Every positive divisor of the order of a finite cyclic group occurs as the order of a subgroup

Given: A finite cyclic group $G=\langle g\rangle$ of order $n$, and positive integers $d,q$ satisfying $n=dq$.

The generator has $\operatorname{ord}(g)=|G|=n$, and $g^k=e$ exactly when $n\mid k$ (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).

Integer powers satisfy $(g^q)^k=g^{qk}$ (Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$, Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

The generated set $\langle g^q\rangle$ is a subgroup, and its cardinality is the order of $g^q$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups, $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).

The canonical embedding preserves products and order; $\mathbb Z$ is an ordered commutative ring and multiplication by a nonzero integer is cancellative (The naturals embed in the integers, The integers form a commutative ring, The integers form a totally ordered ring, The integers have no zero divisors; multiplicative cancellation, Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$).

If a positive natural exponent $r$ satisfies $x^r=e$, then $x$ has finite order, and its order is the least such positive exponent and is at most $r$; if there is no such exponent, its order is $\infty$ (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).