Let $G$ be an infinite group with identity element $e$.

We can consider the first order theory of $G$-spaces i.e. pairs $(X,.)$ where $.$ is an action of $G$ on $X$ as the extension of the usual set theory along with unary symbol $g.$ for each $g\in G$ such that $\forall x (g.(h.x)=(gh).x)$ and $\forall x (e.x=x)$.

Now consider the class $\mathcal C$ of all transitive $G$-spaces. This is a sub-collection of the class of all $G$-spaces (i.e. the models of the theory of $G$-sets) .

Is $\mathcal C$ elementary ? Is the complement of $\mathcal C$ in the collection of all $G$-sets, elementary ?

  • 1
    $\begingroup$ As usual, the compactness theorem is your friend. $\endgroup$ Apr 28, 2018 at 21:21
  • $\begingroup$ @AlexKruckman: Yes I tried that but to no avail ... could you please elaborate ? $\endgroup$
    – user
    Apr 28, 2018 at 21:22

2 Answers 2


Remember that if $G$ acts transitively on $X$, we must have $\vert X\vert \le\vert G\vert$. This sort of size bound runs counter to the compactness theorem:

Since $G$ is infinite, the compactness theorem - in its incarnation as the upwards Lowenheim-Skolem theorem - tells us that in order for the class of transitive $G$-sets to be elementary, there must be transitive $G$-sets of arbitrarily large cardinality. But the point above explains why that can't happen.

There are two differences between this answer and YCor's. First is the absence/presence of ultraproducts; this is really superficial (remember that ultraproducts provide a proof of the compactness theorem!). The second difference, which is meaningful, is that YCor's answer uses an even stronger fact about transitive $G$-sets than the cardinality bound above: that no transitive $G$-set has a proper extension. That is, if $\emptyset\not=X\subsetneq Y$, $G$ acts on $X$ and $Y$, the action of $G$ on $Y$ extends the action of $G$ on $X$, and $G$ acts transitively on $X$, then $G$ does not act transitively on $Y$.

Arguing via this stronger fact gives in some sense a more complete answer to the question, since it really shows just how far off from satisfying compactness the class in question is. By contrast, cardinality arguments like the above are rarely as satisfying, but do have the advantage of being broadly applicable and easy to use.


No, it's not closed under ultraproducts. Just consider a nonprincipal ultrapower of $G$ with countable index set ($G$ being viewed as $G$-set under left multiplication). It contains an orbit given by constants, and is larger, so is not transitive.


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