$|G|=p^2q^2$ where p,q are primes, $p<q$ and $p \nshortmid (q^2-1)$ Show $G$ is Abelian.
($p$ does not divide $q^2-1$)
The first thing to note is that $(q^2-1)=(q-1)(q+1)$ so we can conclude that $p \nshortmid (q-1)$. Therefore if $P \in Sylow_p(G)$ then we must have $P \triangleleft$G by a standard application of Sylow's theorems.
If I can show now that $|Sylow_q(G)|=1$, i.e. theres is a unique sylow-q subgroup and hence this subgroup is normal, we would have $G = P \times Q$ where both $P$ and $Q$ have order of a prime squared and so are both abelian.
To this end, let $Q \in Sylow_q(G)$. Then by the orbit stabilzer theorem, the number of conjugates of $Q$ in $G$ is equal to $x=[G:N_Q(G)] \cong 1$ mod q and also must be $1, p,$ or $p^2$. If $x \cong 1$ we are done. Also, $x \neq p$ because $p<q$. Thus if i can show that $x \neq p^2$ then I believe I will be finished, but I am having trouble doing this.
I want to invoke some sort of element counting arguement, but since $|Q|=q^2$ this means that various conjugations of $Q$ may not have trivial intersection as they do when $|Q|=q$
Insight appreciated!
Let's say $n_q$ is the number of $q$-Sylow subgroups. If $n_q=p^2$ then $p^2\equiv 1$(mod $q$) and hence $q|(p^2-1)$. Since $q$ is prime this means that either $q|(p-1)$ or $q|(p+1)$. Can you finish from here?