Rudiger Gobel and Saharon Shelah- Uniquely Transitive Torsion-free Abelian Groups

8/3/2019 Rudiger Gobel and Saharon Shelah- Uniquely Transitive Torsion-free Abelian Groups

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Uniquely Transitive Torsion-free

Abelian Groups

Rudiger Gobel and Saharon Shelah

Abstract

We will answer a question raised by Emmanuel Dror Farjoun concerning the

existence of torsion-free abelian groups G such that for any ordered pair of pure

elements there is a unique automorphism mapping the first element onto the

second one. We will show the existence of such a group of cardinality for any

successor cardinal = + with = 0.

1 Introduction

We will consider the set pG of all non-zero pure elements of a torsion-free abelian

group G. Recall that g G is pure if the equations xn = g for natural numbers n = 1have no solution x G. Clearly every element of the automorphism group Aut G ofG induces a permutation on the set pG and it is natural to consider groups where theaction of Aut G on pG is transitive: for any pair x, y pG there is an automorphism Aut G such that x = y. In this case we will say that G is transitive, for short G isa T-group. (Transitive groups are A-transitive groups in Dugas, Shelah [5]). This kindof consideration is well-known for abelian p-groups. It was stimulated by Kaplanskyand studied in many papers, see [14, 15, 1] for instance. If G is a free abelian groupwith two pure elements x and y, then there are two sets X and Y of free generators of

This work is supported by the project No. I-706-54.6/2001 of the German-Israeli Foundation for

Scientific Research & DevelopmentAMS subject classification:primary: 13C05, 13C10, 13C13, 20K15,20K25,20K30secondary: 03E05, 03E35Key words and phrases: automorphism groups of torsion-free abelian groups,

GbSh 650 in Shelahs list of publications

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G such that x X and y Y. We can chose a bijection : X Y with x = y which

extends to an automorphism Aut G. Hence free groups are T-groups and a similarargument holds for a wide class of abelian groups. There are T-groups like Z1/Z[1]

with Z[1] the set of all elements in Z1 of countable support, which are 1-free but notseparable, see Dugas, Hausen [4]. The existence of 1-free, indecomposable T-groupsin L for any regular, not weakly compact cardinality was also shown in Dugas, Shelah[5, Theorem (b), p. 192]. This was used to answer a problem in Hausen [11], see also[12] and [5, Theorem (a), p. 192].

Thus we may strengthening the action of Aut G on pG and say that G is a U-groupif any two automorphisms , Aut G with g = g for some g pG must be thesame = . Hence G is a UT-group if and only if G is both a T-group as well as a

U-group. Thus G is a UT-group if and only if Aut G is transitive and (every non-trivialautomorphism acts) fix point-free on pG. In connection with permutation groups suchaction is also called sharply transitive. Note that pG may be empty, if G is divisiblefor instance. In order to avoid trivial cases we also require that 0 = G = Z and G isof type 0, hence G is torsion-free and every element of G is a multiple of an elementin pG. If G is of type 0 and not finitely generated then |pG| = |G| is large and theproblem about the existence of UT-groups becomes really interesting. This question isrelated to problems in homotopy theory and was raised by Dror Farjoun. In responsewe want to show the following

Theorem 1.1 For any successor cardinal = + with = 0 there is an 1-freeabelian UT-group of cardinality .

We will also determine the endomorphism rings of these groups. They are isomorphicto integral group rings R = ZF of groups F freely generated (as a non-abelian group)by elements (with as in the theorem). Since endomorphisms of a group G willact on the right (in accordance with used results from [9, 10]),we will also view G asa right R-module (and as a left or right Z-module). Using classical results on grouprings it will follow that Aut G = F, where 1 Z is scalar multiplication by 1,hence F is a direct product of a group of order 2 and F. Moreover ZF has noidempotents except 0 and 1, hence G in the theorem is indecomposable and obviously

the center ofZF is just Z, hence Z is also the center of End G. Therefore Theorem1.1 strengthens the Theorem in Dugas, Shelah [5, p. 192] substituting T-groups byUT-groups and removing the restriction V = L to the constructible universe. Alsonote that it is straightforward to replace the ground ring Z by a p-reduced domain Sfor some prime p. Hausens [12] problem can be answered also in ordinary set theory.Further applications can be found in Section 5.

