Is a subgroup of a free abelian group free abelian? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:05:43Z http://mathoverflow.net/feeds/question/3405 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3405/is-a-subgroup-of-a-free-abelian-group-free-abelian Is a subgroup of a free abelian group free abelian? Anton Geraschenko 2009-10-30T05:21:46Z 2010-10-17T05:30:03Z <p>It's well-known that that <a href="http://ncatlab.org/nlab/show/Nielsen-Schreier+theorem" rel="nofollow">a subgroup of a free group is free</a>. Is a subgroup of a free <em>abelian</em> group (may not be finitely generated) always a free abelian group?</p> http://mathoverflow.net/questions/3405/is-a-subgroup-of-a-free-abelian-group-free-abelian/3407#3407 Answer by Alon Amit for Is a subgroup of a free abelian group free abelian? Alon Amit 2009-10-30T05:25:03Z 2009-10-30T05:25:03Z <p><a href="http://en.wikipedia.org/wiki/Free%5Fabelian%5Fgroup" rel="nofollow">Yes</a>.</p> <p>(EDIT: If you don't like following links, this is the Wikipedia article on Free abelian groups which, uncharacteristically, contains a complete (and correct) proof of precisely that statement).</p> http://mathoverflow.net/questions/3405/is-a-subgroup-of-a-free-abelian-group-free-abelian/4908#4908 Answer by Danny Calegari for Is a subgroup of a free abelian group free abelian? Danny Calegari 2009-11-10T19:24:32Z 2009-11-10T19:24:32Z <p>A variety of groups \$V\$ is said to have the Schreier property if every subgroup of a free group in the variety is free. It is a classical theorem of Peter Neumann and James Wiegold that the only varieties of groups with the Schreier property are: the (absolutely) free groups, the free abelian groups, and the free exponent \$p\$ abelian groups for \$p\$ prime. </p>