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Math Help - Group problem with Abelian

  1. #1
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    Group problem with Abelian

    Prove that a group G is Abelian iff (ab)^{-1} = a^{-1}b^{-1}.
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    MHF Contributor red_dog's Avatar
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    We have (ab)^{-1}=b^{-1}a^{-1}
    Then the equality becomes a^{-1}b^{-1}=b^{-1}a^{-1}.
    Multiply both sides at left by a:
    aa^{-1}b^{-1}a=ab^{-1}a^{-1}a\Rightarrow b^{-1}a=ab^{-1}
    Now, multiply both sides by b
    bb^{-1}ab=bab^{-1}b\Rightarrow ab=ba


    If G as abelian the identity is obvious:
    (ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}
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  3. #3
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    I don't understand how you get <br />
(ab)^{-1}=b^{-1}a^{-1}<br />
as I have <br />
(ab)^{-1} = a^{-1}b^{-1}<br />
in the problem.

    What allows you to switch them as we are proving the group is communtative, and not assuming it is. (From my understanding)
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    I don't understand how you get <br />
(ab)^{-1}=b^{-1}a^{-1}<br />
as I have <br />
(ab)^{-1} = a^{-1}b^{-1}<br />
in the problem.

    What allows you to switch them as we are proving the group is communtative, and not assuming it is. (From my understanding)
    (ab)^{-1} is an element such that when multiplied by ab produces 1 (identity).

    Since (ab)(b^{-1}a^{-1})=1 it means (ab)^{-1}=b^{-1}a^{-1}.
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  5. #5
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    Thank you, sir, it is clear now.
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