# Equivalence of Definitions of Infinite Order Element

## Contents

## Theorem

The following definitions of the concept of **Infinite 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 infinite order**, or **has infinite order** if and only if there exists no $k \in \Z_{> 0}$ such that $x^k = e_G$:

- $\order x = \infty$

### Definition 2

$x$ **is of infinite order**, or **has infinite order** if and only if the powers $x, x^2, x^3, \ldots$ of $x$ are all distinct:

- $\order x = \infty$

### Definition 3

$x$ **is of infinite order**, or **has infinite order** if and only if the group $\gen x$ generated by $x$ is of infinite order.

- $\order x = \infty \iff \order {\gen x} = \infty$

## Proof

### $(1)$ implies $(2)$

Let $x$ be an infinite order element of $G$ by definition 1.

Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$.

Aiming for a contradiction, suppose not all $x^r$ are distinct for all $r \in \Z_{>0}$.

Then $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$.

Without loss of generality, let $m > n$.

Then:

- $x^{m - n} = e^G$

As $m > n$ we have that $m - n \in \Z_{>0}$.

Let $k = m - n$.

and so:

- $\exists k \in \Z_{>0}$ such that $x^k = e_G$.

But this contradicts the statement that no such $k$ exists.

Hence no such distinct $m$ and $n$ exist.

Thus $x$ is an infinite order element of $G$ by definition 2.

$\Box$

### $(2)$ implies $(1)$

Let $x$ be an infinite order element of $G$ by definition 2.

Then by definition the powers $x^r$ of $x$ are distinct for all $r \in \Z_{>0}$.

That is, there exist no $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$.

Aiming for a contradiction, suppose there exists $k \in \Z_{>0}$ such that $x^k = e^G$.

Consider some $n \in \Z_{>0}$ such that $m = n + k$.

\(\displaystyle x^m\) | \(=\) | \(\displaystyle x^{n + k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^n x^k\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^n e^G\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^n\) |

But this contradicts the statement that no such $m, n \in \Z_{>0}$ exist.

Thus there can be no $k \in \Z_{>0}$ such that $x^k = e_G$.

Thus $x$ is an infinite order element of $G$ by definition 1.

$\Box$

### $(2)$ implies $(3)$

Let $x$ be an infinite order element of $G$ by definition 2.

Then by definition the powers $x^r$ of $x$ are distinct for all $r \in \Z_{>0}$.

Aiming for a contradiction, suppose not all $\gen x = r$ for some $r \in \Z_{>0}$.

Let $s > r + 1$.

Then by the Pigeonhole Principle at least two of the elements of $\gen x$ are such that $x^m = x^n$ for some $m, n \in \closedint 0 r$

But this contradicts the statement that all $x^k$ are different for all $k \in \Z_{>0}$.

Thus $x$ is an infinite order element of $G$ by definition 3.

$\Box$

### $(3)$ implies $(1)$

Let $x$ be an infinite order element of $G$ by definition 3.

Then by definition the group $\gen x$ generated by $x$ is of infinite order:

- $\order {\gen x} = \infty$

Aiming for a contradiction, suppose there exists $k \in \Z_{>0}$ such that $x^k = e^G$.

Then by Order of Cyclic Group equals Order of Generator, $\gen x$ has order $k$

But this contradicts the statement that $\order {\gen x} = \infty$.

Thus there can be no $k \in \Z_{>0}$ such that $x^k = e_G$.

Thus $x$ is an infinite order element of $G$ by definition 1.

$\blacksquare$