Group Action on Subgroup of Symmetric Group

Theorem
Any subgroup of the Symmetric Group $$S_n$$ acts as a group of transformations on any set $$X$$ with $$n$$ elements.

The stabilizer of a given $$i \in \mathbb{N}^*$$ is the set of all those permutations which leave $$i$$ unchanged.

$$\operatorname{Stab} \left({i}\right) = \left\{{\pi \in S_n: i \in \operatorname{Fix} \left({\pi}\right)}\right\}$$

Thus $$\operatorname{Stab} \left({i}\right) \cong S_{n-1}$$.

Proof

 * The identity permutation takes each element of $$X$$ to itself, thus fulfilling GA-1.


 * The product rule in $$S_n$$ ensures fulfilment of GA-2.


 * The stabilizer of $$n$$ in $$S_n$$ is all the permutations of $$S_n$$ which fix $$n$$, which is clearly $$S_{n-1}$$.

A permutation can be applied to $$\mathbb{N}^*_n$$ so that $$i \to n$$ for any $$i$$. Thus one can build an isomorphism to show the result for a general $$i$$.