| Line 9: | Line 9: | ||
<math>g^k=1</math> element g having order of k | <math>g^k=1</math> element g having order of k | ||
| − | <math>(g^k)^-1=(1)^-1</math> | + | <math>(g^k)^{-1}=(1)^{-1}</math> |
<math>g^-k=1</math> | <math>g^-k=1</math> | ||
| − | <math>(g^-1)^k=1</math> inverse of g having order of k | + | <math>(g^{-1})^k=1</math> inverse of g having order of k |
This could be wrong, but it makes sense. | This could be wrong, but it makes sense. | ||
-Daniel | -Daniel | ||
Revision as of 18:24, 16 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