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First we would like to explain why constructing UT-groups is a hard task, much

harder then finding suitable T-groups. Because R+ above is a free abelian group, wecan easily find groups G with R = End G, see [3, 2]. But there are still two obstacleswhich must be taken into consideration. Often |R| < |G| in realization theorems, thusthe units of R which represent Aut G will never act transitive on a bigger group andG cant be a T-group. More importantly, inspecting the constructions in [3, 2], it isclear that they provide no control about the action, thus both U and T for UT-groupsare a problem. Note that in many earlier constructions G has a free dense and pureR-submodule of rank > 1 mostly of rank |G|. But T-groups must obviously be cyclicover their endomorphism ring R, hence [3, 2] do not apply in principle. Inspecting theproof in [5] it is easy to see that G is not torsion-free over its endomorphism ring. This

comes from the list of new variables x,y,... added to R in the construction in order tomake R acting transitive on all pairs of pure elements. Even refining the list of purepairs in [5] it seems hard to avoid clashes of related pairs such that x y for examplehas a proper annihilator. Thus the groups in [5] are T-groups and not UT-groups (evenmodifying the arguments).

Thus a new approach is need, which will be established in Section 3. We willuse a geometric argument choosing carefully new partial automorphisms for making Gtransitive but with very small domain and image in order to preserve the U-propertyfor the new monoid. Then we will feed the partial maps with pushouts to grow themup and become real automorphism without destroying the UT-property. At the end we

will have a suitable subgroup F of automorphisms of some group G, thus G becomesan R-module over the ring ZF =df R.Finally we have to fit these approximations to ideas getting rid of the endomor-

phisms outside R, see Section 4. We need the Strong Black Box as discussed and provenin terms of model theory in Eklof, Mekler [6, Chapter XIV]. Note that this predictionprinciple is stronger then (Shelahs) General Black Box, see [2, Appendix]. The StrongBlack Box is also restricted to those particular cardinals mentioned in the theorem.However, here we will apply a version of the Strong Black Box stated and proven onthe grounds of modules in ordinary, naive set theory, which can be found in a recentpaper by Gobel, Wallutis [10], see also [9]. In order to show End G = R well-knownarguments for realizing rings as endomorphism rings must be modified because the finalring and its action are only given to us at the very end of the construction: We willfirst replace the layers G from the construction by a new filtration only depending onthe norm. Note that the members of the new filtration of the right R-module G mustbe right R-submodules for suitable subrings R of R to have cardinality less than|G|. But they are still good enough to kill unwanted endomorphisms referring to the

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prediction used during the construction. Moreover note that the two tasks indicated

in the last two paragraphs must be intertwined and applied with repetition. Whilethe final G is an 1-free abelian group, hence torsion-free, it is not hard to see thatG is torsion as an R-module: If 0 = g G, then we can choose distinct elementsg, x , y pG such that ng = g, and x y pG. Hence there are distinct unit el-ements rx, ry, rxy U(R) = F such that g

rx = x, gry = y, g

rxy = x + y. Theendomorphism rx + ry does not belong to F, in particular it can not be rxy. Hencer = rx + ry rxy = 0 but g

r = x + y (x + y) = 0 and g is torsion. It is worth notingthat the result can be strengthened under V = L, where we get strongly--free groupsof cardinality as in Theorem 1.1 for each regular, uncountable cardinal which isnot weakly compact. In this case the approximations in Section 3 can be improved,

replacing 1-free by free at all obvious places. The main result of this section willthen be a theorem on free groups G with a free (non-abelian) group F Aut G actinguniquely transitive on G. Also Section 4 must be modified: The Strong Black Box 4.2must be replaced by following arguments similar to [3].

2 Warming up: Construction of a special group

We begin with a particular case of an old theorem and discuss extra properties of theconstructed group. Part of this proposition will be used in Section 3.

Proposition 2.1 Let be a cardinal with 0 = , F be a free (non-abelian) groupof rank < and R = ZF be its integral group ring. Then there is a group G with the following properties.

(i) G is an 1-free abelian group of rank with End G = R.

(ii) G is torsion-free as an R-module.

(iii) Aut G = F

(iv) If End G, then is injective.

Proof. Note that the integral group ring R = ZF has free additive group R+ withbasis F. We can apply a main theorem from [2] showing the existence of an 1-freeabelian group G with End G = R. The free group F is orderable (i.e. has a linearordering which is compatible with multiplication by elements from the right), see Mura,Rhemtulla [16, p. 37]. However note, that torsion-free groups may be non-orderable,

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see [16, pp. 89 - 95, Example 4.3.1.]. The integral group ringZF of any orderable group

F satisfies the unit conjecture, this is to say that the units of R = ZF are the obviousones, hence U(ZF) = F, see Sehgal [17, p. 276, Lemma 45.3]. Moreover, any groupring which satisfies the unit conjecture also satisfies the zero divisor conjecture, henceR has no zero-divisors, see Sehgal [17, p. 276, Lemma 45.2]. Therefore R is torsion-freeas an R-module.

Now the remaining part of the proof is easy: Aut G = U(R) = F and if 0 = g G,then g R G because G is also an 1-free R-module by construction in [2]. Ifwe consider multiplication of g by some r R on a non-trivial component of g in thisfree direct sum, then r = 0 because R is torsion-free as an R-module. Hence G istorsion-free as an R-module. Any End G = R is scalar multiplication by a suitable

r R hence injective because G is a torsion-free R-module.

