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March 7th, 2012, 07:18 AM  #1 
Senior Member Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0  prove that a group is Abelian
I have a solution which I dont get 100% Let G be a finite multiplicative group for which with is an automorphism of G. Show that G is Abelian. the solution is like this this is okej but what has happed after the second equality? what is the name of the ''law'' that allows to do that? and finally what has happened after the last equality, how have g and h been swapped?(we are to show to that G is Abelian...not to use it...) Any help appreciated 
March 7th, 2012, 10:09 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: prove that a group is Abelian Quote:
 
March 7th, 2012, 10:25 AM  #3 
Senior Member Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0  Re: prove that a group is Abelian
Okay then . In German it is very common to write it as ''okej'' I got the idea behind this solution after all. Thank you again 
March 7th, 2012, 11:51 AM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: prove that a group is Abelian
i'm not aware of the name of the "law" you talk about, but it's everywhere in mathematics: "equals to equals are equal". that is, if a = b, and i do the same thing to a that i do to b, the results will be equal. the proof can be rewritten this way: if (gh)^2 = g^2h^2 then: (gh)(gh) = (gh)^2 = g^2h^2 = (gg)(hh), so multplying (gh)(gh) and (gg)(hh) on the right by h^1 we have: (gh)(gh)h^1 = (gg)(hh)h^1 (gh)g(hh^1) = (gg)h(hh^1) <asociativity (gh)ge = (gg)he <rule of inverses (gh)g = (gg)h <rule of identity multiplying on the left by g^1: g^1(gh)g = g^1(gg)h (g^1g)(hg) = (g^1g)(gh) e(hg) = e(gh) hg = gh 

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