Prove that any subgroup of $\mathbb{Z}$ is cyclic Could someone please help me answer this question? Using the subgroup test it was proved that a subgroup $H$ $\le $ $\mathbb{Z}$ is necessarily of the form ${n}$ $\bullet$ $\mathbb{Z}$ for some n $\ge$ $0$.
 A: Assume $a$ is the smallest non-zero element of the subgroup A(we can always choose the smallest one(why?)).
Then, consider any element $x\in A$,then $\exists n,r\in\Bbb Z$ such that $x=na+r$ where $0\leq r\lt a$(division algo.)
But then, $r=x-na\in A$ and $r\lt a$ which contradicts that $a$ is the smallest non-zero element unless $r=0$ itself which gives $x=na$ where $n\in\Bbb Z,x\in A$ 
A: $\Bbb Z$ is cyclic since $\Bbb Z=\langle 1\rangle$, and any subgroup of a cyclic group is cyclic.
A: HINT: Let $G$ be a subgroup of $\Bbb Z$. Let $a$ be the smallest positive element of $G$. Prove that $G=\{na:n\in\Bbb Z\}=\langle a\rangle$, the subgroup of $\Bbb Z$ generated by $a$. You may find the division algorithm useful.
A: Hint $\ $ From Euclid, a nonempty subset $\rm\,S\subseteq \Bbb Z$ closed under subtraction is closed under mod, so closed under gcd. Thus the least positive  $\rm\,d\in S\,$ divides every $\rm\,n\in S\:$ (else $\rm\:gcd(d,n) < d).$
Remark $\ $ One can skip the intermediate "closed under mod" if one is familiar with the "subtractive" form of the Euclidean algorithm, which calculates the gcd by repeatedly subtracting the smaller integer from the larger. Then it is crystal clear that subgroups, being closed under subtraction, are also closed under the gcd operation.
A: $\quad$$\mathbb{G}$ = { $\mathbb {na}$ : $\mathbb{n}$ $\in$ $\mathbb{Z}$} = $\lt$ $\mathbb{a}$$\gt$
Let $g_1$ $\in$ $\mathbb{G}$
Now want to show that $g_1$ is in { $\mathbb {na}$ : $\mathbb{n}$ $\in$ $\mathbb{Z}$}
$\qquad$  $g_1$ = $\mathbb{k\cdot a}$ + $\mathbb{r}$ $\qquad$  $\mathbb{a}$ $\gt$ $\mathbb{r}$ $\ge$ 0
$\qquad$$\qquad$$\mathbb{r}$ =  $g_1$ -  $\mathbb{k\cdot a}$ $\qquad$( $\mathbb{r}$ $\in$ $\mathbb{G}$ )
therefore it was found that $\quad$$\mathbb{r}$ = 0 
$\qquad$$\qquad$   $g_1$ =  $\mathbb{k\cdot a}$
hence this completes the proof
