From: uranus!ikastan@uunet.uu.net (Ilias Kastanas)
Subject: Re: Is G free?
Date: 26 Oct 2000 08:30:16 -0500
Newsgroups: sci.math.research
Summary: [missing]

In article <8snp3h$mqo$1@nnrp1.deja.com>,
Robin Chapman  <rjchapman@my-deja.com> wrote:
@In article <39D5A5AA.AD5C908B@its.caltech.edu>,
@qiang lin <cw12@its.caltech.edu> wrote:
@> Let G be  the subgroup of direct product of (say, countably many) copies
@> of  Z whose elements have uniformly bounded coordinates.
@>
@> Is G free abelian group?
@>
@> I guess not
@>
@
@It is free. I seem to remember this is a theorem of Specker, and
@a proof can be found in one of L. Fuchs's books on Abelian groups.



	Wasn't Specker's theorem  "any subgroup of  G  of cardi-
nality  w_1  is free"?


					Ilias



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