Subgroups of $\mathbb{Z}$ I am trying to understand subgroups.I know a given a group G under a binary operation ∗, a subset H of G is called a subgroup of G if H also forms a group under the operation ∗. 
So if I where to list 5 subgroups of $\mathbb{Z}$ could I simply say, $\mathbb{Z}_2$, $\mathbb{Z}_4$, $\mathbb{Z}_6$, $\mathbb{Z}_7$, $\mathbb{Z}_9$, are all subgroups of $\mathbb{Z}$?
 A: "H of G is called a subgroup of G if H also forms a group under the operation ∗"
!!AND!!! to be a SUBgroup (and not just a group with a different operation on a subset of $H$) we also need that if $a\in H, b\in H$ then $ab \in H$ is equal to $ab \in G$.
So $\mathbb Z_2 = \{0,1|+:0+0=0;0+1=1+0=1;1+1=0\}$ is NOT a subgroup of $\mathbb Z$ because in $\mathbb Z$, we must have $1+1 = 2 \ne 0$.
And $\mathbb Z_9$ is not a subgroup because in $5+6=11 \in \mathbb Z$ but $5+6 = 2 \in \mathbb Z_9$.
Instead though $2\mathbb Z = \{2*a|a \in \mathbb Z|+\}$ (i.e. the even integers) is a subgroup.  
And $k\mathbb Z = \{k*a|a \in \mathbb Z|+\}$ (the multiples of $k$) is a subgroup.
A: The subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ with $n\in\mathbb{N}$.
Proof:
Let $H\neq\{0\}$ (else simply $n=0$) and let $n=\min\{m|m\in H, m>0\}$.
Suppose $H\neq n\mathbb{Z}$, then exists $k\in H\setminus n\mathbb{Z}$.
Then $k=q\cdot n+r$ with $q,r\in\mathbb{Z}$ and $0<r<n$. This is division with remainder.
Hence $0<r=k-qn\in H$. 
But $r<n$ in contradiction to the choice of $n$.
Therefore $H\setminus n\mathbb{Z}=\emptyset$
