Order of Group Element equals Order of Inverse

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

1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(ii)}$
1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Exercise $5.6$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $10$