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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 A p r i l , 1976  ©  Richard J . Dean  In p r e s e n t i n g t h i s t h e 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  for  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 r e f e r e n c e and I f u r t h e r agree t h a t p e r m i s s i o n  for extensive copying of t h i s  study. thesis  f o r s c h o l a r l y purposes may be g r a n t e d by the Head o f my Department or by h i s  representatives.  It  i s understood that c o p y i n g o r p u b l i c a t i o n  o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my written  permission.  Department o f  Mathematics  The U n i v e r s i t y o f B r i t i s h Columbia 2075 Wesbrook P l a c e V a n c o u v e r , Canada V6T 1W5  Date  A p r i l , 1976  ii Abstract  An a b e l i a n group Whitehead's  G i s called a  problem asks which groups are  proved t h a t the answer to cardinality set  W-group i f  w  1  ,  Whitehead's  Ext(G.Z) = 0.  W-groups.  problem,  f o r groups of  i s independent of the axioms of  theory w i t h the axiom of c h o i c e .  Zermelo-Frankel  T h i s t h e s i s g i v e s a complete  and d e t a i l e d p r o o f , based on Shelah's p r o o f , of t h i s result.  Saharon Shelah  independence  I l l  T a b l e of Contents  Introduction  page 1  Whitehead Groups and t h e i r S t r u c t u r e  page 3  (G,Z)-Groups  page 19  V = L and W-groups  page 30  M a r t i n Axiom and W-groups  page 37  The Independence R e s u l t  page 49  Bibliography  page 50  Appendix  page 51  iv Acknowledgment  I would l i k e  to thank Dr. Andrew A d l e r f o r h i s time and guidance  throughout the p a s t y e a r . for  reading  this thesis.  I would a l s o l i k e L a s t l y I would l i k e  h i s h e l p i n t r a n s l a t i n g German.  to thank Dr. Stan Page to thank Ron A i k e n f o r  •1 Introduction  The "The  f o l l o w i n g r e s u l t was  proved by  Saharon Shelah i n ( 1 1 ) . IOJ ,  Whitehead problem, f o r groups of c a r d i n a l i t y  of•and c o n s i s t e n t w i t h  ZFC."  In t h i s t h e s i s we  t h i s r e s u l t based on Shelah's p r o o f .  present  a proof  C e r t a i n a l t e r a t i o n s had  made, as w e l l as a good d e a l of f i l l i n g of some of the a l t e r a t i o n s  i s independent  i s given  of.  to  be  i n of d e t a i l s . A d e s c r i p t i o n  i n the appendix at the end  of  the t h e s i s . A Whitehead group, for  which  or simply  Ext(G,Z) = 0.  w i l l be e x p l a i n e d  a  W-group,  (Ext(G,Z) = 0  i s an A b e l i a n  group  i s a mapping p r o p e r t y  in d e t a i l in this thesis.)  The  which  Whitehead Problem  asks: "Are  The  be  are We  ZFC,  ca^  freely  generated."  Z e r m e l o - F r a n k e l set theory w i t h  the axiom of  the axioms on which a l l c u r r e n t mathematics can be  will  solved..  all  W-groups of c a r d i n a l i t y  axioms of  choice-, upon.  all  We  show t h a t w i t h i n will  do  t h i s by  W-groups are f r e e ,  ZFC  the  built  Whitehead Problem cannot  showing t h a t w i t h i n one  model of  ZFC  and w i t h i n another model t h e r e e x i s t s non  W-groups. Godel e x h i b i t e d a c o n s t r u c t i o n which produced a model of H i s c o n s t r u c t i o n i s r e f e r r e d to as consistent. combinatorial and  use  where  and  so  ZFC  Jensen showed t h a t w i t h i n such a model of property  t h i s property V = L  V = L  holds,  called  'diamond'  holds.  We  + V = L ZFC,  will  W-groups are f r e e l y  generated.  a  define  diamond to show t h a t w i t h i n a modsl of all  ZFC.  ZFC  is  free  2  M a r t i n and Soloway showed t h a t is consistent with  .ZFC.  e x i s t e n c e of non f r e e h  o  l  d  s  .  .  We w i l l d e f i n e and use  W-groups i n . a model of  MA  ZFC  W-groups of c a r d i n a l i t y (i)  ZFC + V = L  (ii)  ZFC + MA + 2  implies W  > a)  1  u>  1  X  Let  X  X  t o show t h e i n which  MA + 2  W  > u)]/  be t h e f o l l o w i n g s t a t e m e n t :  are f r e e l y generated." and so ZFC + X  implies  -i X  and so  consistent. Thus  > ooj-  U  .  And so t h e s i t u a t i o n i s t h i s . "All  MA ( M a r t i n Axiom) + 2  i s c o n s i s t e n t w i t h and independent of  ZFC.  Then:  i s consistent. ZFC + - i X  is  3  Whitehead  Groups  and  their  Structure  In t h i s s e c t i o n we w i l l g i v e some p r e l i m i n a r y f a c t s and d e f i n i t i o n s •.. about Whitehead  groups and w.-free groups.  groups o f c a r d i n a l i t y 0)^ i n t o  three p o s s i b i l i t i e s .  group we w i l l always mean A b e l i a n Definition  Z  that  h:H ->''G  (the i n t e g e r s ) gh:G ->• G  of subgroups  B  (i)  there e x i s t s  (2):  For. a group  and  C  of  B + C = A the form  T h i s i s w r i t t e n as  a d i r e c t sum of a copy of  gh  Let  g  and  h  Clearly  G*  c o u l d not be  such :  A  i s the d i r e c t  sum  where  i s the s e t o f a l l sums of  b  i s in: B  and  c  is in  i s the i d e n t i t y element of J  C. A.  H. as i n the d e f i n i t i o n of a W-group i s Z  and a copy of  G.  be as i n the d e f i n i t i o n of' a W-group.  Since  G* = image(g)  - G. .  G,  - Z.  and k e r n e l ( h ) 1-1.  g:G ->• H  C.  i s the i d e n t i t y map on  By d e f i n i t i o n k e r n e l ( h )  i s . isomorphic .  G.  B + C  where' 0, A  A = B©  h  A i f :  b + c  Lemma ( 1 ) : The group  Proof:  a homomorphism  A. we say  where  ( i i ) B C\ C = 0. . A  group or W-group i f f o r  such t h a t the. k e r n e l o f  i s the i d e n t i t y map on  Definition  In t h i s thesis^ by  group.  ( 1 ) : G i s c a l l e d a Whitehead  every epimorphism to  We w i l l c l a s s i f y the oi^-free  g  i s 1-1,  We show t h a t  have o n l y  Now l e t  h*  be  0 h  so  H = G * © kernel(h)• i n common,  r e s t r i c t e d to  else  gh  G*..  If  • .  xeH,  then  and  x = h* ^ ( x ) + (x - , h * ^h(x))'  Preliminary Definition  Facts  Definition  G  W-groups  '  (4) :  of  Z.  A group i s  u^-free i f every countable  subgroup  i s a W-group t h e n i t i s : (i)  Torsion  (ii)  u>\.  free  co^-free  F r o m now o n  G  will  (2)  p a g e 178  (2)  p a g e 178  be t a k e n t o be t o r s i o n f r e e and o f c a r d i n a l i t y  So we c a n 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 G  are a l l the ordinals  ordinal:  a l l  z€Z,  (5.) : where  i s p u r e i f f o r any  solvable  <. ipi  where  toj  the elements  i s the f i r s t  uncountable  coj = { a : a < t o } } = the s e t of a l l o r d i n a l s l e s s than u ^ .  Definition  B  about  free.  If  for  1  ( 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  of  h * h ( x ) £ G*  x - h* ^h(x)£ k e r n e l ( h ) .  Some  is  and c l e a r l y  in  B  i s a. p u r e s u b g r o u p o f zG = { g £ G : g = z x z£Z,  b£B,  G then i t i s s o l v a b l e  G . i f B fl  f o r some  i f the equation in  B.  zG = zB  x€TG}. zx = b  Equivalent!y i s  4  5  Lemma ( 2 ) : co_'.  Then  G  Let  G  be a t o r s i o n f r e e group of c a r d i n a l i t y  can be w e l l - o r d e r e d as  t h a t f o r any l i m i t o r d i n a l of  5,  ^ a  = {§  :  ^  a  ^  < <$}  : a a  n  s u c  ^  a  w a  Y  i s a pure subgroup  G.  •P r o o f :  We  define  g. by t r a n s f i n i t e i n d u c t i o n . "a  :  l e s s than  co]_  0 < 0 < co],.  The l i m i t  a r e p r e c i s e l y the o r d i n a l s of the form  Suppose  has been d e f i n e d ,  $ < y,  t h a t f o r every  and  -G  how. to extend.the d e f i n i t i o n so t h a t  co$,  and every  i s a pure subgroup of G  where  a. < tog,  G.  i s pure.  ordinals  g^  We w i l l  show  There a r e two  .  coy cases to c o n s i d e r . Case  (i):  Y  i s a limit ordinal.  d e f i n e d f o r every subgroup of G  G,  .  Case  (ii) :  o r d e r type  Y  G  ^  =  CO],  be the f i r s t  G  0<  G.  element of g^  f o r any  B  g^  3 < Y»  i s already G  i s a pure  cop ,  :G „ ,  U  i t is trivial  to v e r i f y  that  w f 3 Y  w i t h the. f i r s t  generated by  torsion free  S i n c e f o r every  i s a successor.  which i s not a of  and  i s a pure subgroup of  COY  g  a < COY •  Then of c o u r s e  G  Take a f i x e d w e l l - o r d e r i n g of  G  element' i n the o r d e r i n g  Let  ^ 0.  w i t h r e s p e c t to t h i s f i x e d a < a n c  ^  has c a r d i n a l i t y  CO(Y 1) -  •  Let  ^  (where  co.  So by  c o n t a i n e d i n a c o u n t a b l e pure subgroup of  B  ( 1 ) , page say  order  be the subgroup  G^ = (j>) .  G,  in  Since 115,  B*.  As  G  B  is  is zg  frG T  for  any  z  in  and so i t may  Z  i t i s c l e a r that  be enumerated  i s a pure subgroup of  G.  as  B*\G  ..  CO(Y-I)  has c a r d i n a l i t y  {g :CO(Y-1) < a < COY).  a  —  Thus  G  co,  .= B* COY  .  OJ(Y~1)  If  G  of l a b e l l i n g it  i s a ( t o r s i o n f r e e ) group of c a r d i n a l i t y the elements o f  G  ls  instead  by the o r d i n a l s • l e s s than  co  i s n o t a t i o n a l l y more c o n v e n i e n t to assume the elements of  are the o r d i n a l s l e s s than  .  i s a pure subgroup o f assumption.  Lemma (2)  .of a^-Free  Groups  Possibility II,  6,  G  admissible.  of  Cardinality  i n this section classify  into three p o s s i b i l i t i e s ,  G  G^ .= {a:a < <5} -  shows t h i s i s . a harmless  C a l l any such naming o f  Classification We w i l l  G.  l5  Whenever such n o t a t i o n 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  oo j  io  lO].  a ^ - f r e e groups of c a r d i n a l i t y  c a l l e d unimaginatively  and P o s s i b i l i t y  III.  P o s s i b i l i t y I,  F i r s t we need a remark and  then some p r e l i m i n a r y group t h e o r e t i c and s e t theoretic- d e f i n i t i o n s .  Remark ( 1 ) :  F o r t o r s i o n f r e e groups the e q u a t i o n  have a t most one s o l u t i o n , x = y.  So i f zx = g  f o r zx = g = zy  i s solvable i n  G,  belongs to a l l pure subgroups c o n t a i n i n g of pure subgroups i s a g a i n pure. definition.  Definition Gwhich. c o n t a i n s  zx = g  implies  can  zx = zy I m p l i e s  then the unique  solution  g and thus the i n t e r s e c t i o n  T h i s a l l o w s us to make the . f o l l o w i n g  .  (6):  Let  L,  L<=LG.  <L,G>^  be the s m a l l e s t pure subgroup o f  Remark (2) :  If  by. Remark ( 1 ) . G  S  i s a pure subgroup o f  We w r i t e  <L>  = <L,G> .  >V  G,  then  <L,S>  jV  = <L,G>.  V  i f i t i s . c l e a r which group  y  we a r e r e f e r r i n g t o .  Definition s u b s e t of if  <SUL>  G,  and  = <S>  •Definition  A  S  be a subgroup o f  an element of  <L>  A  Jk  Let  a  = <S> ®  4  <SLVL'f{a}>  (7):  jV  b u t f o r no  ©<Ll/{b}>.,  (8) :  A subset  G.  