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Find x in Groups

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Let G be a group with identity e, and suppose that a, b, c, x in G.

    Determine x, given that x2a=bxc-1 and acx = xac.

    2. Relevant equations


    3. The attempt at a solution

    I know the three axioms for group.

    G1. Associativity. For all a, b, c in G, (a * b) * c = a * (b * c).
    G2. Identity. a * e = a = e * a.
    G3. Inverses. a * b = e = b * a.

    But how i will use the axioms to find x?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 19, 2012 #2

    morphism

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    Step 1: Multiply the first equation through by c on the right. What do you notice?
     
  4. Feb 19, 2012 #3

    Dick

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    Start by multiplying both sides (x^2)a=bxc^(-1) by c on the right. See where that leads you.
     
  5. Feb 19, 2012 #4
    so it will be :

    (x^2)*a*c=b*x(c^-1)*c

    (x^2)*a*c=b*x

    this is from the first equation.

    what about acx = xac.

    what i have to to next?
     
  6. Feb 19, 2012 #5

    Dick

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    Use acx=xac. (x^2)*a*c=x*x*a*c.
     
  7. Feb 19, 2012 #6
    so
    (x^2)*a*c=b*x(c^-1)*c

    (x^2)*a*c=b*x

    (x^2)*a*c=x*x*a*c=x*(acx)=b*x

    Thus b=acx

    and x=b*a^(-1)*c^(-1) ??
     
  8. Feb 19, 2012 #7

    Dick

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    b=xac, yes. The rest no. Cancel them one at a time. The group isn't given to be commutative.
     
  9. Feb 19, 2012 #8
    so
    (x^2)*a*c=b*x(c^-1)*c

    (x^2)*a*c=b*x

    (x^2)*a*c=x*x*a*c=x*(acx)=b*x

    Thus b=acx

    a^(-1)*b=a^(-1)*a*c*x

    a^(-1)*b=e*c*x

    a^(-1)*b=c*x

    c^(-1)*a^(-1)*b=c^(-1)*c*x

    c^(-1)*a^(-1)*b=e*x

    therefore x= c^(-1)*a^(-1)*b
     
  10. Feb 19, 2012 #9

    Dick

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    Oh heck. I read your answer wrong. b=xac. not b=acx. Try that again. Order counts here. You were almost right. Sorry.
     
  11. Feb 19, 2012 #10
    so
    (x^2)*a*c=b*x(c^-1)*c

    (x^2)*a*c=b*x

    (x^2)*a*c=x*x*a*c=x*(xac)=b*x

    thus b=xac

    a^(-1)*b=a^(-1)*x*a*c

    a^(-1)*b=x*c

    c^(-1)*a^(-1)*b=c^(-1)*x*c

    c^(-1)*a^(-1)*b=x

    Is it correct now?
     
  12. Feb 19, 2012 #11

    Dick

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    No. Why do you think a^(-1)xa=x?? If you have b=xac cancel c on the right first.
     
  13. Feb 19, 2012 #12
    so
    (x^2)*a*c=b*x(c^-1)*c

    (x^2)*a*c=b*x

    (x^2)*a*c=x*x*a*c=x*(xac)=b*x

    thus b=xac

    b*c^(-1)=x*a*c*c^(-1)

    b*c^(-1)=x*a*e

    b*c^(-1)=x*a

    b*c^(-1)*a^(-1)=x*a*a^(-1)

    b*c^(-1)*a^(-1)=x*e

    Hence

    x=b*c^(-1)*a^(-1)

    ok now?
     
  14. Feb 19, 2012 #13

    Dick

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    You forgot to use xac=acx again. But other than that, it's ok.
     
  15. Feb 19, 2012 #14
    where do i have to use xac=acx ??
     
  16. Feb 20, 2012 #15

    Dick

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    On the third line. You want to cancel x on the right.
     
  17. Feb 20, 2012 #16
    cancel x? i dont understand!
     
  18. Feb 20, 2012 #17

    Dick

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    How did you conclude "thus b=xac"? If you multiply one side of your equality by x^(-1) on the right, you have to multiply the other side by x^(-1) on the right. You can't switch sides. The group might not be commutative. That's why they gave you acx = xac.
     
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