Order of Group Element equals Order of Inverse

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

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

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.