We will use Proposition 2.1 in Section 3. We get more out of it if we know that aparticular endomorphism is pure:

Lemma 2.2 LetF be the free (non-abelian) group and End G = ZF be the endomor-phism ring of the 1-free abelian group G given by Proposition 2.1. If ZF \ F,then is a monomorphism and not onto. If 0 = ZF, then is pure inZF+ ifand only if Im is pure in G.

Proof. All endomorphisms of G in Proposition 2.1 are monomorphisms as shown

there. If R = ZF = End G would be onto, then must be an automorphism, thus U(R), which is F; and this was excluded.

We come to the last assertion. We shall write 0 = = r R and suppose thatr = nr (n = 1) is not a pure element of R+. Note that nG = G, hence Gr = Gnr

and we can pick an element g Gr\ Gr which is mapped into Gr under multiplicationby n. Hence Gr is not pure in G. Conversely let r be pure in R and consider any g Gsuch that gp Gr for some prime p. Hence gp = g r and by construction of G (justnote that G is 1-free as R-module) there is h G such that g

hR and hR is a puresubgroup of G. Hence also g hR and we can write g = hrg, g

= hrg which giveshprg = hrgr and prg = rgr because G is R-torsion-free. Using that p cannot divide r

by purity in R and that r, rg are elements of the group ring ZF we can write rg = r

pfor some r R. Finally gp = gr = (hrp)r, hence g = hrr Gr and Gr is purein G.

If we replace [2] in the proof of Proposition 2.1 by [3], then we can strengthenProposition 2.1 in the constructible universe L. We get a

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Lemma 3.1 The set pAut G of all partial automorphism of G with G/Dom and

G/Im both 1-free abelian groups is a submonoid of all partial automorphisms with1 = id G and 1 = id G acting as multiplication by 1 and 1 respectively, which isclosed under taking (weak) inverses.

Moreover, if F pAut G, then F pAut G is the submonoid of all productstaken from the set {1} F F1, where F1 = {1 : F}.

We begin with an observation which allows us to consider induced partial automor-phisms on a factor group.

Observation 3.2 If U G are abelian groups and pAut G with (Im U)1 U and (Dom U) U, then induces a partial automorphism of G whereG = {g = g +U : g G} takingg to g for anyg Dom . MoreoverDom = Dom and Im = Im .

Proof. If g G and g = 0 in G, then g = g U and

g = g1 (Im U)1 U,

hence g = 0 and pAut G. The other assertions are also obvious.

In order to show Proposition 3.9 we relate elements of free (non-abelian) groupsand elements in pAut G. It is important to be able to work with elements of pAut Gacting on a partial free basis of G. To be precise, we will need the following definitionextending freeness from G to pAut G.

Definition 3.3 LetF = {t : t u} pAut G be a finite set of partial automor-phisms. Then(G,F) is called1-free if any countable subset of G belongs to a countablesubgroup X G with basis B and the following properties for any F.

(i) G/X is 1-free.

(ii) induces a partial injection on B, that is, if b B Dom , then also b B.

(iii) X Dom = B Dom .

Passing to weak inverses, it follows from (iii) that also XIm = BIm . MoreoverX Dom and X Im are summands of X with free complements generated byB \ Dom and B \ Im , respectively. It also follows that G is 1-free. We can easearguments in Lemma 3.11 and Lemma 3.12 to note here that we only need a partial

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basis bF (a subset ofB) with the property (ii) and bF Dom = bF Dom ;

see Proposition 3.9.Next we relate basis elements of free non-abelian groups and partial automorphisms

with care. Suppose the set F = {t : t J} generates a free group F and as inDefinition 3.5 there is a map : F pAut G (acting on the left), then this map can beextended to F. The extension is unique if we restrict ourselves to reduced elements = 1 . . . n in F with i F F

1 and define naturally () = (1) . . . (n).However, if1, 2 F are reduced and is the reduced element which coincides withthe formal product 12 in F, then only (1)(2) () holds as a graph and thismeans Dom((1)(2)) Dom () and () Dom((1)(2)) = (1)(2).Thus we have equality if also the formal product 12 is reduced.

Definition 3.4 With : F pAut G as above we say that (or (F) = {() : F}) satisfies the U-property if = for any reduced elements , F withx() = x() and some x Dom() Dom() pG.

The last definition is crucial for this paper because it is the microscopic version ofU-groups discussed in the introduction. We also must pass from groups Gx with thisU-property to suitable extension Gy with the U-property. All this we encode into ourmain definition of quintuples x and their extensions. Normally our maps will act on theright, but we allow three exceptions, the maps , and h below. Also P0(J) denotesall finite subsets of the set J.

Definition 3.5 LetKbe the family of all quintuples x = (G,F, , , h) = (Gx,Fx, x, x, hx)such that the following holds.

