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Determining when a mapping is an isomporphism

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that ##\phi : \mathbb{Z}_n \rightarrow \mathbb{Z}_n##, where the rule is ##\phi([a]_n) = [ka]_n##. Formulate and prove a conjecture that gives necessary and sufficient conditions on the positive integers ##k## and ##n## which would guarantee that ##\phi## is an isomorphism.

    2. Relevant equations


    3. The attempt at a solution
    I have already shown that is function is a homomorphism. After having worked with a few examples, I found that if either ##n## or ##k## can divide the other number, then ##\phi## would not be an isomorphism. When they say that they want me to give necessary and sufficient conditions, does that mean they want to prove that

    "##\phi## is an isomorphism iff ##k## or ##n## do not divide each other?"

    The only thing that troubles me is that there may be more restrictions upon ##k##.
     
  2. jcsd
  3. Nov 15, 2014 #2

    pasmith

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    Consider the case [itex]n = 10[/itex] and [itex]k = 15[/itex]. Neither divides the other, and yet [itex][a]_{10} \mapsto [15a]_{10} = [5a]_{10}[/itex] is not invertible, because [itex][5]_{10}[/itex] does not have a multiplicative inverse.

    Consider the case [itex]k = 1[/itex] and [itex]n \geq 1[/itex]. Then clearly [itex]1[/itex] divides [itex]n[/itex], but [itex][a]_n \mapsto [a]_n[/itex] is trivially invertible.

    You need to think more carefully about the circumstances in which [itex]\phi[/itex] is invertible. To start with, you might consider what [itex]\phi^{-1}[/itex] can possibly be.
     
  4. Nov 15, 2014 #3
    Actually, the group operation is not multiplication, but the addition operator ##\oplus##, which is defined as

    Edit: For someone reason, latex on physicsforums has been very problematic for me. I had the break the following up into three segments of latex code.

    ##[a]_n## ##\oplus## ##[a+b]_n##
     
  5. Nov 15, 2014 #4

    pasmith

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    The reason that [itex][a]_{10} \mapsto [5]_{10}[a]_{10}[/itex] is not an invertible additive homomorphism is that [itex][5]_{10}[/itex] has no multiplicative inverse.

    The operation which takes [itex][a]_n[/itex] to [itex]\phi([a]_n)[/itex] is not addition of k mod n, but multiplication by k mod n, and that is the operation you need to invert.
     
  6. Nov 16, 2014 #5
    I don't quite understand. We have not even discussed multiplication with respect to the group ##\mathbb{Z}_n##, only the addition operator ##\oplus##. Why is ##[a]_{10}## being mapped to ##[5]_{10} [a]_{10}##?
     
  7. Nov 16, 2014 #6

    pasmith

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    This is the homomorphism you are dealing with:

    Note that [itex][ka]_n = [k]_n[a]_n[/itex], because [tex](k + pn)(a + qn) = ka + n(kq + pa + pqn)[/tex].
     
  8. Nov 18, 2014 #7
    I don't understand what this demonstrates.

    This is my problem: at this point in my algebra course, it does not even make sense to write ##[k]_n [a]_n##, unless you wish to define this as ##[k]_n [a]_n := [k]_n \oplus [a]_n##, which I do not believe is your import.

    So, for all intents and purposes, I do not know what it means to talk about multiplication with respect to ##\mathbb{Z}_n##, nor do I know about multiplicative inverses. My point is, I do not believe the problem requires these concepts, otherwise it would have introduced them.
     
  9. Nov 21, 2014 #8
    Okay, this problem is still troubling my mind. Could someone else possibly take a look at it? My professor said I was on the right track with both of these conjectures:

    ##\phi## is an isomorphism iff ##k## is a prime number such that ##k \nmid n##

    and

    ##\phi## is an isomorphism iff ##gcd(k,n) =1##.

    However, I ran into difficulties with both proofs.
     
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