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Math Help - Isomorphism

  1. #1
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    Exclamation Isomorphism

    Let n be a positive integer. Define f : Z ! Zn by f(a) = [a]. Is f a homomorphism? One-to-one? Onto? An isomorphism? Explain.

    Can any one help? Thanks very much.
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  2. #2
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    Quote Originally Posted by suedenation
    Let n be a positive integer. Define f : Z ! Zn by f(a) = [a]. Is f a homomorphism? One-to-one? Onto? An isomorphism? Explain.

    Can any one help? Thanks very much.
    Since [ ] is the greatest integer function we have,
    [\, a \, ]=a for all integers.

    1)Onto: We need to show that for all x\in\mathbb{Z}_n we can find such a y\in \mathbb{Z} such as, [\, y\, ]=x which is true if you take x=y.

    2)One-to-One: We need to show that if [\, x \, ]=[\, y \, ] then, x=y. Based on the first paragraph that x,y are integers, we have that x=y

    3)Homomorphism: We need to show that,
    [\, x+y\, ]=[\, x\, ]+_n[\, y\, ]
    Because, x,y are integers we have,
    x+y=x+_ny. Which is definitely not true.
    -----
    Thus, this map is not an isomorphism. Think about it, how can you have an isomorphsim between a finite and an infinite set? Impossible.
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  3. #3
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    Actually, [a] is the equivalence class of a modulo n; i.e., [a]={k:k in Z and k=a (mod n)}.
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  4. #4
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    :o

    Well, the method is the same as ThePerfectHacker has demonstrated.


    For the homomorphism part, check whether <br />
f(x+y)=[\, x+y\, ] is equal to f(x)+f(y)=[\, x\, ]+[\, y\, ].<br />
Not hard at all

    Ifr f were to be one-to-one, you would have had to show that x\neq y \Rightarrow f(x)\neq f(y). Try testing this, with integers like a and 2a.

    The onto part is also not hard. Just consider an equivalence class [a], and find an integer (obvious!) to map to this.


    And, since an isomorphism needs to be one-to-one, this (is?/is not?) an isomorphism.

    Try it.
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