then mx £ <a. ,-..'.. ,a ,a> for some m, and so x i s i n 40 no t independent of {a-^  a ,a} . Thus I contains exactly n+1 • * . independent elements, and so I i s a free group on n+1 generators. Let I be f r e e l y generated by ^_'""*'bn4._ * A s shown before, for any f i n i t e number of elements i n <a,,...,a ,a>, and not i n <a,,...,a ,a I n " 1 n there e x i s t s an element, say a ., , such that the f i n i t e set of n+1 elements i s i n <an,...,a ,a >. (see G../G, i s i n f i n i t e , page 24). 1 n n+1 5 4 Let B be the set of elements of {b^ , ... . ,b n +^} i n <a^ , . . . ,a^,a>i( and not i n <a •, . . . ,a ,a> . Then for some a , , , B — <a, , . . . , a •> , 1 n n+i 1 n+1 and so { b 1 , . . . , b 1 1 } f = < a 1 , . . . , a J > . Since I = <b,,...,b >, 1 n+1 1 n+1 1 n+1 then I = <a,,,..,a , > and the r e s u l t i s proved. . 1 n+1 . Proof of (b) : Suppose there ex i s t s a set of CJ_ pairwise contradictory elements of P, say {gr:<5 < c o i } . We w i l l derive a contradiction. • o . 6 6 Let the domain of g. be f r e e l y generated by a.,...,a ,.... where o 1 n(.o; n(6) i s a f i n i t e p o s i t i v e integer. We can replace W = (g^ : <5 < w _ } by any subset of W of the same c a r d i n a l i t y without loss of generality. As each n(6) i s f i n i t e and card(W) = co_ , then some n(6) must occur co_ times. So we can assume n(5) = n, for some fi x e d n, for a l l the g 's. That i s without loss of o 6 (5 generality the domain of g r i s generated by ,{a1,...,a } for every • o 1 n g. i n W. Let K = {a,,...,a } be a maximal set of elements of', o 1 m G which f r e e l y generate a pure subgroup and {a , . . . ,a^} S.Dom g^ for co_ 6's. Note that K can be empty. For i f any uncountable family of the Dom (g.)'s has a t r i v i a l i n t e r s e c t i o n , then K i s empty, 6 else K i s non empty. So again without loss of generality we can 41 assume a,,...,a g.Dom g. tor a l l 6 < u>i . Let a, = a, a = a 1 m 6 1 1 1 m m For any 6 we can extend {a,,...,a } to an n-element generating • . 1 m . . set for Dom g. . Thus we can assume Dom g i s generated by o o {a.,...,a ,a ,...,a } for,each g, i n W. 1 m. m+1 n o Now consider Dom g. . Dom g. i s f r e e l y generated by n elements. o o As any homomorphism g . from Dom g. i s uniquely determined by where 6 the generators are mapped and g(a) = (a,z) for any generator a, there can be only countably many d i f f e r e n t homomorphisms from Dom • g into H. So i f there were only countably many d i f f e r e n t domains of. the g 's, there would only be countably many g 's, a contradiction. 6 o Thus there are ix>i d i f f e r e n t domains on which the gg' s a r e defined. Choose one g' on each domain. So without loss of generality we • ° can assume Dom g £ Dom g f o r S ^  a. In other words, i n the o a , o 6 ' ' set l a , , .'