Subgroup of Order p in Group of Order 2p is Normal

Theorem

Let $p$ be an odd prime.

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

Let $a \in G$ be of order $p$.

Let $K = \gen a$ be the subgroup of $G$ generated by $a$.

Then $K$ is normal in $G$.

Corollary

Let $G$ be non-abelian.

Every element of $G \setminus K$ is of order $2$, and:

$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$

Proof

The result Non-Abelian Order 2p Group has Order p Element demonstrates that such an element $a$ exists in $G$.

By definition of generator of cyclic group, $K$ is the cyclic group $C_p$ of order $p$.

By Lagrange's Theorem, the index of $K$ is:

$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$

The result follows from Subgroup of Index 2 is Normal.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $23$