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A Group Problem

  1. Jul 17, 2006 #1
    A Group Problem :)

    So.. I have to demonstrate that those following two groups are isomorfic , that there is an isomorphism between those 2 groups :

    [​IMG]

    Now I know that in such a way that an isomorphism might be , there must also be defined a function in G with values in Z3, and is soooo weird... :frown: So pls help :smile:

    PS: How do I write LaTex ? I have TeXaide and I tried to copy paste the translation in all possible ways into the code tags, but dosen't works :grumpy: .
     

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    Last edited: Jul 17, 2006
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  3. Jul 17, 2006 #2

    HallsofIvy

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    First, what are the elements of G? What are the elements of R3? Of course, an isomorphism must take the identity of one group to the identity of the other so there are not very many choices! (And the choices are irrelevant- there may be more than one isomorphism.)
     
  4. Jul 17, 2006 #3
    The G is a set and is defined in my attachment(made from complex elements with the cube equal with 1) ,and Z3 ... I don't know exactly how to explain it mathematicaly but in school we learn it just as Z3.
    Particular case for Zn which has n elements , from 0 to n-1, and is a group, so if you add 1 to 4 from Z5 you don't get 5, but 0.
    I hope I explained well , that thing... One thing more, pls someone who understood the term I was talking about, write it here? :D THX( NOT a native englishman :) ).
     
  5. Jul 17, 2006 #4

    Office_Shredder

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    So Z3 is the closed additive group of 0-2.

    The elements of the z^3 set, under multiplication (call them a-c, where a is 1 and we go counterclockwise around the circle):

    a*a=a a*b=b a*c=c b*b=c b*c=a c*c=b

    Looking at Z3, calling 0-2 A-C, you can get the exact same formulae, making them equivalent.
     
  6. Jul 17, 2006 #5

    StatusX

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    G is a set of complex numbers, but what are these complex numbers, and how many are there? (hint: one of them is 1). Now, you have to construct an isomorphism between this set and Z3. That is, for each element g in G, associate to it a unique element f(g) in Z3, and verify that for all g,h in G, f(g*h)=f(g)+f(h), where the multiplication g*h is done in G and f(g)+f(h) is done in Z3. Note that this means, for example, the identity of G must map to the identiy of Z3.
     
  7. Jul 18, 2006 #6
    Oh, guys thanks a lot, that enlighted me :).
    So I only have to find element of G, construct a function with the property of f:G->Z3 and f(g*h)=f(g)+f(h)...
    Elements of G are the cubic roots of the unity, and I kinda blocked here, there is 1 and there must be other 2 complex ones... help?!
     
  8. Jul 18, 2006 #7
    Ok, I've kinda find out another root but I'm not sure about this one (dosen't sounds so good :) ) :
    i^(4/3)

    Damn, LaTex still dosen't works for me :((, any help here too ?! .
    Thx.
     
    Last edited: Jul 18, 2006
  9. Jul 18, 2006 #8

    Office_Shredder

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    The easy way to use the cubic roots is to not calculate them. For any z^n=1, the roots can be graphed on the unit circle in the complex plane. For n=3, there are going to be 3 even spaced points on the unit circle, with one of them at z=1. If n=5, there would be 5 evenly spaced points, with one of them at z=1. When you start multiplying them together, it's the same as adding them in a counterclockwise fashion (so call the other four roots z1-z4, with z1 the first point counterclockwise from 1). Then z1*z3=z4. You can call 1 z0 for simplicity's sake

    This is most easily verifiable for z^4=1
     
  10. Jul 18, 2006 #9

    HallsofIvy

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    Remember my original question? What are the elements of G and Z3? There are only three elements of each and I wanted you to actually list them. If you can list the elements of the group (at least in principle for an infinite group) it makes finding an isomorphism much easier.

    The "principal third root of unity", commonly represented by [itex]\omega_3[/itex], is [itex]\frac{-1+ i\sqrt{3}}{2}[/itex]. Another root is just [itex]\omega_3^2[/itex] and, of course, [itex]\omega_3^3= 1[/itex] is the third. [itex]i^\frac{4}{3}[/itex] doesn't have anything to do with your original problem.
    1 is, of course, the identity. [itex]\omega *\omega= \omega^2[/itex], [itex]\omega *\omega^2= \omega^3= 1[/itex] and [itex]\omega^2 *\omega^2= (\omega^3)*(\omega)= \omega[/itex]. Those, together with the fact that multiplication of complex numbers is commutative gives the operation table for the group. It might be good practice to actually write out the table.

    Z3 also has 3 elements. We can take them to be 0, 1, 2. Of course, 0 is the identity. 1+ 1= 2, 1+ 2= 3= 0 (mod 3), and 2+ 2= 4= 1 (mod 3). Again, the group operation is commutative.

    Since 1 is the group identity in G and 0 is the group identity in Z3, any isomorphism must map 1 to 0: f(1)= 0. It seems kind of obvious to take f([itex]\omega[/itex])= 1 and f([itex]\omega^2[/itex])= 2. That that is an isomorphism follows from the fact that multiplying powers of the same base is just adding the exponents.
     
    Last edited: Jul 18, 2006
  11. Jul 18, 2006 #10
    @HallsofIvy : Thanks a lot! Now I can get the things started :).
     
    Last edited: Jul 18, 2006
  12. Jul 18, 2006 #11
    Thanks again to all of you, I just finished the problem, and another one (same type), and I totally understood the tehnique, thx!
     
  13. Jul 18, 2006 #12

    HallsofIvy

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    Actually, it's easy to prove that any two groups with 3 elements are isomorphic (same is true of 1 or 2 element groups but not 4 or more elements). Call the elements of one a,b,c (with a the identity), the other x,y,z (with z the indentity). Any isomorphism must map the identity to identity: a to x. Then there are two choices, b to y and c to z or b to z and c to y. You can show that they are both isomorphisms!
     
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