Does anyone know of any interesting examples of an infinite abelian $p$-group which is uncountable?

By non-interesting here I mean the direct sums of cylic and quasi-cylic groups, and totally projective groups. (The class of totally projective $p$-groups is the smallest class of abelian groups which contains the cyclic group of order $p$, is closed under taking direct sums and summands, and contains a group A if and only if it contains, for every ordinal $x$, both $p^{x}A$ and $A/p^{x}A$. It contains all countable abelian $p$-groups, and all groups in this class can be described by a recursive construction.)

The only example outside of these I know is the torsion subgroup of an unrestricted Cartesian product of cylic groups. Does anyone know of any others?

  • 2
    $\begingroup$ Community wiki? Since there might not be a definitive answer, and you seem to want a list. My own answer would be: ncatlab.org/nlab/show/Pr%C3%BCfer+group $\endgroup$
    – Todd Trimble
    Sep 6, 2012 at 11:53
  • $\begingroup$ Sorry, I missed "uncountable". $\endgroup$
    – Todd Trimble
    Sep 6, 2012 at 13:15
  • $\begingroup$ Todd, the Prufer group is also known as the quasicyclic group. I mention it in the second sentence It's also countable. Any divisible $p$-group is a direct sum of copies of this group. As any divisible subgroup of an abelian group is a direct summand, my question is only really about reduced groups, those with trivial maximal divisible subgroup. I don't want to make this community-wiki yet. I'm sort of looking for a list, but I am not sure if any examples other than the ones I mention are known. $\endgroup$ Sep 6, 2012 at 13:18
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    $\begingroup$ For fixed prime p, the torsion subgroup of $\prod\mathbf{Z}_{p^n}$ is an uncountable abelian p-group all of whose Ulm invariants are equal to 1; it is not isomorphic to a direct sum of cyclic groups (if it were, then the Ulm invariants would guarantee that it equals the direct sum of the $\mathbf{Z}_{p^n}$, which it plainly does not). It is also not a direct sum of indecomposable gruops. (Exercise 10.37 in Rotman's "Introduction to the Theory of Groups" 4th edition, though the text before the exercises incorrectly directs the reader to exercise 10.39). $\endgroup$ Sep 6, 2012 at 15:10
  • $\begingroup$ @Arturo Magidin: I assume this group is what OP meant in the penultimate sentence of the question. $\endgroup$
    – user9072
    Sep 6, 2012 at 15:12


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