# Thread: finding elements of a group

1. ## 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? Also, how do i implement the axioms as asked in the question, pretty confused how to incorporate it into the answer.

2. ## 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.

3. ## 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)

4. ## Re: finding elements of a group

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

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

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