Classify all subgroups of $\Bbb Z.$ Show that all but one of these are isomorphic to $\Bbb Z$ itself. I need to Classify all subgroups of $\Bbb Z$. Show that all but one of these are isomorphic to $\Bbb Z$ itself.
I think subgroups of $\Bbb Z$ is of the form $n\Bbb Z$ for any integer $n.$
But I need to understand how.
Further I think subgroup of identity is only subgroup which is not isomorphic to $\Bbb Z$.
 A: Yes, you are right, all subgroups have this form. Let $H\leq\mathbb{Z}$ be a subgroup. If $H=\{0\}$ then $H=0\mathbb{Z}$ and we are done. Otherwise $H$ must contain non-zero elements, some of them positive. Let $n=\min\{k>0: k\in H\}$. Then it can be shown that $H=n\mathbb{Z}$. We have $n\in H$ and hence obviously $n\mathbb{Z}\leq H$, because $H$ is closed to addition and additive inverses. For the other direction let $k\in H$. We can divide $k$ by $n$ with remainder. We get an expression of the form $k=qn+r$ where $q,r\in\mathbb{Z}, 0\leq r<n$. But then note that $r=k+(-q)n\in H$. Since $n$ is by definition the smallest positive number in $H$ and $0\leq r<n$ we have to conclude that $r=0$, and hence $k\in n\mathbb{Z}$. This shows that $H\leq n\mathbb{Z}$. 
If $n\ne 0$ then $n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}$. You can define an isomorphism $\varphi:\mathbb{Z}\to n\mathbb{Z}$ by $k\to nk$. 
A: First we note that $(n\mathbb{Z},+)$ is an infinite cyclic group. And it is a subgroup of $(\mathbb{Z},+)$ so in particular it's an infinite abelian group. From this we can conclude that $(n\mathbb{Z},+)$ is isomorphic to $(\mathbb{Z},+)$, infact to get isomorphism just send $1$ in $n$.
Let $H$ be a non-trivial subgroup of $(\mathbb{Z},+)$.
Because $H$ is non-trivial: $∃m∈Z:m∈H:m≠0$.
Because $H$ is itself a group:$−m∈H$.
So either $m$ or $−m$ is positive and therefore in $\mathbb{Z}_{>0}$. So let we call, $n$, the smallest element in $H∩\mathbb{Z}_{>0}$. So let's generate $n$ and we obtain that H contains $(n\mathbb{Z},+)$.
To find a contraddiction suppose:
$∃m∈Z:m∈H∖n\mathbb{Z}$.
Then $m≠0$, and also $−m∈H∖n\mathbb{Z}$. Wlog $m>0$, otherwise we consider $−m$. By the Division Theorem:
$$m=qn+r$$
If $r=0$, then $m=qn∈n\mathbb{Z}$, so $0≤r<n$. Now this means $r=m−qn∈H$ and $0≤r<n$.
This would mean n was not the smallest element of H∩Z, and it's a contraddiction. So $H=(n\mathbb{Z},+)$.
