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<math>g^k=1</math> element g having order of k | <math>g^k=1</math> element g having order of k | ||
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<math>(g^k)^-1=(1)^-1</math> | <math>(g^k)^-1=(1)^-1</math> | ||
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<math>g^-k=1</math> | <math>g^-k=1</math> | ||
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<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 | ||
Revision as of 17:23, 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
