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Let (G,*) be a group with identity element e such that a*a=e for all a in G. prove that G is abelian
 2 years ago
 2 years ago
Let (G,*) be a group with identity element e such that a*a=e for all a in G. prove that G is abelian
 2 years ago
 2 years ago

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badreferencesBest ResponseYou've already chosen the best response.2
Abstract algebra?
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
Bleargh I don't know how to solve this. I will call upon @TuringTest and @FoolForMath for help. :P
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
Although presumably if a*a=e, a is an identity element, and a group of identity elements are abelian? I don't know; I might be talking out of my retricehere.
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
If I may suggest a better place to ask advanced questions, http://math.stackexchange.com/ . The average advanced user here is probably a college math student. :P Abstract algebra isn't for everyone.
 2 years ago

anilorapBest ResponseYou've already chosen the best response.1
ok yea that web is another level.. let me see if they could help
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
Okay, this wasn't hard, \( a*a=e \implies a=a^{1} \) similarly \( b=b^{1} \) So, \[ a*b=(a*b)^{1} =b^{1} *a^{1} =b*a. \]
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
M.SE thread: http://math.stackexchange.com/questions/118772/
 2 years ago
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