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.