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Math Help - group theory abstract algebra

  1. #1
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    Question group theory abstract algebra

    consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.
    show that this is a group.
    why is 1 excluded
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  2. #2
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    Verify that \left\langle\mathbb Q\setminus\{1\},\star\right\rangle satisfies all the group axioms. (Do you know what the group axioms are?) Hint for identity: a\star0=0\star a=\ldots\,?

    This group is isomorphic to \left\langle\mathbb Q\setminus\{0\},\times\right\rangle (the group of nonzero rationals under ordinary multiplication) via the mapping a\mapsto\frac1{1-a} for a\in\mathbb Q\setminus\{1\}.
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  3. #3
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    Quote Originally Posted by bettywhit View Post
    consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.
    show that this is a group.
    why is 1 excluded
    thanks i figured out how to show it is a group but i am not sure why 1 is excluded
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  4. #4
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    Quote Originally Posted by bettywhit View Post
    thanks i figured out how to show it is a group but i am not sure why 1 is excluded

    The group's unit is 0, since a*0=a+0-a \cdot 0=a. If 1 were an element of this group, then you'd get a*1=a+1-a\cdot 1=a+1-1=a and you have ANOTHER unit, which cannot be in a group.

    Tonio
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  5. #5
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    Something is wrong here.

    a\star1\ =\ a+1-a\cdot1=a+1-\color{red}a\color{black}=\color{red}1

    You just cannot have two identities, even in a semigroup! Including 1 would make \left\langle\mathbb Q,\star\right\rangle a semigroup rather than a group (the element 1 will not have an inverse). If a semigroup has an identity, then that identity is always unique.
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  6. #6
    Senior Member sfspitfire23's Avatar
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    If 1 was included, the inverse would also be undefined,
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