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

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In what context were you asked this? There are a lot of approaches, depending on how advanced results you have seen. I assume you have seen Sylow's theorems? – Tobias Kildetoft Nov 22 '12 at 20:50
I've seen them, but don't understand them enough to feel comfortable applying them. I like to stay away from "black box" application of theorems if at all possible – user45814 Nov 22 '12 at 20:52
As I saw this fact before, it has a long proof. – Babak S. Nov 22 '12 at 20:53

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 $(n, |G|)$.

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great answer. ++ for Frobenius theorem! Thanks! – user45814 Nov 23 '12 at 3:50
How do you know that if you count the elements when $np \ge q, nq \ge r$ and $nr \ge pq$, you will get more than $pqr$ distinct elements? – Al Jebr Apr 11 '15 at 4:04
So you get $\geq (p-1)q$ elements of order $p$, also $\geq (q-1)r$ elements of order $q$, and $\geq (r-1)pq$ elements of order $r$.. putting all this together gives you at least $pqr$ elements, and then you still have the identity – Mikko Korhonen Apr 11 '15 at 11:07

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 $n_r>r>p>q$.

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I think you need to do more. What if $n_r=pq$ and $pq\equiv1\pmod r$? – Jyrki Lahtonen Jun 24 '14 at 8:52