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Homework Help: -Hmomorphism @:Z_2->Z_4 &Z_2->Z_6

  1. Nov 24, 2008 #1
    Urgent!-Hmomorphism @:Z_2-->Z_4 &Z_2-->Z_6

    1. The problem statement, all variables and given/known data
    THis is probbably very easy, just i am kinda bogged down:

    (a) Show that the mapping [tex]\theta:Z_2-->Z_6[/tex] with [tex] \theta(\bar 0)=\bar 0, \theta(\bar 1)=\bar 3[/tex]

    Is an injective ring homomorphism?
    (b) Show that there is no injective ring homomorphism [tex]\theta:Z_2-->Z_4[/tex]
    Proof:

    (a) Well, i said since [tex]\bar 0 =/=\bar 3[/tex] in Z_6 then such theta is injective.
    Now to establish homomorphism, i proceded

    [tex]\theta(\bar 0 \bar 1)=\theta(\bar 0)=\bar 0[/tex]
    [tex]\theta(\bar 0)\theta(\bar 1)=\bar 0[/tex] so they are equal

    [tex]\theta(\bar 0+\bar 1)=\thta(\bar 1)=\bar 3[/tex]
    [tex]\thetea(\bar 0)+\theta(\bar 1)=\bar 3[/tex]

    So, i concluded that we have a homomorphism

    Well, this didn't cause me any problems, as far as my understanding goes. However, i am having trouble on the second part:

    (b) I started like this:

    In order for [tex]\theta:Z_2-->Z_4[/tex] to be a homomorphism, if [tex]\theta(\bar a)=\bar b, \bar a \in Z_2, \bar b \in Z_4[/tex] then we should have the following [tex]o(\bar b)|o(\bar a)[/tex] ( i think there is a theorem that says this)

    So, looking at the orders of the elements in both rings, we notice that the only such possibility is:

    [tex]\theta(\bar 0)=\bar 0, \theta(\bar 1)=\bar 2[/tex]

    Now, i am failing to show that this is not an injective homomorphism.

    Here is what i am doing:

    [tex]\theta(\bar 0 \bar 1)=\theta(\bar 0)=\bar 0[/tex]

    [tex]\theta(\bar 0)\theta(\bar 1)=\bar 0[/tex]

    [tex]\theta(\bar 0 +\bar 1)=\theta(\bar 1)=\bar 2[/tex]

    [tex] \theta(\bar 0)+\theta(\bar 1)=\bar 2[/tex]

    So, by this reasoning it looks to me that such a mapping is a homomorphism.

    However, now everything that i am trying is also 'showing' that such a theta is also injective.

    SO, how to show that theta is not injective/????????



    Many thanks in return!
     
  2. jcsd
  3. Nov 24, 2008 #2

    Office_Shredder

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    Re: Urgent!-Hmomorphism @:Z_2-->Z_4 &Z_2-->Z_6

    [tex]\theta(1) = 2 [/tex] gives us

    [tex]\theta(1^2) = \theta(1)^2 = 2^2 = 4 = 0[/tex]

    Whoops
     
  4. Nov 24, 2008 #3
    Re: Urgent!-Hmomorphism @:Z_2-->Z_4 &Z_2-->Z_6

    Well since [tex]\theta:\bar 1->\bar 0, and \bar 1-> \bar 2[/tex] this, from my point of view, would mean that theta is not a mapping at all right, since it is violating the definition of a mapping, right? To show that a mapping is not injective wouldn't we need something like this:

    [tex] If, x_1=/=x_2=/>\theta(x_1)=/=\theta(x_2)[/tex]

    Or, if [tex]\theta(a)=\theta(b)=/>a=b[/tex] ??

    [tex]=/>[/tex] stands for "Does not follow"

    So, how would that tell us that theta is not injective? Pardone my ignorance!
     
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