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.
--Dan Castillo 21:42, 4 February 2009 (UTC)
