The setwise stabiliser of a finite set is maximal in Sym(N) 0
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? 
 A: 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 $g\in \langle S_A, f \rangle$.
A: The similar result is in fact true for the symmetric group
$S=\mathrm{Sym}(\Omega)$
of any infinite set $\Omega$: if $A$ is a finite nonempty subset of $\Omega,$ then the setwise
stabilizer $S_{\{A\}}$ of $A$ is a maximal subgroup of $S.$ This fact is due to R.W.Ball and it is contained
in the paper ``Subgroups of infinite symmetric group'' by Macpherson and Neumann [MN].
According to the standard notation, given a subset
$\Gamma$ of $\Omega,$ the setwise stabilizer
of $\Gamma$ is denoted by $S_{\{\Gamma\}}$ and
the pointwise stabilizer of $\Gamma$ by $S_{(\Gamma)}.$
Also, it is convenient to write $\mathrm{Sym}(\Gamma)$ for the
group of all permutations
of $\Omega$ which fix $\Gamma$ setwise and fix $\Omega - \Gamma$
pointwise (and the latter condition clearly implies the former).
$\textbf{Lemma}$ ([MN, Lemma 2.1]). Let $\Gamma_1,\Gamma_2$ be infinite subsets of
$\Omega$ satisfying $|\Gamma_1 \cap \Gamma_2| = |\Gamma_1 \cup \Gamma_2|.$
Then
$$
\langle \mathrm{Sym}(\Gamma_1), \mathrm{Sym}(\Gamma_2) \rangle=\mathrm{Sym}(\Gamma_1 \cup \Gamma_2).
$$
(if this result is unfamiliar it is certainly worth studying the rather short proof which actually is contained in another paper, namely ``Subgroups of small index in infinite symmetric groups'' by Dixon, Neumann, and Thomas).
Let a subgroup $G$ of $S$ strictly contain the setwise stabilizer $S_{\{A\}}$
of a finite nonempty subset $A$ of $\Omega.$ We claim that $G=S.$
We use induction on $|A|.$ Let then $A=\{a\},$ where $a \in\Omega.$ Suppose a
permutation $g \in G$ takes the element $a$ to a different element $b \in \Omega.$
But then, by the lemma,
$$
\begin{aligned}
G &\geqslant  \langle\, \mathrm{Sym}(\Omega - A), g\, \mathrm{Sym}(\Omega - A)\, g^{-1}\,\rangle \\
  &= \langle\, \mathrm{Sym}(\Omega - A), \mathrm{Sym}(\Omega - gA)\, \rangle \\
  &= \mathrm{Sym}(\, (\Omega - A) \cup (\Omega - gA)\,) \\
  &= \mathrm{Sym}(\,\Omega - (A \cap gA)\,) \\
  &= \mathrm{Sym}(\, \Omega - (\{a\} \cap \{b\})\,) \\
  &= \mathrm{Sym}(\Omega)=S.
\end{aligned}
$$
Induction step: if an element $g \in G$ does not
stabilize the set $A,$ that is, if
$$
gA \ne A,
$$
the set
$$
B = A \cap gA
$$
is of cardinality strictly less than $|A|.$
Arguing as before we see that
$$
G \geqslant \langle\, \mathrm{Sym}(\Omega - A), g\, \mathrm{Sym}(\Omega - A)\, g^{-1}\,\rangle
=\mathrm{Sym}(\, \Omega - (A \cap gA)\,)=\mathrm{Sym}(\,\Omega - B\,).
$$
Now as $G$ contains the group $\mathrm{Sym}(A),$
$G$ contains its subgroup $\mathrm{Sym}(B),$ which implies that
$$
G \geqslant S_{\{B\}}.
$$
Since, however, $G = S_{\{B\}}$ is impossible
due to $S_{\{B\}} \not\geqslant S_{\{A\}},$ we obtain that
$G > S_{\{B\}},$ and hence $G=S$ by the induction hypothesis.
