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Math Help - Group Axiom G3

  1. #1
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    Group Axiom G3

    The group

    (Q*,o) where x o y = xy/7

    G1 closure hold

    G2 Idenitiy - I have found the identity to be 7

    I just need to prove G3 inversers and am not sure how to do this.

    Can anyone help?

    Many thanks.

    Arron
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  2. #2
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    We have x\circ x^{-1}=e ( e is the unit), so x\cdot x^{-1}/7=7. Thus, x^{-1}=49/x.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by emakarov View Post
    x\cdot x^{-1}/7=7. Thus, x^{-1}=49/x.
    What?

    \displaystyle x \circ x^{-1} = e \implies \frac{xx^{-1}}{7} = 1

    doesn't it?

    -Dan
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  4. #4
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    The OP has already established that the unit is 7.
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  5. #5
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    No. The OP said that e=7 in his original post.
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  6. #6
    Forum Admin topsquark's Avatar
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    (Ahem!) Missed that. Thanks.

    -Dan
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