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Let $p$ be any prime number and $G=\prod\limits_{n=1}^{\infty}\mathbb Z/(p^n)$ be the direct product of the finite cyclic groups $\mathbb Z/(p^n)$. Let $H=\mathbb Z_p$ be the group of $p$-adic integers with its usual norm $|.|_p$. Then the function $f_n:G\to H$ sending a sequence $(x_k)_{k=1}^{\infty} \in G$ to its $n$-th term $x_n \in \{0, 1, \dots , p^n − 1\} \subset \mathbb Z_p$ is a continuous $p^{−n}$–homomorphism.

As there is a term "continuity" of the map $f_n:G\to H$, there should be topologies on both $G$ and $H$, with respect to which $f_n$ is continuous.

Here, I've two questions.

Question 1 : It seems that the topology on $H=\mathbb Z_p$ is generated by the metric induced by the norm $|.|_p$. Is it true?

Question 2 : What is the topology on $G=\prod\limits_{n=1}^{\infty}\mathbb Z/(p^n)$?

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    $\begingroup$ A1: Yes. A2: Most likely the product topology of discrete topologies on the factors. I don't know of other "natural" topologies for that space. $\endgroup$ – Jyrki Lahtonen Aug 20 at 7:05
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    $\begingroup$ I agree. If you construct the $p$-adic integers more algebraically (not via completions and metrics, but via inverse limits) one will also take $\mathbb{Z}/p^n\mathbb{Z}$ as a discrete space, take the product over $n \geq 1$ and realize the $p$-adic integers as a subspace (which is just the concrete construction of the inverse limit in the category of groups). $\endgroup$ – ThorWittich Aug 20 at 7:11
  • $\begingroup$ @JyrkiLahtonen What is happened if we take other topology instead of discrete topology on the factors of $G$? $\endgroup$ – BijanDatta Aug 20 at 7:56
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    $\begingroup$ Connection with adeles and ideles (mathoverflow.net/q/41253/88984) ? $\endgroup$ – Jean Marie Aug 20 at 8:20
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    $\begingroup$ There is no homomorphism $\Bbb{Z}/(p^n) \to \Bbb{Z}_p$ except the one sending everything to $0$. $\endgroup$ – reuns Aug 20 at 10:02

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