. . ,a ,a , . . . ,a }, m < n. 1 m n+1 n Again without loss of generality we w i l l take a subset of the g P's o of c a r d i n a l i t y ; this -time such that the set {{a^,...,a }. U 6 {a^:m < I n', & < coi}} i s independent i n G. Dom g^ i s generated by {a,,...,a ,a^,.,...,a^} which i s an independent set i n G. 1 m m+1 n Assume for a < 3 < coi we have chosen g 's such that the set 1 a {fa^,. . . ,a^} U {a":m < I i . n, ct <3}} i s an independent set i n G. Now consider < U Dom g ' >', . From the remaining g„'s, toss out ct<3 a x o any g „ such that (< ^ „Dom g <a, ,...,a >) fi Dom g^ i s non empty. / 66 a<3 . a * 1 m 6 5 J We claim that only countably many g^'s w i l l be tossed out. This w i l l follow from the fac t that < U Dom g >. <a, a > = D i s countable. ot<3 a * 1 m If some element i n D, say b. was i n uncountably many Dom g ' s 42 then as b i s not an element of <a, a >, i t i s independent of 1 m the pure subgroup <a , ...,a >,' and so we would get a contradiction to the maximality of the pure subgroup <ar,...,a > . That i s 1 m <a^, . . . ,a j b ^ would be a pure subgroup of order m + 1 contained i n the domain of coi of the g ' s. So for any element b £ D , we toss out only countably many &^Sm Since D i s countable we toss out only countably many g.'s and so there are uncountably many o g 's l e f t . Choose one and c a l l i t g . We claim that the elements • o 3 {{a^,...,a } U {a":m < £ n, a <. 3.}} are independent i n G. This i s e a s i l y seen. We need only check that no l i n e a r combination of the 3 3 ' elements of {a a } i s a l i n e a r combination of the elements of m+1 n •{•{a,,...,a } U (a :m < £ <_ n, a < 3}}- Any l i n e a r combination of the 1 m x. — • elements of (a^,, a^} i s i n Dom g„ <a, , . . . ,a > and any l i n e a r m+1 n 3 1 m • combination of. the elements of'- {{a,,...,a } U (a :m < £ < n, a <' $}} 1 m £ i s i n < i < 3 D 0 m So.* ' B y c h o i c e o f Sg ' (Dom'g -^<a1,...,am>) f) < o f < g D o i n ^ a * * i S e m P ^ y a n d s o t h e s e t ^ a i ' " * * ' a m ^ ^  ^ a £ : m K 1 = n ' a < i s independent. So in d u c t i v e l y we can choose coi of the g 's i n • o such a way that V = {{a. ,. . . ,a } U {a„ :m < £ <_ n, 6 < coi }}• i s an i m £ — 1 independent set i n G. Since P o s s i b i l i t y I f a i l s , there i s an admissible naming of the elements of G such that there does not e x i s t G. so that o conditions (A) and (B) of the d e f i n i t i o n of P o s s i b i l i t y I hold. So assume G has such a naming. Also make sure G contains a.,...,a co 1 m This i s e a s i l y done. Then there i s an uncountable subset of V independent over G/G . If an element g £ G i s a f i n i t e l i n e a r CO CO A3 combination of elements of V, then for any a^ i n the representation 6 6 of g, toss out .a a . Since G i s countable and i f g £ G •.m+1 n co co can be represented as a l i n e a r combination of elements of V, then 6 6 that representation i s unique,' only countably many subsets. ( a m + 1 > • • • > a n^ w i l l be tossed out of V. Let V be the remaining elements of V. ft ft 5 C l e a r l y V i s uncountable. Now l e t J be the set of a l l a ,,'s i n m+i ft ft 5 V , say J = {'a + 1 = <S < ooi). Then: ft (A) J i s an independent family i n G/G co • '(B) <G >, = G c o * . c o • 6 6 Since P o s s i b i l i t y I f a i l s , then there must e x i s t b, i n <G U {a ,,'}>* 1 co m+1 * 5 6 such that <G^U { a m + 1 ^ f t = G w 9 < { b 1 ^ > f t f o r a 1 1 b ut countably many 6's. So <G U {a6J_.}>. = <G. U ( b f } > + = G & <{b?}>. for {b?:6 < O H } . co m+1. * co 1 '< co 1 x 1 1 6-, .r , , , * * . . , . 6 m+2 Now l e t L_'= lb.}, f o r 6 < coi and J be the set of a l l a , „'s 6 1 1 n corresponding to the b.'s. So J = {a,„:6<coi}. Then: 1 m+2 x ft ft (A) J i s an independent family i n G/G . w •(B) <G 0/ L->. = G © <L > for 6 < cox . co 6 * , co 6 * 1 6 Again since P o s s i b i l i t y I f a i l s there must exi s t b„ i n <G \J L. U 2 co 6 t^oKv such that <G U L U fa?