# Why is the index of the stabilizer neccesarily greater than 1?

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.

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

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.
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 $$G_a\ne G$$.