Sylow p-Subgroups of Group of Order 2p/Proof 2

Theorem

Let $p$ be an odd prime.

Let $G$ be a group of order $2 p$.

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 of $G$.

From the First Sylow Theorem, there exists at least $1$ Sylow $p$-subgroup of $G$.

Let $P$ be such a Sylow $p$-subgroup of $G$.

The index of $P$ is $2$.

From Subgroup of Index 2 is Normal, $P$ is normal in $G$.

From Sylow $p$-Subgroup is Unique iff Normal it follows that there is only $1$ Sylow $p$-subgroup of $G$.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$