We call $G$ a torsion group if for every $g\in G$ there exists a natural $n>0$ such that $g^n=1$. We say that $G$ has finite exponent if the inverse quantification holds, i.e. there exists a $N>0$ such that $g^N=1$ for every $g\in G$.
Let $G$ be a torsion group, and assume that every elementary extension $G'\succ G$ is a torsion group. Show that $G$ has finite exponent using the Compactness theorem.
Any hint? I am confused because the Compactness theorem says that every finitely consistent theory is consistent, so I would find a model of the theory, and I cannot say if that model is actually $G$ or something else.
The argument outlined below is the "generic" Compactness Theorem argument, adapted to our particular situation.
Let $L$ be the following language:
(i) $L$ has a binary function symbol (multiplication), and a constant symbol $1$.
(ii) $L$ has a constant symbol $K_g$ for every element $g$ of $G$.
(iii) $L$ has an additional constant symbol $C$.
Let $T$ be the theory over $L$ that has the following axioms:
(A) All the sentences of $L$ that are true in $G$ when the constant symbols $K_g$ are given their natural interpretations.
(B) Sentences $C\ne 1$, $C^2\ne 1$, $C^3\ne 1$, and so on.
Suppose that $G$ does not have finite exponent. Then for any finite subset $F$ of the axioms of $T$, $F$ has a model. For any such finite set contains only finitely many of the sentences in (B). And since $G$ does not have finite exponent, there is for any $n$ an element $g$ of $G$ such that $g^k\ne 1$ for all $k$ between $1$ and $n$. If the "largest" axiom of (B) that is in $F$ is $C^n\ne 1$, then $g$ provides a suitable interpretation for $C$.
So by Compactness, if $G$ does not have finite exponent, then $T$ has a model. Any such model is (isomorphic to) an elementary extension $G'$ of $G$. But $G'$ is not a torsion group, contradicting the assumption that every elementary extension of $G$ is a torsion group.