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  1. What does "finite rank" mean in the context of divisible abelian $q$-group?

  2. A divisible abelian $q$-group of finite rank is always a Prüfer $q$-group or it can be also a finite product of Prüfer $q$-group?

Thanks for all future answers to my questions.

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    anon I was thinking about the [rank](http://en.wikipedia.org/wiki/Rank_of_an_abelian_group)... $A$ Prüfer p-group would have rank at most p-1? And a finite product of Prüfer would have rank at most (p-1)*n? If so, i think this is the right one. AlexanderGruber I don't have to work with the concepts in the [definition](http://en.wikipedia.org/wiki/Pr%C3%$B$Cfer_rank) so this can't be the right one.2012-10-08

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The Prüfer rank of an abelian group is the cardinality of a maximal linearly independent subset over $\mathbb{Z}$.

So if you got your Prüfer group $\mathbb{Z}(p^\infty)$, the maximum size of a linearly independent subset is going to be $1$. Using the representation of a Prüfer $p$-group as $\mathbb{Z}(p^\infty)=\{e^{\frac{2\pi i m}{p^n}}:m,n\in \mathbb{Z}\}$, we take the set $\{e^{2 \pi i a p^{-n}},e^{2 \pi i b p^{-m}}\}$ and try to solve $\left(e^{2 \pi i a p^{-n}}\right)^{x_1}\left(e^{2 \pi i b p^{-m}}\right)^{x_2}=e^{2 i \pi \left(a p^{-n}x_1+b p^{-m} x_2\right)}=1$ So we need to pick nontrivial $x_1$ and $x_2$ to satisfy $a p^{-n}x_1+b p^{-m} x_2=0$, which we can certainly do. Thus $\mathbb{Z}(p^\infty)$ has rank $1$. If you take a finite direct product of Prüfer groups, you'll still have finite rank because rank is additive; each component will just contribute $1$ to the total rank of the group.

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    Oh! This is exactly what I need! At a glance I have done a larger valuation of the Prüfer rank of a Prüfer group, but your answer have cleared all my doubts! Thank you!2012-10-08