Even Order Group has Order 2 Element

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Theorem

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

Let $G$ be of even order.

Then $\exists x \in G: \left|{x}\right| = 2$.

That is, $\exists x \in G: x \ne e_G: x^2 = e$.

In any group $G$, the identity element $e$ is self-inverse with order 1, and is the only such.

That leaves an odd number of elements.

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

The final element which has not been paired off with any of the others must be self-inverse.

The result follows from Self-Inverse Order 2.

$\blacksquare$