(i) G is an 1-free abelian group.

(ii) F = {t : t J} is a set of free generators t indexed by J = Jx of a group F.

(iii) : J {1, 1} is a map.

(iv) : F pAut G is a map which satisfies the U-property. We shall write x(t) =xt and omit x if the meaning is clear from the context.

(v) h : P0(J) Im (h) is a partial function from Dom h P0(J). If u Dom hand U = h(u), F = {t : t u}, then the following conditions must hold.

(a) U is a countable subgroup of G and (Dom x U)x U for all F;hence x induces x pAut (G/U); see Observation 3.2. Let G = G/U andF = {x : F}.

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(b) (G,F) is 1-free; see Definition 3.3.

It follows that G above is 1-free. From Definition 3.5 (iv) and Lemma 3.1 follows

Corollary 3.6 If x K then xt : t Jx pAut Gx.

Hence Gx/Dom x is 1-free for all F. We will carry on this condition induc-tively, just checking the generators in F and using the following simple

Test Lemma 3.7 If U G are groups and any countable subset of G is contained ina countable subgroup X with free generators B1B2 such thatB1 U andUB2 = 0then G/U is 1-free.

The same test lemma will be used inductively for U in Definition 3.5 (v)(b). Wewill pass from groups Gx to larger groups Gy related to x, y K by taking pushoutsor unions. This is reflected in the next definition (in particular condition (iii)) of anordering on K. This is the final step before we can start working.

Definition 3.8 Let x y (x, y K) if the following holds for x = (Gx,Fx, x, x, hx)and y = (Gy,Fy, y, y, hy).

(i) Gx Gy and Gy/Gx is 1-free.

(ii)

y

extends

x

in the weak sense (

x

y

), that is J

x

J

y

(equivalently F

x

Fy

)and also xt y

t extends for all t Jx.

(iii) If t Jx, then one of the following cases holds

(a) x(t) = y(t) and xt = y

t .

(b) Gx Dom yt Im y

t .

(c) x(t) = 1 = y(t) and Gx Dom yt .

(d) y(t) = 1 = x(t) and Gx Im yt .

(iv) hx hy extends (i.e., if hx(u) Gx, then hx(u) = hy(u) Gy).

(v) If u Dom hx and Gx

=df Gx/hx(u) Gy

=df Gy/hx(u), then any basis B of acountable subgroup X G

xas in Definition 3.5 extends to a basis B of some

countable subgroup X of Gy

which also satisfies Definition 3.5.

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Proposition 3.9 Suppose thatx < y inKandu Dom hy, |Gy| > 0, Gy = Dom yt =

Im yt for every t u. If F is the group freely generated by {t : t u} and =iIaii ZF is an element of the integral group ring, ai = 0 for all i I and if the

is are pairwise distinct (reduced) elements of F, such that y =

iIaiy

i is bijective,then I is a singleton and its coefficient is 1 or 1.

Proof. If F = t : t u, then by hypothesis Fy = yt : t u is a free subgroup of

Aut Gy. If also =

iIaii ZF is as above, then y ZFy and we may assume that

ker y = 0. It remains to show that y is surjective if and only if I = {0} is a singletonand a0 = 1. If I = {0}, then it is clear that y is surjective if and only if a0 = 1.Hence we may assume that y is an isomorphism, and |I| > 1 for contradiction. Inorder to apply Definition 3.5 we pass to the quotient G

y= Gy/h(u) and to the induced

maps t, which we rename again as Gy, t. It follows that h(u)

y h(u) and silentlywe assume that h(u) is invariant under (y)1; otherwise we must enlarge h(u) by aback and forth argument such that the quotient satisfies again Definition 3.5 (v). AlsoG

y= Gy/h(u) = 0 because |h(u)| = 0 < |G

y|.If X = 0 is a countable subgroup of Gy, then X is free. We may assume without

restriction that Xy X, XFy X and X(y)1 X and X =df y X End X.

If x X, then x = gy Gyy = Gy, thus g = x(y)1 X and X is also surjective(on X). We can start with some X with a special basis B = as in Definition 3.5 (v)(the weak version mentioned after the Definition 3.3) and let X be its closure as above.Then X will be a summand ofX because G/X is 1-free. Hence B

extends to a basis

B of X: The maps X = y X, ( F) (by hypothesis) are total automorphismsof X, thus all automorphisms of FX = {X : F} are permutations of B

whenrestricted further to B.

