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Math Help - Another Group Theory problem...

  1. #1
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    [Solved] Another Group Theory problem...

    This time, it's about proving 2 elements have the same order:-

    Let G be a group and x be an element of G. The order of x is the least positive number such that x^n = e.. That is, n > 1 satisfies

    x^n = e and x^m =/= e for all 1<m<n

    For x in G prove that x and x-1 have the same order
    .

    So, I've had a fair whack at the question, but I'm not sure if I'm going in the write direction.

    x^{n/2}. x^{n/2} = x^n= e

    x.x^{-1} = e
    x.x^{-1} = x.x^-1
    = x^{1-1}
    = x^0
    =1 ---> e (identity is 1)

    x^n = e
    (x^{n})^{-1} = e^{-1}
    x^{-n} = e^{-1}

    Note: e^{-1} = e

    Therefore,

    x^{-n} = e = x^{n}
    x^{-n} = x^{n}
    x^{n}/x^{-n} = e
    x^{n}.x^{n} = e
    x^{2n} = 1 = e
    e = 1 = x^{0}= x^{-n}= x^{n}= x^{2n}

    THUS. If
    x^{n} = 1 = e
    x^{n}.x^{-n} = 1 = e
    x^{-n} = 1 = e
    x^{n} = x^{-n}

    Therefore they have the same order.

    That is huge im sorry, but yeh. I dunno if it's correct. Any pointers?
    Last edited by exphate; August 27th 2009 at 03:02 PM. Reason: Noting that it's solved now :)
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  2. #2
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    Hi!
    You formulated the problem well, but very few lines of your solution attempt make sense (what is x^{n/2} ??, why two symbols for an identity element?, what does the symbol "/" mean? ). In the language of groups, there are originally no 'powers'. x^n is just our mere shorthand for x \cdot x \cdot \ldots \cdot x with n occurrences of x, n \in \mathbb{N}, with agreement that x^0 = e.
    You included the definition of order of an element, so let's use it to solve our problem.
    Let n \geq 1 be the order of x.

    <br />
(x^{-1})^n \cdot x^n = \underbrace{(x^{-1} \cdot x^{-1} \cdot \ldots \cdot x^{-1})}_{n} \cdot \underbrace{(x \cdot x \cdot \ldots \cdot x)}_{n}=<br />

    <br />
= \underbrace{x^{-1} \cdot x^{-1} \cdot \ldots \cdot (x^{-1}}_{n} \cdot \underbrace{x) \cdot x \cdot \ldots \cdot x}_{n}=<br />

    <br />
  = \underbrace{(x^{-1} \cdot x^{-1} \cdot \ldots \cdot x^{-1})}_{n-1} \cdot \underbrace{(x \cdot x \cdot \ldots \cdot x)}_{n-1}= \ldots = x^{-1} \cdot x = e<br />
    because group operation is associative. This shows that (x^{-1})^n = (x^n)^{-1} but (x^n)^{-1} = e^{-1} = e, so we verified that (x^{-1})^n = e which is the first requirement for n to be the order of x.
    Now let's verify the second requirement. We argue by contradiction. Assume that there is a number m, 1 \leq m < n, such that (x^{-1})^m = e. Then, in the same way as in the first step, we show that (x^{-1})^m \cdot x^m = e. But (x^{-1})^m = e, thus x^m = e. And this contradicts our assumption that n is the order of x.
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  3. #3
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    Thank you for your help. The "x n/2" was to denote 0.5x, and the "/" was to denote fractions (except for when it was used as =/= which was my attempt at "does not equal.).

    Thank you VERY much for the clarification.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by exphate View Post
    Thank you for your help. The "x n/2" was to denote 0.5x, and the "/" was to denote fractions (except for when it was used as =/= which was my attempt at "does not equal.).

    Thank you VERY much for the clarification.
    Well, even if you are using real numbers, x^{n/2} does not mean the same thing as 0.5x.

    But the more serious problem here is that 0.5x has no meaning whatsoever in an arbitrary group G. As for fractions they could be defined in an appropriate way by defining x/y by xy^{-1}, but you should be aware that such a notation is never used in group theory and you would do well to drop it altogether.
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