# Order of a Group acting on a Set vs. Order of the Set

Let the group $$G$$ acting on a finite set $$X$$ with order $$|X|=n$$.

Because each element of $$G$$ permutes the set $$X$$, and there are $$n!$$ permutations of X, does this mean $$|G|=|X|$$?

Remember that an action of a group $$G$$ on a set $$X$$ is just a homomorphism $$G\to\operatorname{Sym}(X)$$ from $$G$$ to the group of permutations of $$X$$. This homomorphism need neither be injective (so $$|G|>n!$$ is possible), nor surjective (so $$|G| is possible), and if the action is not transitive, we may well even have $$|G|.
In fact, any group can act on any set by means of the trivial action so that the mere existence of an action imposes no relation whatsoever between $$G$$ and $$X$$.
If you add the hypothesis that $$\phi : G \times X \rightarrow X$$ is a $$\textit{regular}$$ action, i.e. that $$\forall x,y \in X. \ \exists!\ g \in G.\ y = \phi(g, x),$$ then you will have $$|G|=|X|$$.