Order of Group Element equals Order of Inverse

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Theorem

Let $G$ be a group whose identity is $e$.


Then:

$\forall x \in G: \order x = \order {x^{-1} }$

where $\order x$ denotes the order of $x$.


Proof

By Powers of Group Elements: Negative Index:

$\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$


Hence:

\(\ds x^k\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds \paren {x^{-1} }^k\) \(=\) \(\ds e^{-1}\)
\(\ds \) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds \order {x^{-1} }\) \(\le\) \(\ds \order x\) Definition of Order of Group Element


Similarly:

\(\ds \paren {x^{-1} }^k\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds \paren {\paren {x^{-1} }^{-1} }^k\) \(=\) \(\ds e^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x^k\) \(=\) \(\ds e^{-1}\)
\(\ds \) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds \order x\) \(\le\) \(\ds \order {x^{-1} }\) Definition of Order of Group Element


A similar argument shows that if $x$ is of infinite order, then so must $x^{-1}$ be.

Hence the result.

$\blacksquare$


Sources

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