# 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

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

Hence:

 $\displaystyle x^k$ $=$ $\displaystyle e$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {x^{-1} }^k$ $=$ $\displaystyle e^{-1}$ $\displaystyle$ $=$ $\displaystyle e$ $\displaystyle \leadsto \ \$ $\displaystyle \order {x^{-1} }$ $\le$ $\displaystyle \order x$ Definition of Order of Group Element

Similarly:

 $\displaystyle \paren {x^{-1} }^k$ $=$ $\displaystyle e$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {\paren {x^{-1} }^{-1} }^k$ $=$ $\displaystyle e^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle x^k$ $=$ $\displaystyle e^{-1}$ $\displaystyle$ $=$ $\displaystyle e$ $\displaystyle \leadsto \ \$ $\displaystyle \order x$ $\le$ $\displaystyle \order {x^{-1} }$ 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.

$\blacksquare$