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

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SUMMARY

The discussion centers on proving that an abelian group G of order pq is isomorphic to the cyclic group ##\mathbb{Z}_{pq}##. Participants emphasize that if G contains an element of order pq, it is cyclic. They also explore the scenario where G has elements of order p and q, leveraging Cauchy's theorem to establish the existence of such elements. The conclusion drawn is that G remains cyclic under these conditions, confirming the isomorphism to ##\mathbb{Z}_{pq}##.

PREREQUISITES
  • Understanding of abelian groups and their properties
  • Familiarity with cyclic groups and isomorphisms
  • Knowledge of Lagrange's theorem in group theory
  • Comprehension of Cauchy's theorem regarding group elements
NEXT STEPS
  • Study the implications of Lagrange's theorem in group theory
  • Explore Cauchy's theorem in detail and its applications
  • Investigate the structure of cyclic groups and their isomorphisms
  • Learn about the classification of finite abelian groups
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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and educators looking to deepen their understanding of group isomorphisms and properties of abelian groups.

Mr Davis 97
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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?
 
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Mr Davis 97 said:

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?
 
Dick said:
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
 
Mr Davis 97 said:
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.
 
Dick said:
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
 
Last edited:
Mr Davis 97 said:
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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