9
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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, $P \triangleleft G$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.