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Stuck on Rank(G/H) = Rank(G) - Rank(H) Should be trivial(?)

  1. Dec 19, 2013 #1
    I can't believe I've been stuck on this the whole day... I'd appreciate any help.

    Suppose G is a free abelian group, and even feel free to assume it's finitely generated. H is a subgroup. I'm trying to prove that Rank(G/H) = Rank(G) - Rank(H).

    Also, I'm looking for the most basic proof. (I know there is a proof out there using flatness of the rational numbers as a module etc, but I'm looking for a more direct approach.)

    I've been messing about with elements, but not getting anywhere. Any tips/insights?
     
  2. jcsd
  3. Dec 19, 2013 #2

    jgens

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    Gold Member

    This is essentially the same proof as tensoring with Q, but localization functors are exact, so localizing with respect to the multiplicative set S = Z-0 allows the proof to go through without worrying about flatness.
     
  4. Dec 19, 2013 #3
    That's actually the kind of proof I was trying to avoid. I'm wondering whether I can do it without commutative algebra-type stuff. After all, that kind of proof is very general, for any abelian group G, but I'm willing to assume G is (abelian and) (1) free and (2) finitely generated. So there should be a more basic/direct aproach (?)

    Basically I'm looking for something that just uses straight-forward group theory.
     
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