Groups of order 30 
Dummit&Foote p.143 Let $G$ be a group of order 30. Let $P \in Syl_5(G)$ and let $Q \in Syl_3(G)$. If either $P$ or $Q$ is normal in
  $G$, then both $P$ and $Q$ are characteristic subgroups of $PQ$. .....

I don't understand the above parts, stuck on it for hours. How can it be derived?

Since either $P,Q$ is normal you know that $PQ\leqslant G$. But, since $|P\cap Q|=1$ (necessarily since $|P\cap Q|\mid \text{gcd}(|P|,|Q|)=1$) one has that $\displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{5\cdot 3}{1}=15$. Thus, $PQ$ is a group of order $15$. Now, it's easy to check that since $5\not\equiv1\text{ mod }3$ that $P,Q$ are normal in $PQ$. But, it's a common fact that normal Sylow subgroups are characteristic.
Remark: In fact, all groups of order $15$ are cyclic.
