Prove that all the subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$.

Question

Describe all subgroups of the group $\mathbb{Z}$

The subgroups are $\{ 0 \}$, $\mathbb{Z}$ and $n \mathbb{Z} = \{ n k : k \in \mathbb{Z} \}$ How to prove it?

Answer 1

It is easy to show that $n\mathbb{Z} < \mathbb{Z}$. To show these are all, let $H$ be any subgroup of $\mathbb{Z}$, and then we can pick a least element in $H$, as $\mathbb{Z_{\ge 0}}$ is well ordered. Definitely $H$ will always have positive elements too, as it is a subgroup and thus all inverses will be in it (if $k$ is in there , $-k$ will also be). Suppose $n$ is the least positive integer in $H$ then we claim $H= n\mathbb{Z}$.

Clearly $n\mathbb{Z} \subseteq H$

Now, for reverse containment, let $k \in H$ be any element then use division algorithm, i.e. $k=nq+r$ , where $0 \leq r < n$, then $k-nq =r \in H$, which implies $r=0$, so $k=nq \in \mathbb{Z}$, hence $H \subseteq n\mathbb{Z}$, and so $H= n\mathbb{Z}$

Hence proved.

Answer 2

Clearly, $\{0\}$ is subgroup of $Z$. Let $H$ be a subgroup of $Z$ other than $\{0\}$, then there exist a non-zero integer $a ∈ H$. Since, $H$ is subgroup, $−a ∈ H$. Hence, $H$ contains atleast one positive integer. By Well ordering principle, there exist a smallest positive integer $k$ in $H$.

Claim: $H = kZ$

Since $k ∈ H$ and $H$ is a subgroup implies all integral multiples of $k ∈ H$ which in turn implies that $kZ ⊆ H$. Now, we show that $H −kZ = φ$ which would imply that $H = kZ$. If possible, assume $H − kZ ̸\neq φ$ then there exist $x ∈ H −kZ$, then $x$ is not a multiple of $k$. By division algorithm, there exist integer $q$,$r$ such that $x = kq+r$; $\space0 < r ≤ k−1$, which in turn implies that $x−kq = r ∈ H$, therefore $H$ contains a positive integer $< k$, which is a contradiction as $k$ is the least positive integer in $H$. Hence, $H −kZ = φ \implies H = kZ$. Therefore, subgroups of $Z$ are $\{0\}$ and $nZ$.