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Math Help - group homomorphism challenges

  1. #1
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    group homomorphism challenges

    Here is the question:
    Suppose that ϕ: Z_50 to Z_15 is a group homomorphism with ϕ(7) = 6
    a)determine ϕ(x)
    b) determine the image of ϕ
    c) determine the kernel of ϕ
    d) determine ϕ^(-1)(3). That is, determine the set of all elements that map to 3.


    --- a) doesn't ϕ(x) = (6/7)x
    b) ok, so the image of ϕ is something that i am confused on, we didn't really talk about in class and was barely covered in my book. My professor told me that the image of ϕ is denoted phi(G) and is the set of all elements of G-bar that are 'hit' by ϕ. so does this mean that the image of ϕ is any element in Z_15 such that that element times 6/7 gets me an element in Z_50?
    c)I understand the kernel. It is any element in Z_50 times 6/7 that gets me the element, that is 0 in Z_15 (an element of 15). but there are not many of these, only {0, 35} i think this is right, but i am a little skeptical about ϕ(x)=6/7
    d)so this is similar to the kernel, but instead of mapping to the identity, 0, it maps to three? if that is the case then i know that 21 is an example, but otherwise i am having a difficult time. thank you for any help.
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  2. #2
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    Re: group homomorphism challenges

    a) try not to use "fractions" for elements of Z15. what you want to do is notice that 7 generates Z50, since gcd(7,50) = 1. now, usually we think of elements of Z50 as being "multiples" of 1, that is: x = 1+1+...+1 (x times). note that 7^2 = 49 = -1 (mod 50), so 7^4 = 1 (mod 50). this means "1/7" is 7^3 (mod 50), which is 43.

    so x = 1x = (43)(7)x = 43x(7) (mod 50). since φ is a homomorphism, φ(43x(7)) = 43xφ(7) = 43x(6) (mod 15).

    but mod 15, 43x(6) = 13x(6) = 78x = 3x (mod 15).
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  3. #3
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    Re: group homomorphism challenges

    Ok this helps. But the sentence " this means "1/7" is 7^3 (mod 50), which is 43." is really confusing. I see why 7^3 is 43 but where do you get this 1/7 from? That is what is confusing.
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  4. #4
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    Re: group homomorphism challenges

    well, see, that's why fractions are confusing.

    what i really mean is 7^{-1} under multiplication mod 50.
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  5. #5
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    Re: group homomorphism challenges

    Thank you i understand it now. My last question is i found from spmeone that the answer to b is 0,3,6,9. But why? This i just dont understand. Everything else i get.
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  6. #6
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    Re: group homomorphism challenges

    well, we know that φ(1) = 3, right?

    so φ(2) = φ(1+1) = φ(1) + φ(1) = 3 + 3 = 6,
    φ(3) = φ(1+1+1) = φ(1) + φ(1) + φ(1) = 3 + 3 + 3 = 9
    φ(4) = φ(1+1+1+1) = φ(1) + φ(1) + φ(1) + φ(1) = 3 + 3 + 3 + 3 = 12
    φ(5) = φ(1+1+1+1+1) = φ(1) + φ(1) + φ(1) + φ(1) + φ(1) = 3 + 3 + 3 + 3 + 3 = 15 = 0.

    now φ(k + 5n) = φ(k) + φ(5n) = φ(k) + nφ(5) = φ(k) + n(0) = φ(k), so we're not going to get any more new values for φ(k).
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