Subgroup of integers I know that the only possible subgroups of $\mathbb Z$ are of the form $m\mathbb Z$. But how can I prove that these are the only possible subgroups?
 A: Let $H$ be a nontrivial subgroup of $\mathbb Z$. Then $H$ has a smallest positive element; call it $n$. Now let $k$ be an arbitrary element of $H$. By the division algorithm, we can write $k = nq + r$ where $q,r \in \mathbb Z$ and $0 \leq r < n$. Since $k \in H$ and $n \in H$, we also have $r = k - nq \in H$. Since $0 \leq r < n$ and $n$ is the smallest positive element of $H$, we must have $r = 0$. Therefore $k = nq$. We have shown that an arbitrary element of $H$ is of the form $nq$ where $q \in \mathbb Z$; hence $H = n\mathbb Z$.
A: Prove the following two intermediate results:


*

*If a subgroup of $\Bbb Z$ contains some number $m$, then it contains $m\Bbb Z$

*If a subgroup of $\Bbb Z$ contains two numbers $m,n$, then it contains $\gcd(m,n)$


Now let $k$ be the smallest positive integer in your subgroup. If there is an element $a\notin k\Bbb Z$ in the subgroup, reach a contradiction.
A: Suppose $H$ is a subgroup and $m$ its least positive element. Note that if $H$ contains some $k$ other than a multiple of $m$, then the residue of $k$ modulo $m$ is a positive element of $H$, smaller than $m$. 
