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Consider the group $\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$, where each $\mathbb{Z}_{(i)}$ is a copy of $\mathbb{Z}$. Given $x \in \bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$, let $x^{(k)}$ be the element in $\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$ obtained by shifting every entry of $x$ to the right by $k$ places. Let $x_1, x_2, \ldots, x_n \in \bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$ be some non-zero elements.

I was wondering if there is a "nice description" for the subgroup $$ \langle x_j^{(i)} \mid i \in \mathbb{Z}, 1 \leq j \leq n \rangle. $$ We know that this subgroup is free Abelian (https://mathoverflow.net/q/3405/479955) and infinitely generated. Hence, it is isomorphic to the original group $\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$.

Any reference about this would be really appreciated.

ghc1997
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    You wrote $n$ for the index of the sum. You probably meant $i$. – jjagmath Jun 22 '23 at 17:09
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    What do you mean by "not isomorphic to (...)"? Assuming at least one of $x_i\neq 0$ then your subgroup cannot be finitely generated. And thus it has to be a free abelian group of countable rank. Thus isomorphic to the original group. – freakish Jun 22 '23 at 17:37
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    You are using $i$ for two different things. – jjagmath Jun 22 '23 at 18:07
  • @freakish: the direct product of infinitely many copies of $\mathbb{Z}$ is not free abelian, so is not isomorphic to a direct sum of infinitely many copies of $\mathbb{Z}$. – Arturo Magidin Jun 22 '23 at 20:10
  • @ArturoMagidin that's not a direct product, it is a direct sum. OP refers to "subgroup of free abelian is free abelian" which wouldn't make sense otherwise. – freakish Jun 22 '23 at 20:26
  • @freakish Huh? Yes, the OP is looking at a direct sum, and a subgroup of a direct sum. That does not preclude some subgroups to be isomorphic to direct products. I interpret that particular line as saying that the resulting group, although free abelian, will not in general be isomorphic to a direct product of copies of the infinite cyclic group, (even though it is of course isomorphic to a direct sum of copies of the infinite cyclic group). – Arturo Magidin Jun 22 '23 at 20:30
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    @ArturoMagidin I see, I missed the direct product part in his claim. It's a weird statement though. A random claim that it is not isomorphic to random group, which comes out of nowhere. – freakish Jun 23 '23 at 05:50
  • The subgroup in question is the normal closure of a finitely generated subgroup of the base group in the restricted wreath product ${\mathbb Z} \wr {\mathbb Z}$. – Derek Holt Jun 23 '23 at 07:47

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