Even Order Group has Order 2 Element

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

Proof
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.