[Math] Show that a group of order $p^2q^2$ is solvable

Question

I am trying to prove that a group of order $p^2q^2$ where $p$ and $q$ are primes is solvable, without using Burnside's theorem. Here's what I have for the moment:

• If $p = q$, then $G$ is a $p$-group and therefore it is solvable.

• If $p \neq q$, we shall look at the Sylow $p$-subgroups of $G$. We know from Sylow's theorems that $n_p \equiv 1 \pmod p$ and $n_p \mid q^2$, therefore $n_p \in \{1, q, q^2\}$.

• If $n_p = 1$, it is over, because the Sylow $p$-Subgroup $P$ is normal in $G$ of order $p^2$, and $G/P$ has order $q^2$. Thus both are solvable and $G$ is solvable.

• If $n_p = q^2$, we have $q^2(p^2-1)$ elements of order $p$ or $p^2$ in $G$, and we have $q^2$ elements left to form a unique Sylow $q$-subgroup. By the same argument as before, $G$ is solvable.
• That's where I'm in trouble. I don't know what to do with $n_p = q$. It seems to lead nowhere.

Thanks in advance for any help!

Laurent

Answer 1

You argument works just as well with $p$ and $q$ switched, so the only time you have trouble is if both $n_p=q$ and $n_q = p$. Since $1\equiv n_p \mod p$ and $1\equiv n_q \mod q$ this puts very strong requirements on $p$ and $q$.

Hint 1:

Unless $n_p=1$, $n_p > p$.

Hint 2:

If $n_p=q$, then $q>p$. If $n_q =p$, then $p>q$. Oops.

Fix for OP's argument:

The OP's argument is currently flawed in the case $n_p=q^2$, so this answer is only truly helpful after that flaw is fixed.

A very similar argument to the one given in this answer works. First part of your argument works, and the $p-q$ symmetry helps:

If $n_p=1$ or $n_q=1$, then the group is solvable.

Now we use the Sylow counting again to get some severe restrictions:

If $n_p \neq 1$, then $n_p \in \{q,q^2\}$ and in both cases we have $1 \equiv q^2 \mod p$. Similarly, if $n_q \neq 1$, then $1 \equiv p^2 \mod q$.

Unfortunately now we don't get an easy contradiction, but at least we only get one possibility:

Since $p$ divides $q^2-1 = (q-1)(q+1)$, we must also have $p$ divides $q-1$ or $q+1$, so $p \leq q+1$ and $q \leq p+1$, so $p-1 \leq q \leq p+1$. If $p=2$ is even, then $q$ is trapped between 1 and 3, so $q=3$. If $p$ is odd, then $p-1$ and $p+1$ are both even, so the only possibility for $q \neq p$ is $q=p-1=2$ (so $p=3$) or $q=p+1=2$ (so $p=1$, nope). Hence the only possibility is $p=2$ and $q=3$ (or vice versa).

In this case, we get:

If $p=2$ and $q=3$, then $n_q \in \{2,4\}$. Considering the permutation action of $G$ on its Sylow $q$-subgroups, we know that $n_q=2$ is impossible (Sylow normalizers are never normal) and $n_q=4$ means $G$ has a normal subgroup $K$ so that $G/K$ is isomorphic to a transitive subgroup of $S_4$ containing a non-normal Sylow 3-subgroup and having order a divisor of 36. The only such subgroup is $A_4$, so $K$ has order 3. Hence $G/K\cong A_4$ and $K \cong A_3$ are solvable, so $G$ is solvable.