Cyclic abelian group of order pq

[guildmage]

I'm looking at the exercises of Hungerfod's Algebra. Some looks easy but it seems the proofs are not so obvious. Here's one I'm particularly having a hard time solving:

Let G be an abelian group of order pq with (p,q)=1. Assume that there exists elements a and b in G such that |a|= p and |b| = q. Show that G is cyclic.

Help anyone?

 

[Office_Shredder]

What's the order of ab?

 

[guildmage]

Oh yeah. |ab|=pq because p and q are relatively prime. Whice means ab will generate the whole of G. And hence G is cyclic. Thanks.

 

[Johnson04]

For this problem, in order for the group G to be cyclic, is the abelian condition necessary? In other words, if the problem is restated as: "if a finite group of order pq, where p and q are distinct primes, the the group is cyclic", is it still true?

The reason I asked this question is that in my proof, I didn't see why we need the group to be abelian. Thanks!

 

[Spartan Math]

It's absolutely necessary. Consider the permutation group on three letters (ie. S3), then this is a group of order 6 = 2 *3 and is clearly not cyclic (and definitely not abelian either). However we do have this result:

If G is a group order pq, pq distinct primes say P < q and p does not divide q-1, then G is abelian, hence cyclic. The hard part is proving it's abelian and the cyclic part follows from your initial problem.

There's also a bit more interesting of a problem:

If G is a group of order pq as above and p does q-1, then G is the unique nonabelian group of order pq.