Let $G$ be a discrete abelian group and $G^{\mathbb{N}}$ be the direct product (with the product topology), which consists of sequences $(a_1,a_2,a_3,\dots)$ of elements of $G$.

Let $G^{\infty}$ denote the group of sequences with values in $G^{\mathbb{N}}$, which converge to the identity $(0,0,0,\dots)\in G^{\mathbb{N}}$. Hence, $G^{\infty}$ is naturally a subgroup of $(G^{\mathbb{N}})^{\mathbb{N}}$.

Is it possible to characterize the isomorphism type of $G^{\infty}$ in terms of the group $G$ or, perhaps, in terms of the structural characterization of infinite abelian groups, à la Fuchs, given a structural characterization of $G$?

I am curious about the general problem but am most interested in the case where $G$ is finitely generated. Also, I am aware of several results that help identify the isomorphism type of various subgroups of the Baer-Specker group $\mathbb{Z}^{\mathbb{N}}$ but it's not so clear to me how this applies or if it informs this more general situation.

  • $\begingroup$ I think your meaning is clear, but, just to make sure: a "sequence in $G^{\mathbb N}$" could mean a sequence that is an element of $G^{\mathbb N}$ or that has values in $G^{\mathbb N}$. You mean the latter, right? $\endgroup$
    – LSpice
    Feb 6, 2021 at 17:32
  • $\begingroup$ Correct. I do believe "converges to $(0,0,0,\dots)$" and $G^{\infty}\leq (G^{\mathbb{N}})^{\mathbb{N}}$ make this pretty clear but I guess I can try to clarify more. $\endgroup$
    – J.K.T.
    Feb 6, 2021 at 18:40
  • $\begingroup$ A more robust and standard notation: for an arbitrary set $G^X$ is the group of maps $X\to G$, and $G^{(X)}$ is its subgroup of finitely supported functions (= which converge to 0 as sequences, when $G$ is discrete and $X=\mathbf{N}$). $\endgroup$
    – YCor
    Feb 6, 2021 at 19:29
  • $\begingroup$ @YCor I imagine $G^{(\mathbb{N})}$ is relevant to the problem but I'm having a hard time seeing how it could be used to rephrase the question to be more clear than it is now. $\endgroup$
    – J.K.T.
    Feb 6, 2021 at 20:10
  • $\begingroup$ Don't take so hard time, I didn't suggest this. $\endgroup$
    – YCor
    Feb 7, 2021 at 0:24


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