= Solution
The order of an element $x$ is the least positive integer $d$ such that $x^d=e$, and the order of a finite group is its number of elements.
<Lagrange theorem> states that if $H\leq G$ and $G$ is finite, then
$$
|H|\mid |G|,
\qquad
|G|=[G:H]|H|.
$$
Indeed, the left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection from $H$ to each coset, so every coset has $|H|$ elements. Summing over the $[G:H]$ cosets proves the formula.
The cyclic subgroup $\langle x\rangle$ has exactly $d$ elements when $x$ has order $d$. Applying Lagrange to $\langle x\rangle\leq G$ gives
$$
\boxed{\operatorname{ord}(x)\mid |G|}.
$$
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