# Subgroup of multiples of $m$, $m$-torsion and $p$-subgroup of $\mathbb{Z}_n$

Disclaimer: this terminology might be different from what you're used to and this is why I'm writing down some definitions first.

Let $$A$$ be an abelian group, $$m \in \mathbb{N}^* := \mathbb{N} \setminus \{0\}$$ and let $$p$$ be a prime number. We define:

1. The subgroup of multiples of $$m$$ in $$A$$ as the set $$mA := \{\,ma \, \mid \, a \in A\,\}$$.
2. The $$m$$-torsion of $$A$$ as the set $$A[m] := \{\,a \in A \, \mid \, ma = 0\,\}$$.
3. The $$p$$-subgroup of $$A$$ as the set $$A(p) := \{\,a \in A \, \mid \, p^k a = 0 \hspace{1mm} \text{for some} \hspace{1mm} k \in \mathbb{N}\,\}$$.

I've proven these are all subgroups of $$A$$. It is known that $$A(p) = \{\,a \in A \, \mid \, o(a) = p^k \hspace{1mm} \text{for some} \hspace{1mm} k \in \mathbb{N}\,\}$$. Now we want to prove the following statement considering $$A := \mathbb{Z}_n$$. Square brackets (or a bar if it's a single letter) denote the equivalence class of an element in $$\mathbb{Z}_n$$, whereas the angle brackets denote the subgroup genereated by the element within them and finally the $$\simeq$$ symbol denotes an isomorphism.

1. $$m\mathbb{Z}_n = \gcd(n,m)\mathbb{Z}_n = \langle [\gcd(n,m)] \rangle \simeq \mathbb{Z}_\frac{n}{\gcd(n,m)}$$
2. $$\mathbb{Z}_n[m] = \mathbb{Z}_n[\gcd(n,m)] = \langle [\frac{n}{\gcd(n,m)}] \rangle \simeq \mathbb{Z}_{\gcd(n,m)}$$
3. If $$n = p^r n'$$ and $$p \nmid n'$$ (not divides), then $$\mathbb{Z}_n(p) = \langle n' \rangle \simeq \mathbb{Z}_{p^r}$$

I've tried to solve it in this way. First of all, we have that $$\gcd(n,m) \mid m$$, so there exists $$k \in \mathbb{Z}$$ such that $$m = \gcd(n,m)k$$. Given an element $$\bar{x} \in m\mathbb{Z}_n$$, by definition there exists $$\bar{a} \in \mathbb{Z}_n$$ such that $$\bar{x} = m \bar{a}$$ and therefore $$\bar{x} = \gcd(n,m)k\bar{a} = \gcd(n,m)[ka]$$. This proves that $$\bar{x} \in \gcd(n,m)\mathbb{Z}_n$$, thus $$m\mathbb{Z}_n \subseteq \gcd(n,m)\mathbb{Z}_n$$. I was not able to prove the inclusion $$\supseteq$$ though.

Then I guess I solved the equality $$\gcd(n,m) = \langle [\gcd(n,m)] \rangle$$. In fact (tell me if I'm mistaken somewhere): \begin{align} \gcd(n,m)\mathbb{Z}_n & = \{\,\gcd(n,m)\bar{a} \, \mid \, \bar{a} \in \mathbb{Z}_n\,\} \\ & = \{\,[\gcd(n,m)a] \, \mid \, a \in \mathbb{Z}\,\} \\ & = \{\,a[\gcd(n,m)] \, \mid \, a \in \mathbb{Z}\,\} = \langle [\gcd(n,m)] \rangle \end{align}

For the isomorphism part, let $$f \colon \langle [\gcd(n,m)] \rangle \to \mathbb{Z}_{\frac{n}{\gcd(n,m)}}$$ be the map defined by $$f(k[\gcd(n,m)]) := \bar{k}$$. This is trivially a surjective homomorphism. It is injective, too, since: \begin{align} \ker{f} & = \{\,k[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, f(k[\gcd(n,m)]) = \bar{0}\,\} \\ & = \{\,k[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, \bar{k} = \bar{0}\,\} \\ & = \{\,nh[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, h \in \mathbb{Z}\,\} \\ & = \{\,h\gcd(n,m)\bar{n} \in \langle [\gcd(n,m)] \rangle \, \mid \, h \in \mathbb{Z}\,\} = \{\bar{0}\} \end{align}

Am I on the right track or is there a better way to prove this stuff? And if this is the right way, how can I complete this proof including statement 2. and 3.?

Edit: I found a better way to prove all isomorphisms. I use this result (the order of an element $$a \in G$$ is denoted by $$o(a)$$):

Let $$G$$ be a cyclic group, $$G = \langle a \rangle$$. If $$o(a) = n$$, then $$G \simeq \mathbb{Z}_n$$.

Now one easily verifies the following conditions:

1. $$o([\gcd(n,m)]) = \frac{n}{\gcd(n,m)}$$
2. $$o([\frac{n}{\gcd(n,m)}]) = \gcd(n,m)$$
3. $$o([n']) = p^r$$

So I only need to prove the equalities now if my reasoning is correct.