Let G be the group given by Proposition 2.1. Note that t F (t u) is givenby Proposition 3.9 and ZF = End G, hence any t can be viewed as an automorphismof G. In order to distinguish it from the element in F we will call the automorphismt Aut G and F becomes F

. The mapping extends naturally to all ofZF, (bythe identification ZF = End G), thus =

iIai

i End G, where

i F

. From|I| > 1 and Lemma 2.2 it follows that is not surjective. Let y G \ G whichwill help us showing that X can not be surjective either, in fact we want to show

that B

XX = . Fix an element c B

and define a map : B G such thatc = y. If b B and there is F such that cX = b, then put b = c

. If exists, then it is unique by the U-property. Hence is defined on cFX(= cF

y). Ifb B \ cFX , then let b = 0. Hence is well-defined on B and extends uniquelyto an homomorphism : X G. It follows bX = b, hence bX = b

for all b B, i.e. commuting with replaces the X by. If c XX , then there is

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x =

bB xbb X with xX = c. Thus c = xX =

bB xb(bX) and we apply to

this equation to get the contradiction

y = c =bB

xb(bX) =bB

xb(b) G.

Thus X , hence y is not surjective.

It is convenient to check the U-property by the following simple characterization.

Proposition 3.10 LetF be the group freely generated byF and : F pAut Gx bea map as in Definition 3.5 with(t) =

x

t for allt J. ThenF satisfies the U-property

if and only if any reduced product F with xx = x for some x Dom x pG is = 1 F.

Proof. If we can choose a reduced element F with xx = x = (x1x) for somex Dom x pG, then = 1 follows by the U-property ofF. Conversely, if there arereduced elements , F with xx = xx for some x Dom x Dom xpG, thenwe can write x = xx(x)1. Hence x Dom x(x)1 and we can cancel 1 to geta reduced F with = 1 in F. From x Dom x(x)1 Dom x it followsx = xx. We have = 1 by hypothesis, and = follows.

The last proposition shows that the U-property is a strong restriction on (F). Ifonly xx = x for a reduced and pure x G, then = 1. However note that if F \ {1}, then xx(x)1 = x for some x pG, hence x(x)1 id G but notx(x)1 = id G because

1 is not reduced.The next lemma will be used to make the desired group more transitive. We want

to isolate the argument on the existence ofh(u): Ifx = (G,F, , , h) K, 0 pAut Gwith 0 / J as in Lemma 3.11, then we extend h : P0(J) Im (h) to h

: P0(J)

Im (h) where J = J {0}. If u Dom h, then h(u) = hx(u). If x, y and x are fromLemma 3.11, then let h(u) for u = u {0} be a countable subgroup U G containing{x, y} h(u) such that G = G/U and the induced maps F

ysatisfy Definition 3.5 (v)

(the weak version mentioned after Definition 3.3 will suffice). Note that we only use

extensions of groups Gx as in Lemma 3.11 or Lemma 3.12 and unions of ascendingchains of such groups. By a back and forth argument, and a moments reflection aboutthe action of the extended partial automorphisms by the pushouts, it follows that sucha countable subgroup U exists.

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Lemma 3.11 to get more partial automorphisms. If x = (G,F, , , h) K and

x, y pG such that xx = y for all F, then let y0 : xZ yZ(x y)be the natural isomorphism. IfFy = F {0}, J

y = J {0}, y = {(0, 1)}, y =x {(0,

y

0)} and hy = h as above, then x < y = (G,Fy, y, y, hy) K

Proof. Obviously x y. Also y0 pAut G because xZ, yZ are pure subgroupsof G and G is 1-free, hence G/xZ and G/yZ are 1-free. But it is not clear at thebeginning that y satisfies the U-property. We will check this with Proposition 3.10.

Let be Fy = F, 0 = and F. We write = 1

12 . . . k1k with

0 = i Z and i F reduced and assume that all is are different from 1, exceptpossibly 1, k. Now we assume that z

y = z for some z Dom y pG and want to

show that = 1.However next we claim, that the product must be very special and show first

that i = 1 for all i < k. If this is not the case, then some 2 or 2 is a factor of

. We may assume that 2 appears. Note that Dom (y)2 = (Im (y) Dom y)(y)1,and Im y Dom y = Zy Zx = 0 by the choice of x, y. Hence (y)2 = 0 and y = 0is a contradiction, because 0 = z Dom y, so the first claim follows. Next we showthat

= 112. (3.2)

We look at the path ofz, the set [z] of all consecutive images of z:

z0 = z, z1 = zy

1, z2 = z1(y)1, . . . , z2k1 = z2k2

y

k

In order to apply (y)1 to z1 we must have z1 Dom(y)1, but Dom(y)1 is

either Zx or Zy, hence z1 is one of the four elements of the set V = {x, y} bypurity. Inductively we get zi V for all 0 < i < 2k 1. Suppose that 2 appearsin , then z3 = z2

y

2 V because z4 = z3(y)2 is defined and z3 is pure. However

x Dom y2 or y Dom y

2, respectively. Hence 2 is multiplication by 1 on Zx oron Zy or x2 = y. The last case was excluded by our hypothesis on x, y and the firsttwo cases and the U-property would give 2 = 1 which also was excluded. Hence

(3.2) follows.Our assumption is reduced to z = zy1(y)1y2 for some z pG. We may replace

by 1, hence 1 = 1 without loss of generality and z = zy

1yy2. We consider

the path [z] and have z1 = zy

1 = x from purity of z1 Dom y. It follows z2 = y

and z3 = yy

2 = z from our assumption. Thus yy

2y

1 = x and 21 F, whichcontradicts our choice of x, y. Hence only = 1 is possible and y K follows.

