Application of the Sylow Theorems to groups of order $p^2q$ I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$.  But we also have $n_p\equiv1\mod p$ which rules out $q$.  So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p<q$ however, I run into a problem.. Again we have $n_q|p^2$ (so $n_q\in{1,p,p^2}$). Again, the condition $n_q\equiv1\mod q$ rules out $p$.
Now I am attempting to rule out the case $n_q=p^2$:  Assume $n_q=p^2$.  Then $p^2\equiv1\mod q\implies (p+1)(p-1)\equiv0\mod q\implies q$ must divide $(p+1)$ since $p<q$ and $q$ prime.  But since $p<q$ and $q|(p+1)$ we see $q=p+1$.  For primes this only happens when $p=2, q=3$.
Does this mean I need to check groups of order $2^2\cdot3=12$ or did I miss something along the way that lets me conclude $n_q\neq p^2$ and thus that $n_q=1$.
 A: 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.
A: 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!
A: For the Sylow $p$ subgroups you're not taking into account that these subgroups can intersect at $p^2, p,$ or $1$ element/s. It can't be $p^2$ because then all of the subgroups would be the same. However it could be $p$ or $1$. For $p$ you can use a counting argument that will give a contradiction with the order of the group and then you don't have to do the argument for $1$. 
