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More Abstract-Homomorphism&quotient

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The last parts of the problems were:
    1. prove that for each natural n, the additive quotient group Q/Z contains a one and only subgroup of order n and that sub-group is cyclic.
    2. let G,H be two finite sub-groups of Q/Z. prove that G is contained in H if and only if
    o(G)|o(H).
    3. find all homomorphisms from Z/nZ to Q/Z.
    but I've no clue how to start with the 4th part:
    4. find all homomorphisms from Q/Z to Z...

    TNX for all the helpers...


    2. Relevant equations
    none.
    3. The attempt at a solution
    none...
     
  2. jcsd
  3. Dec 7, 2009 #2

    Dick

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    For number 4, consider a nonzero element of Q/Z. You can represent it by a rational n/m with n and m integers. If you add n/m to itself m times, you get 0 in Q/Z. What does that tell you about the image of n/m under the homomorphism?
     
  4. Dec 7, 2009 #3
    For the first one, try finding a more familiar group that's isomorphic to Q/Z (hint: think complex)
     
  5. Dec 8, 2009 #4
    Let n/m+Z be a nonzero element of the quo. group. the order of this element if m (if we assume that gcd (n,m)=1) . the image of n/m under the homo. has to be an element in Z of order m. But in Z, which element has order m?? how come this is the only homo. possible?

    Help is needed :(

    TNX !
     
  6. Dec 8, 2009 #5

    Dick

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    If phi is your homomorphism, it doesn't really say phi(n/m) has order m. It says phi(n/m)*m=0. The only integer satisfying that is 0. So phi(n/m)=0, for all elements of Q/Z. That's why it's the only homomorphism possible.
     
  7. Dec 8, 2009 #6
    so the homo is:
    each and every element in Q/Z goes to 0?

    TNX
     
  8. Dec 8, 2009 #7

    Dick

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    What else could it be?
     
  9. Dec 8, 2009 #8
    right...Tnx a lot!! Hope you'll be able to help me in my next thread too...
     
  10. Dec 8, 2009 #9
    Can you please give me directions about the 3rd one?I
    I'm pretty lame in all the Homomorphism theorems...

    TNX again
     
  11. Dec 8, 2009 #10

    Dick

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    Just start thinking about it. If you know what the image of the element '1' in Z/nZ is, then you know the rest of the homomorphism. What are the possibilities?
     
  12. Dec 8, 2009 #11
    The image of "1" in Z/nZ (which is of course "n" or "0" ) can be each and every one of the generators of the sub-groups from order n...I've proved that there's only one sub-group of order n in Q/Z and it's cyclic....Each one of it's generators are from the form i/n... Am I on the right track?

    TNX a lot!
     
  13. Dec 8, 2009 #12

    Dick

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    Z/nZ={0,1,2,...n-1}. '1' doesn't mean 0 or n. It means 1. But yes, every element of Z/nZ has order n, so the image of an element i under the homomorphism has to satisfy phi(i)*n=0 in Q/Z. That makes it one of the i/n, sure.
     
  14. Dec 8, 2009 #13
    TNX a lot man...You were very helpful!
     
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