Just by accident I came across something online about the Baer-Specker group, in the Bulletin of the Irish Mathematical Society of all places: http://www.maths.tcd.ie/pub/ims/bulletin/index.php?file=byvol There's a proof in the Easter 1998 issue, which wanders into logic. There's a later article in Winter 2004 issue. Gary McGuire Eugene Salamin wrote:

--- dasimov@earthlink.net wrote:

...

Let P denote the direct product of countably many copies of the integers Z.

E.g., the set {f: Z+ --> Z | f is a function} under addition of functions.

Puzzle: Is P a free abelian group? Prove your answer.

--Dan

This group is known as the Baer-Specker group. The requested proof is said to be in Reinhold Baer, "Abelian groups without elements of finite order", Duke Mathematical Journal, vol. 3 (1937) pp. 68-122. Perhaps someone with online access could look this up and give us a description, hopefully somewhat more concise that of the original paper.

Gene

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Dan asked last week:

...

Let P denote the direct product of countably many copies of the integers Z.

E.g., the set {f: Z+ --> Z | f is a function} under addition of functions.

Puzzle: Is P a free abelian group? Prove your answer.

And Gene just re-expressed interest in the subject. I started writing something about this yesterday, but realized I was on shaky ground. I thought I had a proof that P was not free abelian, but then realized I was making some sort of unwarranted assumptiong about continuous functions on P which depended on some notion of topology of P, which made me uncomfortable. Given a function f : P->Z, just because you know all the f(ei), where ei=(0,0,...,0,1,0,0,...), doesn't mean you know f(a1,a2,a3,a4,...). (The rest of yesterday's argument said that there were only countably many homomorphisms P->Z, since all but finitely many of the ei had to go to zero; but a free abelian group on uncountably many generators -- which P would have to be, it being uncountable itself -- should have uncountably many such homomorphisms. But as I said, this seems hollow to me now.) I second the motion for an answer. --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

Mike Speciner wrote:

Am I missing something? According to mathworld, a free abelian group is an abelian group with no torsion, seemingly meaning that the only element with finite order is the identity (in this case the zero function). Clearly here, no function other than zero has finite order.

Mathword does badly on this one -- its definition is just fine for finitely generated groups, but not in general. An abelian group G is "free abelian" if there exists a set of generators {g_i, i in I}, for some index set I (not necessarily countable, of course), such that every element can be written as a finite sum of generators and their inverses in exactly one way. The group P in question is precisely a counterexample to the hypothesis "That just means the group is torsion-free, right?" --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.