An abelian group whose endomorphism group $Hom(G,G)\cong \mathbb{Z}$ Let $C$ be the category of all abelian groups and $G\in ob(C)$. When the morphism class $Hom(G,G)$ is an infinite cyclic group, I am trying to prove $G\cong \mathbb{Z}$.  If this is true, every additive and fully faithful functor $F: C\rightarrow C$ maps an infinite cyclic group to an infinite cyclic group.
If $G$ is a finitely generated abelian group, then $G$ is a finite diret sum of cyclic groups. Thus $G\cong \mathbb{Z}$.
But I don't know if this is right when $G$ is infinitely generated abelian group.
 A: Maybe there is a much more straightforward counterexample, but denote by $\mathcal{P}$ the set of prime numbers, and for each prime $p$ define $H_{p}=\left\{\frac{n}{\prod_{\substack{p'\in\mathcal{P}\\p'\leqslant p}}p'}\in\mathbb{Q}\mid n\in\mathbb{Z}\right\}$. It is a subgroup of the (additive) abelian group $\mathbb{Q}$, consisting of all fractions whose denominator is a product of primes smaller than $p$ (appearing just once). For instance, $H_{5}$ consists of all rationals of the form $\frac{n}{30}$ for some $n\in\mathbb{N}$.
Put:
\begin{equation*}
G=\bigcup_{p\in\mathcal{P}}H_{p}\text{.}
\end{equation*}
It is also an additive subgroup of $\mathbb{Q}$. Take $\varphi\in\operatorname{Hom}\left(G,G\right)$. Then $\varphi$ is uniquely determined by $\varphi\left(1\right)$. Suppose that $\varphi\left(1\right)\notin\mathbb{Z}$. Let $p$ be the smallest prime such that $\varphi\left(1\right)\in H_{p}$. Then $\varphi\left(\frac{1}{p}\right)$ is of the form $\frac{n}{m}$, with $p^{2}\mid m$ and $n$ prime with $p$. Such numbers are not, by definition, in $G$, so we get a contradiction. Therefore, $\operatorname{Hom}\left(G,G\right)\cong\mathbb{Z}$
