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I believe <math>\scriptstyle g</math> would be the element from the group. <math>\scriptstyle g</math> is arbitrary, so the proof holds for any such <math>\scriptstyle g</math> belonging to the group. | I believe <math>\scriptstyle g</math> would be the element from the group. <math>\scriptstyle g</math> is arbitrary, so the proof holds for any such <math>\scriptstyle g</math> belonging to the group. | ||
:--[[User:Narupley|Nick Rupley]] 05:05, 4 February 2009 (UTC) | :--[[User:Narupley|Nick Rupley]] 05:05, 4 February 2009 (UTC) | ||
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| + | Look at Example 1 of Chapter 3. That helped to explain where g^k = 1 comes from, and how that helps us on this problem. | ||
Revision as of 16:41, 4 February 2009
So is g^k the element for the group? Just trying to see how g^k gives us the answer for any element.
--Jrendall 14:02, 1 February 2009 (UTC)
I believe $ \scriptstyle g $ would be the element from the group. $ \scriptstyle g $ is arbitrary, so the proof holds for any such $ \scriptstyle g $ belonging to the group.
- --Nick Rupley 05:05, 4 February 2009 (UTC)
Look at Example 1 of Chapter 3. That helped to explain where g^k = 1 comes from, and how that helps us on this problem.
