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Math Help - what is [x]?

  1. #1
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    what is [x]?

    I have to prove Y: Z -> Zm defined by Y(x) = [x] is a homomorphism, but I don't remember what [x] means. Any help would be greatly appreciated! Thanks!
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  2. #2
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    Ok, I figured out [x] (it's the congruence class of x, yay!), but I'm given that the mapping in my first post is a homomorphism and I'm told to prove it. In order to prove that, I would have to show [x + y] = [x] + [y].
    I don't see how that is true; for mod 3, [0] = {..., -6, -3, 0, 3, 6, ...} and
    [1] = {..., -5, -2, 1, 4, 7, ...}. According to that equation, [0 + 1] = [0] + [1] = {..., -11, -5, 1, 7, 13, ...} which would be [1] with respect to mod 6.
    Am I on the right track here, or am I completely off?
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  3. #3
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    Recall the definition of homomorphism.

    You need to show that Y(xy)=Y(x)Y(y) for all x,y in Z.

    So, Y(xy)=[xy] =[x][y]=Y(x)Y(y). Since you can always choose x to be the identity or 1, so [xy]=[x][y] will always be true if x,y in Z.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by mdoyle View Post
    Ok, I figured out [x] (it's the congruence class of x, yay!), but I'm given that the mapping in my first post is a homomorphism and I'm told to prove it. In order to prove that, I would have to show [x + y] = [x] + [y].
    I don't see how that is true; for mod 3, [0] = {..., -6, -3, 0, 3, 6, ...} and
    [1] = {..., -5, -2, 1, 4, 7, ...}. According to that equation, [0 + 1] = [0] + [1] = {..., -11, -5, 1, 7, 13, ...} which would be [1] with respect to mod 6.
    Am I on the right track here, or am I completely off?
    You're on the right track but you're not adding the cosets properly. Remember that if A, B \subset G then A+B = \{a+b : a \in A, b \in B\}. For instance 3+7=10 should be in there.
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