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All homomorphisms on Z to Z

  1. Oct 26, 2008 #1
    I've started self-studying algebra. So I want to err on the side of getting guidance so I don't get off on the wrong track. This is problem 2.4.4 in Artin.

    Describe all homomorphisms from Z+ to Z+ (all integers under addition). Determine if they are injective, surjective, or isomorphisms.

    So I need ALL the functions f s.t. f(x+y) = f(x) + f(y) for all integers x,y.

    Clearly any linear function f will do this, and these are all isomorphisms.
    Also f(x) = 0 for all x satisfies the definition of the homomorphism. This is not injective, surjective, nor an isomorphism.

    So far so good?

    I don't know of any way to prove that there are no other such functions that will satisfy the definition of a homomorphism.

    Any hints?

    thanks
     
  2. jcsd
  3. Oct 26, 2008 #2
    Z is the infinite cyclic group generated by 1 with identity 0. to specify a homomorphism, one only needs to specify the action of the generator

    for instance, f(1) = 2000 will define a homomorphism by induction. In fact, there should be infinitely many homomorphism.
     
  4. Oct 27, 2008 #3
    Thanks. I'll look into that. I don't think Artin has presented that result yet.

    Since been doing a lot of analysis lately, and ignored that these are integers here. Hence, my initial comment of all linear functions isn't true. Although all linear functions with integer co-efficients should do it.

    But then the inverse may not map to integers everywhere. Hence these may not all be isomorphisms.

    So now the question makes more sense. I'll dig further.
     
  5. Feb 17, 2011 #4
    oh Obviously there are infinitely many homomorphisms on Z to Z

    But could you suggest the number more precisely by using cardinal numbers?
     
  6. Feb 17, 2011 #5
    As was stated above, a homomorphism from a cyclic group is defined by where it sends one generator. You can send 1 to any integer and get a homomorphism from Z to Z, and so there are as many such maps as there are integers.
     
  7. Feb 18, 2011 #6
    For example, H: Z_12 -> Z_5

    and Z_12 is a cyclic group but it cannot create a homorphism to Z_5 by specifying H(1).

    Then Can I understand it as "we can create a homomorphism by defining each H(1) when the homomorphism is an endomorphism on a cyclic group(like Z)"?
     
  8. Feb 18, 2011 #7

    lavinia

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    Everything is determined by what happens to 1
     
  9. Feb 18, 2011 #8
    Sure we can, and in this case it's unique: H(1)=0, the trivial homomorphism.

    The general result is that the homomorphisms between finite cyclic groups Z_m and Z_n themselves form a cyclic group of order gcd(m,n).
     
  10. Feb 18, 2011 #9

    mathwonk

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    you should be able to prove that everything is determined by where 1 goes. If you cannot, you might try an easier book. you should also try to prove that very few of these maps are isomorphisms.
     
  11. Feb 18, 2011 #10

    Landau

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    It's analogous to what you might be used to from linear algebra: any linear map is uniquely determined by its action on basis vectors. (Z is of course a Z-module of rank 1, so it's exactly the same principle.)
     
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