This statement is not true!

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.

-Proof

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,

S(p)=1,p,q,pq

By checking each one we see that,

p,q,pq do not work in the congruence (because p > q).

Thus, S(p)=1.

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 remainsinvariant under conjugation, i.e. it is a normal subgroup of G, and hence G is not simple.