Re: [math-fun] Abelian group puzzle

A proof can be found in the book "Group Theory" by Eugene Schenkman, page 67: Theorem II.6.e (Baer, Specker) The group (direct product of countably many copies of an infinite cyclic group) is not free abelian; but its subgroups of countable rank are free abelian. Search for "Baer Specker" gives lots of hits and definitions of this group, but I didn't find a webpage containing a proof. I'm too lazy to type in the one from Schenkman's book.