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Math Help - Order a = Order a inverse proof i have no idea how to start

  1. #1
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    Post Order a = Order a inverse proof i have no idea how to start

    Hey everyone im taking a first semester set theory class and im trying to get ahead on hw since im not doing well and i was wondering if anyone could give me some kind of hint or sumthin to start this proof for hw:

    Let G be a group, Prove ord(a) = ord(a)inverse the hint for the problem is to prove 2 cases G in finite order and G is infinte order. We have gone over this kinda stuff im just not sure where to start. If G is finite order then for some integer m a^m = e show that a^m(inverse) = e? thus order(ainverse) = order(a) and for infinite you would show for infinite order for a, ainverse has infinite order?
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  2. #2
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    Quote Originally Posted by ChrisBickle View Post
    Hey everyone im taking a first semester set theory class and im trying to get ahead on hw since im not doing well and i was wondering if anyone could give me some kind of hint or sumthin to start this proof for hw:

    Let G be a group, Prove ord(a) = ord(a)inverse the hint for the problem is to prove 2 cases G in finite order and G is infinte order. We have gone over this kinda stuff im just not sure where to start. If G is finite order then for some integer m a^m = e show that a^m(inverse) = e? thus order(ainverse) = order(a) and for infinite you would show for infinite order for a, ainverse has infinite order?
    In groups it's true the same as with real numbers: \left(g^{-1}\right)^k = g^{-k}=\left(g^k\right)^{-1}

    Tonio
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