Prove, or disprove the statement:
An infinite group $G$ can never act transitively on a finite set $X$.
What i know is that a group action $G \times X \to X$ is transitive if it possesses only a single group orbit, i.e., for every pair of elements $x$ and $y$, there is a group element $g$ such that $gx=y$.
On
Already @Yanko has given a very nice example but I will add something so that you can construct your own example.
Take any infinite group G acting on a finite set X. We know that every orbit is invariant under that action of G, so we can restrict our action on that orbit and that action will be transitive(why?). Of course this orbit is a finite set(as it is a subset of a finite set).
I will ended with a nice example here.
Example-: Consider an infinite group G with a finite index subgroup H. Then consider the action of G on co-sets of H. It will work(check)!
As mentioned by Lord Shark the Unknown (see comments) every infinite group acts on a set of size $1$.
If you require moreover that $X$ has cardinality greater than $1$ then you can still define a transitive action on $X$ by an infinite group.
Example: If $X=\{a,b\}$ we can define the action of $\mathbb{Z}$ on $X$ by
$$n.x = x$$ if $n$ is even, and $$n.x = \cases{ a & x=b \\ b & x=a}$$ if $n$ is odd.
I claim that this is a group action: The trivial element $0$ is even and it acts trivially. It is left to show that $g(hx)=(g+h)x$
If $n,m$ are even then $n+m$ is even and so $n(mx) = (n+m)x = x$.
If $n,m$ are odd then $n+m$ is even and we still have $n(m(x))=(n+m)x=x$.
If $n$ is odd and $m$ is even then $n+m$ is odd. In this case
$$n(m(a))=n(a)=b = (n+m)(a)$$ and $$n(m(b))=n(b) = a = (n+m)(b)$$
Finally if $n$ is even and $m$ is odd then $n+m$ is odd. In this case
$$n(m(a))=n(b)=b = (n+m)(a)$$ $$n(m(b))=n(a)=a=(n+m)(b)$$
Thus in all cases we have that $n(m(x))=(n+m)x$ and so this really defines a group action.
The action is transitive of course because $\{0.x,1.x\} = \{a,b\}$ for all $x\in\{a,b\}$.