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Thread: 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 $\displaystyle a$ has order n. Therefore,
    $\displaystyle a^n = e$
    ii. Likewise, assume that $\displaystyle a^{-1}$ has order m, with m different from n. Say, $\displaystyle m < n$. Therefore,
    $\displaystyle a^{-m} = e$
    iii. Since $\displaystyle a^n = e$ and $\displaystyle a^{-m} = e$
    $\displaystyle a^n = a^{-m}$

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

    $\displaystyle a^n a^{-m} = a^{-m} a^{-m}$

    By laws of exponents,

    $\displaystyle a^{n-m}= (a^{-m})(a^{-m})$

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

    $\displaystyle a^{n-m}= e$
    iv. Since $\displaystyle m < n$, as defined at step ii., it follows that
    $\displaystyle n - m < n$
    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 $\displaystyle a^n = e$ 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 $\displaystyle (a^n)^{-1}=a^{-n}$ for all integer n. Therefore, $\displaystyle a^n=e$ iff $\displaystyle a^{-n}=e$. So, $\displaystyle \{n\in\mathbb{Z}\mid a^n=e\}=\{n\in\mathbb{Z}\mid (a^{-1})^n=e\}$. From here the claim follows.
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