No simple group of order $p^nq^m$, with barely invoking Sylow theorems

Question

It is a well known fact that for two distinct primes $p$ and $q$, and natural numbers $m, n \geq 1$, there can be no simple group of order $p^nq^m$. Most proofs I have seen of this statement either make heavy use of the Sylow theorems, or just treat it as a corollary to Burnside's Theorem, the proof of which is even more involved. I have, however, not seen the following proof; I suspect that this is because it contains an error which I am unable to find.

Let $G$ be a group of order $p^nq^m$, with $p, q, m$ and $n$ as above. Then, by the first Sylow theorem, there exist two Sylow subgroups $P$ and $Q$ such that $|P| = p^n$ and $|Q| = q^m$. We will now show that the intersection of these groups is trivial. From Lagrange's theorem we immediately get $$ |P| = (P:P \cap Q)|P\cap Q| = p^n\\ |Q| = (Q:P \cap Q)|P\cap Q| = q^m $$ So that $|P\cap Q|$ must divide both $p^n$ and $q^m$. But this means that $|P\cap Q| = 1$, and therefore, that $P\cap Q$ contains only the identity. Again applying Lagrange's theorem, we get that there is a one-to-one correspondence between the cosets of $P$ and the elements of $Q$ (and vice versa). But this means that every element of $G$ can be written as a product of elements of $P$ and $Q$, which means that $G = PQ$, which in turn means that at least one of $P$ and $Q$ must be normal, and hence that $G$ cannot be simple.

The final argument that $G$ must be at least a semidirect product of $P$ and $Q$ seems somewhat flimsy to me, but I do not have enough experience in algebra to know where to poke the relevant holes into it. Any help taking this apart would be greatly appreciated.

Answer 1

Everything is correct up to $G = PQ$, but it does not follow that one of $P$ or $Q$ must be normal. In fact, there are examples where neither $P$ nor $Q$ is normal, but there is some other normal subgroup. For example, if $G = S_4$ with order $24 = 2^3 \cdot 3$, then neither a 2-Sylow subgroup (of order 8) nor a 3-Sylow subgroup (of order 3) is normal, but there is a normal subgroup of order 4.

Answer 2

It is false that if $G=PQ$ then at least one of $P$ and $Q$ must be normal.

In general, we know that for subgroups $P$ and $Q$, the set $PQ$ is a subgroup if and only if $PQ=QP$ as sets. this will hold if either $P$ normalizes $Q$ or $Q$ normalizes $P$, but this condition is only sufficient, it is not necessary.

For an example where a product of two subgroups equals the group but neither one is normal, consider $G=A_5$, and let $H$ be the copy of $A_4$ obtained by fixing $5$, and let $K=\langle (1,2,3,4,5)\rangle$, cyclic of order $5$. Then we have that $\gcd(|H|,|K|)=1$, so $H\cap K=\{e\}$; and $|HK|=\frac{1}{2}(4!)5 = \frac{5!}{2} = |A_5|$, so $G=HK$, but neither $H$ nor $K$ are normal in $G$.

So you cannot simply conclude that at least one of $P$ and $Q$ must be normal. You need to prove that this is the case.

(Of course in my example $H$ is not a prime-power order group, but the point still stands. And there are examples where neither Sylow is normal.)