Groups with $|G| = p^2q$. Prove that if $p$ and $q$ are primes, then there are no simple groups of order $p^2q$. Groups with $|G| = p^2q$. 
Prove that if $p$ and $q$ are primes, then there are no simple groups of order $p^2q$.
Also another question, do p and q have to be distinct for this to hold? 
On top of that, I know we have to use the Sylow Theorems but I do not understand them as they are currently. 
 A: First, suppose $p<q$. Then $n_q\equiv 1\pmod{q}$, and $n_q\mid p^2$. So the only possibilities are that $n_q=1,p,p^2$. If $n_q=1$, the group is not simple. But $n_q=p$ is impossible, else $q\mid p-1$. So $n_q=p^2$. Then we have $p^2$ groups of prime order $q$, so they have pairwise trivial intersection, and this forces the $p$-Sylow subgroup to be unique.
The $q$-Sylow subgroups give a total of $p^2(q-1)$ elements of order $q$. This accounts for all but $p^2$ elements of the group, so any $p$-Sylow subgroup must be these remaining elements, hence unique.
The case where $q<p$ is quite quick. Any $p$-Sylow subgroup has index $q$. This is smallest prime dividing $|G|$, so your $p$-sylow subgroup is normal.
A: Either $p^2 < q$ or $q < p^2$.  Assume WLOG the former.  
By Sylow I, there must be at least one Sylow $q$-subgroup.  Now by Sylow III, the total number of $q$-subgroups must be congruent to $1 \pmod{q}$, and further, the total number of such subgroups must divide $|G| = p^2q$.
The only number that satisfies these criteria is $1$, so there must be $1$ Sylow $q$-subgroup.  By Sylow II, all Sylow $p$-subgroups are conjugate to one another for a specific $p$ and there's only $1$ Sylow $q$-subgroup, then it is necessarily normal in $G$.
$\Longrightarrow G$ is not simple.
