I have this following question from my class note on Sylow Theorem:
Let $p$ and $q$ be prime numbers such that $p \nmid (q-1).$ Show that each group of order $pq$ possesses a normal subgroup of order $p.$
For that I know the followings:
(1) Group $G$ is p-group if there exists an integer $e$ such that $|G| = p^e,$
(2) A $p$-subgroup $H$ of $G$ is called Sylow p-subgroup of $G$ if $p \nmid |G/H|,$ and finally
(3) A subgroup $H$ of $G$ is a Sylow p-subgroup if there exist integers $e$ and $m$ such that $|G| = p^em,$ and $(e, m) = 1,$ and also $|H| = p^e,$
but unfortunately I don't know how to assemble them to solve the problem. Any help would therefore be very much appreciated. Thank you for your time and help.
PS. If you have choices of elegant versus dummy down-to-earth solutions, do please give me the latter. I know it would be tedious for you but you have slowpoke turtle over here. Thank again for your patience.
By the Sylow theorems, the number of Sylow $p$-subgroups of the group is congruent to $1$ mod $p$ and divides the order of the group. Suppose $n=kp+1$ and $n\mid q$. Then since $q$ is prime we in fact have $kp+1=q$ or $k=0$. Assume $k\neq 0$. Then $q-1=kp$, meaning that $p\mid q-1$. Since we assume that this is not the case, we must have that $k=0$, hence the number of Sylow $p$-subgroups is $1$, and this Sylow $p$-subgroup is normal because all Sylow $p$-subgroups are conjugate.