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Math Help - Prove that in any group, an element has the same order of its inverse

  1. #1
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    Prove that in any group, an element has the same order of its inverse

    Am I on the right track?







    This will be a proof by contradiction.



    i. Assume a has order n. Therefore,
    a^n = e
    ii. Likewise, assume that a^{-1} has order m, with m different from n. Say, m < n. Therefore,
    a^{-m} = e
    iii. Since a^n = e and a^{-m} = e
    a^n = a^{-m}

    Multiply both sides by a^{-m} on the right. Then,

    a^n a^{-m} = a^{-m} a^{-m}

    By laws of exponents,

    a^{n-m}= (a^{-m})(a^{-m})

    By the definition of order, on the right we have that
    (a^{-m}) (a^{-m}) = e
    So then we have,

    a^{n-m}= e
    iv. Since  m < n, as defined at step ii., it follows that
    n - m < n
    v. But the result of iv. contradicts the claim that a has order n, since by definition this means that n is the least positive integer such that a^n = e holds.




    vi. Therefore an element has the same order of its inverse.







    Thanks for anyone who helps!
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  2. #2
    MHF Contributor
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    Formally, you also need to consider the case when m > n, but I agree with your proof so far.

    It may be easier to just note that (a^n)^{-1}=a^{-n} for all integer n. Therefore, a^n=e iff a^{-n}=e. So, \{n\in\mathbb{Z}\mid a^n=e\}=\{n\in\mathbb{Z}\mid (a^{-1})^n=e\}. From here the claim follows.
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