4
$\begingroup$

I think this is probably a very simple question, but I've been puzzling over it for a while and can't seem to get anywhere.

Suppose $M$ is a structure, $\alpha$ is an automorphism of $M$, and $N$ is an elementary extension of $M$. Does $\alpha$ necessarily extend to an automorphism of $N$?

It seems to me that the answer should be no, but I can't construct a counterexample. In case the answer is no, is there some natural condition on $N$ (or on how $M$ sits inside $N$) that guarantees that each automorphism of $M$ extends to an automorphism of $N$?

$\endgroup$
1
  • $\begingroup$ A friend of mine just showed me a great counterexample - what's the etiquette behind answering one's own question? $\endgroup$ – Noah Schweber Apr 2 '12 at 0:52
3
$\begingroup$

The answer is "no," and there is a straightforward (class of) counterexample(s). Let $A\prec B$ with $A\not\cong B$, and consider the structure $M$ consisting of two copies of $A$ "side-by-side" (say, with an equivalence relation with two equivalence classes, each of which is a copy of $A$). Now let $M'$ be the structure gotten by replacing exactly one of the copies of $A$ with a copy of $B$, and let $\alpha$ be the automorphism of $M$ gotten by swapping the copies of $A$. Then clearly $\alpha$ does not extend to an automorphism of $M'$.

For a more difficult, but possibly more interesting answer to the question: in the paper "Automorphism groups of models of Peano Arithmetic" (Journal of Symbolic Logic vol. 67 no. 4), J. Schmerl shows that every group $G$ which is isomorphic to a subgroup of the automorphism group of some linear order has the property that every $\mathcal{M}\models PA$ has an elementary extension $\mathcal{N}$ with $Aut(\mathcal{N})\cong G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.