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

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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 1

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.


Proof 2

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

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 $2 n$.

This contradicts Lagrange's Theorem (Group Theory).

The result follows.