Automorphisms of recursively saturated models of arithmetic

Kaye, Robert; Kossak, Roman; Kotlarski, Henryk

doi:10.1016/0168-0072(91)90098-7

Cited by 39 publications

(49 citation statements)

References 18 publications

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“…In Section 4 we consider a more refined extendibility question: Suppose M and N are countable and recursively saturated, M ≺ N and f is an automorphism of M which can be extended to an automorphism of N ; can f be extended to an automorphism g such that g moves all points in N \ M? We give an answer to this question in the case when N is an elementary end extension of M. The result generalizes a theorem from [8], which is a special case for f = id.…”

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confidence: 55%

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More on extending automorphisms of models of Peano Arithmetic

Kossak

Kotlarski

2008

Fund. Math.

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Abstract. Continuing the earlier research [Fund. Math. 129 (1988) and 149 (1996)] we give some information about extending automorphisms of models of PA to cofinal extensions.In recent work on automorphisms of recursively saturated models of PA two themes stand out. One is the classification of conjugacy classes of single automorphisms, the other is the classification of subgroups of the automorphism group. Many results in both areas depend on information on how automorphisms of a recursively saturated model can be extended from the model to an elementary extension.Let M be a countable model of Peano Arithmetic, PA, and let N be its cofinal extension. If f is an automorphism of M, g is an automorphism of N and f ⊆ g, then for each a ∈ N , there is b ∈ N such that f (M ∩ a) = M ∩ b. This b is, of course, g(a). Here and elsewhere in the paper we identify elements of models of PA with the sets they code. If for each a ∈ N , there is b ∈ N such that f (M ∩ a) = M ∩ b, we say that f sends coded sets to coded sets. Thus, if f has an extension to an automorphism of N , then both f and f −1 send coded sets to coded ones. We showed in [9] that if M and N are countable and recursively saturated, N is an end elementary extension of M, M = inf{(a) n : n < ω} for all a ∈ N , and f is an automorphism of M such that f and f −1 send coded sets to coded ones, then f can be extended to an automorphism of M. In [10] we showed that the restriction on the type of end extension in the above result is essential. This, more or less, settles the problem of extending automorphisms of recursively saturated models to elementary end extensions. In [10] we considered extendability of automorphisms to cofinal extensions. This is a harder problem.2000 Mathematics Subject Classification: 03C62, 03C50.

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confidence: 55%

“…M[a]. The moving gaps lemma states that only the trivial automorphism of a countable recursively saturated model of PA fixes all gaps setwise (see Section 3 in [8] or Section 5 in [11]). The original proof of the lemma uses a type MG(x, y) with the following two properties.…”

Section: Automorphisms That Do Not Extendmentioning

confidence: 99%

More on extending automorphisms of models of Peano Arithmetic

Kossak

Kotlarski

2008

Fund. Math.

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“…The results presented in this paper suggest the question that is dual to the one asked in [8] and could have just as easily been asked there. The analogous question for countable, arithmetically saturated models is also open.…”

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confidence: 83%

“…There are several ways to prove this. One way is to use [8,Coro. 5.4] that asserts: If M 0 is recursively saturated, then M 0 is arithmetically saturated iff there is g ∈ Aut(M 0 ) and an open H < Aut(M 0 ) such that for all f ∈ Aut(M 0 ), f −1 gf ∈ H. We leave it to the reader to complete this proof.…”

Section: Lemma 411mentioning

confidence: 99%

“…appeared in [8], it has been of interest to determine to what extent (the isomorphism type of) the group Aut(M) of all automorphisms of a countable, recursively saturated model M of Peano Arithmetic determines (the isomorphism type of) M. It was proved in [8] that whenever both M and N are countable, recursively saturated models of PA and exactly one of them is arithmetically saturated, then Aut(M) and Aut(N ) are not isomorphic as topological groups. In 1994, Lascar [16] proved that countable, arithmetically saturated models of PA have the small index property, and that result then implied that Aut(M) ∼ = Aut(N ) as abstract groups.…”

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Automorphism Groups of Countable Arithmetically Saturated Models of Peano Arithmetic

Schmerl¹

2015

J. symb. log.

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Abstract. If M, N are countable, arithmetically saturated models of Peano Arithmetic and Aut(M) ∼ = Aut(N ), then the Turingjumps of Th(M) and Th(N ) are recursively equivalent.Since 1991, when the question Are there countable, recursively saturated models M, N of PA such that Aut(M) ∼ = Aut(N ) (as abstract groups)? appeared in [8], it has been of interest to determine to what extent (the isomorphism type of) the group Aut(M) of all automorphisms of a countable, recursively saturated model M of Peano Arithmetic determines (the isomorphism type of) M. It was proved in [8] that whenever both M and N are countable, recursively saturated models of PA and exactly one of them is arithmetically saturated, then Aut(M) and Aut(N ) are not isomorphic as topological groups. In 1994, Lascar [16] proved that countable, arithmetically saturated models of PA have the small index property, and that result then implied that Aut(M) ∼ = Aut(N ) as abstract groups. This gave the first positive answer to the above question. A neater way, in which the use of the small index property is masked, that automorphism groups distinguish those models that are arithmetically saturated from the other countable recursively saturated ones was obtained the next year in [13, Coro. 3.9] (or see [15, Th. 9.3.10]): If M is countable and recursively saturated, then M is arithmetically saturated iff the cofinality of Aut(M) is uncountable. Finally, we mention that Kaye's Theorem [7] (see §1) characterizing the closed normal subgroups of Aut(M), which appeared in the same volume [9] as did Lascar's Theorem, yields that whenever M, N are countable, arithmetically saturated models and M is a model of True Arithmetic (TA) while N is not, then Aut(M) ∼ = Aut(N ).

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