Let P be a Sylow p-subgroup and Q a Sylow q-subgroup. The number nq of Sylow q-subgroups is congruent to 1 (mod q) and nq divides 1G1, but then p<q implies that the only possibility is nq=1. Thus, Q is a normal subgroup.
Similarly, np must be either 1 or q. However, by our assumption, p does not divide q-1, so this rules out the latter. So again, np=1 and P is a normal subgroup.
Because they are of prime order, P and Q are cyclic, say P=<a> and Q=<b>. Since P and Q have trivial intersection and are normal, a and b commute (prove this by showing a^(-1)*b^(-1)*a*b is in their intersection). Now, 1ab1=1a1*1b1=pq, showing that G must be the cyclic group of order pq.
