Let $G=S_n$ and fix $i \in A=\{1,2,\ldots,n\}$ and let $G_{i}=\{\alpha \in G | \alpha(i)= i\}$. Prove $G_{i} \leq G$ and find $|G_i|$

Question

I am not sure about $|G_i|$ but I was able to show $|G_i|$ as a subgroup as follow:

Define $\; \;$ $\cdot$ $: G\times A \to A$ as $\alpha.i= \alpha(i) \; \; \forall \alpha \in G, \; i\in A$

Then $\cdot$ is a group action of $G$ on $A $ since for $\alpha, \beta \in G, \; i \in A$

$\alpha.(\beta.i) = \alpha(\beta(i)) = (\alpha\circ\beta)(i) = (\alpha \beta).i$ $\; \; $and $ \; \;I.i = I(i)= I$

Now clearly $G_i = stab_G(i) \leq G$

I am not sure about $|G_i|$

if $\alpha \in G_i$ then $\alpha $ must fix $i$. This would be possible if $\alpha$ does not contain $i$ in its cycles. Hence $\alpha \in S_{n-1}$ and therefore, $|G_i| = (n-1)!$ Will this be correct?

EDIT : Can we show $G_i \sim S_{n-1}?$

Answer 1

By definition $G_i =\{\sigma \in S_n|\sigma(i)=i\}$.

Now, define $f:G_n \rightarrow G_i$ by $f(\sigma)=\phi$.

Here $$\phi(x)=\begin{cases} \sigma(i) & x=n \\ \ n & \sigma(x)=i \\ \sigma(x) & \text{otherwise}. \end{cases} $$

This defines an isomorphism between $G_i$ and $G_n$ and since $G_n$ is really $S_{n-1}$, we get the isomorphism between $G_i$ and $S_{n-1}$.