I am trying to understand the proof of the following statement:
If $p$ and $q$ are primes, than a group of order $p^2q$ cannot be simple.
The proof is given as follow:
Assume that G is simple and $ p \neq q$. The number of $p$-Sylow subgroups must be $q$ and hence $q \equiv 1 \mod p$. Hence, $q > p$. The number of $q$-Sylow subgroups is either $p$ or $p^ 2$. It cannot be $p$, because this would imply that $p \equiv 1 \mod q$ implying $p > q$.
Hence, the number of $q$-Sylow subgroups must be $p^2$. But any pair of these subgroups can have only the identity as a common element. Here I got confused. Why is it true? Suppose we have that $p^2$ $q$-Sylow subgroups $Q_1,\dots, Q_{p^2}$. Why we have that $$Q_j \neq Q_i \;\Rightarrow Q_i \cap Q_j = \{e\}$$
Furthermore why we get than that $\vert Q_1 \cup \dots \cup Q_{p^2} \vert=1 +p^2*(q-1)$?
EDIT: In my textbook it uses also the following statement: Let $P_1, \dots, P_q$ be the $p$-Sylow subgroups of $G$. Then $\vert P_1 \cup \dots \cup P_q \vert \geq \vert P_1 \cup P_2 \vert \geq p^2+(p^2-p)$. How we get there the last inequality $\vert P_1 \cup P_2 \vert \geq p^2+(p^2-p)$?
Many thanks for your help!
If $H$ and $K$ are different subgroups of $G$ of order $q$ a prime, then $H \cap K$ is a subgroup of $H$ and by Lagrange $|H \cap K|$ divides $|H|=q$. Since $q$ is prime, $H \cap K=H$, impossible since $H$ and $K$ are different, or $H \cap K=1$.
To finish the argument: hence there are $p^2(q-1)$ elements of order $q$. That leaves $|G|-p^2(q-1)=p^2$ elements of $p$-power order, so these consitute the Sylow $p$-subgroup of order $p^2$ which must be normal, whence $G$ is not simple.