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Thread: Homomorphic function

  1. #1 Homomorphic function 
    Forum Junior anticorncob28's Avatar
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    I'm reading the chapter in my abstract algebra book about homomorphisms between two groups. It states as a theorem:
    Let f : G --> H be a homomorphism. Then,
    (i) The kernel of f is a normal subgroup of G.
    (ii) The range of f is a subgroup of H.
    I don't understand part (ii). How could that not work? The function f is defined to be from G to H, and being a homomorphism the function must be surjective but not necessarily injective. So it seems to me that the range of f has to be exactly the group H.
    My only possible guess is that homomorphisms don't have to be surjective? Say the function from Z6 to Z3:
    0 1 2 3 4 5
    0 1 2 0 1 2
    Here, every element of Z3 has at least one correspondence in Z6. Are they saying this might not always be true for one group that is a homomorphic image of another?


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  3. #2  
    KJW
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    Quote Originally Posted by anticorncob28 View Post
    I'm reading the chapter in my abstract algebra book about homomorphisms between two groups. It states as a theorem:
    Let f : G --> H be a homomorphism. Then,
    (i) The kernel of f is a normal subgroup of G.
    (ii) The range of f is a subgroup of H.
    I don't understand part (ii). How could that not work? The function f is defined to be from G to H, and being a homomorphism the function must be surjective but not necessarily injective. So it seems to me that the range of f has to be exactly the group H.
    My only possible guess is that homomorphisms don't have to be surjective? Say the function from Z6 to Z3:
    0 1 2 3 4 5
    0 1 2 0 1 2
    Here, every element of Z3 has at least one correspondence in Z6. Are they saying this might not always be true for one group that is a homomorphic image of another?
    I'm not sure, but I strongly suspect that it's not about the definition of a homomorphism, but about what the notation "f : G --> H" means. For this, I don't think that H is necessarily the image of G under f, but that the image of G is a subset of H.


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  4. #3  
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    anticorncob I think you are overthinking the theorem

    First note that any group has 2 trivial subgroups - the identity and itself.

    Second note that group homomorphisms may be injective, surjective or bijective (in the latter case they are, of course, called isomorphisms)

    So given your homomorphism the image set (you called it the range - it's the same thing) may be all of or the identity or "anything in between"

    So forget that "sub" in "subgroup" and concentrate on the "group" bit - you need to show, assuming the group operation is preserved, that the identity in maps to the identity in and likewise for inverses.

    I can give proofs for both, but you would do well to try for your own

    Good luck!
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