I've been trying to come up with an example of two models $\mathcal{M,N}\models PA$ such that $\mathcal{M}\not\cong\mathcal{N}$, but $Aut(\mathcal{M})\cong Aut(\mathcal{N})$.

I know that for countable recursively saturated models this is still an open question (Automorphism Groups of Countable Arithmetically Saturated Models of PA, Schmerl, 2014), but I was wondering if there is an example for such models without the limitations of being countable and recursively saturated.

Any help would be appreciated!


1 Answer 1


There are a lot of rigid models of $\textsf{PA}$ -- models with no nontrivial automorphisms. For example, any finitely generated model of arithmetic is rigid (this is an easy application of Ehrenfeucht's Lemma, which says that if $a, b$ are in $\mathcal{M}$ are such that there is a Skolem term $t(x)$ with $\mathcal{M} \models t(a) = b$, then their types are different). So take $\mathbb{N}$ and any finitely generated elementary extension of $\mathbb{N}$ -- they will both be rigid but will not be isomorphic.

  • $\begingroup$ Thank you for your answer! Looking for material about rigid models, I stumbled upon the following result: For every regular cardinal $\kappa$, there is a rigid, recursively-saturated and $\kappa$-like model of Peano Arithmetic (Minimal Satisfaction Classes with an Application to Rigid Models of Peano Arithmetic, by Kossak and Schmerl). So even under the limitation of being recursively saturated there is such an example for non-isomorphic models of PA with trivial (and isomorphic) automorphism groups. $\endgroup$ Sep 5, 2016 at 12:12

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