prove that in any group, an element and its inverse have the same order.
let $\displaystyle a$ be in $\displaystyle G$ and let $\displaystyle n\in \mathbb{Z}^+$ be the order of $\displaystyle a$, i.e. $\displaystyle n$ be the smallest positive integer such that $\displaystyle a^n=e$
then $\displaystyle (a^{-1})^n = e$.
suppose there exists a positive integer $\displaystyle m <n$ such that $\displaystyle (a^{-1})^m= e$. then $\displaystyle \underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{m \mbox{ copies}} = e $ implies that $\displaystyle \underbrace{aa\cdots a}_{m \mbox{ copies}} = a^m = e$ which is a contradiction.
hence, order of $\displaystyle a^{-1}$ is also $\displaystyle n$.