G,  We say  L  a finite  H(a,L,S,G)  b£<SULU{a}>  is  A  holds .  c  C  of  LOJ I s c l o s e d and unbounded  if: (i) '  F o r every non-empty subset T h i s says  (ii)  Definition  that  sup C = a)]..  (9) :  C  S  A  C  sup  S£CL/{toi}.  i s closed.  T h i s says t h a t  .A subset  of  of -u)_  C  i s unbounded.  i s stationary  if  CfiA  ^ <j)  f o r every c l o s e d and unbounded subset of t o _ - .  • Definition co f i n a l that  (10) :  Let X  i n io_ i f for a l l a  be a subset of i n to  t o _ . Then  there e x i s t s  g  X is in X  such  a <_ 3 .  Remark (3) :  No c o u n t a b l e set. i s c o f i n a l w i t h  t o _ . . (3) page 207..  We a r e now ready to d e f i n e the three p o s s i b i l i t i e s . Definition satisfies  ( l l ) : An  Possibility  I  toj-free  and  a n  ct (  a  ) ^  6 < to'].,  a l l . a < ai ,  r  a  i ®  4 ^'  :  <  ( ) } such  n  a  Remark (4) :  Since  of  G  6 <  then  G^  (12):  Possibility  An  toj-free  6 = co' .  G. o  II  i f G  in A  there a r e elements  n(a) i s a f i n i t e o r d i n a l ) ,  Rename  i s countable.  G  of  o  Now rename'  of c a r d i n a l i t y  does n o t s a t i s f y co  G. by  Lemma ( 2 ) .  group  there i s a s t a t i o n a r y subset of a  family  i s c o u n t a b l e and so. we can  u s i n g .the technique o f  Definition  any  that:  1  {ct:0 £. ct < co} which can be done as  and  ordinal)  i s an independent  assume w i t h o u t l o s s of g e n e r a l i t y t h a t  satisfies  G,  h o l d s f o r a l l a < u, .  a  the r e s t  of  G/G, . o  n ( a . . ,L ,G.,G) n(ct) a o  (B)  n(ct) i s a f i n i t e  I j^nta)}  {a^ + G^:a < coi, in  •. CO]  of c a r d i n a l i t y  and t h e r e a r e elements  (where  1  L^' = ^  subsets (A)  o  G  i f f o r any a d m i s s i b l e naming o f G  t h e r e i s some l i m i t o r d i n a l say  group  l5  say A,  ct G,. say a ,  and subsets  co^  Possibility I such t h a t f o r £ <_ n ( a ) ,  (where  Ct  L = {a..:0-<_ i < n(ct) } a Jc —  such  that: Ct  (A)  ^ £ 0 4 ^ < n(ct)}  (B)  n(a°' \ ,L ,G ,G> n(ct) a a  a  :  /  Definition Possibility III  (13):  An  i s an independent  family i n / G  G a  holds.  co^-free group  i f i t doesn't  satisfy  G  of c a r d i n a l i t y  Possibility.I  or  coj  satisfies  Possibility II.  9  Lemma ( 3 ) :  The c l a s s i f i c a t i o n of a g i v e n group  p o s s i b i l i t i e s depends on Proof:  satisfy  Case ( i ) :  G  Suppose  G satisfies.  Then  •G*  I.  (Case ( i i ) :  Suppose  h:G  G*"  where  from  hlG.  Possibility  I.  of P o s s i b i l i t y  satisfies  i s an isomorphism,  i s an isomorphism ui  G  G^  onto  i t will  There are t h r e e cases to c o n s i d e r . Let  h:G -> G*.  can be thought of as a renaming  and so by the d e f i n i t i o n  Possibility  of  to.isomorphism.  the same p o s s i b i l i t y .  isomorphism.  if  o n l y up  to the^ t h r e e  We must show t h a t under any a d m i s s i b l e naming of . G,  always  of  G  G  G*  I,  of the  II.  F i r s t we  then the s e t  C  d e f i n e d by  show t h a t  i s the r e s t r i c t i o n  of  h  to  G..  G  i s closed  i s an isomorphism.  C  a  As  Choose  n < c a define = a  a  = sup  ag < ui],  that  a  such t h a t  i n d u c t i v e l y as  ({h(6):6 <a  then  1  }'{J  a n  n-1  That i s i f we assume i n d u c t i v e l y  coi ,  then  1  a  3 < ct , ,-c  of some  a  t 1  n+1 n  < coi  f o r a l l n < co,  then. 3 < -.  f o r some  Similarly i f  and so  C.  follows:  i s the sup of a c o u n t a b l e s e t and  no c o u n t a b l e s e t i s c o f i n a l w i t h Since  i s an upper bound f o r  (3:h(3) < a . ' } )  n-1 .  < coj f o r a l l n.  then  , < can ,  n-1  a  a < co_  Suppose  n  ag n  r  o  o  s i n c e the u n i o n of a c h a i n of isomorphisms  For  C = {6:h|G  i s a c l o s e d . a n d unbounded subset  o  i s bounded.  elements  satisfies  Possibility  G*}  be an  3 = h(p)  3£G* where  then n,  a n  a* < coi .  < co-i . Let  and so h(3) <  such t h a t p < a ,. .  a n+  3£G -^ then  Thus  h G .  a  i s not an upper bound f o r  1  C  a * = sup a . n<& n such that the. d e f i n i t i o n  3 < OL*  n+1 and so a*€C. Thus unbounded. .  Let  1  since  3 < a i s an  n  for isomorphism  a* and so  ' C  is  10  Now by  we  the d e f i n i t i o n ' of  G*  will  set.  satisfy  CD  A  then  H  C^  Possibility  and  C^ f cj>,  II  and  for l e t  since  n  If  even  C /I.C*,  Thus  (C 11 A) D  Case ( i i i ) : neither  By can  was  satisfies  bounded,  C*)  ^ <j> .  Possibility  I  classification  by  say  into  Possibility :  the  ^  Thus  X/J  i s i n both  C^O C^  a < coj.,  £j = a  E 6C„ . n 2  odd  C^  is  and.  .C^  then  define  to get a c o n t r a d i c t i o n .  i s c l o s e d and C.-fl A  unbounded.  i s stationary  II.  Possibility  nor  A N  n  Actually  c l o s e d unbounded s e t C*,  By. d e f i n i t i o n  for  i s c l o s e d as  as b e f o r e x<rith  C* = A f) (C n  Possibility  The only up  f o r any  and  {E2> ?4 > ?6> • • • }  C^  sets  he an i n c r e a s i n g sequence  E € C, n 1  D  stationary  unbounded set is.  are c l o s e d and unbounded  :  then  as the r e q u i r e d  c l o s e d and  C^ are c l o s e d .  {£J j.,£2> 53> • • • }  Thus  G*  C^  C fl A  using  Clearly  C^H  required  i s a s t a t i o n a r y s e t , and  2  and  unbounded.  are c l o s e d .  so  C^  i s the s t a t i o n a r y s e t  {£^, £J , ?3 v • • }  such t h a t f o r  C^  a sequence  A  II,  tj) = sup. {£]_., f ^ , ^ . , .. } = sup  c l o s e d and  G  Possibility  That i s i f ' C^  of o r d i n a l s  Then  where  i s s t a t i o n a r y because any  stationary.  and  C C\ A,  show t h a t  III II  h o l d s i f and'only i f holds'.  t h r e e p o s s i b i l i t i e s depends on.  G  to isomorphism.  the d e f i n i t i o n of the t h r e e p o s s i b i l i t i e s , satisfy  partition.  The  by a p a r t i c u l a r  only'one, . and  an  tu^-free group  so the t h r e e p o s s i b i l i t i e s form a  f o l l o w i n g lemma shows t h a t each p o s s i b i l i t y ' i s s a t i s f i e d coi-free. group.  11  Lemma ( 4 ) : • Each  Possibility  i s s a t i s f i e d by some  co._-free  group. Proof:  A g a i n t h e r e a r e t h r e e cases t o c o n s i d e r .  Case ( i ) :  We w i l l c o n s t r u c t an  F i r s t we d e f i n e a s e t C of l e n g t h T  co_-fr.ee group s a t i s f y i n g  o f i n c r e a s i n g sequences o f n a t u r a l numbers  co such t h a t t h e c a r d i n a l i t y o f  a r e i n C, .n ^ T, then  n a t u r a l numbers i n common; such a s e t C  P o s s i b i l i t y I.  n  and  T  C  i s co_, and i f " n and.  have a t most f i n i t e l y many  that i s n /i x  e x i s t s we g i v e an example.  i sfinite. Consider  To show t h a t  the f o l l o w i n g  diagram:  The  sequences a r e d e f i n e d by t a k i n g p o s s i b l e p a t h s .  F o r example: .  {1,2,4,8,...} (1,2,4,9  }  {1,2,5,10,...} etc. By t h e n ^  1  row o f t h e diagram  2  n  and  i n the l i m i t there a r e 2  The  i n t e r s e c t i o n o f any two i s f i n i t e f o r they can agree o n l y up to  W  >_ co j  sequences o r paths a r e d e f i n e d ,  the p o i n t where t h e i r c o r r e s p o n d i n g  sequences.  paths  Choose any . co _  separate.  sequences.  12  Let  G  be generated by:  (i)  x,  f o r k < 10  (ii)  m k! x '= , 1 ( r')x , T k=m m. x(k) 00  f o r m < oo and  —  Using the notation of the d e f i n i t i o n of C = {.x(ct):ot < a)].} (i)  G^  be the group freely, generated by the ^ ' x  and so  We must show that  G  {a + G.:'ct < OJI }  1 n  • fixed.  m  .  (A)  and  (B)  i n the  That i s :  i s an independent family i n G/G. .  o  n(ao,cj),G ,G)  .(B)  . tor a < to . and  P o s s i b i l i t y I.  s  = <j>.  s a t i s f i e s conditions  Ct  n  L a  /. . . \ • ct m (ixx) an = x . . . x(ct)  (A)  P o s s i b i l i t y I and  let:.  ( i i ) . n(ct) = 0,  definition.of  xeC  • .  .  o  holds f o r a l l a < oo . L  (A) follows from the f i n i t e i n t e r s e c t i o n property of the elements of  C.  That i s i f . z.x ".  . +...+  1  1  x. (a.)£C, l  and  i ( a1 i )  g£G/ , 6  then  z x™ ' . = g where z , £ Z . z. ± 0, n x(ct ) i i n z , x , .+...+ z x , , i s a f i n i t e lx(ai) nx(a). .  n  l i n e a r 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 l i n e a r combinations of the x, 's , then z,x . . +...+ z x™. . m  k  1 x(cti)  must be an i n f i n i t e l i n e a r combination of the x, 's x™, . x(aj  and  is finite.  x™. . x (a )  agree at only f i n i t e l y many  This i s a contradiction and so  Now we show condition  (B) holds.  L  x 's k  = <L  f o r x(ct..) C\ x(ct ') I J  then  a  © <L > . Choose any a n = x™, . . Then . a * . x(a) / , i m m , . , . .. (m + l ) x , s = x - x . . and so by d e f i n i t i o n of purity x(ct) x(ct) x(a)(m)  <G U L >.=' <G > r  6  ct * s m+1  .6  m+  x(ct)  u  r  /  x \  J  *  N  G<G, U {x™, }>, . 6 x(a) *  w  J  Similarly J  x . . £ < G U { . }>, x(a) 6 x(a) * k  m  r  x  T  since f o r i ^. j  (A) holds.  As  n (ct )  for a l l  k ^ m + 1. Using the f i n i t e i n t e r s e c t i o n property f o r elements  n  13  of  C  <G.V 6  i t i s c l e a r t h a t no o t h e r elements of G ( x . }> . . x(a)  So i f f o r some  m  <G^>,. ©• , v '  x e <G. V • { , }> , , 6 x(a) * ' m  x  then we c a n assume t h a t  < x >  5  w i l l , be thrown  into  <G U {x™, . }> , = 6 x ( a ) *.  <x>, = <x> = t h e group generated 5  {  x = . 2 , z.y. where z.e.Z and each y. i s some i=l i i l i k k o r x . . . C l e a r l y x w i l l cause o n l y f i n i t e l y many o f t h e x > . x(a.) . t(a)  by  x  f o r some  J  t k  to be i n ^g ,'- ® <x> <  n(x  >  A  ;  a n d so t h i s i s i m p o s s i b l e .  , \ A, G.,G) = n ( a , L ,G.,G) x (cO o :. a o 0  (B)  Therefore  h o l d s f o r a l l a < ui x  and so c o n d i t i o n  i s satisfied. Now we show t h a t  for  any g  G  g' £G , n .  i  i s oj_-free.  I t i s s u f f i c i e n t t o show t h a t  t h e pure subgroup generated by  g.,...,g I n  i s f r e e on a f i n i t e number o f g e n e r a t o r s , (4) page 25. Without of g e n e r a l i t y has rank, n  g ,...,g I n  a r e independent and so  loss  <{g.,,...,g }>, 1 n x  b generate <{g,,.••,g }>, , 1 n 1 n { h ^ , . . . ,b^}> . We do an i n d u c t i o n on t h e  (1) page 116.  So l e t b  :<  that i s  '  <  * * •' §  n  ^  >  A  number o f g e n e r a t o r s . i s torsion free.  <  =  For n = 1  c l e a r l y . <{b^}>  Assume any pure subgroup on  i s free since  n - 1  G  generators i s  f r e e and l e t  <{g. , . . ..g }>, be generated by b. b . If I n * I n <{b- ,...,b }> i s n o t f r e e l y generated, by b,,...,b , then f o r I n I n 1  some  z.e Z, n o t a l l z e r o , .? • 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 g e n e r a t e d by g,,...,g r n 1 n-1 x ° I n has rank l e s s than generates  <  n,  ^S-^ > • • • J § ^ n  a contradiction. >  So  freely  and by t h e i n d u c t i o n h y p o t h e s i s any pure  subgroup g e n e r a t e d by a f i n i t e subset o f G to . - f r e e .  