Let G be a group of order $pqr$,where $pI 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.
 A: 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.
