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

Theorem

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

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

$H_2 := \set {g \in G: g^2 = e}$

has precisely two elements.

One of them has to be $e$, since $e^2 = e$.

The result follows.

$\blacksquare$

Aiming for a contradiction, suppose $y$ is another element of order $2$.

Then $x y = y x$ is another element of order $2$.

The subset $H = \set {g \in G: g^2 = e} = \set {e, x, y, x y}$ of $G$ forms a subgroup of $G$.

Thus $\order H = 4$.

But as $n$ is odd, it follows that $\order H$ is not a divisor of $2 n$.

The result follows.

$\blacksquare$