Saharon Shelah- Minimal Bounded Index Subgroup for Dependent Theories

8/3/2019 Saharon Shelah- Minimal Bounded Index Subgroup for Dependent Theories

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arXiv:math/0

603652v1[math.L

O]28Mar2006

MINIMAL BOUNDED INDEX SUBGROUPFOR DEPENDENT THEORIES

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat RamJerusalem 91904, Israel

Department of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Abstract. For a dependent theory T, in CT for every type definable group G,the intersection of type definable subgroups with bounded index is a type definable

subgroup with bounded index.

The author would like to thank the Israel Science Foundation for partial support of this research(Grant No. 242/03), Publication 876I would like to thank Alice Leonhardt for the beautiful typing.First Typed - 05/Mar/28Latest Revision - 06/Jan/11

Typeset by AMS-TEX

1
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2 SAHARON SHELAH

0 Introduction

Assume that T is a dependent (complete first order) theory C is a -saturated

model ofT (a monster) G is a type definable (in C) group in C (of course we consideronly types of cardinality < ).

A type definable subgroup H of G is call bounded if the index (G : H) is < .We prove that there is a minimal bounded definable subgroup. The first theoremon this line for T stable is of Baldwin-Saxe [BaSx76].

Recently Hrushovski, Peterzil and Pillay [HPP0x] investigated definable groups,0-minimality and measure, see also the earlier work on definable subgroups in 0-minimal T in Bezarducci, Otero, Peterzil and Pillay.

Hrushovski in a lecture in the logic seminar in the Hebrew University mentionedthat for every definable group in CT, the intersection of the definable subgroupwith bounded index is of bounded index is proved in [HPP0x] for definable groupswith suitable definable measure on them and T dependent; but it is not clear if theexistence of definable measure is necessary.

Recent works of the author on dependent theories are [Sh 783] (see 3,4 ongroups) [Sh 863] (e.g. the first order theory of the peadics is strongly1 dependentbut not strongly2 dependent, see end of1; on strongly2 dependent fields see 5) and[Sh:F705]. This work is continued in [Sh:F753] (for G abelian and L,-definablesubgroups).


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MINIMAL BOUNDED INDEX SUBGROUP FOR DEPENDENT THEORIES 3

1

1.1 Lemma. For T dependent.

1) If below holds then:

() q(C) is a subgroup of p(C)

() q(C) is of index 2|T|0

() essentially q(x)\p(x) is of cardinality |T|0 (i.e., for some q(x) q(x)of cardinality |T|0 , q(x) is equivalent to p(x) q(x)).

where

(a) p(x) is a type such thatp(C) is a group we callG (under some definableoperation xy,x1 and the identity eG which are constant here;

of course all types are of cardinality < ,C is -saturated)(b) q(x) = p(x)

{r(x) : r(x) R} where

(c) R = {r(x) : r(x) a type such that (p r)(C) is a sub-group of p(C) ofindex < }.

2) There exists some q q over Dom(p), equivalent to q and such that |q| |T| + |Dom(p)|. So (p(C) : q(C)) 2|Dom(p)|+|T|.3) If ri(x) R for i < (|T|

0)+ then for some < (|T|0)+ we have (p(x) {ri(x) : i < })(C) = (p(x)

{ri(x) : i < (|T|

0)+})(C).

Proof. 1) Note

1 R is closed under unions of < hence q q |q| < q R

2 ifr(x) R, r(x) r(x) is countable then there is a countable r(x) r(x)

including r(x) which belongs to R[Why? Let r(x) = {n(x, an) : n < } (can use n = (x = x)). Withoutloss of generality r(x) is closed under conjuctions and also r(x). Now wechoose n(x, bn) =

1n(x, b

1n)

2n(x, b

2n) with

1n(x, b

1n) r(x),

2n(x, b

2n)

p(x) by induction on n < such that n+1(x, bn+1) n+1(y, bn+1) n(xy

1, bn)n(x, an). Such formula exists as (p(x)r(x))(p(y)r(y))

n(xy1

, bn) n(x, an).Now r(x) = {n(x, an), n(x, bn) : n < } is as required.]

Clause () is obvious.Assume toward contradiction that the conclusion () fails. So we can choose

(c, r) by induction on < (|T|0)+ such that


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4 SAHARON SHELAH

3 (a) c (p(x)

{r : < })(C)\q(C)

(b) r = {n(x, b

n) : n < } q and b

n b

n+1

(c) n+1(x, bn+1)

n(x, b

n)

(d) r R

(e) c does not realize r in fact C |= 0 (c, b

0 ).

Without loss of generality

4 n+1(x, y

n+1)

n(x, y

n) and (

n+1(x1, y

n+1)

n+1(x2, y

n+1)) (

n+1(e, y

n+1)

n+1(x1x12 , y

n+1))

5 n(x, y

n) = n(x, yn) and n+1(x, yn+1)) n(x, yn).

6 ca : < (|T|0)+ is an indiscernible sequence over Dom(p) where a =

b0 b

1 b

2 . . . , without loss of generality b

n = a kn

[Why? By Ramsey theorem and compactness.]

