Prove Let G be an abelian group of order mn where m and n ar

Let G be an abelian group of order mn, where m and n are relatively prime. If G has an element of order m and an element of order n, G is cyclic.

Solution

Let G be an abelian group of order mn with gcd(m,n) = 1.
By hypothesis, a and b are elements of G with orders m and n, respectively.
Consider <ab>; I claim that this has order mn. (This shows that G is cyclic.)

Since (ab)^(mn) = (a^m)^n (b^n)^m = e^n e^m = e, we see that the order is a factor of mn.
On the other hand, if (ab)^k = a^k b^k = e, then k must be a multiple of both m and n; hence k is a multiple of mn (since gcd(m,n) = 1). Hence, the order of ab equals mn. hence G is cyclic.