Equivalence of Definitions of Finite Order Element
Theorem
The following definitions of the concept of Finite Order Element are equivalent:
Let $G$ be a group whose identity is $e_G$.
Let $x \in G$ be an element of $G$.
Definition 1
$x$ is of finite order, or has finite order if and only if there exists $k \in \Z_{> 0}$ such that $x^k = e_G$.
Definition 2
$x$ is of finite order, or has finite order if and only if there exist $m, n \in \Z_{> 0}$ such that $m \ne n$ but $x^m = x^n$.
Proof
$(1)$ implies $(2)$
Let $x$ be a finite order element of $G$ by definition 1.
Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$.
Consider some $m, n \in \Z_{>0}$ such that $m = n + k$.
\(\ds x^m\) | \(=\) | \(\ds x^{n + k}\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds x^n x^k\) | Powers of Group Elements/Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds x^n e_G\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds x^n\) | Definition of Identity Element |
Thus $x$ is a finite order element of $G$ by definition 2.
$\Box$
$(2)$ implies $(1)$
Let $x$ be a finite order element of $G$ by definition 2.
That is, there exists $m, n \in \Z_{>0}$ such that $x^m = x^n$ but $m \ne n$.
Without loss of generality, suppose that $m > n$.
Let $k = m - n$.
From $x^m = x^n$ it follows from Powers of Group Elements that:
\(\ds x^k\) | \(=\) | \(\ds x^{m - n}\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds x^{m + \paren{-n } }\) | Definition of Subtraction/Integers | |||||||||||
\(\ds \) | \(=\) | \(\ds x^m x^{-n}\) | Powers of Group Elements/Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds x^n x^{-n}\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren{x x^{-1} } ^ n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren{e_G } ^ n\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds e_G\) | Definition of Identity Element |
Thus there exists $k \in \Z_{>0}$ such that $x^k = e_G$.
Thus $x$ is a finite order element of $G$ by definition 1.
$\blacksquare$