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| + | =[[HW3_MA453Fall2008walther|HW3]], Chapter 3, Problem 4, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]= | ||
| + | ==Problem Statement== | ||
| + | ''Could somebody please state the problem?'' | ||
| + | |||
| + | ---- | ||
| + | ==Discussion== | ||
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. | 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 | -Wooi-Chen | ||
| + | |||
| + | ---- | ||
| + | I thought this worked as a proof. | ||
| + | |||
| + | <math>g^k=1</math> element g having order of k | ||
| + | |||
| + | <math>(g^k)^{-1}=(1)^{-1}</math> | ||
| + | |||
| + | <math>g^{-k}=1</math> | ||
| + | |||
| + | <math>(g^{-1})^k=1</math> 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 | ||
| + | |||
| + | ----- | ||
| + | |||
| + | Daniel, have we defined what it is to take an inverse function? I am not sure if that is ok, to just take the inverse of both sides. And, can we bring the inverse in from the power? I am pretty sure it is ok to have the inverse of g^k is equal to the inverse of one, however I am not sure it is trivial to get that the inverse of g to the k is equal to one from this. | ||
| + | |||
| + | -Allen | ||
| + | ---- | ||
| + | [[HW3_MA453Fall2008walther|Back to HW3]] | ||
| + | |||
| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]] | ||
Latest revision as of 16:12, 22 October 2010
HW3, Chapter 3, Problem 4, MA453, Fall 2008, Prof. Walther
Problem Statement
Could somebody please state the problem?
Discussion
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
Daniel, have we defined what it is to take an inverse function? I am not sure if that is ok, to just take the inverse of both sides. And, can we bring the inverse in from the power? I am pretty sure it is ok to have the inverse of g^k is equal to the inverse of one, however I am not sure it is trivial to get that the inverse of g to the k is equal to one from this.
-Allen
