# Group Action and stabilizer problem.

$n$ is a positive integer the group $S_n$ acts on the set $A={1,2,....,n}$ as :

$s\cdot i = s(i)$ for all $i \in$ {1,2,...,n}.

We need to show that the action is faithful and the stabilizer $G_i$ is isomorphic to $S_{n-1}$.

I tried the first part as follows :

If I somehow show that the Kernel $K$ of the group action only contains the identity element , this would imply the action is faithful.

So , $K = [ s \in S_n | s\cdot i = i]$ for all values of $i$.

=> $s(i)=i$ , which is the cycle that maps each element to itself only , which is an identity element for $S_n$ and hence group action is faithful.

Now for the second part , for stabilizer $G_i$ we know $i$ is fixed , but since $K= (1)$ thus $G_i = (1)$ , but how to show that it is isomorphic to $S_{n-1}$ ? Could anyone help ?

## 1 Answer

By definition, $G_i:= \{s \in S \mid s \cdot i = i\}$, that is, $G_i$ consists of the permutations in $S_n$ that fix $i$. You can view such permutations as just permutations of the other $n-1$ elements of $A$.