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:
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:
\(\ds y x y^{-1}\) | \(\in\) | \(\ds P\) | as $P$ is normal in $G$. | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y x y^{-1}\) | \(=\) | \(\ds x^i\) | for some $i \in \Z_{>0}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y x y^{-1} x\) | \(=\) | \(\ds x^{i + 1}\) | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds y x y x = \paren {y x}^2\) | \(=\) | \(\ds x^{i + 1}\) | $y^2 = e$, 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:
\(\ds y x y^{-1} x\) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y x y^{-1}\) | \(=\) | \(\ds x^{-1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y x\) | \(=\) | \(\ds x^{-1} y\) |
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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \beta$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$