Prove that if $|G| = pqr$ and $p < q < r$, then $G$ has a normal subgroup of order $p$,$q$, or $r$. I proved that $G$ isn't simple but I am having troubles with how to prove it that the normal subgroup has order $p$,$q$, or $r$.
 A: Let $n_s$ be the number of $s$-Sylow subgroups of $G$ for $s \in \{p,q,r\}$. By Sylow's Theorem, for each $s$ we have that $n_s$ divides $|G| = pqr$ and $n_s \equiv 1 \pmod{s}$. In particular, $s$ does not divide $n_s$. One way of proceeding is to show that there is at least one $s$-Sylow subgroup which is normal, that is, $n_s \geq 1$ for some $s \in \{p,q,r\}$.
Suppose not. Then, $n_s \geq s + 1$ for $s \in \{p,q,r\}$. Since $n_p \in \{q,r,qr\}$ and $q < r$ by assumption, we must have $n_p \geq q$. Moreover, since $n_q \in \{p,r,pr\}$ and $p<q$, $r\leq qr$ we get $n_q \geq r$. Finally, since $n_r \in \{p,q,pq\}$ and $p < q < r$, we obtain $n_r = pq$.  Each $s$-Sylow subgroup has prime order and is thus cyclic. Hence, each of these subgroups has $s-1$ elements of order $s$ and so $G$ contains at least 


*

*$q(p - 1)$ elements of order $p$, 

*$r(q - 1)$ elements of order $q$,

*$pq(r - 1)$ elements of order $r$. 


Can you derive a contradiction from here?
A: Once you know that $G$ cannot be simple, it must have a normal subgroup. By Lagrange, it has order $p,q,r$, or $pq,pr$, or $qr$. In the first case we are done. In the second, use the fact that there are no such simple groups of order $pq$ where $p$ and $q$ are distinct primes.
A: A group of square-free order is solvable, so you can consider minimal normal subgroup that is elementary abelian and so has order p,q or r.
