Order is product of three distinct primes implies normal Sylow subgroup

Statement
Suppose $$p < q < r$$ are three distinct prime numbers and $$G$$ is a group of order $$pqr$$. Then, either the $$q$$-Sylow subgroup or the $$r$$-Sylow subgroup of $$G$$ is normal in $$G$$. In particular, $$G$$ has a normal Sylow subgroup.

Facts used

 * 1) uses::Number of elements of prime order for multiplicity-free prime divisor equals Sylow number times one less than the prime
 * 2) uses::Congruence condition on Sylow numbers
 * 3) uses::Divisibility condition on Sylow numbers

Proof
Given: Three distinct primes $$p < q < r$$. A group $$G$$ of order $$pqr$$.

To prove: Either $$n_q = 1$$ or $$n_r = 1$$.

Proof: Let $$n_q, n_r$$ denote the Sylow numbers for the primes $$q,r$$; in other words, $$n_q$$ is the number of $$q$$-Sylow subgroups of $$G$$ and $$n_r$$ is the number of $$r$$-Sylow subgroups of $$G$$.


 * 1) By fact (1), the number of elements of order $$q$$ is $$n_q(q-1)$$ and the number of elements of order $$r$$ is $$n_r(r-1)$$. Since both these sets of elements are disjoint and they are all non-identity elements, we get $$n_q(q-1) + n_r(r-1) \le pqr - 1$$.
 * 2) $$n_r = 1$$ or $$n_r = pq$$: By fact (2), $$n_r|pq$$. Thus, $$n_r = 1, p, q, pq$$. By fact (3), $$n_r \equiv 1 \mod r$$. Thus, if $$n_r \ne 1$$, $$n_r > r$$, so $$n_r = pq$$.
 * 3) $$n_q = 1$$ or $$n_q \ge r$$: By fact (2), $$n_q|pr$$. Thus, $$n_q = 1,p,r,pr$$. By fact (3), $$n_q \equiv 1 \mod q$$. Thus, if $$n_q \ne 1$$, $$n_q > q$$. Thus $$n_q = r, pr$$. In either event, $$n_q \ge r$$.
 * 4) If $$n_q \ne 1$$ and $$n_r \ne 1$$, we have a contradiction: If $$n_q \ne 1$$, $$n_q \ge r$$, and if $$n_r \ne 1$$, $$n_r = pq$$. Thus, $$n_q(q-1) + n_r (r-1) \ge r(q - 1) + pq(r- 1) = pqr + r(q-1) - pq$$. Also, note that $$r(q - 1) = p(q - 1) + (r-p)(q-1) \ge p(q-1) + 2(q-1) > p(q-1) + p = pq$$. Thus, $$n_q(q-1) + n_r(r-1) > pqr$$, a contradiction to the conclusion of step (1).