Proving that additive groups are isomorphic $(n\mathbb{Z}$ and $m\mathbb{Z}$) or not ($\mathbb{Q}$ and $\mathbb{Z}$) I switched my study course and am currently in the process of self-studying to catch up with the first few weeks of my linear algebra lecture which is why I'm still not really confident in how to prove things.
I'm stuck on the following two tasks from the book that I use for the lecture. I'll try to summarize as good as I can as it's not written in English. I'm to prove that that the following groups are not isomorphic:
$\left(\mathbb{Z}, +\right)$ and $\left(\mathbb{Q}, +\right)$
My idea is that the integer group is cyclic while the rationale numbers one is not. So I could probably somehow prove down the line that there's a rational number in $\mathbb{Z}$ which would be a contradiction.
I also have to show that the following groups are isomorphic:
$\left(n\mathbb{Z}, +\right)$ and $\left(m\mathbb{Z}, +\right)$
I'm at a complete loss with this one. I could probably somehow prove that $n$ and $m$ divide each other and thus are identical but I have no idea how I would go about writing that down in proper "math speak".
I think my main problem is that I can't quite wrap my head around the definitions of homomorphisms and isomorphisms which I probably should use for those tasks.
I appreciate any help.
EDIT: Thanks for the help so far. I've now tried for hours to write down something resembling a proper mathematical-looking solution and wrap my head around it all but failed miserably so far.
Cheers
 A: Hint:


*

*You are on the first track in the first problem.

*For the second problem, prove that both these groups are actually isomorphic to $(\Bbb Z,+)$. That is, both are free abelian groups generated by a single element.

A: One could do that as : 
The group $(\mathbb Z,+)$ is not divisible. i.e.; for $1\in\mathbb Z$, and $n\in\mathbb N$ there is no $y\in\mathbb Z$ such that $ny=1$. But the additive $\mathbb Q$ is divisible.
A: For the first question, the trick is to assume they are indeed isomorphic, and to then construct some contradiction. If $h$ was an isomorphism $(\mathbb{Z},+) \to (\mathbb{Q},+)$, then it has to map $1_\mathbb{Z} ($i.e. $1$ as an element of $\mathbb{Z}$) to some element $q \in \mathbb{Q}$. If $q = 0$, you get an instant contradiction if you look at $h(1_\mathbb{Z}+1_\mathbb{Z})$ and use that you may swap the application of $h$ and the addition. If $q \neq 0$, what then would the preimage of $\frac{q}{2}$ be? Whatever you choose, $h^{-1}\left(\frac{q}{2} + \frac{q}{2}\right)$ smeels like trouble...
For the second question, your idea is spot on. Just write your proposed isomorphism down explicitly as a function from $(n\mathbb{Z})$ to $(m\mathbb{Z})$, and verify all the required properties of an isomorphism, i.e. one-to-one, $h(a+b)=h(a)+h(b)$, $\ldots$.
A: An easy way to show that $\mathbb{Q}$ is not cyclic is to take any rational $q=\frac{a}{b}$, with $\gcd(a,b)=1$, and find an element that is not an integer multiple of it. Take a prime $p$ not dividing $b$ (which surely exists); then saying that, for some $m\in\mathbb{Z}$,
$$
m\frac{a}{b}=\frac{1}{p}
$$
means that
$$
pma=b,
$$
which is impossible.
Since a group homomorphism $f\colon\mathbb{Z}\to\mathbb{Q}$ always satisfies
$$
f(m)=mf(1)\qquad(m\in\mathbb{Z}),
$$
no homomorphism $f$ can be surjective.

For the second question, notice that isomorphism is reflexive, symmetric and transitive: $G\cong G$; if $G\cong G'$, then $G'\cong G$; if $G\cong G'$ and $G'\cong G''$, then $G\cong G''$.
Now, if you show $\mathbb{Z}\cong m\mathbb{Z}$, which is a particular case of your statement (for $n=1$), you're done.
A: You are going about the first question correctly. If you can prove that $\mathbb{Z}$ is cyclic whereas $\mathbb{Q}$ is not, then you have automatically shown that they are not isomorphic.
For the second question it is sufficient to merely exhibit an isomorphism between the two groups. Here's a hint. Define a map $\varphi :m\mathbb{Z} \rightarrow n \mathbb{Z}$ that takes integers in $m\mathbb{Z}$ and sends them to integers in $n\mathbb{Z}$. We know that for any $a\in m \mathbb{Z}$ we have $a = mk$ for some $k\in\mathbb{Z}$. Let $\varphi(a) = b$ where $b \in n\mathbb{Z}$ and $b = nk$. Now all that is left is to show that this map is bijective and is a homomorphism, i.e. $\varphi(x+y) = \varphi(x)+\varphi(y)$ for all $x, y \in m \mathbb{Z}$.
Good luck!
