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Abstract Algebra group problem.

  1. Feb 5, 2007 #1
    ProbelmLet p and q be distinct primes. Suppose that H is a porper subset of the integers and H is a group under addition that contains exactly three elements of the set {p, p+q, pq, p^q, q^p}. Determine which of the following are the three elements in H.
    a.pq, p^q, q^p
    b. p+q, pq, p^q
    c. p, p+q, pq
    d. p, p^q, q^p
    e. p, pq, p^q

    2. Relevant equations
    The answer is e, as it's given in the book.

    3. The attempt at a solution
    My basic strategy here was to try and show it's only closed under addition for e. It's easy to see that adding p to itself q times yields pq, however the same cannot be said for p^q. This didn't seem satisfactory then. My next idea, the one that MIGHT be right, but i'm very much not sure, is that e contains the only terms that are integer factors of p. You know p, pq, and p^q are integers when divided by p, and furthermore they are the only ones that are (since p and q are prime). However, I don't know if this is even signficant, and I certainly don't see the signifigance under a group who's operation is addition.

    EDIT - Actually I suppose it's significant in that being an integer multiple it can be seen as "P+P+P+....." an integer number of times. Is that all there is to it? It seems a bit of an odd problem if so.
    Last edited: Feb 5, 2007
  2. jcsd
  3. Feb 5, 2007 #2


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    Hint 1: H has to be a proper subset of Z
    Hint 2: If x and y are coprime, then there exists integers a and b such that ax + by = gcd(x,y)
  4. Feb 5, 2007 #3
    I've been fiddling for a little while and can't seem to come up with anything satisfactory. The crux of the problem seems to be that the items in the group must be integers, however I can't seem to find the relevance of your second hint. Having fiddled with it, it doesn't seem to get me anywhere.

    Is it not important that only the terms in e can have p factored out and have the resulting term still be integer valued?
  5. Feb 5, 2007 #4


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    Yes, that's good. Since p, pq, and pq are integer multiples of p, they can appear in the subgroup of Z consisting of all integer multiples of p, which is indeed a proper subgroup. Now with all the other options, a-d, you can't factor out p from each of the three elements. So p is never a common factor of the elements in a-d. But you need to show that for each of a-d, the three elements have no common factors at all.

    The point of my hint was that if in each of a-d, you can find a pair of elements that are coprime, e.g. in b, pq and p+q are coprime. Therefore, there exist integers a and b such that a(pq) + b(p+q) = 1. But then 1 would be in H, and hence H would be all of Z, contradicting Hint 1.
  6. Feb 6, 2007 #5

    Thanks. That makes perfect sense. I wasn't thinking more along the lines of showing a-d were not groups than were not subgroups. I suppose it then makes sense that I could never explain that they weren't, because they are! Thanks alot, that really helped.
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