b,,...,b 1 n  i s free.  Thus  G is  14  Now Choose  l e t G*  be any admissible naming of the elements of  6 <'ID,  such that  1  {x , . :a < coi } x (a)  x  is in  n  i s uncountable and  m  1  G* 6  G„ 6  for a l l n < co . i s countable,  uncountably many  x™. . 's such that . G • for x (a) T (a) ' 6 {x™^^:3 < c o . } be such a set. By l e t t i n g :  n(3) = 0,  (ii)  and so  L  G.  As  we can find k < co . l e t  = <j> 3  /•••\3 m ( i n ) a. = ( g )  _  T  3  satisfies  Possibility I  x  0  i t follows that way as  G  G*  f o r  was shown to s a t i s f y  <  co.  i n exactly the same  P o s s i b i l i t y I.  Case ( i i ) : We w i l l construct a group s a t i s f y i n g For this example, for  the stationary set  P o s s i b i l i t y II  A  Possibility II.  required by the d e f i n i t i o n  w i l l be the" set of a l l l i m i t ordinals. A = {6 < co_:6 i s a l i m i t ordinal}  we show that this set  First  i s stationary.  This follows from the observation that any closed and unbounded set contains a l i m i t o r d i n a l . i n f i n i t e subset of  C  That i s i f . {a.,a ,a ,... } 2  where  a. < a.,, , 1  ordinal for i f not then  a  then  sup a. = a 1<GJ  has a predessor  i s any countably  3  1+1  a - 1  is a limit  1  which would be an  upper bound to the sequence. Now for of length  6  a"limit o r d i n a l ,  co such that  ordinal of the sequence (i)  l e t x _ be a sequence of ordinals o  sup x „ (n) = 6 n<co  where  x.(n)  o  x. .  i s the  n'th  o  Let  6  for a < co-. a m k\ x = .1 (—r)x ' . m. x ( k ) a 6l i m ik=m t ordinal  G  be generated by:  x  1  00  (ii)  J  for  m <  to  ,  6 < .  C  co 1  n  ,  and  6  15  Using the notation of the d e f i n i t i o n of P o s s i b i l i t y II (i)  x  £ G.  ,  X j i G . ,  for  ct  <  6  ,  m  <  cu  let:  ,  and  6  a  l i m i t ordinal (ii) /  Since G/G  • • • \  &  .  .  .  .  m  (m) ag = x^ , m fixed x™£G_ , then {x™ + G.} o o o o  and so condition  6  and so L = <> J .o  n(<5) = 0 ,  i s an independent family i n  (A) 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 II  n(x.,L-,G.,G) .holds using the same arguement as used for o o o ' P o s s i b i l i t y I , and so condition (B) i s s a t i s f i e d . The oj^-freeness holds.  of  G  i s again similar to P o s s i b i l i t y I.  Lastly we must show that  G  doesn't s a t i s f y  P o s s i b i l i t y I.  It i s s u f f i c i e n t to show that a given admissible naming of G not.satisfy i t . (i) (ii)  does  Let 'G_ be generated by: o x for a < 5 ct ' x^. . for $ < 6 , m < co , 3 a l i m i t ordinal 3  Now define the x s 1  S  <•(!>!  . Then  to be increasing sequences for a l l l i m i t ordinals  It(x™,<j>p ,G)  cannot hold for any other  holds for any m < w .. The  x™' 's ,  3 / <5 , since for  3  "II" condition  3 < 6 , x™ i s  -  3  m Ic in G , and for 3 > 5 , <G,U {x }>. = G ® <x>, where k i s o o p o 3 " the largest element of the increasing sequence x less' than 6 . r  n  r  '  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 i s s a t i s f i e d at only  many places for each G  satisfies  G. . Thus o P o s s i b i l i t y II."  Possibility I  countably  cannot hold and so  Case ( i i i ) : sufficient for  Let  G  be the f r e e group on  to show t h a t  G  does not s a t i s f y  some a d m i s s i b l e naming of  elements 3 <_ £,  a  3  for  cu_ g e n e r a t o r s .  G.  Let' G  ,  3 < coi .  L  not a l i m i t o r d i n a l .  Let  G  Possibility  G  That i s  G _ = „ffi^G^ ,  That i s  G  = »©_G  3  3  by the element Claim  that  where  L  a  .  ,  a  3 <  ,  P  where  G  •  3  L  and  5  where  Let  3  a limit  G _  w5  ordinal.  G  generated  i s an a d m i s s i b l e naming of  G.  does not h o l d f o r any l i m i t o r d i n a l  i s a f i n i t e subset of  suppose f o r some  for  a. .  i s the subgroup of  Clearly this  n(a,L,G^,G )  .  a  G  and  5,  a i s an element o f . G.  II(a,L,G ,G)  that  r  ,  3  S<('  generated by the element  be the group generated by  a  be the group generated by  cot;  CJcj  i s the subgroup of  I or I I  be.generated by the  —  G^  It i s  holds.  I f we  So  can  o  show t h a t o n l y f i n i t e l y many element's a r e i n the group W •= <G. U L U {a}>./<<G U L>  U  p  page 24 "11"  CG 7G^  is infinite i f  5  must f a i l .  If  where  Then the o n l y new the  a 's  z.CZ,  W.  G  is  (1) page 74.  and so  where  then by the r e s u l t  holds)'  since  and i f  a = a  W ='G ,  on /G^ ,  then each  o  a.  i s the g e n e r a t o r of  w i l l be l i n e a r  a/s.  Thus the  t h i s a d m i s s i b l e naming, Also  and  elements i n .W  which make up the  many of. these i n  "n"  L = {a, ,.. . .a } I n  m. a. =.E-,z.a. ,  {a}>,  W  G J .  combinations of  C l e a r l y there are only f i n i t e l y  "II" c o n d i t i o n f a i l s  in  G  under  P o s s i b i l i t y I or I I cannot h o l d .  io_-free s i n c e any subgroup of a f r e e group i s f r e e , Thus  G  must s a t i s f y  Possibility III.  17  Lemma (5) : equivalent Proof:  to  Let G G  Suppose  o)i-free.  Then  be  being G  ooi-free.  the d i r e c t  i s the d i r e c t  Then  Possibility  III • i s  sum o f c o u n t a b l e groups. sum o f c o u n t a b l e groups and  G is  ct ct $ G where each G i s countable. Since a<co ct ct each G i s free,, so each G i s isomorphic to a G =  G  V  1  to T - f r e e ,  is  countable d i r e c t direct  sum o f  sum o f c o p i e s  U[- c o p i e s  Lemma ( 4 ) case  By  Now suppose if 6  C  (iii), G  limit  G  Z  Z.  Thus  and so  G  G  satisfies  satisfies  i s i s o m o r p h i c to a  i s f r e e on  III.  F i r s t we show t h a t  i s a, c l o s e d and unbounded subset of  OJ ,  then  C  1  C* = {6:6 € C  i s a l s o c l o s e d and unbounded.  C*. i s c l o s e d  i s c l o s e d and the sup o f a sequence o f l i m i t o r d i n a l s i s a  ordinal.  o r d i n a l of  C*  C.  i s unbounded s i n c e  Thus  ordinals. . Since  C  C*  i s unbounded and the sup  L,  i sa limit only  <  Possibility  and  C  i s c l o s e d and unbounded and c o n t a i n s  I  and  Possibility  a c l o s e d unbounded s e t C .such t h a t i f a£G  generators.  Possibility III..  of an i n f i n i t e i n c r e a s i n g sequence of o r d i n a l s of  limit  to^  Possibility  i s a l i m i t ordinal}  since  of  of  a finite  subset o f  G,  II  fails,  6£C ,  we can f i n d  t h e r e does not. e x i s t  such t h a t  n(a,L,G.,G)  holds.  6  That i s ,  i f f o r every c l o s e d and unbounded s e t such a  then by t a k i n g  the s e t of these  condition  that  i s a maximal independent f a m i l y i n L,  of the  Possibility L's .  of  Possibility  I I would be s a t i s f i e d  Since  Possibility  I  exists,  we get a s t a t i o n a r y s e t which  satisfies L*  (B)  6's  6  II.  using  fails,  By t a k i n g  then c o n d i t i o n  the L*'s  then  L*£L  such (A)  i n p l a c e of  Possibility  II  would  and  18  hold for  G, a contradiction.  Therefore such a  C exists.  From previous remarks, i n this proof we can assume that only l i m i t ordinals. is  Since  sup C = OJ. ,  rename  G  5 i s a l i m i t ordinal and a as follows: {3:6  Rename  <.' 6 < 6 ci+1  a  Now we can assume that  OJ, . So l e t  C = {6 :a < aii }  a < 3 ^ 6 < 6„ . Now we a 6  as  —  contains  then the c a r d i n a l i t y of C  coi since no countable set i s c o f i n a l with  where each  C  C = {toa:a < to.-},  (3:toa < 3< —  w(a + 1 ) } .  and i t i s clear that we s t i l l  have an admissible naming of G. Now we do an induction to show that for some Let  G  G , = G ffi <b, ,'b„, . . . >. coa+oj ua . 1 2 * b.'s i n G . ^ G . Suppose b,, . . .b have been choosen. i wa+co a>a I n rr  ® <b -,...b >., . Now l e t L = {b,,...b } and l e t I n " . 1 n a = i n f '{6:6 6 G , — G } . As to a i s i n 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  = G  n  n+1  toot ,  *  <G U L V {a}>, = G ® < L U {b ,,}>,= G © <b ,...,b ,b ,,>, . wa ooa n+1 > toa 1 n n+1. * x  c  Since c l e a r l y  n  U G =. G , we get that. G , = G ® <b,,b ,...>, . n toa toa+to , toa+to toa 1 2 H = <b, , b ...>_. . Then G , = G <B H . Thus wa 1 1 > toa+w coa toa n  0  5C  Let  o5  G = G • ffi H (o l_<a<cj toa 1  and so '  G  i s the direct sum of countable groups,  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  about them n e c e s s a r y f o r the c o n s i s t e n c y  Definition set  (14):  A  G x Z = {(a,b):a£G,  For  (i)  (a,b) +  (ii)  The map  a given  G/  i s a group  b£Z}  that:  wi-free  is  a  h:H ->- G  ,  H  d e f i n e d by  w i l l denote.a  Proof:  First  and  is  G^  Let  G^  wi-free.  of  G„ 2  2^ 1  G  G  a.  8  e n e r a t e c  = <{a} U  Thus from the r e s u l t  (i):  <a + G^>  Case  ( i i ) : <a + .G >  to a  Let  where  G^  Wi .  Let  H be 1  (G^,Z)-group.  G  ,  where  G . is a  The r e s u l t  will  the i n d u c t i o n we  will,  be the subgroup a  is in  generated by  z.  i s i s o m o r p h i c to is cyclic  W-groups.  <a + G^>  a + G  G^.  {{a} U G  and  x  order. (ii) H  can be  of  Let }.  Z.  of prime  show t h a t i n cases ( i )  (G^,Z)-group.  G^  noted b e f o r e ,  To s i m p l i f y  be the subgroup of  Case  We w i l l  cases f i r s t .  * by the element  G > j.  of  i s f r e e l y generated s i n c e i t i s c o u n t a b l e  be proved by t r a n s f i n i t e i n d e c t i o n . d e a l w i t h two s p e c i a l  and the c o r r e s p o n d i n g  i s a t most  can be extended to a G^  i s a group homomorphism.  (G^,Z)-group,  be a c o u n t a b l e subgroup  Then  note t h a t  h(a,b) = a  W-group s i n c e f r e e l y generated groups a r e  G  with underlying  h_^ .  and the c a r d i n a l i t y  (G^,Z)-group.  H  (0,c) = (a,b+c) ,  homomorphism w i l l be denoted by  Lemma ( 6 ) :  result.  (G,Z)-group such  and prove some lemmas  extended  20  Proof o f case b £G a Now  (i):  Suppose  <a + G^>  i s isomorphic t o Z.  has a unique r e p r e s e n t a t i o n as  define.for  b  l  = d  +  1  z eZ  and c € G, . 1  the f o l l o w i n g :  (b ,k ) 2  2  (z a + c . k ^ + ( z a H-'c , k ) 2  i ( (  z )a  f  Z  l  +  2  2  c ,k )  +  3  3  (c^,k^) + ( c , k ) = (c^.k  where  where  , b„ i n G and k , k i n Z , z• a 1 /  1  (b ,k )  za + c  Then every  2  2  t h i s n a t u r a l e x t e n s i o n o f H,  )'  in  .  I t i s easy t o check  forms a group.  Call  t h i s group  H . a  1 Then  that  (b,k) + (0,m) = (za + c,k) + (0,m) .=  (za + c,k + m)  ' = (b,k + m) A l s o the mapping homomorphism,  and so  Proof of case Since  h :H  (ii):  -> G H a  d e f i n e d by  is a  Suppose  h^'.H^.