Sum of endomorphisms that is not an endomorphism Consider a group $(G,+)$ and the structure $\operatorname{End}(G)$ of all endomorphisms of $G$. We know that if $G$ is abelian, then $\operatorname{End}(G)$ forms (typically non-commutative) unitary ring together with pointwise addition and function composition. The reason we want $G$ to be commutative is that sum of endomorphisms may fail to be an endomorphism. (here sum means the $+$ group operation of $G$).
Here is the proof that sum of two endomorphism under pointwise commutative addition is an endomorphism again (maybe it will help someone to find the counterexample). Let $(G,+)$ be the abelian group and $(\operatorname{End}(G),\oplus)$ be the additive subgroup of its endomorphism ring. Let $\varphi,\psi$ be two endomorphisms, then 
\begin{align}
(\varphi\oplus\psi)(x+y) &= \varphi(x+y)+\psi(x+y) =\\&= \varphi(x)+\varphi(y)+\psi(x)+\psi(y) =\\&=\varphi(x)+\psi(x)+\varphi(y)+\psi(y) =\\&=
(\varphi \oplus \psi)(x) + (\varphi\oplus\psi)(y)
\end{align}
so $\varphi\oplus\psi$ is an endomorphism.
Is there an example of (necessarily non-commutative) group $G$, s.t. $\operatorname{End}(G)$ fails to be a ring exactly due to the closure of endomorphisms under non-commutative pointwise addition? 
 A: Take $G=S_3$, and consider the endomorphisms $\psi,\theta\colon S_3\to S_3$, both of which map $(1,2,3)$ to the identity. The endomorphism $\psi$ maps every transposition to $(1,2)$; and the endomorphism $\theta$ sends every transposition to $(1,3)$.
The “sum” $\psi\cdot\theta\colon S_3\to S_3$ given by $(\psi\cdot\theta)(x) = \psi(x)\theta(x)$ is not an endomorphism. To see this, simply note that $(1,2)\mapsto (1,2)(1,3)=(1,3,2)$ (I compose permutations right to left), but then $e = (\psi\cdot\theta)( (1,2)^2)\neq (\psi\cdot\theta(1,2))^2 = (1,2,3)$. 
(Or course, in addition this “pointwise operation” is not commutative, so it cannot be the addition of a ring.)
Now, explicitly, what is $\mathrm{End}(S_3)$? You have the automorphisms, which correspond to conjugation by elements of $S_3$; the remaining endomorphisms are (i) the trivial map; and (ii) the maps that factor through $S_3/A_3$; there are three such maps: the two I mention above, plus the one that maps all odd permutations to $(2,3)$. So there are exactly ten endomorphisms for $S_3$. 
Can this be given a ring structure in which composition of endomorphisms is the ring multiplication? No. Because $S_3$ is noncommutative and every element gives a different inner automorphism, the monoid structure of $\mathrm{End}(S_3)$ under composition is noncommutative. But it is a theorem of Eldrige that if the order of a finite ring is cube free, then it is commutative. Since $|\mathrm{End}(S_3)|=10$ is cube free, the ring structure would necessarily be commutative, and so we see that it cannot be given such a ring structure.