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The next lemma will increase domain and image of partial automorphisms, respec-

tively.

Lemma 3.12 growing the partial automorphisms. If x = (G,F, , , h) KwithF = {s : s J} and t J, then there is x y = (G

y,Fy, y, y, h) K such that thefollowing holds.

(i) Fx = Fy, x (J \ {t}) = y (J \ {t}), x(t) = y(t) and Gy = G0 + G1 is apushout with D = G0 G1 and G

x = G0 = G1.

(ii) (a) Ifx(t) = 1, then G0 = Dom y

t , G1 = Im y

t and D = Dom x

t.

(b) If

x

(t) = 1, then G0 = Im

y

t , G1 = Dom

y

t and D = Im

x

t.

Proof. The set J and h do not change when passing from x to y. Thus we considerx next and restrict to x = 1 (the case x = 1 follows if we replace t by

1t ). For

the pushout we let Gy = (G G)/H with H = {(x, x) : x Dom t}. If we alsosay that U0 = (U 0) + H/H G

y and U1 = (0 U) + H/H for any U G, then inparticular Gy = G0 + G1 and D =

df G0 G1 = (Im x

t)0 = (Dom x

t)1 by the pushout.Moreover we identify G0 = G

x, hence Dom xt = (Dom x

t)0 =df D and D = Im xt.

The canonical map

yt : Gy Gy ((x, 0) + H (0, x) + H)

extends xt because ((x, 0) + H)y

t = (0, x) + H = (xx

t, 0) + H for all x Dom x

t.Clearly G0 = Dom

y

t and G1 = Im y

t . Moreover Gy/D = G0/Dom

x

tG1/Im x

t is 1-free, hence also Gy and Gy/Gx = G1/D are 1-free, and the maps

xs =

ys (t = s J)

remain the same. It follows that y : Fy pAut Gy. The existence of a partial basissatisfying Definition 3.3 was discussed before Lemma 3.11. So for x y Kwe onlymust check the U-property for F with the new partial automorphisms from ys (s J)and apply Proposition 3.10:

Consider a reduced product = 11t 2 . . .

n1t n, where 1 = i F \ {t}

except possibly 1 = 1 and n = 1.Suppose that zy = z for some z pGy and let

z0 = z, z1 = z0y

1, t1 = z1(y

t )1 , z2 = t1

y

2, t2 = z2(y

t )2, . . . , zn = tn1

y

n

be the path [z] of z. Thus zn = z0 by assumption on . First we note that y

i = x

i

for all i n with the possible exceptions for 1 = 1 or n = 1. If also all the (y

t )i

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can be replaced by (xt)i , then [z] Gx and by the U-property ofF, zy = zx = z it

follows = 1. We will consider the two cases z0 G0 and z0 G1 \ G0.First we reduce the second case z0 G1 \ G0 to the first case. Since z0 Dom 1 it

follows 1 = 1, hence z1 = z0 G1 \ G0. From z1 Dom(y

t )1 it follows 1 1

and t1 =df z1(

y

t )1 G0. From zn = z0 it follows zn1(

y

t )n1yn = z0 G1 \ G0 and

therefore n = 1 and n1 1. The equation z0y = z0 reduces to

t1(y

t )1+1x2(

y

t )2 xn1 = t

1 with t

1 G0,

which is the first case for a new z = t1 pG0.If z0 G0, then also z0 Dom

x

1 and z1 = z0x

1 G0. We will continue alongthe path step by step having two subcases each time, but one of them will lead to acontradiction. In the first step either z1 D

or z1 / D. In any case 1 = 1 and in the

second case t1 = z1y

t G1 \ D, but t1 Dom y

2 so necessarily 2 = 1 and n = 2. Weget t1

y

2 = z2 = t1, and t1 = z0 G0 contradicting t1 G1 \ D. We arrive at the othercase z1 D

. Hence t1 = z1x

t D and we must have t1 Dom x

2. Therefore alsoz2 = t1

x

2 G0, and 2 = 1. Again we have two cases z2 D and z2 / D

with 2 = 1,where the second case leads to a contradiction. Hence t2 = z2

x

t D and we continueuntil we reach n and all the yt s are replaced by the

x

ts. Now our first remark appliesand = 1 follows from the U-property for F.

Lemma 3.13 Let be a limit ordinal. Then any increasing continuous chain xj =

(Gj,Fj, j, j, hj)j in(K,


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finite character by Pontryagins theorem (speaking about subgroups of finite rank).