- _:}>"•= G 9 ,<L. U {b!?}>, for m+2 5t co 6 m+2 « co ' 6 2 * . 6 uncountably many.of the a , 's. Continuing this f i n i t e process we m+Z 6 6 6 * 6 get that <G (J {a M l . . . , a }>, = G © <b,,...,b >, for uncountably co m+1 n « co 1 n-m * J-many 6's. Choose such a 6,say a, and l e t a ( l ) = a. Note that <G \J {a™ ,aa}>. = <G U Dom g >, . Also note that b" . . . ,b a . co m+1 n * co ct <c 1 n-m are elements of < a a ;a a ,c c. , .>, for some I ' m ' m+1' ' n 1 k(ct) * c™, . . ., c" , N i n G . This i s because b? £ <G U {bf, ...,b? , } U {a0-, . }> , 1 k(a) . co . l co 1 l - l m+i * 44 Now we can repeat t h i s process for any G and choose a(a) wa d i f f e r e n t each time since we are choosing from an uncountable set of which only countably many have been chosen before. In f a c t we could choose o(a) > a(g) for a l l g < a since the a(g)'s are not c o f i n a l in-our uncountable set. Thus we can define a s t r i c t l y increasing sequence of or d i n a l s a(a), a < u i , such that <G U Dom g . .> = 'cua. 0(a) . o(a) . a(a) , ,, a(a) ,a(a), G © <b. , . . . ,b '> . where {b, , . . . ,b ' } <~ <a. , . . . ,a , wa 1 .'. ' n-m * 1 .' ' n-m 1'- ' m a(a) a (a) a (a) a (a) , a (a) . a (a) . ' _ . a ,. ,...,a ,c. c, ; ;>. and c. ,...,c, ; ( € G . Note m+1 ' ' n ' 1 ' ' k(a) * 1 ' ' k(a) wa v , r o"(a) o(a) that by choosing k ( a ) A we can assume that l a , , . . . , a ,a ,...,a . • », 1 m m+1 n a(a) a(a) minimally c, ,...,c, - v} i s an independent set i n G. To sim p l i f y notation 1 k(a) assume a(a) = a. . At t h i s point we must make the observation that the set of ordinals C = {6:6 < w_ and w6 = 6} i s a closed and unbounded subset of OJ_ . . C i s unbounded: By (7), page 108, ordinals of the form w. , for any 3, are i n C. That i s w = w(wW ). Thus i f C was bounded above by say a, a.< w_ , then choose 3 such that w3 J ' 3 a < 3 < co_ . Then co ^ 3 and so w > a.. As w i s i n C we get a contr a d i c t i o n . (b) C i s closed: Let 3 y be a countable sequence of members of C. Then f or each element of the sequence 3 = 'tog . Let a =. lim 3 V V V Then: . a = lim 3 = lim wg = w(lim g ) = wa '. • • • v v v Thus a i s i n C and so C i s closed. 45 For every a < co_ , k(a) is. f i n i t e and so some k(a) , ' say k(a) = occurs uncountably many times. Let A be th i s set under the natural ordering of o r d i n a l s . A = {a:k(a) = t} = {a:k(d.) = t, 6 < co-. , and ' o o a. < a • i f and only i f 6 i < 6o}. Now rename G so that G . = G oi 02 . • coo • coa^ Thus at every o r d i n a l a our set { c c , . G has t elements. 1 k(oi) — coa Now l e t JQ = C = { 6 : 6 < co_ and co6 •= 6}.. i s closed and unbounded and hence stationary. Since k(a) = t for a l l a < co_ (, then k(a) = t for a l l 6 i n . Note also that f o r 1 < i <. t, 6 6 c < co6 as. c € G . Now we can apply a r e s u l t of Fodor to, define I p too J a sequence of stationary sets 2 . . .• 2 J such that for a l l 6 6 € J ^ , c ^ = C £ for some fi x e d c^ . We proceed as follows.. D e f i n i t i o n (22): A function f : J -> A , where J and A are sets of ordi n a l s , i s c a l l e d regressive i f f.