# 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 $m = n + k$.

 $\displaystyle x^m$ $=$ $\displaystyle x^{n + k}$ $\displaystyle$ $=$ $\displaystyle x^n x^k$ $\displaystyle$ $=$ $\displaystyle x^n e^G$ $\displaystyle$ $=$ $\displaystyle x^n$

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:

$x^k = x^{m - n} = x^m x^{-n} = x^n x^{-n} = e_G$

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.

$\blacksquare$