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Math Help - group theory

  1. #1
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    group theory

    In any group G prove that (a^{-1})^{-1} = a

    Let f be a 1-1 mapping of G onto itself such that f(a) = a^{-1} and let f^{-1} be a 1-1 mapping of G onto itself such that f^{-1} (a^{-1}) = a

    then (a^{-1})^{-1} \equiv (f(a))^{-1} = f^{-1} \circ f(a) = f^{-1}(a^{-1}) = a

    Is this a valid argument? I'm just getting started with abstract algebra.
    Last edited by Jonroberts74; May 11th 2014 at 05:35 PM.
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  2. #2
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    Re: group theory

    No, I don't think this is valid. First you have to prove that f is a "good" function. The only way I can see to prove this is to prove that inverses are unique. But if you have this, the desired conclusion $(a^{-1})^{-1}=a$ follows immediately from $a^{-1}a=aa^{-1}=1$; i.e. $a$ satisfies the condition to be the inverse of $a^{-1}$. Secondly, I don't see how to prove f is 1-1 and onto without knowing $(a^{-1})^{-1}=a$.
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  3. #3
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    Re: group theory

    Couldn't I prove first that for a 1-1 and onto mapping of G onto itself that f , f^{-1} are inverses by showing  f^{-1} \circ f (a) = a then state by the property of being 1-1 and onto that this is a sound argument for proving that a has an inverse a^-1 such that the inverse of the inverse is the original element a because the mapping can be argued as being the operation * under which the group is closed. saying a*a^-1 = 1 seems valid but doesn't seem general enough because then e = 1 which is not general.

    ??
    Last edited by Jonroberts74; May 11th 2014 at 08:08 PM.
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  4. #4
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    Re: group theory

    ie * is defined as f(a) = b where b = a^{-1} and by the existence of inverses f^{-1}(b) = a

    ??

    or

    b,c \in G \ni ab=e=ac \Rightarrow b=c  \Rightarrow b=c=a^{-1} and (a^{-1})^{-1} = b^{-1} = c^{-1}


    then
     b^{-1} = c^{-1} = a

    therefore  (a^{-1})^{-1} =a
    something more along these lines?
    Last edited by Jonroberts74; May 11th 2014 at 08:26 PM.
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  5. #5
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    Re: group theory

    In a group, inverses are unique, it's a one line proof:

    Suppose $a\ast b = b\ast a = e$ and $a \ast c = c\ast a = e$. Then $b = e\ast b = (c \ast a)\ast b = c\ast (a \ast b) = c \ast e = c$.

    Now since inverses are unique, ANY $b \in G$ such that $a^{-1} \ast b = b \ast a^{-1} = e$ must be THE inverse of $a^{-1}$.

    Clearly $b = a$ works.
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