proof that all subgroups of a cyclic group are cyclic

The following is a proof that all subgroups of a cyclic group are cyclic.

Proof.

Let $G$ be a cyclic group and $H\le G$. If $G$ is trivial, then $H=G$, and $H$ is cyclic. If $H$ is the trivial subgroup, then $H=\{e_G\}=⟨e_G⟩$, and $H$ is cyclic. Thus, for the of the proof, it will be assumed that both $G$ and $H$ are nontrivial.

Let $g$ be a generator of $G$. Let $n$ be the smallest positive integer such that $g^n\in H$.

Claim: $H=⟨g^n⟩$

Let $a\in ⟨g^n⟩$. Then there exists $z\in \mathbb{Z}$ with $a={(g^n)}^z$. Since $g^n\in H$, we have that ${(g^n)}^z\in H$. Thus, $a\in H$. Hence, $⟨g^n⟩\subseteq H$.

Let $h\in H$. Then $h\in G$. Let $x\in \mathbb{Z}$ with $h=g^x$. By the division algorithm, there exist $q,r\in \mathbb{Z}$ with such that $x=qn+r$. Thus, $h=g^x=g^{qn+r}=g^{qn}{g}^r={(g^n)}^q{g}^r$. Therefore, $g^r=h{(g^n)}^{-q}$. Recall that $h,g^n\in H$. Hence, $g^r\in H$. By choice of $n$, $r$ cannot be positive. Thus, $r=0$. Therefore, $h={(g^n)}^q{g}^0={(g^n)}^q{e}_G={(g^n)}^q\in ⟨g^n⟩$. Hence, $H\subseteq ⟨g^n⟩$.

This proves the claim. It follows that every subgroup of $G$ is cyclic.∎