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Group theory finding order of element and inverse

  1. May 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Let a and b be elements of a group,with a^2=e , b^6=e and a.b=b^4.a find its order and express its inverse in form of a^m.b^n

    2. Relevant equations



    3. The attempt at a solution
    (ab)^2=(ab)(ab)=(ab)(b^4.a)=a(b^5)a
    (ab)^3=a(b^5)a(ab)=a(b^5)(a^2)b=a(b^6)=ae=a
    it implies(ab)^6=e
    thus order(ab) divides 6.It is not 3 hence it is 6.

    Help me in solving second part.I can express inverse in form of b^m.a^n but not in the asked form.
     
  2. jcsd
  3. Jun 1, 2012 #2
    you want (ab)^(-1)?

    try (b^4a)^(-1)
     
  4. Jun 1, 2012 #3
    (ab)^-1=(b^4 a)^-1 as algebrat has said.

    Since you can find (ab)^-1 in the form b^m a^n, you should have no problem expressing (b^4 a)^-1 in the form a^m b^n.
     
  5. Jun 1, 2012 #4
    thanks i have got it
    if my solution to finding the order of ab correct or not
     
  6. Jun 1, 2012 #5
    So you deduced that (ab)^2 = a b^5 a
    If you raise that to the third power, you get a b^15 a.

    So apparently e= (ab)^6 = a b^15 a.
    Also, b^4=e

    Hmmm.......
     
  7. Jun 3, 2012 #6
    a=sqrt(e)

    b=6th root of e => a=b^3

    =>(ab)=b^4=>(b^4)^(-1)=b^(-1/4)=e^(-1/24)
     
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