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It's well-known that that a subgroup of a free group is free. Is a subgroup of a free abelian group (may not be finitely generated) always a free abelian group?

flag	asked Oct 30 2009 at 5:21 Anton Geraschenko♦♦ 13.7k●3●48●108

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I've downvoted this, not because it's a bad question, but because it's answered by the first Google hit for "free abelian group." – JSE Oct 30 2009 at 5:41

@JSE: So it is. – Anton Geraschenko♦ Oct 30 2009 at 5:43

Alon Amit · Accepted Answer · 2009-10-30 05:25:03Z UTC

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Yes.

(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).

answered Oct 30 2009 at 5:25

Alon Amit
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In case the Wikipedia article gets modified, the proof is on page 880 of Lang's Algebra book (3rd ed), and it shows that submodules of free modules over a PID are free: books.google.com/… – Anton Geraschenko♦ Oct 30 2009 at 5:51

Danny Calegari · Answer 2 · 2009-11-10 19:24:32Z UTC

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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.

answered Nov 10 2009 at 19:24

Danny Calegari
1,767●6●12

	A simpler version of this proof is in Neumann & Newman, "On Schreier Varieties of Groups", Math. Z. 98 (1967), 196--199. – Steve D Mar 23 2010 at 10:33
	Grammar nit-pick: you mean, "the only varieties of groups with the Schreier properties are the variety of all groups, the variety of all abelian groups, and the variety of all abelian groups of exponent $p$, for $p$ a prime." – Arturo Magidin Oct 17 2010 at 6:09

Is a subgroup of a free abelian group free abelian?

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