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I have read that for any prime number $p$ the Prüfer $p$-group is countable.

My question is: where can I find a proof of this fact?

Thanks.

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    It would probably be a good idea if you told us what definition you are using of these groups.2012-10-30
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    i just wrote it.2012-10-30
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    You just wrote it *where*? I was suggesting you write it in the body of the question!2012-10-30
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    I was considering the Pruffer $p$-group as the abelian group $\langle x_1,x_2\cdots\,:\, px_1=0,\, px_{i+1}=x_i \rangle$ But @rschwieb already pointed me that they are all subsets of $\mathbb{Q}/\mathbb{Z}.$2012-10-30

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It is can be viewed as a subset of the group $\mathbb{Q}/\mathbb{Z}$, which is obviously countable.

The subset representing the Prüfer group is just the set of all elements with order a power of $p$.

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    Thanks. Now i see that it was an easy question, my problem is that i was watching those groups in the form $\langle x_1,x_2\cdots\, :\, x_{i+1}^p=x_i, x_1^p=1 \rangle.$ I have had forgotten the characterization you wrote.2012-10-30
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    Countably presented groups are countable: this is independent of anything. They are quotients of countably generated free groups, and it is easy to see these are countable.2012-10-30