 Solution

The order of a group element $o(g)$ is the least positive integer $n$ for which $g^n=e$, whose existence follows from part (i). If $g^k=e$, use the Euclidean division to write $k=qn+r$, where $0\le r<n$. ThenMinimality of $n$ forces $r=0$. ThusThe argument also works for negative $k$, using integer powers and the same remainder convention.