Hence G/Dom xt is 1-free, as required. Now it is easy to see that x K: Definition3.3 is easily verified by taking union of partial bases. The U-property carries over tolimit ordinals.

The next main result of this section follows by iterated application of the Lemma 3.11,Lemma 3.12 and Lemma 3.13. Without danger we now can identify the free groups F

and F by the isomorphism in the theorem.

Theorem 3.14 Ifx = (G,F, , , h) is inK, then we can also findx = (G,F, , , h)inKwith x x such that the following holds.

(a) F

is a set of automorphisms of G

which freely generates F

Aut G, hence G

is a module over R = ZF.

(b) F acts uniquely transitive on pG.

(c) |G| + |F| = |G| + |F|

Proof. We will proceed by induction to get a chain x xn K (n ) takingseveral steps each time. Finally x will be the supremum of the xns.

In the first step we apply |G0|-times Lemma 3.11 and Lemma 3.13 to x = x0 takingcare of all appropriate pairs of elements in pG0 and let x0 x1 = (G1,F1, 1, 1, h1) bethe union of this chain such that F1 acts transitive on pG0. In this case G0 = G1 but

the other parameters in x0 increase (along a chain) by Lemma 3.11. In the next stepwe apply |G1|-times Lemma 3.12 and Lemma 3.13 to get x1 x2 = (G2,F2, 2, 2, h2)such that G0 Dom

x2t and G0 Im

x2t for all t J

x2 = Jx1 . We continue this wayfor each n. From Lemma 3.13 follows x x K, and by construction F Aut G

acts uniquely transitive on pG such that also (c) of Theorem 3.14 holds.

Now we can restrict elements in K to triples and say that x = (Gx,Fx, x) belongsto K if and only if Gx is an 1-free abelian group, F = {t : t J

x} freely generates a(non-abelian) group F and x : F Aut Gx is an injective map such that x(F) ={x(t) =

x

t : t Jx} satisfies the U-property. Still we can view K as a subset ofK

and have an induced ordering, compare Definition 3.8: We have x y (x, y K) if (v)and the following holds.

(i) Gx Gy and Gy/Gx is 1-free.

(ii) x y.

The other conditions in Definition 3.8 are vacuous.

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4 Construction of uniquely transitive groups

In this section we will sharpen Theorem 3.14 which shows that a group ring R = ZFfor some free group F can be represented as R End G of some 1-free group G(hence G is an R-module), such that the units U(R) = F act uniquely transitive onG. If F Aut G is not necessarily transitive but satisfies the uniqueness property(( , F and x pG,x = x) = ), then we will also say that thepair (G, F) has the U-property. We will modify G such that equality holds, that isR = End G. Thus we have to kill unwanted endomorphisms, which is done using theStrong Black Box 4.2. We need some preparation to work with this prediction principle.

First we need an S-adic topology. Let S = {qn : n } be an enumeration of

a multiplicatively closed set generated by 1 and at least one more natural numberdifferent from 1 and define s0 = 1, sn+1 = sn qn for all n which obviously definesa Hausdorff S-topology on the groups G under consideration. Let G be the S-adiccompletion of G and G G naturally, where purity is S-purity.

Next we must formulate the Strong Black Box and adjust our notations; see StrongBlack Box 4.2. We rely on the version adjusted to and proven for modules using onlynaive set theory as explained in the introduction, see [10, 9]. Thus we have to fix a fewparameters next. To do so we choose an enumeration by ordinals , with = +

a successor cardinal such that 0 = : Let F be a free non-abelian group with a setof free generators F, the Fs constitute a strictly increasing, continuous chain in < . We will write R = ZF with R =

R and note that R+

and R+ are freeabelian groups. By Theorem 3.14 there is an R-module G of cardinality whichis cyclic as R-module and 1-free as abelian group and F acts sharply transitive onpG. We can choose a free abelian and S-dense subgroup B G B. Also theBs constitute a strictly increasing, continuous chain in < . Passing to + 1 wecan let B+1 = B A with A =

i


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(, i) [ g ]} . Recall that e,i = aie, where ai A, which explains the use

of the notion A-support.Next we define a normon , respectively on B, by {} = + 1 ( ), H =

sup H (H ), hence = , and g = [ g ] (g B), i.e. g = min{

| [ g ] }. Note, [ g ] holds if and only if g B for B =


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We will enumerate C taking care of the order of the norm and can write C = {a :

< } such that [ ] [ ] for all < < . Note that we distinguishbetween F and given by the Strong Black Box 4.2.

The enumeration is now used to find a continuous, ascending chain G ( < )

such that B G B and the corresponding x = (G,F, ) K. Let E obe a fixed stationary subset of o and choose x0 = (G0,F0, 0) to be any triple given byTheorem 3.14 for cardinality . For any < , let P = Dom and suppose that thetriples x = (G,F, ) are constructed for all < . If is a limit ordinal, then bycontinuity we let x = sup


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which ensures that the U-property can be extended when applying the step lemma.

