# Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$ [duplicate]

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

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 29 at 9:01

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