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Show the map is isomorphic

  1. Nov 5, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Suppose H is an infinite cyclic subgroup of Z. Show that H and Z are isomorphic.

    2. Relevant equations

    We know that any infinite cyclic group H isomorphic to Z.
    H = <a> ≠ <0>
    |a| = ∞

    3. The attempt at a solution

    Define f : Z → H | f(k) = ak for all k in Z. We want to show that f is injective, surjective and operation preserving.

    ( Note, I think that map is correct. Since Z is an additive group in this case i think ak is the same as saying ka )

    Case : Injectivity. Suppose that f(k) = f(q) for integers k and q. We want to show k = q. So :

    f(k) = f(q)
    ka = qa
    (k-q)a = 0

    We know that |a| = ∞ and (k-q) is in Z. Hence our equation reduces to k-q = 0 and thus k=q. Therefore f is injective.

    Case : Surjectivity. Pick r in H. Then f(k) = ka = r for some integer k. ( Having a bit of trouble with this one ). So :

    Case : Operation preserving ( Homomorphic ). To show f is a homomorphism, we must show that f(k+q) = f(k) + f(q) for integers k and q. So :

    f(k+q) = ka + kq = f(k) +f(q)

    Hence f is a homomorphism.

    Now if I could get a bit of help showing that f is surjective, that will mean f is a bijective homomorphism aka an isomorphism and thus Z≈H and H≈Z.
     
  2. jcsd
  3. Nov 5, 2012 #2

    Dick

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    That is pretty good. The DEFINITION of <a> is Za. So if r is in <a> then r=ka for some k in Z. So f(k)=r. It's kind of obviously surjective.
     
  4. Nov 5, 2012 #3

    Zondrina

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    Yeah I didnt know whether I had to actually justify it or to say it was trivial. I guess I got my answer then.

    Thanks
     
  5. Nov 5, 2012 #4

    Dick

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    You ALWAYS have to justify it. Quoting a definition and showing why it works is a fine way to do that. Just saying 'it's trivial' is criminally lame. Don't do that! You didn't think it's trivial, why should someone else? Please forget my 'kind of obvious' comment.
     
    Last edited: Nov 5, 2012
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