Finding all subgroups of $\mathbb{Z_n}$

Question

I can't use any theorems regarding cyclic groups or isomorphisms. A hint the exercise provides is: "If $f: G \to H$ is a surjective homomorphism and $N$ is a subgroup of $(G, *)$, then $f(N)$ is a subgroup of $(H, •)$. Consider the surjective homomorphism $f: \mathbb{Z} \to \mathbb{Z_n}$". but I'm having a hard time using that. I'd appreciate any help.

Update: I'm almost certain that the following proposition solves my problem:

Let $K$ be a subgroup of $\mathbb{Z}_n$. Then $K = \{\bar{0}, \overline{m}, \overline{2m}, \cdots,\overline{n-m} \}$, where $m \in \mathbb{N}$ such that $m \mid n$.

I just have to prove it now, which is the current problem.

Update 2: I think I've managed to prove that the only subgroups of $\mathbb{Z}_n$ are those in the form of $x\mathbb{Z}_n = \{\overline{wx} \ \vert \ \overline{w} \in \mathbb{Z}_n \} = \{\overline{0}, \overline{x}, \overline{2x}, \cdots, \overline{(n-1)x}\}$ , where $x \in \mathbb{N} $ is such t hat $x \mid n$. For this, I fist prove that any $x\mathbb{Z}_n$ with $x \mid n$ is a subgroup of $\mathbb{Z}_n$ and after that I prove there are no subgroups not in that form. First, notice that $n = qx$ for some $q \in \mathbb{N}$. Trivially, $x\mathbb{Z}_n \subseteq \mathbb{Z}_n$ and $\overline{0} \in x\mathbb{Z}_n$ is its neutral element. Now I just need to take two elements $a, b$ in $x\mathbb{Z}_n$ and prove that $ab' \in x\mathbb{Z}_n$. For this, let $a = \overline{\alpha x}$ and $b = \overline{\beta x}$. Then $a - b = \overline{\alpha x} + \overline{n - \beta x} = \overline{(\alpha + q - \beta)x} \in x\mathbb{Z}_n $, and so $x\mathbb{Z}_n$ is a subgroup of $\mathbb{Z_n}$.

Now, let $K$ be a subgroup of $\mathbb{Z}_n$. Let $k$ be the smallest positive integer such that $\overline{k} \in K$. By long division, we have that $n = pk + r$ for some $0\leq r < k$. Then $r = n - pk$, and since $K$ is a subgroup, it's trivial that $\overline{r} \in K$. Now, since $0 \leq r < k$ and $k$ is the smallest positive integer such that $\overline{k} \in K$, we have $r = 0$. Then $n =pk $ and $k \mid n$. Trivially, $k\mathbb{Z}_n \subseteq K$. Now, let $y = \overline{m} \in K$. By the exact same argument, $k\mid m$ and therefore $y = \overline{tk} \in k \mathbb{Z}_n$ for some $t \in \mathbb{N}$ and it follows that $K=k\mathbb{Z}_n$. Then we're done.

Answer 1

Hint: Let $k \in \mathbb{Z}$. Then $N = \langle k \rangle \leq \mathbb{Z}$ and $f(N) \leq \mathbb{Z}/n\mathbb{Z}$.

(Possibly helpful:

• We're using $(\mathbb{Z}, +)$ and $(\mathbb{Z}/n\mathbb{Z},+)$.
• $\langle k \rangle = \langle -k \rangle$.

)

Things to show/work out:

• Are there any subgroups of $\mathbb{Z}$ that are not $\langle k \rangle$ for some $k$? (Any such subgroup is a subset which, since $\mathbb{Z}$ is well ordered, has a least positive element...)
• Does the fourth isomorphism/lattice isomorphism/correspondence theorem tell you that you have found all the subgroups of $\mathbb{Z}/n\mathbb{Z}$ by this method?
• What relationship between $k$ and $n$ forces $f(N)$ to be all of $\mathbb{Z}/n\mathbb{Z}$? Then what happens when that relationship does not hold?

Answer 2

A better hint is: the subgroups of $H$ are the images of the subgroups of $G$ that contain $\ker f$.

In your case, this leads to: Which subgroups of $\mathbb Z$ contain $n \mathbb Z$?