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

Proof
By Even Order Group has Order 2 Element, $G$ has an element $x$ of order $2$.

$y$ 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 $2 n$.

This contradicts Lagrange's Theorem (Group Theory).

The result follows.