Why the ring 2Z and 3Z are not isomorphic?

Answer 1

Let's let A=2Z and B=3Z Suppose their is a ring isomorphism from A to B so f:A->B is a ring isomorphism. Then f(2)=3n for some in integer n (1) Now use the ring isomomorphism property that f(a+b)=f(a)+f(b) so f(4)=f(2+2)=f(2)+f(2)=3n+3n using (1) above. Also f(4)=f(2)f(2)=3nx3n=9n^2 But then comparing the two expressions we have for f(4), we obtain 3n+3n=9n^2 but this implies n=0 and we have f(2)=0 (since f(2)=3n for some n as we said above), however, we have f(0)=0 so this is not possible and we must conclude that 2Z is NOT isomorphic to 3Z.