18
$\begingroup$

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.

  • 0
    @BrettFrankel The construction of Tarski Monster groups is not easy - it uses a rather complicated "graded" small cancellation theory (but this is a useful tool for your toolbox, and is not too complicated in itself). It can be found in the book *The Geometry of Defining Relations in Groups* by A. Yu Ol'shanskii (1991 - translated from the Russian in 1994 I think). The book is easier than the original papers, for some value of "easier"...2014-06-16

2 Answers 2

11

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.)

  • 0
    now is this a full answer?2013-10-22
31

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.

  • 0
    Now I see. Under that hypothesis, it was just equate two things which are both 1. Thanks2013-05-02