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Let $\mathbb{Z}(p^{\infty})$ be defined by

$\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$, I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$.

i try several ways but I have not been successful,

some help ??

thank you

YuiTo Cheng
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1 Answers1

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I cite from this duplicate:

Every element of ${\bf Z}_{p^\infty}$ is represented by some $a/p^n$, where $0\leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${\bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

Dietrich Burde
  • 140,055