Subgroups of cyclic groups

Proof. Let $G\subseteq \mathbb{Z}$ be a subgroup. If $G=\{0\}$, then $G=0\mathbb{Z}$. Otherwise, let $b\in G$ be a nonzero element of $G$. With $b\in G$, we alse have $-b\in G$, and so we can say that $G$ contains positive integers. With positive integers in $G$, we can ask for the smallest positive integer in $G$, thanks to the well-ordering principle of $\mathbb{Z}$. Let $d$ be the smallest positive integer in $G$. We will show that $G=d\mathbb{Z}$. First, we have $d\mathbb{Z}\subseteq G$, as $d$ is in $G$ and $d\mathbb{Z}$ is the smallest subgroup of $\mathbb{Z}$ containing $d$. Second, we have $G\subseteq d\mathbb{Z}$. To see this, let $a\in G$. By the division algorithm, we can write $a=qd+r$ for some integers $q$ and $r$ with $0\le r<d$. Since $a$ and $qd$ are in $G$, we have $r=a-qd\in G$. Since $r$ is a positive integer less than $d$, we must have $r=0$, as $d$ is the smallest positive integer in $G$. Thus, $a=qd\in d\mathbb{Z}$. Therefore, $G\subseteq d\mathbb{Z}$, and we have $G=d\mathbb{Z}$.

Proof. Let $G\subseteq \mathbb{Z}/n\mathbb{Z}$ be a subgroup. Let ${\pi }_n:\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ be the quotient map. Then $G^{\prime }={\pi }_n^{-1}(G)$ is a subgroup of $\mathbb{Z}$, and so it is of the form $d\mathbb{Z}$ for some $d\ge 0$. In other words, $G^{\prime }$ is generated by $d$, denoted as $G^{\prime }=⟨d⟩$. Going forward through the quotient map, we have $$G={\pi }_n(G^{\prime })={\pi }_n(⟨d⟩)=⟨[d{]}_n⟩.$$ Thus, $G$ is generated by an element $[d{]}_n$. In addition to this result, notice that $n\in G^{\prime }$, as ${\pi }_n(n)=[0{]}_n\in G$, which means that the generator $d$ of $G^{\prime }$ divides $n$.