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) $?
Here's another way to think about it. Assume $ \scriptstyle\mid a\mid=\sigma $. Then $ \scriptstyle a^\sigma = e $.
$ \textstyle\Rightarrow\scriptstyle\ \underbrace{\scriptstyle aaa\cdots aaa}_{\textstyle\sigma}\ =\ e\ \textstyle\Rightarrow\scriptstyle\ \underbrace{\scriptstyle aaa\cdots aaa}_{\textstyle\sigma}\ \cdot\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma}\ =\ e\ \cdot\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma} $
$ \textstyle\Rightarrow\scriptstyle\ \underbrace{\scriptstyle aaa\cdots aa}_{\textstyle\sigma-1}\ \cdot\ e\ \cdot\ \underbrace{\scriptstyle a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma-1}\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma} $
$ \textstyle\Rightarrow\scriptstyle\ \underbrace{\scriptstyle aaa\cdots a}_{\textstyle\sigma-2}\ \cdot\ e\ \cdot\ e\ \cdot\ \underbrace{\scriptstyle a^{-1}\cdots a^{-1}}_{\textstyle\sigma-2}\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma} $
This reduces iteratively until we reach:
$ \scriptstyle\underbrace{\scriptstyle eee\cdots eee}_{\textstyle\sigma}\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma} $
$ \textstyle\Rightarrow\scriptstyle\ e\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma}\ =\ (a^{-1})^{\sigma} $
Therefore, $ \scriptstyle\mid a^{-1}\mid\ =\ \sigma $. $ \scriptstyle\Box $
- --Nick Rupley 22:04, 4 February 2009 (UTC)
