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Abstract Algebra, order of ab is equal to the order of a times the order of b?

  1. Oct 9, 2012 #1
    Abstract Algebra, order of ab is equal to the order of a times the order of b??

    Hi!
    I am working on some problems in abstract algebra and I am stuck at the moment. I hope some of you guys could help me out a little.

    1. The problem statement, all variables and given/known data
    a and b are two elements in a group G.
    Assume that ab=ba.
    Show that if GCD(o(a),(ob))=1, then o(ab) = o(a)*o(b)

    Where o(a) is the order of a. (i.e. a^(o(a))=1.)


    2. Relevant equations
    -


    3. The attempt at a solution
    I call o(a)=n, o(b)=m, o(ab)=k, then show that k=mn.
    Since SGD(m,n)=1, then m and n are coprime integers, and I have this relation: 1=sn+tm, where s and t are some integers.

    However I am stuck now and I am not sure how to use this or where to start.
    So any suggestions would be very appreciated.

    Thanks in advance
     
  2. jcsd
  3. Oct 9, 2012 #2

    jbunniii

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    Re: Abstract Algebra, order of ab is equal to the order of a times the order of b??

    Clearly [itex](ab)^{mn} = 1[/itex], so [itex]o(ab)[/itex] divides [itex]mn[/itex]. Can you show that [itex]\langle ab\rangle[/itex], the subgroup generated by [itex]ab[/itex], contains subgroups of order [itex]m[/itex] and [itex]n[/itex]? If so, what does that imply?
     
  4. Oct 10, 2012 #3
    Re: Abstract Algebra, order of ab is equal to the order of a times the order of b??

    I am not sure why [itex](ab)^{mn} = 1[/itex] is clear.
    [itex](a)^{n} = 1[/itex] and [itex](b)^{m} = 1[/itex] but does that say anything about [itex](ab)^{mn}[/itex] ? Sorry maybe to early in the morning for this, Ill think about what you wrote during the day and see where I get!
    Thanks for the help
     
  5. Oct 10, 2012 #4

    jbunniii

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    Re: Abstract Algebra, order of ab is equal to the order of a times the order of b??

    You are given that [itex]ab = ba[/itex], so [itex](ab)^{mn} = a^{mn} b^{mn} = \ldots[/itex]
     
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