# Burnside theorem?

Burnsides theorem

If $G$ has $t$ orbits on $\Omega$, then $t=\frac{1}{|G|} \sum_{g \in G} |\operatorname{fix}_{\Omega} (g)|$

It seems to be done by counting in two ways , then saying that the $G$-orbits $\Omega= \Delta_{1} \cup \Delta_{2} \cup\cdots \cup \Delta_{t}$.

However, don't see you can concludes for $g \in G$, $\operatorname{fix}_{\Omega}(g)=\operatorname{fix}_{\Delta_1} \cup \cdots \cup \operatorname{fix}_{\Delta_{t}}(g)$.

I just don't see what $\operatorname{fix}_{\Omega}(g)$ is doing. Can you please help me understand this proof?

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Here's a nice (albeit wordless) proof: en.wikipedia.org/wiki/Burnside's_lemma#Proof, where your $\Omega=X$, your $t=|X/G|$, and $fix_\Omega(g)=X^g$. It's using the orbit-stabilizer theorem (en.wikipedia.org/wiki/Orbit-stabilizer_theorem) and counting the number of pairs $(g,x)\in G\times X$ for which $g(x)=x$. – bgins Jan 20 '12 at 11:13

I will summarize the main idea but leave you some details. Let me know if you'd like those details as well.

The idea is to count the pairs $(g,\omega)$ where $g\in G, \omega\in \Omega$, and $g\omega = \omega$, i.e. the pairs (group element, element of $\Omega$ fixed by this group element), in two different ways. $\mathrm{fix}_{\Omega}(g)$ is notation for the set of elements of $\Omega$ fixed by a particular group element $g$. Here are the two ways of counting involved in the proof:

Way number 1: sum over the elements of the set $\Omega$; organize these elements by orbit to clean up the sum.

When we fix $\omega\in \Omega$, the set of group elements fixing $\omega$ is $\omega$'s stabilizer. So the sum looks like

$$\sum_{\omega\in\Omega} \text{size of}\;\omega\text{'s stabilizer}$$

It is convenient to break this sum down into orbits:

$$\sum_{\omega\in\Delta_1}+\sum_{\omega\in\Delta_2}+\dots+\sum_{\omega\in\Delta_t}$$

When it is broken down this way, it becomes possible to conclude that the total is $t|G|$. Can you see why? This is the proof's key step. Also, it's the only point in the proof where the decomposition $\Omega=\Delta_1\cup\Delta_2\cup\dots\cup\Delta_t$ is used.

Way number 2: sum over the elements of the group.

Now the sum looks like

$$\sum_{g\in G} \text{number of elements of}\;\Omega\;\text{fixed by}\;g$$

But $\mathrm{fix}_{\Omega}(g)$ is just the name for the set of elements of $\Omega$ fixed by $g$, so this sum can be written

$$\sum_{g\in G}|\mathrm{fix}_{\Omega}(g)|$$

The proof can be completed from here.

Now it seems to me since you asked about it that it is actually true that $\mathrm{fix}_{\Omega}(g)=\mathrm{fix}_{\Delta_1}(g)\cup\dots\cup\mathrm{fix}_{\Delta_t}(g)$, because the left side is the set of all elements of $\Omega$ fixed by $g$, and the right side is the disjoint union of the set of elements fixed by $g$ that are in each individual orbit. But the proof I've outlined here doesn't make use of this.

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