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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 $N$?

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A friend of mine just showed me a great counterexample - what's the etiquette behind answering one's own question? –  user28111 Apr 2 '12 at 0:52
You can answer your own question and accept the answer. –  Rachel Apr 2 '12 at 6:12
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1 Answer

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 $Aut(\mathcal{N})\cong G$.

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