## Is it possible to construct (without choice, even?) a non-finitely-generated group with no proper non-finitely-generated subgroup?

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

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The Prufer p-group. en.wikipedia.org/wiki/Pr%C3%BCfer_group – George Lowther Jan 23 2011 at 1:21

## 1 Answer

(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 $H_i$.

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Sweet... didn't know this one. I'm quite charmed. – Todd Trimble Jan 23 2011 at 18:33
@Peter: Thanks. That's just what I would have said. No need to delete this answer, as 6 people have already bothered upvoting it. – George Lowther Jan 24 2011 at 0:24

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Also, every proper subgroup is cyclic, and not just finitely generated. – George Lowther Jan 24 2011 at 0:26
Excellent example and explanation---thanks! 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 clear-cut construction, I suppose.) – Z Norwood Jan 24 2011 at 4:26
This group is isomorphic to the group of p-th power roots of unity in the complex numbers (restrict the isomorphism of Q/Z with the roots of unity to the subgroup of elements with p-power order). This is a more concrete model for the group. That every subgroup of this group is cyclic is related to the fact that any finite subgroup of the nonzero elements of a field is a cyclic group: any proper subgroup is missing some root of unity of order, say, $p^n$ and therefore has no root of unity of order p^n or higher, which means it is a finite subgroup. – KConrad Jan 24 2011 at 9:04
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