Self-Inverse Order 2

From ProofWiki

Jump to: navigation, search

Theorem

Let $\left({S, \circ}\right)$ be a group whose identity is $e$.

An element $x \in \left({S, \circ}\right)$ is self-inverse iff $\left\vert{x}\right\vert = 2$.

Let $x \in G: x \ne e$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left\vert{x}\right\vert$	$=$	$\displaystyle 2$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \iff$	$\displaystyle $	$\displaystyle x \circ x$	$=$	$\displaystyle e$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Order of an Element
$\displaystyle $	$\displaystyle \iff$	$\displaystyle $	$\displaystyle x$	$=$	$\displaystyle x^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Self-Inverse

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{x}\right\vert\)	\(=\)	\(\displaystyle 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x \circ x\)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Order of an Element
\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle x^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Self-Inverse