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I need to show that there exist unique r-Sylow subgroup. I know that the number of Sylow r-groups denoted by $n_r$ is congruent to $1mod \ r$. Also $n_r$ divides $pq$. So $n_r$ can be $1,p,q,pq$. But $n_r$ can not be $p$ or $q$ since $r$ is greater than $p$ and $q$. Now I need to get a contradiction when $n_r$ is $pq$. Could you please help me on this problem.

Shaun
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Rashad Ek
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    If $n_r=pq$, then the number of elements in the Sylow $r$-groups is $pqr$. This implies that either every element of $G$ is in one of these groups (clearly not true), or that two of the Sylow groups overlap (which also isn't possible since $r$ is prime). Herein lies your contradiction – Rushabh Mehta Jul 20 '19 at 15:44
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    if $n_r$=pq , then number of elements in the sylow r -groups is $pq(r-1)+1$. Because identity element is common for all. May I know whether I'm correct? – Rashad Ek Jul 20 '19 at 15:48
  • I suggest you check that if a Sylow-$p$ is normal then its elements commute with the Sylow-$r$ and so a Sylow-$p$ is in the normaliser of a Sylow-$r$, which does not happen in your case; ditto for Sylow-$q$. Now count the number of $p$-elements and $q$-elements and see they can't be fitted into the space you have left. – ancient mathematician Jul 20 '19 at 16:07

1 Answers1

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By Sylow's theorems, $n_r\in\{1,pq\}$. If $n_r=1$ then the Sylow $r$-subgroup of $G$ is normal. Now suppose for contradiction that $n_r=pq$. By Lagrange's theorem, any two Sylow $r$-subgroups of $G$ intersect trivially. In particular, $G$ contains $pq(r-1)$ elements of order $r$. This is not a contradiction yet so we will now analyze the Sylow $q$-subgroups of $G$.

By Sylow's theorems, $n_q\in\{1,r,qr\}$. If $n_q\geq r$ then $G$ contains at least $r(q-1)$ elements of order $q$. This gives a grand total of at least $$pq(r-1)+r(q-1)=pqr+qr-pq-r$$ elements of order $q$ or $r$. This is actually impossible since $$qr-pq-r>qr-pr-r=r(q-p-1)\geq0.$$ Thus, $n_q=1$. Let $Q$ be the normal Sylow $q$-subgroup of $G$. The quotient group $G/Q$ has order $pr$. By Sylow's theorems, $G/Q$ has a normal Sylow $r$-subgroup. By the correspondence theorem, $G$ has a normal subgroup $N$ of order $qr$. By Sylow's theorems, $N$ has a normal Sylow $r$-subgroup $R$. In general, normality is not transitive so we cannot immediately conclude that $R$ is a normal subgroup of $G$. In our case however, the following lemma shows that $R$ is a normal subgroup of $G$ which contradicts our assumption that $n_r=pq$.

Lemma: Let $G$ be a finite group, let $N$ be a normal subgroup of $G$, and let $P$ be a normal Sylow $p$-subgroup of $N$. Then $P$ is a normal subgroup of $G$.

Proof: For any $g\in G$, $P\leq N$ so $gPg^{-1}\leq gNg^{-1}=N$. Since conjugation preserves cardinality, $gPg^{-1}$ is a Sylow $p$-subgroup of $N$. However, Sylow's second theorem shows that $P$ is the unique Sylow $p$-subgroup of $N$. Thus, $gPg^{-1}=P$ which shows that $P$ is a normal subgroup of $G$.

Effectively, we are using the fact that a normal Sylow $p$-subgroup is characteristic.

Thomas Browning
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