Determine the Ring End(G) for each abelian Group G Well here is the confusing question I have gotten from the professor. 
Determine the Ring End(G) for each abelian Group G from the previous exercice.
I found this 
https://crazyproject.wordpress.com/2010/08/15/the-set-of-all-endomorphisms-of-an-abelian-group-is-a-ring-under-pointwise-addition-and-composition/
But i do not know if it helps with my question. Here are all the Groups of the previos exercice. 

 A: To simplify notation, note that your groups are $\mathbb Z_n $, with $n=2,3,4$. For the endomorphisms to form a ring, as your link says, one uses pointwise addition and composition as the operations. Of course, as is customary, here one uses additive notation for the group operation.
A homomorphism  $\phi:\mathbb Z_n\to\mathbb Z_n $ is determined by its action on the generator $1$ (that would be $f $ in the three groups you have). So there are $n $ of them, say $\phi_0,\ldots,\phi_{n-1} $ where $\phi_k (1)=k$. That is, $\phi_k (x)=kx $ with the cyclic addition of $\mathbb Z_n $. 
Now note that $$(\phi_k+\phi_h)(x)=\phi_k (x)+\phi_h (x)=kx+hx=(k+h)x=\phi_{k+h}(x )$$
and
$$
\phi_k\circ\phi_h (x)=k (hx)=(kh)x=\phi_{kh}(x). $$
In other words, the map $\alpha:\mathbb Z_n\to\text {End}(G) $ given by $\alpha(k)=\phi_k$ is a ring homomorphism.  It is clearly onto, and if $\alpha (k)=0$ then $kx=0$ for all $x\in\mathbb Z_n $, so $k=0$; so it is also one-to-one. 
In summary, the ring $\text {End}(\mathbb Z_n) $ is isomorphic to the ring $\mathbb Z_n $.
