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Thread: [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 $\displaystyle \mathbb{Z}_6$ into $\displaystyle \mathbb{Z}_9$.
    Choose a homomorphism $\displaystyle \phi:\mathbb{Z}_6\to\mathbb{Z}_9$ with $\displaystyle a,b\in\{0,1,2,3,4,5\}$. Then $\displaystyle \phi([a+b]_6)=\phi([a]_6)+\phi([b]_6)$. What else can we say about $\displaystyle \phi$? 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 $\displaystyle \phi_3([x]_6)=[3x]_9$, $\displaystyle \phi_6([x]_6)=[6x]_9$, or $\displaystyle \phi_0([x]_6)=[0]_9$, defined for all $\displaystyle [x]_6\in\mathbb{Z}_6$.
    First of all, I am not completely certain I understand the notation, here. Is $\displaystyle \phi_n(\alpha)$ simply another way of writing $\displaystyle [\phi(\alpha)]^n$? If so, then by algebra of integers modulo, isn't $\displaystyle \phi_n([x]_6)=\phi([nx]_6)$ true for all $\displaystyle n$? 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 $\displaystyle \mathbb{Z}_6$ into $\displaystyle \mathbb{Z}_9$.
    If $\displaystyle \phi : \mathbb{Z}_6 \to \mathbb{Z}_9$ is a homomorphism then it is completely determined by $\displaystyle \phi( [1]_6)$ since $\displaystyle [1]_6$ generates the group $\displaystyle \mathbb{Z}_6$.
    If $\displaystyle \phi([1]_6) = [n]_9$ then $\displaystyle \phi([x]_9) = [nx]_9$.

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

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

    Now check that these are indeed all homomorphisms.
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