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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun