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In this question the answer by Alex Kruckman gives a standard argument using the compactness theorem for why a torsion group with elements of arbitrary large order is elementarily equivalent to a nontorsion group. Paraphrasing slightly:

Let $G$ be a torsion group containing elements of arbitrarily large order. Let $c$ be a new constant symbol, and consider the theory $T' = \text{Th}(G)\cup \{c^n \neq e\mid n=1,2,...\}$ (where $c^n$ is an abbreviation for the product of $n$ copies of $c$). By compactness, this theory has a model, which is a non-torsion group elementarily equivalent to $G$ (given a finite subset of $T'$, take $G$ and interpret $c$ as an element of large enough order).

I am having trouble seeing why the resulting group, call it $H$, is elementarily equivalent to $G$ (ignoring the new constant $c$ so that $G$ and $H$ are two structures in the same first order language $\mathcal L$). Clearly every sentence true in $G$ is true in $H$ since $H$ is a model of $\text{Th}(G)$. For the converse, given a sentence $\phi$ in $\mathcal L$ holding in $H$, the constant $c$ is not in the language $\mathcal L$, so does not appear in $\phi$. But why does $\phi$ also hold in $G$?

PatrickR
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1 Answers1

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We can use negation: by definition of $\models$, if $\varphi$ is a sentence true in $H$ but not in $G$ then $\neg\varphi$ is a sentence true in $G$ but not in $H$. You already know that there aren't any of the latter, therefore there can't be any of the former!

Note that this uses almost nothing about first-order logic specifically. The argument above applies to any logic which admits negation, in the sense that for every sentence $\varphi$ of the logic there is a sentence $\psi$ whose class of models is exactly the class of non-models of $\varphi$. Not every logic admits negation - some natural and important examples include existential second-order logic and equational logic (which doesn't even admit "and" and "or"!) - but in my experience naturally-occurring logics tend to.

Noah Schweber
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