= Solution
<Lagrange's theorem> states that for a <subgroup> $H$ of a finite <group> $G$, $|G|=[G:H]|H|$; in particular $|H|$ divides $|G|$. This follows by partitioning $G$ into <cosets>, each in <bijection> with $H$.
If $|G|=p$ is prime, take any $g\ne e$. Its <cyclic subgroup> has order dividing $p$, but its order is not one. Thus it has order $p$ and equals $G$. Hence
$$
\boxed{\text{Every group of order }p\text{ is cyclic and isomorphic to }C_p.}
$$
This describes a single <group isomorphism> type for each prime.
Now let $G=\langle x\rangle$ have order $n$, and let $H$ be any <subgroup>. Choose the least positive integer $d$ with $x^d\in H$; this exists because $x^n=e\in H$, even when $H$ is trivial. Divide $n=qd+r$ with $0\le r<d$. Since $x^r=x^n(x^d)^{-q}\in H$, minimality forces $r=0$, so $d$ divides $n$. Likewise, dividing an exponent $k=qd+r$ for any $x^k\in H$ shows $r=0$. Therefore $H=\langle x^d\rangle$.
Conversely, for each divisor $d$ of $n$, $\langle x^d\rangle$ is a <subgroup> of order $n/d$. The least positive exponent property makes $d$ unique. This proves the complete <subgroups and quotients of a cyclic group> classification:
$$
\boxed{H=\langle x^d\rangle\ (d\mid n),\quad |H|=n/d;
\qquad H_m=\langle x^{n/m}\rangle\text{ is the unique subgroup of order }m\ (m\mid n).}
$$
The choices $d=n$ and $d=1$ include the trivial <subgroup> and the whole <group>.
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