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Burnside's Lemma / Counting Formula says that the number of orbits of an action is equal to the average number of fixed points of the acting permutations. In my case, I'm particularly interested in the sizes of the orbits themselves for a particular action.

Is there a result as general as Burnside's Lemma but that deals with the sizes of orbits? Or, more specifically, does the following setup seem familiar to anyone?

Additional Motivation: The motivation for my question is something like the following.

Suppose we have a set of $m$ sharply transitive permutations $\Pi = \{\pi_1, \ldots, \pi_m\}$ on a set $X$. Here I mean that for $x,y \in X$, there is a unique $\pi \in \Pi$ such that $\pi(x) = y$.

The permutations cooperate "nicely" in the following sense. Let $A$ be an $m \times m$ matrix with off-diagonal entries drawn from $\{1, \ldots, m\}$ such that for $i \neq j$ whenever $\pi_i (x) = \pi_j (y)$, it follows that $\pi_j (x) = \pi_{A(i,j)} (y)$.

Let $G$ be the permutation group generated by $\Pi$. In the general case I'm working on, I really only know $A$, and I'm shooting for a statement of the form: "If $A$ has a certain property, then $|G(x)|$ is divisible by $m$."

The origins of the matrix $A$ while of course essential to proving anything like this statement remain sufficiently messy that I'd rather not get into it.

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Knowing the size of every orbit is the same as knowing the size of every stabilizer. I don't know if one can say more than this in general; if I ever needed to find the sizes of the orbits of an action I would do it "by hand." Is there a specific problem which motivates this question? –  Qiaochu Yuan Jul 26 '10 at 6:02
I've added some motivation in the question. Yes, in specific cases, I'm able to work out the sizes "by hand" but so far the calculations aren't particularly revealing for how the general case should go. –  user2971 Jul 26 '10 at 14:14

1 Answer 1

up vote 2 down vote accepted

Maybe you're after the Orbit-Stabiliser Theorem. Let $G$ be a group that acts on a set $X$ and let $x \in X$. Then $|G|=|G_x||G(x)|$ where $G_x$ is the stabliser of $x$ in $G$ and $G(x)$ is the orbit of $x$.

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I suspect you're right that this is as good as I'm going to get given the generality of my original question. Thanks. –  user2971 Jul 26 '10 at 14:15

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