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I am reading Dummit and Foote, and in Section 4.1: Group Actions and Permutation Representations they give the following example of a group action:

The symmetric group $G = S_n$ acts transitively in its usual action as permutations on $A = \{1, 2, \dots, n\}$. Note that the stabilizer in $G$ of any point $i$ has index $n = | A |$ in $S_n$ (My italics)

I am having trouble seeing why the stabilizer has index $n$ in $S_n$. Is this due to the fact that we have $n$ elements of $A$ and thus $n$ distinct (left) cosets of the stabilizer $G_i$?

And how would I see this if I were to work with the permutation representation of this action? The action is not very clear to me.

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The stabilizer is in effect the group of permutations of the other $n-1$ elements, thus isomorphic to $S_{n-1}$. And $\frac{n!}{(n-1)!}=n$. – Hagen von Eitzen Nov 22 '12 at 20:59
up vote 3 down vote accepted

We have that $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$ for every $i\in \{1,\ldots,n\}$. (Why?)

$S_n$ contains all possible permutations on $n$ vertices. The set of those permutations which fix $i$ are simply the ones that permute the set $\{1,\ldots,n\}\setminus i$ in every possible way. Thus $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$.

Now, $|S_{n-1}|=(n-1)!$, so $[S_n:\mbox{Stab}_{S_n}(i)]=\ldots?$


And how big is $A$?

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Wow thanks for this answer, I like the hidden mouse over too! – Moderat Nov 23 '12 at 5:13

The action of $S_n$ on $A$ is given by $\sigma \cdot (1,2,...,n)=(\sigma(1),\sigma(2),...,\sigma(n))$ for each $\sigma\in S_n$. Remember that $\sigma$ is a bijection $\sigma:A\rightarrow A$.

Recall that for a subgroup $H$ of $G$, the index $[G:H]=|G|/|H|$ when $G$ and $H$ are finite.

If we look at all of the $\sigma\in S_n$ which fix a particular element $i\in A$, there are no restrictions on the $n-1$ remaining components of $\sigma$ (except they cannot map to $i$). Then the stabilizer is isomorphic to $S_{n-1}$.

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$\sigma \in \textrm{ Stab } (i)$ for $1\le i \le n$ iff $\sigma $ fixes $i.$ You just have to count the number of permutations that fix $i$ - working in the usual order, there are $n-1$ choices for the image of the first element in the domain, $n-2$ for the second, and so on, so that $|\textrm{ Stab } (i) | = (n-1)!.$ Apply Lagrange's Theorem.

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