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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?