Groups of order $pqr$ and their normal subgroups

Solution 1:

You know that $G$ has a normal subgroup $N$ of order $p$ or $q$. Suppose $|N| = p$, then you are done, so suppose $|N| = q$. Consider the group $G/N$ which has order $pr$. By Cauchy's theorem, $G/N$ has a subgroup $K$ of order $p$. Since this group has index $r$, which is the smallest prime dividing $|G/N|$, $K \triangleleft G/N$.

Now, let $\pi : G \to G/N$ be the natural quotient map, then $H := \pi^{-1}(K)$ is a normal subgroup of $G$ which has order $|K||N| = pq$.

As before, $H$ has a normal subgroup $P$ of order $p$. Now I claim that $P \triangleleft G$. Suppose $g \in G$, then $gHg^{-1} \subset H$. Hence, $$ gPg^{-1} < H $$ So, $gPg^{-1}$ is a $p$-Sylow subgroup of $H$. But $P$ is the unique $p-$Sylow subgroup of $H$. Hence, $$ gPg^{-1} = P \quad\forall g \in G $$ Hence, $P \triangleleft G$.