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

Click For Summary

Homework Help Overview

The discussion revolves around proving that an abelian group \( G \) of order \( pq \) is isomorphic to \( \mathbb{Z}_{pq} \). The participants explore properties of group elements and their orders, particularly focusing on cyclic groups and the implications of Lagrange's theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessity of showing that \( G \) is cyclic to establish isomorphism. They consider the existence of elements of specific orders and question how to demonstrate the existence of an element whose order is \( pq \). There is also a reference to Cauchy's theorem regarding the existence of elements of certain orders in groups.

Discussion Status

The discussion is ongoing, with participants actively questioning assumptions and exploring the implications of known theorems. Some guidance has been provided regarding theorems that relate to the existence of elements of specific orders, but no consensus or resolution has been reached.

Contextual Notes

Participants note that \( p \) and \( q \) are primes and discuss the implications of this on the orders of elements within the group \( G \). There is an emphasis on the need to clarify all premises related to the problem.

Mr Davis 97
Messages
1,461
Reaction score
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?
 
Physics news on Phys.org
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##?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
9K