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Prove: G is an abelian group

  1. Jan 31, 2007 #1
    Please HELP!!

    So, I have to go about proving the following, but I have no idea where to even start:

    I. Let S = R – {3}. Define a*b = a + b – (ab)/3.
    1. Show < S,*> is a binary operation [show closure].
    2. Show < S,*> is a group.
    3. Find *-inverse of 11/5

    II. Let G be a group with x,y contained in G .
    Prove: (xy)^2 = x^2 y^2 which implies xy = yx.
    xy = yx which implies (xy)^2 = x^2 y^2.

    III. Let G be a group with x • x = e ,  For any x contained in G.
    Prove: G is an abelian group.

    Honestly, any help would be great because I really have no idea where to even start for any of these!
    P.S. - This is not homework, but simply review for a test coming up in Abstract Algebra.
     
    Last edited: Jan 31, 2007
  2. jcsd
  3. Jan 31, 2007 #2

    HallsofIvy

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    If you really have "no idea where to even start", then you should open your textbook and review the definitions.
    R-{3} is all real numbers except 3. To "show closure" you need to show that if a and b are two such numbers then so is a+ b- (ab)/3. The onlything you really need to show is that a+ b- (ab)/3 cannot be equal to 3 if neither a nor b is.

    To show that this is a group, show that each of the "axioms" or requirements for a group are satisfied: associativity of the operation, existence of an identity, inverses for each element.
    If you know what an "inverse" is, then finding the inverse of 11/5 is easy. If you don't, look it up.

    Do you mean "prove that (xy)^2= x^2y^2 implies xy= yx" and it converse?
    The second one is easy- write out what (xy)^2 is.

    What is (xy)(yx)? What is the inverse of xy? But each member of G is its own inverse.

    When you haven't been studying it at all?
     
    Last edited: Feb 1, 2007
  4. Jan 31, 2007 #3

    mathwonk

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    closure is a litle tricky but easy if you know about solutions to quadratic equations.

    i.e. X^2 - (a+b)X + ab can only equl zero if X = a or b.

    your statement has this form for X = 3.
     
  5. Jan 31, 2007 #4
    Actually, the course has no textbook at all
    And secondly, I'm more of a differential equations kind of guy that likes having practical applications to real topics rather than abstract concepts that serve no purpose in the physical world.
     
  6. Jan 31, 2007 #5

    mathwonk

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    bah, what is this "wheel" as you call it good for?
     
  7. Feb 1, 2007 #6
    What determines something having purpose in the physicsal world to you? I'm pretty sure that abstract algebra has applications to quantum physics.
     
  8. Feb 1, 2007 #7

    matt grime

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    Hmm. Right. All thos differential equations you're solving have loads to do with the ones you actually meet in the real world....

    Groups and algebra in general control physics, so you might want to reevaluate your ideas: gauge groups, lorentz transformations, hecke algebras, equivariant coherent sheaves on a variety, A and B models, braid groups, blob algebras.....
     
  9. Feb 1, 2007 #8
    So, I solved the first two, yet still remain a little stuck on the third one, dealing with x • x = e.
     
  10. Feb 1, 2007 #9

    mathwonk

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    thats got to be the easiest one, a calculation.

    oh im sorry thats a hypothesis. its really easy though try it some more.
     
  11. Feb 1, 2007 #10
    So, I'm basically giving up on that last one. If anyone knows how to do it, any advice would be incredible. Thanks.
     
  12. Feb 1, 2007 #11

    Dick

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    x*x=e for any x is the same as saying x^(-1)=x for any x. Now calculate (x*y)*(y*x)^(-1)=e. Which is the same as x*y=y*x.
     
  13. Feb 2, 2007 #12

    HallsofIvy

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    That's what I said, "What is (xy)(yx)? What is the inverse of xy? But each member of G is its own inverse" ten posts back!
     
  14. Feb 2, 2007 #13

    Dick

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    Indeed you did. Sorry for not thoroughly reading the whole thread. Like rbzima.
     
  15. Feb 2, 2007 #14
    Sorry, just a little confusion going on here. LOL. Anyway, I managed through it. Thanks a lot everyone for all the help you've given. It means the world to me. I think now I can see much clearer how to initiate problems like these in the future.
     
  16. Feb 2, 2007 #15

    HallsofIvy

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    I wasn't complaining about you Dick! I was a bit annoyed that rbzima had ignored it!

    If xx= e for every x, then (xy)(yx)= x(yy)x= xex= xx= e. That is, xy and yx are inverses. Since every element is its own inverse (that's what xx= e means), xy= yx.
     
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