= Solution
For every positive integer $m$, the group laws give $(g^{-1})^m=(g^m)^{-1}$. Consequently
$$
g^m=e\quad\Longleftrightarrow\quad(g^{-1})^m=e.
$$
The 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,
$$
\boxed{\operatorname{ord}(g^{-1})=\operatorname{ord}(g).}
$$
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 $\langle g\rangle=\langle g^{-1}\rangle$ as <cyclic subgroups>.
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