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So I'm reading a paper which assumes the following statement but I would like to be able to prove it.

Let $S=Sym(\mathbb{N})$ denote the symmetric group on the set of natural numbers.

If $\emptyset\subset A \subset \mathbb{N}$ then: $S_{A}= \{ q \in S : aq\in A,\;\forall a\in A \}$ is a maximal subgroup of S.

Here is how I would like to prove it. I select $f\in S\setminus S_{{A}}$. I want to show that $\langle S_{{A}}, f \rangle = S$, otherwise we have a contradiction. So i take $g\in S$. If $g\in S_{{A}}$ or $g=f$ we are done so assume $g\in S\setminus (S_{{A}}\cup f )$. How can I show that $g\in \langle S_{{A}}, f \rangle$? I had thought about doing something like finding $h\in\langle S_{{A}}, f \rangle$ such that $gh\in S_{{A}}$ so that $g=ghh^{-1}\in\langle S_{{A}}, f \rangle$ but I can't seem to get it to work. Can anyone help? EDIT: I mean A finite. Why is it enought to show the transposition in the answer is in the group generated by these two?

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    This question was also asked simultaneously on MathOverflow http://mathoverflow.net/questions/59059/the-setwise-stabiliser-of-a-finite-set-is-maximal-in-symn. Please post a link when you're doing this!2011-03-24

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I will asume $A \subset \mathbb N$ is finite and nonempty. Let $f \in S$ with $f \not \in S_A$. Fix a point $a \in A$ with $b := f(a) \in \mathbb N \setminus A$. Let b' \in \mathbb N \setminus A with c := f^{-1}(b') \in \mathbb N \setminus A, which exists beacause $A$ was asumed to be finite. Now (a \; c) = f^{-1} \circ (b\; b') \circ f \in \langle S_A, f \rangle. The rest should be easy.

edit: We now formalize the rest of the proof:

Let $g\in S$. Let $o_1, \dots, o_n$ be all the elements of $A$ for which $g(o_k)\in\mathbb N \setminus A$. Since $g$ is bijective, and therefore $\mathrm{card }g(A) = \mathrm{card }A$ there must be exactly $n$ distinct elements $i_1, \dots, i_n$ of $\mathbb N \setminus A$ with $g(i_k)\in A$. Now consider $(o_k\;i_k) = (o_k\;a)\circ(i_k\;c)\circ(a\;c)\circ(i_k\;c)\circ(o_k\;a) \in \langle S_A, f \rangle$ and observe that $ h(A) := g\circ (o_n\;i_n)\circ\dots \circ(o_1\;i_1) (A) = A$ and therefore $h\in S_A$. This shows $g\in \langle S_A, f \rangle$.

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    I still don't really understand why it is enough just to show that we can move one point of$A$outside.2011-03-21