Group actions: Why do we place the condition that $S$ be finite in the following theorem? Theorem. Let $G$ be a group, $S$ be a $G$-set, and $S$ be finite, then $$|S|= \sum_{a \in A} [G : G_a],$$ where $A$ is a subset of $S$containing exactly one element from each orbit $[a]$.
Here, $G_a$ is the subgroup $\{g \in G \,: \, ga = a\}$ of $G$.
Proof. $S$ can be partioned as the union of orbits $S=\cup_{a \in A}[a]$. We know $[G : G_a] = |[a]|$. Hence, $$S=\sum_{a \in A}|[a]| =\sum_{a \in A}[G : G_a].$$
 A: The basic fact is that there is a bijection
$$
G/{\rm Stab}(a)\longrightarrow{\rm Orb}(a)\qquad
g{\rm Stab}(a)\mapsto g\cdot a.
$$
Given this, there is an obvious bijection
$$
S\longleftrightarrow\bigcup_{a\in\cal O}G/{\rm Stab}(a)
$$
where the disjoint union on the right runs over a complete system $\cal O$ of representatives of the orbits of $G$ in $S$. As usual, a bijection between sets translates into an equality between their cardinalities, so there's no restriction on $S$. The point is that when $|S|=\infty$ the equality just reads
$$
\infty=\infty
$$
which is rather uninteresting except maybe for the observation that either there are infinitely many orbits or at least one of the stabilizers has infinite index in $G$ (of course both things may happen at the same time).
The equality mentioned by the OP is obtained by assuming that $S$ is finite: this gives an a priori non-trivial numerical identity that may be useful to compute some of the summands given the proper information on the action.
