Proof: each subgroup of $\mathbb Z/n\mathbb Z$ is of the form $\{\overline d,\overline{2d},\dots,\overline{n-d},\overline n\}$, where $d$ divides $n$

Question

Show that each subgroup of $\mathbb Z/n\mathbb Z$ is of the form $\{\overline d,\overline{2d},\dots,\overline{n-d},\overline n\}$, where $d$ is a divisor of $n$.

I’m assuming they’re talking about non-trivial subgroups here? Anyhow, I'll line out the proof my book provides:

Let $K=\{a\in\mathbb Z\mid\overline a\in H\}$. They show this is a subgroup, so we have $K=d\mathbb Z$ for some $d\in\mathbb N$. However, they argue that since $n\in K=d\mathbb Z$, it follows that $d\mid n$. That makes sense to me, but I fail to see how this holds in an example.

Say we have $\mathbb Z/7\mathbb Z$ and we take $n=2$. Then our subgroup is simply the entire group, but $2$ doesn’t divide $7$. So I'm missing something, but I don't see what. Could someone help?

Answer 1

Yes $2$ doesn't divide $7$. But as you said the subgroup you get is the entire group, which can be written in the form $\{\bar{1},\bar{2},\dots,\bar{6}\}$ with $1\mid 7$, so that's not a contradiction.

Answer 2

If you take $G = \mathbb Z/7\mathbb Z$, your "theorem" says that the subgroups are of the form $\mathbb Z/d\mathbb Z$, where $d$ divides $7$. That means the subgroups are $\{0\}$ and $G$.

Answer 3

Remark: Every subgroup of a cyclic group is again cyclic.

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Now suppose that $H \leq \mathbb{Z}/ n\mathbb{Z}$ be any aritrary subgroup;

let's denote it's genereator by $t$; now let $d:=\gcd(t,n)$; then we have:

$$H = <t> = <d> .$$

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In your specefic case, we have $d=\gcd(2,7)=1$ :

$$ <2> = <1> .$$