Q:

Direct product decomposition for infinite abelian groups with constrained torsion

Let $g$ be a positive integer, and let $G$ be a commutative group with the following constraint on its torsion subgroup: there is an injection $G[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^{2g}$.  Must there be subgroups $G_1,\ldots,G_g$ of $G$ such that   
(i) $G = G_1 \times \ldots \times G_g$ (internal direct product), and
(ii) For all $1 \leq i \leq g$, there is an injection $G_i[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^2$?  
Motivation: If this is true, then it reduces the "Inverse Mordell-Weil Problem for Abelian Varieties" to the "Inverse Mordell-Weil Problem for Elliptic Curves".
Thus although the given question certain has an affirmative answer in many special cases -- e.g. it is a triviality if $G$ is finitely generated -- I am not really interested in that.  But it would be "lucky for me" if the answer turns out to be affirmative in the general case, so it's worth asking.  
Added: This previous question contains some information on when the torsion subgroup of a commutative group is a direct summand.

A:

The answer is negative.
Define $S=\bigoplus_p C_p$, where $C_p$ is cyclic of order $p$ and $p$ ranges over prime numbers. Fix an integer $n\ge 1$. Define $P=\prod_p C_p$, and consider a maximal $\mathbf{Z}$-free family in $P^n/S^n$, generating a free abelian group (of continuum rank) $L/S^n$. Fix a prime $q$ and let $U/S^n$ be (*) a subgroup of $P^n/S^n$ isomorphic to $\mathbf{Z}_q$ (the group of $q$-adic numbers), with $L\subset U$.
Let's check that for every direct decomposition $U=V\oplus W$, either $V$ or $W$ is finite. Since $T(U)\simeq S^n$, this implies that $U$ answers negatively your question (as soon as $n\ge 3$).
First observe that $\mathbf{Z}_q$ is indecomposable (i.e. has no nontrivial decomposition as direct product, easy exercise (**)). Since $U/T(U)\simeq\mathbf{Z}_q$, it follows that either $V$ or $W$, say $W$, is torsion. So $V/T(V)\simeq\mathbf{Z}_q$.
Write $T(V)=\bigoplus A_p$ and $T(W)=\bigoplus B_p$, so that $A_p\oplus B_p=C_p^n$ for all $p$.
We have $P^n=(\prod A_p)\times (\prod B_p)$. So $P^n/T(V)=(\prod A_p/\bigoplus A_p)\times \prod B_p$. Since there is no nonzero homomorphism $\mathbf{Z}_q\to C_p$ for $p\neq q$ and since the subgroup $V/T(V)$ of $P^n/T(V)$ is isomorphic to $\mathbf{Z}_q$, the projection of $V/T(V)$ into $\prod B_p$ is contained in $B_q$. Thus in the above decomposition, $V\subset (\prod A_p)\times B_q$ and $W\subset (\bigoplus A_p)\times (\bigoplus B_p)$. So $U\subset (\prod A_p)\times (\bigoplus B_p)$. Since $P^n/U$ is torsion (because it is a quotient of $P^n/M$ which is torsion), it follows that $\prod B_p/\bigoplus B_p$ is torsion. This means that $B_p=0$ for large $p$. It follows that $W$ is finite.
Here are the two easy verifications:
(*): pick in $\mathbf{Z}_q$ a maximal $\mathbf{Z}$-free family, generating an abelian free subgroup $Z$; since $P^n/S^n$ is a rational vector space, any isomorphism $Z\to L/S^n$ (it exists since both groups are free abelian of continuum rank) can be extended to an injective homomorphism $\mathbf{Z}_q\to P^n/S^n$.
(**): $\mathbf{Z}_q$ is $p$-divisible for all $p\neq q$ and $\mathbf{Z}_q/q\mathbf{Z}_q$ has order $q$, so for every direct decomposition of $\mathbf{Z}_q$, at least one factor is divisible. But $\mathbf{Z}_q$ does not contain any copy of $\mathbf{Q}$.

