Group of Order p q has Normal Sylow p-Subgroup
Let $p$ and $q$ be prime numbers such that $p > q$.
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.
From the Fourth Sylow Theorem:
- $n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
- $n_p \divides p q$
where $\divides$ denotes divisibility.
The divisors of $p q$ are $1$, $p$, $q$ and $p q$.
- $p$ and $p q$ are $\equiv 0 \pmod p$
and as $p > q$:
- $q \equiv q \pmod p$
So the only possibility for $n_p$ is $1$.