Endomorphism rings and torsion subgroups. Let $G$ be an abelian group and let $T$ be its torsion subgroup, i.e., $T = \{g \in G \hspace{1mm} | \hspace{1mm} g \text{ of finite order}\} $. Is the restriction map
$\phi: \text{End}(G) \rightarrow \text{End}(T), \hspace{1mm} f \mapsto f|_{T}$,
which is a homomorphism of rings, a surjection? 
If $G$ is finitely generated, the answer to this question is positive. This can be seen by invoking the classification of finitely generated abelian groups. I don't think that $\phi$ is a surjection in general, but I haven't been able to construct a counterexample yet. 
 A: Here's a counterexample.  Let $H=\prod_p \mathbb{Z}/p$, where $p$ ranges over all primes.  Let $G$ be the subgroup of $H$ consisting of all sequences $x=(x_p)$ such that there exists $q\in\mathbb{Q}$ such that $x_p$ is the mod $p$ residue of $q$ for all but finitely many $p$ (the mod $p$ residue of $q$ is well-defined for any $p$ that does not divide the denominator of $q$).  In this case, we say that $x$ approximates $q$.  The torsion subgroup of $G$ is $T=\bigoplus_p \mathbb{Z}/p$, since an element of $G$ is torsion iff all but finitely many of its terms are $0$.  The quotient $G/T$ is isomorphic to $\mathbb{Q}$, by sending $x\in G$ to the unique $q\in\mathbb{Q}$ which $x$ approximates.
Now partition the primes into two infinite sets $A$ and $B$ and consider the endomorphism $f:T\to T$ which projects onto $\bigoplus_{p\in A}\mathbb{Z}/p$.  That is, $f(x)_p=x_p$ if $p\in A$ and $f(x)_p=0$ if $p\in B$.  I claim that $f$ cannot be extended to a homomorphism $g:G\to G$.
Indeed, suppose such an extension $g$ exists.  For each $p$, let $e_p\in T$ be the element whose $p$th coordinate is $1$ and all other coordinates are $0$.  Let $1\in G$ be the element all of whose coordinates are $1$.  Note that an element $x\in G$ is divisible by a prime $p$ iff its $p$th coordinate is $0$ (if $x$ approximates $q$, we can choose $y\in G$ which approximates $q/p$ such that $py=x$).  In particular, for any $p$, $1-e_p$ is divisible by $p$, and hence $g(1-e_p)=g(1)-g(e_p)$ is divisible by $p$.  This means that the $p$th coordinate of $g(1)$ is $1$ for all $p\in A$ and $0$ for all $p\in B$.  Since $g(1)\in G$, it approximates some $q\in\mathbb{Q}$.  The numerator of $q$ must then be divisible by $p$ for all but finitely many $p\in B$.  Since $B$ is infinite, this can only happen if $q=0$.  But since the mod $p$ reduction of $q$ is $1$ for all but finitely many $p\in A$ and $A$ is infinite, $q\neq 0$.  This is a contradiction.
