Groups of Order 2p

Theorem
Let $p$ be a prime number.

Let $G$ be a group.

Let the order of $G$ be $2 p$.

Then $G$ is either:
 * the cyclic group $C_{2 p}$

or:
 * the dihedral group $D_p$.

Proof
When $p = 2$, the result follows from Groups of Order 4.

Let $p$ be an odd prime.

From Sylow p-Subgroups of Group of Order 2p, $G$ has exactly $1$ normal subgroup $P$ of order $p$.

$p$ is prime number.

So from Prime Group is Cyclic, $P$ is a cyclic group.

Let $P = \gen x$ for some $x \in G$.

By the First Sylow Theorem there exists at least one subgroup of $G$ of order $2$.

Hence:
 * $\exists y \in G: y^2 = e$

It follows that the elements of $G$ are known:
 * $G = \set {e, x, \ldots, x^{p - 1}, y, y x, y x^{p - 1} }$

Then:

Thus:
 * the even powers of $y x$ are powers of $x$
 * the odd powers of $y x$ are of the form $y x^j$ for some $j \in \Z_{>0}$.

By Order of Element Divides Order of Finite Group:
 * $\order {y x} \divides 2 p$

where:
 * $\order {y x}$ denotes the order of $y x$
 * $\divides$ denotes divisibility.

We have that:
 * $y x \ne e$

and so by Identity is Only Group Element of Order 1:
 * $\order {y x} \ne 1$

Thus:
 * $\order {y x} \in \set {2, p, 2 p}$

Suppose $i \ne -1$ in $(1)$ above.

Then:
 * $\paren {y x}^2 \ne e$

and so:
 * $\order {y x} \ne 2$

Because odd powers of $y x$ are of the form $y x^j$:
 * $\paren {y x}^p \ne e$

and so:
 * $\order {y x} \ne p$

It follows that:
 * $\order {y x} = 2 p$

and from Group whose Order equals Order of Element is Cyclic, $G$ is cyclic.

Thus, when $i \ne -1$:
 * $G = \gen {y x}$

and so is cyclic.

Thus by Cyclic Group is Abelian:
 * $y x = x y$

When $i = -1$ in $(1)$ above, we have that:

leading to the group presentation of $G$:
 * $G = \gen {x, y: x^p = e = y^2, y x = x^{-1} y}$

which is the Group Presentation of Dihedral Group $D_p$.