Is the poset of all precompact group topologies on an abelian group $G$, order-isomorphic to $\operatorname{Sub}(\hat{G})$? 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?
 A: 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.
A: 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.
A: 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.
