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[[Question about Chapter 3, Problem 4]] | [[Question about Chapter 3, Problem 4]] | ||
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| + | [[Category:MA453Spring2009Walther]] | ||
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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 | ||
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| + | ----- | ||
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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>? | ||
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| + | ---- | ||
| + | Here's another way to think about it. Assume <math>\scriptstyle\mid a\mid=\sigma</math>. Then <math>\scriptstyle a^\sigma = e</math>. | ||
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| + | <math>\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}</math> | ||
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| + | <math>\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}</math> | ||
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| + | <math>\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}</math> | ||
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| + | This reduces iteratively until we reach: | ||
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| + | <math>\scriptstyle\underbrace{\scriptstyle eee\cdots eee}_{\textstyle\sigma}\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma}</math> | ||
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| + | <math>\textstyle\Rightarrow\scriptstyle\ e\ =\ \underbrace{\scriptstyle a^{-1}a^{-1}\cdots a^{-1}a^{-1}}_{\textstyle\sigma}\ =\ (a^{-1})^{\sigma}</math> | ||
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| + | Therefore, <math>\scriptstyle\mid a^{-1}\mid\ =\ \sigma</math>. <math>\scriptstyle\Box</math> | ||
| + | :--[[User:Narupley|Nick Rupley]] 22:04, 4 February 2009 (UTC) | ||
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| + | ---- | ||
| + | I agree with Nick's method. That way you aren't assuming which operation the group is using. | ||
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| + | --[[User:Davis29|Davis29]] 08:49, 5 February 2009 (UTC) | ||
Latest revision as of 03:49, 5 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) $?
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)
I agree with Nick's method. That way you aren't assuming which operation the group is using.
--Davis29 08:49, 5 February 2009 (UTC)
