Group action on a subset having trouble with this problem. It's homework. We're given a finite set $S$ on which a finite group $G$ acts on transitively. If $U$ is a subset of $S$, I'm supposed to show that the subsets of $gU$ cover $S$ evenly. By evenly, I mean that each $s\in S$ is in the same number of sets $gU$.
Things I know are that for any $g,g'\in G$, $gU$ and $g'U$ both have order $|U|$. I also know that, since the operation is transitive, there is only one orbit (and I suspect this is important). 
I also noticed that, when $|U| = 1$, this property is just the transitivity of the group action. I'm just having trouble generalizing this to where the sets overlap (ie, $s\in S$ is in more than one set $gU$.)

Let $X_s = \{ gU \;|\; g \in G \;\mathrm{and}\; s \in gU \}$ (this is the set of all "$gU$" with $s \in gU$).
Since the action on $S$ is transitive, if you pick two elements, says $s,t \in S$, there exists some $x \in G$ such that $x \cdot s = t$. 
Now you just need to show that $\varphi:X_s \rightarrow X_t$ defined by $gU \mapsto (xg)U$ is a bijection.
