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Group Theory: Ord(x) = Ord(g^-1xg)

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    If x and g are elements of the group G, prove that |x|=|g-1xg|. Deduce that |ab| =|ba| for all a,b [itex]\in[/itex] G


    2. Relevant equations



    3. The attempt at a solution
    I've written a proof but does not seem quite right:
    if (g-1xg)n= xm then n must be equal m
    g-nxngn = xm then
    xngn =gnxm
    xn =gnxmg-n
    xm =g-nxngn
    xn = gn(g-nxmgn)g-n
    xn = xm
    n = m
     
  2. jcsd
  3. Nov 15, 2011 #2

    Dick

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    There's more than one thing wrong with that. The problems start with (g^(-1)xg)^n=g^(-n)x^ng^n. That's only true if the group is commutative. I'll give you a big hint. (g^(-1)xg)^2=(g^(-1)xg)(g^(-1)xg). What's that?
     
  4. Nov 15, 2011 #3
    (g-1xg)2 = (g-1xg)(g-1xg) => g-1x2g
    so by induction (g-1xg)n = g-1xng
    and?
     
  5. Nov 16, 2011 #4

    Dick

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    Now it's your turn. Think about what order of x and order of g^(-1)xg mean. Show they are the same number.
     
  6. Nov 16, 2011 #5
    g-1xng = e
    xng=ge
    xn = g.g-1
    xn = e

    Thanks.
     
  7. Nov 16, 2011 #6

    Deveno

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    just because g-1xng = e, and xn = e, does NOT mean that |g-1xg| = |x|.

    at best, it only shows that |g-1xg| divides |x| (assuming by n, you mean |x|).

    you need to show that no smaller positive integer will work (this isn't that hard, try proof by contradiction).

    then there's still part 2...
     
  8. Nov 16, 2011 #7

    HallsofIvy

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    It is also, by the way, not true in a general group that is [itex]x^m= x^n[/itex], m= n.
     
  9. Nov 16, 2011 #8

    Deveno

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    right. an obvious example is in the multiplicative group of non-zero real numbers.

    (-1)2 = 12 = 1, but -1 ≠ 1.
     
  10. Nov 17, 2011 #9
    A bit of a blow. I thought I finished but apparently it requires much more careful thinking.

    The choice of n is the smallest positive integer that xn=e. So any m < n would contradict this. What's to prove?
     
  11. Nov 17, 2011 #10
    Prove that [itex](g^{-1} x g)^m = e[/itex] cannot happen for a positive integer [itex]m < n[/itex]. Use the fact mentioned earlier that [itex](g^{-1} x g)^m = g^{-1} x^m g[/itex] for any integer [itex]m[/itex] to derive a contradiction based on the assumption that [itex]1 \leq m < n[/itex] and [itex](g^{-1} x g)^m = e[/itex] both hold.
     
  12. Nov 17, 2011 #11

    Deveno

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    that n is, in fact, the order of g-1xg. we know no smaller m works for x, but g-1xg, isn't x, it's some other element of G, so you need to prove no smaller m will work for g-1xg.

    suppose (as an imaginary example) that g-1xg = x2, and |x| = 4.

    then g-1xg would have order 2, and so would still be e if raised to the 4th power since e2 = e.

    do not confuse the statements xn = e, and |x| = n. this can get you into trouble. the second implies the first, but NOT the other way around.

    you have to get used to this sort of thing, because often you will consider functions (homomorphisms) from one group to another f:G→H. if f is not 1-1 (and you will encounter such functions), than f will map several elements to the identity of H.

    due to the nature of homomorphisms, it will be the case that if |x| = n, (f(x))n will be eH. but this does not mean |f(x)| = n.

    so the map G→G given by x→g-1xg is special, it doesn't display this kind of behavior.
     
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