$|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. $|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK:
Let $|G| = pqr$, and assume $p < q < r$. We have at least one Sylow-$q$ subgroup (call it $Q$) and at least another Sylow-$r$ subgroup (call it $R$). Consider the subgroup $K = $ $<Q, R>$. Now $K$ is not the whole group because there is no element of order $p$ in there. Further, $|K|$ must equal $qr$ because nothing smaller is possible by Lagrange's theorem.
Therefore [G : K] = p, which is the smallest prime dividing G. Hence by a standard theorem (for instance, see corollary 4.5, p. 44, Isaacs' Algebra), K must be normal. Hence $G$ is not simple.
Thanks!
WH
 A: There is nothing stopping $K$ from being equal to $G$ - ie. $K$ might have elements of order $p$. However, I cannot think of a counterexample of the top of my head. Here is an alternate solution though.
You can do this by counting elements of orders $p, q$ and $r$ : Let $n_s$ denote the number of $s$-Sylow subgroups, for $s\in \{p,q,r\}$. Since any two distinct $s-$Sylow subgroups intersect trivially (why?), it follows that

The number of elements in $G$ of order $s$ is $n_s(s-1)$

Now assume $n_s \neq 1$ for all $s\in \{p,q,r\}$. Then


*

*$n_r \equiv 1\pmod{r}, n_r\mid pq \Rightarrow n_r = pq$

*$n_q \equiv 1\pmod{q}, n_q \mid pr \Rightarrow n_q \in\{ r,pr\} \Rightarrow n_q \geq r$

*$n_p \equiv 1\pmod{q}, n_p\mid qr \Rightarrow n_p \geq q$
Hence,
$$
|G| \geq n_r(r-1) + n_q(q-1) + n_p(p-1) + 1
$$
$$
\Rightarrow pqr \geq pq(r-1) + r(q-1) + q(p-1) + 1
$$
Solving this gives you a contradiction, and so $G$ cannot be simple (in fact, one of these Sylow subgroups must be normal)
A: Can you use Feit-Thompson? An odd order simple group must have prime order. So, in this case, the smallest prime must be 2. But if a group has order equal to twice an odd number, it must have a normal subgroup of index two.
