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Math Help - Determining if a group

  1. #1
    Member Maccaman's Avatar
    Joined
    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  (G, *) a semigroup, or group in the following?
    (a)  G = \mathbb{R} \ \ \{\frac{1}{2}\},  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)  (a * b) * c = (a - 2ab + b) * c
    (2)  \ \ \ \ \ \ \ = (a - 2ab + b) - 2(a - 2ab + b)c + c
    (3)  \ \ \ \ \ \ \ = a + b + c - 2ab - 2ac - 2bc + 4abc
    (4)  \ \ \ \ \ \ \ = a - 2a(b - 2bc + c) + (b - 2bc + c)
    (5)  \ \ \ \ \ \ \ = a * (b - 2bc + c)
    (6)  \ \ \ \ \ \ \ = 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:
     = (a - 2ab + b) + c - 2c . I cannot see why (2) is  \ \ \ \ \ \ \ = (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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    To easy confision let us write multiplication as x * y = x - 2xy +y. So when going from line 1 to line 2 we have that x=a*b, y=c. a * b = a - 2ab +b, and the result follows.

    The a and 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
    Joined
    Sep 2008
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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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