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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 $\theta :G \rightarrow G$ with $\theta(g)=g^2$ is an automorphism of G. Show that G is Abelian. the solution is like this $g,h\in G \rightarrow (gh)^2=\theta(gh)=\theta(g)\theta(h)=g^2h^2$ this is okej but $gh=g^{-1}(g^2h^2)h^{-1}=g^{-1}(gh)^2h^{-1}=hg$ 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
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Re: prove that a group is Abelian

Quote:
 Originally Posted by rayman I have a solution which I dont get 100% Let G be a finite multiplicative group for which $\theta :G \rightarrow G$ with $\theta(g)=g^2$ is an automorphism of G. Show that G is Abelian. the solution is like this $g,h\in G \rightarrow (gh)^2=\theta(gh)=\theta(g)\theta(h)=g^2h^2$ this is okej but $gh=g^{-1}(g^2h^2)h^{-1}=g^{-1}(gh)^2h^{-1}=hg$ 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
After the second equality you're using the result above it, the one you said is "okej". (In English, by the way, this is written "okay"; our "j" sound is very different.) For the last equality note that (gh)^2 = ghgh.

 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 re-written 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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