# Similar result to Burnside's lemma

I have an exercise scribbled down, and I am not sure what it is asking. It is somewhat similar to Burnside's lemma.

We have a finite group $G$ acting on a set $X$. For each $g \in G$, let $X^g$ denote the set of elements in $X$ fixed by $g$.

$$\sum_g |X^g|^2 = |G| \cdot \text{(number of orbits of a stabilizer)}$$

I am not sure what it means by "orbit of a stabilizer". I am guessing that it refers to the action of $G$ on cosets of a stabilizer by multiplication. But this really doesn't make sense to me since this action is transitive and the orbit is just the entire set.

Does anyone know of such an exercise and can someone explain what the precise statement of the problem should be?

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Perhaps you mean $\sum_{g} |X_g|^2 = |G| \cdot [G: \text{Stab}(x)]$? –  PEV Feb 20 '11 at 4:28
You define the notation $X^g$, and you use the notation $X_g$. Are the two supposed to be one and the same? –  Arturo Magidin Feb 20 '11 at 4:36

## 2 Answers

Consider $G$ acting on $X^2$ component-wise. The number of elements of $X^2$ fixed by $G$ is exactly $|X_g|^2$. So the average of $|X_g|^2$ is the number of orbits of $G$'s action on $X^2$.

This "exercise" (replacing $2$ with an arbitrary natural number) can be used to show that the distribution of the number of fixed points of a permutation is roughly Poissonian. It has lots of other uses in enumeration theory.

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The "number of orbits of a stabilizer": given $x \in X$, the stabilizer $G_x$ acts on $X$. The number of orbits of the stabilizer means the number of orbits of $G_x$ on $X$.

If you add the extra condition that the action of $G$ is transitive on $X$. If that is so, you first not that $\Sigma_{g \in G_x} |X_g| = \Sigma_{g \in G_y} |X_g|$, for every $x, y \in X$.

Since the action of $G$ is transitive on $X$, then $|G|=|X||G_x|$, implying that $$k|G|=k|X||G_x|=(|X|)(k|G_x|)=\Sigma_{x\in X} \Sigma_{g \in G_x} |X_g| = \Sigma_{g\in G} \Sigma_{x \in X_g} |X_g| = \Sigma_{g\in G}|X_g|^2 ,$$ where $k$ is the number of orbits of $G_x$ on $X$.

Now, if the action of $G$ on $X$ is not transitive, the stabilizers can have different number of orbits, depending on the point that is stabilized. For example if $1, 2, 3, 4$ represent the vertices of a regular square, and $G$ consists of the identity and the reflection $R$ in the line through $1$ and $3$, then $\Sigma_{g\in G}|X_g|^2 = |X_{id}|^2 + |X_R|^2 = 4^2+2^2 =20$. On one hand the stabilizer of 1, $G_1$, has three orbits on the vertices (the one containing 1, the one containing 3 and the one containing 2 and 4), and hence ("number of orbits of the stabilizer")|G|=3*2=6.

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I think you want the number $k$ of orbits of the stabilizer to be the number of orbits of $G_x$ on $X$ rather than on $X \setminus \{ x \}$. –  Derek Holt Feb 20 '11 at 12:14
True! Sorry. I've changed it ;-) –  Isabel Feb 20 '11 at 16:52