Exercise 2.2.8 (Stabilizer of $i$ in $S_n$)

Proof. For the action of $S_n$ on $[[1,n]]=\{1,2,\dots ,n\}$ defined by $\sigma \cdot i=\sigma (i)$ for every $\sigma \in S_n$ and $i\in [[1,n]]$, $G_i$ is the stabilizer of $i$, so $G_i$ is a subgroup of $G$.

Consider the set $X=[[1,n]]\setminus \{i\}$ and the map

where $\sigma {\text{|}}_I$ is the restriction of $\sigma$ to $X=\{1,2,\dots ,i-1,i+1,\dots ,n\}$.

• Note that $\phi$ is well defined: if $j\in X$, then $j\ne i$, therefore $\sigma (j)\ne \sigma (i)=i$, so $\sigma (j)\in X$. This shows that the restriction $\sigma {\text{|}}_I\in S_X$.
• $\phi$ is injective: If $\phi (\sigma )=\phi (\tau )$, where $\sigma ,\tau \in G_i$, then $\sigma (x)=\tau (x)$ for all $x\ne i$, and $\sigma (i)=\tau (i)=i$, so $\sigma =\tau$.
• $\phi$ is surjective: Let $\lambda \in S_X$. If $\sigma :[[1,n]]\to [[1,n]]$ is defined by $\sigma (x)=\lambda (x)$ if $x\ne i$ and $\sigma (i)=i$, then $\sigma \in S_n$ and $\sigma (i)=i$ shows that $\sigma \in G_i$. Moreover $\phi (\sigma )=\sigma {\text{|}}_X=\lambda$. This shows that $\phi$ is surjective.

So $\phi$ is a bijection. Therefore

Since $|X|=n-1$, $S_X\simeq S_{n-1}$, thus $|S_X|=(n-1)!$.

In conclusion,

□