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Thread: 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 $\displaystyle x^n$ = e.. That is, n > 1 satisfies

    $\displaystyle x^n$ = e and $\displaystyle x^m$ =/= 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.

    $\displaystyle x^{n/2}$.$\displaystyle x^{n/2}$ = $\displaystyle x^n$= e

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

    $\displaystyle x^n$ = e
    $\displaystyle (x^{n})^{-1} = e^{-1}$
    $\displaystyle x^{-n} = e^{-1}$

    Note: $\displaystyle e^{-1}$ = $\displaystyle e$

    Therefore,

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

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

    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; Aug 27th 2009 at 02:02 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 $\displaystyle x^{n/2}$ ??, why two symbols for an identity element?, what does the symbol "/" mean? ). In the language of groups, there are originally no 'powers'. $\displaystyle x^n$ is just our mere shorthand for $\displaystyle x \cdot x \cdot \ldots \cdot x$ with $\displaystyle n$ occurrences of $\displaystyle x$, $\displaystyle n \in \mathbb{N}$, with agreement that $\displaystyle x^0 = e$.
    You included the definition of order of an element, so let's use it to solve our problem.
    Let $\displaystyle n \geq 1$ be the order of $\displaystyle x$.

    $\displaystyle
    (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}=
    $

    $\displaystyle
    = \underbrace{x^{-1} \cdot x^{-1} \cdot \ldots \cdot (x^{-1}}_{n} \cdot \underbrace{x) \cdot x \cdot \ldots \cdot x}_{n}=
    $

    $\displaystyle
    = \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
    $
    because group operation is associative. This shows that $\displaystyle (x^{-1})^n = (x^n)^{-1}$ but $\displaystyle (x^n)^{-1} = e^{-1} = e$, so we verified that $\displaystyle (x^{-1})^n = e$ which is the first requirement for $\displaystyle n$ to be the order of $\displaystyle x$.
    Now let's verify the second requirement. We argue by contradiction. Assume that there is a number $\displaystyle m$, $\displaystyle 1 \leq m < n$, such that $\displaystyle (x^{-1})^m = e$. Then, in the same way as in the first step, we show that $\displaystyle (x^{-1})^m \cdot x^m = e$. But $\displaystyle (x^{-1})^m = e$, thus $\displaystyle x^m = e$. And this contradicts our assumption that $\displaystyle n$ is the order of $\displaystyle x$.
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  3. #3
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    Thank you for your help. The "x$\displaystyle n/2$" 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$\displaystyle n/2$" 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, $\displaystyle x^{n/2}$ does not mean the same thing as $\displaystyle 0.5x$.

    But the more serious problem here is that $\displaystyle 0.5x$ has no meaning whatsoever in an arbitrary group $\displaystyle G$. As for fractions they could be defined in an appropriate way by defining $\displaystyle x/y$ by $\displaystyle xy^{-1}$, 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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