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Let [tex]S[/tex] denote the symmetric group on the natural numbers.

If [tex] \emptyset\subset A \subset \mathbb{N}[/tex] then [tex] S_{\{A\}}=\{f\in S:af\in a,\;\forall{a}\in A\}$ [/tex] is a maximal subgroup of [tex] S. [/tex]

Here is how I would like to prove it. I select [tex] f\in S\setminus S_{\{A\}}.[/tex] I want to show that [tex]\langle S_{\{A\}}, f \rangle = S[/tex], otherwise we have a contradiction. So i take [tex]g\in S[/tex]. If [tex]g\in S_{\{A\}}[/tex] or [tex]g=f[/tex] we are done so assume [tex] f\neq g\in S\setminus S_{\{A\}}.[/tex] How can I show that [tex]g\in \langle S_{\{A\}}, f \rangle[/tex]? I had thought about doing something like finding [tex]h\in\langle S_{\{A\}}, f \rangle[/tex] such that [tex]gh\in S_{\{A\}}[/tex] so that then [tex]g=ghh^{-1}\in\langle S_{\{A\}}, f \rangle[/tex] but I can't seem to get it to work. Can anyone help?