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Let $G$ be a group of order $pqr$, where $p,q,r$ are different primes. Let $P,Q,R$ be sylow-$p,q,r$ subgroups respectively. Show that if $P\subseteq C(G)$ and $R\subseteq N(Q)$, then $G\cong P\times QR$

The aproach I tried to take is to prove that $G=PQR$ and then that $P, QR$ satisfy $P\vartriangleleft G, QR\vartriangleleft G$ and $P\cap QR=\{e\}$. However I couldn't find how to prove that $G=PQR$ and $QR\vartriangleleft G$.

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The claim that $G = PQR$ is essentially an order argument. Recall that for subsets $A,B$ of $G$ we have the product formula: $$|AB| = \frac{|A||B|}{|A\cap B|}$$ Since $p,q,r$ are different primes the pairwise intersections of $P,Q,R$ are trivial so we already know that $|PQ| = |P||Q| = pq$. On the other hand, since $P$ is central in $G$ it is a normal subgroup and $PQ$ is a subgroup as well. Using the same argument we find $PQ \cap R = \{ 1\}$ so $|PQR| = pqr = |G|$ so we indeed have $G = PQR$.

To prove that $QR\vartriangleleft G$ we have to prove that $g^{-1}QRg\subset QR$ for all $g\in G$. Since we already know that $G = PQR$ it is enough to prove this for the cases that $g$ belongs to $P$,$Q$ or $R$. Can you finish the argument?

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    Do I also need to prove that $QR$ is a subgroup after I say that $g^{-1} QR g \subseteq QR$ in order to say it is a normal subgroup? Or does it fall between the lines? – 2017-02-12
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    Good point. I guess it doesn't follow automatically from that verification. In general, if $A$ en $B$ are subgroups of $G$, then $AB$ is a subgroup of $G$ if and only if $AB = BA$. So you should indeed prove that $QR=RQ$. – 2017-02-12