I was proving 2Z is non-isomorphic to 3Z I tried it by contradiction, of course. If possible there exists a ring homomorphism f : 2Z ---> 3Z Then, f (x.y) = f(x).f(y) must hold x,y belong to 2Z So x = 2a, f.s. a belonging to Z y = 2b, f.s. b belonging to Z so x.y = 2z, f.s. z belonging to Z Also, f(x) belongs to 3Z, and so does f(y) => f(x).f(y) = 3n , f.s. n belonging to Z So, f(2z) = 3n f(2).f(z) = 3n, (since f is a ring homomorphism) which is a meaningless statement, since f is not defined for all z belonging to Z, but only for elements of the form 2z. Is it correct ? Someone got something neater ?