It's wellknown 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?

5$\begingroup$ 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." $\endgroup$ – JSE Oct 30 '09 at 5:41

1$\begingroup$ @JSE: So it is. $\endgroup$ – Anton Geraschenko Oct 30 '09 at 5:43

1$\begingroup$ More generally every submodule of a free module over a PID is free. $\endgroup$ – YCor Jan 27 '19 at 17:13
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).

3$\begingroup$ 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/… $\endgroup$ – Anton Geraschenko Oct 30 '09 at 5:51

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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.

1$\begingroup$ A simpler version of this proof is in Neumann & Newman, "On Schreier Varieties of Groups", Math. Z. 98 (1967), 196199. $\endgroup$ – Steve D Mar 23 '10 at 10:33

1$\begingroup$ Grammar nitpick: 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." $\endgroup$ – Arturo Magidin Oct 17 '10 at 6:09