# For an infinite group $G$, is the collection of all transitive $G$-sets first order elementary?

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 ?

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

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 $Y$.
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.