I am currently self-studying Aluffi's Algebra book at chapter IV on on Group actions in which he gives the counting formula where $G$ is a group that acts on a finite set $S$ being:

$$\lvert S \rvert = \lvert Z \rvert + \sum_{a \in A} [G: G_a],$$

where $Z$ is the set of fixed point under the action and $A \subseteq S$ contains exactly one element for each nontrivial orbit of the action and $G_a$ is the stabilizer of $a$.

He claimed that each $[G:G_a]$ is a divisor of $\mid G \mid$ and also each $[G:G_a] > 1$ in which I understand the divisor part since it is equal to the cardinality of the orbit of $a$ and orbits partion the set, but I just don't get why $[G:G_a] > 1$ since isn't $1$ is a divisor for every integers?

Since I am self-studying, any information would be helpful.

  • 4
    $\begingroup$ Singleton orbits are accounted for by $|Z|$, so $A$ is a set of representatives of non-singleton orbits. $\endgroup$
    – user870827
    Jan 17, 2021 at 18:13
  • $\begingroup$ Yeah! It was pretty obvious and I missed it! $\endgroup$
    – LamNg.
    Jan 17, 2021 at 18:15

2 Answers 2


If $[G:G_a] = 1$, then $G = G_a$ which would mean that $a$ is fixed by the group action and hence $a \in Z$ and hence $a \notin A$ since it would have a trivial orbit.

  • $\begingroup$ Oh! I know that I am missing something trivial, thanks for that! $\endgroup$
    – LamNg.
    Jan 17, 2021 at 18:14

This follows from the definition of the set $A$. If $a\in A$ then $a$ belongs to a nontrivial orbit. So there must be some element $g\in G$ such that $g.a\ne a$. Hence $G_a\ne G$.


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