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

In summary: 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##?pleximal. The order of ##ab## is q.
  • #1
Mr Davis 97
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44

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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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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:
  • #6
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##?
 

1. How do you prove that an Abelian group of order pq is isomorphic to Z_pq?

To prove that an Abelian group of order pq is isomorphic to Z_pq, we use the Fundamental Theorem of Finite Abelian Groups. This theorem states that any finite Abelian group is isomorphic to a direct product of cyclic groups. In this case, since the order of the group is pq, we can represent it as Z_p x Z_q. Then, by using the Chinese Remainder Theorem, we can show that this is isomorphic to Z_pq.

2. What is the significance of the order pq in this proof?

The order pq is significant because it is a product of two distinct primes, p and q. This allows us to use the Chinese Remainder Theorem, which states that if two integers are relatively prime, then the system of congruences has a unique solution modulo their product. This is crucial in showing that the group is isomorphic to Z_pq.

3. Can you explain the concept of isomorphism in this context?

In mathematics, isomorphism is a relationship between two mathematical objects that preserves their structure. In the case of an Abelian group of order pq and Z_pq, an isomorphism is a bijective function between the two groups that preserves the group operation. This means that if we apply the group operation to two elements in one group and then apply the isomorphism function, the result will be the same as if we applied the group operation to the corresponding elements in the other group.

4. What is the role of the Fundamental Theorem of Finite Abelian Groups in this proof?

The Fundamental Theorem of Finite Abelian Groups plays a crucial role in this proof by allowing us to represent the group of order pq as a direct product of cyclic groups. This representation makes it easier to show that the group is isomorphic to Z_pq, as we can use the Chinese Remainder Theorem to prove the isomorphism.

5. How does this proof relate to the concept of group theory?

This proof is a demonstration of the concepts of group theory, specifically the properties of Abelian groups and the Fundamental Theorem of Finite Abelian Groups. It also illustrates the use of the Chinese Remainder Theorem, a concept from number theory, in the context of group theory. This proof helps us understand the structure and properties of groups and how they can be represented in different ways while preserving their fundamental characteristics.

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