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Math Help - abstract algebra function

  1. #1
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    abstract algebra function

    Let m, n be positive integers with m | n, and let k be any integer. Show that f :
    Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
    is a well-defined function.
    Hint: We know that g : Z
    m ! Z
    m defined by g([x]m) = [x]k
    m is a function if k is a
    positive integer.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sd215 View Post
    Let m, n be positive integers with m | n, and let k be any integer. Show that f :
    Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
    is a well-defined function.
    Hint: We know that g : Z
    m ! Z
    m defined by g([x]m) = [x]k
    m is a function if k is a
    positive integer.
    I'm sorry, I have no idea what this says.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by sd215 View Post
    Let m, n be positive integers with m | n, and let k be any integer. Show that f :
    Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
    is a well-defined function.
    Hint: We know that g : Z
    m ! Z
    m defined by g([x]m) = [x]k
    m is a function if k is a
    positive integer.
    Do you mean that [x]n is the equivalence class of numbers that are equivalent mod n and [x]m^k is the equivalence class of numbers that are equivalent mod m^k? That is, [x]3 is an equivalence class mod 3 so that with n= 6, m= 3 (so that m|n) and k= 2, you are asked to show that the function f that maps any number, x, in Z_6 to x(2^2)= 4x in Z_4[ is "well defined", that is that two numbers in the same equivalence class in Z_n are mapped into numbers in the same equivalence class in Z_m.
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