 Solution

For every positive integer $m$, the group laws give $(g^{-1}{)}^m=(g^m{)}^{-1}$. ConsequentlyThe sets of positive exponents returning the two elements to the identity element are identical. If this set is nonempty, its least member is the order of a group element in both cases. If it is empty, neither element has finite order. Therefore, including the infinite-order case,For a direct divisibility argument in the finite case, if $g$ has order $n$, then $(g^{-1}{)}^n=e$, so its inverse’s order divides $n$. Reversing the roles shows $n$ divides that order, forcing equality. This is the general fact that an element and its inverse have the same order; it also reflects $⟨g⟩=⟨g^{-1}⟩$ as cyclic subgroups.