If an infinite group $G$ acts freely on two sets of same cardinality $> |G|$, then the sets are bijective via an action preserving bijection?

Question

Let $G$ be an infinite group. Let $X,Y$ be two sets on which $G$ acts freely.

(An action of a group $G$ on a set $X$ is called free iff $G_x:=\{g\in G : gx=x\}$ is singleton for every $x\in X$ ).

If $X$ and $Y$ are bijective as sets and $|X| >|G|$ , then is there a bijection $f: X \to Y$ such that $f(gx)=gf(x), \forall g\in G, \forall x\in X$ ?

Answer 1

Still no - consider $\mathbb{Z}$ acting on $\mathbb{Q}$ by translation versus $\mathbb{Z}$ acting on $\mathbb{Z}$ by translation.

The obstacle is the number of orbits. Specifically, if $G$ acts on $X, Y$ freely and there are the same number of orbits in each case, then there is an action-preserving bijection between $X$ and $Y$; but if the number of orbits is different, then there isn't. The point is that the cardinality of the sets isn't enough information in general to determine the number of orbits. (I give more details at my answer to your other question.)

---

The answer to your edited question is yes. If $G$ acts on $X$, and $X$ is infinite and of cardinality $>\vert G\vert$, then the number of orbits is exactly $\vert X\vert$. This follows from a quick exercise in cardinal arithmetic: that $\kappa\times\lambda=\max\{\kappa,\lambda\}$ for $\kappa,\lambda$ infinite cardinals.

Now by the above,this means that if $G$ acts freely on infinite sets of cardinality $>\vert G\vert$, those sets are in action-preserving bijection.

Note that at no point does $G$ need to be infinite!

Answer 2

Here's a counterexample: $$G = \mathbb{Z}$$ $$X = \mathbb{Z}$$ $$Y = \frac{1}{2} \mathbb{Z} = \{\frac{p}{2q} \mid p,q \in \mathbb{Z}, q \ne 0\}$$ and the action of $G$ on both sets is simply addition: $$n \cdot x = n + x \quad\text{for all $x \in X$} $$ $$n \cdot y = n + y \quad\text{for all $y \in Y$} $$ Perhaps you have overlooked the concept of transitivity of the action: $G$ acts transitively on $X$ but not on $Y$.

Answer 3

Let $G$ be the multiplicative group of non-zero reals. Let $X=\Bbb R\setminus\{0\}$ and $Y=\Bbb R^2\setminus\{0\}$. Let $G$ act on $X$ and also on $Y$ by multiplication: $gx=gx$. Both actions are free, but $G$ acts transitively on $X$ but not on $Y$.