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Disproving a set is a group

  1. Nov 6, 2009 #1
    G is a group with respect to o.

    Another operation # is defined by x#y=(xoy)^-1

    Show that G is not a group wrt #

    I've gotten that the operation is closed but I can't figure out how to prove associativity because the inverse is a bit confusing.
  2. jcsd
  3. Nov 6, 2009 #2


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    How else can you write (xoy)-1?
  4. Nov 6, 2009 #3
    o is an operation. (xoy)^-1 is just the inverse of (xoy) if that makes sense
  5. Nov 6, 2009 #4
    I think what office shredder was asking is "Does the inverse distribute over the group operation o?" In other words, does (xoy)-1 = x-1oy-1 or does it equal something else?
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