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Thread: Determining if a group

  1. #1
    Member Maccaman's Avatar
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    Sep 2008
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    85

    Determining if a group

    Hi there,
    I'm doing some revision for my exam and I cannot see how this one particular solution has come about. The question is:

    Is $\displaystyle (G, *) $ a semigroup, or group in the following?
    (a) $\displaystyle G = \mathbb{R} \ $\$\displaystyle \{\frac{1}{2}\}$, $\displaystyle a * b = a - 2ab +b $

    I have solved a lot of questions like this, so I know its a group because i have been able to show that it has an identity and inverse, and i have the solutions to this problem so i have seen the associative property proved, but that is where I am getting stuck

    The solution I have says (I'll number each line)....
    (1) $\displaystyle (a * b) * c = (a - 2ab + b) * c $
    (2) $\displaystyle \ \ \ \ \ \ \ = (a - 2ab + b) - 2(a - 2ab + b)c + c $
    (3) $\displaystyle \ \ \ \ \ \ \ = a + b + c - 2ab - 2ac - 2bc + 4abc $
    (4) $\displaystyle \ \ \ \ \ \ \ = a - 2a(b - 2bc + c) + (b - 2bc + c) $
    (5) $\displaystyle \ \ \ \ \ \ \ = a * (b - 2bc + c) $
    (6) $\displaystyle \ \ \ \ \ \ \ = a * (b * c) $

    Now normally what I like to do when looking to see if a particular operation is associative is do steps (1), (2), (3), and then do (6), (5), (4), (3) and see if i get the same result for (3) both times. However in this case I cannot see how (2) was obtained from (1). Initially I thought that step 2 would be:
    $\displaystyle = (a - 2ab + b) + c - 2c $. I cannot see why (2) is $\displaystyle \ \ \ \ \ \ \ = (a - 2ab + b) - 2(a - 2ab + b)c + c $. Can anyone please explain why this is the case?
    Thanks in advance to those who can help.
    Cheers,
    Maccaman
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  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
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    1,176
    To easy confision let us write multiplication as $\displaystyle x * y = x - 2xy +y$. So when going from line 1 to line 2 we have that $\displaystyle x=a*b, y=c$. $\displaystyle a * b = a - 2ab +b$, and the result follows.

    The $\displaystyle a$ and $\displaystyle b$ in the definition of multiplication and associativity are just arbitrary elements from the group.
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  3. #3
    Member Maccaman's Avatar
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    I see. I must have been tired or something to miss that. Thanks very much Swlabr. You
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