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Proving converse of fundamental theorem of cyclic groups

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    If G is a finite abelian group that has one subgroup of order d for every divisor d of the order of G. Prove that G is cyclic.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2009 #2

    Tom Mattson

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    Post what you've done on this problem please.
     
  4. May 7, 2009 #3

    matt grime

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    Can you check the question. C_2 x C_2 has subgroups of orders 1,2 and 4, but is not cyclic.
     
  5. May 7, 2009 #4

    Dick

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    I think the point is that it is supposed to have ONE subgroup of each order. Your example has several subgroups of order 2.
     
  6. May 8, 2009 #5

    matt grime

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    And that's why we have the word 'exactly'.
     
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