Claim: Let $|G|=pqr$, where $p$, $q$ and $r$ are primes with $p<q<r$. Then $G$ has a normal $Syl_p$ AND a normal $Syl_q$ AND a normal $Syl_r$ subgroup.
Proof: It can be proven that $G$ has a normal Sylow subgroup for either $p$, $q$ or $r$.
Case I: Suppose $Syl_p\lhd G$; then $\overline{G}:=G/Syl_p$ is a group. and $|\overline{G}|=|G/Syl_p|=|G|/|Syl_p|=pqr/p=qr$. Thus $\overline{G}$ contains a normal Sylow $r$-subgroup by the classification of groups of order equaling a product of two distinct primes-call it $\overline{Syl_r}$. By The Fourth Isomorphism Theorem, there must exist a normal subgroup of $G$ of order $r$, which is a Sylow $r$-subgroup in $G$-call it $Syl_r$.
Apply the above argument to $Syl_r$ yields another normal Sylow subgroup in $G$, namely $Syl_q$.
Case II: Suppose $Syl_q\lhd G$. See Case I.
Case III: Suppose $Syl_r\lhd G$. See Case I.
Q.E.D.
My Question: Something doesn't feel right in this proof; but I couldn't tell what it is. Any help would be greatly appreciated.