Non-Abelian Order 2p Group has Order p Element

Theorem

Let $p$ be an odd prime.

Let $G$ be a non-abelian group of order $2 p$.

Then $G$ has at least one element of order $p$.

Proof

By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$.

From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$.

From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$.

If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic group $C_{2 p}$.

Thus $G$ has no element of order $2 p$.

So all elements of $G$ except the identity have orders $2$ or $p$.

Aiming for a contradiction, suppose $G$ has no element of order $p$.

Then all elements are of order $2$.

Then $G$ is a Boolean group.

Then by Boolean Group is Abelian, $G$ is abelian.

This contradicts the assertion that $G$ is non-abelian.

Hence the result by Proof by Contradiction.

$\blacksquare$

Sources

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