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Proof help

  1. Nov 5, 2006 #1
    If G is a group and x, y are in G. Show that o(x) = o(y^-1xy), where o(x) means order of x.

    thanks
     
    Last edited: Nov 5, 2006
  2. jcsd
  3. Nov 5, 2006 #2

    radou

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    Which is your group operation?
     
  4. Nov 5, 2006 #3

    0rthodontist

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    Why not try computing a few powers of y-1xy and see if that gives you any clues?
     
  5. Nov 6, 2006 #4
    isn't y-1xy=x since y-1y=e. so o(y-1xy)=o(x).
     
  6. Nov 6, 2006 #5

    0rthodontist

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    Why would you think that? Not all groups are commutative.

    What is (y-1xy)2 equal to?
     
  7. Nov 6, 2006 #6

    matt grime

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    Do not assume your group is commutative unless expressly told it is. yxy^-1 is most definitely not the same as x in general, and rule number 1 is that you should never assume otherwise unless told you may. I'm sure this was hammered home in your first lesson (and if not it should have been).

    Now, please try to follow the hint given. What is (yxy^-1)^2? (yxy^-1)^n?

    Remember that the order of z is the least positive n with z^n=1, so it obviously suffices to show that the order of z is less than the order of x, since by symmetry the order of x is then less than the order of z, so they are equal.

    You should go through that argument a couple of times to see why it is true, as well.
     
  8. Nov 6, 2006 #7
    (y-1xy)2 =y-2x2y2 =y-2+2x2=x2
     
  9. Nov 6, 2006 #8

    0rthodontist

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    I don't think you're getting it. Could you provide what you think the justification is for each of those steps? This means working directly from the group axioms. You have three equal signs, which means you'll need three justifications.

    There's more than one problem here. First you say that (y-1xy)2 =y-2x2y2. That isn't necessarily true. Then you say that y-2x2y2 =y-2+2x2, which again is not necessarily true.

    To get you started,
    (y-1xy)2 = (y-1xy)(y-1xy). What does that equal? You can use the associative law and the definition of inverses.
     
    Last edited: Nov 6, 2006
  10. Nov 6, 2006 #9

    matt grime

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    For the fourth (?) time: groups are not to be presumed commutative, you cannot do this. xy=/=yx. You cannot just swap round the order of x and y at will.
     
  11. Nov 6, 2006 #10
    i see. how about this..
    (y-1xy)(y-1xy) = y-1xyy-1xy = y-1x2y
     
    Last edited: Nov 6, 2006
  12. Nov 6, 2006 #11

    matt grime

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    Yes. Now what about (yxy^-1) raised to the n'th power?
     
  13. Nov 6, 2006 #12
    that will be...
    (y-1xy)n = y-1xny
     
  14. Nov 6, 2006 #13

    NateTG

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    And, now what happens when [itex]n=o(x)[/itex]?
     
  15. Nov 6, 2006 #14

    mathwonk

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    2015 Award

    1) conjugation is an isomorphism.
    2) isomorphisms preserve order of elements.
     
  16. Nov 7, 2006 #15
    xn= e.what are the next steps in this proof?
     
  17. Nov 7, 2006 #16

    matt grime

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    Can you please try and put two and two together? You're asked to raise something to the power n, and then asked to consider what happens when n is the order of x. Now, please, try to think what that might mean.
     
  18. Nov 7, 2006 #17
    thanks. that was easy.
     
  19. Nov 7, 2006 #18

    0rthodontist

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    You're not done yet, though, because you also must show that no smaller positive power of y-1xy equals e. Pick the smallest power (y-1xy)k that equals e, and show that k = |x|. (the order of x)
     
  20. Nov 8, 2006 #19

    HallsofIvy

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    You just said "xn= e". Use that in your formula (y-1xy)n= y-1xny!

    However, o(x) is defined as the smallest n such that xn= e. Just showing that (y-1xy)n= e is not enough to say that the order of y-1xy is n.
     
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