# Setwise stabilizer of a finite set is a maximal subgroup of Sym(N)

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

Let $$S$$ denote the symmetric group on the natural numbers.

If $$\emptyset\subset A \subset \mathbb{N}$$ then $$S_{\{A\}}=\{f\in S:af\in a,\;\forall{a}\in A\}$$ 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 $$f\neq g\in S\setminus S_{\{A\}}.$$ 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 then $$g=ghh^{-1}\in\langle S_{\{A\}}, f \rangle$$ but I can't seem to get it to work. Can anyone help?

If f isn't in S_{A} then how can af be in S_{A} and for that matter A? Also, I'm a little uncomfortable with your notation. By <S_{A},f> what exactly do you mean? Are you trying to say the subgroup S_{A}U{<f>}, or in words the subgroup S_{A} union the subgroup generated by f?

Also are you talking about N under multiplication? If so I would start by acknowledging that f:N--->S is at least a homomorphism

I am talking about N as a SET. So S=Sym(N) is the set of permutations on a countably infinite number of points enumerated by N.

The notation is all standard. The angled brackets denote "the group generated by" so above I mean the group generated by:
$$S_{\{A\}}$$ and $$f$$.

To avoid the confusion that the use of f has caused, replace the above definition of $$S_{\{A\}}$$ by:

$$S_{\{A\}}=\{q\in S:aq\in A,\;\forall{a}\in A\}$$

(All maps are written on the right).