Zero divisors among endomorphisms Let $(A,+)$ be a nontrivial abelian group (with the neutral element "$0$") and $\mathrm{End}(A)$ the ring of endomorphisms $\phi : A \to A$, defined with the "pointwise function addition", 
$$(f+g)(a) \equiv f(a)+g(a)$$
and the function composition $f \circ g$. If $A$ is finite, say with $|A| = mn$, where $m$ and $n$ are integers strictly greater then $1$, then we can always easily find zero divisors in $\mathrm{End}(A)$. For example, a pair of functions $f(a) = m\cdot a$ and $g(a) = n\cdot a$ (where I'm using abbreviation $m\cdot a = a + \dots + a$, $m$ times).
But what if $(A,+)$ is a finite group of prime order, or if it is not finite at all: Does the ring $\mathrm{End}(A)$ necessarily have zero divisors? 
Note: Here I'm not taking into account the trivial answer, namely a pair of "zero maps" $z(a) \equiv 0$.
 A: Let $A$ be a finite, nontrivial abelian group, of prime order (i.e. $|A|$ is a prime number). If $f : A \to A$ is an endomorphism, then $\ker(f)$ is a subgroup of $A$. As any subgroup of $A$ is either trivial $\{0\}$ or the whole $A$, and we assume that $f$ is not the zero map, then in fact $\ker(f) = \{0\}$ ($f$ has to be an automorphism). Now, let $f,g : A \to A$ be two endomorphisms and $a \in A$, $a \ne 0$. Then $f(a) \ne 0$ and $g(f(a)) \ne 0$, hence this endomorphism ring doesn't have (nontrivial) zero divisors.
What about the infinite groups?
For example, if $A = (\mathbb{Z},+)$, then any endomorphism $f : A \to A$ has to be of the form $f(z) = kz$, where $k \in \mathbb{Z}$, see e.g.
https://proofwiki.org/wiki/Endomorphism_from_Integers_to_Multiples
From here it is easy to show that $\mathrm{End}(A)$ doesn't have any (nontrivial) zero divisors.
On the other hand, if we take $A$ to be the group of sequences of real numbers with "naturally" defined operation of addition, $(a_n) + (b_n) = (a_n + b_n)$, then the associated ring of endomorphisms has zero divisors, see e.g.
https://en.wikipedia.org/wiki/Zero_divisor#Examples
