Let p, q be distinct primes. Suppose G is a group of order pq such that G has a unique subgroup of order p and a unique subgroup of order q. Prove that G is cyclic.

To prove that G is cyclic, we need to show that there exists an element in G that generates the entire group.

Step 1: Let's denote the unique subgroups of order p and q as H and K respectively. By Lagrange's theorem, the order of any subgroup divides the order of the group. Therefore, the orders of H and K are indeed p and q respectively.

Step 2: Since p and q are primes, the subgroups H and K are cyclic. This means there exist elements h in H and k in K such that the cyclic groups generated by h and k are H and K respectively.

Step 3: The intersection of H and K must be the identity element. This is because the order of the intersection of H and K must divide both the order of H (which is p) and the order of K (which is q). Since p and q are distinct primes, the only common divisor is 1. Therefore, the intersection of H and K is a subgroup of order 1, which is the identity element.

Step 4: Consider the element g = hk. We claim that g generates G. To see this, note that the order of g must divide the order of G (which is pq) by Lagrange's theorem.

Step 5: If the order of g were p, then g would belong to H. But g cannot belong to H, because h and k commute (since their intersection is the identity) and k is not in H (since the only element in both H and K is the identity). Similarly, g cannot have order q. Therefore, the order of g must be pq.

Step 6: Since the order of g is pq, the cyclic group generated by g has pq elements, which means it must be G. Therefore, G is a cyclic group.