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

1. Mar 16, 2017

### Mr Davis 97

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. Mar 16, 2017

### Dick

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. Mar 16, 2017

### Mr Davis 97

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. Mar 16, 2017

### Dick

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. Mar 16, 2017

### Mr Davis 97

Yes, I think that that is Cauchy's theorem

Last edited: Mar 16, 2017
6. Mar 16, 2017

### Dick

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 $ab$?