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Question in Group theory.

  1. Feb 16, 2009 #1

    MathematicalPhysicist

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    The question:
    Prove that each group of order 4 is isomorphic to Z/4Z or the Klein Group: (Z/2Z)x(Z/2Z).

    Attempt at solution:
    basically I think that a group of order 4 has e,a,b,c then this group can be characterise by the ordering 0,1,2,3 in the group Z/4Z or (0,0),(0,1),(1,0),(1,1) where addition on Klein Group is defined as (x,y)+(z,w)=(x+z,y+w), well it seems like a really easy question, I don't believe that I am asking this. (-:

    What I mean 0<->e, 1<->a, 2<->b,3<->c
    or (0,0)<->e, (0,1)<->a, (1,0)<->b, (1,1)<->c.

    ofcourse in one group 3+1=0 then ca=e but in the other ca=b.

    The real question is how do I show there aren't more options for this structure, I mean why only these two?

    Thanks in advance.
     
  2. jcsd
  3. Feb 16, 2009 #2

    Hurkyl

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    Well, there are more options. e.g. e,a,b,c could correspond to 0,3,1,2 respectively, and so forth.

    There are only 4,294,967,296 different binary operations GxG --> G. I would just exhaust through all of them, proving each one is either isomorphic to Z/4, isomorphic to Z/2 x Z/2, or is not a group.

    Of course, I would exhaust intelligently, by cleverly applying the group laws so as to handle billions of those "not-a-group" possibilities at a time. :wink:


    I'll get you started. Look for an element 'e' satisfying ex = xe = x. If it doesn't exist, then your set of order 4 is not a group. So, assume 'e' does exist. Label the other elements 'a', 'b', and 'c' in any order you please. So now we're down to 262,144 possibilities.

    Let's consider one of our unknowns: there are only four possibilities for the value of ab. Treat each one as a separate case. What can you do now?
     
    Last edited: Feb 16, 2009
  4. Feb 16, 2009 #3

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    For me e is the indifferent term i.e ae=a=ea be=eb=b etc.
    So the possibilities are ab=ba=e or ac=ca=e, now I understand why people dislike abstract algebra, though it would be nice to prove that a degree five equation doesn't have an algebraic solution.
     
  5. Feb 16, 2009 #4

    Hurkyl

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    Hrm. I can't follow your train of thought here -- how did you get here, and what was your idea? Incidentally, you did miss a possibility, since neither of those would be true for the Klein 4-group, no matter how you labelled its elements as a, b, or c.

    I was thinking more along the lines of the fact you know that ab is either e, a, b, or c, and treat each case separately.

    The idea is that you have 9 unknowns (the values of aa, ab, ac, ba, bb, bc, ca, cb, cc), and the group laws are equations. So your goal now is to try and solve that system, and identify each of the solutions.
     
    Last edited: Feb 16, 2009
  6. Feb 16, 2009 #5

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    Yes I understand now, though it seems quite annoying to check all of the possibilities, you say there are ~200,000 possibilities for legitimate 4-groups, correct?
     
  7. Feb 16, 2009 #6

    Hurkyl

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    No; there are ~200,000 different binary operations on {e, a, b, c} that satisfy ex=xe=x. Very, very few of those will actually be groups: I'd guess less than 10, but probably more like 4.
     
  8. Feb 16, 2009 #7

    Tom Mattson

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    How about using the theorems in your book?

    * Every group of order 5 or less is abelian.

    * Apply the Fundamental Theorem of Finite Abelian Groups.

    Done!
     
  9. Feb 17, 2009 #8

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    Tom, I didn't know this theorem, I am following notes from a lecturer at my univ including my textbook.

    But thanks, I'll try it.
     
  10. Feb 23, 2009 #9

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    Well Tom, I found the proof in Saunders MaClane and Birkhoff book Algebra which is quite comprehensive if I might add.

    He uses there the fact that a finite order group is isomorphic to its non trivial divisors, I can add on on the proof and its simplicity but you can check for yourself. (-:
     
  11. Feb 23, 2009 #10

    dx

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    Lagrange's theorem is useful here.
     
  12. Feb 23, 2009 #11

    HallsofIvy

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    I would try to prove this: If G, a group of order 4, has a non-identity member whose order is 2, then G is isomorphic to the Klein group. If not, it is isomorphic to the rotation group. (I think that is basically what dx is suggesting: by Lagrange's theorem, every non-identity member has order 2 or 4.)
     
  13. Feb 23, 2009 #12

    MathematicalPhysicist

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    Yes, this is what Suanders does there.
     
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