Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You could look up the construction of a Tarski monster. – Brett Frankel Dec 18 '12 at 2:50
wow that answers it! Do you know where I can find such a construction? – Henrique Tyrrell Dec 18 '12 at 2:53
"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? – Code-Guru Dec 18 '12 at 3:20
Tarski Monsters are not abelian groups, so they are not relevant to this problem. – Derek Holt Dec 18 '12 at 8:30
They may not be abelian, but they have the most desired property. So it's a very good partial answer. – Henrique Tyrrell Dec 18 '12 at 12:49
up vote 6 down vote accepted

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.)

share|cite|improve this answer
now is this a full answer? – Henrique Tyrrell Oct 22 '13 at 17:07

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.

share|cite|improve this answer
Just checked. This is a perfect answer. So the problem was trivial after all. – Henrique Tyrrell Dec 18 '12 at 17:43
In retrospect, at least not hard, if one thinks about the kind of tower we need. – André Nicolas Dec 18 '12 at 17:48
Did not go that deep in group theory. just starting the incantations on that realm. – Henrique Tyrrell Dec 18 '12 at 20:47
Now, does anyone know where I can find the construction of them Tarski monsters? – Henrique Tyrrell Dec 18 '12 at 20:47
It is not clear to me how the last sentence holds. – leo May 2 '13 at 20:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.