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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...2012-04-29
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    @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).2012-04-29
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    @t.b.: I think that might as well be an answer.2012-04-29
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    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.2012-04-30

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