Let $G$ be an infinite group with identity element $e$. Consider the first order theory of faithful $G$-sets (sets along with a faithful $G$ action) $(X,.)$ which along with the axioms of set theory, has the following set of axioms : For every $g\in G$, consider the unary function symbol $g.$ and collect the sentences $\forall x (g.(h.x)=(gh).x)$; $\forall x (e.x=x)$, and for every non-identity element $g$ in $G$, collect the sentences $\exists x (g.x\ne x)$.

Is this first order theory of faithful $G$-sets complete ? Is it model complete ?


No, certainly not; faithfulness is a very weak assumption. For instance, if $X$ is any (nonempty) free $G$-set (e.g., $G$ acting on itself by translation), then you can add a fixed point to $X$ to get another faithful $G$-set $Y$. Then $X$ and $Y$ do not have the same theory; for instance, for any $g\neq e$, $X\vDash \forall x (g.x\neq x)$ and $Y\vDash \neg\forall x (g.x\neq x)$

  • $\begingroup$ Note that this answer answers both completeness and model-completeness as $X$ embeds in $Y$ but clearly not elementarily $\endgroup$ – Max Apr 30 '18 at 6:05
  • $\begingroup$ Note also that the OP explicitly included the axioms of set theory as part of the theory under consideration. That gives an alternative proof of incompleteness. $\endgroup$ – Andreas Blass May 14 '18 at 15:14

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