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I am a bit confused about why a finitely generated group has infinite Prufer rank?

A group is said to have finite Prufer rank $r$ if every finitely generated subgroup can be generated by $r$ elements and $r$ is the least such integer. Otherwise it has infinite Prufer rank.

If the subgroups we are considering is finitely generated, surely we can find a max over all these subgroups and find $r$? I am not sure what I am missing here...

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  • $\begingroup$ A finitely generated group $G$ may have a non-finitely generated subgroup $H$. Inside $H$ it could be possible that there is an infinite family of finitely generated subgroups $\{ F_n \}_{n \ge 1}$ where $F_n$ is not generated by less than $n$ elements. Anyway, this is impossible for abelian groups: in fact finitely generated abelian groups have finite Prufer rank. $\endgroup$ – Crostul Mar 14 '16 at 11:46
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A pretty convincing example : take the free group $F_2$ over two elements. It is obviously finitely generated.

Then it is known that every free group $F_n$ is a subgroup of $F_2$, so the Prufer rank of $F_2$ is infinite, enven though it has just two generators.

Actually, you even have that $F_\infty$ is a subgroup, so every countably generated group is a subquotient of $F_2$.

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