(a) < a for a l l a £JV{0} and f (0) = 0 i f 0 £ J. Fodbr's r e s u l t (8), page 141, says that for a regular c a r d i n a l A > co, and J a stationary subset of A, there exists for each. k defined regressive function f on J a stationaty subset J of J •k such that f(a) = 3 for a l l a i n J , and some fi x e d 3 i n A.. Now consider = C, a stationary set, and to_ , a regular 6 c a r d i n a l . As noted for 1 < t < t, c < 'u>6 and so for 6 £ , 6 c.^  < 6 as 6 = co6. Thus we can define a regressive function r :^Q ~* u i 6 6 by f ( 6 ) = c^ since c^ < 6 for a l l 6 £ . By Fodor's result' there e x i s t s , a stationary subset of J , such that f ( 6 ) = c^ for a l l 6 i n . Now repeat t h i s f o r by defining f : J ^ ->- co_ 46 6 by f ( 6 ) = < 6. This w i l l produce a —• ""^1 s u c n that i s stationary and f (6) = c^ for a l l 6 i n . By repeating t h i s t times we can define a nest of stationary sets J ^ J ^ < ~ ... — Jj-6 such that for a l l 6 € , Jl = 1,. . . , t , c^ = c^ for some fixed 6 c . In p a r t i c u l a r for 6 €. J , c = c for a l l I = l , . . . , t . Ay t ' il/ XJ 6 6 That i s the set {c^,. . . >ct}' = {c^,...,c t} for a l l 6 € J ^ . We w i l l , now make one more observation. If A and B are pure * subgroups of G and i f A © B i s a pure subgroup then for A and * A A B , pure subgroups of A and B r e s p e c t i v e l y , A © B i s pure ft ft ft i n G. This i s e a s i l y v e r i f i e d . Let nx = a + b where a i s ft ft ft i n A. and b i s in' B . As A © B i s pure then x = a + b for some a i n A and b i n B. Then: ft ft nx = a .+ b = na +. nb , ft ft so na = a and nb = b , * - A A A so a £ A and b £• B as A and B are pure, so x £ A" © B" , . A A so A © B is, pure i n G. 6 6 Now consider the pure subgroups < a a ,a ,c,,...,c >, 1 m m+1 n 1 t where 6 i s i n . We can extend g^ to this domain. C a l l these 6 new homomorphisms g . By the above observation <a^,...,a^,c , . . . ,ct>. 6 6 6 © <b,,...,b , >, = B i s pure and contained i n Dom g since each 1' . n-m * 1 6 6 6 6 b. i s an element of Dom g . By construction each a ,. i s an 1 m+i ' 6 6 element of B and so Dom g i s contained i n B. Thus Dom g = B. 6 6 5 So Dom g i s f r e e l y generated by {a.,....a , a , a , } where 1 m+t m+t+1 n+t .{a^  , .. . > a m + t ;} f r e e l y generates <a^ , . . . , a^, c^ , . . . , c >^  and. 47 { V t + l ' - ' V t } f r e e l y generates • <bJ,...,b^ _m>A . 6 Since g (a^-) £ { (a , z) : z €• Z} f o r SL = 1 , ...,m+t , there are 6 only countably many d i f f e r e n t images of the g (a )'s, and ,so must 6 appear uncountably many times. So we can assume that g (a ) i s fi x e d for SL = l,...,m+t where 6 i s i n J , an uncountable subset of J . t Now choose a , 3 £ J t such that a < 3 and so the generators ct ct of g £G . Then Dom g i s a pure subgroup of G „ Also top top. . < ^ , , , , . • . , a 3 , > i s pure and equal to . < b 3 , b ^ > , . As m+t+1 . n+t ^ H m+1 . n * 3 3 G . © <b b >.• i s pure then by the observation on the l a s t top m+1 n page Dom g a $ <- a m+t+l' '' ''am+t> = < D o m ^ ^  D o m 8 ^ i s pure i n G. F i n a l l y we have the extension needed to produce the contradiction. <Dom g™ U Dom g3> i s f r e e l y generated by »•••' a m + t' a^+ t+l''''' an+t' am+t+l'-'-' an+t } a n d g " ^ ) = g 3 U A ) . for = l , . . . > m 4 - t . Thus a 3 ct 3 g:<Dom g U Dom g > -> H defined by g(a + b) = g (a) + g (b) where > -pi a • " U / • T N 3 . ' „ . r: oi , 3 a t Dom g , b fc Dom g i s a common extension of g and g ct 3 Now g and g are themselves extensions of some g " and g A 0 r e s p e c t i v e l y i n our o r i g i n a l set assumed to be pairwise contradictory. So g i s a common extension of some g^and g. and g £ P since A . 