Cyclic abelian group of order pq

Click For Summary

Discussion Overview

The discussion revolves around the properties of a finite abelian group G of order pq, where p and q are distinct primes. Participants explore the implications of G being abelian and whether this condition is necessary for G to be cyclic.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant presents a problem from Hungerford's Algebra regarding proving that an abelian group G of order pq is cyclic given elements a and b with orders p and q, respectively.
  • Another participant claims that the order of the product ab is pq, leading to the conclusion that G is cyclic.
  • A question is raised about the necessity of the abelian condition for G to be cyclic, suggesting a restatement of the problem without the abelian requirement.
  • In response, a participant argues that the abelian condition is indeed necessary, citing the example of the permutation group S3, which is of order 6 and not cyclic.
  • This participant also mentions a related result about groups of order pq and conditions under which they are abelian and cyclic.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of the abelian condition for the cyclic nature of the group, with some asserting it is essential while others question this requirement.

Contextual Notes

The discussion includes references to specific group properties and examples, but does not resolve the question of whether the abelian condition is strictly necessary for the cyclicity of groups of order pq.

guildmage
Messages
25
Reaction score
0
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?
 
Physics news on Phys.org
What's the order of ab?
 
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.
 
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!
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
820
  • · Replies 13 ·
Replies
13
Views
3K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 26 ·
Replies
26
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K