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Q: If $ H = \{ \sigma \in S_n : \sigma(n) = n \} $ is a subgroup of $S_n$, then show that $H \simeq S_{n-1}$.

I know any group is isomorphic to a subgroup of the symmetric group. But I don't know how to proceed.

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Think of $S_n$ as the group of bijections from $\{1,\dots,n\}$ to itself with composition as the group multiplication. What does a bijection fixing $n$ do? It must give a bijection from $\{1,\dots,n-1\}$ to itself and, conversely, every bijection of $\{1,\dots,n-1\}$ gives a bijection of $\{1,\dots,n\}$ fixing $n$, so... –  t.b. Apr 29 '12 at 10:45
    
@Faisal: Any group is isomorphic to a subgroup of a symmetric group. It is an important distinction: not every group is isomorphic to a subgroup of a given symmetric group (obviously, because there exist infinite groups and $S_n$ is finite, for example). –  Tara B Apr 29 '12 at 10:58
    
@t.b.: I think that might as well be an answer. –  Tara B Apr 29 '12 at 10:59
    
If you need further elaborations please tell me what is unclear or what you'd like to have clarified. I'd be happy to add to my answer. –  t.b. Apr 30 '12 at 13:45

1 Answer 1

Following Tara B's suggestion I'm posting my comment as an answer:

Think of $S_n$ as the group of bijections from $\{1,\dots,n\}$ to itself with composition as the group multiplication. What does a bijection fixing $n$ do? It must give a bijection from $\{1,\dots,n-1\}$ to itself. Conversely, every bijection of $\{1,\dots,n-1\}$ gives a bijection of $\{1,\dots,n\}$ fixing $n$, so...

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