I was wondering if the following proof looks okay.
Proof. Let $g$ be any element of a group $G$. Then by definition, the order of $g$ is the least positive integer $n$ such that $g^{n}=e$. We define the order of the element $g$ as $|g|=n$. Now, we wish to show that in any group, an element and its inverse have the same order.
For the inverse of the group element g,
$(g^{-1})^{n}=g^{-n}=(g^{n})^{-1}=e^{-1}=e$.
Since the order of $g^{-1}$ is the least positive integer $m$ such that $(g^{-1})^{m}=e$, it follows that $|g^{-1}|\le |g|=n$.
Likewise, $((g^{-1})^{-1})^{m})=(g^{-1})^{-m}=(g^{-m})^{-1}=((g^{m})^{-1})^{-1}=(e^{-1})^{-1}=e$
Then since $(g^{-1})^{-1}=g$, $|g|\le |g^{-1}|.$
Therefore, the order of $g$ is equal to the order of its inverse. QED
Best Answer
It's okay for me.
Maybe it could be shortened observing it is enough to prove that $$ \forall n,\quad g^n=1\implies \bigl(g^{-1}\bigr){}^n=1. $$