Is there a nonfinitelygenerated group each of whose proper subgroups is finitely generated? If so, what form of choice (if any) is required to construct such a group?

14$\begingroup$ The Prufer pgroup. en.wikipedia.org/wiki/Pr%C3%BCfer_group $\endgroup$ – George Lowther Jan 23 '11 at 1:21
(CW since this is just expanding on George Lowther’s comment to the question, which could really have been an answer in the first place; if George L wants to convert his answer to a comment himself, I can delete this one.)
For any prime $p$, the Prüfer $p$group is as desired.
There are several constructions of this; a good one for present purposes is $$\mathbb{Z}[1/p]\ /\ \mathbb{Z}$$ i.e. rationals with denominator a power of $p$, modulo the integers.
To see that this works, note that it is the union of the linearly ordered chain of finitely generated (indeed, cyclic) subgroups $H_i := \{ [a / p^i]\ \ 0 \leq a < p^i \}$, over $i \in \mathbb{N}$.
Now any element of $H_{i+1}$ not in $H_{i}$ must be of the form $[a/p^{i+1}]$ with $a$ coprime to $p$, and hence generates the whole of $H_{i+1}$. So any subgroup is either equal to some $H_i$, or else contains them all and is the whole group.
On the other hand, the entire group is clearly not finitely generated since any finite set of elements is contained in some $H_i$.

1$\begingroup$ Sweet... didn't know this one. I'm quite charmed. $\endgroup$ – Todd Trimble♦ Jan 23 '11 at 18:33

$\begingroup$ @Peter: Thanks. That's just what I would have said. No need to delete this answer, as 6 people have already bothered upvoting it. $\endgroup$ – George Lowther Jan 24 '11 at 0:24

1$\begingroup$ Also, every proper subgroup is cyclic, and not just finitely generated. $\endgroup$ – George Lowther Jan 24 '11 at 0:26

$\begingroup$ Excellent example and explanationthanks! It's still not clear to me to what extent choice is necessary for this example, but I'll look through the argument more closely later. (I was expecting a less clearcut construction, I suppose.) $\endgroup$ – Zach N Jan 24 '11 at 4:26

1$\begingroup$ The pth power roots of unity can be viewed as a (dense) subset  even subgroup  of the unit circle, thus providing a natural context where the group of interest lives. That's why I consider the realization of this group as the pth power roots of unity to be more concrete. I also like this point of view because it gives you another way to understand why the finite subgroups of this group are cyclic, as I wrote in my previous comment. $\endgroup$ – KConrad Jan 25 '11 at 5:51