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thus <math> a^{n}=1</math> by the def of order | thus <math> a^{n}=1</math> by the def of order | ||
| − | then <math>\phi_{a}^{n}(x)=a^{n}xa^{- | + | then <math>\phi_{a}^{n}(x)=a^{n}xa^{-n}=1x1=x</math> making x the identity |
since <math>\phi_{a}^{n}(x)=</math>identity then <math>|\phi_{a}(x)|=min(n, c) </math> were c is a division of n since any multiple of c, including n, will also give the identity | since <math>\phi_{a}^{n}(x)=</math>identity then <math>|\phi_{a}(x)|=min(n, c) </math> were c is a division of n since any multiple of c, including n, will also give the identity | ||
Latest revision as of 02:35, 25 September 2008
Any body have any ideas???? I'm lost.
I don't know either. I looked in the back of the book, but I don't see how what they're saying has anything to do with the problem. The back of the book is talking about how $ \phi_{a^{n}}=1 $, but I thought that we basically needed to show that $ \phi_{a}^{n}=1 $. All I can show is that $ \phi_{a}^{n}(x)=ax^{n}a^{-1} $, and that $ \phi_{a}^{n}(x)=x $ if G is abelian.
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What the book is saying:
let $ |a|=n $
thus $ a^{n}=1 $ by the def of order
then $ \phi_{a}^{n}(x)=a^{n}xa^{-n}=1x1=x $ making x the identity
since $ \phi_{a}^{n}(x)= $identity then $ |\phi_{a}(x)|=min(n, c) $ were c is a division of n since any multiple of c, including n, will also give the identity
hope this helps
-zach
Yeah, I just figured that out and was posting about it when you put that up. Thanks.
