This is an abstract algebra related question I pulled this f

This is an abstract algebra related question. I pulled this from wiki.

Consider the group with addition modulo 6: G = {0, 1, 2, 3, 4, 5}. Consider the subgroup N = {0, 3}, which is normal because G is abelian. Then the set of (left) cosets is of size three:

G/N = { aN : a G } = { {0, 3}, {1, 4}, {2, 5} } = { 0+N , 1+N, 2+N }.

Thank you!

Solution

Each value in G/N should be element of G because G/N is defined as:
G/N = { aN : a G }
But 6 or 7 or 8 is not an element of G
So, {3,6} or {4,7} or {5,8} can\'t be element of G/N