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Suppose I take an infinite direct power $\prod G$ of some (not necessarily finite) group $G$. I want to know about the subgroups of $\prod G$ that are maximal subject to having trivial intersection with $\oplus G$. Is there a general description of such subgroups (in terms of ultrafilters maybe), without delving into the structure of $G$? If not, are there at least some interesting general constructions of 'large' subgroups of $\prod G$ that intersect trivially with $\oplus G$?

One way to construct subgroups intersecting trivially with $\oplus G$ is as follows: take a family $\mathcal{P}$ of partitions of the indexing set $I$ that is closed under coarsest common refinement, such that all the parts in any given $P \in \mathcal{P}$ are infinite. (For instance $I = \mathbb{Z}$ and $\mathcal{P}$ is congruence classes modulo $n$.) Now take all those $g \in \prod G$ for which there is some $P$ in $\mathcal{P}$ for which the $i$-th entry of $g$ is determined by which part of $P$ contains $i$.

Are the groups I just described ever maximal?

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I know just a particular construction. Take an abelian G, and call the Pontryagin dual P. Now, Pontryagin dual of the direct sum of P is isomorphic to direct product of G, and quotients of the direct sum correspond to closed subgroups of the direct product. It might be that it's of no use to you, because I can't find an example such that the resulting closed subgroup doesn't intersect the direct sum (maybe it always intersects it?) –  Łukasz Grabowski Jan 27 '11 at 17:46
    
In general, I have no idea. But if is the additive group of a vector space, then the map $\prod G\to \prod G/\oplus G$ will split, and the image of a splitting should be maximal with the above property. –  Donu Arapura Jan 27 '11 at 18:17
    
@Lukasz Grabowski: I think $\oplus G$ will be dense in the product topology, so closed subgroups may not be all that useful as you say. –  Colin Reid Jan 27 '11 at 18:51
    
...although this doesn't clearly rule out large closed subgroups, come to think of it. –  Colin Reid Jan 27 '11 at 19:18
    
In fact subgroups you describe in the revised question are closed. –  Łukasz Grabowski Jan 28 '11 at 10:15
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Here is an example showing that the particular subgroups you describe need not be maximal.

Consider the case of $\Pi_{n\in\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$, with the subgroup $H$ consisting of the periodic elements of the product, which I think is the same as the subgroup you describe in your modulo $n$ example. This is a subgroup and has trivial intersection with the direct sum.

But I claim that it is not maximal. The reason is that this $H$ is countable, but I claim any subgroup of this product that is maximal with respect to having trivial intersection with the direct sum must be uncountable (and in fact must have size continuum). To see this, suppose that $H$ is a countable subgroup having trivial intersection with the direct sum. Thus also, the set of elements of the product that differ at most finitely from an element in $H$ is also countable. Since the direct product is uncountable, there is an element $h$ in the direct product that does not differ finitely from any element of $H$. It follows that the group generated by $H$ and $h$ continues to have trivial intersection with the direct sum. So $H$ is not maximal.

More generally, the same argument shows in this case that any maximal group with the property must have size continuum, for if not, then the set of elements differing finitely from $H$ will also have size less than continuum, and we will be able to find an $h$ that can be added.

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Well, in the case of a product of cyclic grousp of order two, we have vector spaces, so maximal subgroups with the property of having trivial intersection with the direct sum are simply complements of the latter subspace. Since the direct sum is countable and the direct product is not, a complement must be uncountable. –  Mariano Suárez-Alvarez Jan 28 '11 at 18:45
    
Is it possible for the set $\mathcal{P}$ of partitions to be uncountable and/or maximal subject to the given conditions? (If $\mathcal{P}$ can be extended, so can the subgroup.) –  Colin Reid Jan 28 '11 at 23:12
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