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Proving that an Abelian group of order pq is isomorphic to Z_pq

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Given that G is an abelian group of order pq, I need to show that G is isomorphic to ##\mathbb{Z}_{pq}##

    2. Relevant equations


    3. The attempt at a solution
    I am trying to do this by showing that G is always cyclic, and hence that isomorphism holds. If there is an element of order pq, then we immediately see that G is cyclic.

    If there is an element x of order p, I want to show that the there is an element y not in the cyclic subgroup generated by x, such that the order of xy is pq, which would mean that G is cyclic, right. How could I go about doing this?
     
  2. jcsd
  3. Mar 16, 2017 #2

    Dick

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    p and q aren't just any old numbers, right? You should state all of the premises explicitly. Can you say whether there are elements of order p and q?
     
  4. Mar 16, 2017 #3
    Sorry. p and q are primes. By Lagrange, all of the element of G must be of order 1, p, q, or pq. THe identity is the only element of order 1. And if there is an element of order pq then G is autmatically cyclic. Now I want to show that if there are only elements of order p and q, that G is still cyclic
     
  5. Mar 16, 2017 #4

    Dick

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    Do you know the theorem that tells you that if a prime p divides the order of G, then G has an element of order p? This would be a great help for your other thread as well.
     
  6. Mar 16, 2017 #5
    Yes, I think that that is Cauchy's theorem
     
    Last edited: Mar 16, 2017
  7. Mar 16, 2017 #6

    Dick

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    Yes, it is. That means you have an element ##a## of order p and another element ##b## of order q. What's the order of ##ab##?
     
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