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Math Help - [SOLVED] homomorphism issue on Z_m \to Z_n

  1. #1
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    [SOLVED] homomorphism issue on Z_m \to Z_n

    My textbook asks:

    Write down the formulas for all homomorphisms from \mathbb{Z}_6 into \mathbb{Z}_9.
    Choose a homomorphism \phi:\mathbb{Z}_6\to\mathbb{Z}_9 with a,b\in\{0,1,2,3,4,5\}. Then \phi([a+b]_6)=\phi([a]_6)+\phi([b]_6). What else can we say about \phi? Nothing significant, as far as I can tell.

    But my textbook has another story. According to it, there are only

    3 different homomorphisms, given by the formulas \phi_3([x]_6)=[3x]_9, \phi_6([x]_6)=[6x]_9, or \phi_0([x]_6)=[0]_9, defined for all [x]_6\in\mathbb{Z}_6.
    First of all, I am not completely certain I understand the notation, here. Is \phi_n(\alpha) simply another way of writing [\phi(\alpha)]^n? If so, then by algebra of integers modulo, isn't \phi_n([x]_6)=\phi([nx]_6) true for all n? If not, then what does the notation mean?

    As you can see, I'm a little bit befuddled by this. I would welcome any assistance.

    Thanks!
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  2. #2
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    Write down the formulas for all homomorphisms from \mathbb{Z}_6 into \mathbb{Z}_9.
    If \phi : \mathbb{Z}_6 \to \mathbb{Z}_9 is a homomorphism then it is completely determined by  \phi( [1]_6) since [1]_6 generates the group \mathbb{Z}_6.
    If \phi([1]_6) = [n]_9 then \phi([x]_9) = [nx]_9.

    But if \phi is is homomorphism it means the order of [1]_6 must be dividsibly by order of \phi ([1]_9) = [n]_9. The order of [1]_6 is six. Therefore, the order of [n]_9 in \mathbb{Z}_9 must be have order 1,2,3,6. It cannot have order 2,6 since 2,6\not | 9 and so only two possibilites remain, that the order of [n]_9 must be 1,3. For order 1 we get [n]_9 = [0]_9, for order 3 we get [3]_9,[6]_9 have order 3.

    The the only possibily homomorphism are:
    \phi ([x]_6) = [0]_9, \phi([x]_6) = [3x]_9, \phi([x]_6) = [6x]_9.

    Now check that these are indeed all homomorphisms.
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