MINIMAL BOUNDED INDEX SUBGROUP FOR DEPENDENT THEORIES

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 3, March 2008, Pages S (07) Article electronically published on November 9, 2007 MINIMAL BOUNDED INDEX SUBGROUP FOR DEPENDENT THEORIES SAHARON SHELAH (Communicated by Julia Knight) Abstract. For a dependent theory T,inC T for every type definable group G, the intersection of type definable subgroups with bounded index is a type definable subgroup with bounded index. 0. Introduction Assume that T is a dependent (complete first-order) theory, C is a κ-saturated model of T (a monster), G is a type definable (in C) groupinc (of course we consider only types of cardinality < κ). A type definable subgroup H of G is call bounded if the index (G : H) is< κ. We prove that there is a minimal bounded definable subgroup. The first theorem on this line for T stable is due to Baldwin and Saxl [BaSx76]. Recently Hrushovski, Peterzil and Pillay [HPP0x] investigated definable groups, o-minimality and measure. In an earlier work on definable subgroups in o-minimal T in Berarducci, Otero, Peterzil and Pillay [BOPP05] the minimal type-definable bounded index theorem and more results are proved for o-minimal theories. Hrushovski, in a lecture at the Hebrew University, mentioned that he, Peterzil and Pillay had observed the main result of the current paper, but assuming in addition the existence of an invariant measure on the group in question, and Hrushovski asked if the measure assumption could be removed. So we answer it positively. The current version of their paper [HPP0x] includes an exposition of our proof. Recent works of the author on dependent theories are [Sh:783] (see 3, 4 on groups) [Sh:863] (e.g., the first-order theory of the p-adics is strongly 1 dependent but not strongly 2 dependent, see end of 1; on strongly 2 dependent fields, see 5) and [Sh:F705]. This work is continued in [Sh:F753] (getting mainly a parallel result for G abelian and L, κ -definable subgroups). I thank Aviv Tatarski and Ilay Kaplan for their very careful checking. Received by the editors January 22, 2006 and, in revised form, July 19, 2006, July 30, 2006, and October 19, Mathematics Subject Classification. Primary 03C45; Secondary 03C60. The author would like to thank the Israel Science Foundation for partial support of this research (Grant No. 242/03), Publication 876. I would like to thank Alice Leonhardt for the beautiful typing c 2007 American Mathematical Society

2 1088 SAHARON SHELAH Theorem. Assume T is a dependent (complete first-order) theory, C a κsaturated model for T. We consider types of cardinality < κ. 1) If below holds, then: () q(c) isasubgroupofp(c), () q(c) is of index < κ, (γ) essentially q(x)\p(x) is of cardinality λ := T ℵ 0 (i.e., for some q (x) q(x) of cardinality λ, q(x) is equivalent to p(x) q (x)), (δ) we can strengthen (), () to q(c) isasubgroupofindex 2 λ, where (a) p(x) is a type such that p(c) is a group which we call G (with some definable operations xy, x 1 and the identity e G which is constant here), (b) q(x) =p(x) {r(x) :r(x) R}, where (c) R = {r(x) :r(x) atypesuchthat(p r)(c) isasubgroupofp(c) of index < κ}. 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 r i (x) R for i<λ +, then for some <( T ℵ 0 ) + we have (p(x) {r i (x) : i<})(c) =(p(x) {r i (x) :i<λ + })(C). Proof. 1) Note 1 R is closed under unions of < κ. 2 (a) If r(x) R,r (x) r(x) is countable, then there is a countable r (x) r(x) including r (x) which belongs to R. (b) If p(x) r(x) R and r(x) is closed under conjunctions and r (x) r(x) is countable, then we can find ψ n (x, b n )forn<ωsuch that () ψ n (x, b n ) r(x), () ψ n+1 (x, b n+1 ) ψ n (x, b n ), (γ) ψ n+1 (x, b n+1 ),ψ n+1 (y, b n+1 ) ψ n (x 1 y, b n ) ψ n (x 1, b n ) ψ n (xy, b n ), (δ) ψ n (x, b m+1 ) ϕ n (x, ā n ), where {ϕ n (x, ā n ): n<ω} list r, (ε) C = ψ n (e G, b n ), actually follows from clause (). [Why? Let r (x) ={ϕ n (x, ā n ):n<ω}(can use ϕ n =(x = x)). Without loss of generality, r(x) is closed under conjunctions and also r (x) is. Nowwechooseψ n (x, b n ) by induction on n<ωsuch that ψ n+1 (x, b n+1 ) ψ n+1 (y, b n+1 ) ψ n (xy 1, b n ) ϕ n (x, ā n ) ψ n (xy, b n ); notice that trivially e G ϕ n (C, ā n ) ψ n (C, b n ). Such a formula exists as (p(x) r(x)) (p(y) r(y)) ψ n (xy 1, b n ) ϕ n (x, ā n ) ψ n (xy, b n ). Now r (x) ={ϕ n (x, ā n ),ψ n (x, b n ):n<ω} is as required in clause (a), ψ n (x, b n ):n<ω in (b).] In the conclusion of Theorem 1.1, Clause () isobvious. Assume toward a contradiction that the conclusion () +(γ) fails. So we can choose (c,r ) by induction on <λ + such that 3 (a) c (p(x) {r : <})(C), (b) r = {ψn(x, b n):n<ω} q and b n b n+1, (c) r R, andψ n+1 (x, b n+1 ) ψ n(x, b n), (d) c does not realize r,infactc = ψ0 (c, b 0 ), hence c / q(c).