-*- G^  , since i n  (c,k) + (0,m)  h (b,k) = b  = (c,k + m).  i s clearly a  (G ,Z)-group. a  <a + G^>  i s c y c l i c o f prime  has k e r n e l i s o m o r p h i c t o Z  and G^  order  is a  p.  W-group,  then t h e r e e x i s t s for  c£G^.  where  g :G -> H such t h a t 'h g = 1 . L e t g (c) = (c,m(c)) 1 i 1 1 1 Gi I Every b £ G^ has a unique, r e p r e s e n t a t i o n as za + c  0 <_ z < p ,  c £ G ' . Now f o r  b, , b„  1  ~~•  1  in G  2  and k, , k-  a  1 . 2  define: (b ,k ) l  +  1  (b ,k ) 2  (z a + c ^ k ^  2  + ( z a +c , k )  x  2  def ^ ' ( ( z ^ + z > a + c^ + '^i c  2  . wiere  + f(z  0 <.-z •, z  otherwise;  +  l  2  + 2 ~ ^ k  m  c  p  - m(c ) 2  + m(c^ + c ) 2  z )) 2  < p ,  where  2  M  and f (n) = 0  when  n < p  and f (h) ,=. M € Z  i s an a r b i t r a r y c o n s t a n t which once  chosen  in Z  remains the same f o r a l l such d e f i n e d call  i t H  show  H  ,  a  forms a  i s a group,  of i n v e r s e s . i n v e r s e of  Let c  sums.  (G ,Z)-group under t h i s a t h e o n l y non t r i v i a l  (b,k) = (za + c,k)  in •G  We w i l l show t h a t t h i s s e t , defined  operation.  To  t h i n g to show the e x i s t e n c e  be i n  H  and  a  - c be the  . ; 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 i n v e r s e o f a group.  Now l e t  (b,k) i s  ((p - z ) a - c, -k - M)  (b,k) = (za + c,k)  (0,t)  and  ,  and so  be i n  H  3-  is  H,. Then: a  ' (b,k) + (0,t)  Also so  =  (b,k + t - m(c) - m(0)  + m(c) + f ( z > )  =  (b,k + t )  z < p .  the mapping H  a  is a  Since and  , as  G^  h :H G, , a a 1 (G ,Z)-group. a  d e f i n e d by  i s .coj-free and  so i t i s f r e e l y  generated.  Now we do t h e - i n d u c t i o n .  G^  h (b,k) = b i s a homomorphism, a  i s countable,  Thus  G  a  is a  F i r s t we f i n d  say  G„., 2  and such t h a t i f J . = <G, U {a : p < 6}> o 1 p  i s countable  W-group.  such that  G^U  for a l l  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 o r d e r .  A i s d e f i n e d as f o l l o w s :  G^  a sequence o f elements, o f  G^ ,  <a^ + J^>  A = {a^:6 < a,a an o r d i n a l } ,  then  A  generates  6 < a,  then  The sequence  22  Assume  a„ has been d e f i n e d f o r a l l g < 6. L e t b = inf{a:a£ G.\ g 2 <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 o r d e r , l e t a. = b.  If  o ,  I f not, where  •  then p  <b + J > o  i s prime.  .  •  •  i s c y c l i c of non prime o r d e r ,  Then l e t  a. = nb. 6  I t i s c l e a r that c a r d (A) < oj; ,  So ' <a G^V  since  6  A  o  say of order  '+ J > 6  np  has prime o r d e r  generates  G^  and  cardCG^) <_ o)_ . Let  =  (J,Z)-group  3  (a)  •  ,  a  for a l l  •  (J^,Z)-group.  to be a  g <_ a.  .  ~  .3 i s not a l i m i t o r d i n a l :  has a p r e d e s s o r .  We w i l l d e f i n e  So we  Since  3  can suppose a  i s not a l i m i t o r d i n a l ,  (J  1  ,Z)-group,  K  g-1  defined.  ii-  Then by c o n s t r u c t i o n of the sequence  A,  <a  ,  L  it  has been  . + J  > is 3-1 is infinite  .3-1 infinite cyclic cyclic  ,  then case  can be extended order, it  then  is a  or i s c y c l i c of prime o r d e r . (i)  to a  as  can be a p p l i e d d i r e c t l y (J^,Z)-group ^  W-group.  s  K  .  countable,  to  K  i t i s freely  Then t h e r e e x i s t s  to show i t  g-1  I f i t i s c y c l i c of prime  g  :J p-1  generated  K 3-1  (b) that  *j  and  so  such t h a t g-1 and  where as u s u a l h ^ i K ^ r> J ^ _ h(a,b) .= a. g-1 Thus we can a p p l y case ( i i ) u s i n g g as the r e q u i r e d map, and so •p-1 extend K to a ( J , Z ) - g r o u p K . h  g_l§3_l  If i t  =  1  3 i s a limit ordinal: • K  is a  3  r>  r  I t i s easy  to check  R  Gj f  (G ,Z)-group n  I  (J,Z)-group.  So i n d u c t i v e l y we the s e t  K. •==V K . g 6<g 6  Define  A and  can d e f i n e a  generates  G^  ,  (J^,Z)-group.  then  the lemma i s proved.  = G^  ,  Call i t and  so  H  J.} 6  . is a  As.  p.  23  Lemma(7):  Let  homomorphisms, h g  i  n  :  be a ^  H  G  -[  a n <  (G^,Z)-group. ^  h (a,b) = a,  G-L .  Let  n(a,A,G^,G2,)  G^  be  extends Proof:  holds.  Then  can be extended g :G  <G  (i) '  A = {a, ,...,a } 1 m  (ii)  G  (ill)  G.  =  (iv)  G  = <G  H. 4  be a  H-^' ,  be  such t h a t  g  b = c + za >  a n c {  the  ^  x  a^'s  £  ng(x) = y. = a.  As  3  4 5  = <G  <  g  3  As  ^ *  '  >  x  e  <  U  ^- *  h g •= 1 4 G^  *.Z  g(x)  and  So each  = <G > 1  show t h a t  make some o b s e r v a t i o n s about i n t o two  possibilities.  where  gz  5V  i<  mo v C o n s i d e r the homorphisms  and  .  Any  and  z£Z.  such  a. So  g  g:G.  -> H, 4 4  i s uniquely  That i s i f .  b € G. , ' 4  b = d + x + za where  nx i s a l i n e a r  Thus the  g(a_^)'s  combination  determine  g(nx) = ng(x) ,  s i n c e t h e r e i s a unique s o l u t i o n to g(d)  i s a l r e a d y determined by  h. (b,z) = b, 4 g(a^)  be. o n l y c o u n t a b l y many such . g's.. we w i l l  = 1 Z  0 <A>  A  then -  G /G, 5 r  g^  .  Let  g(a.)£{(a.,z):z£Z} l l  can be d e f i n e d i n o n l y c o u n t a b l y  many ways and s i n c e t h e r e a r e o n l y f i n i t e l y many  Now  Suppose  {a}>  implies  >  ,  ?v  a.,...,a 1 m  i = ,l,...,m.  A  does :h g  U A U {a}>  c £ G^  d'£<G^>  i = 0,...,m.  A>  h g = 1 4  maps  where  for  U  1  <G„  and such t h a t  1  .  h^Ca.b) = a.  (G.,Z)-group. 4  and so they determine  for  g^  to a (G2>Z)-group  Z  Let:  determined by where  d £ ^ ,'  2  and as u s u a l  t h a t extend  then  -> H„  N  g^  Let  n  g^:G  1  2  U  and  t o i - f r e e and card(G2) 4 to  such t h a t f o r no homomorphism  a  h^  = 1  11  of  Let  Call  them  a  -^'  {g :n n  must be i n f i n i t e . 4 '  the s t r u c t u r e of  G^/G^  s  »  t h e r e can  < io) = R. Then we  a n <  ^  will  classify  it  24  G '/G 5  i s infinite:  4  then there that  Since x £ <G^  exists  nx = g + c + ka  and  k ^ 0.  G  Let n  = <G U A U {a}>^ $  5  A V {a}>  where f o r some  g, k, and  k  f o r i f not then say m  c.  Then the g r e a t e s t  • m((—)x - (—)a) = g + e, m m there and  g , 6 <G >. • 1 . 1 *  exists  so  and s i n c e and  consider  exists integers  divides  ®  V  <  A  {a}> -. ,  u  A  © <A U {a}>^  A  g£<G^>  ,  A  such  c 6 <A> , 4  common d i v i s o r of n  and  k.  <G. f A>. = <G > 1 * 1 * such t h a t  and  z,  such t h a t  n and  Then then  J  (—)'x - (—)a = g, + c • , m m' 1 1 ' 6  c o n t r a d i c t i n g the m i n i m a l i t y •  w  there  ® <A> , .* '  u  c,€ <A> . 1 *  (—)x = g + c, + (—)a, m l l m  Thus t h e r e  n ^ 1,  .  > 1  be t h e s m a l l e s t p o s i t i v e i n t e g e r f o r .which  i s such a i s 1,  , ;x^<G >  5V  < G  of n.  nw + kz = I. Now  wa + zx: nx = g + c + ka ,  so and  wg + wc = wnx - wka , so  wg + wc + k(wa + zx) = wnx -wka + kwa +kzx = wnx 4- kzx ' = (wn + k z ) x =. x  Similarly:' -zg - zc + n(wa + zx) = -znx + zka + nwa + nxz = (zk + nw)a = a So i f x  was t h e o n l y new element i n <G^ C A L/ {a}>.,. ,  <G\^U A U ( a } > = G >,, © <A  {wa +zx}>  <  A  II (a,A,G^ ,0 ) ,  f o r some  y u  {  since  s  must be a  1  such.that Z.  <G U A U {a}>^  wa + zx  A  ,  i s i n <G^U  a contra  or  U s i n g the same method we can f i n d such t h a t  y  and  wa + zx  d i c t i o n to  A U {a}>  my = g + c + sa + t x ,  then  A  •  So t h e r e  my = g^ + c^ + u(wa + zx) an element  are i n  b in  ® <A (/. {b}>^ .  S i n c e t h i s p r o c e s s can be r e p e a t e d f o r any f i n i t e number of such elements,  i t f o l l o w s that t h e r e must be i n f i n i t e l y many of them e l s e  we get a c o n t r a d i c t i o n to the • countably i n f i n i t e Definition G the  Thus  .G /G. b 4  isa  r  t o r s i o n group.  (15):  and every i n t e g e r equation  II c o n d i t i o n .  A group n,  G  i s d i v i s i b l e i f f o r every  t h e r e e x i s t s elements i n G  that  x in satisfy  ny = x.  From Kaplansky ( 5 ) , (a)  Any a b e l i a n group  and  G = M ® N  (b)  Any d i v i s i b l e group i s a d i r e c t  where  G  we have the f o l l o w i n g two r e s u l t s :  N  has a unique l a r g e s t d i v i s i b l e has no d i v i s i b l e subgroups. sum of groups,  subgroup  M,  (5) page 9. each i s o m o r p h i c  oo to  the . a d d i t i v e group of r a t i o n a l s  all  p ^  Q.  G^/G^  So i f G^/G^  the group o f  (5) page 10.  has a non t r i v i a l G^/G^  d i v i s i b l e subgroup,  c o n t a i n s a copy of  then by  Z(p ) f o r some  CO  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. Definition  divisible  (16):  A group  G  i s reduced i f i t has no non t r i v i a l  subgroups.  From Kaplansky we have the f o l l o w i n g (c)  Z(p ) ,  i s a t o r s i o n group i t cannot have a subgroup i s o m o r p h i c  Kaplansky s two r e s u l t s , prime  or to  r o o t s of u n i t y f o r v a r i o u s primes p.  As to  Q,  If  G  i s a reduced group which i s not t o r s i o n f r e e ,  has. a f i n i t e c y c l i c summand. Since a finite  result".  G^/G^  (5)  Now a p p l y  G  page 21.  i s a reduced t o r s i o n group,  c y c l i c summand.  then  then by (c) G  (c) to the other summand.  has Repeated  a p p l i c a t i o n of (c) to the i n f i n i t e that  G^/G^  r e m a i n i n g summand of  c o n t a i n s an i n f i n i t e d i r e c t  sum o f f i n i t e c y c l i c  From each of these choose an element of prime o r d e r . assume t h a t t h e r e 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 for  n < co, .  such t h a t  p a € G, n n 4  G^/G^. shows  where each  p  n  Thus we can  elements,  say a^  i s prime.  Now we can say t h a t one of the f o l l o w i n g p o s s i b i l i t i e s in  groups.  occurs  G./G.: D 4 (I)  Gj_/G^  c o n t a i n s i n f i n i t e l y many elements of prime  order.  CO  or  ( I I ) 'G /G^ c o n t a i n s a copy of let  (I) h o l d i n G /G. . 5 4 r  prime o r d e r . Let  '*  Let a  That i s t h e element  , n .  f o r some  n < co,  a + G, n 4  .  G n  Z(p )  ft  (G  n+1  *  extends  ,Z)-group  H  ,  n+1.  be the elements of  has o r d e r  be generated by G. U { a , , a ^ 4 O n - 1  define a  p.  i n G /G, . 5 4  p n  We w i l l  n  inductively '  ft  using  Lemma (6) so t h a t  H  *  '  n+1  •  .  First  use Lemma (6)  to extend  to a  (G^,Z)-group  ft which we w i l l  call  .  Clearly  t h i s can be done as  meet a l l the c o n d i t i o n s of  Lemma ( 6 ) .  