Abelian Group of Order Twice Odd has Exactly One Order 2 Element

Theorem
Let $G$ be an abelian group whose identity element is $e$.

Let the order of $G$ be $2 n$ such that $n$ is odd.

Then there exists exactly one $g \in G$ such that $g = g^{-1}$.

Proof
By definition, $G$ is of even order.

From Even Order Group has Order 2 Element it follows that $G$ has at least one element $g$ of order $2$.

It remains to show that $g$ is the only such element.

Suppose that:
 * $\exists h \in G: h^2 = e = g^2$