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Let $G$ be a countable group. Consider the subgroup $$\widetilde{\Pi}G = \left\{(g_n) \in \prod\nolimits_{\mathbb{N}} G\,:\,g_n \neq e\text{ only for finitely many }n\right\}$$ of the countable power of $G$ with itself. Then $\tilde{\Pi}G$ is a countable group which one might call the reduced power of $G$. This leads me to the first question:

First question: is there an official name for $\widetilde{\Pi}G$?

For some bizarre reasons, I came to be interested in groups $G$ that are isomorphic to $\widetilde{\Pi}G$ and lacking any information about the name of this construction it is hard for me to start trying to find information on them. Rather vaguely my question is:

What can be said about the properties of $\widetilde{\Pi}G$? What properties of $G$ are reflected in $\widetilde{\Pi}G$? Is there a place where groups of the form $\widetilde{\Pi}G$ are studied?

Two simple observations about this construction:

  • If $G \cong \widetilde{\Pi}G$ then $G$ cannot be finitely generated because every finite set $S \subset \widetilde{\Pi}G$ generates a subgroup having only finitely many nontrivial coordinates. This means that many basic group theoretic properties aren't preserved in a straighforward way by this construction. (This makes me a bit doubtful about the value of this construction.)

  • About the only mildly interesting positive result I could find is: $\widetilde{\Pi}G$ is amenable if and only if $G$ is amenable. If $G$ is amenable then $\widetilde{\Pi}G$ is the union of the amenable groups $G^n$, and if $\widetilde{\Pi}G$ is amenable, then so is its subgroup $G$.

Since this site seems to favor specific questions, here's one:

Suppose $\widetilde{\Pi}G \cong \widetilde{\Pi} H$. Is it true that either $G^n \cong H^m$ for some $n,m$ or $G \cong \widetilde{\Pi} H$ or $\widetilde{\Pi}G \cong H$?

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$\widetilde{\Pi}G$ is the direct sum of $\omega$ copies of $G$. – Brian M. Scott Mar 1 '12 at 10:13
For a counterexample to your final question (buried at the end of my answer), take $G=\mathbb{Z}^2\oplus (\mathbb{Z}/2\mathbb{Z})$ and $H=\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$. No finite power of $G$ is isomorphic to a finite power of $H$, nor are $G$ or $H$ isomorphic to the other's "reduced power" (restricted direct power), but $\displaystyle\widetilde{\prod}G \cong \widetilde{\prod}H$. – Arturo Magidin Mar 2 '12 at 16:51

As Brian Scott notes, what you have is called the direct sum of $\omega$ copies of $G$, or the restricted direct product (or perhaps, power) of $\omega$ copies of $G$. ("Direct sum" is sometimes reserved for abelian groups, in which case it will be called a "restricted direct product"). No need to restrict to index set $\mathbb{N}$, or all groups to be the same: you can consider the same construction for any family of groups, finite, countable, or uncountable.

"Reduced" is a bad name in any case, since it already has several meanings within group theory. A "reduced abelian group" is one with no nontrivial divisible subgroups; and a "reduced word" has specific meaning in group theory as well.

About your observations: the fact that the restricted direct power of an amenable group is amenable is a special case of the fact that a restricted direct product of any family of amenable groups is amenable. This follows from the following observations:

  1. The direct product of two (hence, of finitely many) amenable groups is amenable;

  2. Subgroups of amenable groups are amenable; and

  3. A group is amenable if and only if all its finitely generated subgroups are amenable.

Since a finitely generated subgroup of a restricted product is isomorphic to a direct product of finitely many of the factors, the result follows. In particular, it follows when all the factors are the same amenable group, then the restricted direct power (over any nonempty index set) is amenable.

Restricted direct powers are useful; they show up when considering restricted wreath products, for example.

As to your final question... The answer is no. Let $G=\mathbb{Z}^2\oplus\mathbb{Z}/2\mathbb{Z}$, and $H=\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. Then $G^n\cong \mathbb{Z}^{2n}\oplus(\mathbb{Z}/2\mathbb{Z})^n$, while $H^m\cong \mathbb{Z}^m\oplus(\mathbb{Z}/2\mathbb{Z})^m$. We cannot have $G^n\cong H^m$ for any $n,m\gt 0$, since the rank of $G^n$ is always twice the $2$-rank, but the rank of $H^m$ is always equal to its $2$-rank. Nor is $G$ isomorphic to $\oplus H$, nor $H$ to $\oplus G$. However, the restricted direct power of $G$ and of $H$ are isomorphic.

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