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.


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