Groups of Order 2p

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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:

\(\displaystyle y x y^{-1}\) \(\in\) \(\displaystyle P\) as $P$ is normal in $G$.
\(\displaystyle \leadsto \ \ \) \(\displaystyle y x y^{-1}\) \(=\) \(\displaystyle x^i\) for some $i \in \Z_{>0}$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y x y^{-1} x\) \(=\) \(\displaystyle x^{i + 1}\)
\((1):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle y x y x = \paren {y x}^2\) \(=\) \(\displaystyle x^{i + 1}\) $y^2 = e$, so $y = y^{-1}$


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:

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

$\Box$


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:

$y x = x y$

$\Box$


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

\(\displaystyle y x y^{-1} x\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y x y^{-1}\) \(=\) \(\displaystyle x^{-1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y x\) \(=\) \(\displaystyle x^{-1} y\)

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 $D_p$.

$\blacksquare$


Sources