Any group of 3 elements is isomorphic to Z3

  1. 1. The problem statement, all variables and given/known data
    Prove that any group with three elements is isomorphic to [tex]Z_{3}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Let G be the group of three elements
    We have an isomorphism if given c:G--->[tex]Z_{3}[/tex],
    if c is one-to -one and onto and c(ab)=c(a)c(b)

    First, we check one-to-one
    We want c(a)=c(b) to imply a=b
    My problem here is how to define c(a), c(b).
    Onto:
    We want c(a)=x and want to solve for a?
    c(ab):
    Same problem with not knowing what c(ab) is
     
  2. jcsd
  3. micromass

    micromass 18,924
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    Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=a². Thus the group is {e,a,a²}.

    Define the map G --> Z3 by
    e ---> 0
    a ---> 1
    b ---> 2

    It is easily checked that this is indeed an iso.
     
  4. So c(a)=c(b)
    c(a)=1
    c(b)=2
    1=2 not true, but that means it's not an isomorphism
     
  5. micromass

    micromass 18,924
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    Uh what? c(a) doesnt equal c(b)??? does it?
     
  6. I thought for the one to one part, you assume c(a)=c(b)
     
  7. micromass

    micromass 18,924
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    Yes... never mind...
     
  8. but I assumed that but that amounts to 1=2. How does that work?
     
  9. micromass

    micromass 18,924
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    You assumed that c(a)=c(b), and from that assimption followed that 1=2. So your assumption is wrong, and thus [tex]c(a)\neq c(b) [/tex]
     
  10. Then I guess I don't see how to show 1-1

    onto
    y=c(x)
    Do I just take any element, say a
    y=c(a)=1
    y=1, but we want to solve for x I thought

    c(a)c(b)
    1*2
    2=c(ab)
     
  11. micromass

    micromass 18,924
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    You show 1-1 by a simple proof by contradiction. I'm sorry, but a math major really should be able to do such a thing...
     
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