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Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

  • $S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

  • $S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\not\cong G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\not\cong G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

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Doesn't $S_\infty$ itself contain a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$? Say the subgroup preserving the odd part. –  Aaron Meyerowitz Sep 19 '11 at 16:59
    
Yes it does; as noted above one such subgroup can be embedded in $G$. –  ARupinski Sep 19 '11 at 18:26
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So how does the sentence before your question establish that $S_\infty$ is not isomorphic to $G$? Not that I am saying it is. –  Aaron Meyerowitz Sep 19 '11 at 18:51
    
@Aaron: Now I see the issue you have. The important claim I am making there is that $G$ is not isomorphic to $S_\infty$, but you are correct, my given reasoning does not prove this claim. I will edit accordingly. –  ARupinski Sep 19 '11 at 21:45
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Well, consider $\mathbb{\bar N}=\mathbb{N}\cup\{\infty\}$ with the topology where the open sets $S$ are either finite and excluding $\infty$, or include infinity and $\sum\_{n\not\in S}w(n) < \infty$. Then, $G$ is the set of self-homeomorphisms of $\mathbb{\bar N}$ fixing a neighborhood of $\infty$. I suppose the answer depends on what a base of neighborhoods of $\infty$ looks like (I guess the space is not secodn countable). –  George Lowther Sep 20 '11 at 1:19
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1 Answer

up vote 5 down vote accepted

Not an answer, just a series of observations which will hopefully be of some use, and too long for a comment. The last observation might be relevant to your question about whether $G$ is the direct sum of countably many copies of $S_\infty$, unless I've over-simplified something.

  1. For each weight $w$, we may associate the collection $$I_w = \left\{A \subseteq \omega\ |\ \sum_{n \in A}w(n) < \infty\right\}$$
    Then $I_w = P(\omega)$ iff $\sum_{n \in \omega} w(n) < \infty$. If the series is divergent, then $I_w$ is a non-principal ideal extending the ideal $\mathrm{Fin}$ of finite sets, and moreover this ideal is not maximal, i.e. its dual filter is not an ultrafilter.
  2. $S_{\infty}(w) = \bigcup _{A \in I_w}S_A$ where $S_A$ is the subgroup of $S_{\infty}$ which fixes $\omega \setminus A$ pointwise, i.e. it's the set of permutations of $A$.
  3. The map $(I_w,\subseteq) \to ($subgroups of $S_\infty,\leq)$ defined by $A \mapsto S_A$ is an injective homomorphism of posets. We're essentially interested in the union of the range of this map, but unfortunately this is not equal to the image of the union of the domain of this map.
  4. Also note that $(I_w,\subseteq)$ is a lattice and a directed set, as is the range of the above map.
  5. If $\lim_{n\to\infty}w(n) = 0$ but $\sum_{n\in\omega}w(n) = \infty$, then there is no partition of $\omega$ into countably many infinite sets $A_n (n \in \omega)$ such that $S_\infty(w)$ is isomorphic to the direct sum of the $S_{A_n}$ in the "obvious" way. That is, $S_\infty(w)$ doesn't consist merely of those permutations which can be expressed as a finite product of elements from finitely many of the $S_{A_n}$. [Proof: For each $n$, pick some $m_n \in A_n$ such that $w(m_n) < 2^{-n}$. Set $A = \{m_n : n \in \omega\}$. Then $S_A \leq S_{\infty}(w)$ but there are elements of $S_A$ which can't be written as a finite product of elements of the $S_{A_n}$.]
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Your point 5 seems to confirm what I suspected but could not rigorously prove: that there must be more to $G$ than just a bunch of copies of $S_\infty$. –  ARupinski Sep 20 '11 at 16:12
    
@ARupinski: $G$ is not a countable union of subgroups isomorphic to images of $S_\infty$. –  George Lowther Sep 20 '11 at 19:34
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