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Math Help - Prove

  1. #1
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    Unhappy Prove

    Prove that:
    a) In a group (ab)^-1=b^-1 * a^-1
    b) A group G is Abelian iff (ab)^-1=a^-1 * b^-1
    c) In a group (a^-1)^-1=a for all a
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    Prove that:
    a) In a group (ab)^-1=b^-1 * a^-1
    b^{-1}*a^{-1}
    Multiply it to the right by (ab)

    b^{-1}*a^{-1}*a*b

    Since it is a group, * is associative.
    So we can write :

    b^{-1}*a^{-1}*a*b=b^{-1}*\underbrace{(a^{-1}*a)}_{1}*b
    =b^{-1}*b=1

    So we have (b^{-1}*a^{-1})*(ab)=1
    Therefore...

    b) A group G is Abelian iff (ab)^-1=a^-1 * b^-1
    Abelian means that * is commutative.

    If it is Abelian, (ab)^{-1}=(ba)^{-1}=a^{-1}*b^{-1} by the previous relation.

    If (ab)^{-1}=a^{-1}*b^{-1}, then since (ab)^{-1}=b^{-1}*a^{-1}, we have b^{-1}*a^{-1}=a^{-1}*b^{-1}
    Thus it is commutative --> Abelian

    The equivalence is proved.

    c) In a group (a^-1)^-1=a for all a
    (a^{-1})^{-1}=a \Longleftrightarrow \underbrace{(a^{-1})^{-1}*(a^{-1})}_{1}=\underbrace{a*(a^{-1})}_{1}

    Sorry for the wording & the method... just trying myself to explain
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  3. #3
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    thanks

    Thank you so much!!! This helps me a lot!!
    I like your animated cow by the way!!!
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  4. #4
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    Quote Originally Posted by mandy123 View Post
    c) In a group (a^-1)^-1=a for all a
    a\cdot a^{-1} = e therefore inverse of a^{-1} is a.
    Thus, (a^{-1})^{-1} = a.
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