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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.

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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, $S_p(G) \neq q^2$.

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Since $S_p(G)$ divides $q^2$ then $S_p(G)=1,q,$ or $q^2$. Note if that $S_p(G)=q$ then that implies that $S_p(G)=q\equiv 1\mod p$ and so $p$ divides $q-1$, a contradiction. If $S_p(G)=q^2$, then $p$ divides $q^2-1=(q-1)(q+1)$. Since $p$ is prime $p$ divides either $q-1$ or $q+1$, but neither can happen. Therefore, $S_p(G)=1$.

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