Even Order Group has Odd Number of Order 2 Elements

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

Let $G$ be of even order.

Then $G$ has an odd number of elements of order $2$.

Proof
Let $S = \set {x \in G: \order x > 2}$ be the set of all elements of $G$ whose order is strictly greater than $2$.

Let $h \in S$.

Then from Order of Group Element equals Order of Inverse:
 * $h^{-1} \in S$.

Because $\order h > 2$ it follows that $h^{-1} \ne h$ by definition of self-inverse.

Thus every element in $S$ can be paired off with its inverse which is also in $S$.

It follows that $S$ has an even number of elements.

Thus $G \setminus S$ also contains an even number of elements.

From Identity is Only Group Element of Order 1, there is precisely $1$ element of $G$ whose order is less than $2$.

This leaves $G$ with an odd number of elements of order $2$.