Solution
= Solution
{c}
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 $0\leq r<n$. Then
$$
e=g^k=(g^n)^qg^r=g^r.
$$
Minimality of $n$ forces $r=0$. Thus
$$
\boxed{o(g)\mid k.}
$$
The argument also works for negative $k$, using integer powers and the same remainder convention.