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Really Easy Group Calculations which i cant do

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    let a,b[tex]\in[/tex]G Assume that a has order 5 and that [tex]a^{3}b=ba^{3}[/tex]

    Prove that ab=ba



    3. The attempt at a solution

    I know there is some magical thing you can multiply by that will make things nice... i just cant figure this out.

    if tried left and right multiplying by a2 and a-2 with no prevail.
     
  2. jcsd
  3. Oct 12, 2008 #2

    Mark44

    Staff: Mentor

    Some questions to help you get going here.
    What does it mean that a has order 5?
    Why did you pick a2 to multiply by?
    What did you get when you multiplied by a2?

    One other thing: at some point you'll want to use the fact that a3b = ba3.
     
  4. Oct 13, 2008 #3
    a has order 5 means that a5=1

    when i multiply by a2 i get b=a2ba3
     
  5. Oct 13, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you multiplied on the other side you must have gotten b=a^3ba^2 as well. So that's a^3ba^2=a^2ba^3. Now what? Can't you see an 'ab=ba' in there somewhere?
     
  6. Oct 13, 2008 #5
    its not fair dick is too good at math.


    Thanks.. feel free to lock
     
  7. Oct 13, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Dick practiced a lot. The rest of us are trying to catch up.
     
  8. Oct 13, 2008 #7
    Well, I find it is often possible to prove it by a series of equation as ab=.....=....=ba, which is clear and forthright.
    e.g. ab=aeb=aa^5b= a^3(a^3b)=a^3(ba^3)=(a^3b)a^3=(ba^3)a^3=baa^5=bae=ba
    But it may still need magics to prove more challenging problem....
     
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