Application of the Sylow Theorems to groups of order $p^2q$

Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.

This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.

Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.

To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!

Consider the case of $q > p$. Then,

$n_q = p^2$ or $1$ and

$n_p = q$ or $1$.

Let G have no non-trivial normal subgroup. Then,

$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following , \begin{align} p^2q &\geq p^2(q-1) + q(p^2-1) + 1 \\ & \geq p^2q+p^2q-p^2-q+1 \\ & \geq p^2q + (p^2-1) (q-1) \end{align} $$\implies (p^2-1)(q-1) = 0 \implies p=1,-1 \text{ or } q=1$$ But $p$ and $q$ are primes.

Hence $G$ has a non-trivial normal subgroup.