Prove that $n\mathbb{Z}$ for $n\in\mathbb{Z}^+$ are the only subgroups of $\mathbb{Z}$ I want to prove that $n\mathbb{Z}$ for $n\in\mathbb{Z}^+$ are the only subgroups of $\mathbb{Z}$, under the binary operation of addition, where, $n\mathbb{Z}$ is defined as $\{nk:k\in\mathbb{Z}\}$. 
Naturally, I have to prove that $n\mathbb{Z}$ is a subgroup first.
We hence have to prove that the subgroups are closed under addition, and preserve identity and invertibility. We have:


*

*For all $n\in\mathbb{Z}^+$, we have that $n\cdot 0=0$, and so the additive identity is preserved.

*For $a,b\in n\mathbb{Z}$, we have that $a=nk$, and $b=n\ell$, for $k,\ell\in\mathbb{Z}$. Then, $a+b=n(k+\ell)$, but because $k,\ell\in\mathbb{Z}$, $k+\ell\in\mathbb{Z}$, such that $n(k+\ell)\in n\mathbb{Z}$.

*For $a\in n\mathbb{Z}$, we have that $a=nk$, with $k\in\mathbb{Z}$. For $k\in\mathbb{Z}$ however, there exists a $k^{-1}\in\mathbb{Z}$, such that there exists a $nk^{-1}\in n\mathbb{Z}$, such that $(nk)(nk^{-1})=n^2kk^{-1}=0$. Here, $k^{-1}=-k$, such that $nk^{-1}=-nk$.


I am having some trouble proving that these are the only allowed subgroups of $\mathbb{Z}$. A hint or nudge in the right direction would be appreciated. I'm thinking a proof by contradiction would be the best way to go, but I'm not able to formulate a good beginning.
 A: Given a subgroup $G$ of $\mathbb{Z}$, consider the set
$$
G_+ := \{x \in G \mid x>0\}.
$$
Can $G_+$ be empty? If so, what can you say about $G$ in case it is empty. If $G_+$ is nonempty, consider it's smallest element $n$. What can you say about $G$ in terms of $n$?
Also, in light of yanko's comments, have a second look at your proof in part 3. You should show that if $a\in n\mathbb{Z}$, then so is $-a$.
A: Naturally, you have to prove that $\Bbb Z$ is cyclic first. Because subgroups of a cyclic group are cyclic, now take an element $n$ of $\Bbb Z$, and generate a cyclic subgroup with it. You get $n\Bbb Z$. If $n<0$, obviously, $n\Bbb Z=(-n)\Bbb Z$
A: I wouldn't call this a proof per say, but it was my approach in ultimately solving the problem.
Suppose there exists some subgroup $H$ not of the form $n\mathbb{Z}$. We first note that it's impossible for this group to trivially be the identity element, as $\{0\}=0\mathbb{Z}$. Suppose instead we have some element $a$ in addition to the identity element. Then, $-a$ must also be in this subgroup. Then, we have that $H=\{-a,0,a\}$. But as $H$ is a subgroup, we must have that $H$ exhibits closure. Then, we must have that $na\in H$, for all $n\in\mathbb{Z}$. However, here, $a$ just takes on the role of $n$. That is, $H=a\mathbb{Z}$.
Essentially, it is impossible to construct a subgroup of $\mathbb{Z}$, which is not of the form, $n\mathbb{Z}$.
