Theorem – Orbit-Counting theorem

Contents
1. Theorem
2. Proof
3. Consequences of the theorem.

Theorem

Let $\langle G, * \rangle$ be a finite group. Let $X$ be a set. Consider the group action of $G$ on $X$ . Let the set $S$ be equal to the set

$S = \{\text{orb}(x) ~|~ x\in X\}$

Then, $|S| = \frac{1}{|G|} \sum_{g\in G} |\text{fix}(g)|$ .

Proof

Let $O_i$ be some element of $S$ (that is, $O_i$ is an orbit of $g$ ). Then, we have that

$\sum_{x\in O_i} |\text{stab}(x)| = |\text{orb}(x)| \times \frac{|G|}{|\text{orb}(x)|} = |G|$

Thus, as the $n$ orbits of $X$ partition $X$ , we have $O_1 \cup O_2 \cup ... \cup O_n = X$ , (with each pair $O_i, O_j$ disjoint for $i\neq j$ ). This means that

$\begin{aligned} \sum_{x\in O_1} |\text{stab}(x)| + \sum_{x\in O_2} |\text{stab}(x)| + ... + \sum_{x\in O_n} |\text{stab}(x)| &= \sum_{x\in X} |\text{stab}(x)| \\ &= n\times |G| \end{aligned}$

And so, dividing both sides by $|G|$ (assuming $G$ to be non-empty),

$n = \frac{1}{|G|} \sum_{x\in X} |\text{stab}(x)|$ .

Finally, we use the fact that $\sum_{g\in G} |\text{fix}(g)| = \sum_{x\in X} |\text{stab}(x)|$ ,

$n = \frac{1}{|G|} \sum_{x\in X} |\text{stab}(x)| = \frac{1}{|G|} \sum_{g\in G} |\text{fix}(x)| \quad \square$