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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!

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  • $\begingroup$ It is quite possible for an intersection of non-subgroups to be a subgroup, so that argument does not work. I think the way to use group actions is to note that this is a general thing: The stabilizer of an element is a subgroup (for any action). $\endgroup$ – Tobias Kildetoft Oct 27 '16 at 12:02
  • $\begingroup$ @TobiasKildetoft That's another thing that confused me. They proved this fact (that the stabilizer is always a subgroup) in the chapter preceding this exercise, so the exercise becomes trivial. I suppose the purpose of the exercise was to identify the group action that induces this collection of stabilizers (is that how one would phrase that?), which C. Falcon did in his answer. $\endgroup$ – user193319 Oct 27 '16 at 12:21
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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

  1. id $\cdot s$ = $s$, $\text{ }$ for all $s \in S$, and
  2. $\sigma \cdot \left( \tau \cdot s \right) = (\sigma \tau) \cdot s$, $\text{ }$ for all $\sigma, \tau \in S_n$, $s \in S$

equivalently we need to show

  1. id $ ( s ) = s$, $\text{ }$ for all $s \in S$, and
  2. $\sigma \left( \tau (s) \right) = (\sigma \circ \tau) (s)$, $\text{ }$ 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.

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$G$ 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 $G$.

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