Timeline for Is there an easy description of the structure of this infinite group?
Current License: CC BY-SA 3.0
20 events
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| Sep 22, 2011 at 12:15 | comment | added | François Brunault | @ARupinski : It turns out that $G$ isn't isomorphic to a direct product (finite or infinite) of copies of $S_\infty$. Indeed in $G$ no element has an abelian centralizer, while in the direct product some elements have abelian centralizers (take an element of the form $(x,x,\ldots)$ where $x$ has trivial centralizer). I don't see how to rule out infinite direct sum, but I share your impression that $G$ is more complicated than that. | |
| Sep 20, 2011 at 16:25 | comment | added | ARupinski | @Francois: Although your argument doesn't work, it did make me think some more and I realized that I was too quick to rule out infinite direct products. At any rate, the more I think about it and about Amit's answer, I am fairly convinced $G$ is not any sort of product. | |
| Sep 20, 2011 at 16:12 | vote | accept | ARupinski | ||
| Sep 20, 2011 at 11:28 | comment | added | François Brunault | Sorry, the argument in my comment doesn't work, also in the direct sum every element has uncountable normalizer... | |
| Sep 20, 2011 at 11:22 | comment | added | François Brunault | In the last sentence of your question, do you mean direct sum or direct product ? Because in a countable direct sum $\oplus S_\infty$, there are elements with countable centralizers, while in $G$ every element has uncountable centralizer. So $G$ cannot be isomorphic to the direct sum. | |
| Sep 20, 2011 at 7:47 | answer | added | Amit Kumar Gupta | timeline score: 5 | |
| Sep 20, 2011 at 3:09 | comment | added | ARupinski | @David: No problem; I honestly never realized there was a \not command I could use in place of all the backspaces. Good to learn a simplifying trick like that. | |
| Sep 20, 2011 at 2:38 | comment | added | David Roberts♦ | I just couldn't help myself. I simplified some of the LaTeX, which was I'm sorry to say, a little bit dreadful. | |
| Sep 20, 2011 at 2:37 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
Tidied up LaTeX
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| Sep 20, 2011 at 2:19 | comment | added | George Lowther | I should have said - the open sets $S$ are those excluding $\infty$ (no finiteness required) or those including $\infty$ with $\sum\_{n\not\in S}w(n) < \infty$. And, it is not second countable. | |
| Sep 20, 2011 at 1:19 | comment | added | George Lowther | 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). | |
| Sep 20, 2011 at 1:15 | comment | added | Aaron Meyerowitz | OK that works! I suppose that you could just say that in $S_\infty$ there are fixed point free elements but not in $G$. Consider the subgroup of $S_\infty$ consisting of elements fixing all but finitely many even integers (or powers of 2 or members of $\lbrace 2,2^2,2^{2^2},\cdots\rbrace$). Can we rule out $G$ being isomorphic to that? | |
| Sep 19, 2011 at 22:03 | history | edited | ARupinski | CC BY-SA 3.0 |
added 124 characters in body
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| Sep 19, 2011 at 21:47 | history | edited | ARupinski | CC BY-SA 3.0 |
added 30 characters in body
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| Sep 19, 2011 at 21:45 | comment | added | ARupinski | @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. | |
| Sep 19, 2011 at 18:51 | comment | added | Aaron Meyerowitz | So how does the sentence before your question establish that $S_\infty$ is not isomorphic to $G$? Not that I am saying it is. | |
| Sep 19, 2011 at 18:26 | comment | added | ARupinski | Yes it does; as noted above one such subgroup can be embedded in $G$. | |
| Sep 19, 2011 at 16:59 | comment | added | Aaron Meyerowitz | 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. | |
| Sep 19, 2011 at 16:24 | history | edited | ARupinski | CC BY-SA 3.0 |
edited title
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| Sep 19, 2011 at 15:35 | history | asked | ARupinski | CC BY-SA 3.0 |