Prove that no group of order $p^2q$ is simple where $p$ and $q$ are prime Can we argue: There is a Sylow $p$-subgroup $K$ of order $p^2$ which should be abelian and thus normal and so $K$ is a non trivial normal subgroup
 A: If $p =q$, then G is a p-group and its center is non-trivial and so we are done in that case. Suppose further that $p \neq 2$ [otherwise you can easily do a counting argument in which the number of elements of order q would be $4q -4$ so that the remaining four elements would then form a unique 2-sylow subgroup]
Suppose $p \neq q$, then by Sylow Theorems,  $n_{p} \neq 1 \implies n_{p} = q$ so that $p|n_{p}-1 \implies p \leq q-1 < q$ ( *).
Note also that $n_{q} \neq 1 \implies n_{q} \in \{p, p^{2}\}$. If $n_{q} = p$, then $q|n_{q} - 1 \implies q \leq p-1 < p$ contradicting (*). 
Hence $n_{q} = p^{2}$, but then $q|n_{q} -1 = p^{2} -1 = (p-1)(p+1) \implies q|(p-1)$ or $q|(p+1)$. 
But once again, $q|(p-1)$ contradicts (*). Hence $q|(p + 1)\implies q < p +1$ strict inequality because one number is a prime while the other one is not [this is where I am using the assumption that $p \neq 2$].
We have $q -1 < p < q$, but this is a contradiction since a whole number cannot sit between two consecutive numbers. It must be the case that either $n_{p} = 1$ or $n_{q} =1$.  
A: Say first that $q<p$, then consider the map
$$\varphi: G\to S_q$$
given by sending $g\in G$ to the action it has when conjugating a $p$-Sylow subgroup. Since $p>q$, it must be that the map is trivial on the elements of a $p$-Sylow subgroup, hence one such is contained in the kernel of such a map. Also the map is clearly non-trivial, hence the existence follows. (Incidentally, this is the root of the argument that says that a subgroup with index the smallest prime dividing the order of a group is always normal).
One can also simply note that the usual congruence  relation says $n_p\equiv 1\mod p and n_p | q\implies n_p=1$ since $q<p$ as Joan Pemo says in the comments. I opted for the former approach in my original answer to illustrate a very useful and common technique.
Now if $p<q$, we note that $\#\mathcal{Q}=n_q\equiv 1\mod q$ and $n_q\big | p^2$. If $n_q\ne 1$ then $n_q = p^2$ since $p<q$ makes $p$ impossible. But then since $\mathcal{Q}$ is cyclic of prime order, this means there are $p^2(q-1)$ elements of order $q$, which means that $n_p=1$.