0 a - 3 • <Dom g [J Dom g > i s a f i n i t e l y generated pure subgroup. This contradicts our o r i g i n a l assumption that g. and g. have no A o upper bound i n P. Thus there doesn't e x i s t any subset of to_ pairwise contradictory elements of P. 48 We can now complete the proof of Theorem- ( 2 ) . Since under our assumptions, co_ < .2^° , by Martin's Axiom there e x i s t s a (generic) subset B- of P such that B fl D = (j) for a l l a < coi , and such a T - l that any two members of B have a common upper bound i n B. . Let * IJ * g = g. Since B i s generic i t i s easy to v e r i f y that g i s B a function from G to H. Since each g6 B i s a homomorphism, so i s g . Since hg i s the i d e n t i t y map on the domain of g, we ft ft have. hg = 1 . Thus there e x i s t s a homomorphism g :G •->- H such . C J ft that hg = 1 , and so G is. a W-group. 49 The Independence Result Theorem (3) : The statement: "Every W-group of c a r d i n a l i t y u>n i s free " i s independent of ZFC (Zermelo-Frankel set theory plus the axiom of choice). Proof: Since W-groups are oii-free, then by Theorem (1) i f V = L, any W-group must s a t i s f y P o s s i b i l i t y I I I . By Lemma (5) G = © G where each G i s countable. As G i s co-i-free, each a<oii a a G i s free and so G i s free. Thus ZFC + V = L implies that every . W-group of c a r d i n a l i t y oil i s free. But by Godel (8) , ZFC + V = L i s consistent i f ZFC i s consistent. By Martin and Solovay (10), ZFC + MA + 2^° > coi i s consistent i f ZFC i s . But by Theorem (2) , i n the presence of MA + 2^° > , any group s a t i s f y i n g P o s s i b i l i t y II i s a W-group. By Lemma (4) there are groups s a t i s f y i n g P o s s i b i l i t y I I . So i t i s consistent with ZFC to assume that there are W-groups of' c a r d i n a l i t y co^  which are not free. Thus the statement: "Every W-group of c a r d i n a l i t y u>i i s free " i s independent of ZFC. Bibliography 1. L. Fuchs, I n f i n i t e Abelian Groups, Vol.1, Academic Press, N.Y. and London, 1970. 2. L. Fuchs, I n f i n i t e Abelian Groups, Vol.11, Academic Press, N.Y. and London, 1973. 3. Pinter, Set Theory, Addison-WesTey Publishing Company, London, 1971 4. Jon Barwise, Back and Forth Thru I n f i n i t a r y Logic, i n Studies i n ' Model Theory, M.A.A., 1973,pp. 5-34. 5. I. Kaplansky, I n f i n i t e Abelian Groups, University of Michigan Press Ann Arbor, 1954. 6. R. B. Jensen, The Fine Structure of the Constructible Hierarchy, Annals of Mathematical Logic (1972), Vol.4, pp.229-308. 7. J . Monk, Introduction to Set Theory,. McGraw-Hill Book Company, Toronto, 1969. 8. G. Fodor, Eine Bemerdeeng zur Theorie der Regressiven Funktionen, Acta. S c i . Math. (1956), Vol.17, pp. 139-142. 9. K. Godel, The Consistency of the Axiom of Choice and of the Generalized Continuum-Hypothesis with the Axioms of Set .Theory, Princeton University Press, Princeton N.J., 1940. 10. D. M. Martin, R. M. Solovay, Internal Cohen Extensions, Annals of Mathematical Logic (1970), Vol.2, 143-178. 