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Comunitative Group

  1. Mar 17, 2012 #1
    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?
     
  2. jcsd
  3. Mar 17, 2012 #2
    Hmm, and how do you know it's associative?

    Give us more precise name of your starting structure.
     
  4. Mar 17, 2012 #3
    No, if you have some set H, with a binary operation HxH → H, such that all (a,a) maps to same element (that we may call 1), then this will in general not be a group. You have that every element is its own inverse wrt to the element 1, but 1 is not a unit element (which requires that a*1=1*a=a), and you do not have associativity as there are absolutely no conditions set to the value of a*b with a≠b.

    I didnt understand the word "comunitative", but you may have meant commutative, which btw is not taking you a lot further to making this into a group.
     
  5. Mar 17, 2012 #4
    So I have a set <a in H : a^2=1> where H is a commutative group,
    1 being the identity, and aa means a*a, sorry should have specified before

    I have been told it is a group but I just want to see if I can justify it myself, having no clear binary operation confused me.

    So I know each element is it's own inverse, and it does contain an identity element, 1

    To show associativity , for a,b and c in the set

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

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

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

    so all boxes ticked?
     
    Last edited: Mar 17, 2012
  6. Mar 18, 2012 #5
    Maybe you want to show that subset of H <a in H : a^2=1> is a subgroup of H?

    In this case, you don't need to worry about associativity etc., only under closure of operations. There are three operation you need to verify, one nullary - taking unit, one unary - taking inverse and one binary - multiplication.
     
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