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In this page, in abstract, it is claimed that the poset of all Hausdorff precompact group topologies on an abelian group $G$, is order-isomorphic to the the subgroup lattice of $\hat{G}$, the character group of $G$.

What is or where, on the internet, can I find a proof for this claim?

If this is the correct version of the claim, for an infinite abelian group $G$, we can have $|\hat{G}|=2^{2^{|G|}}$. Is this correct?

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Pre-compact group topologies on $G$ are described by isomorphism classes of epimorphisms $G \to K$, where $K$ is a compact abelian group. (This follows from the observation that a homomorphism is an epimorphism if and only if it has dense range.) By Pontryagin duality, such isomorphism classes of maps $G \to K$ are in bijection with isomorphism classes of monomorphisms $\hat{K} \to \hat{G}$, where $\hat K$ is now treated as a discrete group. Note that a homomorphism is a monomorphism if and only if it is injective.

Hence, the pre-compact group topologies on $G$ are in bijection with the subgroups of $\hat{G}$. It is easy to see that the lattice operations are preserved under this bijection.

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  • $\begingroup$ thanks. I will try to complete/expand your sketch of proof in coming days so I may understand/know the correct version of the claim. $\endgroup$ Feb 15, 2015 at 19:58
  • $\begingroup$ What is the order-isomorphism? $\endgroup$ Feb 27, 2015 at 10:38
  • $\begingroup$ The compactification $G \to K$ goes to $\hat K$, which is naturally embedded in $G$. $\endgroup$ Feb 27, 2015 at 11:22
  • $\begingroup$ A precompact group topology on a group $G$ is describe by an injective homomorphism $G\to K$ where $K$ is compact group. I do not know why do you use epimorphisms instead. $\endgroup$ Mar 20, 2015 at 20:02
  • $\begingroup$ @user47958: Because only the closure of G matters (and a map with dense image is an epimorphism). $\endgroup$ Apr 3, 2015 at 14:00
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I didn't think about the precompact group topologies, but the statement $|\hat G|=2^{2^{|G|}}$ is incorrect. A character of $G$ is a map from $G$ to the complex numbers. Hence there are no more than $(2^{\aleph_0})^{|G|}$ characters. If $G$ is infinite, then $(2^{\aleph_0})^{|G|}=2^{|G|}$ and not $2^{2^{|G|}}$.

But I don't understand how you get the equation $|\hat G|=2^{2^{|G|}}$ from the claim you mention. $2^{2^{|G|}}$ is an upper bound for the number of Hausdorff precompact group topologies on $G$. The character group is of size at most $(2^{\aleph_0})^{|G|}$, i.e., $2^{|G|}$ for infinite $G$. But $\hat G$ can have up to $2^{|\hat G|}$ subgroups. So for infinite $G$ we arrive at an upper bound of $2^{2^{|G|}}$ for both the number of subgroups of $\hat G$ and the number of Hausdorff precompact group topologies on $G$. Nothing wrong there.


Edit: I went through the literature concerning this question. The claim is proved in [W. W. Comfort and Kenneth A. Ross, Topologies induced by groups of characters, Fundamenta Math., 55 (1964), 283-291]. Another good read is [Bernahu, Comfort, Reid, COUNTING SUBGROUPS AND TOPOLOGICAL GROUP TOPOLOGIES, Pacific Journal of Mathematics, Vol. 116, No. 2, 1985], where it is shown, among other things, that every uncountable abelian group $G$ has $2^{|G|}$ subgroups.

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  • $\begingroup$ 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. $\endgroup$ Feb 14, 2015 at 4:50
  • $\begingroup$ 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? $\endgroup$ Feb 14, 2015 at 5:00
  • $\begingroup$ 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. $\endgroup$ Feb 14, 2015 at 8:43
  • $\begingroup$ 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. $\endgroup$ Feb 14, 2015 at 8:53
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    $\begingroup$ 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 $\tau$. $\endgroup$ Feb 16, 2015 at 8:16
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There is a more elementary complete proof for this claim in:

D. Dikranjan; L. Stoyanov. An elementary approach to Haar integration and Pontryagin duality in locally compact abelian groups. Topology Appl. 158 (2011), no. 15, 1942--1961.

Precompact in this claim, which is called Comfort-Ross Theorem, is not going to include Hausdorffness. So if $PC(G)$ is the set of all precompact (not necessarily Hausdorff) group topologies on $G$, ordered by $\subseteq$, then it is order-isomorphic to the lattice of all subgroups of $\hat{G}$ (also ordered by $\subseteq$). Here $\hat{G}$ is the group of all homomorphisms $f:G\to \mathbb T$ where $\Bbb T$ is the circle group. No continuity (or any other topological) conditions is assumed for such homomorphisms.

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