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?
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6$\begingroup$ 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." $\endgroup$– JSECommented Oct 30, 2009 at 5:41
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1$\begingroup$ @JSE: So it is. $\endgroup$– Anton GeraschenkoCommented Oct 30, 2009 at 5:43
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2$\begingroup$ More generally every submodule of a free module over a PID is free. $\endgroup$– YCorCommented Jan 27, 2019 at 17:13
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$\begingroup$ I've upvoted this because it is now the first Google hit for "subgroup of free abelian group is free" $\endgroup$– Alex GroundsCommented Apr 10, 2023 at 16:03
2 Answers
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).
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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/… $\endgroup$ Commented Oct 30, 2009 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.
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1$\begingroup$ A simpler version of this proof is in Neumann & Newman, "On Schreier Varieties of Groups", Math. Z. 98 (1967), 196--199. $\endgroup$– Steve DCommented Mar 23, 2010 at 10:33
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2$\begingroup$ 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." $\endgroup$ Commented Oct 17, 2010 at 6:09