Abstract algebra (another question about isomorphic and direct product )

Prove that Z_m × Z_n is cyclic if and only if (m,n)=1. Moreover,
if (m,n)=1, then prove that Z_m × Z_n ≃ Z_mn.
Note: ≃ is isomorphism

Notation: Let's use [a,b] for elements of Z_m × Z_n, and (m,n) for gcd of m and n.
Since the group operation is addition, we'll use ka instead of k^a to indicate a operating on itself k times.

Part 1: Prove that Z_m × Z_n is cyclic if and only if (m,n)=1

Order of this group is mn. If it's cyclic, then there are elements a and b in Z_m and Z_n such that the order of [a,b] = mn. Let k = (a,b). Assume k >1, i.e., that m and n are not relatively prime. Then the lcm of m and n is mn/k < m. Let's say r = mn/k. Consider any [a,b] in Z_m × Z_n, and r[a,b] = [ra,rb], with multiplication mod m or n respectively, in accordance with the definition of Z_m × Z_n. Since r is a multiple of n and m, [ra,rb] = [0,0], the identity element of Z_m × Z_n, so the order of [a,b] is <= r < mn. Therefore there is no element of Z_m × Z_n with order mn. Therefore k cannot be >1, so (m,n) = 1.

To go the other way, assume (m,n) = 1. Suppose that k[1,1] = [0,0] for some k > 0. This means k = 0 mod m and k = 0 mod n, so k = sm =tn, so K is a common multiple of m and n. Since (m,n)=1, the lcm of m and n is mn, and k must be a multiple mn. So the order of [1,1] must be >= mn. Since it can't be greater than the order of Z_m × Z_n, we must have the order of [1,1] = mn, so Z_m × Z_n is cyclic.

Part 2: We've just shown that Z_m × Z_n is cyclic, generated by [1,1]. That means that for any [a,b] in Z_m × Z_n, [a,b] is a multiple (or power) of [1,1]. Obviously an isomorphism must map a generator of the domain to a generator of the range. Let f([1,1]) = 1 in Z_mn, and f(k[1,1]) = k

Pick [a,b] and [c,d] in Z_m × Z_n. Since [1,1] is a generator, these must = s[1,1] and t[1,1] for some s and t. Therefore f([a,b] + [c,d]) + f(s[1,1] + t[1,1]) = f( (s+t)[1,1] ) = s+t = f(s[1,1]) + f(t[1,1]) + f(a) + f(b).