Using Group Actions to Show a Subset is a Subgroup I find this question, which comes from section 2.2 of Dummit and Foote's algebra text, to be somewhat confusing:

Let $G = S_n$, fix $i \in \{1,...,n\}$ and let $G_i = \{\sigma \in G ~|~ \sigma(i) = i\}$ (the stabilizer of $i$ in $G$). Use group actions to prove that $G_i$ is a subgroup of $G$. Find $|G_i|$.

Here is what I came up with, but it hardly uses group actions. Let $\ker(\cdot)$ denote the kernel of $G$ acting on $\{1,...,n\}$ (?). It is easy to show that $\ker(\cdot)$ is the intersection of all stabilizers of elements in $G$, i.e., $\ker(\cdot) = \bigcap_{i=1}^n G_i$. But since $\ker(\cdot)$ is a subgroup of $G$, and since $\bigcap_{i=1}^n G_i$ is a subgroup if and only if each $G_i$ is, then the stabilizer $G_i$ is a subgroup. 
That is the best I could come up with; as I mentioned, it really doesn't use many ideas of group actions. Also, I am not 100% certain $G$ is acting on $\{1,...,n\}$ in this case; perhaps it is acting on $n$-tuples of elements in $\{1,...,n\}$.
PS What is the standard notation for the kernel of a group action? Dummit and Foote offers no convenient notation for it---in fact, they haven't yet offered any notation for it!
 A: $G$ acts on $\{1,\cdots,n\}$ through $(\sigma,i)\mapsto\sigma(i)$ (check that the axioms are satisfied i.e. $\textrm{id}$ acts trivially and compatibility with the group structure), $G_i$ is the stabilizer of $i$ under this action, whence a subgroup of $G$.
A: On page 51 of Dummit and Foot the following is shown

If $G$ is a group acting on a set $S$, then for all $s \in S$, the stabilizer of $G$ in $s$ is a subgroup of $G$, i.e. $G_s \leq G$.

This gives us a way of showing $G_i \leq G$ without using the subgroup criteria. Instead we need only show that $S_n$ acts on the set $\{1 , \ldots , n \}$ via $\sigma \cdot s = \sigma(s)$, specifically we need to show 


*

*id $\cdot s$ = $s$, $\text{ }$ for all $s \in S$, and

*$\sigma \cdot \left( \tau \cdot s \right) = (\sigma \tau) \cdot s$, $\text{ }$ for all $\sigma, \tau \in S_n$, $s \in S$


equivalently we need to show 


*

*id $ ( s ) = s$, $\text{ }$ for all $s \in S$, and

*$\sigma \left( \tau (s) \right) = (\sigma \circ \tau) (s)$, $\text{ }$ for all $\sigma, \tau \in S_n$, $s \in S$.


Now 1. is true by the definition of the identity and 2. is true by the definition of the composition of maps. 
