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I found this question in Arhangel'skii and Tkachenko's book Topological Groups and Related Structures. The first chapter of the book is devoted to algebraic preliminaries.

The question actually reads:

Give an example of an infinite abelian group all proper subgroups of which are finite.

What I have done is: Every element of this group has finite order, else we could find an infinite proper subgroup, namely the group generated by $x²$ if $x$ has infinite order.

I think this can be strengthened: every element should have a prime order. Although I haven't proved this.

Intuitively this group cannot be and infinite product of smaller groups, because you could take the product of the even group factors and find an infinite proper subgroup.

Well, this is it, a highly non-trivial problem. Thanks in advance.

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    $\begingroup$ You could look up the construction of a Tarski monster. $\endgroup$ – Brett Frankel Dec 18 '12 at 2:50
  • $\begingroup$ wow that answers it! Do you know where I can find such a construction? $\endgroup$ – Henrique Tyrrell Dec 18 '12 at 2:53
  • $\begingroup$ "Every element of this group has finite order, else we could find an infinite proper subgroup, namely the group generated by $x^2$ if $x$ has infinite order." How can you be certain that $<x^2>$ is a proper subgroup? Could it be the original group itself? $\endgroup$ – Code-Guru Dec 18 '12 at 3:20
  • $\begingroup$ Tarski Monsters are not abelian groups, so they are not relevant to this problem. $\endgroup$ – Derek Holt Dec 18 '12 at 8:30
  • $\begingroup$ They may not be abelian, but they have the most desired property. So it's a very good partial answer. $\endgroup$ – Henrique Tyrrell Dec 18 '12 at 12:49
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More generally, you can show that the abelian groups whose proper subgroups are finite are precisely the Prüfer groups $\mathbb{Z}[p^{\infty}]$.

(Mentioned in Kaplansky's book, Infinite abelian groups, exercice 23.)

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  • $\begingroup$ now is this a full answer? $\endgroup$ – Henrique Tyrrell Oct 22 '13 at 17:07
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Consider the set of all $2^n$-th roots of unity, as $n$ ranges over the non-negative integers. An infinite subgroup involves elements of arbitrarily high order, which generate everything below them.

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  • $\begingroup$ Just checked. This is a perfect answer. So the problem was trivial after all. $\endgroup$ – Henrique Tyrrell Dec 18 '12 at 17:43
  • $\begingroup$ In retrospect, at least not hard, if one thinks about the kind of tower we need. $\endgroup$ – André Nicolas Dec 18 '12 at 17:48
  • $\begingroup$ Did not go that deep in group theory. just starting the incantations on that realm. $\endgroup$ – Henrique Tyrrell Dec 18 '12 at 20:47
  • $\begingroup$ Now, does anyone know where I can find the construction of them Tarski monsters? $\endgroup$ – Henrique Tyrrell Dec 18 '12 at 20:47
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    $\begingroup$ @leo: There are $m$ $m$-th roots of unity, so for any infinite subgroup $A$, there must be a sequence $2{n_1}\lt 2^{n_2}\lt \cdots$ such that $A$ contains a primitive $2^{n_i}$-th root of unity. But if $k\lt n_i$, and $g$ is a primitive $2^{n_i}$-th root of unity, and $a$ is a $2^k$-th root of unity, then $a$ is a power of $g$. $\endgroup$ – André Nicolas May 2 '13 at 21:29

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