Classify all groups of order $p^2q^2$ up to isomorphism Let $p,q \in \mathbb{N}$ be prime numbers with the properties 
$2 < p < q$  and $q - 1 , q + 1 \notin \left\langle p \right\rangle$
Classify all groups of the order $p^2q^2$ up to isomorphism.

This was a question given by my algebra professor and quite frankly I am stumped.
My initial thought was that this question is referring to p-Sylow subgroups and one would need to apply the Sylow-theorems. If this is true, how would you apply them? Then what does "up to isomorphism" exactly mean?
I also thought to try and break it down and look at different possible cases. For example something like this:
Since 
$$q - 1 \notin \left\langle p \right\rangle \Rightarrow p \nmid q - 1 $$
$\Rightarrow \exists! $ subgroup of order $p$  $\Rightarrow \exists p - 1 $elements of order $p$ and $q-1$ elements of order $q$.
But honestly I am not sure how to answer this question. I would really appreciate if someone could try and explain this to me. Thank you in advance!
 A: Let $G$ be a finite group of order $p^2q^2$ where $2<p<q$ and where $q\not\equiv\pm1\pmod{p}$. Then $p\nmid q-1$ and $p\nmid q+1$ so $p\nmid q^2-1$. Thus, $q\not\equiv1\pmod{p}$ and $q^2\not\equiv1\pmod{p}$. By Sylow's theorems, $G$ has a unique Sylow $p$-subgroup $P$.
Since $2<p<q$, we have that $q\geq p+2$. Then $q\nmid p-1$ and $q\nmid p+1$ so $q\nmid p^2-1$. Thus, $p\not\equiv1\pmod{q}$ and $p^2\not\equiv1\pmod{q}$. By Sylow's theorems $G$ has a unique Sylow $q$-subgroup $Q$.
Thus, $G$ factors as the direct product $P\times Q$. In particular, there are four possibilities for $G$: $C_{p^2}\times C_{q^2}$, $C_{p^2}\times C_q\times C_q$, $C_p\times C_p\times C_{q^2}$, and $C_p\times C_p\times C_q\times C_q$.
A: There are only two groups up to isomorphism of order $p^2$: $C_{p^2}$ and $C_p\times C_p$. Sylow $q$-subgroup of $G$ is normal, and moreover, $G$ is a semi-direct product $Q:P$. Hence, there is a homomorphism $\varphi:P\to\mathrm{Aut}(Q)$.
A more general case: (without your assumption on $p$ and $q$)

If $Q\cong C_{q^2}$ then $\mathrm{Aut}(Q)\cong C_{q^2-q}$.
If $\gcd(p^2,q^2-q) = 1$, then the homomorphism $\varphi$ must be trivial. Thus, $G = Q\times P$.
If $p\mid q-1$ but $p^2\nmid q-1$. Then there is a non-trivial $\varphi$ such that $\mathrm{Im}(\varphi)\cong C_p$. If $P\cong C_p\times C_p$ then the group $G$ becomes $(C_p\times C_{q^2}):C_p\cong C_p\times (C_{q^2}:C_p)$. If $P\cong C_{p^2}$ then $G\cong C_{q^2}:C_{p^2}$ (with $\mathrm{Im}(\varphi)\cong C_p$).
If $p^2\mid q-1$ then $\mathrm{Im}(\varphi)$ can be larger. In this case if $\mathrm{Im}(\varphi)\cong C_p$ then it is the same as the previous case. Now if $\mathrm{Im}(\varphi)$ has order $p^2$, then $\mathrm{Im}(\varphi)\cong P$ (it is an inclusion). The group $G$ becomes $C_{q^2}:C_{p^2}$ (with $\mathrm{Im}(\varphi)\cong C_{p^2}$) or $C_{q^2}:(C_p\times C_p)$ (with $\mathrm{Im}(\varphi)\cong C_p\times C_p$).

If $Q\cong C_q\times C_q$ then $\mathrm{Aut}(Q)\cong \mathrm{GL}(2,q)$, which has order $(q^2-1)(q^2-q) = q(q-1)^2(q+1)$. You can try also to discuss as above to find all.

Now by your assumption, $p\nmid q-1$ and $p\nmid q+1$, and in particular $p\nmid q$. In both cases $\varphi$ must be trivial. Therefore, it becomes a direct product $G\cong Q\times P$, and there are totally four groups up to isomorphism.
