If g^k=1 then dividing by g^k you get g^(-k)=1. So ord(1/g) is less than or equal to ord(g). This goes both ways, so equality must hold.
--Awika 14:42, 30 January 2009 (UTC)
Question about Chapter 3, Problem 4
I'm not sure why this problem took me awhile to think through...something about doing things algebraically with representations of the order didn't seem to fit well with me...but I think this is definitely a good approach to the problem
Let $ ord(g)=k $. So $ g^k = 1 $ and we divide by $ g^k $ to get $ 1=g^{-k} $. Can we just say $ 1=(g^{-1})^k $, and therefore $ ord(g^{-1})=k=ord(g) $?
