# Divisible abelian $q$-group of finite rank

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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link The Prüfer p-group is divisible. I don't think the Rank of an abelian group fits... I have to find a definition that take care of the case of a product of Prufer q-group... – W4cc0 Oct 8 '12 at 22:24
You're right. But what about rank doesn't fit? – anon Oct 8 '12 at 22:27
Do you mean Prüfer rank? – Alexander Gruber Oct 8 '12 at 22:29
anon I was thinking about the rank... 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 so this can't be the right one. – W4cc0 Oct 8 '12 at 22:38

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.