subgroup  i s coj-free  of  G^ ,  and' G^  That i s G^  G^  and  G^  i s a countable  as i t i s a subgroup  of the  c o i - f r e e group G^ . A l s o i s g i v e n to be a (G^,Z)-group. •* >v * .ft H„ exists. Assume i n d u c t i v e l y H i s defined. G and G ,, 0 n n n+1  Thus  ft  satisfy  the c o n d i t i o n s of  ft ft ft  Lemma. (6)  using  as the r e q u i r e d  (G ,Z)-group. So by Lemma (6) can be extended to a ( G ^ ^ , Z ) ~ ft . f t group, H . As a. + G. has o r d e r p , a + G has o r d e r p • n+1 n 4 n' n n n n  Let  +  M  n  be the c o n s t a n t used i n Lemma (6)  I n d u c t i v e l y we d e f i n e to  a (G,_,Z)-group,  ft. ft H . Again apply CO  say . H,. ,  to extend  Lemma'(6)  which extends a l l the  ft  H n  to  to extend H  ' s  n  -  ft  H ,, . n+1 H  CO  27  If h.gr  g :G,.  = l  where  H  •' i s a homomprphism extending  then f o r some  r  5  g £R  n,  g  ->  g  ;  n  as d e f i n e d e a r l i e r i n the proof..  n  such t h a t  t h a t i s g| / = g 0 c  n  T  Now we w i l l  ft  i t . * '  using  extends  g^  show  • < '  .  g l * § as the r e q u i r e d map i n extending to H _ > n (see Lemma ( 6 ) , case ( i i ) ) , that c o n s t a n t s M can be chosen •n i ft such that gr-L* has no e x t e n s i o n to G f o r each n < to. T h i s 5 ii n+1 n • * n w i l l show that g^ and thus g has no e x t e n s i o n to G^ and so =  5  such a As  H  n  +  gj. does not e x i s t . a + G. n 4  i s of o r d e r  n g (b ) = (b ,k ) n n n g  ft  n  xi  extends  g  p , n  then l e t  ft g (a ) = (a ,c ) . n n n n  and  Since  n  we have  g (b ) = g (b ) n n n  g ( b ) = (b ,k ) • n ' n n  p a = b 6 G, . L e t n n n 4 b i s i n G, n 4  and so:  by d e f i n i t i o n  n  = g (b ) n n .  as  b € G. n 4  = g (p a ) n n n  as  p a -= b .n n n  P g (a ) n n n  as  g n  X  ft  =  and  ft ft  i s a homomorphism 1  = p (a ,c ) n n n  by d e f i n i t i o n  =  by d e f i n i t i o n of "+" i n H" n+1  (b ,p c + M ) n n n n  as  as d e f i n e d i n Lemma ( 6 ) , case ( i i ) . So i n H ,,: n+1 p  n  (a c ) = (a ,c ) +.....+(a ,c ) n n n n. n n =  (p n  (a + a ,c + c + f ( l + 1)) + (a ,c ) + n n n n n n (2a ,2c ) + (a ,c ) + + (a ,c ) n n n n n n  =  = ( ( P - . D a , ( P - D O + (a ,c ) n  =  =  (  V n ' (b ,p  n  n  P  n  n  c  n  C  n  +  f  ( P  + M )  n  n  n  n  } 0  .  n  times)  + (a ,c ) n n  28  R e c a l l t h a t the c o n s t a n t was k  n  arbitrary. = p c + M n n n  n  and so  k  suppose  f o r some  elements, (a)  pa  (b)  pa  That i s  p.  say  a  Q  n  a -  0  Q  u n i t y such that  •  n  g G  (mod  Lemma (6) case ( i i )  p ). n  By c h o o s i n g  & ,, n+1  and so  That i s  g  n  p (pa  n < OJ,  = k  n G  +1,  n  and  n+1  cannot be extended -  G^/G^  so  to  G 5  c o n t a i n s a copy of .  C  O  Z(p )  t h e r e are  such t h a t :  n  6 G  '•  4  r o o t of u n i t y and  t b  - a  n  a  ,) = 0(mod G,). n-i 4  i s the ( p ) ^  r o o t of  n  n  *  Again l e t  *  by  G  be generated •  n  *  G. U {a ,...,a }. and l e t H be a (G ,Z)-group c o n s t r u c t e d 4 0 n-1 n n . i n d u c t i v e l y as b e f o r e u s i n g the c o n s t a n t s M . We w i l l a g a i n show n t h a t by proper c h o i c e of the  G  * n +  l  a n  and  d  n  n  n  n  G  .  g £ R n  has no e x t e n s i o n to  As b e f o r e l e t  *  g (a_) = (a ,c ) n n  Then-:  ,  0  =: p  ( a  o' o c  = ( o> o b  k  n  that  -  = pg^(a )  so  ' » s  n  ="g*(b ) = ( b , k ) - -  =  And  M  thus no e x t e n s i o n to  g (b )  p c  = pc -+ M Q  Q  )  V  +  or  k  Q  = M (mod p ) . Q  Also: g (b n  n  ) = (b ,k ) n n = g\b ) n  n  4  = b  n-1  M  cannot be extended to  n  holds.  for  b €G  - a  EM  Then from the s t r u c t u r e of  n  is a  n  to  (II)  /  00  Z(p )  n  and so  cannot be extended  Now  as chosen i n  n  By the c a l c u l a t i o n on the p r e v i o u s page we have t h a t  t h i s i s impossible,g  M  •  by ^  definition b £G n  4  29  V  =  p  n " Vl>  a  b  y  ( b )  ft ft ft  n-1-j) P8n ( n ) ~ Sn (a  =  as  a  gn  = p ( a ,c ) - (a ,c ) n n n-1 n-1  by d e f i n i t i o n .  = (b + a ,pc. .+ M ) - (a . ,c .) n n-1. n n n-1' n-1  "+" i n H  = (b.,pc + M . - c ) n n n. n-1  "+" i n that  And  so  k = pc + M - c , n n n n-1  Thus we have:  or  cannot be extended F i n a l l y use  g  2  that  I.  n g  As Thus  g  2  to  k's  Then  ^  * n +  ^  + 1  and f o r (2) choose  M  ft exist.  g  2  to extend g~:G  2  l t :  does n o t e x i s t . .  2  H,.  ->- H  extends  cannot be extended to  of t h e lemma.  H n  n  Thus  g^ ,  and so  to a  (G^,Z)-group,  = k + c , + n n-1 n g ,  .  Lemma (6)  g^ .  extends  ft  a , n-1  were chosen a r b i t r a r i l y we c a n do t h e n  be any. homomorphism ,  extends  fact  k + c , = M (mod p ) . n n-1 n  M,. = k D O  F o r (1) choose  C l e a r l y I n b o t h cases no such  g„  using  n+1  k + c = M (mod p) n n-1 n  Keeping i n mind the M^'s  Let  n+1  (1) k^ = M^mod p) (2)  following.  i s a homomorphism  some  * g ,, n+1  such that g 6 R. n  such that  h g  2 2  — 1'  G2  and  We have j u s t  * ft h •, ,e n+1. n+1  H  2  shown  = 1 * +1 G  n  n g cannot be extended to  follows that Therefore  2  say  §2 '  s a t i s f i e s the requirements  30  V = L  In  this  s e c t i o n we  groups s a t i s f y i n g  If  W-groups  w i l l show that under the  Possibility  Lemma ( 8 ) :  and  G  I  or  satisfies  assumption  Possibility  Possibility  I  V =  II  are. not  or  II  L,  W-groups.  then  G  can 6  be and  named so  that for  a f i n i t e subset  Proof: Case ( i ) :  any  limit ordinal  6,  L. o  such that  ,L ,  There are Let  that f o r any  G  II(a  two  cases to  satisfy  Possibility 6 < ui'-i  G  ,  i s the  a  '  holds.  0+0)  0  I.  G  Thus  by  and  (B)  i n the  :ct < a}  , I  4 n(ct)}  G  i s named such  i s a pure subgroup and o  g,  p  element  prove.  some l i m i t • o r d i n a l (A)  G.,G_, )  r  0  limit ordinal  conditions  t h e r e e x i s t s an  particular  pure subgroup  definition  of  for  required  Possibility  I..  That i s : (A)  {a  (B) Let  Jo  + G  (a  $ = {6:6  conditions  p  , . ,L',GJG)  holds for a l l  is a limit ordinal, (A)  and  i s an  (B)}.  6 <  That i s  independent f a m i l y i n  a < wi  where  ,  G--  and  t h e r e do  not  L  =  {a„:£  does not  exist  G/G  p  < n(ct)}.  satisfy  suitable  a  £' > s  a < cu 1 , such t h a t G^ could replace G^ i n the above. We c l a i m that $ i s bounded i n oo . If. $ i s unbounded then c a r d ( $ ) = 10j , n  say  $ = {6  that. so  {6:6  ct  :a < aii } 1  <e'<6 ,.} ct+1  err  :  that- under the  new  where  6 < 6 • cti ' 012 •  becomes  {9:CJ  ordering,  a if  <  0  —  6  if  a-i < a?  < oj(ct+l)}.  .  Rename  G  T h i s renames  i s a limit ordinal,  then  so G  G  R  0  cannot s a t i s f y  conditions  Possibility  under any  of  I  Possibility  I,  this  (A)  and  (B).  As  G  must  naming that i s a d m i s s i b l e , is a contradiction.  Thus  by $  satisfy the  definition  i s bounded,  ..  by  3i  say G  p < oi.  where  . becomes  G co  p.  p  and  i s a limit ordinal.  {8:p+toa <_ 9 < p+co(a+l)} . becomes —  Under this admissable naming i f s a t i s f y conditions  (A) and  Now consider  G  . As  a  G  Now rename  and L  such that  G  so that  G  (A) and  (B),  Choose a  Suppose  n(a,L,G ,G , ) holds.  {9:co  6  6  (A) and  <_ 6 < co+co} }  for  Now we do the  CO+CO  II(a ,L„,G  o  II(a,L,G ,G) CO  has been named so that for a l l  and L. such that  s a t i s f i e s conditions  . Thus  {9:co+co+coa 4 9 < co+co+co (a+1)  becomes  G  r  0  becomes  <^ 9 < 6}  CO  6  will  a < oy_ .  *  {9:<o  x  a  G^  II(a,L,G ,G) holds.  (0  a < coi . Under this naming  there exists  a ' s,  <G U L U {a}> £ G  so that  =  induction step.  {8:co+coa <_ 6 < co+co(a+l)}.  co  {9:6+coa. < 9 < 6+co(a+l)}  and  so that  i s a l i m i t ordinal then  s a t i s f i e s conditions  x  holds.  G  CO  6 < coi  l i m i t ordinal  6  (B) for some  CO  then there exists  Now rename  .„,G  coo  6 <g  „, ) holds.:  Since  coo+co  (B), then there exists  a  and L  cop  such that  II(a,L,G ,G)  holds.  n  p < coi  As before choose a l i m i t ordinal  cop  such that  <G  L  {a}>. G and so (a.L.G „,G ) holds. Rename * p cog p remains unchanged, { 9 : cog ^ 9 < p} becomes {9:cog <_ 8 < cog+co},  cog  G  so that  G cog  '  ,  {9:p+coa <; 9 < p+to(a+l)}  and  Thus  n(a,L,G „,G , )  {8 : to g+co+coa ^ 8 < cog+co+co (a+1)  }.  holds.  n  cog  — becomes  cog+co 6  Thus we can assume  G  can be named so that  II (a ,L G.,G r)  6  o  for any l i m i t ordinal  )  holds  o+co  6  6 < coi  and suitable  a 's and  L.'s. o •  Case ( i i ) : same.  Let G  Let A  satisfy  Possibility II.  The proof i s almost the  be the required stationary set 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 I . Let :  A = {6 : a < coi }, and of course the 6 's are a a 6 < 6 i f a i < a ? . From condition (B) in 1  l i m i t ordinals with  a \  a2  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 ,  there exists  a  and L  such that  •n(a-,L,G. ,G) 6  holds.  Choose  6„ 6  0  Rename  G  so t h a t  {0:to 4 6<(o+u)}_, for  G._ 6  a l l a < oil .  chosen from the  Then  and so  co^  {a:X fi a = S } ct  { 0 : 6 n < 9 < 6„} — B  becomes.  u  co+co  )  becomes  {0:to+co+toa < 6 < oo+co+oo (a+l) }  h o l d s . . T h e i n d u c t i o n step i s o r d i n a l s w i l l always be  A.  An i n f i n i t e K  cardinal  K  i s cofinal i n  K.  (18): L e t K.  V = L  K  be an i n f i n i t e  i s r e g u l a r i f no s e t  coi ,  c a r d i n a l number and l e t  IE there i s a sequence  a,  and f o r each subset  i s stationary i n  From Jensen  tr  and  i s regular.  i s a subset of  O (A) K  ,  Remark ( 3 ) , no c o u n t a b l e s e t i s c o f i n a l i n  be a subset of  "Assume  co  ( i ) e x c e p t . t h a t the l i m i t  (17) :  in  Definition  S a  CO  c a r d i n a l i t y l e s s than As. noted  G  II(a,L,G ,G  6 Vs .of a  Definition  A  becomes  t  0  and. {0:6' +uia < 0 < 6^+co(a+l)}  the same a s . case  of  •  0  L U {a}>, £ G . » 6g  <G„ u 6.  so t h a t  (6) page 293> and l e t K  K,  X  ,  a  A,  of  K  the s e t  <X (A) K  then we say -  7  be a r e g u l a r i n f i n i t e A  of  cardinal.-  K  i s the r e g u l a r c a r d i n a l  Now c o n s i d e r a group  G  of c a r d i n a l i t y  stationary set .of•limit ordinals. any s t a t i o n a r y s e t to i t s l i m i t  co . 1  Then  K."  So t h i s r e s u l t h o l d s when  A's.  holds.  we get the f o l l o w i n g r e s u l t :  h o l d s f o r every s t a t i o n a r y subset  are many such  such that  co .  Let  1  A  be a  As shown b e f o r e the r e s t r i c t i o n of  o r d i n a l s i s again s t a t i o n a t y ,  C o n s i d e r the s e t  so t h e r e  G x Z = {(a,z):a £ G,z€Z}.  33  Name the elements of .(a)  Name  G x Z  Suppose  i s countable. ( 3 : 3 < co6},  { ( a , z ) : a < co6} has been named as  then, name t h e elements of  J , , = {(a,z):to6 <_ a < co(5+l)} o+l — which i s e a s i l y done as J is. o+l  {3:w6 < 3 < co(6+1) } —  as  ( 3 : 3 < co} which i s e a s i l y  = { ( a , z ) : a < co} as  done as (b)  as f o l l o w s :  r  r  i  l  countable. So by t h i s i n d u c t i v e naming p r o c e s s the elements of {a:a < coi }  ordinals {a:a < 6}  G • x Z has been named as the o r d i n a l s o f o r a l l l i m i t o r d i n a l s l e s s than co_ . L e t H be such a  H = co. = {ot:a < co.