We want to prove that a group, say $G$, of order $pqr$ where $p \gt q \gt r$ has a normal Sylow $p$-subgroup and deduce that it has a normal subgroup of order $pq$.

I know how to show $G$ has a normal Sylow $p$-subgroup or normal Sylow $q$-subgroup, but not the rest. Thanks for your advices.


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, $P \triangleleft G$.

  • $\begingroup$ First of all i want to thank for your sincere help. i am concerned with this matter,i think we should say K is a subgroup of G/N. Am i right? $\endgroup$ – Aref Nov 15 '13 at 15:46
  • 1
    $\begingroup$ @Aref : Yes, that was a typo. Thanks. $\endgroup$ – Prahlad Vaidyanathan Nov 15 '13 at 16:24

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