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I have this result which I do not understand.

This is the statement of the result:

Suppose G is a group acting on a finite set X with fixed point set A (i.e. elements of X which are mapped to themselves for all G). Let X1,...Xm be the (disjoint) orbits of G in X with |Xi| Greater than or equal to 2, and let Hi be the stabilizer of a point in Xi for each i ∈ {1,...m}. Then each Hi has finite index [G : Hi] in G and m |X| =|A| + sum over i ∈ {1,...m} of [G : Hi].

So here are my thoughts. I think that this is equal to the size of X because they are summing over all the orbits.

|A| Is the sum of orbits of size 1 (as if all elements of G stabilise x, there is one element in the orbit, x).

For the second term in the sum, I believe that this is summing over all orbits of size greater than 2. So as there are m orbits in total, then this means that [G:Hi] is the size of each orbit (greater than or equal to 2). This is the bit I don’t get.

We know from the orbit stabilizer theorem that each orbit Xi is isomorphic to all the left cosets of Hi in G. But does this imply the thing I do not understand? I am struggling to visualise this. I understand that in an orbit, each element of the orbit uniquely defines a left coset of the stabilizer of an element.

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Let $X_i$ be one of the orbits mentioned in your question.

Moreover, let $p \in X_i$ and $H_i$ be the stabilizer of $p$, as in your question.

Since $p \in X_i$, we have $X_i = Gp$.

Now, the orbit-stabilizer theorem says that the map

$$ \phi: G/H_i \rightarrow X_i, \quad gH_i \mapsto gp $$

is a bijection between the set of cosets $G/H_i$ and the orbit $X_i = Gp$.

From the assumption that $X$ is finite, we get that $X_i \subseteq X$ is finite, and also

$$ |G/H_i| = [G : H_i ] < \infty. $$

(You already mentioned in your question that $H_i$ has finite index in $G$.)

But then $\phi$ is a bijection between two finite sets. So those sets must have the same size. So we have

$$ [G : H_i] = |G/H_i| = X_i. $$

By the way, this argument also works for orbits of size $1$.

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  • $\begingroup$ Thanks, I get it now. I wasn’t using OST correctly I think. $\endgroup$
    – jacob
    Nov 6, 2018 at 7:42

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