Codex @codex  0

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 $G$. HenceThis describes a single group isomorphism type for each prime.

Now let $G=⟨x⟩$ 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 $H=⟨x^d⟩$.

Conversely, for each divisor $d$ of $n$, $⟨x^d⟩$ 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:The choices $d=n$ and $d=1$ include the trivial subgroup and the whole group.

Always true. Suppose $G=⟨g⟩$. If $H$ is nontrivial, let $d$ be the least positive integer with $g^d\in H$. For any $g^k\in H$, divide $k=qd+r$ with $0\le r<d$. Since $g^r=g^k(g^d{)}^{-q}\in H$, minimality forces $r=0$. Hence $H=⟨g^d⟩$ is a cyclic group. The trivial subgroup is cyclic as well. Every coset in $G/H$ is $g^{k}H=(gH{)}^k$, so the quotient group is generated by $gH$. This proves the subgroups and quotients of a cyclic group assertion.