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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 | 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 | ||
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| + | Let <math>ord(g)=k</math>. So <math>g^k = 1</math> and we divide by <math>g^k</math> to get <math>1=g^{-k}</math>. Can we just say <math>1=(g^{-1})^k</math>, and therefore <math>ord(g^{-1})=k=ord(g)</math>? | ||
Revision as of 17:23, 4 February 2009
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) $?