7 if < < then cc1 r(C).

[Why? Without loss of generality is infinite, as (p r)(C) is a subgroupof p(C) of index < . If , ci : i < is a sequence of indiscerniblesover Dom(p) b of elements ofp(C) pairwise non-equivalent modulo G =(p r)(C), then extend it to ci : i < a sequence of indiscerniblesover Dom(p) b and arrive at < cc

1 / G cc

1 / G so

cG : < pairwise distinct (equivalently Gc : < pairwisedistinct) contradiction.]

8 c r(C) iff = .[Why? Let

c = c2+1 (c2)1

r = r2.

So:

(i) if < , c (p r)(C) as c2+1, c2 belong to the subgroup (p

r2+1)(C

) by clause (a) of3(ii) if > , c belongs to (p r

)(C) by 7

(iii) if = then c does not belong to (p r)(C) as it is a subgroup,

c2+1 belongs to it and c2 does not belong to it by clause (e) of3.


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Let a = a2+1, b

n = b2n retaining the same s. So we have gotten an example

as required in 8 (not losing the other demands).]

9 if d1, d2 (p r)(C) then d1cd2 / 1(C, b1 ).

[Why? Fix , if this holds for some n(, bn) by indiscernibility renamingthe is this is O.K. Otherwise for each n < there are d

n1 , d

n2 (p r)(C)

such that C |= n(dn1 cd

n2 , b

n). By compactness for some d

1, d

2 (p

r)(C) we have |= n[d1cd

2, b

n] for every n < . So d

1cd

2 belongs to

the subgroup (p r)(C) but also d1, d

2 belongs to it hence c belongs,

contradiction.]

10 if w = {i1, . . . , in}, i1 < . . . < in < (2|T|)+, and dw := ci1ci2 . . . cin then

|= 1[dw, b1] / w.

[Why? If w let k be such that = ik, so (ci1 , . . . , cik1) (pr)(C) by7 and cik+1 . . . cin (p r)(C) hence dw = (ci1 . . . cik)cik(cik+1 . . . cin) /

(p r)(C) by 9.Second, if / w by 8 as {ci : < n} is included in the subgroup(p r)(C).]

So we get a contradiction to T dependent hence clause () holds. Also clause ()follows by the following observation:

Observation. If r(x) R and |r(x)| then (p(C) : (p r)(C)) 2 (except finitewhen is finite).

Proof. 1) If is finite then by compactness. If is infinite then without loss ofgenerality r is closed under conjunctions. Let r = {i(x, b) : i < }, b is possiblyinfinite.

For each i < let u ord be such that > |u| > (p(C) : (p r)(C)) leti,u = {p(x) : u} {i(xx

1 , b) : < from u}. So for some finite

ui u, i,ui is contradictory so i,ni is contradictory when ni = |ui|. It suffices to

use (2)+ (. . . ni . . . )i


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By definition of P and c1, for each (x, e) tp(c1, Dom(p)) there exists somep(x,e) P such that (x, e) p(x,e) and therefore some c(x,e) q(C) realizes(x, e). So tp(c1, dom(p)) q(x) is finitely satisfiable and is therefore realized bysome c

2. Thus tp(c

1, Dom(p)) = tp(c

2, Dom(p)), but c

1/

q(C) and c2

q(C) acontradiction.3) By the proof of part (1). 1.1


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2

Claim. [T dependent] Assume

(a) G is a A-definable semi-group with cancellation

(b) q(x, a) is a type, q(C, a) a sub-semi group.

Then we can find q and ai : i < and B such that

() < (|T|0)+

() tp(ai, A) = tp(a, A)

() q = {qi(x, ai) : i < }

() B {qi(C, ai) : i < } and |B| ||

() if a

realizes tp(a, A

) and B q(C, a

) then q

(C) q(C, a

).

Proof. We try to choose a, b by induction on < (|T|+0)+ such that

(a) a realizes tp(a, A)

(b) b / q(C, a)

(c) b realizes q(x, a) for <

(d) b realizes q(x, a) for < .

If we succeed we get contradiction as in the proof in 1. If we are stuck at some < (|T|0)+ then take ai : i < , B = {bi : i < }.


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8 SAHARON SHELAH

REFERENCES.

[BaSx76] John T. Baldwin and Jan Saxl. Logical stability in group theory. J.Austral. Math. Soc. Ser. A, 21:267276, 1976.

[HPP0x] Ehud Hrushovski, Yaacov Peterzil, and Anand Pillay. Definable groups,o-minimality and measure. preprint.

[Sh:F753] Saharon Shelah. Definable Groups for Dependent Theories.

[Sh 783] Saharon Shelah. Dependent first order theories, continued. Israel Journalof Mathematic, accepted. math.LO/0406440.

[Sh 863] Saharon Shelah. Strongly dependent theories. Israel Journal of Mathe-matics, submitted. math.LO/0504197.

[Sh:F705] Shelah, Saharon. Representation over orders of elementary classes.

Saharon Shelah- Minimal Bounded Index Subgroup for Dependent Theories

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