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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.

A group homomorphism preserves multiplication, $\varphi (ab)=\varphi (a)\varphi (b)$. A group isomorphism is a bijective homomorphism. An isomorphism from a group to itself is a group automorphism.

The automorphism group contains the identity map. Composition is associative, and if $\varphi ,\psi$ are automorphisms, $(\varphi \circ \psi )(ab)=\varphi (\psi (a)\psi (b))=(\varphi \circ \psi )(a)(\varphi \circ \psi )(b)$; composition is also bijective. For $u=\varphi (a)$ and $v=\varphi (b)$, ${\varphi }^{-1}(uv)={\varphi }^{-1}(\varphi (ab))=ab={\varphi }^{-1}(u){\varphi }^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.

A finite cyclic group is $G=⟨g⟩=\{e,g,\dots ,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the roots of unity, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.

For prime $n$, Lagrange's theorem leaves only subgroup orders one and $n$: there is no nontrivial proper subgroup. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of groups with no proper nontrivial subgroups makes the distinction explicit.

For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},H_1=⟨g^p⟩}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The subgroups and quotients of a cyclic group are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.

For the intended finite nontrivial case, let $G=⟨g⟩$ have composite order $N$. Choose a prime divisor $p$ of $N$, with $1<p<N$. Thenis a nontrivial proper cyclic subgroup of order $p$. Indeed $(g^{N/p}{)}^p=e$, and no smaller positive power is the identity because $g$ has order $N$. Thus $1<|H|=p<N$.

If the source also allows an infinite cyclic group, $⟨g^2⟩$ is a proper cyclic subgroup: its powers have even exponents and do not include $g$. The trivial group, of order one, is an exception to the literal formulation “not of prime order”: it has no proper subgroup at all. The nontriviality assumption is therefore necessary for that general wording. These facts are instances of subgroups and quotients of a cyclic group.