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Proving a subgroup is equivalent to Z

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    for [itex] n \in N, n \geq 1 [/itex] Prove that [itex] (n^{3} +2n)Z + (n^{4}+3n^{2}+1)Z= Z[/itex]
    2. Relevant equations
    I know subgroups of Z are of the form aZ for some a in Z and also that aZ+bZ= dZ, where d=gcd(a,b)

    3. The attempt at a solution

    So I was thinking if I could prove that the gcd of (n^3+2n) and (n^4+3n^2+1) was 1, then I could make the proof, but I'm struggling to figure out how to find a gcd of two polynomials... I also tried factoring to see if that led anywhere, but it didn't really...

    Then I was thinking that if I could show that 1 was in the group, and since 1 generates Z, that would prove that the group was equivalent to Z... but then I wasn't actually sure that logic was sound.

    Any help or some guidance in the right direction would be appreciated. Thanks!
     
  2. jcsd
  3. Feb 6, 2013 #2
    Your first idea is correct. Do you know the Euclidean algorithm and polynomial division? So that you can check your work, you should find that

    [tex] -(n^3+2n)(n^3+2n) + (1+n^2)(n^4+3n^2+1) = 1[/tex]
     
  4. Feb 7, 2013 #3
    Thank you! For some reason I didn't think polynomial long division was the way to go... I'm always so unconfident when I do these sorts of proofs.
     
  5. Feb 8, 2013 #4
    There is something to be said for confidence (or perhaps intuition?). Unfortunately, it seems like a lot of the time the only way to find that confidence/intuition is to fail/succeed at a tonne of questions.

    Also, something you may notice is that (like this question) there are always many ways to attack a problem. We often cannot determine which way is correct until we have followed a path through to its conclusion and found that it is a dead end. Luckily, even dead-ends often provide insight and understanding into the structure of a problem: Even through failure, we are constantly learning.
     
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