Proposition 17

Let $\mathrm{H}$ be a subgroup of $\mathbf{Z}$. There exists one and only one integer $a \geqslant 0$ such that $\mathrm{H} = a\mathbf{Z}$.

If $\mathrm{H} = \{0\}$, then $\mathrm{H} = 0\mathbf{Z}$. Suppose that $\mathrm{H} \neq \{0\}$. The subgroup $\mathrm{H}$ has an element $x \neq 0$. Then $x > 0$ or $-x > 0$, and therefore $\mathrm{H}$ has elements $> 0$. Let $a$ be the smallest element $> 0$ in $\mathrm{H}$. The subgroup $a\mathbf{Z}$ generated by $a$ is contained in $\mathrm{H}$; we show that $\mathrm{H} \subset a\mathbf{Z}$. Let $y \in \mathrm{H}$. By Proposition 16, there exists an integer $r$ such that $y \equiv r \pmod{a}$ and $0 \leqslant r < a$. A fortiori $y \equiv r \pmod{\mathrm{H}}$, whence $r \in \mathrm{H}$. But this is only possible if $r = 0$ and therefore $y \in a\mathbf{Z}$. The integer $a$ is unique: if $\mathrm{H} = \{0\}$, then necessarily $a = 0$, and if $\mathrm{H} \neq \{0\}$, the integer $a$ is the order of $\mathbf{Z}/\mathrm{H}$.