Abstract Algebra: define an operation

  • Thread starter murmillo
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  • #1
murmillo
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Homework Statement


Does the rule g*x = xg^-1 define an operation of G on G?


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The Attempt at a Solution


I don't even know what this means. Could someone just tell me what it means for a rule to define an operation of one group on itself? I should be able to figure it out from there.
 

Answers and Replies

  • #2
spamiam
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  • #3
murmillo
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Oh, I see. OK, thanks, I can take it from here.
 
  • #4
Delta2
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So you have to prove that (gh)*x=g*(h*x)
[tex](gh)*x=x(gh)^{-1}=x(h^{-1}g^{-1})=(xh^{-1})g^{-1}=(h*x)g^{-1}=g*(h*x)[/tex]
 

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