}. A.  that  a r e the  and each  naming of the elements of  and  G x Z  Thus  Assume  0 (A) rl  V = L  holds.  T1  g(a) = (a,z^)  G x Z, and so the s e t A and a p p l y  L e t g:G  where the  z a  '  Jensen's r e s u l t to H  G x Z  be a f u n c t i o n  a r e i n Z.  s  i s stationary, i n  Then  g  such  can be  say L = {(a,z ) :a < COT and z e Z } , a a can be viewed as a.subset of H, say Y, where 66 Y i f  viewed as a s e t of ordered p a i r s , and so  g  and o n l y i f 6  i s t h e name i n H  of some  (a,z ) i n L. By Jensen's a r e s u l t t h e r e e x i s t S , f o r a£ A, such that S S a and f o r any a a • X £. H, the s e t {a:Xfio r - S_^} i s s t a t i o n a r y . In p a r t i c u l a r ,  ft A :  = {a:Yfta  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 a a and Y = { 3 £ H : 3 = ( 6 , z ) f o r &€G }, then f o r a e A\ Y 0 a = o a ( 3 6 H: 3 = (6,z.) f o r i e G } . Thus Y fi a = S = g L can be viewed o a a G r  1  Z S = a as  a  a function arbitrary,  S :G ->• G x Z. . a a. ct  Let S  then f o r aiy f u n c t i o n  a  = g  a  '*  fora l l a£ A .  g:G ->- G. x Z  where  As  g  was  g(a). has shape  34  (ct,z),the s e t  '{<5 < u>i:g  f u n c t i o n so i s  g.:G. ;0 o i s a function.from G  Lr  = g 6 x Z  o into  r  6  '= S } 0  i s s t a t i o n a r y and s i n c e  0  = g„ o o Thus we can make the f o l l o w i n g  G. x Z}. 6'  . " I f V = L,  there  are functions  G a  r  i s stationary.  = g } 6  Possibility  Assume  II,  Suppose  G  G  Then i f  i s not a  satisfies  where  G  and  L. 6  Possibility  I or I I . • By  holds.  Let  s t a t i o n a r y s e t c o n s i s t i n g of a l l l i m i t o r d i n a l s . since  V = L,  so f o r Let  K  5^A  we can assume  ft ft ,  g  o  (J)  as above.  i s a f u n c t i o n from  Lemma (8)  G.  e x i s t a map  g:G -> H .. oil  I f we can c o n s t r u c t  such a  such t h a t H  ,  do the c o n s t r u c t i o n by t r a n s f i n i t e  then  exists  A. be the  into  H  and  r  6  and  g(a) = (a,z)  • * * • ' •  K={g.:5£A }. o (G,Z)-group, H , such that u  not  there  L e t . H. = G ' x Z, 6 6  be the s e t of these f u n c t i o n s , a  l  hg = 1„ G G  .  and as u s u a l  cannot be a  induction.  t h e r e does  .  Define  a  h(a,z) = a. '  W-group.  , arbitrarily. Suppose we have d e f i n e d a oi • f o r a l l a < 5, such that H extends H „ oia 0)3  We  (G ,Z)-group, 0)  H  I  Thus by Jensen,  o  We w i l l now c o n s t r u c t  Possibility  W-grbup.  6 II(a ,L.,G.,G.., ) 6 6 6+0)  such that  g(a) = ( a , z ) ,  satisfies  can be named so t h a t f o r any l i m i t o r d i n a l ' 6 < coj , 6  A,  6  V = L..  then  6 £ A'°" £  r  the s e t  Theorem. (1) :  Proof:  g :G -> G x Z, 6 6  g:G -> G x Z  G  or  •  such t h a t f o r any f u n c t i o n {5 < toj_:g |  is a  ft* So l e t A = {6€A:S.  a function.  statement: (J)  g  (G ,Z)-group, H o>a • oia f o r a l l 3 < a.  35  Define  H' coo  as f o l l o w s :  Case ( i ) : H  ,„ co(6-i)  6  i s not a l i m i t o r d i n a l ,  i s well defined.  1 N  with  Suppose  so we can assume  As b e f o r e l e t H = G x Z  and  h:H -> G  h(a,z) = a. (a)  If co(6-l)^A  function  o r i f .to (6-1) 6 A  g in K co (6-1) / r  i s not a homomorphism  1 N  , . = 1 , then extend co(6-l; G co ( 6 - 1 ; a r b i t r a r i l y u s i n g Lemma (6)..  hg  H co(6-l)  1 N  (b)  I f co(6-l)£A  , g , t h e n  .  in K  .' co(6-l) g  satisfy  •k-k  '  co(6 1) " '  a  ,  ufi  w  L  co(6-l)  ' <o(6-l) G  n  d  G  (  6  -  1  }  co6  Lemma ( 7 ) . That i s :  and  cor-free  i s a ( G ^ ^ ,Z)-group.  i K a ^ " - * ,L ,G ..,G J to(6-l) co(6-l) co6 6  '  a  i s a c o u n t a b l e subgroup o f the  1 X  (ii)  to H co6  i s a homomorphism such t h a t hg . „ ,. = 1_  G .„ co(6-l)  group" 6  such t h a t  and the c o r r e s p o n d i n g f u n c t i o n  the c o n d i t i o n s of  (i)  and the c o r r e s p o n d i n g  1  1 N  h o l d s by  Lemma. (8)  as s t a t e d i n the b e g i n n i n g of the proof.. (iii)  g : G , ..,.->- H . i s a homomorphism co (6-1) to(6-l)co(6-l) such t h a t hg ••. . = 1 • (o(6-l) So we can apply Lemma (7) to extend- H .. '. t o H „ co(6-l) co6 s  t 0 ( < 5 _ 1 )  so t h a t  § (,5__)  cannot  u  be extended  :G  :  a < 6.  S  U  C  h  t  h  a  t  ^6  to a homomorphism  G , • co6 Case ( i i ) : Suppose 6 i s a limit ordinal. Define H .= I JH , — coo , toa ' a<6 and as b e f o r e H .. i s a (G .,Z)-group extending H for a l l co6 co6 • coa g  co6 co6 * \&  G  H  =  X  Let Suppose  H  be the  g:G -> H  (G,Z)-group  constructed  by t h i s  i s a homomorphism such t h a t ui\  A  = {6:g  .  u .  '  .  - g.}  i s stationary  0  6  non empty, g„  say  6  is in .A .  cannot be extended hg., = 1^ o+o) G.,  .  6+0)  ft  .  hg = 1^, ,  and so  Thus no such  g  g  g. o  g„, :G. 0+OJ ,  .  A  G  i s not a  g. ,  ft  •  is H  0)1 ->• H., ; such  0+0)  cannot be extended  i s not an e x t e n s i o n of  e x i s t s and so  .  Then •  Thus by the c o n s t r u c t i o n of  to a homomorphism Thus  hg = 1_ . G .  by ( J ) . In p a r t i c u l a r  0  that  induction.  ,  0+0)  to  g  such t h a t  contradiction.  W-group.  37  Martin ; Axiom  and W-groups  In this section we w i l l show that under the assumption Martin Axiom and is a  2  > oi] ,. any group s a t i s f y i n g  W  Possibility I I  W-group.  D e f i n i t i o n (19) : .Let P and l e t  a,b£P.  We say a  be a poset ( p a r t i a l l y ordered s e t ) ,  and b  no common upper bound i n the poset  D e f i n i t i o n (20): We say that is a  of the  b  D  in D  Let P  are contradictory i f they. have, P.  be a poset and l e t  D  i s a dense subset of P' i f for any such that  D e f i n i t i o n (21) :  be a subset of P. a  in P  there-  a < b.  Let A  be a cardinal number..  Let MA  be  . A  the following assertion: "Let  P  be any poset of c a r d i n a l i t y  A.  Suppose i n P  there i s  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  two.members of B Such a set the  B  BHD  ^ <j>. for a l l a < A,  and such that any.  have a common upper bound i n B."  i s called a generic subset of P (with respect to  D ' s ) . MA (Martin Axiom)  says that  MA,  holds for any  A < 2 . U  38  Theorem (2) : has  cardinality  then  G  Proof: the  u)_ ,  direct  h:H -> G  G  G  satisfies H  satisfies  Possibility III.  i s free.  Thus  As . G  G  subgroups. •I belong to  P,  be a group whose s e t o f elements i s  P  a r e homomorphisms of. G write,  into  H  g  is  P  hg = 1 extends-  OJ_ . d i f f e r e n t co_  these pure subgroups  I  finitely  co_  into If  finite  H  g:I ->- H  many c h o i c e s f o r the image of each b ''s,-'  .  P.  The  If  g^ .  g^  &2  and  We w i l l  now  generated pure G  subgroups  i s t h e i r union,  generated pure generated pure s u b s e t s of  G.  there  subgroups. subgroups Now each s e t as  I , with generators  G  b , ... ,b  a r e u n i q u e l y determined by the  b_^ ,g(b )£{ (b ,z) : z £ Z) .  f i n i t e l y many  and l e t  generated pure  i s f r e e l y generated by a f i n i t e  images of the g e n e r a t o r s . generator  finitely  F o r a g i v e n pure subgroup  the homomorphisms of  G x Z,  i s u)_ .  i s c o u n t a b l e and  s i n c e there a r e o n l y  oj_-free.  Gis  So we can assume  from f i n i t e l y  i f g^  of  There cannot be more than  of  Lemma (5)  Now we d e f i n e a poset  such that  g^ <. g^  As each subgroup  must be a t l e a s t  G  P o s s i b i l i t y I.  i s u ^ - f r e e then each summand  F i r s t we compute the number of f i n i t e l y  of  If G  PossibilityII.  show t h a t t h e c a r d i n a l i t y  of. . G.  By  i s a W-group.  be d e f i n e d by. h(a,b.) = a.  elements of  >. t6_ .  W  and does not s a t i s f y  sum of c o u n t a b l e groups.  and so  Let  i s oj.-free,  2  W-group.  Suppose-  free, G  isa  Assume the M a r t i n Axiom and  i s i n P,  then f o r each  Thus there a r e o n l y c o u n t a b l y b_^ ,  and s i n c e t h e r e a r e o n l y  t h e r e a r e o n l y c o u n t a b l y many d i s t i n c t mappings  ,  of the  b_/s.  Thus t h e r e a r e o n l y c o u n t a b l y many  g's  f o r each  S i n c e the number o f f i n i t e l y generated pure subgroups i s the c a r d i n a l i t y of  P  is  then  ooi .  We now d e f i n e s u b s e t s = {g£.P:a  uii. ,  I.  D  of  P  for  a < •u>.y  i s i n the domain o f  as f o l l o w s :  g}. '  •.  We now show: (a)  Each  , (b)  D  a < u>i ,  There do not e x i s t  Proof of (a) : Then  ,  a  I  OJ  be i n  i s pure and f r e e l y generated by say  (domain o f  g^) ,  So suppose now  then  g  in  such that  g^eD^  a<^Dom g^  and  g  T  4 g  and c o n s i d e r  i s a f r e e group generated by  P.  p a i r w i s e c o n t r a d i c t o r y elements of  n  L e t a< ui^ and l e t g  show t h a t t h e r e e x i s t s  I  i s dense i n  P  where  a^,...,a g' <_ g.  T  I  ,  P  . If  We must a"6Dom g^  so we can l e t  = <a^,. . ..,a a> n>  a^,...,a^,a^ ^  A  then d e f i n e  +  g = g. . If g  as  follows: (i)  Let  g ( a ) = g (a )  (ii)  Let  g(a  »  l>--«»n+l,  As the  a  £  1  ±  =  for  ±  n + 1  ) = (a  ^z.)  i =  l,...,n.  f o r any  z  Z.  generate the f r e e group  I ,  then t h i s , it  mapping of the g e n e r a t o r s can be extended to a homomorphism  g:I  -> H.  is  Clearly  g  extends  g^. and  a £ Dom g = I •.  Thus  g£D^  and  ^ g.  •k I t remains to show that f o r some  a n  +  ^  •  Since  I  i s f r e e l y generated'by  a ^ I , - then  a  a,,...,a ,a 1 n n+1 i s independent of {a^,....,a }  i<  since  I  i s pure. *  If  •  x £ I "^1,  So  I  contains at l e a s t  n+1  independent  elements  '  then  mx £ <a. ,-..'.. ,a ,a>  i  n  f o r some  m,  and so  x  is  40  no t independent  of  {a-^  a ,a} . •  independent  elements,  and so  Thus  *  I  contains exactly  .  I  i s a f r e e group  Let I be f r e e l y generated by ^_'""*' 4._ * any f i n i t e number of elements i n <a,,...,a ,a>, b  A  on  I  elements Let  B  is in  and so I  elements  1 1  of  J  some  n(6)  W  1  •> ,  n+1  = <b,,...,b >, 1 n+1  pairwise contradictory  derive a contradiction. 6 6 a.,...,a ,.... 1 n(.o;  can r e p l a c e  where  W = (g^ 5 < w _ } :<  i s f i n i t e and card(W) = co_ ,  n(6)  must occur n,  i(  of the same c a r d i n a l i t y w i t h o u t l o s s of  As each  some f i x e d  <a^ , . . . ,a^,a>  B — <a, , . . . , a  CJ_  be f r e e l y generated by We  I  page 2 4 ) .  i s proved. .  r  g. o  in  n  s e t of  is infinite,  a , , , n+i  Since  and the r e s u l t  i s a f i n i t e positive integer.  