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Let $G$ is an abelian infinte group such that for all nontrivial subgroups $H$ $$\forall H\leq G, \left|\frac{G}{H}\right|<\infty$$ Prove that $G\cong\mathbb Z$.

What I have done:

Clearly, it is enough to show that $G$ is a cyclic group. Moreover, I see as $G/H$ is a finite group for all subgroups $H$, then $G$ cannot have any elements with the finite order. This means to me that if $H \leq G$ and $H$ is cyclic then $H$ cannot be finite. Help your friend for the rest. Thanks. :)

Martin Sleziak
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Mikasa
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2 Answers2

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Your group is torsion-free (otherwise it has a finite subgroup of finite index, contradicting $G$ being infinite). Now let $g\in G$, and consider $H=\langle g\rangle$, which is of some finite index $n=[G:H]$. The map $G\rightarrow H$, $g\mapsto g^n$, then maps $G$ into an infinite cyclic group. The kernel, if non-trivial, would have finite index, impossible because $H$ is torsion-free. Thus the kernel is trivial and $G$ embeds in a cyclic group, so is itself cyclic.

  • Thanks Steve for the answer. – Mikasa Aug 26 '12 at 17:30
  • This is indeed the better answer. William's answer is incomplete because your group is not necessarily finitely-generated. – Chris Gerig Aug 26 '12 at 19:55
  • @ChrisGerig: thanks for the kind words, but the group will be finitely generated, because it is virtually cyclic (any subgroup generated by one element has finite index). I also prefer my answer :), because it avoids the classification. –  Aug 26 '12 at 20:52
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    This is beautiful in its simplicity. – David Wheeler Aug 27 '12 at 22:12
  • $n$ is fixed so how $g \to g^n$ maps $G$ into $H$ ? – Souvik Dey Aug 11 '14 at 03:59
  • @SouvikDey since $G$ is abelian, every subgroup $H$ is normal, so $G/H$ is a group. Being a group of order $n$, $g^n \equiv e \bmod H$ for all $g \in G$, which is another way of saying $g^n \in H$ for all $g \in G$. – KCd Aug 29 '20 at 04:33
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Show that such a group must be finitely generated. Using the fundamental theorem of finitely generated abelian groups, you have that

$G \simeq \mathbb{Z}^r \times \mathbb{Z} / n_1 \mathbb{Z} \times ... \times \mathbb{Z} / n_k \mathbb{Z}$

where $n_i \in \mathbb{Z}$. There is an isomorphic copy of each factor of the direct product in $G$. Quotienting $G$ by $\mathbb{Z}^s$ for various $1 \leq s \leq r$ and the $\mathbb{Z} / n_i \mathbb{Z}$, you see that $r = 1$ ($r \geq 1$ for the group to be infinite and $r \leq 1$ for the finite quotient property) and $k = 0$ (for the finite quotient property).

William
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