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

  • #1
So I'm reading a paper which assumes the following statement but I would like to be able to prove it.

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?
 

Answers and Replies

  • #2
36
0
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?
 
  • #3
36
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Also are you talking about N under multiplication? If so I would start by acknowledging that f:N--->S is at least a homomorphism
 
  • #4
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:
[tex]S_{\{A\}}[/tex] and [tex] f [/tex].

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

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

(All maps are written on the right).
 

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