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Math Help - Comunitative Group

  1. #1
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    Comunitative Group

    Hey,


    I have a small question about groups,


    If you have a comunitative 'group' H = <a in H : a2=1>,


    Is that enough information to show that it is a group, without knowing the binary operation?


    say b is also in H


    then a*b=b*a


    (a*b)*(b*a) = (a*a)*(b*b) = 1 (since its comunitative)


    So that shows there is an identity, and each element is it's own inverse


    It's also associative so everything is satisfied for H to be a group,


    So only knowing these two properties of this group can show that it is indeed a group?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Comunitative Group

    If x^2=1 then, x=x^{-1}. On the other hand, (ab)^2=1\Rightarrow abab=1\Rightarrow bab=a^{-1}\Rightarrow\ldots Could you continue?
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  3. #3
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    Re: Comunitative Group

    if H is commutative, then H is indeed a group if every element of H is of order 1 or 2. the commutativity is essential, since otherwise there is no guarantee that H is closed under multiplication. for example, the subset of S3 consisting of {e, (1 2), (1 3), (2 3)} is NOT a group.

    as Fernando's post and mine taken together show, for such a subset H of some group G (that is, for all h in H, h^2 = e), closure under multiplication is equivalent to commutativity.
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  4. #4
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    Re: Comunitative Group

    Quote Originally Posted by Angela11 View Post
    Hey,


    I have a small question about groups,


    If you have a comunitative 'group' H = <a in H : a2=1>,


    Is that enough information to show that it is a group, without knowing the binary operation?
    Strictly speaking, no. However, since the condition to be in the group is " a^2= 1", which involves multiplcation, we would be pretty much assumed that the group operation is multiplication. Specifically, if a^2= 1 and b^2= 1, then [tex](ab)^2= (a^2)(b^2)= (1)(1)= 1[/itex] while (a+ b)^2= a^2+ 2ab+ b^2= 2(1+ ab) which would not generally be equal to 1.

    say b is also in H


    then a*b=b*a


    (a*b)*(b*a) = (a*a)*(b*b) = 1 (since its comunitative)


    So that shows there is an identity, and each element is it's own inverse


    It's also associative so everything is satisfied for H to be a group,


    So only knowing these two properties of this group can show that it is indeed a group?
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  5. #5
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    Re: Comunitative Group

    abab = aa bb = 1 a^2 = b^-2

    b^-2 = 1
    b^2 = 1

    Is that what you wanted me to continue?

    Thanks for your replies guys,

    I've been questioning the associativity a little more, I kind of just guessed it was associative,

    Would this be sufficient to show associativity?

    (a*b)*c=a*(b*c)

    (a*c)*b= b*(a*c)


    (a*c)*b= (a*c)*b

    EDIT:

    I'm sure the way my lecturer writes binary operations is a*b=ab (to save writing the star every time, so really ab is a*b), sorry I should of specified that, I thought it might have been the way everyone writes it
    Last edited by Angela11; March 17th 2012 at 05:18 PM.
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