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Subgroup of Finitely Generated Abelian Group

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that any subgroup of a finitely generated abelian group is finitely generated.

    2. Relevant equations


    3. The attempt at a solution
    I've attempted a proof by induction on the number of generators. The case n=1 corresponds to a cyclic group, and any subgroup of a cyclic group is cyclic, and so generated by one element. Then for the inductive step, I supposed that G was finitely generated, say [itex]G = \mathbb{Z} a_{1} + ... + \mathbb{Z} a_{n}[/itex], and that the result holds for groups generated by fewer than n elements. I've then let [itex]H \le G[/itex], and considered the quotient group [itex]G/\mathbb{Z}a_{n}[/itex], and then hoped that the correspondence theorem would help me out, but so far I can't seem to make it work.

    Am I even attacking this problem correctly?
     
  2. jcsd
  3. Nov 22, 2012 #2

    micromass

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    Try to prove the following general result, it might help:

    Let H be an abelian group. If there exists a finitely generated subgroup K of H such that H/K is finitely generated, then H is finitely generated.
     
  4. Nov 24, 2012 #3
    Does this work?

    Let [itex]K \le H[/itex] be finitely generated where [itex]H/K[/itex] is finitely generated, say [itex]K = \mathbb{Z}a_{1} + ... + \mathbb{Z}a_{n}[/itex] and [itex]H/K = \mathbb{Z}(b_{1} + K) + ... + \mathbb{Z}(b_{m} + K) = \mathbb{Z}b_{1} + ... + \mathbb{Z}b_{m} + K[/itex], where the [itex]b_{i} \in H[/itex]. Let [itex]h \in H[/itex], and consider [itex]h + K \in H/K[/itex]. Then [itex]h + K = y_{1}b_{1} + ... + y_{m}b_{m}[/itex] for some [itex]y_{i} \in \mathbb{Z}[/itex], so [itex]y_{1}b_{1} + ... + y_{m}b_{m} - h \in K[/itex]. But then [itex]y_{1}b_{1} + ... + y_{m}b_{m} - h = z_{1}a_{1} + ... + z_{n}a_{n}[/itex] for some [itex]z_{i} \in \mathbb{Z}[/itex], and so [itex]h = y_{1}b_{1} + ... + y_{m}b_{m} - z_{1}a_{1} - ... - z_{n}a_{n}[/itex]. Thus, [itex]H[/itex] is finitely generated.
     
  5. Nov 24, 2012 #4

    micromass

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    Yes, that works!
     
  6. Nov 24, 2012 #5
    Sorry, I'm going to have to ask for a hint as to where to go from there. My ideas for proofs tend to break down when I want to assume [itex]a_{i} \in H[/itex] for some [itex]a_{i}[/itex]when [itex]G = \mathbb{Z}a_{1} + ... + \mathbb{Z}a_{n}[/itex].
     
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