cyclic groups homomorphism I have the following task: 
"Determine the homomorphism between two cyclic groups. Which are injective, surjective or bijective?" 
I already found this for the cyclic group of integers: http://users.math.yale.edu/~auel/courses/370f06/docs/solutions3.pdf page 2, 4.4.
But what about the cyclic groups of Integers modulo n?
We can suppose the cyclic groups are $\mathbf Z/m\mathbf Z$ and $\mathbf Z/n\mathbf Z$ respectively. A homomorphism from the first to the second is determined by the choice of the image $\bar x$ of $\bar 1$, subject to the condition $m \bar x=0$, i.e.
$$\DeclareMathOperator\Hom{Hom}\Hom(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z) ≃ \operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m).$$
Let $d=\gcd(m,n)$, $m'=\dfrac md$, $n'=\dfrac nd$. Since $m'$ and $n'$ are coprime,
$$\operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m)=n'\mathbf Z/n\mathbf Z ≃ \mathbf Z/d\mathbf Z.$$
Furthermore, 


the surjectivity of such a homomorphism means its image, which is contained in $n'\mathbf Z/n\mathbf Z$, is equal to $\mathbf Z/n\mathbf Z$ . This can happen only if $n'=1$, i.e. $n∣m$.
If $\gcd(m,n)=1$, the only homomorphism is the zero homomorphism.
Injectivity means the image of the homomorphism is isomorphic to $\mathbf Z/m\mathbf Z$. Hence, by Lagrange’s theorem, $ m $ has to be a divisor of $ n $. We’ll suppose this is indeed the case. Then the image $ \bar x $ of $\bar 1$ in $ \mathbf Z/n\mathbf Z $ has to be or order $ m $. As the order of $ \bar x$ is $ \dfrac n{\gcd(n, x)} $,  this means $ \gcd(n, x) = \dfrac nm$.
Finally, if a homomorphism is an isomorphism, the above considerations show it implies $ m = n $. The image of $\bar 1 $ is another generator of $\mathbf Z/m\mathbf Z$, i. e. an element $\bar x,\; x &lt; m$, coprime to $ m $.
Property: let $f: G \rightarrow H$ a finite group morphism then
for all $ x\in G$  the order of $f(x)$ divides both, order of $ x$
and order of $H$.

So, if $n$ and $m$ are  coprime, then there is no nonzero morphism
groups from $\Bbb{Z}/n\Bbb{Z}$ to $\Bbb{Z}/m\Bbb{Z}$. else $n=1$ or $m=1$ this last three cases are evidente

If $n$ divides $m$, the number of  choosing the injection from
$\Bbb{Z}/n\Bbb{Z}$  into $\Bbb{Z}/m\Bbb{Z}$ is the number of pairs
$(i, j)$ where $order (i) = oorder(j) = n$. so  in total there are
$\phi(n)^2$;  these injections are surjections therefore
bijections if and only if $n = m$.

If $d = gcd (n, m)$ with $n \neq d \neq m$, there is  no
injections, no surjections, no bijections.