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Math Help - Element with odd order

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    Element with odd order

    Suppose a and b are in a group, a has odd order, and aba^{-1} = b^{-1} . Show that b^{2} = e.
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    In the relation aba^{-1} = b^{-1} , take the inverse of both sides, to get ab^{-1}a^{-1} = b . It follows that a^2ba^{-2} = ab^{-1}a^{-1} = b . By the same calculation, a^3ba^{-3} = b^{-1}, and so on. In fact, a^kba^{-k} is equal to b^{-1} if k is odd, and b if k is even.

    Now take k to be the order of a, which is odd. Then a^k=e (the identity element of the group), and a^kba^{-k} = ebe = b. But since k is odd, this is also equal to b^{-1}. So b=b^{-1} and therefore b^2=e.
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