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Algbra Proof

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data

    A finite,commutative grp is not cyclic


    it has a subgrp that is isomorphic to (Z mod p) x (Z mod p) , for a prime p.

    2. Relevant equations

    I am having trouble and I would appreciate a hint.

    3. The attempt at a solution

    Let G be a finite commutative grp with order n.

    Assume that G is not cyclic. This means that there are no elements in G which are relatively prime with n and n is not a prime otherwise it would be cyclic.
    Last edited: Nov 3, 2008
  2. jcsd
  3. Nov 3, 2008 #2


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    This doesn't make sense. Instead, you want to notice that n must have a repeated prime factor. The fact that G is commutative is vital here.
  4. Nov 5, 2008 #3
    Could you give me an example of this? I am having trouble without an example and I had the misconception that "comm. iff cyclic". I now see the theorem in the book and it's only one way.

    There is also a note in the book that, A subgrp of a comm. grp has the same left and right cosets.
  5. Nov 5, 2008 #4


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    Example of what? A commutative noncyclic group? (Z mod p) x (Z mod p) is one!
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