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Show that this is a homomorphism

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that [itex] (\mathbb{Z} _4 , \oplus ) \approx ( \mathbb{Z} ^{0}_{5} [/itex] [itex] , \odot ) [/itex]
    Meaning, they are isomorphic. The 0 means the zero is deleted from the set. We are using circle plus and circle dot because we are not allowed to think of the operations as addition and multiplication yet. My trouble is with defining the operation [itex] \theta [/itex].


    2. Relevant equations

    I know [itex] \theta [/itex] must be operation preserving, that is, [itex] \theta (x \oplus y) = \theta (x) \odot \theta (y) [/itex]

    3. The attempt at a solution

    I tried defining [itex] \theta [/itex] as [itex] \theta ([x])= [x+1] [/itex]
    so, [itex] \theta ([0]) = [1] [/itex] up to [itex] \theta ([3]) = [4] [/itex]
    And now, [itex] \theta ([x] \oplus [y]) = \theta ([x \oplus y]) = [x \oplus y +1] [/itex]
    Did I make the wrong steps there? I'm not sure how to arrive at [itex] \theta ([x]) \odot \theta ([y]) [/itex]
     
    Last edited: Nov 11, 2011
  2. jcsd
  3. Nov 11, 2011 #2

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    Hi Arcana!

    θ needs to map each element.
    So you can define it by defining θ(0), θ(1), θ(2), and θ(3).

    Start with θ(0) and then make an arbitrary choice for θ(1).
    Use the preservation of the operation to deduce what the other values must be.
     
  4. Nov 11, 2011 #3
    [itex] \theta ([0])=[1] [/itex]
    [itex] \theta ([1])=[2] [/itex]
    [itex] \theta ([2])=[3] [/itex]
    [itex] \theta ([3])=[4] [/itex]

    But how do I show it's operation preserving? Do I have to show it using each number, or is there some way to do it using arbitrary elements?
     
  5. Nov 11, 2011 #4

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    Sorry, but this won't work.
    Consider θ(1+1).
     
  6. Nov 11, 2011 #5

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    Basically what you would need to do is the show that each combination of a and b has its operation preserved.
    That is, set up the entire multiplication table, map it, and verify that it matches the other entire multiplication table.

    With the size of your current groups this is very doable.
     
  7. Nov 11, 2011 #6
    Okay, if that's the only way I can certainly do that. thanks for the tip-off about my mapping being wrong. I think with this information I'll be okay here. :)
     
  8. Nov 11, 2011 #7

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    It's not the only way.
    That is, you can make a couple of shortcuts.
    But I guess that is better left for a later exercise.

    Note that the definition of your isomorphism says: "for any elements a and b".
    This means you have to show it "for any elements a and b".
     
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