Group with order $|G|=p^2q^2$ and $p\not\mid q-1, p\not\mid q+1$ Group with order $|G|=p^2q^2$ and $p\not\mid q-1, p\not\mid q+1$. $q\not=p$ both prime.
I want to show that there is only one $p$-Sylow subgroup. Let $S_p(G)$ the number of  $p$-Sylow subgroups. I know that $S_p(G) \equiv 1 \mod p$ and $S_p(G) \mid q^2$.
If $S_p(G) > 1$, then $S_p(G)= q = k \cdot p + 1$ or $S_p(G) = q^2 = k \cdot p +1$ with $k \in \mathbb{N}_0$.
If $q = k \cdot p + 1 \Rightarrow q-1=k \cdot p \Rightarrow p \mid q-1$ and this contradict to our assumption. So I have to know if $q^2 = k \cdot p +1$, but I don't know how? 
Thank for your help.

Note that from $q^2 = kp + 1$ we get that $$kp = q^2 - 1 = (q - 1)(q+1).$$ So $p \mid (q-1)(q+1)$, but as $p$ is prime we must have $p \mid q - 1$ or $p \mid q + 1$; either way, we have a contradiction. Therefore, $S_p(G) \neq q^2$.