11. S. Shelah, I n f i n i t e Abelian Groups, "Whitehead Problem, and Some Constructions, I s r a e l J. Math.(1974), Vol.18, pp. 243-256. 51 Appendix We w i l l describe the s i g n i f i c a n t a l t e r a t i o n s that we have made to Shelah's paper (11). The d e f i n i t i o n of P o s s i b i l i t y I was changed from the existence of a G. , under some admissible ordering, which o , s a t i s f i e s conditions (A) and (B), to the existence of a G. , • o under every admissible ordering, which s a t i s f i e s conditions (A) and (B). Under the o r i g i n a l d e f i n i t i o n there appeared to be no way of c l a s s i f y i n g , up to isomorphism, o)_-free groups into the three p o s s i b i l i t i e s . This al t e r e d d e f i n i t i o n of P o s s i b i l i t y I allowed us to si m p l i f y some of the proofs. In p a r t i c u l a r we were able to come up with a lemma (Lemma (8)) which allowed.us to deal with both P o s s i b i l i t y I. and II i n a uniform way v i a Theorem (1). Shelah had used a complicated group theoretic argument i n dealing with P o s s i b i l i t y I. Thus our set theoretic Lemma (8) eliminated the more d i f f i c u l t group theoretic theorem of Shelah's, (see (11) 3.3). Shelah used a rather complicated combinatorial argument, (see (11) 3.1(2) and 3.1(3)), to show every P o s s i b i l i t y I I I group i s a d i r e c t sum of countable groups. Our Lemma (5) gives a simpler and more d i r e c t proof of t h i s f a c t witout using the complicated combinatorial technique of (11) 3.1(2). In the proof of Theorem (2), the method indicated by, Shelah 6 6 for producing the elements c ,...,c .. . appeared to be in c o r r e c t . 1 k. (, o) So a completely different.arguement had to be used, see pages 41-44. 52 In general the set theoretic and group theoretic d e t a i l s were f i l l e d i n to the point where someone with only a l i m i t e d knowledge of set theory and group theory could read the thesi s . This involved much work i n places for Shelah assumed a knowledge of group theory at a l e v e l of Fuch's books (1) and (2). In many cases only the broad o u t l i n e of an argument was given, and so there had to be a s i g n i f i c a n t f i l l i n g i n of d e t a i l . As an example, i n Lemma (7) G^/G^ has to be shown to be i n f i n i t e and then i t has to be shown that t h i s implies / C O e i t h e r G^IG^ contains a copy of Z(p ) or i t contains i n f i n i t e l y many elements of prime order, see pages 24-26. Another example was working out a l l the d e t a i l s i n showing that the examples i n Lemma (4) ac t u a l l y s a t i s f y ' t h e respective p o s s i b i l i t i e s , see pages 11-16. The. main d i f f i c u l t i e s with the set theory, other than redefining P o s s i b i l i t y I and the subsequent c l a s s i f i c a t i o n into the three p o s s i b i l i t i e s , was in showing how the r e s u l t s of Jensen (6) and Fodor (8) applied.to our problem. So again here there had to be su b s t a n t i a l f i l l i n g i n of d e t a i l , see pages 32-34 and 44-46. Also" we had to show that G could be well-ordered such that for a l l l i m i t ordinals 6, G„ i s o pure, see Lemma (2) on page 5. 

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