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March 7th, 2012, 07:18 AM   #1
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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
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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 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
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.
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March 7th, 2012, 10:25 AM   #3
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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
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March 7th, 2012, 11:51 AM   #4
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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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