Can a finite group act transitively on an infinite set? I had a question on my algebra exam, asking me to show that in case of a transitive action of a finite group $G$ on a set $X$, $|X|$ divides $|G|$. In case of a finite set, this is trivial, so I tried to see if I could come up with a possible counterexample with $X$ infinite (I failed to do so). Intuitively I feel like an infinite $X$ is "too big" for a finite group to act transitively on, but how would one show this formally?
 A: Let $y$ be a fixed element of $X$. For any element $x\in X$, transitivity implies there exists some $g\in G$ such that $g\cdot y=x$. This induces a injective map from $X\to G$, which is a contradiction.
A: Actually, if $X$ is empty, any group acts transitively on $X$, but $\lvert X \rvert = 0$ cannot divide $\lvert G \rvert$, so the claim might actually be false (depending on whether or not you allow $G$-sets to be empty).
Anyway, suppose $G$ is a finite group acting transitively on a nonempty set $X$. Since $X$ is nonempty, we may pick some $x \in X$. Then define $a : G \to X$ by $a(g) = g \cdot x$. By definition of transitive action, $a$ is surjective. Thus, $\lvert G \rvert \geq \lvert X \rvert$, so $X$ is finite.
A: Given a fixed, arbitrary $x\in X$, transitivity means that the set
$$
Gx = \{gx\mid g\in G\}\subseteq X
$$
is equal to $X$. However, this set clearly cannot be larger than $G$.
A: For a sort of a sledgehammer here, it falls out of the orbit-stabilizer theorem.  For the "size" of the orbit of any element is the index of the stabilizer: $[G:G_x]=|Gx|$.  But $[G:G_x]\le|G|$.  That'll crack this nut.
A: If $G$ acts transitively on $X$, then, given $x\in X$, for every $y\in X$ there is some $g\in G$ such that $y=g\cdot x$. This is precisely the condition for a surjection $f_x\colon G\longrightarrow X$ to exist. Therefore the cardinality of $G$ is at least equal to the one of $X$.
(Incidentally, since by group action's axioms distinct $x$'s induce distinct $f_x$'s, there are $|X|$ of such surjections.)
