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Inverses of elements of a group

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the inverse for each element in the group {1, 9, 16, 22, 29, 53, 74, 79, 81}, which is under modulo91

    2. Relevant equations

    3. The attempt at a solution

    From notes (again):
    1 = 1
    9 = 81
    16 = 74
    22 = 29
    53 = 79

    How were these numbers found? Sorry for so many questions, I'm just realy lost on this group stuff. Thanks :)
     
  2. jcsd
  3. Mar 5, 2009 #2
    Well, again, an element of V_91, call it a, is invertible if and only if gcd(a,91)=1.

    Find such elements first.
     
  4. Mar 5, 2009 #3
    Does that hold true for all the elements? All of them have gcd(a,91) = 1
     
  5. Mar 5, 2009 #4
    I ndeed to correct myself on my previous post. It should have read: an element of Z_91, call it a, is invertible iff gcd(a,91)=1. Because it doesn't make sense to say an elment of V_91, since we know that V_91 is the set of all invertibles of Z_91.

    Now the task of findiing their inverses is another issue. What you need to do is find two elements a,b of V_91 such that

    [a]=[1] that is, two elements such that when you multiply them and then take mod 91 they should give u 1.

    Say 9 and 81=> 9*81=729=> 729=8*91+1=> so the remainder is 1, which means that

    729=1(mod 91) so 9, and 81 are multiplicative inverses of each other.
     
  6. Mar 5, 2009 #5
    Excellent, thanks! That makes perfect sense now.

    If it's not too much trouble, there's another (3rd) post I made that has worked its way down the board, also on group theory. Could you maybe give me a hand on that one?
     
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