We will use obvious simplified notations:

Proposition 4.3 Let G =

n Gn, xn G1n be pure (Rn-torsion-free) elements,

where Gn+1 = Gn G1n Gn+1 and Gn+1 is obtained from G

n+1 by Theorem 3.14. If

b G0, x = n xnsn1 + b and G = G,xR, thenU(R) = F for the free groupF =

n Fn, the pair (G

, F) satisfies the U-property and G is 1-free.

Proof. It is well-known that G is 1-free, see [2] or [6, 9] for instance. We wantto show that the pair (G, F) satisfies the U-property.

If r R and F, then r Rn and Fn for some n large enough. Hence

there is no restriction to write any 0 = y G

as y = xk

r+g with xk

= n>ksn1

sk xn+bk

,r Rn, b bk G0, g = g0 + g1, g0 G0, 0 = g1 Gn and [G0] [g1] = [xkr] [g] = .

Suppose y = y, hence (xkr) + g = xkr + g and by continuity

m>k

sm1sk

(xmr) + g =m>k

sm1sk

(xmr) + g.

We consider any m > n , k and note that xm is a torsion-free element of the Rn-module Gm and also U(Rn), (hence Gm is invariant under application of , r) and0 = xmr,xm Gm. Restricting the support [xkr] of the last displayed equation toxm shows that

sm1sk

(xmr) =sm1sk

(xmr). Hence 0 = (xnr) = (xnr) and passing to

h pGm with hq = xnr we get h = h, hence = 1 by the U-property for (Gm, Fm).So acts as id Gm for all m large enough. Thus is the identity on G, and thesame holds for G by continuity (G is dense in G in the S-topology).

Next we prove the step lemma.

Step Lemma 4.4 LetP =

(,i)IZe,i for some I and let H be a subgroup

of B with P H B which is 1-free and an R-module, where R =df Rx = I

R

with I =df [ I] = { < : i < , (, i) I}. Assume that x = (H, F, ) K (thus

Rx = ZF). Also suppose that there is a set I = {(n, in) : n < } [P] = I

suchthat 0 < 1 < < n < (n < ) are discrete ordinals and

(i) H =n


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If : P H is a homomorphism which is not multiplication by an element from

R, then there exists an element y P and x x = (H,F, ) K such thatH H

B, y H and y H.Proof. By assumption H is an R-module and hence the S-completion H is anR-module. Let Z be the S-completion ofZ and let x =

n

snen,in P. We will usey = x P or y = x + b P for some b P to construct two new groups H Hy:

Let Hy = H,yR B. By Proposition 4.3 and Theorem 3.14 we obtain xy =(Hy,Fy, y) K such that x xy and y Hy. Clearly y H. Put z = y andy = x. If z Hx, then we choose H

= Hx and have xx K with End Hx. Also

Rx = ZFx.Ifz Hx, then also z Hx by an easy limit argument and there exist integers k andn such that skx = grn+xr

n for some rn, r

n R and g H. It follows x(skr

n) = grn

and rn, rn Rn for some n

. Since (sk rn) = 0 there is b

P such thatb =df b(sk r

n) = 0. Note that b

has finite support. Moreover, by the cotorsion-

freeness of H there exists Z such that b H. Let y = x + b. We claim that End(Hy). By way of contradiction assume that sl(x+b) = grm+(x+b

)rm forsome integer l k and elements rm, r

m Rm for some m

and g H. Withoutloss of generality, we may assume n = m, hence sl(x + b

) = grn + (x + b)rn and

if s = sl/sk, then

sl(b) = sl(x + b) sskx = grn + (x + b)rn s(grn + xrn)

= (grn sgrn) + brn + x(r

n sr

n).

Since [b] = [b], [b] = [b] and g, g H an easy support argument shows thatrn = sr

n, hence ssk(b

) = (grn sgrn) + sbrn and thus s(skb

brn) =(grn sgrn) H. By purity we get (skb

brn) = b H - a contradiction.

Proof of Theorem 1.1. IfR = End G, then also Aut G = F, because U(R) = Fand any r R \ U(R) viewed as endomorphism of G is not bijective by Proposition3.9. Hence it remains to show that End G R.

First we choose a new filtration of G and define

G =df {g G, g < } ( < ).

Lemma 4.5 The new filtration on G has the following properties.

(a) G P G+1 for all < ;20


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(b) {G, < } is a -filtration of G;

(c) If , < are ordinals such that = then G G.

Proof. Inspection of the definitions, or [10, 9].

Assume that End G\R and let = B, hence R. Let I = {(n, in)| n

Documents

Rudiger Gobel and Saharon Shelah- Uniquely Transitive Torsion-free Abelian Groups