= Solution
{c}
<Lagrange's theorem> states that for a <subgroup> $H$ of a finite <group> $G$,
$$
|G|=[G:H]|H|,
$$
where $[G:H]$ is the <index of a subgroup>, the number of its <left cosets>. In particular, $|H|$ divides $|G|$.
The <cyclic subgroup> $\langle g\rangle$ consists of the distinct elements $e,g,\ldots,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the <Euclidean division>, and a repetition among them would contradict minimality of $n$. Thus $|\langle g\rangle|=o(g)$. Applying <Lagrange's theorem> gives
$$
\boxed{o(g)\mid|G|.}
$$
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