Group of Order p q has Normal Sylow p-Subgroup

Theorem

Let $p$ and $q$ be prime numbers such that $p > q$.

Let $G$ be a group of order $p q$.

Then $G$ has exactly one Sylow $p$-subgroup.

This Sylow $p$-subgroup is normal.

Proof

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 $p q$.

Of these:

$p$ 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 $1$.

Let $P$ denote this unique Sylow $p$-subgroups.

From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \alpha$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$: Proposition $12.1$