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-Matt
 
-Matt
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in this case, there are two cases where g has finite/infinite order of k.
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- Panida

Revision as of 02:12, 18 September 2008

How do you prove that an element and its inverse have the same order? I understand the idea but do not know how to prove it.

-Wooi-Chen



I thought this worked as a proof.

$ g^k=1 $ element g having order of k

$ (g^k)^{-1}=(1)^{-1} $

$ g^{-k}=1 $

$ (g^{-1})^k=1 $ inverse of g having order of k

This could be wrong, but it makes sense.

-Daniel


That actually makes sense to me as well. It is kind of playing with the order which power comes, that's the idea I get.

-Wooi-Chen


I think if you prove its cyclic the inverse will always be the same

-Matt


in this case, there are two cases where g has finite/infinite order of k.

- Panida

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009