Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 16H / e

A model of $T_G$ consists of a model $M\models T$ and automorphismssuch that ${\vartheta }_g{\vartheta }_h={\vartheta }_{gh}$ and ${\vartheta }_g\ne {\vartheta }_h$ whenever $g\ne h$. Thus $g\mapsto {\vartheta }_g$ is an injective homomorphism $G\hookrightarrow \mathrm{Aut}(M)$. Conversely, any such embedding supplies interpretations of the $f_g$ and hence a model of $T_G$.

Therefore $G$ is $T$-good exactly when $T_G$ has a model. By part (b), this is equivalent to consistency of $T_G$, and hence

Suppose $T_G$ is inconsistent. By the compactness theorem, some finite subset $\Delta \subseteq T_G$ is inconsistent. Only finitely many added symbols $f_g$ occur in $\Delta$; let $X\subseteq G$ be the finite set of their indices. Every sentence of $\Delta$ belongs to the sublanguage $L_X$, soConsequently $T_G\cap L_X$ is inconsistent.

If $T_G\cap L_X$ is inconsistent for some finite $X\subseteq G$, then its superset $T_G$ is inconsistent. Together with parts (i) and (ii), this proves the equivalence of all three conditions.