by any subset of  for  Then f o r some  {g :<5 < c o i } . We w i l l • o .  say  1  n+  Suppose t h e r e e x i s t s a s e t of P,  generality.  G../G, 5 4  generators.  the f i n i t e  {b^ , ... . ,b ^}  1  = <a,,,..,a , > . 1 n+1  the domain of  n(6)  (see  n  1  "  such that  {b ,...,b }f=<a ,...,a >. 1 n+1 1 n+1  P r o o f of (b) :  Let  <a ,...,a ,a >. 1 n n+1  <a •, . . . ,a ,a> . 1 n  n  a ., , n+1  be the s e t of elements of  and not i n  then  say  n+1  shown b e f o r e , f o r and not i n <a,,...,a ,a  s  n  t h e r e e x i s t s an element,  n+1  co_  times.  f o r a l l the  So we  g 's. o  can assume  then  n(5) = n,  That i s w i t h o u t l o s s of  6 (5 g e n e r a l i t y the domain of g i s generated by ,{a ,...,a } f o r every • o 1 n g. i n W. L e t K = {a,,...,a } be a maximal s e t of elements o f ' , o 1 m r  G  which co_  for of  the  else  f r e e l y generate a pure subgroup 6's. Dom  K  1  Note that  K  and  can be empty.  {a  , . . . ,a^}  S.Dom g^  For i f any uncountable f a m i l y  (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 i s non empty. So a g a i n w i t h o u t l o s s of g e n e r a l i t y we can  41  assume For  a,,...,a g.Dom g. t o r a l l 1 m 6  any  6  •  set  we can extend  Thus we can assume  {a.,...,a ,a ,...,a } 1 m. m+1 n Now c o n s i d e r  1  {a,,...,a } 1 m  .  f o r Dom g. . o  6 < u>i .  for,each  Dom g. . o  As any homomorphism  g . from  Dom g o in  a, = a, 1 1  a  H.  .  .  i s generated by  W. n elements.  i s u n i q u e l y determined by where  g(a) = (a,z)  f o r any generator  t h e r e can be o n l y c o u n t a b l y many d i f f e r e n t homomorphisms from into  = a m m  n-element g e n e r a t i n g  i s f r e e l y generated by  Dom g. 6  the g e n e r a t o r s a r e mapped and  to an  g, o  Dom g. o  Let  a, Dom • g  So i f t h e r e were o n l y c o u n t a b l y many d i f f e r e n t domains of.  the  g 's, t h e r e would o n l y be c o u n t a b l y many g ' s , a c o n t r a d i c t i o n . 6 o Thus t h e r e a r e ix>i d i f f e r e n t domains on which the g g ' defined. s  Choose one  g'  on each domain.  can assume  Dom g  a  r  e  So w i t h o u t l o s s of g e n e r a l i t y we  •°  £ Dom g f o r S ^ a. I n other words, i n the o a , o 6 ' ' l a , , .'. . ,a ,a , . . . ,a }, m < n . 1 m n+1 n A g a i n w i t h o u t l o s s of g e n e r a l i t y we w i l l take a subset of the g ' s  set  P  o of  cardinality  ;  t h i s -time such that the s e t  {{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 s e t i n G. 1 m m+1 n Assume f o r a < 3 < coi  we have chosen  1  { f a ^ , . . . ,a^} U {a":m < I i . n, ct <3}} Now c o n s i d e r  < U Dom g ' >', . ct<3 a x  From the remaining  any g „ such t h a t (< ^ „ D o m g / 6 a<3 . a * We c l a i m t h a t o n l y c o u n t a b l y many 6  f o l l o w from the f a c t I f some element  that  i n D,  g 's such that the s e t a i s an independent s e t i n G.  <a, ,...,a >) fi Dom g^ i s non empty. 6 J 1 m 5 g^'s w i l l be t o s s e d out. T h i s w i l l  < U Dom g >.  ot<3 say b.  g„'s, t o s s out o  <a,  a > = D  a * 1 m was i n uncountably many  i s countable. Dom g ' s  42  then as  b  i s n o t an element  of  <a,  a >, i t i s independent of m and so we would get a c o n t r a d i c t i o n 1  the to  pure subgroup  <a , ...,a >,'  the m a x i m a l i t y of the pure subgroup  <a^, . . . ,a j b ^ would in  of the  g ' s.  t o s s out o n l y c o u n t a b l y many out  &^  That i s  r  be a pure subgroup  coi  the domain of  <a ,...,a > . 1 m of order  m + 1  contained  So f o r any element Since  Sm  D  b£D,  we  i s c o u n t a b l e we t o s s  o n l y c o u n t a b l y many  g 's •o  g.'s and so t h e r e a r e uncountably many o Choose one and c a l l i t g . We c l a i m that the elements 3  left.  {{a^,...,a  } U {a":m < £  i s e a s i l y seen.  a r e independent i n G.  This  We need o n l y check t h a t no l i n e a r combination of the  3  elements of  a <. 3.}}  n,  {a m+1  a } n  3  '  i s a l i n e a r combination o f the elements of a < 3}}-  •{•{a,,...,a } U (a :m < £ <_ n, 1 m x. —  Any l i n e a r combination o f the •  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  <  of<g  i<3  D o i n  o.*  D 0 m  S  ^a**  i  S  i s independent.  e m  '  B  P^y  a  y  n  c  d  h  s  o  i  o  c  e  t  h  o  e  Sg '  f  s  e  So i n d u c t i v e l y  t  (Dom'g  ^ i'"**' a  a  we can choose  1  ^ ^ ^ £  V = {{a. ,. . . ,a } U {a„ :m < £ <_ n, i m £ — independent s e t i n G.  the  Possibility I  elements of  conditions So assume  (A) G  G  fails,  (B)  = '  1  n  •  a  g 's i n o i s an  1  A l s o make sure  G. o  so t h a t  Possibility I G  contains  co  T h i s i s e a s i l y done. independent over  Then t h e r e i s an uncountable subset of  G/G  . CO  <  t h e r e i s an a d m i s s i b l e naming of  of the d e f i n i t i o n of  has such a naming.  m K  6 < coi }}•  such t h a t t h e r e does not e x i s t  and  :  coi o f the  such a way t h a t  Since  m  a  m  f)  -^<a ,...,a >)  I f an element  g£G CO  i s a finite  hold. a.,...,a 1 m  V linear  A3  combination o f elements of  of  6  g,  V,  then f o r any  6  t o s s out .a •.m+1  a . n  can be r e p r e s e n t e d as a l i n e a r  Since  G  co  a^  i n the r e p r e s e n t a t i o n  i s c o u n t a b l e and i f g £ G  co  combination of elements  of  V,  then  t h a t r e p r e s e n t a t i o n i s unique,'  o n l y c o u n t a b l y many s u b s e t s . (  w i l l be t o s s e d out o f  V  Clearly V  ft  ,  V  say  ft  V.  Let  ft  i s uncountable.  ft  J  = {'  5  a  +  1  be the remaining elements  Now l e t  = <S < ooi).  6  6  a  >••• > ^ a  m + 1  of  n  V.  5  J  be the s e t of a l l a ,,'s i n m+i  Then:  ft (A) '(B) Since  i s an independent  So  family i n  G/G  •  co  <G >, = G  co*.co  Possibility I  such t h a t 6's.  J  <G^U  { a  5 m + 1  fails, ^ft  =  G  9  < { b  w  1  6  ^ ft >  f  o  r  a  1  1  •  6  then t h e r e must e x i s t  b, 1  in  <G  6  U {a ,,'}>* m+1 *  co  b  ut c o u n t a b l y many {a _.}>. = <G. U ( b f } > = G & <{b?}>. f o r {b?:6 < O H } . co 1 ' co. , 1 1 6-,m+1. . * , co , , <* * .be . 6 L_'= lb.}, fro r 6 < coi and J the s e t of a l l a , „'s 6 1 nm+2 <G  U  6  J  +  x  Now l e t  1  c o r r e s p o n d i n g to the J  •(B)  <G  i s an independent .  0/ L->. = G  6*,  co Again since 5t  b . ' s . So 1  ft ft  (A)  t^oKv m+2  <G  co  many  that  <G  6's.  co  w  f o r 6 < cox . 1  6  6  6  b„ 2  in  <G  \J L . U 6  co  U fa?- _:}>"•= G 9 ,<L. U {b!?}>, f o r m+2 « co ' 6 2 * .  6  6 * 6  M l  J  6,say  a,  and l e t  ,a }>. = <G U Dom g >, c . n * co ct < a  of  c™, . . ., c" , in 1 k(a) N  G/G  a , ' s . C o n t i n u i n g t h i s f i n i t e process we m+Z  Choose such a  elements  family i n  f a i l s t h e r e must e x i s t  U L  Then:  x  (J {a . . . , a }>, = G © <b,,...,b >, f o r uncountably m+1 n « co 1 n-m * -  <G \J {a™ . co m+1 are  6  = {a,„:6<coi}. m+2  6*  Possibility I  such that  J  © <L >  co  uncountably many.of the get  1  < G  a a ;a I ' m ' m+1' .  . co .  A l s o note that  a ,c ' n 1  T h i s i s because  a ( l ) = a.  b? £ <G l  c. , .>, k(ct) *  Note t h a t b" . . . ,b 1 n-m a  f o r some  U {bf, ...,b? , } U {a , . }> , co 1 l - l m+i * 0-  44  Now we can r e p e a t t h i s p r o c e s s f o r any  G wa  and choose  a(a)  d i f f e r e n t each time s i n c e we a r e c h o o s i n g from an uncountable s e t of which  o n l y c o u n t a b l y many have been chosen b e f o r e .  c o u l d choose  o(a) > a(g)  fora l l  c o f i n a l i n - o u r uncountable s e t . sequence  g < a  s i n c e the  a(g)'s  a r e not  Thus we can d e f i n e a s t r i c t l y  a(a), a < u i ,  of o r d i n a l s  I n f a c t we  such t h a t  <G  increasing  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) t h a t by choosing k ( a ) we can assume t h a t l a , , . . . , a ,a ,...,a . • », 1 m m+1 n a(a) a(a) minimally c, ,...,c, - } i s an independent s e t i n G. To s i m p l i f y n o t a t i o n 1 k(a) A  v  a(a) = a. .  assume  At t h i s p o i n t we must make the o b s e r v a t i o n t h a t ordinals  C = {6:6 < w_  subset of C  w6 = 6}  i s a c l o s e d and unbounded  OJ_ . .  i s unbounded:  w. ,  and  the s e t of  f o r any  3,  By  ( 7 ) , page 108,  a r e i n C.  was bounded above by say a < 3 < co_ .  Then  w co  3  a,  ^ 3  That i s w a.< w_  ,  and so  w  o r d i n a l s of the form = w(w  W  ). 3  then choose  J '  Thus i f C such  that  3  > a.. As  w  is in C  we get a c o n t r a d i c t i o n . (b)  C  i s closed:  Let  Then f o r each element  3  be a c o u n t a b l e sequence  y  of the sequence  3  = 'tog V  .  V  Then: . a = l i m 3 = l i m wg = w(lim g ) = wa '. • • • v v v Thus  a  is in C  and so  C  i s closed.  of members of  Let  a =. l i m 3 V  C.  45  For every  a < co_ ,  k(a)  o c c u r s uncountably many times. ordering a.  i s . f i n i t e and so some Let  < a •  a  our s e t  JQ = C = { 6 : 6 < co_  unbounded and hence s t a t i o n a r y .  c  I  be t h i s s e t under the n a t u r a l  A = {a:k(a) = t} = { a : k ( d . ) = t , 6 < co-. , and ' o o i f and o n l y i f 6 i < 6o}. Now rename G so t h a t G . = G  Now l e t  6  k(a) =  of o r d i n a l s .  oi 02 Thus a t every o r d i n a l  then  A  k(a) , ' say  k(a) = t  < co6  6  fora l l  as. c € G . p too  6  {  1  c  c  co6 •= 6}..  and Since  in  .  i s c l o s e d and  k(a) = t  for a l l  a < co_ , (  Note a l s o that f o r 1 < i <. t ,  Now we can a p p l y  a r e s u l t of  J  2  a sequence of s t a t i o n a r y s e t s  . • coo • coa^ has t elements.  , . G k(oi) — coa  . . .• 2 J  Fodor to, d e f i n e  such t h a t f o r a l l  6  6€J^  ,  c  ^  =  C  £  f o r some f i x e d  Definition  (22):  s e t s of o r d i n a l s , and  f (0) = 0  f : J -> A ,  A function  ( 8 ) , page 141,  and  defined  regressive function  c.^ < 6  as  f(a) = 3  f  fora l l = C,  As noted  for  6 = co6. 6  by  f ( 6 ) = c^  there e x i s t s for a l l 6 i n  and  A are  for a l l  says that f o r a r e g u l a r  a  on  J  in  A,  there  a £J {0} V  J  •k  ,  and some f i x e d  a stationary set, 1 < t < t,  c  and  < 'u>6  cardinal  e x i s t s f o r each.  a s t a t i o n a t y subset  6 6  J  i f f.(a) < a  a s t a t i o n a r y subset of  Now c o n s i d e r cardinal.  where  i f 0 £ J.  A > co,  such t h a t  We proceed as f o l l o w s . .  i s called regressive  Fodbr's r e s u l t J  c^ .  to_ ,  and so f o r  J  k  3  of  J  i n A..  a regular 6£  Thus we can d e f i n e a r e g r e s s i v e f u n c t i o n  , r:  ^Q  ~*  u  6  since ,  c^ < 6  fora l l  6£  a s t a t i o n a r y subset o f . Now repeat t h i s f o r  . J,  By  Fodor's r e s u l t '  such that f ( 6 ) = c^ by d e f i n i n g f : J ^ ->- co_  i  46  by  6  f(6) =  < 6.  