About minimal subgroup of a finite group. Let $G$ be a finite group such that $G=PQ$ where $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is a normal Sylow $q$-subgroup of $G$. I need to prove the following:
(1) If $H/Q$ is a subgroup of $PQ/Q$ of order $p$, then $H/Q=\langle x\rangle Q/Q$ where $x \in P$ and the order of $x$ is $p$.
(2) Suppose that $p=2$. If $H/Q$ is a cyclic subgroup of $PQ/Q$ of order $4$, then $H/Q=\langle x\rangle Q/Q$ where $x \in P$ and the order of $x$ is $4$.
I think that $H/Q=\langle xQ\rangle$ which is isomorphic to $\langle x \rangle / (\langle x \rangle \cap Q)$, but how to prove that $\langle x \rangle \cap Q=\{1\}$. 
Thanks.

Use the isomorphism theorems:
$$(1)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\langle x\rangle Q/Q\cong \langle x\rangle/\left(\langle x\rangle\cap Q\right)\cong\langle x\rangle$$
since assuming $\,p\neq q\,$ , we get that $\,x\in P\Longrightarrow \langle x\rangle\cap Q=\{1\}$
To end this, just observe that $\,H/Q\,$ is a subgroup of order $\,p\,$ then $\,H/Q=\langle hQ\rangle=\langle h\rangle Q\,$ , with $\,\mathcal Ord(x)=p\,$ , and since any $\,p-$subgroup of $\,G\,$ is contained in some Sylow $\,p-$subgroup...
Now point (2) must follow at once from (1) above, introducing some little changes...
