Abelian Group of Order Twice Odd has Exactly One Order 2 Element/Proof 1

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 $g \in G$ with $g \ne e$ such that $g = g^{-1}$.

Proof

By Abelian Group Factored by Prime, the subgroup $H_2$ defined as:

$H_2 := \set {g \in G: g^2 = e}$

has precisely two elements.

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

The result follows.

$\blacksquare$