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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 $gU$.)

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    Have you proved the orbit-stabilizer theorem? This says that as a set with $G$-action, $S$ is identified with the coset space $G/H$ given the action by translations, where $H$ is the stabilizer of a point in $S$. So you really only have to think about the case of $G/H$.2011-11-14
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    Let $s, s' \in S$. Now, suppose that I have $gU$ containing $s$. Can you hit this with something in $G$ to obtain a translate of $U$ containing $s'$? Is this process reversible?2011-11-14
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    Yeah, right as I posted that I had the insight I needed on this mapping here! Thanks for putting it in writing, though.2011-11-14

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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.