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anilorap

  • 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

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  1. badreferences
    • 2 years ago
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    Abstract algebra?

  2. anilorap
    • 2 years ago
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    yep

  3. badreferences
    • 2 years ago
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    Bleargh I don't know how to solve this. I will call upon @TuringTest and @FoolForMath for help. :P

  4. badreferences
    • 2 years ago
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    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.

  5. badreferences
    • 2 years ago
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    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.

  6. anilorap
    • 2 years ago
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    ok yea that web is another level.. let me see if they could help

  7. FoolForMath
    • 2 years ago
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    Okay, this wasn't hard, \( a*a=e \implies a=a^{-1} \) similarly \( b=b^{-1} \) So, \[ a*b=(a*b)^{-1} =b^{-1} *a^{-1} =b*a. \]

  8. FoolForMath
    • 2 years ago
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    [QED]

  9. FoolForMath
    • 2 years ago
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    M.SE thread: http://math.stackexchange.com/questions/118772/

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