= An element and its inverse have the same order
{title2=$\operatorname{ord}(g^{-1})=\operatorname{ord}(g)$}
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>.
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