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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 17:41, 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.

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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