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

Theorem
Let $p$ be an odd prime.

Let $G$ be a group of order $2 p$ whose identity element is $e$.

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

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

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
By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$.

$1$ is not possible because Identity is Only Group Element of Order 1.

Then we have that $G$ is non-abelian.

Hence from Cyclic Group is Abelian, $G$ is not cyclic.

Thus $\order b \ne 2 p$.

It remains to investigate $2$ and $p$.

Let $b \in G \setminus K$.

By Subgroup of Index 2 contains all Squares of Group Elements:
 * $b^2 \in K$

$b$ has order $p$.

Then by Intersection of Subgroups of Prime Order:
 * $K \cap \gen b = e$

which contradicts $b^2 \in K$.

Thus by Proof by Contradiction $\order b \ne p$

Hence it must follow that $\order b = 2$.

We have that:
 * $b a \in G \setminus K$

Thus:
 * $\paren {b a}^2 = e$