= Solution
{c}
<Lagrange's theorem> states that for a <subgroup> $K$ of a <finite group> $G$,
$$
|G|=[G:K]|K|;
$$
in particular, $|K|$ divides $|G|$. Here $[G:K]$ is the <subgroup index>, the number of left <cosets> of $K$.
The <order of a group element> $g\in G$ is the least positive integer $k$ such that $g^k=e$. It exists because $G$ is finite. The <cyclic subgroup> $\langle g\rangle$ consists of exactly $e,g,\ldots,g^{k-1}$, so has order $k$. Applying <Lagrange's theorem> to this <subgroup> gives
$$
\boxed{o(g)\mid|G|.}
$$
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