Given 3 distinct primes {$p,q,r$}, then $|G|=pqr \implies G$ not simple Here's a question I've been asked;
Given distinct primes $p,q,r$, show that any group $G$ of order $pqr$ is not simple. 
So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only $$\dbinom{3}{2} + \dbinom{3}{1} + 1$$ of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of $n$ distinct primes.
Is there some other approach? Thanks for any insights
 A: Let $G$ be group of order $pqr$, where $p, q$ and $r$ are primes and $p < q < r$. 
You can show that there is a normal Sylow subgroup as follows. Let $n_p, n_q$ and $n_r$ be the number of $p$-Sylow, $q$-Sylow and $r$-Sylow subgroups, respectively. Assuming $n_p, n_q, n_r > 1$ implies by Sylow's theorem that $n_p \geq q$, $n_q \geq r$ and $n_r \geq pq$. Counting elements in Sylow subgroups then gives you more than $pqr$ distinct elements, which is a contradiction. Hence there exists a unique Sylow subgroup for some prime dividing $|G|$, which has to be normal. 
Actually more is true: using the fact that there is a normal Sylow subgroup, you can show that the $r$-Sylow subgroup has to be normal. This result generalizes to groups of squarefree order. In a group of squarefree order the Sylow subgroup corresponding to the largest prime divisor of $|G|$ is normal. This fact can be shown with Burnside's Transfer Theorem. A more elementary proof, but not so "well-known" these days can be given using Frobenius theorem, which states that number of solutions to $x^n = 1$ in $G$ is a multiple of $(n, |G|)$.
A: Can't you get away with: Say $|G|=pqr$ where $p$, $q$ and $r$ are distinct primes, then without loss of generality $p<q<r$.  Then by Sylow, $n_r\equiv 1\pmod{r}$.  For sake of a contradiction suppose that $G$ is not simple.  Then $n_r\in\{1+r,1+2r,\ldots,\}$.  ($n_r\neq 1$)  At the very least, $n_r>r$.  Also by Sylow, $n_r\mid pq\Rightarrow (n_r\mid p\lor n_r\mid q)$ which is impossible since $n_r>r>p>q$.
