 An element and its inverse have the same order

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.