s t a t i o n a r y and  T h i s w i l l produce a  f (6) = c^  for a l l 6  —• ""^1 in  s  .  c  Ay  .  ,  6 €. J t 6 6 {c^,. . . > }'  That i s the s e t  c  6 c = c  ,  '  il/  J  ^  J  is  ^~  this ... — Jj-  <  f o r some f i x e d  for a l l I  XJ  {c^,...,c }  =  t  that  n  6 c^ = c^  Jl = 1,. . . , t ,  In p a r t i c u l a r f o r  c  By r e p e a t i n g  t times we can d e f i n e a n e s t of s t a t i o n a r y s e t s such t h a t f o r a l l 6 €  u  =  l,...,t.  for a l l 6 € J ^ .  t  We w i l l , now make one more o b s e r v a t i o n .  If  A  and  B  a r e pure  * subgroups of B  *  ,  G  and i f  pure subgroups of  in  G.  in  A.  ft ft ft and  a  b  i s in' B  in  A  and  ft ft nx = a .+ b  B  *  so  x £ A" © B" , A  Now  and  © B  A  consider  6  in  b £• B  A  B.  g  6  6 6 <b,,...,b , >, = B 1' . n-m * i s an element of  So  Dom  g  6  .{a^ , .. . >  B  and so  A  © B  i s pure then  m+t;  }  i s pure a  is  x = a + b  , A  A  We .  and  B  A  are pure,  G.  can extend  < g^  1  a  a  6 m+1  m  to t h i s domain.  By the above o b s e r v a t i o n  i s pure and c o n t a i n e d 1  Dom  g  6  .  Dom  g  6  6 ,a , c , , . . . , c >, n 1 t  in  t  Dom  By c o n s t r u c t i o n each i s contained  in  f r e e l y generates  C a l l these  <a^,...,a^,c , . . . ,c >.  B.  6  g  6  s i n c e each  6 a ,. i s an m+i ' 6 Thus Dom g = B.  6 i s f r e e l y generated by {a.,....a , a , 1 m+t m+t+1  a  A  ft ft ft + b where  1  element of  A  and  Then:  the pure subgroups .  nx = a  A © B  as  is, pure i n  is in  new homomorphisms  Let  As  nb = b -  a£A  . A  .  respectively,  A  = na +. nb ,  so  so  6 b.  and  b  ft ft na = a and  so  where  A  i s a pure subgroup then f o r  This i s e a s i l y v e r i f i e d .  f o r some  ©  A © B  a  5  , } n+t  <a^ , . . . , a^, c^ , . . . , c >^  where and.  47  {  Vt+l'-'Vt  f r e e l y generates • <bJ,...,b^_> .  }  m  A  6 g (a^-) £ { (a , z) : z €• Z}  f o r SL = 1 , ...,m+t , t h e r e a r e 6 only c o u n t a b l y many d i f f e r e n t images o f the g (a ) ' s , and ,so must Since  6 appear uncountably many times. SL = l,...,m+t  for of  of <  J . t Now choose  ct  g £G  .  top  a,3£J Then  ^,,,,.•.,a , > m+t+1 . n+t  3  Dom g  t  6  is in J  such that  ct  Dom g  a < 3  i s a pure subgroup o f G „  . As  3  a  $ - + t + l ' '' '' m+t <  a  a  > =  <  D  o  m  m  a  ^ ^  }  a  "U  D  o  n  d  TN  g " ^ ) = g U ) . for 3  3  8 ^  i  s  »•••'  A  / •  m  pure i n G.  to produce the c o n t r a d i c t i o n .  i s f r e e l y generated by  a 3 g:<Dom g U Dom g > -> H  a  ' ^+ +l''''' n+t' a  m + t  > m  Thus  ct 3 d e f i n e d by g ( a + b) = g (a) + g (b) where  .  „  '  .  r:  ,  oi  g " and g A 0  s e t assumed to be p a i r w i s e  So  o f some  i s a common e x t e n s i o n  g^and A  3  and g  r e s p e c t i v e l y i n our o r i g i n a l  a -  a  t  = l,... 4-t.  a t Dom g , bfcDom g i s a common e x t e n s i o n o f g ct 3 Now g and g a r e themselves e x t e n s i o n s o f some  g  Also  b >.• i s pure then by the o b s e r v a t i o n on the l a s t n  m+t+l'-'-' n+t  •  top. .  H  3  a  and so the g e n e r a t o r s 3  <Dom g™ U Dom g >  > -pi  i s fixed  , an uncountable subset  F i n a l l y we have the e x t e n s i o n needed  a  g (a )  i s pure and equal to . < b , b ^ > , ^ m+1 . n*  3  G . © <b top m+1 page  where  So we can assume t h a t  contradictory.  g . and g £ P s i n c e .0  3•  <Dom g [J Dom g > i s a f i n i t e l y generated pure subgroup. This c o n t r a d i c t s our o r i g i n a l assumption that g. and g . have no A o upper bound i n P.  Thus t h e r e doesn't e x i s t  p a i r w i s e c o n t r a d i c t o r y elements o f P.  any subset of  to_  48  We  can now  c o _ < .2^° ,  assumptions, subset  B-  complete the proof of  of  P  by  such t h a t  M a r t i n ' s Axiom = (j)  B fl D a  t h a t any two members of g  *  IJ g. B  =  Since  B  a f u n c t i o n from  G  is  hg  g  have.  . hg  Since ft ft = 1 .  .  B  Theorem- ( 2 ) .  f o r a l l a < coi , -  T  S i n c e each  i s the i d e n t i t y map  g6 B  hg  = 1  ,  and so  G  is. a  W-group.  such  B. . L e t g  *  is  i s a homomorphism,  Thus t h e r e e x i s t s a homomorphism  ft that  that  on the domain of  CJ  and  l  have a common upper bound i n  H.  our  there e x i s t s a (generic)  i s g e n e r i c i t i s easy to v e r i f y to  S i n c e under  g,  g :G •->- H  so  we such  49 The  Theorem is  (3) :  independent of  Independence  The statement:  Result  "Every  W-group of c a r d i n a l i t y  ZFC ( Z e r m e l o - F r a n k e l s e t theory p l u s  u> i s f r e e " n  the axiom of  choice).  Proof: V = L,  any  G = © G a<oii a G  Since  W-groups a r e o i i - f r e e ,  W-group must s a t i s f y  where each  i s free  and so  G a  G  Possibility III.  i s countable.  i s free.  Thus  As  oil i s f r e e .  ZFC  ZFC  By  i s consistent i f  M a r t i n and Solovay  if  ZFC  i s . But by  any  group s a t i s f y i n g  Theorem  ZFC  Theorem By  Lemma (5)  i s co-i-free,  But by  implies Godel  ZFC + MA + 2^° > coi  (2) ,  (1) i f  each  that (8) ,  i s consistent. i s consistent  i n the p r e s e n c e of  Possibility II  t h e r e a r e groups s a t i s f y i n g with  (10),  G  ZFC + V = L  every . W-group o f c a r d i n a l i t y + V = L  then by  isa  Possibility II.  W-group.  MA + 2^° >  By  ,  Lemma (4)  So i t i s c o n s i s t e n t  to assume t h a t t h e r e a r e W-groups of' c a r d i n a l i t y  co^  which a r e n o t f r e e . Thus the statement: is  independent o f ZFC.  "Every  W-group of c a r d i n a l i t y  u>i  i s free"  Bibliography  1.  L. Fuchs, I n f i n i t e A b e l i a n Groups, London,  2.  V o l . 1 , Academic P r e s s , N.Y.  and  1970.  L. Fuchs, I n f i n i t e A b e l i a n Groups,  V o l . 1 1 , Academic P r e s s , N.Y.  and  3.  P i n t e r , Set Theory, Addison-WesTey P u b l i s h i n g Company, London,  1971  4.  Jon Barwise, Back and F o r t h Thru I n f i n i t a r y  London,  ' 5.  1973.  Model Theory, M.A.A., 1973,pp. I . Kaplansky, Ann A r b o r ,  6.  Logic, i n Studies i n  5-34.  I n f i n i t e A b e l i a n Groups,  U n i v e r s i t y of M i c h i g a n P r e s s  1954.  R. B. Jensen, The F i n e S t r u c t u r e of the C o n s t r u c t i b l e H i e r a r c h y , Annals of Mathematical L o g i c (1972), V o l . 4 , pp.229-308.  7.  J . Monk, I n t r o d u c t i o n to Set Theory,. McGraw-Hill Book Company, Toronto,  8.  1969.  G. Fodor, E i n e Bemerdeeng zur T h e o r i e der R e g r e s s i v e n F u n k t i o n e n , A c t a . S c i . Math. (1956), V o l . 1 7 , pp.  9.  139-142.  K. Godel, The C o n s i s t e n c y of the Axiom of Choice and of the G e n e r a l i z e d Continuum-Hypothesis  w i t h the Axioms of Set .Theory,  P r i n c e t o n U n i v e r s i t y P r e s s , P r i n c e t o n N.J., 10. D. M. M a r t i n , R. M.  1940.  Solovay, I n t e r n a l Cohen E x t e n s i o n s , Annals  of Mathematical L o g i c (1970), V o l . 2 , 143-178. 11. S. Shelah, I n f i n i t e A b e l i a n Groups, Constructions, Israel  "Whitehead  J . Math.(1974),  Problem,  Vol.18, pp.  and Some  243-256.  51  Appendix  We w i l l d e s c r i b e the s i g n i f i c a n t a l t e r a t i o n s to Shelah's from  paper  (11).  the e x i s t e n c e of a  s a t i s f i e s conditions  The d e f i n i t i o n of G. ,  Possibility I  o  ,  (A) and  ( B ) , to the e x i s t e n c e of a  up to isomorphism,  possibilities. us to s i m p l i f y  T h i s a l t e r e d d e f i n i t i o n of some o f the p r o o f s .  had  I. and I I  (A) and  to be no way of  o)_-free groups i n t o the t h r e e Possibility  In p a r t i c u l a r  I  to d e a l w i t h  i n a u n i f o r m way v i a Theorem ( 1 ) .  used a c o m p l i c a t e d group t h e o r e t i c argument  allowed  we were a b l e to  come up w i t h a lemma (Lemma (8)) which allowed.us Possibility  which  G. , • o  which s a t i s f i e s c o n d i t i o n s  Under the o r i g i n a l d e f i n i t i o n t h e r e appeared  classifying,  was changed  under some a d m i s s i b l e o r d e r i n g ,  under every a d m i s s i b l e o r d e r i n g , (B).  t h a t we have made  both  Shelah  i n dealing with  P o s s i b i l i t y I . Thus our s e t t h e o r e t i c Lemma (8) e l i m i n a t e d the more difficult  group t h e o r e t i c  Shelah used  theorem  of Shelah's,  ( s e e (11) 3.3).  a r a t h e r c o m p l i c a t e d c o m b i n a t o r i a l argument,  (see (11) 3.1(2) and 3 . 1 ( 3 ) ) ,  to show every  i s a d i r e c t sum of c o u n t a b l e groups.  Our  Possibility  I I I group  Lemma (5) g i v e s a s i m p l e r  and more d i r e c t proof of t h i s f a c t w i t o u t u s i n g the c o m p l i c a t e d c o m b i n a t o r i a l technique of (11) 3.1(2). In the proof of Theorem ( 2 ) , t h e method i n d i c a t e d by, Shelah 6 6 f o r p r o d u c i n g t h e elements c ,...,c .. . appeared to be i n c o r r e c t . 1 k. (, o) So a c o m p l e t e l y d i f f e r e n t . a r g u e m e n t had to be used, see pages 41-44.  52  In g e n e r a l t h e s e t t h e o r e t i c and group t h e o r e t i c d e t a i l s were filled of  i n to the p o i n t where someone w i t h o n l y a l i m i t e d knowledge  s e t theory and group theory c o u l d read the t h e s i s .  This involved  much work i n p l a c e s f o r Shelah assumed a knowledge o f group at  a l e v e l of Fuch's books (1) and ( 2 ) .  o u t l i n e of an argument was g i v e n , filling  i n of d e t a i l .  theory  I n many cases o n l y the broad  and so t h e r e had to be a s i g n i f i c a n t i n Lemma (7) G^/G^  As an example,  has to  be shown to be i n f i n i t e and then i t has to be shown t h a t t h i s i m p l i e s /  either  CO  G^IG^  c o n t a i n s a copy of  many elements of prime o r d e r ,  Z(p )  or i t c o n t a i n s  infinitely  see pages 24-26. Another example was  working out a l l the d e t a i l s i n showing t h a t t h e examples i n Lemma (4) actually satisfy'the respective p o s s i b i l i t i e s , main d i f f i c u l t i e s w i t h t h e s e t t h e o r y , and in  than r e d e f i n i n g P o s s i b i l i t y I  the subsequent c l a s s i f i c a t i o n i n t o t h e t h r e e p o s s i b i l i t i e s , showing how the r e s u l t s o f Jensen  our problem. of  other  see pages 11-16. The.  detail,  So a g a i n here  (8) a p p l i e d . t o  t h e r e had t o be s u b s t a n t i a l f i l l i n g i n  see pages 32-34 and 44-46. Also" we had to show t h a t  c o u l d be w e l l - o r d e r e d such pure,  (6) and Fodor  was  that f o r a l l l i m i t o r d i n a l s  see Lemma (2) on page 5.  G  6, G„ i s o  

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