If $P$ or $Q$ is normal then both $P$ and $Q$ are characteristic in $PQ$ Abstract Algebra, Dummit and Foote e3 states that:
"If $G$ is a group of order 30 and $P\in Syl_5(G)$ and $Q\in Syl_3(G)$ with one of them being normal, then both are characteristic subgroups of $PQ$; this then implies that since $PQ$ is normal, both $P$ and $Q$ have to be normal."
Why is it that if either $P$ or $Q$ is normal in $G$ then both are characteristic in $PQ$? I realize that, by Sylow theorem, the normal Sylow subgroup has to be characteristic, but I do not get why the other one has to be also.
 A: First note that since $P$ or $Q$ is normal, $PQ \leq G$. We have $|P|$ and $|Q|$ are coprime, so they must have trivial intersection. Thus $|PQ| = \frac{|P||Q|}{|P \cap Q|} = \frac{15}{1}$. Thus $PQ$ is normal in $G$ since it has index $2$. If $Q$ is normal in $G$ then $P$ has index $3$ in $PQ$. Since $3$ is the smallest prime dividing $|PQ|$, $P$ is necessarily normal in $PQ$ and thus characteristic. If $P$ is normal in $G$ then consider the number of possible subgroups of order $3$ in $PQ$. We know from the Sylow Theorems that it must be congruent to 1 modulo three. Further it must divide 5. The only divisor of $5$ congruent to 1 modulo 3 is 1. Thus there is one Sylow 3 subgroup of $PQ$, namely $Q$. 
A: The normality of $P$ or $Q$ is needed only to ensure that $PQ$ is a subgroup. Once $PQ$ is a subgroup, then $P\in\operatorname{Syl}_5(PQ)$ and $Q\in\operatorname{Syl}_3(PQ)$. By Sylow's theorem, $n_5(PQ)\mid 3$ and $n_5(PQ)\equiv1\pmod{5}$, so $n_5(PQ)=1$, hence $P$ is characteristic in $PQ$. Similarly, $n_3(PQ)\mid 5$ and $n_3(PQ)\equiv1\pmod{3}$, so $n_3(PQ)=1$, hence $Q$ is characteristic in $PQ$.
