Equivalence of Definitions of Infinite Order Element
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 $m = n + k$.
| \(\ds x^m\) | \(=\) | \(\ds x^{n + k}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x^n x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x^n e^G\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x^n\) |
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.
$\blacksquare$