Non-Abelian Order 10 Group has Order 5 Element

Theorem

Let $G$ be a non-abelian group of order $10$.

Then $G$ has at least one element of order $5$.

Proof 1

By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $5$ or $10$.

From Identity is Only Group Element of Order 1, $9$ elements of $G$ have orders greater than $1$.

From Cyclic Group is Abelian, $G$ is not cyclic.

If $g \in G$ was of order $10$ then $g$ would generate the cyclic group $C_{10}$.

Thus $G$ has no element of order $10$.

So all elements of $G$ except the identity have orders $2$ or $5$.

Aiming for a contradiction, suppose $G$ has no element of order $5$.

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.

$\blacksquare$

Proof 2

As $10 = 2 \times 5$, $G$ is a non-abelian group of order $2 p$, where $p$ is an odd prime.

Hence this is an instance of the result Non-Abelian Order 2p Group has Order p Element.

$\blacksquare$