Proof that there exist no finite axiomatic system with Compactness Theorem

Question

Say we would like to prove that the class of all infinite groups $(G, \circ, e)$ is not finitely axiomatizable by making use of the compactness theorem. We normally prove this by contradiction since we know the axiomatic system is defined by the union of $\Phi_{group}$ (the standard FO axiomatic system for groups) and $\Phi_{\infty}$ (in FO logic, there exist $x_1$ to $x_n$ and they are all pairwise non-identical and that for all $n \in \mathbb{N}$).

Assuming there exist a $\Phi_{infGrp} \subseteq FO(\tau)$ that finitely axiomatizes this class. Therefore, we can find a $\psi \in FO(\tau)$ that axiomatizes this class too, since the axiomatic system is finite.

This implies that $\Phi_{infGrp} \cup \{\neg\psi\}$ is not satisfiable. We then go on to find a arbitrary, finite subset $\Phi_0 \subseteq \Phi_{infGrp} \cup \{\neg\psi\}$, and show that we can find a structure that is a model of this subset by making the said structure to have more elements than the formula with a largest $m \in \mathbb{N}$ in $\Phi_0$. We then finally say that since structure $\mathbb{A} \models \Phi_0$, by virtue of compactness theory, $\Phi_{infGrp} \cup \{\neg\psi\}$ is satsifiable.

The question is... how can $\Phi_{infGrp} \cup \{\neg\psi\}$ be satisfiable when $\neg\psi$ contains statements that negate the properties of a group. Wouldn't every structure (that is a group) not be a model for $\Phi_{infGrp} \cup \{\neg\psi\}$?

Answer 1

There is nothing in $\lnot\psi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $\alpha_1,\dots,\alpha_m$, the usual axioms of group theory, plus additional special axioms $\beta_1,\dots,\beta_n$, which in conjunction with the group axioms force the group to be infinite. Then $\psi$ is the sentence $$\alpha_1\land\cdots\land \alpha_m \land \beta_1\land\dots\land \beta_n.$$ The sentence $\lnot \psi$ is then the sentence $$\lnot\alpha_1\lor\cdots\lor \lnot\alpha_m \lor \lnot\beta_1\lor\dots\lor \lnot\beta_n.$$ This is a disjunction. No group property needs to fail.

Answer 2

The following may be a more intuitive proof:

Let $\mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.

If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $\psi$ such that $K$ is the set of the all $\mathcal{L}$-structures $M$ such that $M \models \psi$. Since the group axioms are finite, let $\Phi$ be the conjunction of all the group axioms. Then $ \Phi \wedge \neg \psi$ axiomatizes the class of all finite groups. Let $\mathcal{L}' = \mathcal{L} \cup \{c_i : i < \omega\}$. $\mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = \{\Phi \wedge \neg \psi\} \cup \{c_i \neq c_j : i \neq j\}$. If $\Delta \subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $\Delta$. Take any group $G$ of size $n$. Make $G$ into a $\mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $\mathcal{L}'$-structure satisfies $\Delta$. So $\Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G \models T$. Clearly $G$ must be an infinite $\mathcal{L}'$-structure which satisfies $\Phi \wedge \neg \psi$. The reduct of $G$ back to $\mathcal{L}$ is then an infinite group satisfying $\Phi \wedge \neg \psi$. This is a contradiction since $\Phi \wedge \neg \psi$ was suppose to be an axiomatization of finite groups.