The theorem says if |G|=pq with q>p, and q is not congruent to 1 modulo p then |G| is a cyclic group.
However, when it is congruent to 1 modulo p we cannot conlude that.
However we can show that |G| is not a simple group.
Let |G|=pq and p>q.
Now by Sylow's First Theorem there exists a Sylow p-subgroup. Let S(p) denote the number of Sylow p-subgroups. Then we know that by Sylow's Third Theorem:
S(p) = 1 (mod p) and S(p)|(pq).
Thus, the statement S(p)|(pq) states that,
By checking each one we see that,
p,q,pq do not work in the congruence (because p > q).
Now, any inner automorphism under any element maps a Sylow p-subgroup into a Sylow p-subgroup. Since the number of Sylow p-subgroups is fixed at one, it means that the Sylow p-subgroup remains invariant under conjugation, i.e. it is a normal subgroup of G, and hence G is not simple.