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Consider the group $G = S_n$ acting on $X = \left\{1,2,...n \right\}$. Let $x=n$. How many elements does the stabilizer $\operatorname{Stab_G(x)}$ have? To what group is this stabilizer isomorphic?

Attempt: There is $(n-1)!$ elements in $\operatorname{Stab_G(x)}$. There exists an isomorphism $\phi$ such that $\phi : \operatorname{Stab_G(x)} \rightarrow S_{n-1}$, however, what is the simplest way to prove this? Clearly, the sets have the same size so that is not an issue (in terms of bijectivity) and I have checked some cases, but I can't see how to prove it in general, which I think would be equivalent to finding an explicit homomorphism $\operatorname{Stab_G(x)} \rightarrow S_{n-1}$

Is it correct to say that $\operatorname{Stab_G(x)} \leq G$ acts trivially on the set $X$? If so, I thought about considering the bijection $f_g : X \rightarrow X$, given by $f_g(x) = g \cdot x = x,\,\, g \in \operatorname{Stab_G(x)}$ which I think is a homomorphism from $\operatorname{Stab_G(x)}$ to some symmetric group (since this group is finite and $f_g \in \operatorname{bij}(X) = S_{|X|}),$ but I am unsure

Many thanks.

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Maybe it's a good idea to think permutations as bijections of finite sets. In other words, $S_n$ is the group of bijections of the set $\{1,\ldots,n\}$ with composition as operation.

Let $\eta\in Stab_G(x)\subseteq S_n$ be such a bijection. Note that the map $$ \begin{array}{rccc} \Phi(\eta):&\{1,\ldots,n-1\}&\longrightarrow&\{1,\ldots,n-1\}\\ &a&\longmapsto&\eta(a) \end{array} $$ is a bijection and therefore $\Phi(\eta)\in S_{n-1}$. Then $\Phi:Stab_G(x)\rightarrow S_{n-1}$ is the isomorphism you are looking for. To see that it indeed is an isomorphism you can try to construct its inverse, which should send $\psi$ to the map $\widetilde{\psi}$ given by $\widetilde{\psi}(a)=\psi(a)$ if $a\in\{1,\ldots,n-1\}$ and $\widetilde{\psi}(a)=n$ if $a=n$.

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  • $\begingroup$ How would you prove that this is a homomorphism though? Also, in the edit, what is $\eta$? Thank you. $\endgroup$ – CAF May 3 '13 at 9:42
  • $\begingroup$ I'll edit again the whole think to see if I can make it clearer. $\endgroup$ – A. Bellmunt May 3 '13 at 9:45
  • $\begingroup$ Thanks, can I also construct a homomorphism from what I wrote in my post starting from 'Is it correct..' I don't know how to show there that the symmetric group is explicitly $S_{n-1}$. $\endgroup$ – CAF May 3 '13 at 10:16
  • $\begingroup$ I think you want to say "transitively" instead of "trivially". Anyway, $f_g$ is an element of $S_n$, not $S_{n-1}$. You should send $g$ to $f_g$ restricted to the subset $\{1,\ldots,n-1\}\subseteq X$ (this is exactly the same thing that $\Phi$ makes to $\eta$ in my post). $\endgroup$ – A. Bellmunt May 3 '13 at 10:47
  • $\begingroup$ Why would the stabilizer act transitively? I.e if we consider $Stab_G(x)$ then all it's elements fix x, so there is no element g in this stabilizer that maps x to another element $y \in X$? $\endgroup$ – CAF May 3 '13 at 11:19
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Or; you can use the Orbit-Stablizer Equation. Moerover to @A. Bellmunt's way, when you want to show two permutation groups are isomorphic, you have to consider additional points there. See 1.

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  • $\begingroup$ How would the Orbit Stabilizer help? Sure, I can see why it helps to see that the two groups are indeed the same size, but how else does it help? $\endgroup$ – CAF May 3 '13 at 9:52
  • $\begingroup$ No! This is not about having the same size! When you want to use the Bellmunt approach, you have to show that additional proprty. Because you are working with permutation groups. See the link I appended, and see what prof. Derek nicely and elegantly noted me there. $\endgroup$ – mrs May 3 '13 at 9:57
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If you write the elements that fix $n$ in cycle notation, you will notice that these are precisely the elements that do not contain the symbol $n$ in any of your cycles (when simplified of course).

So the isomorphism should be clear.

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  • $\begingroup$ Do you mean like, if in $S_3$ for example $1 \rightarrow 1,\,2 \rightarrow 3,\,3 \rightarrow 2$, then in cyclic notation we have $(23)$. But surely this is still a permutation in $S_3$? The absence of $1$ is just implicitly saying that $1$ is mapped to $1$. $\endgroup$ – CAF May 3 '13 at 9:47

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