Timeline for Is the poset of all precompact group topologies on an abelian group $G$, order-isomorphic to $\operatorname{Sub}(\hat{G})$?
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| Feb 16, 2015 at 8:16 | comment | added | Stefan Geschke | You will see that one slightly surprizing and crucial fact used in the proof of the claim is that for every Hausdorff precompact group topology $\tau$ the elements of the group are separated by characters that are continuous wrt to $\tau$. | |
| Feb 16, 2015 at 8:10 | comment | added | Stefan Geschke | Note that the collection of all precompact Hausdorff group topologies is actually not a topological structure. You don't fix a specific topology on the group, but rather look at all topologies of a certain type. Put in another way, you look at all topologies of a certain kind that are compatible with the algebraic structure of the group. So that is an algebraic object, in some sense. | |
| Feb 16, 2015 at 8:03 | history | edited | Stefan Geschke | CC BY-SA 3.0 |
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| Feb 15, 2015 at 3:24 | comment | added | Minimus Heximus | thanks. I just need to be sure that that claim is exactly what I wrote above, with the same definition of character group as linked and considering Hausdorff precompact group topologies. That claim seems odd to me because it's connecting a topological structure to an algebraic one, so I worry if I have understood correctly. | |
| Feb 14, 2015 at 8:53 | comment | added | Stefan Geschke | I just found the reference projecteuclid.org/download/pdf_1/euclid.pjm/1102707061 where it is proved that every uncountable abelian group $G$ has $2^{|G|}$ subgroups. The paper also talks about totally bounded group topologies, but at my first glance doesn't seem to prove the claim that you are after. | |
| Feb 14, 2015 at 8:43 | comment | added | Stefan Geschke | The number of subgroups is in general not the same as the size of the group itself. Consider a free abelian group with a basis of some infinite size $\kappa$. Then the size of the group is $\kappa$ as well, but any two distinct subsets of the basis generate two different subgroups. Hence there are $2^{|G|}$ subgroups. | |
| Feb 14, 2015 at 5:00 | comment | added | Minimus Heximus | Yes you are right, an abelian group $G$ can have more subgroups than $|G|$. But in this case, it must have more. Is this correct for any character group? | |
| Feb 14, 2015 at 4:50 | comment | added | Minimus Heximus | I remember there are $2^{2^{|G|}}$ Hausdorff precompact group topologies on an abelian group (see this link page 2 of the paper). So the number of subgroups of $\hat{G}$ must be $2^{2^{|G|}}$. I think for an infinite abelian group (e.g. $\Bbb Z$) the number of subgroups has the same cardinality as the group itself. | |
| Feb 14, 2015 at 4:43 | history | edited | Stefan Geschke | CC BY-SA 3.0 |
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| Feb 14, 2015 at 4:30 | history | answered | Stefan Geschke | CC BY-SA 3.0 |