# Group of order $pqr$

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?

• It is immediate that $n_p=1$ since $P$ is normal. – Eric Wofsey Feb 3 at 21:06
• @EricWofsey Oh yes indeed, but I can't say much about $n_q$ and $n_r$, I think... I'll edit my question. – Yuval Gat Feb 3 at 21:07

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 $$pr$$.