Assume $G$ is a group of order $pqr$, with $p, q, r$ distinct primes. Let $P, Q, R$ be their corresponding Sylow subgroups. In addition, assume $P\subseteq C(G)$ and $R\subseteq N(Q)$ where $C$ and $N$ denote the centralizer and the normalizer, correspondingly. Show that $G\cong P\times QR$ and that if $G$ isn't abelian, it has exactly $q$ subgroups of order $pr$.
I have shown that $G\cong P\times QR$ quite easily; we can show $QR$ is normal in $G$ and so is $P$. We can also easily show $o(G)=o(QR)o(P)$, $P\cap QR=\{1\}$, which completes the proof.
However, I'm having trouble with the second part. All I can deduce is that $QR$ is non-abelian, and playing around with the number of Sylow subgroups doesn't seem to be of much help since I can't say much more than $n_p=1, n_q\in\{1, p, r, pr\}, n_r\in\{1, p, q, pr\}$ (or at least, I can't contradict any of the options).
How should I proceed?
Best Answer
Let us identify $G$ with $P\times QR$. Note that the order of an element $(x,y)\in P\times QR$ is the LCM of the orders of $x$ and $y$. Thus an element of order $p$ has the form $(x,1)$ and an element of order $r$ has the form $(1,y)$, where $x$ has order $p$ and $y$ has order $r$. It follows that a subgroup of $G$ of order $pr$ must have the form $P\times H$ where $H\subset QR$ is a subgroup of order $r$, since it must be generated by an element of order $p$ and an element of order $r$.
In other words, the number of subgroups of $G$ of order $pr$ is the same as the number of $r$-Sylow subgroups of $QR$, which can only be $1$ or $q$. If it is $1$, then $R$ is normal in $QR$. But by assumption $Q$ is also normal in $QR$. If $R$ and $Q$ were both normal then we would have $QR\cong Q\times R$ and $G\cong P\times Q\times R$ would be abelian. So, if $G$ is nonabelian, $R$ must not be normal and so there are $q$ subgroups of $G$ of order $pr$.