Can $G$ of order $pqr$ be simple if it's generated by elements of orders $p,q$? Let $G$ be a group of order $pqr$, where $p$, $q$ and $r$ be three distinct primes. By Cauchy's theorem there exist three elements, $a$, $b$ and $c$, whose orders are $p$, $q$ and $r$, respectively. If the subgroup generated by $a$ and $b$ is the whole group, then I wonder if it is possible that there exists a proper normal subgroup of $G$.
 A: More generally, let $G$ be a finite group with order $p_1p_2\cdots p_n$, where $p_1< p_2< \cdots < p_n$ are distinct primes.  Then by the normalizer-centralizer theorem, the Sylow $p_1$-subgroup is central in its normalizer, so there is a normal $p_1$-complement in $G$ by Burnside's transfer theorem.  Inductively, we can then show the Sylow $p_n$-subgroup is normal in $G$.
A: No, a group of order $pqr$ cannot be simple, regardless of whether you can find elements $a$ and $b$ of orders $p$ and $q$ that generate $G$ or not.
Suppose $p\lt q\lt r$, and let $\ell$, $m$, and $n$ be the number of $p$-Sylow, $q$-Sylow, and $r$-Sylow subgroups of $G$, respectively. Any two distinct Sylow subgroups will intersect trivially, because they are all cyclic of prime order.
If $G$ were simple, then $\ell,m,n\gt 1$. Also, $n|pq$, $m|pr$, and $\ell|qr$. Since $n$ must be congruent to $1$ modulo $r$ and $p,q\lt r$, then we must have $n=pq$. Since $m$ must be congruent to $1$ modulo $q$ and $p\lt q$, the smallest value possible for $m$ is $r$. So the smallest values they can have are $n=pq$, $m=r$, and $\ell=q$. 
So the distinct $p$-Sylow subgroups give at least $q(p-1)$ nonidentity elements. The distinct $q$-Sylow subgroups give at least $r(q-1)$ nonidentity elements; and the distinct $r$-Sylow subgroups give at least $pq(r-1)$ nonidentity elements. So $G$ will have at least
$$\begin{array}{rcl} q(p-1) + r(q-1) + pq(r-1) + 1 &=& 
pqr + pq + rq - q - r - pq + 1\\
& = &pqr + rq +1 - q-r
\end{array}$$
distinct elements. But this is strictly larger than $pqr$, since $r$ and $q$ are greater than 1, so $qr\geq q+r$. This is impossible.
The contradiction arises from the assumption that $G$ is simple; hence $G$ is not simple.
