Equivalence of Definitions of Finite Order Element

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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.

$\blacksquare$

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