3 MINIMAL BOUNDED INDEX SUBGROUP FOR DEPENDENT THEORIES 1089 [Why? Arriving to, wetrytoletq = p(x) {r : <}, soq q and by cardinality considerations, q (C) is a subgroup of G of index < κ. So by our assumption toward a contradiction, q (x) q(x); hence there is r(x) R such that q (x) r,soq (x) ϑ (x, d )forsomeϑ (x, d ) r.letr (x) ={ψn(x, b n):n<ω} r(x) belongtorandbesuchthat ϑ (x, d )=ψ0 (x, b 0 ); it exists by 2() above. Without loss of generality, we can assume 4 8 : 4 e G is an individual constant, ȳn ȳn+1 and ψn+1(x, ȳn+1) ψn(x, ȳn)and (ψn+1(x 1, ȳn+1) ψn+1(x 2, ȳn+1)) (ψn(e G, ȳn) ψn(x 1 x 1 2, ȳ n)). [Why? As in the proof of 2 above, i.e., during the induction in the proof of 3,weuse 2 () andgetψn(x, ȳn), b n and without loss of generality b n b n+1, ȳ n ȳn+1 (as we can change the order and name the free variable and add dummy variables). Now we define ψn, (x, ȳn) by induction on n by ψ, 0 (x, ȳ0 )=ψ0 (x, ȳ0 ),ψ, n+1 (x, ȳ n+1) =ψn+1(x, ȳn+1) ( z)[ψm+1(z, ȳm+1) ψm(z, ȳm)] m n ( z 1,z 2 )[ψm+1(z 1, ȳm+1) ψm+1(z 2, ȳm+1) So clearly ψ, n m n ψ m(e G, ȳ m) ψ m(z 1 z 1 2, ȳ m)]. (x, ȳn):n<ω satisfies 4 and C = ( x)[ψn, (x, b n) ψn[x, b n)].sorenamingwearedone.] 5 ψn (x, ȳ n )=ψ n(x, ȳ n )andψ n+1 (x, ȳ n+1 )) ψ n (x, ȳ n ). [Why? By the pigeon hole principle.] 6 c ā : <λ + is an indiscernible sequence over Dom(p) whereā = 7 {b n : n < ω} b 0 ˆ b 1 ˆ b 2 ˆ...; note that by clause (b) of 3 we have b n = ā k n. [Why? By the Ramsey theorem and compactness.] If <<γ,thenc c 1 r γ(c). [Why? By the indiscernibility, without loss of generality, γ is infinite, so γ ω and c i : i<γ is an indiscernible sequence over Dom(p) b γ of elements of p(c) pairwise nonequivalent modulo the subgroup G γ =(p r γ )(C). Then we can extend it to c i : i< κ, an indiscernible sequence over Dom(p) b γ and arrive at < c c 1 / G γ c c 1 / G γ so c G γ : < κ are pairwise distinct (equivalently G γ c : < κ are pairwise distinct), a contradiction.] 8 c r (C) iff. [Why? Let c = c 2+1 (c 2 ) 1, r = r 2. So: (i) If <, then c (p r )(C), because c 2+1,c 2 belong to the subgroup (p r 2 )(C) by clause (a) of 3. (ii) If >,thenc belongs to (p r )(C) by 7. (iii) If =, thenc does not belong to (p r )(C) because:

4 1090 SAHARON SHELAH () it is a subgroup, () c 2+1 belongs to it by clause (a) of 3 and (γ) c 2 does not belong to it by clause (e) of 3. Let ā, 2 = ā 2+1, b n = b n retaining the same ψ s. Sowehaveobtainedan example as required in 8 (not losing the other demands).] 9 For some n<ωfor every, wehaveifd 1,d 2 (p r )(C), then d 1 c d 2 / ψ n (C, b n); without loss of generality, n =1. [Why? Fix. If this holds for some ψ n (, b n), by indiscernibility, renaming the ϕ i s, this is O.K. Otherwise for each n<ωthere are d n 1,dn 2 (p r )(C) such that C = ψ n (d n 1 c d n 2, b n ). By compactness for some d 1,d 2 (p r )(C) wehave = ψ n [d 1c d 2, b n] for every n<ω.sod 1c d 2 belongs to the subgroup (p r )(C), but also d 1,d 2 belongs to it; hence c belongs, a contradiction. Alternatively, note that n = 2 is O.K.: let c = d 1 cd 2 and assume toward a contradiction that c ψ 2 (C, b )) and let d 1 =(d 1 ) 1,d 2 =(d 2 ) 1, so clearly d 1,d 2 (p r )(C) ψ 2 (C, b 2 ). Now by 4 as d 1,c ψ 2 (C, b 1 )it follows that d 2 (c ) 1 ψ 1 (C, b 1 ). As d 1,d 2 (c ) 1 ψ 1 (C, b 1 )by 4 we have d 1(d 2 (c ) 1 ) 1 ψ 0 (C, b 0 ). But c = d 1 c d 2 ; hence c = d 1((d 2) 1 (c 1 ) 1 ) 1 = d 1(d 2 (c ) 1 ) 1, but d 1(d 2 (c ) 1 ) 1 ψ 0 (C, b 0 ) by the previous sentence, whereas c / ψ 0 (C, b 1 )by 3 (e), a contradiction.] 10 If w = {i 1,...,i n },i 1 <... < i n <λ +,andd w := c i1 c i2...c in G and <λ +,then = ϕ 1 [d w, b 1 ] / w. [Why? If w, letk be such that = i k,soc i1,...,c ik 1 (p r )(C) by 8 and similarly c ik+1...c in (p r )(C); hence d w =(c i1...c ik )c ik (c ik+1...c in ) / (p r )(C) by 9. Second, if / w, this holds by 8 as {c il : l<n} is included in the subgroup (p r )(C).] So we get a contradiction to T is dependent ; hence clauses (), (γ) hold. Also clause (δ) follows by the following observation: Observation: If r(x) R and r(x) θ, then(p(c) :(p r)(c)) 2 θ (except for being just finite when θ is finite). Proof. If θ is finite, then the proof follows by compactness. If θ is infinite, then without loss of generality, r is closed under conjunctions. Let r = {ϕ i (x, b) :i<θ}, where b is possibly infinite. Let u be a set of ordinals (< κ) such that κ > u > (p(c) :(p r)(c)). Now for each i<θ,letγ i,u = {p(x ): u} { ϕ i (x x 1, b) :<from u}. So for some finite u i u, Γ i,u is contradictory, so Γ i i,ni is contradictory, when n i = u i. It suffices to use (2 θ ) + (...n i...) i<θ (why? let c : <(2 θ ) + exemplify the failure and let ζ, = Min{i : = ϕ i (c c 1, b)}). This finishes the proof of part (1). We still need to prove 2), 3). 2) Let q (x) q(x) have cardinality T ℵ 0 andbesuchthatq(c) =(p q )(C); q (x) exists by part (1). Observe that every automorphism of C fixing Dom(p) maps p(c) onto itself and therefore maps q(c) ontoitself.

5 MINIMAL BOUNDED INDEX SUBGROUP FOR DEPENDENT THEORIES 1091 It follows that if c 1,c 2 p(c) are such that tp(c 1, Dom(p)) = tp(c 2, Dom(p)), then c 1 q(c) if and only if c 2 q(c). Let P := {tp(b, Dom(p)) : b q(c)}, P(C) := {r(c) :r P}. Then by the above explanation, P(C) q(c). By definition, q(c) p(c), so they are equal. Let q = {r : r P}, sowehaveq(c) q (C). If they are equal, then we are done. Otherwise take c 1 q (C)\q(C). Without loss of generality, let ψ(x, d) q be such that = ψ(c 1, d). By definition of P and c 1,foreachθ(x, ē) tp(c 1, Dom(p)) there exists some p θ(x,ē) P such that θ(x, ē) p (x,ē) and therefore some c θ(x,ē) q(c) satisfies θ(x, ē). So tp(c 1, dom(p)) q (x) is finitely satisfiable and is therefore realized by some c 2. Thus tp(c 1, Dom(p)) = tp(c 2, Dom(p)), but c 1 / q (C) =q(c) and c 2 (p q )(C), a contradiction. 3) By the proof of part (1). 1.1 References [BaSx76] John T. Baldwin and Jan Saxl. Logical stability in group theory. J. Austral. Math. Soc. Ser. A, 21: , MR (53:10934) [BOPP05] Alessandro Berarducci, Margarita Otero, Yaa cov Peterzil, and Anand Pillay. A descending chain condition for groups definable in o-minimal structures. Annals of Pure and Applied Logic, 134: , MR (2006a:03052) [HPP0x] Ehud Hrushovski, Ya acov Peterzil, and Anand Pillay. Groups, measures, and the NIP. preprint, [Sh:F753] Saharon Shelah. Definable Groups for Dependent Theories. [Sh:783] Saharon Shelah. Dependent first order theories, continued. Israel Journal of Mathematic, accepted. math.lo/ [Sh:863] Saharon Shelah. Strongly dependent theories. Israel Journal of Mathematics, submitted. math.lo/ [Sh:F705] Shelah, Saharon. Representation over orders of elementary classes. The Hebrew University of Jerusalem, Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, Jerusalem 91904, Israel Current address: Department of Mathematics, Hill Center-Busch Campus, Rutgers, The State University of New Jersey, 110 Frelinghuysen Road, Piscataway, New Jersey

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