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Any body have any ideas???? I'm lost. | Any body have any ideas???? I'm lost. | ||
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| − | 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 <math>\phi_{a^{n}}=1</math>, but I thought that we basically needed to show that <math>\phi_{a}^{n}=1</math>. | + | 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 <math>\phi_{a^{n}}=1</math>, but I thought that we basically needed to show that <math>\phi_{a}^{n}=1</math>. All I can show is that <math>\phi_{a}^{n}(x)=ax^{n}a^{-1}</math>, and that <math>\phi_{a}^{n}(x)=x</math> if G is abelian. |
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| + | --- | ||
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| + | What the book is saying: | ||
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| + | let <math> |a|=n</math> | ||
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| + | thus <math> a^{n}=1</math> by the def of order | ||
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| + | then <math>\phi_{a}^{n}(x)=a^{n}xa^{-n}=1x1=x</math> making x the identity | ||
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| + | 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 | ||
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| + | hope this helps | ||
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| + | -zach | ||
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| + | ---- | ||
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| + | Yeah, I just figured that out and was posting about it when you put that up. Thanks. | ||
Latest revision as of 01:35, 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.
