non-abelian groups of order $p^2q^2$. Let $p<q$ be prime numbers and let $G$ be a group of order $p^2q^2$.
I wish to determine up to isomorphism how many groups $G$ are there.
What I know:
The abelian case is very clear.
Moreover, let assume that $pq\neq 6$ then it can be shown that 
$$G=Q\rtimes P,$$
where $P,Q$ are the corresponding Sylow subgroups.
For some $p,q$ the only groups $G$ are abelian, but lets focus on these $p,q$ such that $G$ is non-abelian.
I believe that if $Q$ is cyclic, then for any $P$ (cyclic or of rank $2$) there exist exactly one isomorphism class.
However, in the case where $Q=C_q\times C_q$ I am not sure about the number of isomorphism classes.
Any help will be appreciated. 
 A: Let's study the case $p=3$, $q=19$. Let $P \in {\rm Syl}_{19}(Q)$, $Q \in {\rm Syl}_3(G)$. (Sorry, I have managed to swap $P$ and $Q$!)
Case 1. $P,Q$ cyclic. $Q$ can induce an automorphism of order $3$ or $9$ on $Q$, giving $2$ groups.
Case 2. $P$ cyclic, $Q$ non-cyclic. $Q$ must induce automorphism of order $3$ of $P$, giving $1$ group.
Case 3. $P$ non-cyclic, $Q$ cyclic. Let $\omega$ be an element of order $9$ in ${\mathbb F}_{19}^*$; for example $\lambda=4$.
a) If $Q$ induces automorphism of order $3$ of $P$, then there are $3$ groups, in which the eigenvalues of the action of $P$ on $Q$ are respectively $(1, \omega^3)$, $(\omega^3,\omega^3)$, and $(\omega^3,\omega^6)$.
b) If $Q$ induces automorphism of order $9$ of $P$, then there are $7$ groups, in which the eigenvalues of the action of $P$ on $Q$ are respectively $(1, \omega)$, $(\omega,\omega)$, $(\omega,\omega^2)$, $(\omega,\omega^3)$. $(\omega,\omega^4)$, $(\omega,\omega^6)$, $(\omega,\omega^8)$.
(Note that $(\omega,\omega^5)$ would give a group isomorphic to $(\omega,\omega^2)$ and $(\omega,\omega^7)$ isomorphic to $(\omega,\omega^4)$.)
Case 4. $Q$ and $P$ both non-cyclic.
a) If $Q$ induces automorphism of order $3$ of $P$, then there are $3$ groups, just as in Casse 3 a).
b) If $Q$ acts faithfully on $P$, then there is a unique group.
So we get $17$ nonabelian groups altogether which, together with the $4$ abeliabn groups, makes $21$ groups of this order. This agrees with the number given by GAP.
