Order of Group Element equals Order of Inverse

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

Then:
 * $\forall x \in G: \left|{x}\right| = \left|{x^{-1}}\right|$

where $\left|{x}\right|$ denotes the order of $x$.

Proof
By Index Laws for Monoids: Negative Index, $\left({x^k}\right)^{-1} = x^{-k} = \left({x^{-1}}\right)^k$.


 * Suppose $x^k = e$. Then $\left({x^{-1}}\right)^k = e$.

So $\left|{x^{-1}}\right| \le \left|{x}\right|$.

Similarly, suppose $\left({x^{-1}}\right)^k = e$.

Then $x^{-k} = e$, and so $\left({x^{-k}}\right)^{-1} = x^k = e$.

So $\left|{x}\right| \le \left|{x^{-1}}\right|$.

Thus $\left|{x}\right| = \left|{x^{-1}}\right|$.


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

Hence the result.