{0,3}+{0,3} = {(0+0) (mod 6), (0+3) (mod 6), (3+0) (mod 6), (3+3) (mod 6)} = {0,3,3,0} = {0,3}

{0,3}+{1,4} = {(0+1) (mod 6), (0+4) (mod 6), (3+1) (mod 6), (3+4) (mod 6)} = {1,4,4,1} = {1,4}

{0,3}+{2,5} = {(0+2) (mod 6), (0+5) (mod 6), (3+2) (mod 6), (3+5) (mod 6)} = {2,5,5,2} = {2,5}

Hence {0,3} is the identity.

{1,4}+{1,4} = {(1+1) (mod 6), (1+4) (mod 6), (4+1) (mod 6), (4+4) (mod 6)} = {2,5,5,2} = {2,5}

{1,4}+{2,5} = {(1+2) (mod 6), (1+5) (mod 6), (4+2) (mod 6), (4+5) (mod 6)} = {3,0,0,3} = {0,3}

{2,5}+{2,5} = {(2+2) (mod 6), (2+5) (mod 6), (5+2) (mod 6), (5+5) (mod 6)} = {4,1,1,4} = {1,4}

So, the set G/N is closed under addition, it has an identity element, every element has a unique inverse: -{0,3} = {0,3}, -{1,4} = {2,5}, and -{2,5} = {1,4}. Associativity is inherited from G. Hence, it is a group.

Another way to think about it: for any a in G/N, you can write a=g+N for some g in G. So, {0,3} becomes 0+N=3+N. {1,4} becomes 1+N=4+N. {2,5} becomes 2+N=5+N. Then, the addition becomes:

(0+N) + (0+N) = 0+N

(0+N) + (1+N) = 1+N

(0+N) + (2+N) = 2+N

(1+N) + (1+N) = 1+N

(1+N) + (2+N) = 3+N = 0+N

(2+N) + (2+N) = 4+N = 1+N

This addition table matches the one above.