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

  • #1
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Homework Statement


Given that G is an abelian group of order pq, I need to show that G is isomorphic to ##\mathbb{Z}_{pq}##

Homework Equations




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?
 

Answers and Replies

  • #2
Dick
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Homework Statement


Given that G is an abelian group of order pq, I need to show that G is isomorphic to ##\mathbb{Z}_{pq}##

Homework Equations




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?

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?
 
  • #3
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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?
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
 
  • #4
Dick
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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

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.
 
  • #5
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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.
Yes, I think that that is Cauchy's theorem
 
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  • #6
Dick
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Yes, I think that that is Cauchy's theorem

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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