# Even Order Group has Order 2 Element/Proof 1

## Theorem

Let $G$ be a group whose identity is $e$.

Let $G$ be of even order.

Then:

$\exists x \in G: \order x = 2$

That is:

$\exists x \in G: x \ne e: x^2 = e$

## Proof

In any group $G$, the identity element $e$ is self-inverse with Identity is Only Group Element of Order 1, and is the only such.

That leaves an odd number of elements.

Each element in $x \in G: \order x > 2$ can be paired off with its inverse, as $\order {x^{-1} } = \order x > 2$ from Order of Group Element equals Order of Inverse.

Hence there must be at least one element which has not been paired off with any of the others which is therefore self-inverse.

The result follows from Group Element is Self-Inverse iff Order 2.

$\blacksquare$