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Thread: finding elements of a group

  1. #1
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    finding elements of a group

    i know that the elements in the group is the set {1,2,4,7,8,10,11,13,14} for 2(c), but how do i find the elements so that there is a y such that xy is congruent to 1 mod 15?finding elements of a group-screenshot-77-.png Also, how do i implement the axioms as asked in the question, pretty confused how to incorporate it into the answer.
    Last edited by noobpronoobpro; Oct 17th 2017 at 09:28 AM.
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  2. #2
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    Re: finding elements of a group

    "1" in "Z mod 15" corresponds to 1, 15+ 1= 16, etc.

    To find xy= 1 (mod 15), since the only numbers, x, y, from 0 to 14, such that xy= 1 are x= y= 1, you want to find x, y, such that xy= 16. xy= 16 gives x=2, y= 8, x= 4, y= 4. (There are no numbers, x and y, from 1 to 14, such that xy= 2(15)+ 1= 31, xy= 3(15)+ 1= 46, etc.) The set of numbers that have multiplicative inverses is {1, 2, 4. 8}. Those are "the elements in this group", not {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}. The set, with this multiplication is not a group because, for example 3*5= 0 (mod 15). 3 and 5 do NOT have inverses.

    "Implement group axioms to prove" simply means "use" the group axioms- above, specifically using the axiom that every member of a group has an inverse.
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  3. #3
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    Re: finding elements of a group

    why isn't 7,11,13 and 14 included? 7(13)= 91, 11(11)=121 and 14(14)=196 also gives 1 (modulo 15)
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  4. #4
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    Re: finding elements of a group

    \mathbb{Z}_{15}^*=\{1,2,4,7,8,11,13,14\}

    These are all the elements in \mathbb{Z}_{15} that are relatively prime to 15.

    10 is not included since gcd(10,15)>1
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