Let G be a finite group with $\displaystyle |G|=pqr$, where $\displaystyle p, q, r$ are primes with $\displaystyle p<q<r$. Prove that G has a normal Sylow subgroup for either p, q, or r.
suppose to the contrary that neither of Sylow subgroups of G are normal. let $\displaystyle n_p, n_q, n_r$ be the number of Sylow p, q and r subgroups of G respectively. so $\displaystyle n_p > 1, \ n_q > 1, n_r > 1.$ now $\displaystyle n_r \mid pqr,$
which gives us $\displaystyle n_r \mid pq$ because $\displaystyle \gcd(n_r,r)=1.$ but since $\displaystyle n_r > r > q > p,$ we have $\displaystyle n_r \neq p, \ n_r \neq q.$ thus $\displaystyle n_r=pq.$ now since $\displaystyle n_q \mid pr,$ and $\displaystyle n_q > q > p,$ we have $\displaystyle n_q = r, \text{or} \ pr.$ in either case $\displaystyle n_q \geq r.$
similarly $\displaystyle n_p \geq q.$ so we've proved that G has exactly pq subgroups of order r, at least r subgroups of order q and at least q subgroups of order p. since every two distinct cyclic subgroups of the
same prime order have trivial intersection and also every two cyclic subgroups of distinct prime orders have trivial intersection, a simple counting of elements of G shows that we must have:
$\displaystyle |G| \geq pq(r-1) + (q-1)r + (p-1)q + 1=pqr + (q-1)r - q + 1 > pqr =|G|,$ which is the contradiction we needed! Q.E.D.