So, going through my qual prep classes, there's a two part question: For $p,q$ distinct primes, prove that a group of order $p^2q$ has a nontrivial proper normal subgroup, then prove that it's solvable. Part 2 is pretty trivial given the first, but I've been struggling with it.
Thoughts so far: Attacking with Sylow theorems, if $n_p=1$ or $n_q=1$ we'd be done, so assume neither are one (either for contradiction or to find a group of order $pq$, since I'll show later that such a group would have to be normal)
So, given that we need $n_p=q$ and $n_q=p$ or $n_q=p^2$, but since we also need $n_p>p$ and $n_q>q$, we must have $p<q<p^2$, and $n_q=p^2$. We also have $q=pk+1$ and $p^2=qj+1$ for some natural numbers $k,j$ due to congruency, but I couldn't seem to find a contradiction out of here.
Now, since $p$ is the smallest prime, if I could find a subgroup of order $pq$, it would have index $p$ and thus be normal by the theorem that groups of the smallest prime index of a group are normal. However, I can't seem to figure out why either that would have to be the case, or why the above statements are a contradiction to force a normal $p^2$ group or a normal $q$ group.
Best Answer
For the contradiction: Since $p^2-1=qj$ then $p^2-1$ is a multiple of $q$. Hence, $q$ divides $p-1$ or $p+1$. Since $q>p$ then $q=p+1$, which implies $q=3$ and $p=2$.
This paper lists all possible groups of order $12$ (and includes a proof that none of them are simple).