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

Jump to navigation
Jump to search

## 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.

$\blacksquare$

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Exercise $5.7$