TITLE: Is there a particularly elegant proof that if $|G|=pqr$, then $G$ is solvable?
QUESTION [8 upvotes]: I know for a fact that if a group $G$ has order $pqr$ with $p,q,r$ distinct primes, then $G$ is solvable. 
Most proofs I see of this are very ugly, and require a lot of case checking to show most cases lead to contradictions. 
So is there a relatively "nice" proof of this fact, or am I too optimistic is asking for such thing?

REPLY [2 votes]: Assume that $G$ is a simple group with $p<q<r$ and show that $n_r = pq$, $n_q \geq r$ and $n_p \geq q$. Then count the elements to get a contradiction.