= Solution
\b[No <finite axiomatization> exists.] Write $\Gamma$ for the <axioms> just constructed and suppose a <finite set> $\Delta$ of <first-order sentences> had exactly the same <first-order models>. Its <logical conjunction> $\delta$ is a single <first-order sentence>, and $\Gamma\models\delta$. By the <compactness theorem> there is a finite <subset> $\Gamma_0\subseteq\Gamma$ with $\Gamma_0\models\delta$: otherwise $\Gamma\cup\{\neg\delta\}$ would be finitely satisfiable and hence satisfiable, a contradiction.
Only finitely many of the integers $n$ in the torsion-free scheme occur in $\Gamma_0$. Choose a <prime number> $p$ larger than all of them. The <cyclic group> $\mathbb Z/p\mathbb Z$ satisfies all the <group axioms> and each of these finitely many conditions from the <torsion-free abelian group axiom scheme>: <multiplication> by an <integer> $1\le n<p$ is invertible modulo $p$, so $nx=0$ implies $x=0$. It consequently satisfies $\Gamma_0$, then $\delta$, and therefore $\Delta$.
But its nonzero elements have <finite order> $p$, so it is not a <torsion-free group>, contradicting the assumed description of the <first-order models> of $\Delta$. This establishes the <nonfinite axiomatizability of torsion-free abelian groups>, ruling out every <finite set> of defining <first-order sentences>, not just finite <subsets> of the particular scheme in part (c).
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