Product of three distinct primes is solvability-forcing

Statement
Suppose $$p,q,r$$ are three distinct prime numbers. Let $$G$$ be a group of order $$pqr$$. Then $$G$$ is a solvable group. In particular, $$G$$ is not a simple non-Abelian group.

Facts used

 * 1) uses::Product of two distinct primes is solvability-forcing
 * 2) uses::Congruence condition on Sylow numbers
 * 3) uses::Divisibility condition on Sylow numbers
 * 4) uses::Solvability is extension-closed

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

To prove: $$G$$ is solvable. In particular, $$G$$ is not a simple non-Abelian group.

Proof: Without loss of generality, assume $$p < q < r$$.

Let $$n_p, n_q, n_r$$ denote the Sylow numbers, i.e., the number of $$p$$-Sylow subgroups, $$q$$-Sylow subgroups, and $$r$$-Sylow subgroups.


 * 1) $$n_p(p - 1) + n_q(q - 1) + n_r(r - 1) \le pqr - 1$$: An element of $$G$$ has order $$p$$ if and only if it is a non-identity element in a $$p$$-Sylow subgroup. Further, any two $$p$$-Sylow subgroups intersect in the identity, so the total number of elements of order $$p$$ is $$n_p(p-1)$$. Similarly, the total number of elements of order $$q$$ is $$n_q(q-1)$$, and the total number of elements of order $$r$$ is $$n_r(r-1)$$. Thus, the number of non-identity elements is at least $$n_p(p-1) + n_q(q-1) + n_r(r-1)$$.
 * 2) One of the $$n_p, n_q, n_r$$ must equal $$1$$: If not, we derive a contradiction:
 * 3) By facts (2) and (3), $$n_r \equiv 1 \mod r$$ and $$n_r | pq$$: Thus, $$n_r = 1,p,q,pq$$. Since $$n_r \ne 1$$, we have $$n_r > r$$, so $$n_r = pq$$.
 * 4) By facts (2) and (3) we have $$n_q \equiv 1 \mod q$$ and $$n_q | pr$$: Thus, $$n_q = 1, p, r, pr$$. Since $$n_q \ne 1$$, $$n_q > q$$, so $$n_q \ge q$$.
 * 5) By facts (2) and (3), we have $$n_p \equiv 1 \mod p$$ and $$n_p | qr$$: Thus, $$n_p = 1, q, r, qr$$. Since $$n_p \ne 1$$, $$n_p \ge q$$.
 * 6) We have $$n_p(p-1) +n_q(q-1) + n_r(r-1) \ge q(p-1) + r(q-1) + pq(r-1) = pqr + qr - q - r > pqr - 1$$. This contradicts the conclusion of step (1). Thus, we have the required contradiction.
 * 7) Thus, one of $$n_p, n_q, n_r$$ is $$1$$, so there is a normal subgroup $$N$$ of prime order. Any group of prime order is cyclic of prime order, hence solvable. The quotient $$G/N$$ has order equal to the product of the other two primes, so by fact (1), this is solvable. Thus, by fact (4), $$G$$ is solvable.