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Math Help - Prove that G has a subgroup of order p^n

  1. #1
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    Prove that G has a subgroup of order p^n

    If G is a finite abelian group and p is a prime such that p^n divides |G|, then prove that G has a subgroup of order p^n.

    Note: This proof is very similar to the First Sylow Theorem, but in this one, G is abelian.

    Attempt at the proof:

    If p^n||G|, then |G| = p^nm, m \in \mathbb{Z}, (p,m) = 1.

    Consider a group H.
    Claim: H \subset G is a group such that |H| = p^n.

    I know that I have to use Lagrange's Theorem to show that [G:H] = m. I just don't know how to do this. Can anyone help?
    Last edited by crushingyen; February 24th 2010 at 06:28 PM.
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  2. #2
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    Quote Originally Posted by crushingyen View Post
    If G is a finite abelian group and p is a prime such that p^n divides |G|, then prove that G has a subgroup of order p^n.

    Note: This proof is very similar to the First Sylow Theorem, but in this one, G is abelian.

    Attempt at the proof:

    If p^n|G, then G = p^nm, m \in \mathbb{Z}, (p,m) = 1.


    This well may be false: you're only given p^n\mid |G| , not that n is the maximal power of p dividing the order of G.

    I'd rather go: let p^m\mid |G|\,\,\,s.t.\,\,\,p^{m+1}\nmid |G|\Longrightarrow by Sylow theorems there exists H\leq G\,\,\,s.t.\,\,|H|=p^m . Now, it's easy to prove that any p-group of

    order p^m has a normal sbgp. of order p^k for any 0\leq k\leq m-1 (by induction, say) . In the present case normality is for free since G is abelian, and still

    we're done.

    Tonio


    Consider a group H.
    Claim: H \subset G is a group such that |H| = p^n.

    I know that I have to use Lagrange's Theorem to show that [G:H] = m. I just don't know how to do this. Can anyone help?
    .
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  3. #3
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    Aw man, that would've been perfect. I forgot to say that you're not allowed to use the Sylow Theorem to answer this.
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