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Cyclic abelian group of order pq

  1. Jul 3, 2009 #1
    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?
  2. jcsd
  3. Jul 4, 2009 #2


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    What's the order of ab?
  4. Jul 5, 2009 #3
    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.
  5. Nov 11, 2009 #4
    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!
  6. Nov 11, 2009 #5
    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.
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